WHY centre at a? Because near x=a the term (x−a) is tiny, so higher powers (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).
We assume f equals its power series and is differentiable as many times as we want. Our job: solve for each cn.
Step 0 — find c0. Set x=a directly:
f(a)=c0+c1⋅0+c2⋅0+⋯=c0⇒c0=f(a)Why this step? Every term except the constant has a factor (x−a), which is 0 at x=a.
Step 1 — find c1. Differentiate once:
f′(x)=c1+2c2(x−a)+3c3(x−a)2+⋯
Now set x=a: f′(a)=c1⇒c1=f′(a).
Why this step? Differentiation killed c0 (constant → 0) and shifted c1 into the constant slot.
Step 2 — find c2. Differentiate again:
f′′(x)=2c2+(3⋅2)c3(x−a)+(4⋅3)c4(x−a)2+⋯
Set x=a: f′′(a)=2c2⇒c2=2f′′(a).
Why this step? The power 2 on (x−a)2 came down as a factor when we differentiated twice: 2⋅1=2!.
Step n — the pattern. Differentiate n times. The term cn(x−a)n becomes
n(n−1)(n−2)⋯1⋅cn=n!cn,
and all lower terms have vanished while higher terms still carry a factor (x−a). Set x=a:
f(n)(a)=n!cn⇒cn=n!f(n)(a)
Differentiating (x−a)n once gives n(x−a)n−1, again gives n(n−1)(x−a)n−2… Each derivative peels off the current exponent as a multiplying factor. After n derivatives you've multiplied by n⋅(n−1)⋯2⋅1=n!. Dividing by n! in the coefficient exactly cancels this, so cn cleanly equals f(n)(a)/n!.
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)). Then match how steep it is (that's f′(a)). Then match how it curves (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! keeps the pieces from blowing up too big.
Socho tumhare paas ek curvy function 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(x−a)+c2(x−a)2+⋯ hai, toh yeh coefficients cn kya honge? Taylor series isi ka jawaab hai.
Trick bahut clever hai. Jab tum x=a rakhte ho, toh saare (x−a) wale terms zero ho jaate hain, sirf c0 bachta hai — toh c0=f(a). Phir ek baar differentiate karo, dobara x=a rakho, toh c1=f′(a) mil jaata hai. n baar differentiate karne par us term ka exponent n! banke neeche aa jaata hai, isliye cn=f(n)(a)/n!. Bas yahi poora derivation hai — "derive n times, divide by n!, plug in a".
Yeh n! wala factor sabse zaroori cheez hai jise log bhool jaate hain. Har differentiation exponent ko neeche le aati hai, aur n differentiations milke n! bana deti hain. Isiliye divide karna zaroori hai, warna coefficient galat ho jayega. Aur dhyaan rakho — agar centre a=0 nahi hai (jaise lnx ka a=1), toh powers (x−1) ke honge, x 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). ex≈1+x, sinx≈x — yeh sab Taylor series ke chhote versions hi hain. Lekin yaad rakho: series sirf radius of convergence ke andar hi f ke barabar hoti hai, har jagah nahi.