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The New Calculus is the only correct formulation of calculus.
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Jew Lover
Jan 10, 2019, 7:59:49 AM1/10/19
to
Mainstream mathematics academics are incorrigibly stupid creatures. They will tell you that a straight line function has a derivative and change the definition of derivative in terms of their bullshit "analytic derivative".
Let's see how the orangutans fail!
The slope of a straight line is given by:
[ f(x+h)-f(x) ] / h [A]
The derivative of a straight line is given by:
Lim (h -> 0) [ f(x+h)-f(x) ] / h [B]
Therefore, since a straight line's derivative is its slope, we have:
Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
Can you get any dumber? I have produced a New Calculus which has no ill-formed concepts and have solved the tangent line problem and removed ALL ill-formed concepts such as infinity, infinitesimals and limit theory from differential and integral calculus.
A derivative can be found geometrically using Ancient Greek mathematics: https://lnkd.in/dhN6Gzw
Download the most important mathematics book ever written here (it's free!!!):
https://lnkd.in/dEkeshd
Let's see how the orangutans fail!
The slope of a straight line is given by:
[ f(x+h)-f(x) ] / h [A]
The derivative of a straight line is given by:
Lim (h -> 0) [ f(x+h)-f(x) ] / h [B]
Therefore, since a straight line's derivative is its slope, we have:
Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
Can you get any dumber? I have produced a New Calculus which has no ill-formed concepts and have solved the tangent line problem and removed ALL ill-formed concepts such as infinity, infinitesimals and limit theory from differential and integral calculus.
A derivative can be found geometrically using Ancient Greek mathematics: https://lnkd.in/dhN6Gzw
Download the most important mathematics book ever written here (it's free!!!):
https://lnkd.in/dEkeshd
Alan Mackenzie
Jan 10, 2019, 8:57:15 AM1/10/19
to
Jew Lover <thenewc...@gmail.com> wrote:
> Mainstream mathematics academics are incorrigibly stupid creatures.
That is false and borders on defamation.
> Mainstream mathematics academics are incorrigibly stupid creatures.
> They will tell you that a straight line function has a derivative and
> change the definition of derivative in terms of their bullshit
> "analytic derivative".
derivative.
> The slope of a straight line is given by:
> [ f(x+h)-f(x) ] / h [A]
function of one variable.
> The derivative of a straight line is given by:
> Lim (h -> 0) [ f(x+h)-f(x) ] / h [B]
> Therefore, since a straight line's derivative is its slope, we have:
> Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
variable, x, on the right of the = is a function of two variables, x and
h.
> Can you get any dumber?
> I have produced a New Calculus which has no ill-formed concepts and
> have solved the tangent line problem and removed ALL ill-formed
> concepts such as infinity, infinitesimals and limit theory from
> differential and integral calculus.
is overwhelmingly successful (with applications in science, engineering
and statistics, amongst others), and is rigorously well defined, what's
the point of "New Calculus", whatever it might be?
> A derivative can be found geometrically using Ancient Greek
> mathematics: https://lnkd.in/dhN6Gzw
> Download the most important mathematics book ever written here (it's
> free!!!):
> https://lnkd.in/dEkeshd
--
Alan Mackenzie (Nuremberg, Germany).
Jew Lover
Jan 10, 2019, 10:28:44 AM1/10/19
to
On Thursday, 10 January 2019 08:57:15 UTC-5, Alan Mackenzie wrote:
> Jew Lover <thenewc...@gmail.com> wrote:
> > Mainstream mathematics academics are incorrigibly stupid creatures.
>
> That is false and borders on defamation.
It is indisputably true.
> Jew Lover <thenewc...@gmail.com> wrote:
> > Mainstream mathematics academics are incorrigibly stupid creatures.
>
> That is false and borders on defamation.
>
> > They will tell you that a straight line function has a derivative and
> > change the definition of derivative in terms of their bullshit
> > "analytic derivative".
>
> As you note below, a straight line function does indeed have a
> derivative.
>
> > The slope of a straight line is given by:
>
> > [ f(x+h)-f(x) ] / h [A]
>
> In fact, that is the slope of a chord through (x, f(x)) of any real
> function of one variable.
>
> > The derivative of a straight line is given by:
>
> > Lim (h -> 0) [ f(x+h)-f(x) ] / h [B]
>
> Correct. How well do you understand this definition?
>
> > Therefore, since a straight line's derivative is its slope, we have:
>
> > Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
>
> That is plain nonsense. On the left of the = sign is a function of one
> variable, x, on the right of the = is a function of two variables, x and
> h.
>
> > Can you get any dumber?
>
> I suspect not.
>
> > I have produced a New Calculus which has no ill-formed concepts and
> > have solved the tangent line problem and removed ALL ill-formed
> > concepts such as infinity, infinitesimals and limit theory from
> > differential and integral calculus.
>
> What is "New Calculus" and what can be done with it?
> Given that calculus
> is overwhelmingly successful (with applications in science, engineering
> and statistics, amongst others), and is rigorously well defined, what's
> the point of "New Calculus", whatever it might be?
>
> > A derivative can be found geometrically using Ancient Greek
> > mathematics: https://lnkd.in/dhN6Gzw
>
> > Download the most important mathematics book ever written here (it's
> > free!!!):
>
> > https://lnkd.in/dEkeshd
>
> Is that Euclid's Elements?
j4n bur53
Jan 10, 2019, 11:07:56 AM1/10/19
to
So you are using Eulers blunder, S = Lim S?
j4n bur53
Jan 10, 2019, 11:32:45 AM1/10/19
to
Its rather John Garbageiels blunder.
What is this nonsense moron?
Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
On the LHS there is a value and on the right hand
side there is a function in h. Do you always compare
values with functions?
Such a comparison fails usually. R and R -> R
are distinct spaces. Maybe in Dans DC Proof they
are the same, but usually, in ZFC:
there is no real number r,
which is also a function f : R -> R
and there is no function f : R -> R,
which is also a real number r
How stupid are you bird brain?
What is this nonsense moron?
Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
side there is a function in h. Do you always compare
values with functions?
Such a comparison fails usually. R and R -> R
are distinct spaces. Maybe in Dans DC Proof they
are the same, but usually, in ZFC:
there is no real number r,
which is also a function f : R -> R
and there is no function f : R -> R,
which is also a real number r
How stupid are you bird brain?
Me
Jan 10, 2019, 4:38:28 PM1/10/19
to
On Thursday, January 10, 2019 at 1:59:49 PM UTC+1, Jew Lover wrote:
> The slope of a straight line is given by:
>
> [ f(x+c)-f(x) ] / c [A]
> The slope of a straight line is given by:
>
for some c e IR, c > 0.
> The derivative of a straight line is given by:
>
Right.
> Therefore, since a straight line's derivative is its slope, we have:
>
for some c e IR, c > 0.
Right.
> A derivative can be found geometrically [in this case].
Right.
Well, done, troll boy.
Jew Lover
Jan 10, 2019, 7:11:09 PM1/10/19
to
On Thursday, 10 January 2019 16:38:28 UTC-5, Me wrote:
> On Thursday, January 10, 2019 at 1:59:49 PM UTC+1, Jew Lover wrote:
>
> > The slope of a straight line is given by:
> >
> > [ f(x+c)-f(x) ] / c [A]
>
> for some c e IR, c > 0.
>
> > The derivative of a straight line is given by:
> >
> > lim_(h -> 0) [ f(x+h)-f(x) ] / h [B]
>
> Right.
>
> > Therefore, since a straight line's derivative is its slope, we have:
> >
> > lim_(h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+c)-f(x) ] / c
Changing the variable does not solve the problem, you imbecile.
> On Thursday, January 10, 2019 at 1:59:49 PM UTC+1, Jew Lover wrote:
>
> > The slope of a straight line is given by:
> >
> > [ f(x+c)-f(x) ] / c [A]
>
> for some c e IR, c > 0.
>
> > The derivative of a straight line is given by:
> >
> > lim_(h -> 0) [ f(x+h)-f(x) ] / h [B]
>
> Right.
>
> > Therefore, since a straight line's derivative is its slope, we have:
> >
> > lim_(h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+c)-f(x) ] / c
Not even a good try. Fef.
Dan Christensen
Jan 10, 2019, 8:43:26 PM1/10/19
to
On Thursday, January 10, 2019 at 7:59:49 AM UTC-5, Jew Lover wrote:
> Mainstream mathematics academics are incorrigibly stupid creatures. They will tell you that a straight line function has a derivative and change the definition of derivative in terms of their bullshit "analytic derivative".
>
> Let's see how the orangutans fail!
>
Hey, Troll Boy, did you ever figure out the derivative of the y=x? No??? Hint: It is NOT undefined as you previously claimed. What else have you got?
> Mainstream mathematics academics are incorrigibly stupid creatures. They will tell you that a straight line function has a derivative and change the definition of derivative in terms of their bullshit "analytic derivative".
>
> Let's see how the orangutans fail!
>
Interested readers should see: “About the spamming troll John Gabriel in his own words (December 2018)” at https://groups.google.com/forum/#!topic/sci.math/PcpAzX5pDeY
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Me
Jan 11, 2019, 2:13:50 AM1/11/19
to
On Friday, January 11, 2019 at 1:11:09 AM UTC+1, Jew Lover wrote:
> On Thursday, 10 January 2019 16:38:28 UTC-5, Me wrote:
> > On Thursday, January 10, 2019 at 1:59:49 PM UTC+1, Jew Lover wrote:
> > >
> > > The slope of a straight line is given by:
> > >
> > > [ f(x+c)-f(x) ] / c [A]
> > >
> > for some c e IR, c > 0.
> > >
> > > The derivative of a straight line is given by:
> > >
> > > lim_(h -> 0) [ f(x+h)-f(x) ] / h [B]
> >
> > Right.
> >
> > > Therefore, since a straight line's derivative is its slope, we have:
> > >
> > > lim_(h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+c)-f(x) ] / c
> > >
for some c e IR, c > 0.
Actually, ANY c e IR, c > 0 will do.
> On Thursday, 10 January 2019 16:38:28 UTC-5, Me wrote:
> > On Thursday, January 10, 2019 at 1:59:49 PM UTC+1, Jew Lover wrote:
> > >
> > > The slope of a straight line is given by:
> > >
> > > [ f(x+c)-f(x) ] / c [A]
> > >
> > for some c e IR, c > 0.
> > >
> > > The derivative of a straight line is given by:
> > >
> > > lim_(h -> 0) [ f(x+h)-f(x) ] / h [B]
> >
> > Right.
> >
> > > Therefore, since a straight line's derivative is its slope, we have:
> > >
> > > lim_(h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+c)-f(x) ] / c
> > >
for some c e IR, c > 0.
Well done, troll boy!
Jew Lover
Jan 11, 2019, 6:04:13 AM1/11/19
to
lim_(h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+c)-f(x) ] / c
Me
Jan 11, 2019, 7:37:09 AM1/11/19
to
Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> >
> > Actually, ANY c e IR, c > 0 will do.
> >
> Except 0.
> On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> >
> > Actually, ANY c e IR, c > 0 will do.
> >
Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
Now you claim
> lim_(h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+c)-f(x) ] / c
>
> is NOT true in general.
Hint: For all x e IR and for any c e IR, c > 0:
lim_(h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+c)-f(x) ] / c ,
where f is a function "decribing" a "straight line". With other words:
f(x) = m*x + d for all x e IR (with m e IR and d e IR) .
Zelos Malum
Jan 11, 2019, 8:14:16 AM1/11/19
to
>Better than you or anyone else will ever be able to understand it.
Not really considering you cannot see the differens between a two variable function and one variable function and then proclaim they are the same.
>What you write is nonsense. Both are functions in two variables - x and h.
Jew Lover
Jan 11, 2019, 10:47:37 AM1/11/19
to
On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > >
> > > Actually, ANY c e IR, c > 0 will do.
