4.3.16Calculus III — Sequences & Series

Taylor series — derivation from power series

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WHAT is a power series?

WHY centre at aa? Because near x=ax=a the term (xa)(x-a) is tiny, so higher powers (xa)n(x-a)^n shrink fast. The series is most accurate near the centre, and the first few terms dominate (this is the 80/20 of approximation).


HOW to find the coefficients — the derivation

We assume ff equals its power series and is differentiable as many times as we want. Our job: solve for each cnc_n.

Step 0 — find c0c_0. Set x=ax=a directly: f(a)=c0+c10+c20+=c0c0=f(a)f(a) = c_0 + c_1\cdot 0 + c_2 \cdot 0 + \cdots = c_0 \quad\Rightarrow\quad c_0 = f(a) Why this step? Every term except the constant has a factor (xa)(x-a), which is 00 at x=ax=a.

Step 1 — find c1c_1. Differentiate once: f(x)=c1+2c2(xa)+3c3(xa)2+f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + \cdots Now set x=ax=a:   f(a)=c1c1=f(a)\; f'(a) = c_1 \Rightarrow c_1 = f'(a). Why this step? Differentiation killed c0c_0 (constant → 0) and shifted c1c_1 into the constant slot.

Step 2 — find c2c_2. Differentiate again: f(x)=2c2+(32)c3(xa)+(43)c4(xa)2+f''(x) = 2c_2 + (3\cdot 2)c_3(x-a) + (4\cdot3)c_4(x-a)^2 + \cdots Set x=ax=a:   f(a)=2c2c2=f(a)2\; f''(a) = 2c_2 \Rightarrow c_2 = \dfrac{f''(a)}{2}. Why this step? The power 22 on (xa)2(x-a)^2 came down as a factor when we differentiated twice: 21=2!2\cdot1 = 2!.

Step nn — the pattern. Differentiate nn times. The term cn(xa)nc_n(x-a)^n becomes n(n1)(n2)1cn=n!cn,n(n-1)(n-2)\cdots 1 \cdot c_n = n!\,c_n, and all lower terms have vanished while higher terms still carry a factor (xa)(x-a). Set x=ax=a: f(n)(a)=n!cncn=f(n)(a)n!f^{(n)}(a) = n!\,c_n \quad\Rightarrow\quad \boxed{c_n = \dfrac{f^{(n)}(a)}{n!}}

Figure — Taylor series — derivation from power series

WHY n!n! appears (the engine of the derivation)

Differentiating (xa)n(x-a)^n once gives n(xa)n1n(x-a)^{n-1}, again gives n(n1)(xa)n2n(n-1)(x-a)^{n-2}… Each derivative peels off the current exponent as a multiplying factor. After nn derivatives you've multiplied by n(n1)21=n!n\cdot(n-1)\cdots 2\cdot 1 = n!. Dividing by n!n! in the coefficient exactly cancels this, so cnc_n cleanly equals f(n)(a)/n!f^{(n)}(a)/n!.


Worked examples


Forecast-then-Verify


Common mistakes


Recall Feynman: explain to a 12-year-old

Imagine you want to copy a curvy slide using only straight LEGO pieces that bend a little. First you match where the slide starts (that's f(a)f(a)). Then match how steep it is (that's f(a)f'(a)). Then match how it curves (f(a)f''(a)), then how the curve changes, and so on. Each new piece of information makes your LEGO copy hug the real slide a little better near the starting point. The Taylor series is just the recipe that says exactly how big each LEGO piece must be — and the magic number n!n! keeps the pieces from blowing up too big.


Connections


What forces the coefficients of a power series equal to ff to be unique?
Differentiating nn times and setting x=ax=a isolates exactly one term, forcing cn=f(n)(a)/n!c_n=f^{(n)}(a)/n!.
General formula for the nn-th Taylor coefficient about aa?
cn=f(n)(a)n!c_n = \dfrac{f^{(n)}(a)}{n!}
Why does n!n! appear in Taylor coefficients?
Each of the nn differentiations peels off the current exponent as a factor; together they multiply to n!n!, which the division cancels.
What is the trick that kills all but one term when solving for cnc_n?
Setting x=ax=a, so every (xa)k(x-a)^k with k1k\ge1 becomes 00.
What is a Maclaurin series?
A Taylor series centred at a=0a=0: f(n)(0)xn/n!\sum f^{(n)}(0)x^n/n!.
Maclaurin series of exe^x?
n=0xn/n!\sum_{n=0}^\infty x^n/n!
Maclaurin series of sinx\sin x?
n=0(1)nx2n+1/(2n+1)!\sum_{n=0}^\infty (-1)^n x^{2n+1}/(2n+1)!
For a series centred at a=1a=1, what powers appear?
Powers of (x1)(x-1), not powers of xx.
Does the Taylor derivation guarantee the series equals ff everywhere?
No — only the coefficients are determined; equality needs convergence and Rn0R_n\to0.
First three nonzero terms of lnx\ln x about a=1a=1?
(x1)12(x1)2+13(x1)3(x-1) - \tfrac12(x-1)^2 + \tfrac13(x-1)^3

Concept Map

centred at a

goal

killer move

then set x=a

constant survivor

peels exponents

rearrange

substitute back

special case a=0

near centre

Assume f equals power series

Power series sum c_n x-a ^n

Solve for coefficients c_n

Differentiate n times

Kills all but one term

f^n a equals n! c_n

Factor n! appears

c_n = f^n a over n!

Taylor series

Maclaurin series

Terms shrink, most accurate

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho tumhare paas ek curvy function f(x)f(x) hai, aur tum chahte ho ki use ek polynomial se replace kar do — kyunki polynomials ke saath kaam karna easy hai (add, differentiate, integrate sab simple). Sawaal yeh hai: agar f(x)=c0+c1(xa)+c2(xa)2+f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots hai, toh yeh coefficients cnc_n kya honge? Taylor series isi ka jawaab hai.

Trick bahut clever hai. Jab tum x=ax=a rakhte ho, toh saare (xa)(x-a) wale terms zero ho jaate hain, sirf c0c_0 bachta hai — toh c0=f(a)c_0 = f(a). Phir ek baar differentiate karo, dobara x=ax=a rakho, toh c1=f(a)c_1 = f'(a) mil jaata hai. nn baar differentiate karne par us term ka exponent n!n! banke neeche aa jaata hai, isliye cn=f(n)(a)/n!c_n = f^{(n)}(a)/n!. Bas yahi poora derivation hai — "derive nn times, divide by n!n!, plug in aa".

Yeh n!n! wala factor sabse zaroori cheez hai jise log bhool jaate hain. Har differentiation exponent ko neeche le aati hai, aur nn differentiations milke n!n! bana deti hain. Isiliye divide karna zaroori hai, warna coefficient galat ho jayega. Aur dhyaan rakho — agar centre a=0a=0 nahi hai (jaise lnx\ln x ka a=1a=1), toh powers (x1)(x-1) ke honge, xx ke nahi.

Yeh kyun matter karta hai? Kyunki physics aur engineering mein hum complicated functions ko unke pehle 2-3 terms se approximate karte hain (yeh hai 80/20 — thode terms se bahut accuracy, centre ke paas). ex1+xe^x \approx 1+x, sinxx\sin x \approx x — yeh sab Taylor series ke chhote versions hi hain. Lekin yaad rakho: series sirf radius of convergence ke andar hi ff ke barabar hoti hai, har jagah nahi.

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections