GR8767 Question 20
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
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
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 nonconstant functions that would satisfy the premises in the problem?
Does anyone know of any nonconstant functions that would satisfy the premises in the problem?

 Posts: 8
 Joined: Tue Mar 04, 2008 1:02 am
suppose that the degree of f is n (n>=1)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
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
This is wrong. Fundamental theorem of algebra does not guarantee the existence of roots over field of rational numbers.suppose that the degree of f is n (n>=1)
then there exists a is a root of f(x) then f(a)=0
Consider for example p(x) = (x5)^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 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) = (x5)^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
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.
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.
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.
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.
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
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 nonzero 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
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 nonzero 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.
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.
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 n1 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.
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 n1 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
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
Yes, thats it! f(x+1)  f(x) is exactly one degree smaller. So f(x) must be a constant.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.