Why does X = SIN X for very small X in Radians?
tendim
website/concept about how I can figure it out.
Why is it that for very small values of X in radians, sin(X) equals X?
Tks,
patrick
--
t e n d i m
agent orange as the casual ganster.
calculus is the root of all evil.
Michael
around the point (1,0)
Now, the angle, x, is equal to the length of the curve from a given point on
the circle to the x-axis. Sin x is equal to the vertical distance from that
point to the x-axis. As x becomes smaller, shape of the curve of which you
are measuring the length of becomes extremely close to a vertical line,
which is the same as sin x. This is the concept to which you are referring.
However, they do not become equal until x actually is zero. I assume you
are making your calculations on a calculator, and the reason why you say
they actually are equal is because of the rounding error of the calculator;
there is a point at which the difference between x and sin x is less than
1x10^-93, just to pick a number.
In all actuality, there is nothing amazing about this, atleast nothing more
amazing than any other functions that intersect. y=sin x and y=x intersect
at x=0. So for very small x, x and sin x approach each other in degrees,
gradients, or any other angle measuring system you want to invent in your
spare time. It's just like saying that for very small x, x^40 = x. They
don't, but they come very close. It's just because they intersect at x=0.
Michael
"tendim" <pat...@tendim.cjb.net> wrote in message
news:1ejsot8.18y7c6z3nnzr4N%pat...@tendim.cjb.net...
Zdislav V. Kovarik
tendim <pat...@tendim.cjb.net> wrote:
:Wondering if someone could either tell me *why* this is so, or to a
:website/concept about how I can figure it out.
:
:Why is it that for very small values of X in radians,
:sin(X) equals X?
It isn't (unless x is already 0), as others reminded you. Not even
"practically" - an expression intentionally left vague.
If x is practically like sin(x) for small x, then 1/x should be
practically like 1/sin(x), right? Well, the limit of their
difference is zero. Emboldened by this, let's check how close
(1/sin(x))^2 is to 1/x^2.
I leave it to the reader to find out that the limit of the
difference:
lim[as x -> 0] (((1/sin(x)))^2 - 1/x^2)
is no longer zero. What is it?
Cheers, ZVK(Slavek).
tendim
> In <1ejsot8.18y7c6z3nnzr4N%pat...@tendim.cjb.net> pat...@tendim.cjb.net
> (tendim) writes:
>
> >Wondering if someone could either tell me *why* this is so, or to a
> >website/concept about how I can figure it out.
>
> >Why is it that for very small values of X in radians, sin(X) equals X?
>
> It doesn't. No website/concept can help you prove it.
> But you can examine the series for sin(x) to see why it is
> approximately true. Consult a textbook. Visit your library.
Consulted three text books, turned up nothing.
However, we were told in one class to try different values with a
calculator, and that lead to something else, read my next followup.
> >Tks,
>
> What's *tks*? Perhaps some shorthand in your middle school?
Short hand for "thanks". In the same way that "ppl" is the shorthand
for "people", TIA for "thanks in advance", "dept." for department (as in
your .sig (".sig" for "signature file", even though it may not *be* a
signature file, it conforms to what is usually appeneded to the end of
news articles, eMail's ("eMail" for "electronic mail"), and other forms
of writing on the 'net ("'net" for "internet")), etc.
> >agent orange as the casual ganster.
>
> What's a *ganster*?
Opposite of an ungangster.
>
> >calculus is the root of all evil.
>
> Probably very witty in some circles.
It is at 4:00am when you and all your friends have been wrestling with
problems all night.
> --
> =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
> John E. Prussing
> Dept. of Aeronautical & Astronautical Engineering
> University of Illinois at Urbana-Champaign
> http://www.uiuc.edu/~prussing
> =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
tendim
> Draw a picture of the unit circle. Zoom in extremely closely on the area
> around the point (1,0)
> Now, the angle, x, is equal to the length of the curve from a given point on
> the circle to the x-axis. Sin x is equal to the vertical distance from that
> point to the x-axis. As x becomes smaller, shape of the curve of which you
> are measuring the length of becomes extremely close to a vertical line,
> which is the same as sin x. This is the concept to which you are referring.
