Taylor series — derivation from power series: Question bank

4.3.16 · D5 · Maths › Calculus III — Sequences & Series › Taylor series — derivation from power series

True or false — justify karo

A power series jo

f

ke centre ke paas

f

ke barabar hai, woh hamesha har real

x

ke liye

f

ki taraf converge karta hai.

False. Woh

f

ki taraf sirf radius of convergence ke andar converge karta hai, aur wahan bhi tabhi jab remainder

R_{n} \to 0

ho. Bahar, terms bilkul blow up ho sakti hain.

Agar do power series (same

(x - a)^{n}

mein) ek interval pe equal hain, to unke coefficients term by term equal hone chahiye.

True.

n

baar differentiate karke

x = a

set karne se

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

uniquely isolate ho jata hai, isliye kisi bhi function ka

a

ke baare mein zyada se zyada ek hi power-series representation hota hai.

Taylor coefficient

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

hota hai.

False. Tumhe

n!

se divide karna zaroori hai:

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

. Har differentiation ek exponent peel off karta hai, aur woh factors milke

n!

bana dete hain, jise division cancel kar deta hai.

Har infinitely differentiable function centre ke paas apni Taylor series ke barabar hoti hai.

False. Classic counterexample

f (x) = e^{- 1/ x^{2}}

hai (jahan

f (0) = 0

) — yeh smooth hai aur

0

pe iske saare derivatives zero hain, isliye iska Maclaurin series

0

hai — phir bhi

f

khud zero nahi hai. Coefficients exist karte hain; equality phir bhi fail ho jaati hai.

f

ke liye jo

a = 1

pe centred hai, woh series

x

ki powers mein likhi jaati hai.

False. Building blocks

(x - a)^{n} = (x - 1)^{n}

hote hain.

x

ki powers likhna ek Maclaurin aadat hai jo sirf tab kaam karti hai jab

a = 0

ho.

Maclaurin series Taylor series se ek alag rule hai.

False. Maclaurin sirf usi formula ka special case

a = 0

hai; koi nayi cheez define nahi ho rahi.

Ek geometric series

\sum x^{n} = \frac{1}{1 - x}

ek Taylor series hai.

True. Yeh

\frac{1}{1 - x}

0

ke baare mein Taylor (Maclaurin) series hai — har coefficient

c_{n} = 1

hai kyunki

f^{(n)} (0) / n! = n! / n! = 1

Agar

f^{'} (a) = 0

hai, to linear (degree-1) Taylor approximation sirf constant

f (a)

hai.

True. Degree-1 term

f^{'} (a) (x - a)

hota hai; jab

f^{'} (a) = 0

ho to yeh vanish ho jaata hai, isliye best line flat hai — geometrically tangent horizontal hai.

Zyada Taylor terms add karne se har

x

pe accuracy hamesha improve hoti hai.

False. Zyada terms accuracy ko centre ke paas aur convergence ke andar sharp karte hain, lekin radius of convergence ke bahar terms add karne se sum diverge ho jaata hai — accuracy better nahi, balki worse ho jaati hai.

Galti dhundho

e^{x} = 1 + x + x^{2} + x^{3} + \dots

, isliye yeh geometric series jaisa hi hai."

e^{x}

series mein har term ko

\frac{1}{n !}

se divide kiya jaata hai:

1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \dots

. Factorials drop karne se yeh

\frac{1}{1 - x}

ban jaata hai, jo bilkul alag function hai.

ln x

ke liye

a = 1

ke baare mein:

ln x = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots

Powers

(x - 1)

ki honi chahiye,

x

ki nahi:

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

x^{n}

use karna silently

0

pe re-centre kar deta hai, jahan

ln

define bhi nahi hai.

sin x

ki series mein ek

c_{0}

term hai kyunki

sin

ek achhi smooth function hai."

c_{0} = f (a) = sin 0 = 0

, isliye koi constant term nahi hai. Sirf odd powers survive karte hain kyunki

sin

ke even-order derivatives origin pe

0

hote hain.

cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{4} - \dots

Denominators bare exponents nahi, factorials hain:

cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots

. Kyunki

2! = 2

hai, pehli galti chhup jaati hai, lekin

\frac{x ^{4}}{4}

asal mein

\frac{x ^{4}}{24}

hona chahiye.

"Taylor derivation prove karta hai ki

\sum \frac{f ^{(n)} ( a )}{n !} (x - a)^{n}

f (x)

ke barabar hai."

Derivation sirf yeh dikhata hai ki agar

f

ek power series ke barabar ho to coefficients kya honge. Actual equality prove karne ke liye alag remainder argument chahiye — dekho Taylor's theorem with remainder.

c_{3}

find karne ke liye main teen baar differentiate karta hun aur

(x - a)^{3}

ka coefficient padhta hun."

