Simplifying $\frac1{1+x}+\frac2{1+x^2}+\frac4{1+x^4}+\frac8{1+x^8}+\frac{16}{x^{16}-1}$

Question

We need to simplify $$\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{x^{16}-1}$$

The last denominator can be factored and we can get all the other denominators as factors of $x^{16}-1$. I tried handling the expressions in pairs,starting from the right.I also tried to take a common factor of two out of the numerators to help simplify,but that has yielded nothing.I then tried multiplying all the fractions to get $x^{16}-1$ in the denominator but that worsens things(I think so anyway).

So after doing the above things(and much more),I feel like I am running out of ideas.A really small hint will be appreciated.

If you split that last factor into two fractions with denominators $(x^8\pm 1)$ — Kal S., May 02 '14 at 05:50

Kal S. · Accepted Answer · 2017-02-14T11:17:22.067

12

$$\frac{16}{x^{16}-1}=\frac{8}{x^8-1}+\frac{-8}{x^8+1}$$

So the $4^{th}$ term of the original sum and the $2^{nd}$ part of the decomposition above are canceled. You are left with: $$ \dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{x^{8}-1} $$ Continue similarly. In the end you will get $\frac{1}{x-1}$.

edited Feb 14 '17 at 11:17

answered May 02 '14 at 05:56

Kal S.

3,781

Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help. – rah4927 May 02 '14 at 05:58
@rah4927 This is what I suggested originally in my comment. – Kal S. May 02 '14 at 05:59
I thought you were asking me to factor the denominator(sorry I didn't read your comment properly). – rah4927 May 02 '14 at 06:00
@Test123. I think that you will get $\frac{1}{x-1}$ – Claude Leibovici May 02 '14 at 06:50
@ClaudeLeibovici Yes, it was a typos there. Thank you – Kal S. May 02 '14 at 06:52
2

If you knew the number of typo's I can make ! – Claude Leibovici May 02 '14 at 06:53
Another way is to add $\dfrac{1}{1-x}$ to our expression and see how the sum telescopes. – rah4927 May 04 '14 at 12:57

Robert Israel · Answer 2 · 2014-05-02T15:15:59.293

4

More generally, $$ \sum_{n=0}^N \dfrac{2^n}{1+x^{2^n}} = \dfrac{2^{N+1}}{1-x^{2^{N+1}}} - \dfrac{1}{1-x}$$ as can be proven by induction.

edited May 02 '14 at 15:15

answered May 02 '14 at 06:16

Robert Israel

448,999

Similarly, $$\sum_{n=0}^N \dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 \cdot 3^n}} = \dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - \dfrac{1}{1 - x}$$ – Robert Israel May 02 '14 at 15:14

score 0 · Answer 3 · answered May 13 '14 at 11:21

0

Hint: Add $\dfrac{1}{1-x}$ to the given expression and see the sum telescope.

answered May 13 '14 at 11:21

rah4927

3,864

Simplifying $\frac1{1+x}+\frac2{1+x^2}+\frac4{1+x^4}+\frac8{1+x^8}+\frac{16}{x^{16}-1}$

3 Answers3

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