MHB How Can You Convert \( \frac{1}{1+x^{2m}} \) from a Product to a Series?

[Kiwi1]

I would like to convert:

\frac{1}{1+x^{2m}}

into a sum of terms. Preferably m terms but 2m terms would be OK.

I start off with:

\frac{1}{1+x^{2m}}=\frac{1}{\Pi^k_1(x-e^{i\frac{2k-1}{2m}\pi})(x-e^{-i\frac{2k-1}{2m}\pi})}=\frac{1}{\Pi^k_1 (x^2-2x \cdot cos \frac{2k-1}{2m}\pi +1)} where k = 1 to m.

And I hope that this may be converted from a product of m terms into a sum of m terms?

I don't know for certain that this is possible. But I expect it probably is because the integral of this entity is a sum of k terms.

 

[Prove It]

I would like to convert:

\frac{1}{1+x^{2m}}

into a sum of terms. Preferably m terms but 2m terms would be OK.

I start off with:

\frac{1}{1+x^{2m}}=\frac{1}{\Pi^k_1(x-e^{i\frac{2k-1}{2m}\pi})(x-e^{-i\frac{2k-1}{2m}\pi})}=\frac{1}{\Pi^k_1 (x^2-2x \cdot cos \frac{2k-1}{2m}\pi +1)} where k = 1 to m.

And I hope that this may be converted from a product of m terms into a sum of m terms?

I don't know for certain that this is possible. But I expect it probably is because the integral of this entity is a sum of k terms.

$\displaystyle \begin{align*} \frac{1}{1 + x^{2m}} = \frac{1}{1 - \left( -x^{2m} \right) } \end{align*}$

Now notice that a geometric series $\displaystyle \begin{align*} \sum_{n = 0}^{\infty} r^n = \frac{1}{1 - r} \end{align*}$, which is convergent where $\displaystyle \begin{align*} |r| < 1 \end{align*}$. What you have is simply the closed form of a geometric series with $\displaystyle \begin{align*} r = -x^{2m} \end{align*}$. So...

$\displaystyle \begin{align*} \frac{1}{1 - \left( - x^{2m} \right) } &= \sum_{n = 0}^{\infty} \left( -x^{2m} \right) ^n \\ &= \sum_{n = 0}^{\infty} \left[ \left( -1 \right) ^n \, x^{2\,m\,n} \right] \end{align*}$

and this is convergent where $\displaystyle \begin{align*} \left| -x^{2m} \right| < 1 \implies \left| x \right| ^{2m} < 1 \implies |x| < 1^{\frac{1}{2m}} \implies |x| < 1 \end{align*}$.

 

[chisigma]

I would like to convert:

\frac{1}{1+x^{2m}}

into a sum of terms. Preferably m terms but 2m terms would be OK.

I start off with:

\frac{1}{1+x^{2m}}=\frac{1}{\Pi^k_1(x-e^{i\frac{2k-1}{2m}\pi})(x-e^{-i\frac{2k-1}{2m}\pi})}=\frac{1}{\Pi^k_1 (x^2-2x \cdot cos \frac{2k-1}{2m}\pi +1)} where k = 1 to m.

And I hope that this may be converted from a product of m terms into a sum of m terms?

I don't know for certain that this is possible. But I expect it probably is because the integral of this entity is a sum of k terms.

In...

http://mathhelpboards.com/analysis-50/too-difficult-integral-1842.html#post8616

... it is reported the general expansion...

$\displaystyle\frac{1}{1+t^{n}}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{w_{n,k}^{1-n}}{t -w_{n,k}}\ ,\ w_{n,k}=e^{\ i\ \frac{2k+1}{n}\ \pi}\ (1)$

Kind regards

$\chi$ $\sigma$

 

[Kiwi1]

Thanks Chisigma, I am actually trying to solve a similar "too difficult" integral.

 

[Opalg]

I would like to convert:

\frac{1}{1+x^{2m}}

into a sum of terms. Preferably m terms but 2m terms would be OK.

I start off with:

\frac{1}{1+x^{2m}}=\frac{1}{\Pi^k_1(x-e^{i\frac{2k-1}{2m}\pi})(x-e^{-i\frac{2k-1}{2m}\pi})}=\frac{1}{\Pi^k_1 (x^2-2x \cdot cos \frac{2k-1}{2m}\pi +1)} where k = 1 to m.

And I hope that this may be converted from a product of m terms into a sum of m terms?

I don't know for certain that this is possible. But I expect it probably is because the integral of this entity is a sum of k terms.

What you want to do here is to express $\frac{1}{1+x^{2m}}$ as a sum of partial fractions. There is a neat trick for doing this.

Suppose that $x-\alpha$ is a factor of the polynomial $p(x)$. The cover-up rule says that the coefficient $A$ in the fraction $\frac A{x-\alpha}$ that occurs in the partial fraction decomposition of $\frac1{p(x)}$ is obtained by omitting the factor $x-\alpha$ from the denominator $p(x)$ (in effect, dividing $p(x)$ by $x-\alpha$) and then putting $x=\alpha$. You can express this in terms of a limit by saying that $$A = \lim_{x\to\alpha}\frac{x-\alpha}{p(x)}.$$ But that limit is $$\frac1{p'(\alpha)}.$$

Now apply that trick to the polynomial $1+x^{2m}$ and let $\omega = e^{i\frac{2k-1}{2m}\pi}.$ You find that the coefficient of the term $\frac1{x-\omega}$ in the partial fraction decomposition of $\frac1{1+x^{2m}}$ is $\frac1{2m\omega^{2m-1}}.$ But $\omega^{2m} = 1$, so the coefficient is $\frac{\omega}{2m}.$ Therefore $$\frac1{1+x^{2m}} = \frac1{2m}\sum_\omega \frac\omega{x-\omega},$$ where the sum is taken over all the $(2m)$th roots of $-1.$

If you combine the terms involving $\omega$ and $\overline{\omega}$, where $\omega = e^{i\frac{2k-1}{2m}\pi}$, then you find that $$\frac1{1+x^{2m}} = \frac1m \sum_{k=1}^m \frac{x\cos\bigl( \frac{2k-1}{2m}\pi\bigr) -1}{x^2 - 2x\cos\bigl( \frac{2k-1}{2m}\pi\bigr) + 1}.$$

Edit. If I had read it more carefully, I would have seen that chisigma's comment above already gives that same partial fraction decomposition.