Theory

Theory 1

Suppose that we have a power series function:

$$f(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$$

Consider the successive derivatives of $f$:

$$\begin{array}{ccccccccccccc}f(x)& =& a_0& +& a_{1}x& +& a_2{x}^2& +& a_3{x}^3& +& a_4{x}^4& +& \cdots \\ f^{\prime }(x)& =& 0& +& a_1& +& 2\cdot a_2{x}^1& +& 3\cdot a_3{x}^2& +& 4\cdot a_4{x}^3& +& \cdots \\ f^{\prime \prime }(x)& =& 0& +& 0& +& 2\cdot a_2& +& 3\cdot 2\cdot a_3{x}^1& +& 4\cdot 3\cdot a_4{x}^2& +& \cdots \\ f^{\prime \prime \prime }(x)& =& 0& +& 0& +& 0& +& 3\cdot 2\cdot 1\cdot a_3& +& 4\cdot 3\cdot 2\cdot a_4{x}^1& +& \dots \\ ⋮& & & ⋮& & & ⋮& & & ⋮& & & \\ f^{(n)}(x)& =& 0& +& 0& +& 0& +& 0& +& \cdots +n!\cdot a_n& +& \cdots \end{array}$$

When these functions are evaluated at $x=0$, all terms with a positive $x$-power become zero:

$$\begin{array}{ccccc}f(0)& =& a_0& =& a_0\\ f^{\prime }(0)& =& a_1& =& a_1\\ f^{\prime \prime }(0)& =& 2\cdot a_2& =& 2!\cdot a_2\\ f^{\prime \prime \prime }(0)& =& 3\cdot 2\cdot a_3& =& 3!\cdot a_3\\ ⋮& =& ⋮& =& ⋮\\ f^{(n)}(0)& =& n\cdot (n-1)\cdots 2\cdot 1\cdot a_n& =& n!\cdot a_n\end{array}$$

This last formula is the basis for Taylor and Maclaurin series:

Power series: Derivative-Coefficient Identity

$$f^{(n)}(0)=n!\cdot a_n$$

This identity holds for a power series function $f(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$ which has a nonzero radius of convergence.

We can apply the identity in both directions:

• Know $f(x)$? $⇝$ Calculate $a_n$ for any $n$.
• Know $a_n$? $⇝$ Calculate $f^{(n)}(0)$ for large $n$. (Faster than differentiating.)

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Many functions can be ‘expressed’ or ‘represented’ near $x=c$ (i.e. for small enough $|x-c|$) as convergent power series. (This is true for almost all the functions encountered in pre-calculus and calculus.)

Such a power series representation is called a Taylor series. When $c=0$, the Taylor series is also called the Maclaurin series.

One power series representation we have already studied:

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

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Whenever a function has a power series (Taylor or Maclaurin), the Derivative-Coefficient Identity may be applied to calculate the coefficients of that series.

Conversely, sometimes a series can be interpreted as an evaluated power series coming from $x=c$ for some $c$. If the closed form function format can be obtained for this power series, the total sum of the original series may be discovered by putting $x=c$ in the argument of the function.

Theory 2

Study these!

• Memorize all of these series!
• Recognize all of these series!
• Recognize all of these summation formulas!

$$\begin{array}{cccc}\frac{1}{1-x}& =1+x+x^2+\cdots & =& \sum _{n=0}^{\infty }{x}^n,R=1,\text{interval: }(-\mathrm{1,1})\\ \ln (1-x)& =-\frac{x}{1}-\frac{x^2}{2}-\frac{x^3}{3}-\cdots & =& \sum _{n=0}^{\infty }-\frac{x^{n+1}}{n+1},R=1,\text{interval: }[-\mathrm{1,1})\\ {\tan }^{-1}x& =x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots & =& \sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{2n+1},R=1,\text{interval: }[-\mathrm{1,1}]\\ e^x& =1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots & =& \sum _{n=0}^{\infty }\frac{x^n}{n!},R=\infty \\ \cos x& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots & =& \sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n}}{(2n)!},R=\infty \\ \sin x& =x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots & =& \sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!},R=\infty \end{array}$$