> > >
> > Except 0.
>
> Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
You can't refute the New Calculus, for in order to do so, you would have to reject sound geometry.
> Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > >
> > > Actually, ANY c e IR, c > 0 will do.
> > >
> > Except 0.
>
> Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
In your bogus calculus, you do set h = 0 even though you have produced a whole lot of hand waving arguments that do you no good whatsoever.
f(x,h)= [ f(x+h)-f(x) ] / h
f'(x) = f'(x) + Q(x,h) which is only possible if Q(x,h)=0 meaning h=0.
Therefore, you have NO systematic way of finding the derivative. Your first principles method is a load of crap.
<gibberish>
Me
Jan 11, 2019, 11:09:09 AM1/11/19
to
On Friday, January 11, 2019 at 4:47:37 PM UTC+1, Jew Lover wrote:
> On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> > Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > > >
> > > > Actually, ANY c e IR, c > 0 will do.
> > > >
> > > Except 0.
> > >
> > Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
> >
> You can't <bla>
> On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> > Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > > >
> > > > Actually, ANY c e IR, c > 0 will do.
> > > >
> > > Except 0.
> > >
> > Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
> >
Yeah, whatever.
> In calculus, you do set h = 0
Nope! h = 0 is explicitly excluded when calculating the limit
lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
What a moron!
Alan Mackenzie
Jan 11, 2019, 2:58:03 PM1/11/19
to
Jew Lover <thenewc...@gmail.com> wrote:
> You can't refute the New Calculus, for in order to do so, you would
> have to reject sound geometry.
Seeing as how you won't or can't even say what this "New Calculus" is, or
> You can't refute the New Calculus, for in order to do so, you would
> have to reject sound geometry.
what it's for, there doesn't seem to be anything worth refuting. Don't
try going into advertising for a living; it's not your strong point
(assuming you've got one).
> In your bogus calculus, you do set h = 0 even though you have produced
> a whole lot of hand waving arguments that do you no good whatsoever.
derivative as a limit, and you said "better than me". It's now clear you
barely understand it at all, and haven't a clue with respect to its
rigorous foundation. You clearly belong to the mathematically less
sophisticated part of the world's population.
[ .... ]
> Therefore, you have NO systematic way of finding the derivative. Your
> first principles method is a load of crap.
all this sweary language.
Jew Lover
Jan 11, 2019, 4:59:18 PM1/11/19
to
On Friday, 11 January 2019 14:58:03 UTC-5, Alan Mackenzie wrote:
> Jew Lover <thenewc...@gmail.com> wrote:
>
> > You can't refute the New Calculus, for in order to do so, you would
> > have to reject sound geometry.
>
> Seeing as how you won't or can't even say what this "New Calculus" is
<crap>
> Jew Lover <thenewc...@gmail.com> wrote:
>
> > You can't refute the New Calculus, for in order to do so, you would
> > have to reject sound geometry.
>
> Seeing as how you won't or can't even say what this "New Calculus" is
What a crank you are. I referred you to a publication and you said "No thanks."
Yes, you are a fucking moron. Shut up already.
Jew Lover
Jan 11, 2019, 5:04:09 PM1/11/19
to
On Friday, 11 January 2019 11:09:09 UTC-5, Me wrote:
> On Friday, January 11, 2019 at 4:47:37 PM UTC+1, Jew Lover wrote:
> > On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> > > Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > > > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > > > >
> > > > > Actually, ANY c e IR, c > 0 will do.
> > > > >
> > > > Except 0.
> > > >
> > > Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
> > >
> > You can't <bla>
>
> Yeah, whatever.
>
> > In calculus, you do set h = 0
>
> Nope! h = 0 is explicitly excluded when calculating the limit
>
> lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
>
> idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
>
Quite telling how you failed to address any of the following. Chuckle.
> On Friday, January 11, 2019 at 4:47:37 PM UTC+1, Jew Lover wrote:
> > On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> > > Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > > > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > > > >
> > > > > Actually, ANY c e IR, c > 0 will do.
> > > > >
> > > > Except 0.
> > > >
> > > Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
> > >
> > You can't <bla>
>
> Yeah, whatever.
>
> > In calculus, you do set h = 0
>
> Nope! h = 0 is explicitly excluded when calculating the limit
>
> lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
>
> idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
>
You can't refute the New Calculus, for in order to do so, you would have to reject sound geometry.
In your bogus calculus, you do set h = 0 even though you have produced a whole lot of hand waving arguments that do you no good whatsoever.
Jew Lover
Jan 11, 2019, 5:17:20 PM1/11/19
to
On Friday, 11 January 2019 11:09:09 UTC-5, Me wrote:
> On Friday, January 11, 2019 at 4:47:37 PM UTC+1, Jew Lover wrote:
> > On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> > > Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > > > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > > > >
> > > > > Actually, ANY c e IR, c > 0 will do.
> > > > >
> > > > Except 0.
> > > >
> > > Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
> > >
> > You can't <bla>
>
> Yeah, whatever.
>
> > In calculus, you do set h = 0
>
> Nope! h = 0 is explicitly excluded when calculating the limit
>
> lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
>
> idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
> > On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> > > Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > > > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > > > >
> > > > > Actually, ANY c e IR, c > 0 will do.
> > > > >
> > > > Except 0.
> > > >
> > > Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
> > >
> > You can't <bla>
>
> Yeah, whatever.
>
> > In calculus, you do set h = 0
>
> Nope! h = 0 is explicitly excluded when calculating the limit
>
> lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
>
> idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
In your bogus calculus, you do set h = 0 even though you have produced a whole lot of hand waving arguments that do you no good whatsoever.
Me
Jan 11, 2019, 5:56:09 PM1/11/19
to
On Friday, January 11, 2019 at 11:04:09 PM UTC+1, Jew Lover wrote:
> On Friday, 11 January 2019 11:09:09 UTC-5, Me wrote:
> > On Friday, January 11, 2019 at 4:47:37 PM UTC+1, Jew Lover wrote:
> > > On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> > > > Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > > > > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > > > > >
> > > > > > Actually, ANY c e IR, c > 0 will do.
> > > > > >
> > > > > Except 0.
> > > > >
> > > > Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
> > > >
> > > You can't <bla>
> >
> > Yeah, whatever.
> >
> > > In calculus, you do set h = 0
> >
> > Nope! h = 0 is explicitly excluded when calculating the limit
> >
> > lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
> >
> > idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
> >
>
> Quite telling <bla>
> On Friday, 11 January 2019 11:09:09 UTC-5, Me wrote:
> > On Friday, January 11, 2019 at 4:47:37 PM UTC+1, Jew Lover wrote:
> > > On Friday, 11 January 2019 07:37:09 UTC-5, Me wrote:
> > > > Am Freitag, 11. Januar 2019 12:04:13 UTC+1 schrieb Jew Lover:
> > > > > On Friday, 11 January 2019 02:13:50 UTC-5, Me wrote:
> > > > > >
> > > > > > Actually, ANY c e IR, c > 0 will do.
> > > > > >
> > > > > Except 0.
> > > > >
> > > > Oh, so c > 0 allows for c = 0 in your "New Calculus"? Fascinating!
> > > >
> > > You can't <bla>
> >
> > Yeah, whatever.
> >
> > > In calculus, you do set h = 0
> >
> > Nope! h = 0 is explicitly excluded when calculating the limit
> >
> > lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
> >
> > idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
> >
>
If you say so.
> In calculus, you do set h = 0 <bla>
No. See above.
Me
Jan 11, 2019, 5:56:52 PM1/11/19
to
On Friday, January 11, 2019 at 11:17:20 PM UTC+1, Jew Lover wrote:
> In calculus, you do set h = 0
No.
> In calculus, you do set h = 0
Jew Lover
Jan 11, 2019, 7:46:55 PM1/11/19
to
On Friday, 11 January 2019 17:56:52 UTC-5, Me wrote:
> On Friday, January 11, 2019 at 11:17:20 PM UTC+1, Jew Lover wrote:
>
> > In calculus, <crap>
> On Friday, January 11, 2019 at 11:17:20 PM UTC+1, Jew Lover wrote:
>
In your bogus calculus, you DO set h = 0 even though you have produced a whole lot of hand waving arguments that do you no good whatsoever.
Me
Jan 11, 2019, 7:53:02 PM1/11/19
to
On Saturday, January 12, 2019 at 1:46:55 AM UTC+1, Jew Lover wrote:
> In calculus, you DO set h = 0
No, idiot.
> In calculus, you DO set h = 0
No, idiot.
Jew Lover
Jan 12, 2019, 2:22:09 AM1/12/19
to
f(x,h)= [ f(x+h)-f(x) ] / h
f'(x) = f'(x) + Q(x,h) which is only possible if Q(x,h)=0 meaning h=0.
Therefore, you have NO systematic way of finding the derivative. Your first principles method is a load of crap.
Chuckle. Yes, you are stupid beyond belief.
Alan Mackenzie
Jan 12, 2019, 8:19:22 AM1/12/19
to
Jew Lover <thenewc...@gmail.com> wrote:
> On Friday, 11 January 2019 14:58:03 UTC-5, Alan Mackenzie wrote:
>> Jew Lover <thenewc...@gmail.com> wrote:
>> > You can't refute the New Calculus, for in order to do so, you would
>> > have to reject sound geometry.
>> Seeing as how you won't or can't even say what this "New Calculus" is
> <crap>
> What a crank you are. I referred you to a publication and you said "No
> thanks."
Yes. I'm inviting you to try to persuade me to look at it. Right at the
> On Friday, 11 January 2019 14:58:03 UTC-5, Alan Mackenzie wrote:
>> Jew Lover <thenewc...@gmail.com> wrote:
>> > You can't refute the New Calculus, for in order to do so, you would
>> > have to reject sound geometry.
>> Seeing as how you won't or can't even say what this "New Calculus" is
> <crap>
> What a crank you are. I referred you to a publication and you said "No
> thanks."
moment, it just seems like a waste of time. The said publication was
written by somebody who doesn't even understand the concept of a limit in
maths, and likely contains offensive material.
Like I said, it seems like it has nothing to offer, and you can't even
describe in a paragraph what "New Calculus" is. Why should anybody waste
their time with it?
> Yes, you are a fucking moron. Shut up already.
Have a good Saturday.
Me
Jan 12, 2019, 8:47:36 AM1/12/19
to
On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
> f(x,h) = [ f(x+h)-f(x) ] / h
No, you can't "do" that. Since at the rhs of the equation f is a function with just ONE argument while at the lhs of the equation f is a function with TWO arguments. That just NONSENSE.
Actually, we have:
f'(x) = lim_(h->0) [ f(x+h)-f(x) ] / h
Hint: The variable "h" is BOUND in the formula "lim_(h->0) [ f(x+h)-f(x) ] / h", hence it's not a "free variable" in it. With other words, you can't "access" "h" from the outside of this formula (as if it were a FREE variable).
See: https://en.wikipedia.org/wiki/Free_variables_and_bound_variables
READ IT, IDIOT!
It's ALWAYS THE SAME:
"Cranks who contradict some mainstream opinion in some highly technical field, (e.g. mathematics [...]) frequently:
1. exhibit a marked lack of technical ability,
2. misunderstand or fail to use standard notation and terminology,
3. ignore fine distinctions which are essential to correctly
understand mainstream belief."
(Here you are on par with Mückenheim.)
Now you considert the equation:
> f'(x) = f'(x) + Q which is only possible if Q = 0
Yeah, that's right. :-)
For all r,c e IR: if r + c = r, then c = 0.
> f(x,h) = [ f(x+h)-f(x) ] / h
No, you can't "do" that. Since at the rhs of the equation f is a function with just ONE argument while at the lhs of the equation f is a function with TWO arguments. That just NONSENSE.
