Exercises — Taylor series — derivation from power series

(a) Every power is $(x-3{)}^n$, so the displacement from the centre is $(x-3)$. That means $a=3$. (b) The coefficient sitting on $(x-3{)}^2$ is $c_2=7$. (c) Rearrange $c_2=\frac{f^{\prime \prime }(a)}{2!}$ into $f^{\prime \prime }(a)=2!c_2=2\cdot 7=\boxed{14}$. Why: $2!=2$, and the coefficient already had the factorial divided out, so we multiply it back to recover the raw derivative.

A Maclaurin series is a Taylor series with centre $a=0$. Series A has powers $x^n=(x-0{)}^n$, so $a=0$ — that is the Maclaurin one. Series B has $(x+2{)}^n=(x-(-2){)}^n$, so its centre is $a=-2$. Careful: $(x+2)$ hides a minus a negative.

Differentiate repeatedly; each derivative pulls down a factor of $2$ (chain rule): $f=e^{2x},f^{\prime }=2e^{2x},f^{\prime \prime }=4e^{2x},f^{\prime \prime \prime }=8e^{2x}.$ Plug in $a=0$ (so $e^0=1$): $f(0)=1,f^{\prime }(0)=2,f^{\prime \prime }(0)=4,f^{\prime \prime \prime }(0)=8$. Apply $c_n=f^{(n)}(0)/n!$: $c_0=\frac{1}{1}=1,c_1=\frac{2}{1!}=2,c_2=\frac{4}{2!}=2,c_3=\frac{8}{3!}=\frac{4}{3}.$ $\boxed{e^{2x}=1+2x+2x^2+\frac{4}{3}{x}^3+\cdots }$ Check against the shortcut: replacing $x\to 2x$ in $e^x=\sum x^n/n!$ gives $\sum (2x{)}^n/n!=\sum 2^n{x}^n/n!$, i.e. $c_n=2^n/n!$ — matching $1,2,2,\frac{4}{3}$. ✓

Derivatives cycle every four steps: $\cos ,-\sin ,-\cos ,\sin ,\cos ,\dots$ At $x=0$: $\cos 0=1,-\sin 0=0,-\cos 0=-1,\sin 0=0,\cos 0=1$. So $f(0),f^{\prime }(0),f^{\prime \prime }(0),f^{\prime \prime \prime }(0),f^{(4)}(0)=1,0,-1,0,1$. Divide by $n!$: $c_0=1,c_1=0,c_2=\frac{-1}{2!}=-\frac{1}{2},c_3=0,c_4=\frac{1}{4!}=\frac{1}{24}.$ $\boxed{\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots }$ Only even powers survive because the odd-order derivatives all pass through $\sin 0=0$.

$f=x^{-1},f^{\prime }=-x^{-2},f^{\prime \prime }=2x^{-3}.$ At $a=2$: $f(2)=\frac{1}{2},f^{\prime }(2)=-\frac{1}{4},f^{\prime \prime }(2)=\frac{2}{8}=\frac{1}{4}$. Coefficients: $c_0=\frac{1}{2},c_1=-\frac{1}{4},c_2=\frac{1/4}{2!}=\frac{1}{8}$. $\boxed{\frac{1}{x}=\frac{1}{2}-\frac{1}{4}(x-2)+\frac{1}{8}(x-2{)}^2-\cdots }$ Why powers of $(x-2)$? The centre is $a=2$, so the natural building block is the displacement $(x-2)$, not $x$.

Suppose $\cos x=\sum c_n{x}^n$. Replace $x\to -x$: the left side is unchanged (even function), while $x^n\to (-1{)}^n{x}^n$. Matching: $\sum c_n{x}^n=\sum c_n(-1{)}^n{x}^n\Rightarrow c_n=(-1{)}^n{c}_n\text{ for every }n.$ If $n$ is odd, $(-1{)}^n=-1$, forcing $c_n=-c_n$, i.e. $2c_n=0$, so $c_n=0$. Thus all odd coefficients vanish — only even powers remain. This matches L2.2 without any calculus, just symmetry.

About $a=0$: $f$ is already a polynomial in $x$, so its Maclaurin series is itself: $3x^2-x+4$. Check: $f(0)=4,f^{\prime }(0)=-1,f^{\prime \prime }(0)=6\Rightarrow c_2=6/2=3$. ✓ All coefficients $c_n=0$ for $n\ge 3$ because $f^{\prime \prime \prime }\equiv 0$. About $a=1$: compute $f(1)=3-1+4=6$; $f^{\prime }(x)=6x-1\Rightarrow f^{\prime }(1)=5$; $f^{\prime \prime }(x)=6\Rightarrow f^{\prime \prime }(1)=6$, so $c_2=6/2=3$; $f^{\prime \prime \prime }=0$. $\boxed{f(x)=6+5(x-1)+3(x-1{)}^2}.$ Special feature: the series terminates — a degree-$d$ polynomial has a Taylor series with at most $d+1$ terms about any centre, because all derivatives beyond order $d$ are zero. No convergence worries; equality is exact everywhere.