> However, they do not become equal until x actually is zero. I assume you
> are making your calculations on a calculator, and the reason why you say
> they actually are equal is because of the rounding error of the calculator;
> there is a point at which the difference between x and sin x is less than
> 1x10^-93, just to pick a number.
I discussed this method with a friend, but I was trying to find a less,
graphical, solution.
In one class we were told to take small values of X, using radians on
our calculators, and see what SIN X turned up. The point was to show us
that for very small values of X, SIN X == X. The TA told us that this
was because SIN could be expressed as a "power series" (the name Taylor
comes to mind). I took a look at the material I had, but I couldn't
find anything that was (directly) useful -- although at that point I was
just skimming the text, and not looking for an in depth solution.
>
> In all actuality, there is nothing amazing about this, atleast nothing more
> amazing than any other functions that intersect. y=sin x and y=x intersect
> at x=0. So for very small x, x and sin x approach each other in degrees,
> gradients, or any other angle measuring system you want to invent in your
> spare time. It's just like saying that for very small x, x^40 = x. They
> don't, but they come very close. It's just because they intersect at x=0.
>
> Michael
Tks,
patrick
Randy Poe
wrote:
>Wondering if someone could either tell me *why* this is so, or to a
>website/concept about how I can figure it out.
>
>Why is it that for very small values of X in radians, sin(X) equals X?
Not sure on what level you want this "why" answered... there are a few
different answers.
(1) Taylor-series argument: sin(x) can be written as a series x -
x^3/6 + x^5/120 - ...
(2) Geometric argument: For angles less than 90 degrees, sin is the
ratio of the opposite to the hypotenuse of a right triangle. When the
angle is very small, the picture looks very close to two radii of a
circle with a small chord between them. The length of that chord is
approximately r*x, where the hypotenuse is x.
(3) Graphical argument: If you draw the line y=x and the function
y=sin(x) and blow it up, looking very close to the origin near x=0,
you will see that the straight line falls very close to the curve.
(4) Theological argument...?? Philosophical argument...?
- Randy
Michael
(x-->0) ((sin x)/x)=1. I know that L'hopital's rule makes this an easy
problem, but my HS calc teacher presented us with this problem before he
taught us L'hopital's rule, so we had to do it using the squeeze theorem
(I've also seen it called the sandwich theorem, but whatever). That was a
long proof, the graph was logical to me, so that was how I remembered it.
Michael
"Michael" <jel...@yahoo.com> wrote in message
news:8ud1e3$qhr$1...@mozo.cc.purdue.edu...
> Draw a picture of the unit circle. Zoom in extremely closely on the area
> around the point (1,0)
> Now, the angle, x, is equal to the length of the curve from a given point
on
> the circle to the x-axis. Sin x is equal to the vertical distance from
that
> point to the x-axis. As x becomes smaller, shape of the curve of which
you
> are measuring the length of becomes extremely close to a vertical line,
> which is the same as sin x. This is the concept to which you are
referring.
> However, they do not become equal until x actually is zero. I assume you
> are making your calculations on a calculator, and the reason why you say
> they actually are equal is because of the rounding error of the
calculator;
> there is a point at which the difference between x and sin x is less than
> 1x10^-93, just to pick a number.
>
> In all actuality, there is nothing amazing about this, atleast nothing
more
> amazing than any other functions that intersect. y=sin x and y=x
intersect
> at x=0. So for very small x, x and sin x approach each other in degrees,
> gradients, or any other angle measuring system you want to invent in your
> spare time. It's just like saying that for very small x, x^40 = x. They
> don't, but they come very close. It's just because they intersect at x=0.
