2^y - 3^x
rpg16
2^y - 3^x, where x is max. for a given y and 2^y - 3^x > 0, x,y are
positive integers
Find a formula for y in terms of x???
bert
x is max for y also means y is min. for x.
2^y > 3^x means y * log(2) > x * log(3).
Therefore y = ceil(x * log(3) / log(2)).
--
William Elliot
It's necessary that 1 < y, otherwise the maximum x is zero.
2^y > 3^x; y.log 2 > x.log 3; y(log 2)/log 3 > x; x = [y(log 2)/log 3]
MeAmI.org
Options/directions:
Find the derivative 1. y = 3x + 6x 2.
So we set the derivative equal to 0.
y = 0. 6x(x + 1). 2. (x - 1). 2. = 0 x =-1,0,1. These are the only
critical points. 3.
To find the global max and ...
2^y - 3^x, where x is max. for a given y and 2^y - 3^x > 0, x,y are
positive integers. Find a formula for y in terms of x:
Well, The graph of y = x3 - 10x2 +12x + 23 has a maximum point between
x ...x= - 8/3+10/3 = 2/3 plug these values in original formula and you
will find the min and max ...
y=x^3-10x^2+12x+23=> y'=3x^2-20x+12=> y''=6x-20 y'=0=> (x-6)(3x-2)
=0=> ...
So, the coordinates of the max.pt is (0.67,26.85) approximately.
--
Martin Musatov
Los Angeles, CA
rpg16
Thanks bert... but thats an obvious reply... I think the mistake is
mine...
I should have said:
"Find a formula for y in terms of a polynomial or exponential function
of x"
bert
That's the perpetual difficulty of answering
questions posted on sci.math; trying to decide
where the questioner is starting from. One
can get it quite wrong, in either direction!
I can't see that any polynomial or exponential
modification would convert the linear function
y = x * k, where k is log(3)/log(2), into the
required stepwise function y = ceil(x * k).
But I'll watch for any other replies.
--
Gerry
x * log 3 / log 2 is a polynomial function of x. If it's
the "ceil" that's bothering you, ceil(Q) just means Q rounded
down to the nearest integer. If it still bothers you, you're
probably out of luck - given that there's a formula using
ceil, you're unlikely to find one that doesn't.
--
GM
rpg16
> On Jul 5, 8:31 pm,rpg16<roupam.gh...@gmail.com> wrote:
>
>
>
> > On Jul 5, 1:19 pm, bert <bert.hutchi...@btinternet.com> wrote:
>
> > > On 5 July, 08:58,rpg16<roupam.gh...@gmail.com> wrote:
>
> > > > Given,
> > > >2^y-3^x, wherexis max. for a givenyand2^y-3^x> 0,x,yare
> > > > positive integers
>
> > > > Find a formula foryin terms ofx???
>
> > >xis max foryalso meansyis min. forx.
> > >2^y>3^xmeansy* log(2) >x* log(3).
> > > --
>
> > Thanks bert... but thats an obvious reply... I think the mistake is
> > mine...
> > I should have said:
>
> > ofx"
>
> x* log3/ log2is a polynomial function ofx. If it's
> the "ceil" that's bothering you, ceil(Q) just means Q rounded
> down to the nearest integer. If it still bothers you, you're
> probably out of luck - given that there's a formula using
> ceil, you're unlikely to find one that doesn't.
> --
> GM
I think Gerry is right...
The problem lies with ceil[] which apparently cannot be expressed as a
polynomial or exponential function...
And hence very little chance of finding such a function...
Robert Israel
> I think Gerry is right...
> The problem lies with ceil[] which apparently cannot be expressed as a
> polynomial or exponential function...
> And hence very little chance of finding such a function...
ceil(x) = 1 + x - arccot(cot(pi x))/pi (if x is a non-integer)
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Musatov
<isr...@math.MyUniversitysInitials.ca> wrote:
ceil(x) = 1 + x - arccos(cos(pi x))/pi (if x is a non-integer) is
available by typing pi or PI . See constants.
acos(x) and arccos(x) return the inverse cosine of x in the range 0°
to +180° (0 to pi).
Then, if the result is non-zero, the value of expr1 is computed and
returned.
ceil(x) (ceiling) returns the smallest integer which is no less than
x.
--Martin Musatov
http://MeAmI.org
"Search me."
bert
wrote:
Wow! There you are, then. log(3)/log(2) is known to be
irrational, hence x * log(3)/log(2) is a non-integer for
all integer x, hence Robert's ceil() function will work
for y in all of your cases.
--
David W. Cantrell
> On 5 July, 18:29, Robert Israel <isr...@math.MyUniversitysInitials.ca>
> wrote:
> > rpg16 <roupam.gh...@gmail.com> writes:
> > > I think Gerry is right...
> > > The problem lies with ceil[] which apparently cannot be expressed as
> > > a polynomial or exponential function...
> > > And hence very little chance of finding such a function...
> >
> > ceil(x) = 1 + x - arccot(cot(pi x))/pi (if x is a non-integer)
> Wow! There you are, then. log(3)/log(2) is known to be
> irrational, hence x * log(3)/log(2) is a non-integer for
> all integer x, hence Robert's ceil() function will work
> for y in all of your cases.
Yes, the formula mentioned by Robert takes care of the matter in this case,
unless the result must literally "be expressed as a polynomial or
exponential function", which is impossible, of course.
And the Wolfram Functions site shows a similar formula at
<http://functions.wolfram.com/IntegerFunctions/Ceiling/27/02/>:
Ceiling[z] == z + ArcTan[Cot[Pi z]]/Pi + 1/2
/; Element[z, Reals] && !Element[z, Integers]
Those two formulae for ceiling do not work when its argument is an integer.
That is, they fail precisely at the most interesting places, where the
ceiling function is discontinuous.
But formulae in terms of "elementary functions" and valid for _all_ reals
can be devised for functions such as ceiling. More than two years ago, I
sent several such formulae to those responsible for the Wolfram Functions
site, but my formulae have yet to appear there. :-( Here's the formula
which I sent them for the ceiling function:
Ceiling[x] == x + 1/2 + I/(2 Pi) Log[-Exp[2 Pi I x]]
/; Element[x, Reals]
David W. Cantrell
rpg16
But x is a integer here...
bert
> > ceil(x) = 1 + x - arccot(cot(pi x))/pi (if x is a non-integer)
>
> But x is a integer here...
No, it's not. Your x was an integer, but
Robert's x is your x multiplied by the
irrational quantity log(3)/log(2), which
therefore can NOT be an an integer, so
the formula will always work.
--
jillbones
> On Jul 5, 8:31 pm, rpg16 <roupam.gh...@gmail.com> wrote:
>
>
>
> > On Jul 5, 1:19 pm, bert <bert.hutchi...@btinternet.com> wrote:
>
> > > On 5 July, 08:58, rpg16 <roupam.gh...@gmail.com> wrote:
>
> > > > Given,
> > > > 2^y - 3^x, where x is max. for a given y and 2^y - 3^x > 0, x,y are
> > > > positive integers
>
> > > > Find a formula for y in terms of x???
IMMHO. there is insufficient information. Don't I need the value
of "K" in "2^y - 3^x = K"?
Bill J