- Thread starter
- #1

#### lfdahl

##### Well-known member

- Nov 26, 2013

- 719

\[\frac{2x-1}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+ ...\]

- Thread starter lfdahl
- Start date

- Thread starter
- #1

- Nov 26, 2013

- 719

\[\frac{2x-1}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+ ...\]

- Moderator
- #2

- Feb 7, 2012

- 2,681

\[= \frac{\frac d{dx}(1-x+x^2)}{1-x+x^2} + \frac{\frac d{dx}(1-x^2+x^4)}{1-x^2+x^4}+\frac{\frac d{dx}(1-x^4+x^8)}{1-x^4+x^8}+ \ldots\]

\[= \frac d{dx}\bigl(\ln(1-x+x^2)\bigr) + \frac d{dx}\bigl(\ln(1-x^2+x^4)\bigr) + \frac d{dx}\bigl(\ln(1-x^4+x^8)\bigr) + \ldots\]

\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\right) + \ln\left(\frac{1+x^6}{1+x^3}\right) + \ln\left(\frac{1+x^{12}}{1+x^6}\right) + \ldots\right)\]

\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^3}\frac{1+x^{12}}{1+x^6} \cdots\right)\right)\]

\[= \lim_{n\to\infty} \frac d{dx}\left(\ln\left(\frac{1+x^{3\cdot2^n}}{1+x}\right)\right)\]

\[ = \frac d{dx}\left(\ln\left(\frac{1}{1+x}\right)\right)\]

\[= \frac d{dx}\bigl(-\ln(1+x)\bigr) = - \frac1{1+x}\]

Some of those steps may need heavy machinery to justify them (differentiating a series term by term, interchanging limits, ...). But that should all work satisfactorily for $0<x<1$.

- Thread starter
- #3

- Nov 26, 2013

- 719

Hi, Opalg

\[= \frac{\frac d{dx}(1-x+x^2)}{1-x+x^2} + \frac{\frac d{dx}(1-x^2+x^4)}{1-x^2+x^4}+\frac{\frac d{dx}(1-x^4+x^8)}{1-x^4+x^8}+ \ldots\]

\[= \frac d{dx}\bigl(\ln(1-x+x^2)\bigr) + \frac d{dx}\bigl(\ln(1-x^2+x^4)\bigr) + \frac d{dx}\bigl(\ln(1-x^4+x^8)\bigr) + \ldots\]

\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\right) + \ln\left(\frac{1+x^6}{1+x^3}\right) + \ln\left(\frac{1+x^{12}}{1+x^6}\right) + \ldots\right)\]

\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^3}\frac{1+x^{12}}{1+x^6} \cdots\right)\right)\]

\[= \lim_{n\to\infty} \frac d{dx}\left(\ln\left(\frac{1+x^{3\cdot2^n}}{1+x}\right)\right)\]

\[ = \frac d{dx}\left(\ln\left(\frac{1}{1+x}\right)\right)\]

\[= \frac d{dx}\bigl(-\ln(1+x)\bigr) = - \frac1{1+x}\]

Some of those steps may need heavy machinery to justify them (differentiating a series term by term, interchanging limits, ...). But that should all work satisfactorily for $0<x<1$.

I´m afraid, there is an error in your calculus

There is not a telescoping product, because:

\[1-x+x^2 = \frac{1+x^3}{1+x} \\\\ 1-x^2+x^4 = \frac{1+x^6}{1+x^2} \\\\ 1-x^4+x^8 = \frac{1+x^{12}}{1+x^4} \\\\ 1-x^8+x^{16} = \frac{1+x^{24}}{1+x^8} \;\;...\]

- Apr 22, 2018

- 251

Some of those steps may need heavy machinery to justify them (differentiating a series term by term, interchanging limits, ...). But that should all work satisfactorily for $0<x<1$.

$$\sum_{n=1}^\infty\ln x_n\ =\ \ln\left(\prod_{n=1}^\infty x_n\right)$$

(which in turn assumes that the sum of the left and the product on the right converge).

- Moderator
- #5

- Feb 7, 2012

- 2,681

Hi, Opalg

I´m afraid, there is an error in your calculus

There is not a telescoping product, because:

\[1-x+x^2 = \frac{1+x^3}{1+x} \\\\ 1-x^2+x^4 = \frac{1+x^6}{1+x^2} \\\\ 1-x^4+x^8 = \frac{1+x^{12}}{1+x^4} \\\\ 1-x^8+x^{16} = \frac{1+x^{24}}{1+x^8} \;\;...\]

\[\frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^2}\frac{1+x^{12}}{1+x^4}\frac{1+x^{24}}{1+x^8} \cdots\right)\right)\]

The denominators then form the product

\[(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots = 1+x + x^2 + x^3 + x^4 + x^5 + x^6 + \ldots = \frac1{1-x}.\]

The numerators are the same, with $x^3$ instead of $x$, so their product is $\dfrac1{1-x^3}.$ Therefore we need to find

\[\frac d{dx}\left(\ln\left(\frac{1-x}{1-x^3}\right)\right) = \frac d{dx}\left(\ln\left(\frac1{1+x+x^2}\right)\right) = \frac d{dx}\bigl(-\ln(1+x+x^2)\bigr) = -\frac{2x+1}{1+x+x^2}.\]

I hope that works better than my first attempt.

- Thread starter
- #6

- Nov 26, 2013

- 719

Thankyou for participating in this challenge, Opalg ! Yes sure, your 2nd attempt works perfect!

\[\frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^2}\frac{1+x^{12}}{1+x^4}\frac{1+x^{24}}{1+x^8} \cdots\right)\right)\]

The denominators then form the product

\[(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots = 1+x + x^2 + x^3 + x^4 + x^5 + x^6 + \ldots = \frac1{1-x}.\]

The numerators are the same, with $x^3$ instead of $x$, so their product is $\dfrac1{1-x^3}.$ Therefore we need to find

\[\frac d{dx}\left(\ln\left(\frac{1-x}{1-x^3}\right)\right) = \frac d{dx}\left(\ln\left(\frac1{1+x+x^2}\right)\right) = \frac d{dx}\bigl(-\ln(1+x+x^2)\bigr) = -\frac{2x+1}{1+x+x^2}.\]

I hope that works better than my first attempt.

- Thread starter
- #7

- Nov 26, 2013

- 719

Hi, Olinguito

$$\sum_{n=1}^\infty\ln x_n\ =\ \ln\left(\prod_{n=1}^\infty x_n\right)$$

(which in turn assumes that the sum of the left and the product on the right converge).

Thankyou for your sharp observation in Opalg ´s solution concerning convergence.

Do you perhaps have any idea of how to prove convergence for both sides of the equation? Or is it sufficient to show, that we´re dealing with two geometric series, and that $0<x<1$?