Examples

Geometric series: algebra meets calculus

Consider the geometric series as a power series functions:

$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$

Take the derivative of both sides of the function:

$$\frac{d}{dx}\left(\frac{1}{1-x}\right)\gg \gg \frac{1}{(1-x{)}^2}\gg \gg {\left(\frac{1}{1-x}\right)}^2$$

This means $f$ satisfies the identity:

$$f^{\prime }=f^2$$

Now compute the derivative of the series:

$$1+x+x^2+x^3+\cdots \stackrel{\frac{d}{dx}}{\gg \gg }1+2x+3x^2+4x^3+\cdots$$

On the other hand, compute the square of the series:

$${\left(1+x+x^2+x^3+\cdots \right)}^2\gg \gg 1+2x+3x^2+4x^3+\cdots$$

So we find that the same relationship holds, namely $f^{\prime }=f^2$, for the closed formula and the series formula for this function.

Manipulating geometric series: algebra

Find power series that represent the following functions:

(a) $\frac{1}{1+x}$ (b) $\frac{1}{1+x^2}$ (c) $\frac{x^3}{x+2}$ (d) $\frac{3x}{2-5x}$

Solution

(a) $\frac{1}{1+x}$

Rewrite in format $\frac{1}{1-u}$:

$$\frac{1}{1+x}\gg \gg \frac{1}{1-(-x)}$$

Choose $u=-x$. Plug into geometric series:

$$\begin{array}{c}\frac{1}{1-u}=1+u+u^2+u^3+\cdots \\ \\ \frac{1}{1-(-x)}=1+(-x)+(-x{)}^2+(-x{)}^3+\cdots \\ \\ \gg \gg 1-x+x^2-x^3+\cdots \end{array}$$

Therefore:

$$\frac{1}{1+x}=1-x+x^2-x^3+\cdots =\sum _{n=0}^{\infty }(-1{)}^n{x}^n$$

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(b) $\frac{1}{1+x^2}$

Rewrite in format $\frac{1}{1-u}$:

$$\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}$$

Choose $u=-x^2$. Plug into geometric series:

$$\begin{array}{c}\frac{1}{1-u}=1+u+u^2+u^3+\cdots \\ \\ \frac{1}{1-(-x^2)}=1+(-x^2)+(-x^2{)}^2+(-x^2{)}^3+\cdots \\ \\ \gg \gg 1-x^2+x^4-x^6+\cdots \\ \end{array}$$

Therefore:

$$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\cdots =\sum _{n=0}^{\infty }(-1{)}^n{x}^{2n}$$

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(c) $\frac{x^3}{x+2}$

Rewrite in format $A{x}^3\cdot \frac{1}{1-u}$:

$$\begin{array}{c}\frac{x^3}{x+2}\gg \gg x^3\cdot \frac{1}{2+x}\gg \gg x^3\cdot \frac{1}{2\left(1+\frac{x}{2}\right)}\\ \\ \gg \gg \frac{1}{2}{x}^3\cdot \frac{1}{1+\frac{x}{2}}\gg \gg \frac{1}{2}{x}^3\cdot \frac{1}{1-\left(-\frac{x}{2}\right)}\end{array}$$

Choose $u=-\frac{x}{2}$. Here $A{x}^3=\frac{1}{2}{x}^3$. Plug into geometric series:

$$\begin{array}{c}\frac{1}{1-u}=1+u+u^2+u^3+\cdots \\ \\ \frac{1}{1-\left(-\frac{x}{2}\right)}=\sum _{n=0}^{\infty }{\left(-\frac{x}{2}\right)}^n=1+(-\frac{x}{2})+(-\frac{x}{2}{)}^2+(-\frac{x}{2}{)}^3+\cdots \\ \\ \gg \gg \sum _{n=0}^{\infty }(-1{)}^n\frac{1}{2^n}{x}^n=1-\frac{1}{2}x+\frac{1}{4}{x}^2-\frac{1}{8}{x}^3+\cdots \end{array}$$

Therefore:

$$\begin{array}{c}\frac{x^3}{x+2}\gg \gg \frac{1}{2}{x}^3\cdot \frac{1}{1-\left(-\frac{x}{2}\right)}\\ \\ \gg \gg \frac{1}{2}{x}^3\left(\sum _{n=0}^{\infty }(-1{)}^n\frac{1}{2^n}{x}^n\right)\\ \\ \gg \gg \sum _{n=0}^{\infty }(-1{)}^n\frac{1}{2^{n+1}}{x}^{n+3}=\frac{1}{2}{x}^3-\frac{1}{4}{x}^4+\frac{1}{8}{x}^5-\frac{1}{16}{x}^6+\cdots \end{array}$$

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(d) $\frac{3x}{2-5x}$

Rewrite in format $Ax\cdot \frac{1}{1-u}$:

$$\begin{array}{rl}\frac{3x}{2-5x}\gg \gg 3x\cdot \frac{1}{2-5x}& \\ & \\ \gg \gg 3x\cdot \frac{1}{2\left(1-\frac{5x}{2}\right)}\gg \gg & \frac{3}{2}x\cdot \frac{1}{1-\frac{5}{2}x}\end{array}$$

Choose $u=\frac{5}{2}x$. Here $Ax=\frac{3}{2}x$. Plug into geometric series:

$$\begin{array}{c}\frac{1}{1-u}=1+u+u^2+u^3+\cdots \\ \\ \frac{1}{1-\left(\frac{5}{2}x\right)}=\sum _{n=0}^{\infty }{\left(\frac{5}{2}x\right)}^n=1+(\frac{5x}{2})+(\frac{5x}{2}{)}^2+(\frac{5x}{2}{)}^3+\cdots \\ \\ \gg \gg \sum _{n=0}^{\infty }\frac{5^n}{2^n}{x}^n=1+\frac{5}{2}x+\frac{25}{4}{x}^2+\frac{125}{8}{x}^3+\cdots \end{array}$$

Therefore:

$$\begin{array}{c}\frac{3x}{2-5x}\gg \gg \frac{3}{2}x\cdot \frac{1}{1-\frac{5}{2}x}\\ \\ \gg \gg \frac{3}{2}x\left(\sum _{n=0}^{\infty }\frac{5^n}{2^n}{x}^n\right)\\ \\ \gg \gg \sum _{n=0}^{\infty }\frac{3\cdot 5^n}{2^{n+1}}{x}^{n+1}=\frac{3}{2}x+\frac{15}{4}{x}^2+\frac{75}{8}{x}^3+\frac{375}{16}{x}^4+\cdots \end{array}$$

Manipulating geometric series: calculus

Find a power series that represents $\ln (1+x)$.

Solution

Differentiate to obtain similarity to geometric sum formula:

$$\begin{array}{c}\frac{d}{dx}\ln (1+x)\gg \gg \frac{1}{1+x}\\ \\ \gg \gg \frac{1}{1-(-x)}\gg \gg 1-x+x^2-x^3+x^4-\cdots \end{array}$$

Integrate series to find original function:

$$\begin{array}{c}\int \frac{1}{1-(-x)}dx=\int 1-x+x^2-x^3+x^4-\cdots dx\\ \\ \gg \gg \ln (1+x)=D+x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+\cdots \end{array}$$

Use known point to solve for $D$:

$$\begin{array}{c}\ln (1+0)=D+0+0+\cdots \gg \gg 0=D\\ \\ \gg \gg \ln (1+x)=x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+\cdots \end{array}$$

Recognizing and manipulating geometric series: Part I

(a) Evaluate $\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{1}{n}$. (Hint: consider the series of $\ln (1-x)$.)

(b) Find a series approximation for $\ln (2/3)$.

Solution

(a)

(1) Follow hint, study series of $\ln (1-x)$:

Notice:

$$\frac{d}{dx}\ln (1-x)=\frac{-1}{1-x}\gg \gg -(1+x+x^2+x^3+\cdots )$$

Integrate the series:

$$\begin{array}{c}\int \frac{-1}{1-x}dx=\int -1-x-x^2-\cdots dx\\ \\ \gg \gg \ln (1-x)=C-x-\frac{1}{2}{x}^2-\frac{1}{3}{x}^3-\cdots \end{array}$$

Solve for $C$ using $\ln (1-0)=0$ which (plugging above) implies $0=C-0-0-\cdots$ and thus $C=0$. So:

$$\ln (1-x)=-x-\frac{1}{2}{x}^2-\frac{1}{3}{x}^3-\cdots =\sum _{n=1}^{\infty }-\frac{1}{n}{x}^n$$

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(2) Relate to the given series:

Notice that $x^n=(-1{)}^n$ if we set $x=-1$.
Also, $-(-1{)}^n=(-1{)}^{n-1}$.
Therefore:

$$\ln (1-(-1))\gg \gg \sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{1}{n}$$

So the answer is $\ln 2$.

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(b) Find a series approximation for $\ln (2/3)$:

Observe that $\ln (2/3)=\ln (1-1/3)$.

Plug $x=1/3$ into the series: $\ln (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots$

$$\begin{array}{c}\ln (1-1/3)\gg \gg -1/3-\frac{(1/3{)}^2}{2}-\frac{(1/3{)}^3}{3}-\cdots \\ \\ \gg \gg -\frac{1}{3}-\frac{1}{3^2\cdot 2}-\frac{1}{3^3\cdot 3}-\cdots \\ \\ \gg \gg \sum _{n=1}^{\infty }-\frac{1}{n{3}^n}\end{array}$$

Recognizing and manipulating geometric series: Part II

(a) Find a series representing ${\tan }^{-1}(x)$ using differentiation.

(b) Find a series representing $\int \frac{dx}{1+x^4}$.

Solution

(a)

Notice that $\frac{d}{dx}{\tan }^{-1}(x)=\frac{1}{1+x^2}$.

What is the series for $\frac{1}{1+x^2}$?

Let $u=-x^2$:

$$\begin{array}{c}\frac{1}{1+x^2}\gg \gg \frac{1}{1-u}=1+u+u^2+\cdots \\ \\ \gg \gg 1-x^2+x^4-x^6+x^8-\cdots \end{array}$$

Now integrate this by terms:

$$\begin{array}{c}\int \frac{1}{1+x^2}dx=\int 1-x^2+x^4-x^6+x^8-\cdots dx\\ \\ \gg \gg C+x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots \end{array}$$

Conclude:

$${\tan }^{-1}(x)=C+x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$

Plug in $0$ to solve for $C$:

$$\begin{array}{c}{\tan }^{-1}(0)=C+0+\cdots +0\gg \gg C=0\end{array}$$

Final answer:

$${\tan }^{-1}(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$

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(b)

Rewrite integrand in format of geometric series sum:

$$\frac{1}{1+x^4}\gg \gg \frac{1}{1-(-x^4)}\gg \gg \frac{1}{1-u},u=-x^4$$

Therefore:

$$\begin{array}{c}\frac{1}{1-u}=1+u+u^2+u^3+\cdots \\ \\ \gg \gg 1-x^4+x^8-x^{12}+x^{16}-\cdots =\sum _{n=0}^{\infty }(-1{)}^n{x}^{4n}\end{array}$$

Integrate the series by terms to obtain the answer:

$$\begin{array}{c}\int 1-x^4+x^8-x^{12}+x^{16}-\cdots dx\\ \\ \gg \gg C+x-\frac{x^5}{5}+\frac{x^9}{9}-\frac{x^{13}}{13}+\frac{x^{17}}{17}-\cdots \end{array}$$