Taylor series — derivation from power series

4.3.16 · Maths › Calculus III — Sequences & Series

Power series KYA hoti hai?

$a$ par centre kyun? Kyunki $x = a$ ke paas term $(x - a)$ bahut chota hota hai, toh zyada powers $(x - a)^{n}$ tezi se shrink karte hain. Series centre ke paas sabse accurate hoti hai, aur pehle kuch terms dominate karte hain (yeh approximation ka 80/20 rule hai).

Coefficients KAISE nikaalein — derivation

Hum assume karte hain ki $f$ apni power series ke barabar hai aur jitni baar chaaho differentiate kar sakte ho. Humara kaam: har $c_{n}$ solve karna.

Step 0 — $c_{0}$ nikaalein. Seedha $x = a$ set karo: $f (a) = c_{0} + c_{1} \cdot 0 + c_{2} \cdot 0 + \dots = c_{0} \Rightarrow c_{0} = f (a)$ Yeh step kyun? Constant ke alaawa har term mein $(x - a)$ ka factor hai, jo $x = a$ par $0$ hota hai.

Step 1 — $c_{1}$ nikaalein. Ek baar differentiate karo: $f^{'} (x) = c_{1} + 2 c_{2} (x - a) + 3 c_{3} (x - a)^{2} + \dots$ Ab $x = a$ set karo: $f^{'} (a) = c_{1} \Rightarrow c_{1} = f^{'} (a)$ . Yeh step kyun? Differentiation ne $c_{0}$ ko khatam kar diya (constant → 0) aur $c_{1}$ ko constant slot mein shift kar diya.

Step 2 — $c_{2}$ nikaalein. Dobara differentiate karo: $f^{''} (x) = 2 c_{2} + (3 \cdot 2) c_{3} (x - a) + (4 \cdot 3) c_{4} (x - a)^{2} + \dots$ $x = a$ set karo: $f^{''} (a) = 2 c_{2} \Rightarrow c_{2} = \frac{f ^{''} ( a )}{2}$ . Yeh step kyun? $(x - a)^{2}$ par power $2$ do baar differentiate karne par factor ke roop mein neeche aa gayi: $2 \cdot 1 = 2!$ .

Step $n$ — pattern. $n$ baar differentiate karo. Term $c_{n} (x - a)^{n}$ ban jaata hai $n (n - 1) (n - 2) \dots 1 \cdot c_{n} = n! c_{n},$ aur saare neeche ke terms khatam ho jaate hain jabki upar ke terms mein abhi bhi $(x - a)$ ka factor hota hai. $x = a$ set karo: $f^{(n)} (a) = n! c_{n} \Rightarrow c_{n} = \frac{f ^{(n)} ( a )}{n !}$

Figure — Taylor series — derivation from power series

$n!$ KYUN aata hai (derivation ka engine)

$(x - a)^{n}$ ko ek baar differentiate karne par $n (x - a)^{n - 1}$ milta hai, dobara karne par $n (n - 1) (x - a)^{n - 2}$ milta hai… Har derivative current exponent ko ek multiplying factor ke roop mein neeche kheench leta hai. $n$ derivatives ke baad tumne $n \cdot (n - 1) \dots 2 \cdot 1 = n!$ se multiply kar diya hota hai. Coefficient mein $n!$ se divide karna ise exactly cancel kar deta hai, toh $c_{n}$ saaf tarike se $f^{(n)} (a) / n!$ ke barabar ho jaata hai.

Worked examples

Forecast-then-Verify

Common mistakes

Recall Feynman: 12 saal ke bachche ko samjhao

Socho tum ek curvy slide ko sirf seedhe LEGO pieces se copy karna chahte ho jo thoda moda ja sake. Pehle tum match karte ho ki slide kahan shuru hoti hai (woh hai $f (a)$ ). Phir match karo yeh kitni steep hai (woh hai $f^{'} (a)$ ). Phir match karo yeh kaise curve karti hai ( $f^{''} (a)$ ), phir curve kaise change hoti hai, aur aise aage. Har nayi information tumhari LEGO copy ko starting point ke paas real slide se thoda aur acche se chipkaati hai. Taylor series bilkul woh recipe hai jo batati hai ki har LEGO piece kitni badi honi chahiye — aur magic number $n!$ pieces ko bahut bada hone se rokta hai.

Connections

Power series — radius & interval of convergence (series actually $f$ ke barabar kab hoti hai?)
Taylor's theorem with remainder (error term $R_{n}$ jo derivation ignore karta hai)
Maclaurin series — common expansions ( $e^{x}, sin, cos, \frac{1}{1 - x}$ )
Linear approximation & differentials ( $n = 1$ truncation)
Geometric series (ek power series jiske saare coefficients $1$ hain)
L'Hôpital's rule (aksar leading Taylor terms se replace kiya ja sakta hai)

Kya cheez ek power series ke coefficients ko jo

f

ke barabar ho unique hone par majboor karti hai?

n

baar differentiate karna aur

x = a

set karna exactly ek term ko isolate karta hai,

c_{n} = f^{(n)} (a) / n!

force karta hai.

a

ke baare mein

n

-th Taylor coefficient ka general formula kya hai?

c_{n} = \frac{f ^{(n)} ( a )}{n !}

Taylor coefficients mein

n!

kyun aata hai?

n

differentiations mein se har ek current exponent ko factor ke roop mein neeche kheenchta hai; milke woh

n!

multiply karte hain, jise division cancel kar deta hai.

c_{n}

solve karte waqt sirf ek term ko chhodkar baaki sab ko khatam karne ki trick kya hai?

x = a

set karna, toh

k \geq 1

wale har

(x - a)^{k}

ka maan

0

ho jaata hai.

Maclaurin series kya hoti hai?

a = 0

par centred ek Taylor series:

\sum f^{(n)} (0) x^{n} / n!

e^{x}

ki Maclaurin series?

\sum_{n = 0}^{\infty} x^{n} / n!

sin x

ki Maclaurin series?

\sum_{n = 0}^{\infty} (- 1)^{n} x^{2 n + 1} / (2 n + 1)!

a = 1

par centred series mein kaisi powers aati hain?

(x - 1)

ki powers,

x

ki powers nahi.

Kya Taylor derivation guarantee karta hai ki series har jagah

f

ke barabar hai?

Nahi — sirf coefficients determine hote hain; equality ke liye convergence aur

R_{n} \to 0

chahiye.

a = 1

ke baare mein

ln x

ke pehle teen nonzero terms?

(x - 1) - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3}