Applications — approximation, evaluating limits

WHY this works

WHY the $\frac{f^{(n)}(a)}{n!}$ ? Derive it. Suppose $f(x)=\sum c_n (x-a)^n$ actually equals a power series. Differentiate $k$ times and plug in $x=a$ :

$f^{(k)}(x)=\sum_{n\ge k} c_n\,\frac{n!}{(n-k)!}(x-a)^{n-k}.$

Every term with $n>k$ still has a factor $(x-a)$ , so at $x=a$ it vanishes. Only $n=k$ survives: $f^{(k)}(a)=c_k\,k!\quad\Rightarrow\quad c_k=\frac{f^{(k)}(a)}{k!}.$

That's the only choice of coefficients that makes the polynomial match the function's value and all its slopes/curvatures at $a$ . That's WHY it's the best polynomial fit.

HOW to approximate — and HOW to trust it

WHY a bound? We don't know $c$ , but if $|f^{(N+1)}|\le M$ on the interval, then $|R_N(x)|\le \frac{M}{(N+1)!}|x-a|^{N+1}.$ This is your guarantee: "my answer is correct to within this much."

Figure — Applications — approximation, evaluating limits

Worked example Approximate

$e^{0.2}$ with error $<10^{-4}$ Step 1. $P_N(0.2)=\sum_{n=0}^N \frac{0.2^n}{n!}$ . Step 2. Bound: on $[0,0.2]$ , $|f^{(N+1)}|=e^c\le e^{0.2}<1.3$ . So $|R_N|\le \frac{1.3}{(N+1)!}0.2^{N+1}$ . Why? Pick $M=1.3$ as a safe upper bound for $e^c$ . Step 3. Try $N=3$ : $\frac{1.3}{24}(0.2)^4=\frac{1.3}{24}(0.0016)=8.7\times10^{-5}<10^{-4}$ . ✓ Step 4. $1+0.2+\frac{0.04}{2}+\frac{0.008}{6}=1+0.2+0.02+0.001333=1.221333$ . (true $1.221403$ ).

HOW to kill limits (the powerful trick)

Common mistake Steel-manning the classic errors

(1) "Just plug $x=a$ into the series for a limit." Feels right because the series equals $f$ . Why wrong: the WHOLE point is the lowest powers cancel — you must keep enough terms so the leading surviving power has the same degree as the denominator. Fix: expand to at least the order of the denominator (one safe term beyond).

(2) "Two terms is always accurate enough." Feels right for small $x$ . Why wrong: accuracy depends on $|x-a|^{N+1}$ and on $M$ ; for $x$ far from $a$ the bound can be huge. Fix: always check the remainder bound.

(3) Using $\ln(1+x)$ series for $x=2$ . Feels right — it's a formula. Why wrong: it only converges for $|x|<1$ . Fix: respect the radius of convergence; rewrite or recenter.

(4) Dropping the $x^5/120$ but keeping a stray sign. Fix: line up powers in a column; cancel by degree.

Recall Feynman: explain to a 12-year-old

A wiggly function is hard to compute. But Newton found a magic trick: near one point, the function behaves almost exactly like a simple polynomial — start with its height, then add a piece for its slope, then a tinier piece for how it bends, and so on. The further pieces get tiny super fast (divided by $1,2,6,24,\dots$ ). To get $\sin(0.1)$ you just add a couple of these tiny pieces. And when a fraction looks like $\frac00$ , you swap both top and bottom for their polynomial twins, cross out the equal small bits, and the answer pops out.

Flashcards

What is the Maclaurin coefficient

c_n

and why?

c_n=\frac{f^{(n)}(0)}{n!}

; it's the unique value making the polynomial match

f

n

-th derivative at

0

Lagrange remainder of a Taylor polynomial

P_N

R_N(x)=\frac{f^{(N+1)}(c)}{(N+1)!}(x-a)^{N+1}

for some

c

between

a

and

x

Error bound for an alternating series?

\le

the absolute value of the first omitted term.

\lim_{x\to0}\frac{\sin x - x}{x^3}

-\frac16

(from

\sin x - x=-x^3/6+\cdots

\lim_{x\to0}\frac{e^x-1-x}{x^2}

\frac12

Why must you keep enough series terms in a limit?

Lowest powers cancel; you need the term that matches the denominator's degree to survive.

Maclaurin series of

\cos x

1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots

Convergence interval of

\ln(1+x)

series?

-1<x\le 1

General error bound if

|f^{(N+1)}|\le M

|R_N(x)|\le\frac{M}{(N+1)!}|x-a|^{N+1}

Connections

Taylor & Maclaurin Series
Lagrange Remainder & Error Estimation
Alternating Series Test
Radius & Interval of Convergence
L'Hôpital's Rule (alternative method for $\frac00$ )
Power Series — Operations (add, multiply, differentiate)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, idea bahut simple hai: koi bhi smooth function ko hum uske ek point ke aas-paas ek polynomial se replace kar sakte hain — isko Taylor series bolte hain. Polynomial isliye chahiye kyunki sirf jod (add), guna (multiply) aur bhag (divide) hum haath se kar sakte hain. Har term ka coefficient $\frac{f^{(n)}(a)}{n!}$ hota hai, aur yeh isliye nikalta hai kyunki yahi ek choice hai jisse polynomial ki value, slope, curvature sab original function se match karein point $a$ par.

Approximation ke liye: jaise $\sin(0.1)$ . $\sin x \approx x - \frac{x^3}{6}$ use karo, arithmetic karo, ho gaya. Aur kitna error hai? Lagrange remainder $R_N$ ya alternating series ka rule (error $\le$ first chhoda hua term) se bound nikaalo — yeh guarantee deta hai ki answer kitna sahi hai. Yaad rakho: error $|x-a|^{N+1}$ par depend karta hai, isliye $x$ point ke paas hona chahiye warna do term kaafi nahi honge.

Limits ke liye toh yeh trick magic hai. Jab $\frac00$ form aaye, top aur bottom dono ki series likho aur lowest power cancel kar do. Jaise $\frac{\sin x - x}{x^3}$ : $\sin x - x = -\frac{x^3}{6}+\cdots$ , divide by $x^3$ , $x\to0$ — answer $-\frac16$ . L'Hopital se kabhi-kabhi yeh fast hota hai kyunki baar-baar differentiate nahi karna padta.

Ek warning: limit me itne terms zaroor rakho ki denominator ki degree wala term bach jaaye, warna galat answer milega. Aur $\ln(1+x)$ wali series sirf $|x|<1$ tak valid hai — convergence ka radius hamesha check karo. Mnemonic yaad rakho: Plug, Cancel, Crawl, Conclude.

Applications — approximation, evaluating limits

WHY this works

HOW to approximate — and HOW to trust it

HOW to kill limits (the powerful trick)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections