6.2 Properties of power series  (Page 8/10)

Evaluate ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{3^n}}$ as $f^{\prime }\left(\frac{1}{3}\right)$ where $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{x}^n}.$

Evaluate ${\displaystyle \sum _{n=2}^{\infty }\frac{n\left(n-1\right)}{2^n}}$ as $f\text{″}\left(\frac{1}{2}\right)$ where $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{x}^n}.$

Evaluate ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\mathrm{-1}\right)}^n}{n+1}}$ as ${\displaystyle {\int }_0^{1}f\left(t\right)dt}$ where $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{\left(\mathrm{-1}\right)}^n{x}^{2n}}=\frac{1}{1+x^2}.$

$f\left(x\right)=\text{ln}x$ centered at $x=1$ ( Hint: $x=1-\left(1-x\right))$

$\text{ln}\left(1-x\right)$ at $x=0$

$\text{ln}\left(1-x^2\right)$ at $x=0$

$f\left(x\right)=\frac{2x}{{\left(1-x^2\right)}^2}$ at $x=0$

$f\left(x\right)={\text{tan}}^{\mathrm{-1}}\left(x^2\right)$ at $x=0$

$f\left(x\right)=\text{ln}\left(1+x^2\right)$ at $x=0$

$f\left(x\right)={\displaystyle {\int }_0^x\text{ln}tdt}$ where $\text{ln}\left(x\right)={\displaystyle \sum _{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n-1}\frac{{\left(x-1\right)}^n}{n}}$

[T] Evaluate the power series expansion $\text{ln}\left(1+x\right)={\displaystyle \sum _{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n-1}\frac{x^n}{n}}$ at $x=1$ to show that $\text{ln}\left(2\right)$ is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate $\text{ln}\left(2\right)$ accurate to within 0.001, and find such an approximation.

[T] Subtract the infinite series of $\text{ln}\left(1-x\right)$ from $\text{ln}\left(1+x\right)$ to get a power series for $\text{ln}\left(\frac{1+x}{1-x}\right).$ Evaluate at $x=\frac{1}{3}.$ What is the smallest N such that the N th partial sum of this series approximates $\text{ln}\left(2\right)$ with an error less than 0.001?

${\displaystyle \sum _{k=0}^{\infty }\left(x^k-x^{2k+1}\right)}$

${\displaystyle \sum _{k=1}^{\infty }\frac{x^{3k}}{6k}}$

${\displaystyle \sum _{k=1}^{\infty }{\left(1+x^2\right)}^{\text{−}k}}$ using $y=\frac{1}{1+x^2}$

${\displaystyle \sum _{k=1}^{\infty }{2}^{\text{−}kx}}$ using $y=2^{\text{−}x}$

Show that, up to powers x ³ and y ³ , $E\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{x^n}{n\text{!}}}$ satisfies $E\left(x+y\right)=E\left(x\right)E\left(y\right).$

Differentiate the series $E\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{x^n}{n\text{!}}}$ term-by-term to show that $E(x)$ is equal to its derivative.

Show that if $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ is a sum of even powers, that is, $a_n=0$ if n is odd, then $F={\displaystyle {\int }_0^{x}f\left(t\right)dt}$ is a sum of odd powers, while if f is a sum of odd powers, then F is a sum of even powers.

[T] Suppose that the coefficients a _n of the series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ are defined by the recurrence relation $a_n=\frac{a_{n-1}}{n}+\frac{a_{n-2}}{n\left(n-1\right)}.$ For $a_0=0$ and $a_1=1,$ compute and plot the sums $S_N={\displaystyle \sum _{n=0}^N{a}_n{x}^n}$ for $N=2,3,4,5$ on $\left[\mathrm{-1},1\right].$

[T] Suppose that the coefficients a _n of the series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ are defined by the recurrence relation $a_n=\frac{a_{n-1}}{\sqrt{n}}-\frac{a_{n-2}}{\sqrt{n\left(n-1\right)}}.$ For $a_0=1$ and $a_1=0,$ compute and plot the sums $S_N={\displaystyle \sum _{n=0}^N{a}_n{x}^n}$ for $N=2,3,4,5$ on $\left[\mathrm{-1},1\right].$

[T] Given the power series expansion $\text{ln}\left(1+x\right)={\displaystyle \sum _{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n-1}\frac{x^n}{n}},$ determine how many terms N of the sum evaluated at $x=\mathrm{-1}\text{/}2$ are needed to approximate $\text{ln}\left(2\right)$ accurate to within 1/1000. Evaluate the corresponding partial sum ${\displaystyle \sum _{n=1}^N{\left(\mathrm{-1}\right)}^{n-1}\frac{x^n}{n}}.$

[T] Given the power series expansion ${\text{tan}}^{\mathrm{-1}}\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{\left(\mathrm{-1}\right)}^k\frac{x^{2k+1}}{2k+1}},$ use the alternating series test to determine how many terms N of the sum evaluated at $x=1$ are needed to approximate ${\text{tan}}^{\mathrm{-1}}\left(1\right)=\frac{\pi }{4}$ accurate to within 1/1000. Evaluate the corresponding partial sum ${\displaystyle \sum _{k=0}^N{\left(\mathrm{-1}\right)}^k\frac{x^{2k+1}}{2k+1}}.$

[T] Recall that ${\text{tan}}^{\mathrm{-1}}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}.$ Assuming an exact value of $\left(\frac{1}{\sqrt{3}}\right),$ estimate $\frac{\pi }{6}$ by evaluating partial sums $S_N\left(\frac{1}{\sqrt{3}}\right)$ of the power series expansion ${\text{tan}}^{\mathrm{-1}}\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{\left(\mathrm{-1}\right)}^k\frac{x^{2k+1}}{2k+1}}$ at $x=\frac{1}{\sqrt{3}}.$ What is the smallest number N such that $6S_N\left(\frac{1}{\sqrt{3}}\right)$ approximates π accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?