>
> Michael
>
> "tendim" <pat...@tendim.cjb.net> wrote in message
> news:1ejsot8.18y7c6z3nnzr4N%pat...@tendim.cjb.net...
> > Wondering if someone could either tell me *why* this is so, or to a
> > website/concept about how I can figure it out.
> >
> > Why is it that for very small values of X in radians, sin(X) equals X?
> >
John Prussing
>Wondering if someone could either tell me *why* this is so, or to a
>website/concept about how I can figure it out.
>Why is it that for very small values of X in radians, sin(X) equals X?
It doesn't. No website/concept can help you prove it.
But you can examine the series for sin(x) to see why it is
approximately true. Consult a textbook. Visit your library.
>Tks,
What's *tks*? Perhaps some shorthand in your middle school?
>patrick
>--
>t e n d i m
>agent orange as the casual ganster.
What's a *ganster*?
>calculus is the root of all evil.
Probably very witty in some circles.
rancid moth
> Michael <jel...@yahoo.com> wrote:
>
> > Draw a picture of the unit circle. Zoom in extremely closely on the
area
> > around the point (1,0)
> > Now, the angle, x, is equal to the length of the curve from a given
point on
> > the circle to the x-axis. Sin x is equal to the vertical distance from
that
> > point to the x-axis. As x becomes smaller, shape of the curve of which
you
> > are measuring the length of becomes extremely close to a vertical line,
> > which is the same as sin x. This is the concept to which you are
referring.
> > However, they do not become equal until x actually is zero. I assume
you
> > are making your calculations on a calculator, and the reason why you say
> > they actually are equal is because of the rounding error of the
calculator;
> > there is a point at which the difference between x and sin x is less
than
> > 1x10^-93, just to pick a number.
>
> graphical, solution.
>
> In one class we were told to take small values of X, using radians on
> our calculators, and see what SIN X turned up. The point was to show us
> that for very small values of X, SIN X == X. The TA told us that this
> was because SIN could be expressed as a "power series" (the name Taylor
> comes to mind). I took a look at the material I had, but I couldn't
> find anything that was (directly) useful -- although at that point I was
> just skimming the text, and not looking for an in depth solution.
>
you can take the intuitativly approach it like this...
sin(x) = x - x^3/3! + x^5/5! -......
and that for x<<1 (i.e. very much less than one, or very small) then x^3 is
going to be even smaller, and hence not contribute much to the summation,
and thus x^5 would do so even less and so on and so on, and in fact what you
could say is that for x<<1 we will approximate it by taking anything of the
order of x^2 or higher approximately equal to zero (in relation to the
summation) and hence ignore them to get...
sin(x) is approximately equal to x, which is the only remaining term in the
summation. lets have an example
lets say we choose
x = .0001, which is reasonable enough to say that its <<1 or very small.
then
sin(.0001) = .0001 - (.0001)^3/3! + (.0001)^5/5! -......
= .0001 - .000000000001 / 6 + .00000000000000000001 /
120 -....
now whats the dominating term of the summation...its .0001, simply because
for small x, the other higher powered terms basically become "insignificant"
to the summation, and thus can be approximated out to leave
sin(x) ~ x, x<<1
i dont think i can explain it any more simply with out just writing down the
asymptotic derivation.
regards
moth
> >
> > In all actuality, there is nothing amazing about this, atleast nothing
more
> > amazing than any other functions that intersect. y=sin x and y=x
intersect
> > at x=0. So for very small x, x and sin x approach each other in
degrees,
> > gradients, or any other angle measuring system you want to invent in
your
> > spare time. It's just like saying that for very small x, x^40 = x.
They
> > don't, but they come very close. It's just because they intersect at
x=0.
> >
> > Michael
>
> Tks,
> patrick
>
> --
> t e n d i m
> agent orange as the casual ganster.
George Jefferson
class I'm sure.
