Opalg said:[sp]\[\frac{2x-1}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+ \ldots\]
\[= \frac{\frac d{dx}(1-x+x^2)}{1-x+x^2} + \frac{\frac d{dx}(1-x^2+x^4)}{1-x^2+x^4}+\frac{\frac d{dx}(1-x^4+x^8)}{1-x^4+x^8}+ \ldots\]
\[= \frac d{dx}\bigl(\ln(1-x+x^2)\bigr) + \frac d{dx}\bigl(\ln(1-x^2+x^4)\bigr) + \frac d{dx}\bigl(\ln(1-x^4+x^8)\bigr) + \ldots\]
\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\right) + \ln\left(\frac{1+x^6}{1+x^3}\right) + \ln\left(\frac{1+x^{12}}{1+x^6}\right) + \ldots\right)\]
\[= \frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^3}\frac{1+x^{12}}{1+x^6} \cdots\right)\right)\]
\[= \lim_{n\to\infty} \frac d{dx}\left(\ln\left(\frac{1+x^{3\cdot2^n}}{1+x}\right)\right)\]
\[ = \frac d{dx}\left(\ln\left(\frac{1}{1+x}\right)\right)\]
\[= \frac d{dx}\bigl(-\ln(1+x)\bigr) = - \frac1{1+x}\]
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 $0<x<1$.
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Opalg said:[sp]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 $0<x<1$.
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[sp]Oops, you're right, the product does not telescope at all! It should belfdahl said:Hi, Opalg
I´m afraid, there is an error in your calculus :(
There is not a telescoping product, because:
\[1-x+x^2 = \frac{1+x^3}{1+x} \\\\ 1-x^2+x^4 = \frac{1+x^6}{1+x^2} \\\\ 1-x^4+x^8 = \frac{1+x^{12}}{1+x^4} \\\\ 1-x^8+x^{16} = \frac{1+x^{24}}{1+x^8} \;\;...\]
Opalg said:[sp]Oops, you're right, the product does not telescope at all! It should be
\[\frac d{dx}\left(\ln\left(\frac{1+x^3}{1+x}\frac{1+x^6}{1+x^2}\frac{1+x^{12}}{1+x^4}\frac{1+x^{24}}{1+x^8} \cdots\right)\right)\]
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.
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Olinguito said:IMHO the part that needs most justification is going from the fourth to the fifth line, where it’s assumed that
$$\sum_{n=1}^\infty\ln x_n\ =\ \ln\left(\prod_{n=1}^\infty x_n\right)$$
(which in turn assumes that the sum of the left and the product on the right converge).
An infinite sum is a mathematical concept that involves adding an infinite number of terms together. It is often denoted by the symbol ∑ and can be written as ∑n=1∞ an, where n represents the index or position of the term and an represents the value of each term.
To evaluate an infinite sum, you can use various methods such as the geometric series test, telescoping series, or partial sums. The method used depends on the type of series and its convergence or divergence.
The given infinite sum can be written as ∑n=1∞ (2x-1)/(1-x+x^2)+(4x^3-2x)/(1-x^2+x^4)+(8x^7-4x^3)/(1-x^4+x^8). This can also be written in a more compact form as ∑n=1∞ (2xn-1)/(1-xn+x2n).
To simplify the given infinite sum, you can use partial fractions and algebraic manipulation to rewrite the fraction terms into a form that can be summed using known series. In this case, you can use the formula for the sum of a geometric series to simplify each term.
The given infinite sum will converge for values of x that make the individual terms converge. This can be determined by using convergence tests such as the ratio test or root test. In this case, the sum will converge when |x| < 1.