Exercises — Applications — approximation, evaluating limits

Straight from the toolkit: $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots =1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots$ Notice only even powers appear, alternating $+,-,+$. That happens because $\cos$ is an even function (a mirror image across the $y$-axis), so it cannot contain odd powers like $x$ or $x^3$.

From $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$, the $x^5$ term is $+\frac{x^5}{120}$, so $c_5=\frac{1}{5!}=\frac{1}{120}.$

Every derivative of $e^x$ is $e^x$ again. So $f^{(7)}(x)=e^x$, and at $x=0$: $f^{(7)}(0)=e^0=1.$ Since $c_n=\frac{f^{(n)}(0)}{n!}=\frac{1}{n!}$ for all $n$, that is exactly why $e^x=\sum \frac{x^n}{n!}$.

$\cos (0.2)\approx 1-\frac{(0.2{)}^2}{2}+\frac{(0.2{)}^4}{24}.$ Compute each piece:

• $\frac{(0.2{)}^2}{2}=\frac{0.04}{2}=0.02$
• $\frac{(0.2{)}^4}{24}=\frac{0.0016}{24}\approx 0.0000667$

So $\cos (0.2)\approx 1-0.02+0.0000667=0.9800667$. (True value $0.9800666\dots$)

Here $\ln (1.1)=\ln (1+0.1)$, so we set $x=0.1$ (safely inside $|x|<1$): $\ln (1+0.1)\approx 0.1-\frac{(0.1{)}^2}{2}=0.1-0.005=\mathrm{0.095.}$ (True value $0.0953102\dots$ — two terms already good to three decimals.)

Substitute $x=-0.1$ directly into $e^x=1+x+\frac{x^2}{2}+\cdots$: $e^{-0.1}\approx 1+(-0.1)+\frac{(-0.1{)}^2}{2}=1-0.1+0.005=\mathrm{0.905.}$ (True value $0.9048374\dots$) The sign of $x$ carries through automatically — no separate rule needed.

Value. $0.3-\frac{(0.3{)}^3}{6}=0.3-\frac{0.027}{6}=0.3-0.0045=0.2955$.

Error. The $\sin$ series is alternating with terms shrinking in size, so by the Alternating Series Test the error is at most the first omitted term: $\left|R\right|\le \frac{(0.3{)}^5}{5!}=\frac{0.00243}{120}\approx 2.03\times 10^{-5}.$ So $\sin (0.3)=0.2955$ correct to within about $2\times 10^{-5}$. (True $0.2955202\dots$ ✓)

The Lagrange remainder is $R_N(x)=\frac{f^{(N+1)}(c)}{(N+1)!}{x}^{N+1}$ with $c\in (0,0.5)$. For $f=e^x$, $f^{(N+1)}(c)=e^c\le e^{0.5}<1.65$. Take $M=1.65$: $|R_N(0.5)|\le \frac{1.65}{(N+1)!}(0.5{)}^{N+1}.$ Test values:

• $N=3$: $\frac{1.65}{24}(0.5{)}^4=\frac{1.65}{24}(0.0625)\approx 4.30\times 10^{-3}$ — too big.
• $N=4$: $\frac{1.65}{120}(0.5{)}^5=\frac{1.65}{120}(0.03125)\approx 4.30\times 10^{-4}<10^{-3}$ ✓

So $N=4$ is the smallest that guarantees the accuracy. See how the bound collapses as $N$ climbs.

Why invalid. The series $\ln (1+x)=x-\frac{x^2}{2}+\cdots$ only converges for $-1<x\le 1$ — its radius of convergence is $1$. At $x=2$ the terms $2,-2,\frac{8}{3},\dots$ grow, so the sum is meaningless. Valid rewrite. Write $\ln 3=\ln \left(\frac{1+t}{1-t}\right)$ where $\frac{1+t}{1-t}=3\Rightarrow t=\frac{1}{2}$ (inside the disc). Then use $\ln \frac{1+t}{1-t}=2\left(t+\frac{t^3}{3}+\frac{t^5}{5}+\cdots \right),$ which converges quickly for $t=\frac{1}{2}$. This keeps every argument inside the legal range.

