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I don't really get how any of this works. I tried looking at this power function example already to figure it out. But I got lost when this part comes up. $$\sum_{n=0}^{\infty}x^n=x^0+\sum_{n=1}^{\infty}x^n=1+\sum_{n=1}^{\infty}x^n$$ Why is this entire step happening and why is the $n$ index moving? And do you just distribute the $\sum_{n=1}^{\infty}x^n$ into $(1+x)$?

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  • $\begingroup$ Are you asking why you would want to do that, or why you are permitted to do that? $\endgroup$ – Ian Dec 5 '15 at 17:14
  • $\begingroup$ @Ian Kind of both I guess? I'm trying to use this to be a guide to find the power series representation of $\frac{8+x}{x-1}$ $\endgroup$ – England Dec 5 '15 at 17:17
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Sigma summation is simply a collection of terms that are added together, so the step you are looking at is basically $(x^0+x^1+x^2+\cdots+x^n)=x^0+(x^1+x^2+\cdots+x^n)=1+(x^1+x^2+\cdots+x^n)$

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  • $\begingroup$ Thank you! You're explanation made more sense with what i'm trying to solve with my homework. It's like adding the two sums together right? So it'd be: $$1 + 2\sum_{n=1}^{\infty}x^n$$ because we're just adding the like terms of the series right? Or am I totally wrong? Just a quick side question, how would I find the radius of $8+9\sum_{n=1}^{\infty}x^n$? Do I just use the $x^n$? $\endgroup$ – England Dec 5 '15 at 17:36
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You have a sum of two geometric series. \begin{align} \frac 1 {1-x} & = 1 + x + x^2 + x^3 + \cdots \\[10pt] \frac x {1-x} & = x + x^2 + x^3 + x^4 + \cdots \end{align} You're looking for the sum of the two above. \begin{align} \sum_{n=0}^\infty x^n & = x^0 + x^1 + x^2 + x^3 + x^4 + \cdots \\[10pt] & = x^0 + \Big( x^1 + x^2 + x^3 + x^4 + \cdots \Big) \\[10pt] & = x_0 + \sum_{n=1}^\infty x^n \\[10pt] & = 1+\sum_{n=1}^{\infty}x^n \end{align}

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As you know:

$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$

You are trying to find the power series representation of $\frac{1+x}{1-x}$.

First thing to do is to represent the fraction in a form similar seen at the beginning:

$$\frac{1+x}{1-x} = \frac{1}{1-x} + \frac{x}{1-x} = \sum_{n=0}^{\infty}x^n + \frac{x}{1-x}$$

Now we have to deal with the term $\frac{x}{1-x}$:

If we divide by $x$ on the top and bottom we get:

$$\frac{x}{1-x} = \frac{1}{\frac{1}{x} - 1} = -\frac{1}{1 -\frac{1}{x}} = -\sum_{n=0}^{\infty}\frac{1}{x^n}$$

Now we can represent our sum as:

$$\frac{1+x}{1-x} = \sum_{n=0}^{\infty}\left(x^n - \frac{1}{x^n}\right)$$

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    $\begingroup$ This is not a power series in $x$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 5 '15 at 17:35
  • $\begingroup$ @MichaelHardy How so? Is my math incorrect? $\endgroup$ – Varun Iyer Dec 5 '15 at 17:36
  • $\begingroup$ "Power series in $x$" means something of the form $\displaystyle\sum_{n=0}^\infty a_n x^n$, in which the powers of $x$ are $x^0,x^1,x^2,x^3,\ldots$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 5 '15 at 17:37
  • $\begingroup$ Remark: Note that your $\frac{1}{x}$-move spoils the possibility to evaluate your sum representation for $\frac{1+x}{1-x}$ at $x=0$. Similat story for $\lim_{x\to\infty}$. Simulatnously, the naive termwise substraction of infinite sums makes each $x^n-\frac{1}{x^n}$ well defined at $x=1$, whereas $\frac{1+x}{1-x}$ is not. $\endgroup$ – Nikolaj-K Dec 5 '15 at 17:39
  • $\begingroup$ @MichaelHardy I never realized that! Thanks for the math lesson. $\endgroup$ – Varun Iyer Dec 5 '15 at 17:46

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