How to expand this ratio of polynomials?

[Gourab_chill]

Homework Statement

I've got the question in the attachments. I've got doubt regarding the expansion of 1+x^n

Relevant Equations

i don't know the expansions

I could simplify the expressions in the numerator and denominator to (1+x^n)/(1+x) as they are in geometric series and I used the geometric sum formula to reduce it. Now for what value of n will it be a polynomial?
I do get the idea for some value of n the simplified numerator will contain the (1+x) factor in it. What sort of expansion or formula should be used here?

 

[etotheipi]

You want to be able to state $(x^n + 1) = (x+1)P(x)$, with $P(x)$ as a polynomial. Try and fill in the gaps, i.e. work out the form of $P(x)$, and see if that gives you any clues.

 

[archaic]

I'd suggest that you first perform a long division for two or three terms, and figure out an expression for the quotient and rest as a function of $i=$ the number of operations performed.

 

[epenguin]

Homework Statement:: I've got the question in the attachments. I've got doubt regarding the expansion of 1+x^n
Relevant Equations:: i don't know the expansions

I could simplify the expressions in the numerator and denominator to (1+x^n)/(1+x) as they are in geometric series and I used the geometric sum formula to reduce it. Now for what value of n will it be a polynomial?
I do get the idea for some value of n the simplified numerator will contain the (1+x) factor in it. What sort of expansion or formula should be used here?

Probably rather than things like (1 + x) you should be looking at (1 - x) but you have and advantage over me - you know what the question is.

 

[Mark44]

Probably rather than things like (1 + x) you should be looking at (1 - x) but you have and advantage over me - you know what the question is.

I have edited post #1 so that the question is now an inline image.

 

[epenguin]

I have edited post #1 so that the question is now an inline image.

I see the same as what I saw before but I can't see a question.

(I think I see the answer though, we get used to this sort of thing. )

 

[Mark44]

I see the same as what I saw before but I can't see a question.

From the image
If p(x) = <rational function in image> then n can be
It's missing a question mark.

 

[epenguin]

OK I simplified the polynomial ratio only to then notice the student had already done the same. Otherwise have no idea what the question is looking for - could it be one of these Indian examiner trap questions ?

 

[etotheipi]

OK I simplified the polynomial ratio only to then notice the student had already

l, done the same. Otherwise have no idea what the question is looking for - could it be one of these Indian examiner trap questions ?

We can only write it in the form $(x^n + 1) = (x+1)P(x)$ if $n$ takes a certain set of values. The easiest way to do this IMO is to think what all of the terms in $P(x)$ have to be, but @archaic's suggestion is also equivalent.

 

[archaic]

l, done the same. Otherwise have no idea what the question is looking for - could it be one of these Indian examiner trap questions ?

Try dividing $x^n+1$ by $x+1$, you'll see that for some $n$s the rest is $0$.

 

[ehild]

Homework Statement:: I've got the question in the attachments. I've got doubt regarding the expansion of 1+x^nI could simplify the expressions in the numerator and denominator to (1+x^n)/(1+x) as they are in geometric series and I used the geometric sum formula to reduce it. Now for what value of n will it be a polynomial?

Consider the geometric series $1-x+x^2-x^3+...+(-x)^{n-1}$. What is the sum if n is odd/even?

 

[haruspex]

In case @ehild's hint looks too hard to have thought of for yourself, think what P(x) being divisible by 1+x says about P(-1).

 

[Gourab_chill]

In case @ehild's hint looks too hard to have thought of for yourself, think what P(x) being divisible by 1+x says about P(-1).

Yes, I never thought about this! this actually makes it way easier!

I tried dividing the 1+x^n by 1+x also as others said in this section and I did find n can take only odd values.

@ehild I'm not sure about the sum you said, it is different from my question; it would work for 1+x+x^2+...+x^(n-1)?

 

[ehild]

Yes, I never thought about this! this actually makes it way easier!

I tried dividing the 1+x^n by 1+x also as others said in this section and I did find n can take only odd values.

@ehild I'm not sure about the sum you said, it is different from my question; it would work for 1+x+x^2+...+x^(n-1)?

It is a different sum but also a geometric series, a polynomial of x. It helps you to decide when (x^n+1)/(x+1) is a polynomial, without the long division. The sum of the series is $$1-x+x^2-x^3+...+.(-x)^{n-1}=\frac{(-x)^{n}-1}{-x-1}$$, For what n-s is it equal to $$\frac {x^n+1}{x+1}$$?

 

[Gourab_chill]

It is a different sum but also a geometric series, a polynomial of x. It helps you to decide when (x^n+1)/(x+1) is a polynomial, without the long division. The sum of the series is $$1-x+x^2-x^3+...+.(-x)^{n-1}=\frac{(-x)^{n}-1}{-x-1}$$, For what n-s is it equal to $$\frac {x^n+1}{x+1}$$?

I had made a silly error in calculating the sum of the series before which you mentioned; you are right indeed! I didn't visualize the expression could be like this!

 

[Charles Link]

To perhaps simplify things, $x^n+1$ will have $x+1$ as a factor, if and only if $x^n+1=0$ has $x=-1$ as a root. Plug in $x=-1$ to $x^n+1 =0$ and see what you get.

 

[Gourab_chill]

To perhaps simplify things, $x^n+1$ will have $x+1$ as a factor, if and only if $x^n+1=0$ has $x=-1$ as a root. Plug in $x=-1$ to $x^n+1 =0$ and see what you get.

yes I did that as @haruspex has mentioned already :)