:I discussed this method with a friend, but I was trying to find a less,
:graphical, solution.
plot y=sin x and y=x, you will see that they are equal and tangent
(same slope) at x = 0.
--
george jefferson : geo...@sol1.lrsm.upenn.edu
to reply simply press "r"
-- I hate editing addresses more than I hate the spam!
G. A. Edgar
> graphical, solution.
Actually, they are not precisely equal, just very nearly equal. In
fact, their difference is close to (x^3)/6.
sin(0.01) = 0.009999833334
sin(0.01) - 0.01 = -.000000166666
0.01^3/6 = .000000166667
--
Gerald A. Edgar ed...@math.ohio-state.edu
John Prussing
>"tendim" <pat...@tendim.cjb.net> wrote in message
>news:1ejsvwf.1mn75ob57yx8gN%pat...@tendim.cjb.net...
>> Michael <jel...@yahoo.com> wrote:
>>
>>
>> I discussed this method with a friend, but I was trying to find a less,
>> graphical, solution.
>>
>> In one class we were told to take small values of X, using radians on
>> our calculators, and see what SIN X turned up. The point was to show us
>> that for very small values of X, SIN X == X. The TA told us that this
>> was because SIN could be expressed as a "power series" (the name Taylor
>> comes to mind). I took a look at the material I had, but I couldn't
>> find anything that was (directly) useful -- although at that point I was
>> just skimming the text, and not looking for an in depth solution.
>>
>you can take the intuitativly approach it like this...
>sin(x) = x - x^3/3! + x^5/5! -......
One can either accept this as the definition of sin(x) or interpret it
as a Taylor series expansion about x = 0. (Is an expansion about x =
0 still called a McLaurin (sp?) series?). His name is included on a
list that is chiseled on a building at MIT and I always wondered if
his fame was due solely to the special case of a Taylor series about 0
argument. My guess is that's not the case. Perhaps his
result predates Taylor's and perhaps he discovered other things.
For the Taylor/McLaurin expansion of f(x) = sin(x), f(0) = sin(0) =
0, f'(0) = cos(0) = 1, f''(0) = - sin(0) = 0, f'''(0) = - cos(0) = -1,
etc., so the series expansion is sin(x) = 0 + x + 0 - x^3/3! + .......
>and that for x<<1 (i.e. very much less than one, or very small) then x^3 is
>going to be even smaller, and hence not contribute much to the summation,
For any x < 1, x^3 will be smaller, and x^3/3! even smaller yet.
Depending on your accuracy requirements, x may not be such
a bad approximation to sin(x) for values of x up to 0.5 radian or so.
True, 0.5 is not << 1 but the x^3/3! term is only about an 8% correction
(to the value 0.5).
John Savard
>Wondering if someone could either tell me *why* this is so, or to a
>website/concept about how I can figure it out.
>Why is it that for very small values of X in radians, sin(X) equals X?
My website, at the page
http://home.ecn.ab.ca/~jsavard/other/ide01.htm
actually has a picture in it that helps to explain why this is so, as
part of the reason why mathematicians use radians.
John Savard
http://home.ecn.ab.ca/~jsavard/crypto.htm
Dave L. Renfro
[sci.math Wed, 8 Nov 2000 21:12:56 -0500]
<http://forum.swarthmore.edu/epigone/sci.math/palsmeldphoi/8ud1e3$qhr$1...@mozo.cc.purdue.edu>
wrote
> Draw a picture of the unit circle. Zoom in extremely closely on the
> area around the point (1,0)
> Now, the angle, x, is equal to the length of the curve from a
> given point on the circle to the x-axis. Sin x is equal to the
> vertical distance from that point to the x-axis. As x becomes
> smaller, shape of the curve of which you are measuring the length
> of becomes extremely close to a vertical line, which is the same
> as sin x. This is the concept to which you are referring. However,
> they do not become equal until x actually is zero. I assume you
> are making your calculations on a calculator, and the reason why
> you say they actually are equal is because of the rounding error
> of the calculator; there is a point at which the difference between
> x and sin x is less than 1x10^-93, just to pick a number.
