GR8767 Question 20

Forum for the GRE subject test in mathematics.
phcooh
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GR8767 Question 20

Postby phcooh » Mon Oct 13, 2008 7:19 pm

f(x)=f(x+1) for all real x, if f is a polynomial and f(5)=11,
then f(15/2) is ?
A. -11
B. 0
C. 11
D. 33/2
E. not uniquely determined

I chose E. but the answer C, why?
Thanks a lot

JcraigMSU
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Postby JcraigMSU » Mon Oct 13, 2008 7:26 pm

f(x) = f(x+1) for all real x, not just integers

EmLasker
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Postby EmLasker » Mon Oct 13, 2008 11:27 pm

I think the temptation here is to define a piecewise function such that f(13/2)=2 or some other number than 5.... The reason you can't do this is because f is a polynomial and is therefore continuous for all real numbers.

Does anyone know of any non-constant functions that would satisfy the premises in the problem?

ralphhumacho
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Postby ralphhumacho » Tue Oct 14, 2008 3:04 am

im pretty sure a polynomial can only have a finite number of roots. thus, in order for f(x)=5 at infinitely many points, f(x) must be constant. more over, f(x) = 5. hence C is the correct answer.

please - correct me if i'm wrong.

EmLasker
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Postby EmLasker » Tue Oct 14, 2008 3:17 am

Ralph,

You are right a polynomial of order n has exactly n roots in in the set of complex numbers...

Nameless
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Postby Nameless » Tue Oct 14, 2008 8:28 am

f(x)=f(x+1) for all real x, if f is a polynomial and f(5)=11,
then f(15/2) is ?
A. -11
B. 0
C. 11
D. 33/2
E. not uniquely determined


suppose that the degree of f is n (n>=1)
then there exists a is a root of f(x) then f(a)=0
then f(a+1)=f(a)=0
f(a+2)=f(a+1)=0

then f(a+n+1)=f(a+n)=....=f(a)=0
then f(x)=o has infinite solutions so f(x)=0 for all x , then n= - infinity <0<1 . This a contradiction :D
Note : the degree of the ZERO polynomial is -infinity

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lime
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Postby lime » Tue Oct 14, 2008 9:45 am

suppose that the degree of f is n (n>=1)
then there exists a is a root of f(x) then f(a)=0

This is wrong. Fundamental theorem of algebra does not guarantee the existence of roots over field of rational numbers.
Consider for example p(x) = (x-5)^2+11.

Let degree n>0, then p(x) should go to +inf as x goes to +inf which cause contradiction. Therefore, p(x) must be a constant function.

CoCoA
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Postby CoCoA » Tue Oct 14, 2008 12:26 pm

There is a very recent thread on this question, pleasee see it also.

Nameless
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Postby Nameless » Tue Oct 14, 2008 12:58 pm

Posted: Tue Oct 14, 2008 1:45 pm Post subject:
Quote:
suppose that the degree of f is n (n>=1)
then there exists a is a root of f(x) then f(a)=0

This is wrong. Fundamental theorem of algebra does not guarantee the existence of roots over field of rational numbers.
Consider for example p(x) = (x-5)^2+11.

Let degree n>0, then p(x) should go to +inf as x goes to +inf which cause contradiction. Therefore, p(x) must be a constant function.



thanks lime for correcting me. Your solution is absolutely good

octave145
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Postby octave145 » Fri Oct 24, 2008 8:18 pm

My thought process was similar to the infinite number of roots argument but on
f'(x) rather than f(x). Since f(x) = f(x+1) there must be a point
in the interval (x,x+1) where the derivative is zero. Only when f(x) is constant does this get satisfied for all x.

amateur
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Postby amateur » Fri Oct 24, 2008 11:06 pm

What happens if the condition "f(x) is a polynomial" is relaxed?

