## GR8767 Question 20

Forum for the GRE subject test in mathematics.
phcooh
Posts: 4
Joined: Mon Oct 13, 2008 7:15 pm

### GR8767 Question 20

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
Posts: 13
Joined: Wed Aug 06, 2008 9:25 am
f(x) = f(x+1) for all real x, not just integers

Posts: 24
Joined: Sun Oct 05, 2008 1:41 am
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
Posts: 8
Joined: Tue Mar 04, 2008 1:02 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.

Posts: 24
Joined: Sun Oct 05, 2008 1:41 am
Ralph,

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

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 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

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
Note : the degree of the ZERO polynomial is -infinity

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 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
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
There is a very recent thread on this question, pleasee see it also.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 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
Posts: 4
Joined: Thu Oct 23, 2008 11:27 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
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am
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
Posts: 36
Joined: Mon Oct 06, 2008 7:33 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
Posts: 1
Joined: Fri Nov 02, 2012 2:04 pm

### Re: GR8767 Question 20

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
Posts: 217
Joined: Wed Jun 03, 2009 1:05 am

### Re: GR8767 Question 20

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.

redcar777
Posts: 34
Joined: Sun Nov 04, 2012 11:59 am

### Re: GR8767 Question 20

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 ).

Last edited by redcar777 on Sun Nov 04, 2012 7:07 pm, edited 2 times in total.

vonLipwig
Posts: 51
Joined: Sat Mar 17, 2012 9:58 am

### Re: GR8767 Question 20

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.

redcar777
Posts: 34
Joined: Sun Nov 04, 2012 11:59 am

### Re: GR8767 Question 20

@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
Posts: 51
Joined: Sat Mar 17, 2012 9:58 am

### Re: GR8767 Question 20

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
Posts: 217
Joined: Wed Jun 03, 2009 1:05 am

### Re: GR8767 Question 20

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.