Teen baar differentiate karne ke baad, jo

c_{3}

term bachti hai woh

(x - a)^{3}

term nahi balki constant

3! c_{3}

hoti hai. Phir

x = a

set karo aur

3!

se divide karo.

Why questions

x = a

set karne se ek single coefficient isolate kyun hota hai?

Current constant ke alawa har term mein

(x - a)^{k}

ka factor hota hai jahan

k \geq 1

ho, aur

x = a

(x - a) = 0

hota hai. Isliye woh saari terms zero ho jaati hain, sirf ek survivor bachta hai.

Factorial

n!

kyun aata hai,

n

kyun nahi?

(x - a)^{n}

ko differentiate karne se exponents ek ek karke peel hote hain:

n

, phir

n - 1

, phir

n - 2

, ...

1

tak. Unka product

n \cdot (n - 1) \dots 1 = n!

hota hai.

Power series apne centre ke paas sabse accurate kyun hoti hai?

x = a

ke paas displacement

(x - a)

bahut chota hota hai, isliye high powers

(x - a)^{n}

bahut tezi se shrink karte hain aur pehle kuch terms dominate karte hain — tail almost kuch contribute nahi karta.

Hum

0

se door centre kyun rakhte hain (jaise

ln x

a = 1

pe)?

Kyunki Maclaurin series ke liye

0

pe saare derivatives chahiye, lekin

ln x

wahan defined hi nahi hai. Ek aisa centre choose karna jahan

f

well-behaved ho, expansion ko exist karne deta hai.

Degree-1 Taylor polynomial linear approximation se same kyun hai?

Series ko

n = 1

ke baad truncate karne se

f (a) + f^{'} (a) (x - a)

milta hai — exactly tangent line, jo standard linear/differential estimate hai.

L'Hôpital ko aksar leading Taylor terms se replace kyun kiya ja sakta hai?

Centre ke paas ek function apna leading term hi hota hai (jaise

sin x \approx x

), isliye

0/0

ratio leading powers ke ratio mein reduce ho jaata hai, aur limit seedha mil jaati hai bina baar baar differentiate kiye.

Edge cases

Constant function

f (x) = 5

ki Taylor series kya hai?

Sirf

5

. Har derivative

0

hai, isliye

c_{0} = 5

aur baaki saare

c_{n} = 0

hain — "series" sirf ek term hai.

Polynomial

f (x) = 3 x^{2} - x + 4

a = 0

ke baare mein Taylor series kya hai?

Polynomial khud hi:

4 - x + 3 x^{2}

. Order 2 ke baad derivatives zero ho jaate hain, isliye series terminate ho jaati hai aur

f

ko exactly har jagah reproduce karti hai.

Agar

f

, centre pe differentiable nahi hai (jaise

∣ x ∣

a = 0

) to series ka kya hoga?

Coefficients defined nahi hote —

f^{'} (0)

exist nahi karta — isliye us centre ke baare mein koi Taylor series exist hi nahi karta.

a

pe smoothness ek prerequisite hai.

"Radius of convergence

= 0

" case ka kya matlab hai?

Series sirf

x = a

pe converge karti hai aur kahin nahi, isliye centre se door approximation ke liye yeh useless hai — coefficients itni tezi se badhte hain ki koi bhi nonzero displacement kaam nahi karta.

Agar

a

pe saare derivatives zero hain lekin

f

zero function nahi hai, to Taylor series kya hai?

Woh identically

0

hai, phir bhi

f \neq = 0

hai. Yeh dikhata hai ki derivation coefficients determine karta hai lekin series ko

f

ke barabar force nahi kar sakta (

e^{- 1/ x^{2}}

wali situation).

Degree-

0

Taylor "series" kya hai aur yeh geometrically kya represent karta hai?

Sirf

f (a)

— centre pe function ki height pe ek horizontal line. Yeh sirf starting value match karta hai, slope ya curvature nahi.

Recall Har trap ki ek-line summary

Derivation coefficients deta hai, equality nahi; powers $(x - a)^{n}$ hoti hain, $x^{n}$ nahi jab tak $a = 0$ na ho; aur $n!$ negotiable nahi hai.

Connections

Parent: Taylor series derivation
Power series — radius & interval of convergence (jahan equality hold bhi ho sakti hai)
Taylor's theorem with remainder (kyun coefficients ≠ equality)
Maclaurin series — common expansions ( $a = 0$ wala case)
Linear approximation & differentials ( $n = 1$ truncation)
Geometric series (all-ones coefficients)
L'Hôpital's rule (leading-term shortcut)