Exercises — Maclaurin series of eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ — derive all

WHAT: just read off $a_n=\frac{1}{n!}$. $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$ WHY these values: every derivative of $e^x$ is $e^x$, and $e^0=1$, so all numerators are $1$. WHY it's valid for all $x$ — the ratio test, made concrete. Look at how big each term is compared to the one before it. Term $n$ is $\frac{x^n}{n!}$, term $n+1$ is $\frac{x^{n+1}}{(n+1)!}$, so the step ratio is $\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right|=\frac{|x|}{n+1}.$ Fix any $x$. As $n$ grows, $\frac{|x|}{n+1}\to 0$. So eventually each new term is a tiny fraction of the previous one — the terms crash toward $0$ no matter how big $x$ is. That is the "factorial beats power" statement, now as an honest inequality, not a slogan. The figure below shows the terms of $e^5$ rising at first (while $n<5$) then collapsing once $n>5$.

Answer: (B). WHY: $\sin$ is an odd function ($\sin (-x)=-\sin x$), so only odd powers of $x$ may appear. That kills (C) immediately (it has even powers — it is $\cos x$). (A) has odd and even powers ($x^2$), and its denominators are plain integers, not factorials — that is the $\ln (1+x)$ pattern with a wrong sign. So (A) is wrong twice over. The correct series has factorial denominators and alternating signs: $x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots$

WHAT: plug $x=0.2$ into $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$. $1-\frac{0.04}{2}+\frac{0.0016}{24}=1-0.02+0.0000\overline{6}=0.98006\overline{6}$ WHY it works: the $x^6$ term is about $\frac{(0.2{)}^6}{720}\approx 9\times 10^{-8}$, far below the accuracy we need. True: $\cos (0.2)=0.980066\dots$ ✓ The figure below shows how fast the truncated cosine polynomial locks onto the true curve as you add the $x^2$ then $x^4$ term.

WHAT / WHY: don't differentiate from scratch — substitute $x\mapsto -2x$ into the known $e^x$ series. This is legal because $e^x$ converges for all inputs, so $-2x$ is always a valid input. $e^{-2x}=1+(-2x)+\frac{(-2x{)}^2}{2!}+\frac{(-2x{)}^3}{3!}+\cdots$ $=1-2x+\frac{4x^2}{2}-\frac{8x^3}{6}+\cdots =1-2x+2x^2-\frac{4}{3}{x}^3+\cdots$

WHAT: use $(1+x{)}^m$ with $m=\frac{1}{2}$. $\left(\genfrac{}{}{0ex}{}{1/2}{0}\right)=1,\left(\genfrac{}{}{0ex}{}{1/2}{1}\right)=\frac{1}{2},\left(\genfrac{}{}{0ex}{}{1/2}{2}\right)=\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}=\frac{\frac{1}{2}\cdot (-\frac{1}{2})}{2}=-\frac{1}{8}$ $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\cdots (|x|<1)$ Check at $x=0.21$: $1+0.105-0.0055125=1.0994875\approx \sqrt{1.21}=1.1$ ✓

WHAT: differentiate $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ term by term. $\frac{d}{dx}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \right)=1-\frac{3x^2}{3!}+\frac{5x^4}{5!}-\cdots$ WHY the denominators collapse: $\frac{3}{3!}=\frac{3}{6}=\frac{1}{2!}$ and $\frac{5}{5!}=\frac{5}{120}=\frac{1}{4!}$ — the numerator cancels the top factor of the factorial. $=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots =\cos x✓$ This is why $\sin \to \cos$ under differentiation even at the level of the polynomials.

At $x=1$: the series becomes $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$, the alternating harmonic series, which converges (to $\ln 2$). Its terms shrink to $0$ and alternate — that is enough for convergence. At $x=-1$: substitute $x=-1$: ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n-1}(-1{)}^n}{n}={\sum }_{n=1}^{\infty }\frac{(-1{)}^{2n-1}}{n}={\sum }_{n=1}^{\infty }\frac{-1}{n}=-(1+\frac{1}{2}+\frac{1}{3}+\cdots )$ This is $-\infty$ (the harmonic series diverges). WHY it matches the function: $\ln (1+(-1))=\ln 0=-\infty$. The series honestly reports the singularity by diverging. What the figure shows. The figure below plots the true curve $\ln (1+x)$ (white) against the partial sums using $3$, $8$ and $20$ terms (blue, yellow, green). Watch two things: (1) inside $(-1,1]$ every partial sum hugs the white curve, and the more terms you add the tighter the hug; (2) as $x\to -1^{+}$ the true curve dives to $-\infty$, and the partial sums fail to follow smoothly near that wall — no finite number of terms can reach $-\infty$, which is the visual signature of the series diverging exactly at the singularity marked by the red dashed line. At the right edge $x=1$ (blue dotted line) the sums instead settle onto $\ln 2$, the visual signature of convergence.

WHAT: the raw expression is $\frac{0}{0}$ at $x=0$ — undefined by plain arithmetic. But divide the series by $x$: $\frac{\sin x}{x}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots }{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots$ WHY this is legal: every term of $\sin x$ has at least one factor of $x$, so dividing by $x$ leaves an honest polynomial with no division by zero. Putting $x=0$ now gives $1$. So the removable hole should be filled with $\frac{\sin x}{x}{|}_0=1$.

WHAT: multiply $(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots )(x-\frac{x^3}{6}+\cdots )$ and collect powers of $x$.

• $x^1$: $1\cdot x=x$
• $x^2$: $x\cdot x=x^2$
• $x^3$: $\frac{x^2}{2}\cdot x+1\cdot (-\frac{x^3}{6})=\frac{1}{2}{x}^3-\frac{1}{6}{x}^3=\frac{1}{3}{x}^3$ $e^x\sin x=x+x^2+\frac{x^3}{3}+\cdots$ WHY multiply and not differentiate the product: matching coefficients from two known series avoids computing derivatives of a product (product rule chains) entirely.

