§8.5 Finding and using Taylor series

Introduction

Using substitution and algebra to find new Taylor series expressions

Differentiating and integrating Taylor series

From $\frac{1}{1-x} = 1+x+x^2+\cdots$ and $\frac{1}{(1-x)^2} = 1+2x+3x^2+\cdots$, differentiating $h(x) = \frac{1}{1+x^2}$ via the quotient rule gives

To build the series for $x^4\cos(x^3)$, substitute $x^3$ into the cosine series and multiply by $x^4$. The result begins

To build the series for $g(x) = x^3\sin(x^2)$, substitute $x^2$ into the sine series and multiply by $x^3$, giving

For $p(x) = \frac{1}{1+5x}$, the series $1-5x+25x^2-125x^3+\cdots$ converges exactly when

To expand $e^{-x^4}$, substitute $-x^4$ into the series for $e^x$. The result begins

Since $\frac{1}{1+x} = 1-x+x^2-x^3+\cdots$ for $|x| \lt 1$, integrating term by term from $0$ to $x$ produces the series for

Integrating the cosine series $1-\frac{t^2}{2!}+\frac{t^4}{4!}-\cdots$ term by term from $0$ to $x$ yields

From $\frac{1}{1+x^2} = 1-x^2+x^4-\cdots$, integrating term by term gives the arctangent series

To expand the Gaussian $e^{-t^2}$, substitute $-t^2$ into the exponential series, giving

Multiplying the integrated Gaussian series by $\frac{2}{\sqrt{\pi}}$ gives the error function series

For $q(x) = x^2\ln(1+x^4)$, the series begins $x^6-\frac{1}{2}x^{10}+\frac{1}{3}x^{14}-\cdots$. Report the degree of the first nonzero term.

For $r(x) = \frac{e^{3x}-1}{3x}$, the series is $1+\frac{3x}{2!}+\frac{(3x)^2}{3!}+\cdots$. Report the value $r(0)$ that the series assigns.

From $\int_0^xe^{-t^2}dt = x-\frac{x^3}{3}+\frac{x^5}{5\cdot2!}-\cdots$, the coefficient of $x^5$ is $\frac{1}{5\cdot2!}$. Report it as a decimal.

Using $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}(x-\frac{x^3}{3}+\frac{x^5}{5\cdot2!}-\frac{x^7}{7\cdot3!}+\cdots)$, estimate $\operatorname{erf}(0.5)$, rounded to four decimals.

The Taylor series of $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$ converges for every real $x$, inherited from the everywhere convergence of the $e^x$ series.