Plug the series: $\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots$, so $1-\cos x=\frac{x^2}{2}-\frac{x^4}{24}+\cdots .$ Cancel / divide by $x^2$: $\frac{1-\cos x}{x^2}=\frac{1}{2}-\frac{x^2}{24}+\cdots .$ Crawl $x\to 0$: the $x^2$-term dies, leaving $\boxed{\frac{1}{2}}$.

Substitute $x^2$ into the $e^u$ series ($u=x^2$): $e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\cdots .$ Subtract $1+x^2$: $e^{x^2}-1-x^2=\frac{x^4}{2}+\frac{x^6}{6}+\cdots$. Divide by $x^4$: $\frac{1}{2}+\frac{x^2}{6}+\cdots \to \boxed{\frac{1}{2}}.$ The lowest surviving power was $x^4$, matching the denominator exactly — that is why we needed the series out to $x^4$.

Numerator: $\ln (1+x)-x=\left(x-\frac{x^2}{2}+\cdots \right)-x=-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$ Denominator: $x\sin x=x\left(x-\frac{x^3}{6}+\cdots \right)=x^2-\frac{x^4}{6}+\cdots$ Both start at $x^2$. Divide top and bottom by $x^2$: $\frac{-\frac{1}{2}+\frac{x}{3}-\cdots }{1-\frac{x^2}{6}+\cdots }\stackrel{x\to 0}{\to }\frac{-\frac{1}{2}}{1}=\boxed{-\frac{1}{2}}.$

We need $\tan x=\frac{\sin x}{\cos x}$ as a series. Use series division: $\tan x=\frac{x-\frac{x^3}{6}+\cdots }{1-\frac{x^2}{2}+\cdots }.$ Multiply numerator by ${\left(1-\frac{x^2}{2}\right)}^{-1}=1+\frac{x^2}{2}+\cdots$: $\tan x=\left(x-\frac{x^3}{6}\right)\left(1+\frac{x^2}{2}\right)+\cdots =x+\frac{x^3}{2}-\frac{x^3}{6}+\cdots =x+\frac{x^3}{3}+\cdots$ Therefore $\tan x-x=\frac{x^3}{3}+\cdots$, and $\frac{\tan x-x}{x^3}=\frac{1}{3}+\cdots \to \boxed{\frac{1}{3}}.$

Series. $e^{-x^2}=1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\cdots$ (substitute $-x^2$ into $e^u$). Because this is an alternating, shrinking series in $x$, we can integrate term by term: ${\int }_0^{0.5}\left(1-x^2+\frac{x^4}{2}\right)dx={\left[x-\frac{x^3}{3}+\frac{x^5}{10}\right]}_0^{0.5}.$ Evaluate at $0.5$:

• $0.5$
• $-\frac{(0.5{)}^3}{3}=-\frac{0.125}{3}=-0.0416667$
• $+\frac{(0.5{)}^5}{10}=\frac{0.03125}{10}=0.003125$

Sum: $0.5-0.0416667+0.003125=0.4614583$.

Error bound. First omitted integrated term comes from $-\frac{x^6}{6}$: $\left|{\int }_0^{0.5}\frac{x^6}{6}dx\right|=\frac{(0.5{)}^7}{7\cdot 6}=\frac{0.0078125}{42}\approx 1.86\times 10^{-4}.$ So ${\int }_0^{0.5}{e}^{-x^2}dx\approx 0.46146$ with error $<2\times 10^{-4}$. (True $0.4612810\dots$ ✓)

Two expansions. $\sin x=x-\frac{x^3}{6}+\cdots ,x\cos x=x\left(1-\frac{x^2}{2}+\cdots \right)=x-\frac{x^3}{2}+\cdots .$ Subtract, aligning by degree: $\sin x-x\cos x=\left(x-\frac{x^3}{6}\right)-\left(x-\frac{x^3}{2}\right)+\cdots =-\frac{x^3}{6}+\frac{x^3}{2}+\cdots =\frac{x^3}{3}+\cdots$ Divide by $x^3$: $\frac{1}{3}+\cdots \to \boxed{\frac{1}{3}}$.