>
> In all actuality, there is nothing amazing about this, atleast
> nothing more amazing than any other functions that intersect.
> y=sin x and y=x intersect at x=0. So for very small x, x and
> sin x approach each other in degrees, gradients, or any other
> angle measuring system you want to invent in your spare time.
> It's just like saying that for very small x, x^40 = x. They
> don't, but they come very close. It's just because they intersect
> at x=0.
I think the original poster meant that the percentage error between
x and sin(x) goes to zero as x --> 0 (i.e. [x - sin(x)]/x --> 0 as
x --> 0 or, equivalently, [sin(x)]/x --> 1 as x --> 0), not just
that the difference between x and sin(x) goes to zero as x --> 0.
And to nitpick, saying that the two graphs intersect when x=0 isn't
sufficient. Let f(x) = 5 if x isn't 0 and f(0) = 0, and let
g(x) = 12 if x isn't 0 and g(0) = 0. Then the graphs of f(x) and
g(x) intersect at x=0, but f(x) doesn't approach g(x) as x --> 0.
Dave L. Renfro
Bob Silverman
pat...@tendim.cjb.net (tendim) wrote:
> Wondering if someone could either tell me *why* this is so, or to a
> website/concept about how I can figure it out.
>
> Why is it that for very small values of X in radians, sin(X) equals X?
They are not equal. They are very close, but not equal.
The "reason" follows.
We know that sin(x) = x - x^3/3! + x^5/5! - ...
Now when x is close to zero, x^3/3! is *very* small (and the other
terms are even smaller) thus sin(x) = x + something very small
when x is near 0.
--
Bob Silverman
"You can lead a horse's ass to knowledge, but you can't make him think"
Sent via Deja.com http://www.deja.com/
Before you buy.
Dave L. Renfro
[sci.math Wed, 8 Nov 2000 23:31:23 -0500]
<http://forum.swarthmore.edu/epigone/sci.math/palsmeldphoi/8ud9hm$2o$1...@mozo.cc.purdue.edu>
wrote
> Just a side note, this graphical analysis is a logical reasoning
> for why lim (x-->0) ((sin x)/x)=1. I know that L'hopital's rule
> makes this an easy problem, but my HS calc teacher presented us
> with this problem before he taught us L'hopital's rule, so we had
> to do it using the squeeze theorem (I've also seen it called the
> sandwich theorem, but whatever). That was a long proof, the
> graph was logical to me, so that was how I remembered it.
Something to consider ... In using L'Hopital's rule, you have to
know what the derivative of sin(x) is at 0. Easy, right?? [It's
cos(0) = 1.] But how do we know that the derivative of sin(x) is
cos(x)? The definition of the derivative requires us to evaluate
the following limit:
(d/dx)(sin x) = LIMIT as h --> 0 of [sin(x+h) - sin(x)]/h
Expanding sin(x+h) using the addition angle formula gives
LIMIT as h --> 0 of [(sin x)*(cos h) + (cos x)*(sin h) - (sin x)] / h
LIMIT as h --> 0 of { (sin x)*[(cos h) - 1]/h + (cos x)*(sin h)/h }
(sin x)*{ LIMIT as h --> 0 of [(cos h) - 1]/h }
+ (cos x)*{ LIMIT as h --> 0 of (sin h)/h }
[Both (sin x) and (cos x) are constant as far as the operation
"LIMIT as h --> 0" is concerned, so they can be factored out
of the limits.],
and now we're confronted with limit as h --> 0 of (sin h)/h, the same
limit this thread was about! So some sort of argument independent of
L'Hopital's rule (or Taylor series, etc.) is needed for why
(sin x)/x --> 1 as x --> 0, unless a different way of evaluating
the original limit that (d/dx)(sin x) represents is found (and I
know of no other way, off-hand).