Actually, we have:
f'(x) = lim_(h->0) [ f(x+h)-f(x) ] / h
Hint: The variable "h" is BOUND in the formula "lim_(h->0) [ f(x+h)-f(x) ] / h", hence it's not a "free variable" in it. With other words, you can't "access" "h" from the outside of this formula (as if it were a FREE variable).
See: https://en.wikipedia.org/wiki/Free_variables_and_bound_variables
READ IT, IDIOT!
It's ALWAYS THE SAME:
"Cranks who contradict some mainstream opinion in some highly technical field, (e.g. mathematics [...]) frequently:
1. exhibit a marked lack of technical ability,
2. misunderstand or fail to use standard notation and terminology,
3. ignore fine distinctions which are essential to correctly
understand mainstream belief."
(Here you are on par with Mückenheim.)
Now you considert the equation:
> f'(x) = f'(x) + Q which is only possible if Q = 0
Yeah, that's right. :-)
For all r,c e IR: if r + c = r, then c = 0.
Jew Lover
Jan 12, 2019, 9:24:55 AM1/12/19
to
On Saturday, 12 January 2019 08:47:36 UTC-5, Me wrote:
> On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
>
> > f(x,h) = [ f(x+h)-f(x) ] / h
>
> No, you can't "do" that.
Look moron. I can and did do that and what's more, it's completely correct.
> On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
>
> > f(x,h) = [ f(x+h)-f(x) ] / h
>
> No, you can't "do" that.
> Since at the rhs of the equation f is a function with just ONE argument while at the lhs of the equation f is a function with TWO arguments. That just NONSENSE.
>
> Actually, we have:
>
> f'(x) = lim_(h->0) [ f(x+h)-f(x) ] / h
>
> Hint: The variable "h" is BOUND in the formula "lim_(h->0) [ f(x+h)-f(x) ] / h", hence it's not a "free variable" in it. With other words, you can't "access" "h" from the outside of this formula (as if it were a FREE variable).
>
> See: http
<mind boggling shit>
Jew Lover
Jan 12, 2019, 9:30:19 AM1/12/19
to
On Saturday, 12 January 2019 08:19:22 UTC-5, Alan Mackenzie wrote:
> Jew Lover <thenewc...@gmail.com> wrote:
> > On Friday, 11 January 2019 14:58:03 UTC-5, Alan Mackenzie wrote:
> >> Jew Lover <thenewc...@gmail.com> wrote:
>
> >> > You can't refute the New Calculus, for in order to do so, you would
> >> > have to reject sound geometry.
>
> >> Seeing as how you won't or can't even say what this "New Calculus" is
>
> > <crap>
>
> > What a crank you are. I referred you to a publication and you said "No
> > thanks."
>
> Yes. I'm inviting you to try to persuade me to look at it.
Considering that you are a nobody, I simply treat you with the disdain that you so deserve. Chuckle.
> Jew Lover <thenewc...@gmail.com> wrote:
> > On Friday, 11 January 2019 14:58:03 UTC-5, Alan Mackenzie wrote:
> >> Jew Lover <thenewc...@gmail.com> wrote:
>
> >> > You can't refute the New Calculus, for in order to do so, you would
> >> > have to reject sound geometry.
>
> >> Seeing as how you won't or can't even say what this "New Calculus" is
>
> > <crap>
>
> > What a crank you are. I referred you to a publication and you said "No
> > thanks."
>
> Yes. I'm inviting you to try to persuade me to look at it.
<too much nonsense to read>
Jew Lover
Jan 12, 2019, 9:39:55 AM1/12/19
to
On Saturday, 12 January 2019 08:19:22 UTC-5, Alan Mackenzie wrote:
> Jew Lover <thenewc...@gmail.com> wrote:
> > On Friday, 11 January 2019 14:58:03 UTC-5, Alan Mackenzie wrote:
> >> Jew Lover <thenewc...@gmail.com> wrote:
>
> >> > You can't refute the New Calculus, for in order to do so, you would
> >> > have to reject sound geometry.
>
> >> Seeing as how you won't or can't even say what this "New Calculus" is
>
> > <crap>
>
> > What a crank you are. I referred you to a publication and you said "No
> > thanks."
>
> Yes. I'm inviting you to try to persuade me to look at it.
Chuckle. I couldn't care less whether you looked at it or not. You are a nobody.
> Jew Lover <thenewc...@gmail.com> wrote:
> > On Friday, 11 January 2019 14:58:03 UTC-5, Alan Mackenzie wrote:
> >> Jew Lover <thenewc...@gmail.com> wrote:
>
> >> > You can't refute the New Calculus, for in order to do so, you would
> >> > have to reject sound geometry.
>
> >> Seeing as how you won't or can't even say what this "New Calculus" is
>
> > <crap>
>
> > What a crank you are. I referred you to a publication and you said "No
> > thanks."
>
> Yes. I'm inviting you to try to persuade me to look at it.
<No time to read crap>
Me
Jan 12, 2019, 10:13:36 AM1/12/19
to
On Saturday, January 12, 2019 at 3:24:55 PM UTC+1, Jew Lover wrote:
> On Saturday, 12 January 2019 08:47:36 UTC-5, Me wrote:
> > On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
> > >
> > > f(x,h) = [ f(x+h)-f(x) ] / h
> > >
> > No, you can't "do" that [since it isn't correct].
> On Saturday, 12 January 2019 08:47:36 UTC-5, Me wrote:
> > On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
> > >
> > > f(x,h) = [ f(x+h)-f(x) ] / h
> > >
> >
> Look moron [...] it's completely correct.
Nope, it isn't. Hint:
> > at the rhs of the equation f is a function with ONE argument,
> > at the lhs of the equation f is a function with TWO arguments.
> > That's just NONSENSE.
> >
> Both sides take two arguments: x and h.
I didn't talk about the rhs/lhs formulas, but about "f" (occurring in the rhs formula and occurring in the lfh formula).
> Both sides take two arguments: x and h.
You claimed that you are/were a programmer. So do you think that the strings "foo(x)" and "foo(x, y)" refer to the same C function? Sure?
Ever heard anything about the "signatur of a function"?
> It does not matter that one of them remain constant.
Now you mention that
> f'(x) = lim_(x->h) (f(x+h)-f(x)) / h doesn't even include the h on the left
> hand side.
Right. I already told you the reason for this fact:
> > The variable "h" is BOUND in the formula "lim_(h->0) [ f(x+h)-f(x) ] / h",
> > hence it's not a "free variable" in it. With other words, you can't
> > "access" "h" from the outside of this formula (as if it were a FREE
> > variable).
Hint: Same situation her:
c := lim_(h->0) h + 1
Now c is a constant, actually c = 1. To write "c(h)" at the lhs would be nonsense. c does not "depend" on "h".
See: https://en.wikipedia.org/wiki/Free_variables_and_bound_variables
Jew Lover
Jan 12, 2019, 10:27:33 AM1/12/19
to
On Saturday, 12 January 2019 10:13:36 UTC-5, Me wrote:
> On Saturday, January 12, 2019 at 3:24:55 PM UTC+1, Jew Lover wrote:
> > On Saturday, 12 January 2019 08:47:36 UTC-5, Me wrote:
> > > On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
> > > >
> > > > f(x,h) = [ f(x+h)-f(x) ] / h
> > > >
> > > No, you can't <shit>
> On Saturday, January 12, 2019 at 3:24:55 PM UTC+1, Jew Lover wrote:
> > On Saturday, 12 January 2019 08:47:36 UTC-5, Me wrote:
> > > On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
> > > >
> > > > f(x,h) = [ f(x+h)-f(x) ] / h
> > > >
Afraid that conjuring a new ill-formed concept such as "free variable" does not help your cause. Chuckle. You just keep digging yourself in deeper and deeper.
Jew Lover
Jan 12, 2019, 10:29:34 AM1/12/19
to
On Saturday, 12 January 2019 10:13:36 UTC-5, Me wrote:
> On Saturday, January 12, 2019 at 3:24:55 PM UTC+1, Jew Lover wrote:
> > On Saturday, 12 January 2019 08:47:36 UTC-5, Me wrote:
> > > On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
> > > >
> > > > f(x,h) = [ f(x+h)-f(x) ] / h
> > > >
> > > No, you can't <shit>
> On Saturday, January 12, 2019 at 3:24:55 PM UTC+1, Jew Lover wrote:
> > On Saturday, 12 January 2019 08:47:36 UTC-5, Me wrote:
> > > On Saturday, January 12, 2019 at 8:22:09 AM UTC+1, Jew Lover wrote:
> > > >
> > > > f(x,h) = [ f(x+h)-f(x) ] / h
> > > >
I can actually provide numerous reasons for why everything you write is bullshit, but I am concerned you might create a new Wikipedia entry to give some credibility to your rot. Chuckle.
So no. I will refrain from showing you what a moron you are.
This time: Answer not a fool according to his folly, lest he ...
Jew Lover
Jan 12, 2019, 10:37:51 AM1/12/19
to
" f'(x) = Lim (h->0) [f(x+h)-f(x)] / h
x is a free variable and h is a bound variable; consequently the value of this expression depends on the value of x, but there is nothing called h on which it could depend."
1. Does not the limit depend on h? If h approached anything else BUT ZERO, would that expression still be true? Ha, ha.
2. Does not the limit depend on EVERY possible value of h (except the one that actually matters, i.e. ZERO!) because it is in fact the limit of the FINITE DIFFERENCE quotients that is f '(x)?
Chuckle.
x is a free variable and h is a bound variable; consequently the value of this expression depends on the value of x, but there is nothing called h on which it could depend."
1. Does not the limit depend on h? If h approached anything else BUT ZERO, would that expression still be true? Ha, ha.
2. Does not the limit depend on EVERY possible value of h (except the one that actually matters, i.e. ZERO!) because it is in fact the limit of the FINITE DIFFERENCE quotients that is f '(x)?
Chuckle.
Jew Lover
Jan 12, 2019, 10:43:29 AM1/12/19
to
"A bound variable is a variable that was previously free, but has been bound to a specific value or set of values called domain of discourse or universe."
- http://en.wikipedia.org/wiki/Free_variables_and_bound_variables#CITEREFThompson1991
Gee, so if h is bound, it means that it can be anything except 0?
Time to create a new Wikipedia Moronica entry:
A bound variable is one that has certain restrictions.
See, I am not such a bad guy after all, am I?
The truth is that you fucking morons are so wrong that no matter how many edits, updates and new articles you produce on the Moronica, you will ALWAYS be wrong. Why? Because you are morons.
Me
Jan 12, 2019, 10:48:22 AM1/12/19
to
On Saturday, January 12, 2019 at 4:27:33 PM UTC+1, Jew Lover wrote:
>
> You can't deny it, you moron:
>
>
> You can't deny it, you moron:
>
> f(x,h) = [ f(x+h)-f(x) ] / h
I already told you, that this in nonsense. Hence I won't explain it again.
> [bullshit]
Jew Lover
Jan 12, 2019, 10:52:00 AM1/12/19
to
Chuckle.
Message has been deleted
Me
Jan 12, 2019, 10:54:05 AM1/12/19
to
Jew Lover
Jan 12, 2019, 10:59:20 AM1/12/19
to
On Saturday, 12 January 2019 10:52:30 UTC-5, Me wrote:
> On Saturday, January 12, 2019 at 4:37:51 PM UTC+1, Jew Lover wrote:
>
> > "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
>
> > 1. [...] If h approached anything else BUT ZERO ...
>
> Then we would not get the limit any more (by definition).
Therefore, the limit depends on h, doesn't it, you ignoramus? Not even an attempt at answering all the other questions, eh? Chuckle.
You are inferior to me in all respects. Never, ever forget this. You are privileged that I even bother responding to your shit. You are no doubt incorrigibly stupid with an IQ no greater than 40, if that!
