if f(x) be a polynomial function satisfying
f(x).f(1/x) =f(x) + f(1/x) and f(4)=65
then find f(6)

Question

if f(x) be a polynomial function satisfying
f(x).f(1/x) =f(x) + f(1/x)   and f(4)=65 
then find f(6)

asnaseer · Answer

the polynomial can be written down as:$$f(x)=a_0+a_1x+a_2x^2+a_3x^2+...$$and$$f(\frac{1}{x})=a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\frac{a_3}{x^3}+...$$we are given:$$f(x).f(\frac{1}{x})=f(x)+f(\frac{1}{x})$$therefore:$$f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$$we are given f(4)=65, so:$$f(\frac{1}{4})=\frac{65}{65-1}=\frac{65}{64}$$and$$f(4)=65=a_0+4a_1+16a_2+64a_3+...$$$$f(\frac{1}{4})=\frac{65}{64}=a_0+\frac{a_1}{4}+\frac{a_2}{16}+\frac{a_3}{64}+...$$therefore, multiplying both sides of the last equation by 64, we get:$$65=64a_0+16a_1+4a_2+a_3+...$$
that's as far as I have been able to take this. is there any other information given with this question or are you able to continue from here?

anonymous · Answer

i  am trying from here....thanks

asnaseer · Answer

yw - do let me know if you happen to solve it - I'd be very interested in the solution.

mr.math · Answer

I think I found a function that satisfies the given conditions. Take $f(x)=x^2+1$!

asnaseer · Answer

is that through trial and error or analysis?

asnaseer · Answer

but f(4) != 65?

mr.math · Answer

I meant f(x)=x^3+1 :)

asnaseer · Answer

f(x)=1+16x will also work, as will many others (I think)
but which one is correct?

mr.math · Answer

Actually any polynomial of the form $x^n+1$, satisfies $f(x)f(\frac{1}{x})=f(x)+f(\frac{1}{x})$.

mr.math · Answer

Right?

asnaseer · Answer

I don't think so - e.g. $1+x^3$ does not work

asnaseer · Answer

I mean $1+x$

asnaseer · Answer

there must be another condition to pin down the solution to just one polynomial

mr.math · Answer

(1+x)(1/x+1)=1/x+1+1+x, right?

asnaseer · Answer

yes but it must also fit f(4)=65

mr.math · Answer

I was talking about the statement I said about polynomial of the form x^n+1.
x^3+1 satisfies the conditions in the problem, but as you said there are many other polynomials that do, and they would give different values for f(6). So, I think there's something missing in the problem.

asnaseer · Answer

actually Mr.Math I think you may have something there with $x^n+1$

anonymous · Answer

note that if you plug in x = 1, you get:$$f(1)^2 = 2f(1)$$which says f(1) = 0 or f(1) = 2. not that i know where to go, but that may help.

mr.math · Answer

Wait! Could you give me another example of such a polynomial?

mr.math · Answer

If this was in an exam, I would say $f(6)=6^3+1$. :-)

asnaseer · Answer

I think you are right

asnaseer · Answer

because with f(x)=16x+1, we get f(4)=65, but it does not satisfy the original conditions.

mr.math · Answer

Thank you! :-)

asnaseer · Answer

Although I would be a lot happier if we could prove this

mr.math · Answer

So am I.

asnaseer · Answer

i.e. prove that there are no other polynomials that could satisfy the original condition.

mr.math · Answer

I don't think that we have to do that. The question doesn't say that in anyway. We have only to show that f(6) of any such polynomial is 6^3+1.

mr.math · Answer

any way*

mr.math · Answer

Do you agree?

asnaseer · Answer

I think you have found the answer, but I personally would be interested in a proof that this is indeed the only such polynomial (which I believe it must be for the question to make any sense).

asnaseer · Answer

ok, I think I have a proof. from above I proved that:$$f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$$and we can write:$$\begin{align}
f(\frac{1}{x})=a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\frac{a_3}{x^3}+...+\frac{a_n}{x^n}\
=\frac{1}{x^n}(a_0x^n+a_1x^{n-1}+a_2x^{n-2}+a_3x^{n-3}+...+a_n)\
=\frac{f(x)}{f(x)-1}\quad\text{from above}\
\therefore (f(x)-1)(a_0x^n+a_1x^{n-1}+a_2x^{n-2}+a_3x^{n-3}+...+a_n)=x^nf(x)\
\therefore (a_0-1+a_1x+...+a_nx^n)(a_0x^n+a_1x^{n-1}+...+a_n)=x^n(a_0+a_1x+...+a_nx^n)\
=a_0x^n+a_1x^{n+1}+...+a_nx^{2n}
\end{align}$$
from inspection, we can infer:$$a_n(a_0-1)=0\implies a_n=0\quad\text{or }a_0=1$$
1. if we pursue the case of $a_n=0$ then we will successively get $a_{n-1}=0$, $a_{n-2}=0$, ..., $a_1=0$ and will end up with $a_0=2$.
this gives us $f(x)=f(\frac{1}{x})=2$ and satisfies the condition $f(x).f(\frac{1}{x})=f(x)+f(\frac{1}{x})$ but does not satisfy the condition $f(4)=65$. so we can reject this case.

2. if we pursue the case $a_0=1$ we will successively be able to prove that $a_1=0$, $a_2=0$, ...$a_{n-1}=0$, and will end up with:$$a_nx^n(x^n+a_n)-x^n+a_nx^{2n}$$from which we can prove $a_n=1$.
this gives us $f(x)=1+x^n$ and satisfies the condition $f(x).f(\frac{1}{x})=f(x)+f(\frac{1}{x})$. in order to satisfy the secondary condition $f(4)=65$, we need $n=3$. so we have proved that the only solution is:$$f(x)=1+x^3$$therefore:$$f(6)=1+6^3=1+216=217$$
In conclusion - you were right all along Mr.Math :-)

mr.math · Answer

Great! Nice work. If I might add that even the case where f(x)=2 is of the form 1+x^n, with n=0. Thus we can conclude (based on your proof) that ONLY polynomials of the form 1+x^n (n>=0) satisfies that condition.

asnaseer · Answer

good point - well spotted and thx for your insights :-)

anonymous · Answer

thanks guys..