## Representation Of Functions By Power Series Homework Answers

Recall that we were able to analyze all geometric series "simultaneously'' to discover that $$\sum_{n=0}^\infty kx^n = {k\over 1-x},$$ if $|x|< 1$, and that the series diverges when $|x|\ge 1$. At the time, we thought of $x$ as an unspecified constant, but we could just as well think of it as a variable, in which case the series $$\sum_{n=0}^\infty kx^n$$ is a function, namely, the function $k/(1-x)$, as long as $|x|< 1$. While $k/(1-x)$ is a reasonably easy function to deal with, the more complicated $\sum kx^n$ does have its attractions: it appears to be an infinite version of one of the simplest function types—a polynomial. This leads naturally to the questions: Do other functions have representations as series? Is there an advantage to viewing them in this way?

The geometric series has a special feature that makes it unlike a typical polynomial—the coefficients of the powers of $x$ are the same, namely $k$. We will need to allow more general coefficients if we are to get anything other than the geometric series.

Definition 11.8.1 A power series has the form $$\ds\sum_{n=0}^\infty a_nx^n,$$ with the understanding that $\ds a_n$ may depend on $n$ but not on $x$.

Example 11.8.2 $\ds\sum_{n=1}^\infty {x^n\over n}$ is a power series. We can investigate convergence using the ratio test: $$ \lim_{n\to\infty} {|x|^{n+1}\over n+1}{n\over |x|^n} =\lim_{n\to\infty} |x|{n\over n+1} =|x|. $$ Thus when $|x|< 1$ the series converges and when $|x|>1$ it diverges, leaving only two values in doubt. When $x=1$ the series is the harmonic series and diverges; when $x=-1$ it is the alternating harmonic series (actually the negative of the usual alternating harmonic series) and converges. Thus, we may think of $\ds\sum_{n=1}^\infty {x^n\over n}$ as a function from the interval $[-1,1)$ to the real numbers.

A bit of thought reveals that the ratio test applied to a power series will always have the same nice form. In general, we will compute $$ \lim_{n\to\infty} {|a_{n+1}||x|^{n+1}\over |a_n||x|^n} =\lim_{n\to\infty} |x|{|a_{n+1}|\over |a_n|} = |x|\lim_{n\to\infty} {|a_{n+1}|\over |a_n|} =L|x|, $$ assuming that $\ds \lim |a_{n+1}|/|a_n|$ exists. Then the series converges if $L|x|< 1$, that is, if $|x|< 1/L$, and diverges if $|x|>1/L$. Only the two values $x=\pm1/L$ require further investigation. Thus the series will definitely define a function on the interval $(-1/L,1/L)$, and perhaps will extend to one or both endpoints as well. Two special cases deserve mention: if $L=0$ the limit is $0$ no matter what value $x$ takes, so the series converges for all $x$ and the function is defined for all real numbers. If $L=\infty$, then no matter what value $x$ takes the limit is infinite and the series converges only when $x=0$. The value $1/L$ is called the **radius of convergence** of the series, and the interval on which the series converges is the **interval of convergence** .

Consider again the geometric series, $$\sum_{n=0}^\infty x^n={1\over 1-x}.$$ Whatever benefits there might be in using the series form of this function are only available to us when $x$ is between $-1$ and $1$. Frequently we can address this shortcoming by modifying the power series slightly. Consider this series: $$ \sum_{n=0}^\infty {(x+2)^n\over 3^n}= \sum_{n=0}^\infty \left({x+2\over 3}\right)^n={1\over 1-{x+2\over 3}}= {3\over 1-x}, $$ because this is just a geometric series with $x$ replaced by $(x+2)/3$. Multiplying both sides by $1/3$ gives $$\sum_{n=0}^\infty {(x+2)^n\over 3^{n+1}}={1\over 1-x},$$ the same function as before. For what values of $x$ does this series converge? Since it is a geometric series, we know that it converges when $$\eqalign{ |x+2|/3&< 1\cr |x+2|&< 3\cr -3 < x+2 &< 3\cr -5< x&< 1.\cr }$$ So we have a series representation for $1/(1-x)$ that works on a larger interval than before, at the expense of a somewhat more complicated series. The endpoints of the interval of convergence now are $-5$ and $1$, but note that they can be more compactly described as $-2\pm3$. We say that $3$ is the radius of convergence, and we now say that the series is centered at $-2$.

Definition 11.8.3 A power series centered at $a$ has the form $$\ds\sum_{n=0}^\infty a_n(x-a)^n,$$ with the understanding that $\ds a_n$ may depend on $n$ but not on $x$.

## Exercises 11.8

Find the radius and interval of convergence for each series. In exercises 3 and 4, do not attempt to determine whether the endpoints are in the interval of convergence.

**Ex 11.8.1** $\ds\sum_{n=0}^\infty n x^n$ (answer)

**Ex 11.8.2** $\ds\sum_{n=0}^\infty {x^n\over n!}$ (answer)

**Ex 11.8.3** $\ds\sum_{n=1}^\infty {n!\over n^n}x^n$ (answer)

**Ex 11.8.4** $\ds\sum_{n=1}^\infty {n!\over n^n}(x-2)^n$ (answer)

**Ex 11.8.5** $\ds\sum_{n=1}^\infty {(n!)^2\over n^n}(x-2)^n$ (answer)

**Ex 11.8.6** $\ds\sum_{n=1}^\infty {(x+5)^n\over n(n+1)}$ (answer)

- Абсолютно. Скажи папе, что все в порядке». Но нутром он чувствовал, что это далеко не. Интуиция подсказывала ему, что в глубинах дешифровального чудовища происходит что-то необычное.

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