> On Saturday, January 12, 2019 at 4:37:51 PM UTC+1, Jew Lover wrote:
>
> > "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
> >
> > x is a free variable and h is a bound variable; consequently the value of
> > this expression depends on the value of x, but there is nothing called h
> > on which it could depend."
>
> Exactly.
> > x is a free variable and h is a bound variable; consequently the value of
> > this expression depends on the value of x, but there is nothing called h
> > on which it could depend."
>
>
> > 1. [...] If h approached anything else BUT ZERO ...
>
> Then we would not get the limit any more (by definition).
Therefore, the limit depends on h, doesn't it, you ignoramus? Not even an attempt at answering all the other questions, eh? Chuckle.
You are inferior to me in all respects. Never, ever forget this. You are privileged that I even bother responding to your shit. You are no doubt incorrigibly stupid with an IQ no greater than 40, if that!
Me
Jan 12, 2019, 11:00:24 AM1/12/19
to
On Saturday, January 12, 2019 at 4:43:29 PM UTC+1, Jew Lover wrote:
> On Saturday, 12 January 2019 10:37:51 UTC-5, Jew Lover wrote:
> >
> > "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
> On Saturday, 12 January 2019 10:37:51 UTC-5, Jew Lover wrote:
> >
> >
> > x is a free variable and h is a bound variable; consequently the value of
> > this expression depends on the value of x, but there is nothing called h
> > on which it could depend."
Right.
> > x is a free variable and h is a bound variable; consequently the value of
> > this expression depends on the value of x, but there is nothing called h
> > on which it could depend."
> "A bound variable is a variable that was previously free, but has been bound
> to a specific value or set of values called domain of discourse or universe."
> Gee, so if h is bound
Me
Jan 12, 2019, 11:16:56 AM1/12/19
to
On Saturday, January 12, 2019 at 4:59:20 PM UTC+1, Jew Lover wrote:
> Therefore, the limit depends on h, doesn't it?
I don't think so.
Consider:
c := lim_(h->0) h + 1
Now you may claim that c is the limit of h + 1 as h approaches 0 (or something like that).
Can's the how this (constant) value "depends on h". Sorry.
Now we could introduce a PARAMETER, say "p", with p e IR. Then we might consider
c(p) := lim_(h->p) h + 1
and claim that c(p) is the limit of h + 1 as h approaches p.
Now here I can see (and hence say) that "lim_(h->p) h + 1" or "c(p)" depends on p. No? Actually, it DOES depend on p.
Hint: c(0) differs from c(1).
> Therefore, the limit depends on h, doesn't it?
I don't think so.
Consider:
c := lim_(h->0) h + 1
Can's the how this (constant) value "depends on h". Sorry.
Now we could introduce a PARAMETER, say "p", with p e IR. Then we might consider
c(p) := lim_(h->p) h + 1
and claim that c(p) is the limit of h + 1 as h approaches p.
Now here I can see (and hence say) that "lim_(h->p) h + 1" or "c(p)" depends on p. No? Actually, it DOES depend on p.
Hint: c(0) differs from c(1).
gabriel...@gmail.com
Jan 12, 2019, 11:16:56 AM1/12/19
to
On Saturday, 12 January 2019 11:00:24 UTC-5, Me wrote:
> On Saturday, January 12, 2019 at 4:43:29 PM UTC+1, Jew Lover wrote:
> > On Saturday, 12 January 2019 10:37:51 UTC-5, Jew Lover wrote:
> > >
> > > "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
> > >
> > > x is a free variable and h is a bound variable; consequently the value of
> > > this expression depends on the value of x, but there is nothing called h
> > > on which it could depend."
>
> Right.
>
> > "A bound variable is a variable that was previously free, but has been bound
> > to a specific value or set of values called domain of discourse or universe."
Bwaaa haaaaa haaaaaa. No idiot. h can take on ANY value close to x, except 0, which means it very much depends on EVERY value in your bullshit "universe". Chuckle.
> On Saturday, January 12, 2019 at 4:43:29 PM UTC+1, Jew Lover wrote:
> > On Saturday, 12 January 2019 10:37:51 UTC-5, Jew Lover wrote:
> > >
> > > "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
> > >
> > > x is a free variable and h is a bound variable; consequently the value of
> > > this expression depends on the value of x, but there is nothing called h
> > > on which it could depend."
>
> Right.
>
> > "A bound variable is a variable that was previously free, but has been bound
> > to a specific value or set of values called domain of discourse or universe."
>
> Well, actually, it is "bound" by a certain "expression". In this case, "h" in the formula "[f(x+h)-f(x)] / h" is bound by the expression "lim_(h->0)".
>
> > Gee, so if h is bound
>
> Right. That's why it does not occure in the lhs formula, i.e. in "f'(x)".
In your bogus calculus, you DO set h = 0 even though you have produced a whole lot of hand waving arguments that do you no good whatsoever.
Me
Jan 12, 2019, 11:20:19 AM1/12/19
to
On Saturday, January 12, 2019 at 4:37:51 PM UTC+1, Jew Lover wrote:
> "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
> "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
>
> x is a free variable and h is a bound variable; consequently the value of
> this expression depends on the value of x, but there is nothing called h
> on which it could depend."
> x is a free variable and h is a bound variable; consequently the value of
> this expression depends on the value of x, but there is nothing called h
> on which it could depend."
Exactly.
> 1. [...] If h approached anything else BUT ZERO ...
Then we would not get the derivation of f any more (by definition).
> 1. [...] If h approached anything else BUT ZERO ...
Me
Jan 12, 2019, 11:24:08 AM1/12/19
to
On Saturday, January 12, 2019 at 4:59:20 PM UTC+1, Jew Lover wrote:
> On Saturday, 12 January 2019 10:52:30 UTC-5, Me wrote:
> >
> > On Saturday, January 12, 2019 at 4:37:51 PM UTC+1, Jew Lover wrote:
> > >
> > > "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
> > >
> > > x is a free variable and h is a bound variable; consequently the value of
> > > this expression depends on the value of x, but there is nothing called h
> > > on which it could depend."
> > >
> > Exactly.
> > >
> > > 1. [...] If h approached anything else BUT ZERO ...
> > >
> > Then we would not get the limit any more (by definition).
Sorry, about that. Sould read:
> On Saturday, 12 January 2019 10:52:30 UTC-5, Me wrote:
> >
> > On Saturday, January 12, 2019 at 4:37:51 PM UTC+1, Jew Lover wrote:
> > >
> > > "f'(x) = lim_(h->0) [f(x+h)-f(x)] / h
> > >
> > > x is a free variable and h is a bound variable; consequently the value of
> > > this expression depends on the value of x, but there is nothing called h
> > > on which it could depend."
> > >
> > Exactly.
> > >
> > > 1. [...] If h approached anything else BUT ZERO ...
> > >
> > Then we would not get the limit any more (by definition).
> > Then we would not get the derivative of f any more (by definition).
It seems that you actually have learned something concerning the (essential) difference between free and bound variables. Well done, man!
Me
Jan 12, 2019, 11:44:28 AM1/12/19
to
On Saturday, January 12, 2019 at 5:16:56 PM UTC+1, gabriel...@gmail.com wrote:
> h can take on ANY value close to x, except 0
Exactly. :-)
> h can take on ANY value close to x, except 0
Hint: This contradicts your former claim:
"In your bogus calculus, you do set h = 0"
I'm glad that we agree now: No, we don't set h = 0 (in this context).
Now...
> > Actually, [a bound variable] is "bound" by a certain "expression".
> > In this case, "h" in the formula "[f(x+h)-f(x)] / h" is bound by
> > the expression "lim_(h->0)".
> >
> h cannot be 0.
> > the expression "lim_(h->0)".
> >
That's a DEFFERENT matter, but still true. Right!
> A variable is a variable.
https://en.wikipedia.org/wiki/Free_variables_and_bound_variables
> In the expression "sum_{k=1}^n", only "n" is a variable.
Nope. We do have two variables here: "k" and "n (if "n" actually is a variable).
> "k" is NOT a variable.
Hint: Consider the expression "k" where "k" is just a variable which occures free in it.
Now if we consider the expression
sum_{k=1}^10 k
"k" is STILL a variable. As you can see, it now occures at two "locations". Moreover NOW it is bound (in this expression) by the expression "sum_{k=1}^10" in front of the expression "k".
Hint: The result, again, is a certain NUMBER. This number does not "depend on k". But "the result" of this summation may, say, depend on "n" with n e IN:
sum_{k=1}^n k .
HERE (in this expression) "n" is a free variable.
HENCE we may as well define/introduce the term "s(n)" and write:
s(n) := sum_{k=1}^n k .
Me
Jan 12, 2019, 11:54:33 AM1/12/19
to
On Saturday, January 12, 2019 at 5:44:28 PM UTC+1, Me wrote:
> On Saturday, January 12, 2019 at 5:16:56 PM UTC+1, gabriel...@gmail.com wrote:
> >
> > In the expression "sum_{k=1}^n", only "n" is a variable.
> >
> Nope. We do have two variables here: "k" and "n (if "n" actually is a
> variable).
> >
> > "k" is NOT a variable.
> >
> Sure it is. :-)
>
> Hint: Consider the expression "k" where "k" is just a variable which occures
> free in it.
>
> Now if we consider the expression
>
> sum_{k=1}^10 k
>
> "k" is STILL a variable. As you can see, it now occures at two "locations".
> Moreover NOW it is bound (in this expression) by the expression
> "sum_{k=1}^10" in front of the expression "k".
Check the first example here:
> On Saturday, January 12, 2019 at 5:16:56 PM UTC+1, gabriel...@gmail.com wrote:
> >
> > In the expression "sum_{k=1}^n", only "n" is a variable.
> >
> Nope. We do have two variables here: "k" and "n (if "n" actually is a
> variable).
> >
> > "k" is NOT a variable.
> >
> Sure it is. :-)
>
> Hint: Consider the expression "k" where "k" is just a variable which occures
> free in it.
>
> Now if we consider the expression
>
> sum_{k=1}^10 k
>
> "k" is STILL a variable. As you can see, it now occures at two "locations".
> Moreover NOW it is bound (in this expression) by the expression
> "sum_{k=1}^10" in front of the expression "k".
https://en.wikipedia.org/wiki/Free_variables_and_bound_variables#Examples
Me
Jan 12, 2019, 12:10:25 PM1/12/19
to
On Saturday, January 12, 2019 at 5:44:28 PM UTC+1, Me wrote:
> On Saturday, January 12, 2019 at 5:16:56 PM UTC+1, gabriel...@gmail.com wrote:
> >
> > h can take on ANY value close to x, except 0
> >
> Exactly. :-)
> >
> > h cannot be 0.
> >
> > h can take on ANY value close to x, except 0
> >
> Exactly. :-)
> >
Right. I'm glad we could clarify this misconception on your part.
Peter Percival
Jan 12, 2019, 3:17:52 PM1/12/19
to
Jew Lover wrote:
> " f'(x) = Lim (h->0) [f(x+h)-f(x)] / h
>
> x is a free variable and h is a bound variable; consequently the value of this expression depends on the value of x, but there is nothing called h on which it could depend."
One way of seeing that it doesn't depend on h is to change 'h' to
> " f'(x) = Lim (h->0) [f(x+h)-f(x)] / h
>
> x is a free variable and h is a bound variable; consequently the value of this expression depends on the value of x, but there is nothing called h on which it could depend."
something else. The above statement is the same as this one:
f'(x) = Lim (k->0) [f(x+k)-f(x)] / k
in which 'h' has been replaced by 'k'. [One cannot change things ad
lib, e.g., 'h' cannot be changed to 'x'.]
>
> 1. Does not the limit depend on h? If h approached anything else BUT ZERO
> , would that expression still be true? Ha, ha.