Note that f(x) = 11*[sin (2*x*PI + PI/2) ] satisfies

f(x) = f(x + 1), f is defined for all real x,

f is not a polynomial (OR you may consider it a polynomial if you consider the series expansion of sin(x), I am not sure about this :?: ),

and finally f(15/2) = -11.

moo5003
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Postby moo5003 » Mon Oct 27, 2008 1:54 pm

If you do not restrict the functions to polynomials then there are an infinite host of examples of non-constant functions that would satisfy the conditions. Namely any wave with period 1.

jlt10
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Re: GR8767 Question 20

Postby jlt10 » Fri Nov 02, 2012 2:09 pm

The way that I looked at that question was that since f(x) = f(1+x), f(n) = f(n+1) = f(n+2) = .... for all n in Z. So you just need to realize that at some point continually adding one to 15/2 will produce an integer. So, f(15/2) = f(n) for all n in Z, and f(15/2) = 11.

Legendre
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Re: GR8767 Question 20

Postby Legendre » Fri Nov 02, 2012 9:09 pm

Polynomials are differentiable.

By the mean value theorem, f(x+1) - f(x) = f'(c) = 0 for some c in (x,x+1). This holds for all x in the domain of f.

Therefore, f is a constant.

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redcar777
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Re: GR8767 Question 20

Postby redcar777 » Sun Nov 04, 2012 5:10 pm

Note that, for any polynomial p(x), p(x+a) - p(x) is a polynomial with strictly lower degree than p.

The hypothesis is that p(x+1)-p(x) is identically zero. The only polynomial for which this is true is the zero polynomial (since a non-zero polynomial has at most a finite number of zeros, bounded by the degree of the polynomial).

So p(x) must be constant, and hence f(15/2) = 11.

The mean value theorem doesn't quite get you there-- you need the polynomial hypothesis, as the example f(x) = A sin( 2 \pi x + k ) shows (for some choice of A and k, f(5) = 11 ).

See also: http://en.wikipedia.org/wiki/Newton_polynomial
Last edited by redcar777 on Sun Nov 04, 2012 7:07 pm, edited 2 times in total.

vonLipwig
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Re: GR8767 Question 20

Postby vonLipwig » Sun Nov 04, 2012 5:50 pm

redcar777: You go from saying that p(x+1) - p(x) is the zero polynomial, which is true, to saying that this means that p(x) is constant. This is really assuming the result of the question.

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redcar777
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Re: GR8767 Question 20

Postby redcar777 » Sun Nov 04, 2012 6:48 pm

@vonLipwig:

I guess the fact you really need to know is that p(x+a)-p(x) is exactly one degree smaller than p(x) for polynomials over the real numbers. (Here you need to interpret "one degree smaller" than a degree 0 polynomial as the zero polynomial). In particular, p(x+a) - p(x) is a polynomial of some degree, and therefore has a finite number of zeros (possibly none), unless p(x) is a constant.

This is true for polynomials of the form x^k by the binomial theorem, so a little head scratching shows its true for all polynomials since only the leading term of p(x) can give a term of degree n-1 in p(x+a)-p(x).

The point is if you have p(x), p(x+a), p(x+2a),...,p(x+ma)
and you take the differences, then the differences of the differences, then the differences of the differences of the differences, etc., eventually you get all zeros. That happens exactly when you've applied this procedure n+1 times.

vonLipwig
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Re: GR8767 Question 20

Postby vonLipwig » Sun Nov 04, 2012 7:58 pm

Yes, it's that "exactly one smaller" which is important here. If you use that fact, then you don't need to consider the zeros of anything. You can just say that f(x+1) - f(x) is zero, so f(x) is degree 0, so constant.

Legendre
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Re: GR8767 Question 20

Postby Legendre » Sun Nov 04, 2012 10:45 pm

vonLipwig wrote:Yes, it's that "exactly one smaller" which is important here. If you use that fact, then you don't need to consider the zeros of anything. You can just say that f(x+1) - f(x) is zero, so f(x) is degree 0, so constant.


Yes, thats it! f(x+1) - f(x) is exactly one degree smaller. So f(x) must be a constant.




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