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GR8767 Question 20

Posted: Mon Oct 13, 2008 7:19 pm
by phcooh
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

Posted: Mon Oct 13, 2008 7:26 pm
by JcraigMSU
f(x) = f(x+1) for all real x, not just integers

Posted: Mon Oct 13, 2008 11:27 pm
by EmLasker
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?

Posted: Tue Oct 14, 2008 3:04 am
by ralphhumacho
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.

Posted: Tue Oct 14, 2008 3:17 am
by EmLasker
Ralph,

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

Posted: Tue Oct 14, 2008 8:28 am
by Nameless
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

Posted: Tue Oct 14, 2008 9:45 am
by lime
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.

Posted: Tue Oct 14, 2008 12:26 pm
by CoCoA
There is a very recent thread on this question, pleasee see it also.

Posted: Tue Oct 14, 2008 12:58 pm
by Nameless
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

Posted: Fri Oct 24, 2008 8:18 pm
by octave145
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.

Posted: Fri Oct 24, 2008 11:06 pm
by amateur
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.

Posted: Mon Oct 27, 2008 1:54 pm
by moo5003
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.

Re: GR8767 Question 20

Posted: Fri Nov 02, 2012 2:09 pm
by jlt10
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.

Re: GR8767 Question 20

Posted: Fri Nov 02, 2012 9:09 pm
by Legendre
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.

Re: GR8767 Question 20

Posted: Sun Nov 04, 2012 5:10 pm
by redcar777
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

Re: GR8767 Question 20

Posted: Sun Nov 04, 2012 5:50 pm
by vonLipwig
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.

Re: GR8767 Question 20

Posted: Sun Nov 04, 2012 6:48 pm
by redcar777
@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.

Re: GR8767 Question 20

Posted: Sun Nov 04, 2012 7:58 pm
by vonLipwig
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.

Re: GR8767 Question 20

Posted: Sun Nov 04, 2012 10:45 pm
by Legendre
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.