>
> 2. Does not the limit depend on EVERY possible value of h (except the one that actually matters, i.e. ZERO!) because it is in fact the limit of the FINITE DIFFERENCE quotients that is f '(x)?
>
> Chuckle.
>
"He who will not reason is a bigot;
he who cannot is a fool;
he who dares not is a slave."
- Sir William Drummond
Dan Christensen
Jan 12, 2019, 3:57:51 PM1/12/19
to
Geez, Troll Boy, if you couldn't get any converts to your Wacky New Calclueless at YT, I would think your prospects at sci.math are 100 times worse. Everyone here but BKK is wise to you, and he is totally f---ing insane!
Dan
Dan
Jew Lover
Jan 12, 2019, 9:45:04 PM1/12/19
to
On Saturday, 12 January 2019 15:17:52 UTC-5, Peter Percival wrote:
> Jew Lover wrote:
> > " f'(x) = Lim (h->0) [f(x+h)-f(x)] / h
> >
> > x is a free variable and h is a bound variable; consequently the value of this expression depends on the value of x, but there is nothing called h on which it could depend."
>
> One way of seeing that it doesn't depend on h is to change 'h' to
> something else. The above statement is the same as this one:
>
> f'(x) = Lim (k->0) [f(x+k)-f(x)] / k
>
> in which 'h' has been replaced by 'k'. [One cannot change things ad
> lib, e.g., 'h' cannot be changed to 'x'.]
It is not talking about changing the name of h, but rather its value. Since the value of h changes in such a way that the finite difference quotient approaches a certain limit, it is pretty clear that the limit depends on h for it (the limit) cannot be realised if h is not changing in a certain way, that is, getting smaller.
> Jew Lover wrote:
> > " f'(x) = Lim (h->0) [f(x+h)-f(x)] / h
> >
> > x is a free variable and h is a bound variable; consequently the value of this expression depends on the value of x, but there is nothing called h on which it could depend."
>
> One way of seeing that it doesn't depend on h is to change 'h' to
> something else. The above statement is the same as this one:
>
> f'(x) = Lim (k->0) [f(x+k)-f(x)] / k
>
> in which 'h' has been replaced by 'k'. [One cannot change things ad
> lib, e.g., 'h' cannot be changed to 'x'.]
>
> >
> > 1. Does not the limit depend on h? If h approached anything else BUT ZERO
>
> That's right. It *does* depend on 0 (which isn't h).
f(x,h)= [ f(x+h)-f(x) ] / h
f'(x) = f'(x) + Q(x,h) which is only possible if Q(x,h)=0 meaning h=0.
Therefore, you have NO systematic way of finding the derivative. Your first principles method is a load of crap.
Epsilon-delta arguments are "verifinitions", not definitions, and what is more, they are circular because you need to know the limit before you can show that your first principles guess is true.
>
>
> > , would that expression still be true? Ha, ha.
> >
> > 2. Does not the limit depend on EVERY possible value of h (except the one that actually matters, i.e. ZERO!) because it is in fact the limit of the FINITE DIFFERENCE quotients that is f '(x)?
> >
> > Chuckle.
> >
>
>
> --
> "He who will not reason is a bigot;
> he who cannot is a fool;
> he who dares not is a slave."
> - Sir William Drummond
Perhaps a more serious attempt? Chuckle.
> >
> > 2. Does not the limit depend on EVERY possible value of h (except the one that actually matters, i.e. ZERO!) because it is in fact the limit of the FINITE DIFFERENCE quotients that is f '(x)?
> >
> > Chuckle.
> >
>
>
> --
> "He who will not reason is a bigot;
> he who cannot is a fool;
> he who dares not is a slave."
> - Sir William Drummond
Jew Lover
Jan 12, 2019, 9:49:53 PM1/12/19
to
On Thursday, 10 January 2019 07:59:49 UTC-5, Jew Lover wrote:
> Mainstream mathematics academics are incorrigibly stupid creatures. They will tell you that a straight line function has a derivative and change the definition of derivative in terms of their bullshit "analytic derivative".
>
> Let's see how the orangutans fail!
>
> The slope of a straight line is given by:
>
> [ f(x+h)-f(x) ] / h [A]
>
> The derivative of a straight line is given by:
>
> Lim (h -> 0) [ f(x+h)-f(x) ] / h [B]
>
> Therefore, since a straight line's derivative is its slope, we have:
>
> Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
>
> Can you get any dumber? I have produced a New Calculus which has no ill-formed concepts and have solved the tangent line problem and removed ALL ill-formed concepts such as infinity, infinitesimals and limit theory from differential and integral calculus.
>
> A derivative can be found geometrically using Ancient Greek mathematics: https://lnkd.in/dhN6Gzw
>
> Download the most important mathematics book ever written here (it's free!!!):
>
> https://lnkd.in/dEkeshd
See, it's not possible to refute facts. You can make up rules/decrees, but these have no place in mathematics. Only facts and logic are allowed in mathematics. Ill-formed definitions are rejected.
1: f(x,h)= [ f(x+h)-f(x) ] / h
2: f'(x) = f'(x) + Q(x,h) which is only possible if Q(x,h)=0 meaning h=0.
> Mainstream mathematics academics are incorrigibly stupid creatures. They will tell you that a straight line function has a derivative and change the definition of derivative in terms of their bullshit "analytic derivative".
>
> Let's see how the orangutans fail!
>
> The slope of a straight line is given by:
>
> [ f(x+h)-f(x) ] / h [A]
>
> The derivative of a straight line is given by:
>
> Lim (h -> 0) [ f(x+h)-f(x) ] / h [B]
>
> Therefore, since a straight line's derivative is its slope, we have:
>
> Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
>
> Can you get any dumber? I have produced a New Calculus which has no ill-formed concepts and have solved the tangent line problem and removed ALL ill-formed concepts such as infinity, infinitesimals and limit theory from differential and integral calculus.
>
> A derivative can be found geometrically using Ancient Greek mathematics: https://lnkd.in/dhN6Gzw
>
> Download the most important mathematics book ever written here (it's free!!!):
>
> https://lnkd.in/dEkeshd
See, it's not possible to refute facts. You can make up rules/decrees, but these have no place in mathematics. Only facts and logic are allowed in mathematics. Ill-formed definitions are rejected.
1: f(x,h)= [ f(x+h)-f(x) ] / h
2: f'(x) = f'(x) + Q(x,h) which is only possible if Q(x,h)=0 meaning h=0.
666
Jan 13, 2019, 1:35:02 AM1/13/19
to
perjantai 11. tammikuuta 2019 18.09.09 UTC+2 Me kirjoitti:
> On Friday, January 11, 2019 at 4:47:37 PM UTC+1, Jew Lover wrote:
> > In calculus, you do set h = 0
>
> Nope! h = 0 is explicitly excluded when calculating the limit
>
> lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
>
> idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
lim_(h -> 0) [ f(x+h)-f(x) ] / h = 0/0
if
Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
we have
[ f(x+h)-f(x) ] / h = 0/0
and is possible only if you set h = 0
> On Friday, January 11, 2019 at 4:47:37 PM UTC+1, Jew Lover wrote:
> > In calculus, you do set h = 0
>
> Nope! h = 0 is explicitly excluded when calculating the limit
>
> lim_(h -> 0) [ f(x+h)-f(x) ] / h ,
>
> idiot! (Hint: f(x+h)-f(x) ] / h is not defined for h = 0.)
lim_(h -> 0) [ f(x+h)-f(x) ] / h = 0/0
if
Lim (h -> 0) [ f(x+h)-f(x) ] / h = [ f(x+h)-f(x) ] / h
[ f(x+h)-f(x) ] / h = 0/0
and is possible only if you set h = 0
Me
Jan 13, 2019, 5:13:55 AM1/13/19
to
On Sunday, January 13, 2019 at 3:45:04 AM UTC+1, Jew Lover wrote:
> the only way you have of finding the derivative, you set h equal to 0.
Nonsense. Back at square A, crank?
> the only way you have of finding the derivative, you set h equal to 0.
Me
Jan 13, 2019, 5:21:36 AM1/13/19
to
On Sunday, January 13, 2019 at 3:49:53 AM UTC+1, Jew Lover wrote:
> The slope of a straight line is given by:
>
> [ f(x+c)-f(x) ] / c [A]
> The slope of a straight line is given by:
>
for some c e IR, c > 0.
> The derivative of a straight line is given by:
>
Right.
> Therefore, since a straight line's derivative is its slope, we have:
>
for any c e IR, c > 0.
Right.
> A derivative can be found geometrically [in this case].
Right.
> Ill-formed definitions are rejected [in mathematics].
Indeed! That's the reason why we have to reject the following blunder:
> 1: f(x,h) = [ f(x+h)-f(x) ] / h
Nope. Since at the rhs of the equation, f is a function with just ONE argument while at the lhs of the equation, f is a function with TWO arguments. That just NONSENSE.
Actually, we have:
f'(x) = lim_(h->0) [ f(x+h)-f(x) ] / h
See: https://en.wikipedia.org/wiki/Free_variables_and_bound_variables
Now you considert the trivial equation:
> f'(x) = f'(x) + Q which is only possible if Q = 0
Yeah, that's right. :-)
For all r,c e IR: if r + c = r, then c = 0.
Me
Jan 13, 2019, 5:36:40 AM1/13/19
to
On Sunday, January 13, 2019 at 7:35:02 AM UTC+1, 666 wrote:
> [nonsense]
Running low on medication?
> [nonsense]
Running low on medication?
Me
Jan 13, 2019, 5:48:58 AM1/13/19
to
> On Saturday, January 12, 2019 at 5:16:56 PM UTC+1, gabriel...@gmail.com
> wrote:
> >
> > h can take on ANY value close to x, except 0
> >
> >
> > h cannot be 0.
666
Jan 13, 2019, 6:10:13 AM1/13/19
to
Jew Lover
Jan 13, 2019, 8:14:46 AM1/13/19
to
1: f(x,h)= [ f(x+h)-f(x) ] / h
which is only possible if Q(x,h)=0 meaning h=0. Through sheer luck, Newton knew he was finding the general derivative even though his method is kludgy.
Today's academic morons know this too and that's why they added a whole lot of limit theory in an attempt to make "rigorous" their bogus calculus.
You seem to know that the mainstream definition is shit. But you still have a problem with infinity being a junk concept.
The mainstream orangutans don't claim that h becomes 0, but in order to find f'(x), they have to set h=0. They have no systematic way of finding f'(x) as I have shown is possible in the New Calculus, the first and only rigorous formulation in human history.
Jew Lover
Jan 13, 2019, 8:17:51 AM1/13/19
to
They hate me because I instantly dismiss their bullshit. It would actually have been much easier for them to admit they are wrong first and then they might stand a chance of learning.
Jew Lover
Jan 13, 2019, 8:27:03 AM1/13/19
to
On Sunday, 13 January 2019 05:21:36 UTC-5, Me wrote:
<much whining and bla, bla>
> Now you considert the trivial equation:
>
> > f'(x) = f'(x) + Q which is only possible if Q = 0
>
> Yeah, that's right. :-)
Now think moron: What is the only way that Q(x,h) can be 0? Yes! h = 0.
It doesn't help your cause that you call h bound or free at all. Hand waving is as transparent as glass.
See, it's not possible to refute facts. You can make up rules/decrees, but these have no place in mathematics. Only facts and logic are allowed in mathematics. Ill-formed definitions are rejected.
1: f(x,h)= [ f(x+h)-f(x) ] / h
2: f'(x) = f'(x) + Q(x,h) which is only possible if Q(x,h)=0 meaning h=0.
Therefore, you have NO systematic way of finding the derivative. So far, you haven't even attempted to refute this fact. Your first principles method is a load of crap.