WHAT: put $x\mapsto ix$ in $e^x=\sum \frac{x^n}{n!}$, where $i^2=-1$. $e^{ix}={\sum }_{n=0}^{\infty }\frac{(ix{)}^n}{n!}=1+ix+\frac{(ix{)}^2}{2!}+\frac{(ix{)}^3}{3!}+\cdots$ The powers of $i$ cycle: $i^0=1,i^1=i,i^2=-1,i^3=-i,i^4=1,\dots$ So: $=\underset{\cos x}{\underbrace{(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots )}}+i\underset{\sin x}{\underbrace{(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots )}}$ WHY it works: the real terms (even powers) exactly assemble $\cos x$; the imaginary terms (odd powers) exactly assemble $\sin x$. Three series fuse into one identity. ✓

WHAT: $(1-x^2{)}^{-1/2}$. Use binomial with $m=-\frac{1}{2}$ and input $u=-x^2$. $(1+u{)}^{-1/2}=1-\frac{1}{2}u+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}{u}^2+\cdots =1-\frac{1}{2}u+\frac{3}{8}{u}^2+\cdots$ Substitute $u=-x^2$: $\frac{1}{\sqrt{1-x^2}}=1+\frac{1}{2}{x}^2+\frac{3}{8}{x}^4+\cdots$ WHY the connection: integrating gives $\arcsin x=x+\frac{x^3}{6}+\frac{3x^5}{40}+\cdots$, since $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}$.

WHAT: substitute the $\cos x$ series into the numerator. $\cos x-1+\frac{x^2}{2}=(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots )-1+\frac{x^2}{2}=\frac{x^4}{24}-\frac{x^6}{720}+\cdots$ WHY the low powers cancel: the $1$ and $-\frac{x^2}{2}$ are deliberately subtracted, leaving the first survivor at $x^4$. $\frac{\text{numerator}}{x^4}=\frac{1}{24}-\frac{x^2}{720}+\cdots \stackrel{x\to 0}{\to }\frac{1}{24}$ Answer: $\frac{1}{24}$. (L'Hôpital would need four differentiations — series does it in one line.)

WHAT — the radius: the coefficients are $\left(\genfrac{}{}{0ex}{}{1/3}{k}\right)$. Apply the ratio test — compare consecutive terms: $\left|\frac{a_{k+1}{x}^{k+1}}{a_k{x}^k}\right|=\left|\frac{\left(\genfrac{}{}{0ex}{}{1/3}{k+1}\right)}{\left(\genfrac{}{}{0ex}{}{1/3}{k}\right)}\right||x|=\left|\frac{\frac{1}{3}-k}{k+1}\right||x|\stackrel{k\to \infty }{\to }|x|$ WHY the ratio $\to |x|$: $\frac{\frac{1}{3}-k}{k+1}\to -1$ in magnitude as $k\to \infty$. Convergence needs this limit $<1$, i.e. $|x|<1$. Radius $=1$. Endpoint $x=1$ — converges. For large $k$ the coefficients $\left(\genfrac{}{}{0ex}{}{1/3}{k}\right)$ behave like $\frac{C}{k^{4/3}}$ in size (each new factor $\left|\frac{\frac{1}{3}-k}{k+1}\right|=1-\frac{4/3}{k+1}+\cdots$ shaves the term by an amount that accumulates to a $k^{-4/3}$ decay), and the signs alternate from some point on. A series whose terms shrink like $k^{-4/3}$ converges absolutely (since $\sum k^{-4/3}$ converges, being a $p$-series with $p=\frac{4}{3}>1$). So at $x=1$ the series converges — to the honest value $2^{1/3}$. Endpoint $x=-1$ — converges too. At $x=-1$ the alternating signs are removed, so every term has size $\approx C{k}^{-4/3}$ and they all add with the same sign. But $\sum k^{-4/3}$ still converges (same $p$-series argument), so the series converges at $x=-1$ as well — to $0^{1/3}=0$, matching $(1+(-1){)}^{1/3}=0$. (The function is finite there, and the series behaves.) So the full interval of convergence is $[-1,1]$. Contrast this with $(1+x{)}^{-1/2}$, where the coefficients decay only like $k^{-1/2}$ and the endpoint $x=-1$ diverges — the decay rate of the coefficients is what decides each endpoint. Just past $|x|=1$, the step ratio exceeds $1$ and terms grow, so the sum diverges even though $(1+x{)}^{1/3}$ is a perfectly finite number there.

WHAT: fix any $x$ and pick an integer $N>2|x|$. For $n>N$, each new factor multiplying the previous term is $\frac{|x|}{n}<\frac{|x|}{N}<\frac{1}{2}$. WHY that finishes it: past $n=N$, every step at least halves the size: $\frac{|x{|}^n}{n!}<\frac{|x{|}^N}{N!}{\left(\frac{1}{2}\right)}^{n-N}\stackrel{n\to \infty }{\to }0$ A quantity bounded by a shrinking geometric ratio (Geometric series) must go to $0$. Hence the factorial always wins over the power, confirming the "valid for all $x$" claim — the same halving picture you saw in the Exercise 1.1 figure. ✓

Can you, from memory: (1) write all five series, (2) get $e^{-2x}$ by substitution, (3) explain the $\ln$ endpoint failure, (4) multiply two series for $e^x\sin x$, (5) do the $x^4$ limit, (6) state each radius of convergence and both endpoints of $(1+x{)}^{1/3}$? If any is shaky, redo that level.