$f^{\prime }=(1-x{)}^{-2}$, $f^{\prime \prime }=2(1-x{)}^{-3}$, …, $f^{(n)}(0)=n!$. So $c_n=n!/n!=1$: $\frac{1}{1-x}={\sum }_{n=0}^{\infty }{x}^n=1+x+x^2+\cdots$ This is exactly a geometric series with ratio $x$; it converges only for $|x|<1$. At $x=2$, say, the true value is $\frac{1}{1-2}=-1$, but $1+2+4+\cdots$ diverges — the coefficients are still "correct," yet the series fails to equal $f$ outside the interval. See Power series — radius & interval of convergence.

Start from $e^u=1+u+\frac{u^2}{2!}+\cdots$ and substitute $u=-x^2$: $e^{-x^2}=1+(-x^2)+\frac{(-x^2{)}^2}{2}+\cdots =1-x^2+\frac{x^4}{2}-\cdots$ $\boxed{e^{-x^2}=1-x^2+\frac{1}{2}{x}^4-\cdots }$ Why this is legal: substitution into a convergent power series just relabels the variable; it reproduces exactly what repeated differentiation would give, far more cheaply. (Differentiating $e^{-x^2}$ four times by hand is painful — this is the payoff.)

Use $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$ and $\sin x=x-\frac{x^3}{6}+\cdots$. Multiply, keeping terms up to $x^3$: $(1+x+\frac{x^2}{2}+\frac{x^3}{6})(x-\frac{x^3}{6}).$ Collect by power:

• $x^1$: $1\cdot x=x$.
• $x^2$: $x\cdot x=x^2$.
• $x^3$: $\frac{x^2}{2}\cdot x=\frac{x^3}{2}$, plus $1\cdot (-\frac{x^3}{6})=-\frac{x^3}{6}$. Sum $=\frac{3}{6}-\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$. $\boxed{e^x\sin x=x+x^2+\frac{1}{3}{x}^3+\cdots }$

Expand the numerator: $e^x-1-x=(1+x+\frac{x^2}{2}+\cdots )-1-x=\frac{x^2}{2}+\frac{x^3}{6}+\cdots$ Divide by $x^2$: $\frac{e^x-1-x}{x^2}=\frac{1}{2}+\frac{x}{6}+\cdots \underset{x\to 0}{\to }\boxed{\frac{1}{2}}.$ Why series beat L'Hôpital here: the $\frac{0}{0}$ form would need L'Hôpital twice; the series shows the answer is just the coefficient of $x^2$ in the numerator, read off in one step. See Linear approximation & differentials for why the leading surviving term controls the limit.

Differentiate: $f^{\prime }(x)=\frac{1}{1+x}$. This is a geometric series (ratio $-x$): $\frac{1}{1+x}={\sum }_{k=0}^{\infty }(-x{)}^k=1-x+x^2-x^3+\cdots ,|x|<1.$ Since $f(0)=\ln 1=0$, integrate term by term from $0$ to $x$: $\ln (1+x)={\int }_0^x(1-t+t^2-\cdots )dt=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots ={\sum }_{n=1}^{\infty }\frac{(-1{)}^{n-1}}{n}{x}^n.$ This matches parent Example 3 with the substitution $x\to x-1$ (i.e. $\ln x$ about $1$). Convergence: $|x|<1$ (and, as a bonus, $x=1$ gives the alternating harmonic series $=\ln 2$).

From $\cos x={\sum }_{k=0}^{\infty }\frac{(-1{)}^k}{(2k)!}{x}^{2k}$, the coefficient of $x^6$ (take $2k=6\Rightarrow k=3$) is $c_6=\frac{(-1{)}^3}{6!}=-\frac{1}{720}.$ Reverse the coefficient rule $c_6=f^{(6)}(0)/6!$: $f^{(6)}(0)=6!c_6=720\cdot \left(-\frac{1}{720}\right)=\boxed{-1}.$ Idea: the series is an encyclopedia of all derivatives at the centre — reading off a coefficient and multiplying by $n!$ recovers any derivative instantly.

Let $h(x)=\sum (b_n-d_n)(x-a{)}^n$, which equals $0$ for all $x$ near $a$. Set $e_n=b_n-d_n$.

• Put $x=a$: every term with $(x-a{)}^k$, $k\ge 1$, dies, leaving $e_0=0$.
• Differentiate once and put $x=a$: the same "killer move" from the parent note leaves $1!e_1=0$, so $e_1=0$.
• Differentiate $n$ times and set $x=a$: $n!e_n=h^{(n)}(a)=0$, so $e_n=0$. Since $n!\ne 0$, each $e_n=0$, hence $b_n=d_n$ for all $n$. This is exactly why the Taylor derivation gives THE series, not one of many — the representation is unique.

Read the centre off any series (find $a$)? ::: Force into $(x-a{)}^n$; the number subtracted is $a$. Recover a raw derivative from a coefficient? ::: $f^{(n)}(a)=n!c_n$. Get a series without differentiating? ::: Substitute or multiply known series (L4). Know when a correct series still fails to equal $f$? ::: Outside the radius of convergence, or if $R_n\nrightarrow 0$.