<much whining and bla, bla>
> Now you considert the trivial equation:
>
> > f'(x) = f'(x) + Q which is only possible if Q = 0
>
> Yeah, that's right. :-)
It doesn't help your cause that you call h bound or free at all. Hand waving is as transparent as glass.
See, it's not possible to refute facts. You can make up rules/decrees, but these have no place in mathematics. Only facts and logic are allowed in mathematics. Ill-formed definitions are rejected.
1: f(x,h)= [ f(x+h)-f(x) ] / h
2: f'(x) = f'(x) + Q(x,h) which is only possible if Q(x,h)=0 meaning h=0.
j4n bur53
Jan 13, 2019, 8:46:09 AM1/13/19
to
Q(x,h) := f[x,x+h] - f'(x)
Where: f(x) - f(y)
/ ----------- x <> y
f[x,y] = < x - y
\
f'(x) x = y
The operator "divdiff" f[x,y] is well knowen it, has
a lot of laws that are very similar to derivative.
See for example:
Symbolic Computation of Divided Differences
W. Kahan and Richard J. Fateman - 1999
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.9483&rank=1
But its much older.
Where: f(x) - f(y)
/ ----------- x <> y
f[x,y] = < x - y
\
f'(x) x = y
The operator "divdiff" f[x,y] is well knowen it, has
a lot of laws that are very similar to derivative.
See for example:
Symbolic Computation of Divided Differences
W. Kahan and Richard J. Fateman - 1999
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.9483&rank=1
But its much older.
j4n bur53
Jan 13, 2019, 8:54:31 AM1/13/19
to
Corr:
D(x,h) := f[x,x+h]
Then:
f(x+h) = f(x) + D(x,h)*h
D(x,0) = f'(x)
D(x,h) := f[x,x+h]
Then:
f(x+h) = f(x) + D(x,h)*h
D(x,0) = f'(x)
j4n bur53
Jan 13, 2019, 9:01:16 AM1/13/19
to
divdiff satisfies a form of chain rule:
f(g)[x, y] = f[g(x), g(y)] g[x, y]
See section 1.4 Rules of the paper.
For x = y we get the special case:
f(g)'(x) = f'(g(x)) g'(x)
https://en.wikipedia.org/wiki/Chain_rule
f(g)[x, y] = f[g(x), g(y)] g[x, y]
See section 1.4 Rules of the paper.
For x = y we get the special case:
f(g)'(x) = f'(g(x)) g'(x)
https://en.wikipedia.org/wiki/Chain_rule
666
Jan 13, 2019, 9:13:39 AM1/13/19
to
sunnuntai 13. tammikuuta 2019 15.14.46 UTC+2 Jew Lover kirjoitti:
> On Sunday, 13 January 2019 06:10:13 UTC-5, 666 wrote:
> > sunnuntai 13. tammikuuta 2019 12.36.40 UTC+2 Me kirjoitti:
> > > On Sunday, January 13, 2019 at 7:35:02 AM UTC+1, 666 wrote:
> > >
> > > > [nonsense]
> > >
> > > Running low on medication?
> >
> > maybe you have never heard of limits:
> >
> > lim_(h -> 0) [ f(x+h)-f(x) ] / h = 0/0
>
> They don't do it that way. They do it this way:
>
> 1: f(x,h)= [ f(x+h)-f(x) ] / h
no, it makes no sense.[f(x+h)-f(x)]/h is a finite difference quotient:
> On Sunday, 13 January 2019 06:10:13 UTC-5, 666 wrote:
> > sunnuntai 13. tammikuuta 2019 12.36.40 UTC+2 Me kirjoitti:
> > > On Sunday, January 13, 2019 at 7:35:02 AM UTC+1, 666 wrote:
> > >
> > > > [nonsense]
> > >
> > > Running low on medication?
> >
> > maybe you have never heard of limits:
> >
> > lim_(h -> 0) [ f(x+h)-f(x) ] / h = 0/0
>
> They don't do it that way. They do it this way:
>
> 1: f(x,h)= [ f(x+h)-f(x) ] / h
Delta f/ Delta h = [f(x+h)-f(x)]/h
>
>
> 2: f'(x) = f'(x) + Q(x,h)
>
> which is only possible if Q(x,h)=0 meaning h=0. Through sheer luck, Newton knew he was finding the general derivative even though his method is kludgy.
so what?
>
> Today's academic morons know this too and that's why they added a whole lot of limit theory in an attempt to make "rigorous" their bogus calculus.
>
> You seem to know that the mainstream definition is shit.
lim_(h -> 0) [f(x+h)-f(x)]/h = 0/0
> But you still have a problem with infinity being a junk concept.
Until you learn to handle it, you can't call yourself a mathematician.
> The mainstream orangutans don't claim that h becomes 0, but in order to find f'(x), they have to set h=0.
Lim (h->0) [f(x+h)-f(x)]/h = [f(x+h)-f(x)]/h
we have
[f(x+h)-f(x)]/h = 0/0
and is possible only if you set h = 0
>They have no systematic way of finding f'(x) as I have shown is possible in the New Calculus, the first and only rigorous formulation in human history.
You never calculate the slope of the tangent line. You end up calculating
the slope of a non-parallel secant line. The reason is:
your rejection of infinitesimals.
Me
Jan 13, 2019, 9:41:49 AM1/13/19
to
On Sunday, January 13, 2019 at 2:14:46 PM UTC+1, Jew Lover wrote:
We started with
f(x) := m*x + d
where m,d e IR. (Then f'(x) is defined for all x e IR.)
Now you consider the following equation:
> f'(x) = f'(x) + Q(x,h) [for all x,h e IR]
>
> [This] is only possible if Q(x,h) = 0 [for all x,h e IR].
Indeed!
> meaning h = 0.
Of course not, idiot.
Actually, it suffices that Q(x,h) = 0 for all x,h e IR.
Hence Q(x,h) may be
x*(h - h)
for example. Then the equation
f'(x) = f'(x) + Q(x,h)
holds for ALL x e IR and _all_ h e IR. Hence h = 0 does NOT follow (contrary to your claim).
We started with
f(x) := m*x + d
where m,d e IR. (Then f'(x) is defined for all x e IR.)
Now you consider the following equation:
> f'(x) = f'(x) + Q(x,h) [for all x,h e IR]
>
> [This] is only possible if Q(x,h) = 0 [for all x,h e IR].
Indeed!
> meaning h = 0.
Of course not, idiot.
Actually, it suffices that Q(x,h) = 0 for all x,h e IR.
Hence Q(x,h) may be
x*(h - h)
for example. Then the equation
f'(x) = f'(x) + Q(x,h)
j4n bur53
Jan 13, 2019, 9:42:21 AM1/13/19
to
But divdiff could be also helpful with your Geogebra
tricks bird brain John Garbageiel. Lets say we
want a secant parallel to f'(x). Then we have to
solve the following:
f[x-m,x+n] = f'(x)
Which has a solution sometimes, but not always
for lets say a given m. For example inflection points
cause problems with your secant nonsense.
But if we have solved the above equation, then
obviously, setting m,n=0,0 in divdiff, gives
the folllowing:
f[x-m,x+n]m,n=0,0 = f[x,x] = f'(x)
This is independent whether you have solved
the equation or not. Because the equation is
obviously satisfied for m,n=0,0.
Lets say you have solved the equation:
f[x-m,x+n] = f'(x)
Then your secant would trivially read:
s(x_) = f[x-m,x+n]*(x_-(x-m))+f(x-m)
s(x-m) = f(x-m)
s(x+n) = f(x+n)
s'(x_) = f[x-m,x+n] = f'(x)
But how solve the equation?
tricks bird brain John Garbageiel. Lets say we
want a secant parallel to f'(x). Then we have to
solve the following:
f[x-m,x+n] = f'(x)
Which has a solution sometimes, but not always
for lets say a given m. For example inflection points
cause problems with your secant nonsense.
But if we have solved the above equation, then
obviously, setting m,n=0,0 in divdiff, gives
the folllowing:
f[x-m,x+n]m,n=0,0 = f[x,x] = f'(x)
This is independent whether you have solved
the equation or not. Because the equation is
obviously satisfied for m,n=0,0.
Lets say you have solved the equation:
f[x-m,x+n] = f'(x)
Then your secant would trivially read:
s(x_) = f[x-m,x+n]*(x_-(x-m))+f(x-m)
s(x-m) = f(x-m)
s(x+n) = f(x+n)
s'(x_) = f[x-m,x+n] = f'(x)
But how solve the equation?
Message has been deleted
Me
Jan 13, 2019, 9:50:52 AM1/13/19
to
On Sunday, January 13, 2019 at 2:27:03 PM UTC+1, Jew Lover wrote:
> On Sunday, 13 January 2019 05:21:36 UTC-5, Me wrote:
> >
> > Now you considert the trivial equation:
> > >
> > > f'(x) = f'(x) + Q which is only possible if Q = 0
> > >
> > Yeah, that's right. :-)
> >
> Now think moron: What is the only way that Q(x,h) can be 0?
Depends. If Q(x,h) = x*(h - h) there are INFINTELY MANY ways. :-)
> On Sunday, 13 January 2019 05:21:36 UTC-5, Me wrote:
> >
> > Now you considert the trivial equation:
> > >
> > > f'(x) = f'(x) + Q which is only possible if Q = 0
> > >
> > Yeah, that's right. :-)
> >
> Now think moron: What is the only way that Q(x,h) can be 0?
Hint: In this case
f'(x) = f'(x) + Q(x,h)
> It doesn't help your cause that you call h bound or free at all.
j4n bur53
Jan 13, 2019, 9:52:01 AM1/13/19
to
But how play your Geogebra demonstration tricks?
You need to solve:
f[x-m,x+n] = (f(x-m)-f(x+n))/(x-m-x-n)
= (f(x+n)-f(x-m))/(m+n) = f'(x)
Hence:
f(x+n)-f(x-m) = f'(x)*(m+n)
Hence:
f(x+n)-f'(x)*n = f(x-m)+f'(x)*m
So you can look at this function:
g(n) = f(x+n) - f'(x)*n
If you can solve it, take a root different
of -m, in case it exists:
n = g^(-1)(g(m))
You need to solve:
f[x-m,x+n] = (f(x-m)-f(x+n))/(x-m-x-n)
= (f(x+n)-f(x-m))/(m+n) = f'(x)
Hence:
f(x+n)-f(x-m) = f'(x)*(m+n)
Hence:
f(x+n)-f'(x)*n = f(x-m)+f'(x)*m
So you can look at this function:
g(n) = f(x+n) - f'(x)*n
If you can solve it, take a root different
of -m, in case it exists:
n = g^(-1)(g(m))
j4n bur53
Jan 13, 2019, 9:53:02 AM1/13/19
to
Corr.:
n = g^(-1)(g(-m))
n = g^(-1)(g(-m))
Message has been deleted
Me
Jan 13, 2019, 9:58:35 AM1/13/19
to
On Sunday, January 13, 2019 at 2:14:46 PM UTC+1, Jew Lover wrote:
> 1: f(x,h)= [ f(x+h)-f(x) ] / h
You know, there's a difference, troll boy.
No function which takes two arguments can be _identical_ with a function that only takes one argument.
Hint: "A function that takes a single argument as input (such as f(x) = x^2) is called a unary function. A function of two or more variables is considered to have a domain consisting of ordered pairs or tuples of argument values. "
See: https://en.wikipedia.org/wiki/Argument_of_a_function
Me
Jan 13, 2019, 10:02:08 AM1/13/19
to
On Sunday, January 13, 2019 at 3:13:39 PM UTC+1, 666 wrote:
> no-one seems to be able to take the limit:
>
> [...] = 0/0
> no-one seems to be able to take the limit:
>
The reason for this is, that "0/0" is considered UNDEFINED in the context of real analysis. Hence there's no such limit.
j4n bur53
Jan 13, 2019, 10:03:56 AM1/13/19
to
Example this trivial function often seen
in John Garbageiels demonstration Geogebra:
f(x) = x^2
Now we have:
g(-m) = (x-m)^2 + 2*x*m
= x^2 - 2*x*m + m^2 + 2*x*n
= x^2 + m^2.