Once you know (sin h)/h --> 1 as h --> 0, finding the other limit
above is easy when you multiply the numerator and denominator of
[(cos h) - 1]/h by (cos h) + 1.
Dave L. Renfro
P.S. Yes, I know many of the respondants in this thread know this,
and I'm sure I've read posts by several of you that actually
make this same point. But since it hasn't been made explicit
yet in this thread, and I happen to be up a bit early this
morning, I thought I'd jump in.
Michael Bienstein
problems.
If we define sin(x) as [exp(-ix)-exp(ix)]i/2 where i=sqrt(-1) then we can
find the derivative to be:
[-iexp(-ix)-iexp(ix)]i/2=[exp(ix)+exp(-ix)]/2 = cos(x).
If we define sin(x) as x-x^3/6+x^5/120-x^7/7!+x^9/9!...
then the derivative is 1-x^2/2+x^4/24-x^6/6!+x^8/8!...=cos(x)
How do you want it?
Michael
Dave L. Renfro wrote in message ...
>Michael <jel...@yahoo.com>
>[sci.math Wed, 8 Nov 2000 23:31:23 -0500]
><http://forum.swarthmore.edu/epigone/sci.math/palsmeldphoi/8ud9hm$2o$1@mozo
Dave L. Renfro
[sci.math Tue, 14 Nov 2000 16:54:34 +0200]
<http://forum.swarthmore.edu/epigone/sci.math/palsmeldphoi/8urirg$kjv$1...@license1.unx.sas.com>
wrote
> How do you want to define sin(x)? Since this involves chicked and
> egg type problems.
>
> If we define sin(x) as [exp(-ix)-exp(ix)]i/2 where i=sqrt(-1) then
> we can find the derivative to be:
> [-iexp(-ix)-iexp(ix)]i/2=[exp(ix)+exp(-ix)]/2 = cos(x).
>
> If we define sin(x) as x-x^3/6+x^5/120-x^7/7!+x^9/9!...
> then the derivative is 1-x^2/2+x^4/24-x^6/6!+x^8/8!...=cos(x)
>
> How do you want it?
Since the original poster is in a Calculus I class (this
seemed clear to me from his three Nov. 8 posts at
<http://forum.swarthmore.edu/epigone/sci.math/palsmeldphoi>,
but perhaps I'm mistaken), neither the complex variable formulation
nor the power series formulation are appropriate. Also, to justify
the power series differentiation requires a proof for term-by-term
differentiation of power series, and to justify the complex variable
differentiation probably relies on this as well.
Given this context I would expect that sin(x) is defined in the usual
way one finds in in trig., precalculus, and calculus classes--the
y-coordinate of the point on the unit circle x^2 + y^2 = 1 associated
with the terminal ray of the angle 'x radians' in standard position.
Dave L. Renfro
Hop David
tendim wrote:
> Wondering if someone could either tell me *why* this is so, or to a
> website/concept about how I can figure it out.
>
> Why is it that for very small values of X in radians, sin(X) equals X?
>
> Tks,
> patrick
>
> --
> t e n d i m
> agent orange as the casual ganster.
> calculus is the root of all evil.
I drew a pic. Hope it helps: http://clowder.net/hop/sin(a)overa.html
-- Hop
http://clowder.net/hop/index.html
Erik Max Francis
> tendim wrote:
>
> > Wondering if someone could either tell me *why* this is so, or to a
> > website/concept about how I can figure it out.
> >
> > Why is it that for very small values of X in radians, sin(X) equals
> > X?
>
> I drew a pic. Hope it helps: http://clowder.net/hop/sin(a)overa.html
Since you're just graphically showing that lim{x -> 0} (sin x/x) = 1,
why not go the easy route?
gnuplot> plot sin(x)/x
http://charmaine.alcyone.com/max/graphs/sinxoverx.gif
--
Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/
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