Which can be easily inverted:
/ sqrt(v - x^2)
g^(-1)(v) = <
\ -sqrt(v - x^2)
So that for this function we have:
f[x-m,x+m] = f'(x)
Right?
Lets check:
(x-m)^2 - (x+m)^2 - 4*x*m
----------------- = ------- = 2*x
x-m - x - m - 2*m
Yep!
in John Garbageiels demonstration Geogebra:
f(x) = x^2
Now we have:
g(-m) = (x-m)^2 + 2*x*m
= x^2 - 2*x*m + m^2 + 2*x*n
= x^2 + m^2.
Which can be easily inverted:
/ sqrt(v - x^2)
g^(-1)(v) = <
\ -sqrt(v - x^2)
So that for this function we have:
f[x-m,x+m] = f'(x)
Right?
Lets check:
(x-m)^2 - (x+m)^2 - 4*x*m
----------------- = ------- = 2*x
x-m - x - m - 2*m
Yep!
j4n bur53
Jan 13, 2019, 10:10:09 AM1/13/19
to
See page 98 ff of John Gabriels compiled nonsense:
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
Message has been deleted
Me
Jan 13, 2019, 10:27:08 AM1/13/19
to
On Sunday, January 13, 2019 at 2:14:46 PM UTC+1, Jew Lover wrote:
> [Mathematicians] don't claim that h becomes 0
Yeah, since this case is EXPLICITLY excluded in this context.
> but in order to find f'(x), they have to set h=0.
Nope. Where did get that nonsensical idea from?
> They have no systematic way of finding f'(x)
If course, we have. Want to see how we get f' for f defined with
f(x) = m*x + d (x e IR)
where m,d are some (fixed) elements in IR?
Here it is:
f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(m*(x + h) + d) - (m*x + d)] / h} = lim_(h->0) {[m*x + m*h + d - m*x - d] / h} = lim_(h->0) {m*h / h} = lim_(h->0) {m} = m.
So f' is just a constant function with f'(x) = m for all x e IR. m is the slope of the straight line "described" by the equation
m*x + d ,
for x e IR (and with fixed m,d e IR).
Rememeber, you wrote:
> Therefore, since a straight line's derivative is its slope, we have:
>
> lim_(h -> 0) {[ f(x+h)-f(x) ] / h} = [ f(x+c)-f(x) ] / c
for any c e IR, c > 0.
We may check this here. Let c e IR, c > 0, then
[ f(x+c)-f(x) ] / c = [(m*(x + c) + d) - (m*x + d)] / c = [m*x + m*c + d - m*x - d] / c = m*c / c = m (since c =/= 0).
With other words, f'(x) = [ f(x+c)-f(x) ] / c for any c e IR, c > 0, and all x e IR.
qed
> [Mathematicians] don't claim that h becomes 0
Yeah, since this case is EXPLICITLY excluded in this context.
> but in order to find f'(x), they have to set h=0.
> They have no systematic way of finding f'(x)
f(x) = m*x + d (x e IR)
where m,d are some (fixed) elements in IR?
Here it is:
f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(m*(x + h) + d) - (m*x + d)] / h} = lim_(h->0) {[m*x + m*h + d - m*x - d] / h} = lim_(h->0) {m*h / h} = lim_(h->0) {m} = m.
So f' is just a constant function with f'(x) = m for all x e IR. m is the slope of the straight line "described" by the equation
m*x + d ,
for x e IR (and with fixed m,d e IR).
Rememeber, you wrote:
> Therefore, since a straight line's derivative is its slope, we have:
>
for any c e IR, c > 0.
[ f(x+c)-f(x) ] / c = [(m*(x + c) + d) - (m*x + d)] / c = [m*x + m*c + d - m*x - d] / c = m*c / c = m (since c =/= 0).
With other words, f'(x) = [ f(x+c)-f(x) ] / c for any c e IR, c > 0, and all x e IR.
qed
Me
Jan 13, 2019, 10:41:47 AM1/13/19
to
On Sunday, January 13, 2019 at 4:10:09 PM UTC+1, j4n bur53 wrote:
> See page 98 ff of John Gabriels compiled nonsense:
> https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
Mind blowing stuff. Not even a single reasonable (or "rational") line.
> See page 98 ff of John Gabriels compiled nonsense:
> https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
Jew Lover
Jan 13, 2019, 12:02:07 PM1/13/19
to
On Sunday, 13 January 2019 09:41:49 UTC-5, Me wrote:
<crap>
> Now you consider the following equation:
>
> > f'(x) = f'(x) + Q(x,h) [for all x,h e IR]
> >
> > [This] is only possible if Q(x,h) = 0 [for all x,h e IR].
>
> Indeed!
>
> > meaning h = 0.
>
> Of course not...
Of course YES. Look you mega moron, the only way Q(x,h) can be 0 is if h=0. Otherwise f'(x) =/= f'(x) + Q(x,h) EVER!
<crap>
> Now you consider the following equation:
>
> > f'(x) = f'(x) + Q(x,h) [for all x,h e IR]
> >
> > [This] is only possible if Q(x,h) = 0 [for all x,h e IR].
>
> Indeed!
>
> > meaning h = 0.
>
Of course YES. Look you mega moron, the only way Q(x,h) can be 0 is if h=0. Otherwise f'(x) =/= f'(x) + Q(x,h) EVER!
Jew Lover
Jan 13, 2019, 12:02:47 PM1/13/19
to
Jew Lover
Jan 13, 2019, 12:08:50 PM1/13/19
to
On Sunday, 13 January 2019 10:27:08 UTC-5, Me wrote:
> Yeah, since this case is EXPLICIT...
<rot>
f'(x)=lim (h->oo) [f(x+h)-f(x)]/h is very embarrassing to you.
It's as if you produced a massive turd that stinks more than anything on the planet and so you are trying to hide it by spraying cologne thereon. Tsk, tsk. That strategy hasn't worked for you that well, has it? Chuckle.
f'(x)=f'(x)+Q(x,h) MEANS h = 0, for if not, then you don't have any way of finding that L (limit) so you can pretend that you are verifying it using your delusional limit theory:
0 < |x - c| < delta => | [f(c+h)-f(c)] / h - L|<epsilon
See, everything you touch is SHIT because you have shit for brains. Chuckle.
> Yeah, since this case is EXPLICIT...
<rot>
f'(x)=lim (h->oo) [f(x+h)-f(x)]/h is very embarrassing to you.
It's as if you produced a massive turd that stinks more than anything on the planet and so you are trying to hide it by spraying cologne thereon. Tsk, tsk. That strategy hasn't worked for you that well, has it? Chuckle.
f'(x)=f'(x)+Q(x,h) MEANS h = 0, for if not, then you don't have any way of finding that L (limit) so you can pretend that you are verifying it using your delusional limit theory:
0 < |x - c| < delta => | [f(c+h)-f(c)] / h - L|<epsilon
See, everything you touch is SHIT because you have shit for brains. Chuckle.
Me
Jan 13, 2019, 12:15:23 PM1/13/19
to
On Sunday, January 13, 2019 at 6:02:07 PM UTC+1, Jew Lover wrote:
> On Sunday, 13 January 2019 09:41:49 UTC-5, Me wrote:
> <crap>
>
> > Now you consider the following equation:
> >
> > > f'(x) = f'(x) + Q(x,h) [for all x,h e IR]
> > >
> > > [This] is only possible if Q(x,h) = 0 [for all x,h e IR].
> >
> > Indeed!
> > >
> > > meaning h = 0.
> > >
> > Of course not...
> >
> Of course YES.
No.
> On Sunday, 13 January 2019 09:41:49 UTC-5, Me wrote:
> <crap>
>
> > Now you consider the following equation:
> >
> > > f'(x) = f'(x) + Q(x,h) [for all x,h e IR]
> > >
> > > [This] is only possible if Q(x,h) = 0 [for all x,h e IR].
> >
> > Indeed!
> > >
> > > meaning h = 0.
> > >
> > Of course not...
> >
> Of course YES.
> Look, the only way Q(x,h) can be 0 is if h = 0.
No.
How about Q(x,h) = x*(h - h)?
Clearly Q(x,h) = 0 for all x,h e IR.
Hence
f'(x) = f'(x) + Q(x,h) for _all_ h e IR and x e D(f')
in this case.
Me
Jan 13, 2019, 12:17:52 PM1/13/19
to
On Sunday, January 13, 2019 at 6:08:50 PM UTC+1, Jew Lover wrote:
> then you don't have any way of finding that L (limit) so <bla>
> then you don't have any way of finding that L (limit) so <bla>
Message has been deleted
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Me
Jan 13, 2019, 2:15:52 PM1/13/19
to
On Sunday, January 13, 2019 at 6:08:50 PM UTC+1, Jew Lover wrote:
f(x) = m*x + d (for all x e IR)
We say that
lim_(h->0) [f(x+h) - f(x)] / h = L
if for every epsilon > 0 there exists a delta > 0 such that, for all h =/= 0,
> if 0 < |h| < delta, then |[f(x+h) - f(x)] / h - L| < epsilon
Let's see, if L = m will satisfy this condition. :-)
We had that for all h e IR, h =/= 0: [f(x+h) - f(x)] / h = ... = m.
Hence if L = m, the condition becomes:
> for every epsilon > 0 there exists a delta such that, for all h =/= 0,
> if 0 < |h| < delta, then |0| < epsilon
Since for every epsilon > 0, |0| < epsilon holds, the condition is satisfied.
With other words,
lim_(h->0) [f(x+h) - f(x)] / h = m .
So where's your problem, troll boy?
Note that in the conditions above h = 0 is explicitly excluded.
f(x) = m*x + d (for all x e IR)
We say that
lim_(h->0) [f(x+h) - f(x)] / h = L
if for every epsilon > 0 there exists a delta > 0 such that, for all h =/= 0,
> if 0 < |h| < delta, then |[f(x+h) - f(x)] / h - L| < epsilon
Let's see, if L = m will satisfy this condition. :-)
We had that for all h e IR, h =/= 0: [f(x+h) - f(x)] / h = ... = m.
Hence if L = m, the condition becomes:
> for every epsilon > 0 there exists a delta such that, for all h =/= 0,
> if 0 < |h| < delta, then |0| < epsilon
Since for every epsilon > 0, |0| < epsilon holds, the condition is satisfied.
With other words,
lim_(h->0) [f(x+h) - f(x)] / h = m .
So where's your problem, troll boy?
Note that in the conditions above h = 0 is explicitly excluded.
Jew Lover
Jan 13, 2019, 2:22:03 PM1/13/19
to
On Sunday, 13 January 2019 14:15:52 UTC-5, Me wrote:
> We baboons, say that
>
> if for every epsilon > 0 there exists a delta > 0 such that, for all h =/= 0,
>
> > if 0 < |h| < delta, then |[f(x+h) - f(x)] / h - L| < epsilon
No moron. This is circular. You use that faux "definition" to find f'(x) which is the L you are using. Moron! You have NO systematic way of finding L. Get it ballsack?
<mind numbing stupidity>
> We baboons, say that
>
> lim_(h->0) [f(x+h) - f(x)] / h = L
Fail. f'(x)=L is what you need to determine first, you dumb cunt.
> lim_(h->0) [f(x+h) - f(x)] / h = L
>
> if for every epsilon > 0 there exists a delta > 0 such that, for all h =/= 0,
>
> > if 0 < |h| < delta, then |[f(x+h) - f(x)] / h - L| < epsilon
<mind numbing stupidity>
Jew Lover
Jan 13, 2019, 2:23:13 PM1/13/19
to
On Sunday, 13 January 2019 14:15:52 UTC-5, Me wrote:
<crapola...>
Me
Jan 13, 2019, 2:50:04 PM1/13/19
to
On Sunday, January 13, 2019 at 8:22:03 PM UTC+1, Jew Lover wrote:
> On Sunday, 13 January 2019 14:15:52 UTC-5, Me wrote:
> >
> > We say that
> On Sunday, 13 January 2019 14:15:52 UTC-5, Me wrote:
> >
> >
> > lim_(h->0) [f(x+h) - f(x)] / h = L
> >
> > [etc.]
> > lim_(h->0) [f(x+h) - f(x)] / h = L
> >
> >
> f'(x) = L is what you need to determine
Exactly. Since f'(x) = lim_(h->0) [f(x+h) - f(x)] / h BY DEFINITION, all I need to do is to check if for some suitable L:
lim_(h->0) [f(x+h) - f(x)] / h = L .
With other words,
> > if for every epsilon > 0 there exists a delta > 0 such that, for all h =/= 0,
> >
> > > if 0 < |h| < delta, then |[f(x+h) - f(x)] / h - L| < epsilon
lim_(h->0) [f(x+h) - f(x)] / h = m
for f(x) = m*x + d, and hence
f'(x) = m .
To prove that
lim_(h->0) [f(x+h) - f(x)] / h = m
(and Hence f'(x) = m) I only have to show that
> for every epsilon > 0 there exists a delta > 0 such that, for all h =/= 0,
That's all. And that's indeed the case.
> You have NO systematic way of finding L.
Usually we don't proceed this way, but by applying "limit laws".
Want to see how we get f' for f defined with
f(x) = m*x + d (x e IR)
where m,d are some (fixed) elements in IR?
Here it is:
f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(m*(x + h) + d) - (m*x + d)] / h} = lim_(h->0) {[m*x + m*h + d - m*x - d] / h} = lim_(h->0) {m*h / h} = lim_(h->0) {m} = m.
Me
Jan 13, 2019, 2:59:10 PM1/13/19
to
On Sunday, January 13, 2019 at 8:50:04 PM UTC+1, Me wrote:
> Usually we don't proceed this way, but by applying "limit laws".
>
> Want to see how we get f' for f defined with
>
> f(x) = m*x + d (x e IR)
>
> where m,d are some (fixed) elements in IR?
>
> Here it is:
>
> f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(m*(x + h) + d) -
> (m*x + d)] / h} = lim_(h->0) {[m*x + m*h + d - m*x - d] / h} = lim_(h->0)
> {m*h / h} = lim_(h->0) {m} = m.
>
> Hint: We *do* determine f' in a _systematic way_ here.
Here's another example:
> Usually we don't proceed this way, but by applying "limit laws".
>
> Want to see how we get f' for f defined with
>
> f(x) = m*x + d (x e IR)
>
> where m,d are some (fixed) elements in IR?
>
> Here it is:
>
> f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(m*(x + h) + d) -
> (m*x + d)] / h} = lim_(h->0) {[m*x + m*h + d - m*x - d] / h} = lim_(h->0)
> {m*h / h} = lim_(h->0) {m} = m.
>
> Hint: We *do* determine f' in a _systematic way_ here.
f(x) = x^2 (x e IR)
Then f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(x + h)^2 - x^2] / h} = lim_(h->0) {[x^2 + 2xh + h^2 - x^2] / h} = lim_(h->0) {[2xh + h^2] / h} = lim_(h->0) {2x + h} = lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
With other words, f'(x) = 2x for all x e IR.
Where's your problem troll boy?
Jew Lover
Jan 13, 2019, 3:20:09 PM1/13/19
to
On Sunday, 13 January 2019 14:59:10 UTC-5, Me wrote:
> f(x) = x^2 (x e IR)
>
> Then f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(x + h)^2 - x^2] / h} = lim_(h->0) {[x^2 + 2xh + h^2 - x^2] / h} = lim_(h->0) {[2xh + h^2] / h} = lim_(h->0) {2x + h}
Right here you big monkey!
> f(x) = x^2 (x e IR)
>
> Then f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(x + h)^2 - x^2] / h} = lim_(h->0) {[x^2 + 2xh + h^2 - x^2] / h} = lim_(h->0) {[2xh + h^2] / h} = lim_(h->0) {2x + h}
> = lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
YES, YOU SET h = 0, you fucking ape!!
Thanks.
> With other ...
Shut the fuck up idiot. You don't know shit. Learn some English you dumb Kraut. It's not "with other", but "in other".
I don't know why I even bother to waste my time with you.
Once again, YOU HAVE NO SYSTEMATIC WAY OF DETERMINING f'(x). You can't have h =/= 0 before f'(x)+Q(x,h) and then h = 0 afterwards. Unless you set h=0, you NEVER have f'(x), only f'(x)+Q(x,h) which is NOT equal to f'(x).
Now fuck off kindly.
Me
Jan 13, 2019, 4:16:30 PM1/13/19
to
On Sunday, January 13, 2019 at 9:20:09 PM UTC+1, Jew Lover wrote:
> On Sunday, 13 January 2019 14:59:10 UTC-5, Me wrote:
> >
> > f(x) = x^2 (x e IR)
> >
> > Then f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(x + h)^2 -
> > x^2] / h} = lim_(h->0) {[x^2 + 2xh + h^2 - x^2] / h} = lim_(h->0) {[2xh +
> > h^2] / h} = lim_(h->0) {2x + h}
It seems that you have some question.
> On Sunday, 13 January 2019 14:59:10 UTC-5, Me wrote:
> >
> > f(x) = x^2 (x e IR)
> >
> > Then f'(x) = lim_(h->0) {[f(x+h) - f(x)] / h} = lim_(h->0) {[(x + h)^2 -
> > x^2] / h} = lim_(h->0) {[x^2 + 2xh + h^2 - x^2] / h} = lim_(h->0) {[2xh +
> > h^2] / h} = lim_(h->0) {2x + h}
> Right here
> >
> > = lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
> >
> What did you do to h [?]
> > = lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
> >
Let's see. Actually, it's quite obvous, isn't it?
We have
lim_(h->0) {h} = 0 ,
didn't you know that?
Proof:
You know that (by definition) we have
lim_(x->0) f(x) = L
if for every epsilon > 0 there exists a delta > 0 such that, for all x e D(f),
if 0 < |x| < delta, then |f(x) - L| < epsilon .
Now, I claim that for f(x) = x, L = 0. Let's see (and check). If L = 0 the condition becomes
for every epsilon > 0 there exists a delta > 0 such that, for all x e D(f),
if 0 < |x| < delta, then |x| < epsilon
And that is indeed the case, just take delta = epsilon/2 for any epsilon > 0.
So we have PROVED that
lim_(h->0) {h} = 0 .
Got that, you moron?
Hint: In practice, lim_(x->0) x = 0 is just another "limit law".
From THIS we get
lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
Great.
> Once again,
you have seen that we *do* have
> A SYSTEMATIC WAY OF DETERMINING f'(x).
Actually, you may have heard about additional systematic rules for determining f':
https://en.wikipedia.org/wiki/Differentiation_rules
Message has been deleted
Me
Jan 13, 2019, 4:28:59 PM1/13/19
to
On Sunday, January 13, 2019 at 10:16:30 PM UTC+1, Me wrote:
> Hint: In practice, lim_(x->0) x = 0 is just another "limit law".
Actually, it is
> Hint: In practice, lim_(x->0) x = 0 is just another "limit law".
lim_(x->c) x = c .
Another one is
lim_(x->c) a = a (for any constant a).
You really should learn some math, man.
You might start with this text her: "The Limit Laws"
http://www.oxfordmathcenter.com/drupal7/node/95
Jew Lover
Jan 13, 2019, 4:52:24 PM1/13/19
to
On Sunday, 13 January 2019 16:16:30 UTC-5, Me wrote:
<crapola>
> From THIS we get
>
> lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
No moron. h = 0 gives you f'(x) = 2x. The effect is the same and you are simply taking the limit after you reduce the quotient, because if you tried it before, you have no way of finding f'(x)=2x systematically. You can only find f'(x) by using your kludgy circular bullshit definition lim_(h->0) [ f(x+h)-f(x)] / h and breaking arithmetic in the process, which forbids you from dividing by 0!
So, to summarise, you have NO systematic way. What does systematic mean? It means that you perform a sequence of logical steps and arrive at the derivative without breaking any arithmetic.
<crap>
<crapola>
> From THIS we get
>
> lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
So, to summarise, you have NO systematic way. What does systematic mean? It means that you perform a sequence of logical steps and arrive at the derivative without breaking any arithmetic.
<crap>
Jew Lover
Jan 13, 2019, 4:56:50 PM1/13/19
to
To get from lim_(h->0) {2x} + lim_(h->0) {h}
to
= 2x + 0 = 2x
BUT L = f'(x) which is a problem because you still haven't found f'(x). You need to apply your limit process, which can't be applied because you don't know f'(x) or L, the very thing your limit process requires.
What a complete idiot you are.
j4n bur53
Jan 13, 2019, 5:00:22 PM1/13/19
to
May I make a comment? To find f'(x), we do not
always use if 0 < |h| < delta, then |[f(x+h) - f(x)] / h - m| < epsilon .
Since we are not anymore apes and have evolved, our
arsenal is much wider, Chain Rule, Hopital Rule, etc...
This rules in turn can be reduced to some lim definitions.
always use if 0 < |h| < delta, then |[f(x+h) - f(x)] / h - m| < epsilon .
Since we are not anymore apes and have evolved, our
arsenal is much wider, Chain Rule, Hopital Rule, etc...
This rules in turn can be reduced to some lim definitions.
Me
Jan 13, 2019, 5:05:58 PM1/13/19
to
On Sunday, January 13, 2019 at 10:52:24 PM UTC+1, Jew Lover wrote:
> On Sunday, 13 January 2019 16:16:30 UTC-5, Me wrote:
> >
> > From THIS we get
> >
> > lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
> >
> No
Yes. Didn't you see the proof for lim_(h->0) {h} = 0?
> On Sunday, 13 January 2019 16:16:30 UTC-5, Me wrote:
> >
> > From THIS we get
> >
> > lim_(h->0) {2x} + lim_(h->0) {h} = 2x + 0 = 2x.
> >
> No
> So, to summarise, we have a systematic way.
Right.
> What does systematic mean? It means that you perform a sequence of logical
> steps and arrive at the derivative without breaking any arithmetic.
Thanks for your valuable contribution!
Jew Lover
Jan 13, 2019, 5:07:37 PM1/13/19
to
On Sunday, 13 January 2019 17:00:22 UTC-5, j4n bur53 wrote:
> May I make a comment? To find f'(x), we do not
> always use if 0 < |x-c| < delta, then |[f(c+h) - f(c)] / h - L| < epsilon .
> May I make a comment? To find f'(x), we do not
Yes, you do!
>
> Since we are not anymore apes and have evolved, our
> arsenal is much wider, Chain Rule, Hopital Rule, etc...
>
> This rules in turn can be reduced to some lim definitions.
Me
Jan 13, 2019, 5:08:36 PM1/13/19
to
On Sunday, January 13, 2019 at 11:00:22 PM UTC+1, j4n bur53 wrote:
> May I make a comment?
Sure.
> May I make a comment?
> To find f'(x), we do not
> always use if 0 < |h| < delta, then |[f(x+h) - f(x)] / h - m| < epsilon .
>
> Since we are not anymore apes and have evolved,
> our arsenal is much wider, Chain Rule, Hopital Rule, etc...
>
> This rules in turn can be reduced to some lim definitions.
Very nice text: "The Limit Laws"
http://www.oxfordmathcenter.com/drupal7/node/95
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