Intuition The big pictureA Maclaurin series is just a polynomial pretending to be a complicated function near x = 0 x=0 x = 0 WHY does this work? Because a smooth function is completely pinned down at one point by its value, slope, curvature, rate-of-change-of-curvature, ... If you match ALL the derivatives at x = 0 x=0 x = 0 WHAT we build: an infinite polynomial a 0 + a 1 x + a 2 x 2 + … a_0 + a_1x + a_2x^2 + \dots a 0 + a 1 x + a 2 x 2 + … 0 0 0 0 0 0
Definition Maclaurin seriesThe Maclaurin series of f f f = = 0 = = ==0== == 0 == f ( x ) = ∑ n = 0 ∞ f ( n ) ( 0 ) n ! x n f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}\,x^n f ( x ) = ∑ n = 0 ∞ n ! f ( n ) ( 0 ) x n
Derivation from scratch. Assume f f f f ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ⋯ f(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots f ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ⋯
Find a 0 a_0 a 0 put x = 0 x=0 x = 0 x x x f ( 0 ) = a 0 f(0)=a_0 f ( 0 ) = a 0 Why? 0 k = 0 0^k=0 0 k = 0 Find a 1 a_1 a 1 differentiate once: f ′ ( x ) = a 1 + 2 a 2 x + 3 a 3 x 2 + ⋯ f'(x)=a_1+2a_2x+3a_3x^2+\cdots f ′ ( x ) = a 1 + 2 a 2 x + 3 a 3 x 2 + ⋯ x = 0 x=0 x = 0 f ′ ( 0 ) = a 1 f'(0)=a_1 f ′ ( 0 ) = a 1 Find a 2 a_2 a 2 differentiate again: f ′ ′ ( x ) = 2 a 2 + 6 a 3 x + ⋯ f''(x)=2a_2+6a_3x+\cdots f ′′ ( x ) = 2 a 2 + 6 a 3 x + ⋯ x = 0 x=0 x = 0 f ′ ′ ( 0 ) = 2 a 2 f''(0)=2a_2 f ′′ ( 0 ) = 2 a 2 a 2 = f ′ ′ ( 0 ) 2 a_2=\dfrac{f''(0)}{2} a 2 = 2 f ′′ ( 0 ) General pattern: differentiating n n n x n x^n x n n ! a n n!\,a_n n ! a n x = 0 x=0 x = 0
Intuition WHY it's the cleanestd d x e x = e x \dfrac{d}{dx}e^x=e^x d x d e x = e x every derivative equals e x e^x e x e 0 = 1 e^0=1 e 0 = 1 1 1 1
HOW: f ( n ) ( 0 ) = e 0 = 1 f^{(n)}(0)=e^0=1 f ( n ) ( 0 ) = e 0 = 1 n n n a n = 1 n ! a_n=\dfrac{1}{n!} a n = n ! 1 e x = ∑ n = 0 ∞ x n n ! = 1 + x + x 2 2 ! + x 3 3 ! + ⋯ ( all x ) \boxed{e^x=\sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots}\quad(\text{all }x) e x = n = 0 ∑ ∞ n ! x n = 1 + x + 2 ! x 2 + 3 ! x 3 + ⋯ ( all x )
e 0.1 e^{0.1} e 0.1 1 + 0.1 + 0.01 2 = 1.105 1+0.1+\frac{0.01}{2}=1.105 1 + 0.1 + 2 0.01 = 1.105 Why this step? We just plug x = 0.1 x=0.1 x = 0.1 x n n ! \frac{x^n}{n!} n ! x n 1.10517 … 1.10517\ldots 1.10517 …
Intuition WHY the pattern alternatesDerivatives of sin \sin sin sin → cos → − sin → − cos → sin \sin\to\cos\to-\sin\to-\cos\to\sin sin → cos → − sin → − cos → sin x = 0 x=0 x = 0 0 , 1 , 0 , − 1 , … 0,1,0,-1,\dots 0 , 1 , 0 , − 1 , … only odd powers survive with alternating signs.
HOW for sin x \sin x sin x f ( 0 ) = 0 , f ′ ( 0 ) = 1 , f ′ ′ ( 0 ) = 0 , f ′ ′ ′ ( 0 ) = − 1 , … f(0)=0,\;f'(0)=1,\;f''(0)=0,\;f'''(0)=-1,\dots f ( 0 ) = 0 , f ′ ( 0 ) = 1 , f ′′ ( 0 ) = 0 , f ′′′ ( 0 ) = − 1 , … sin x = x − x 3 3 ! + x 5 5 ! − ⋯ = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! \boxed{\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{(2n+1)!}} sin x = x − 3 ! x 3 + 5 ! x 5 − ⋯ = n = 0 ∑ ∞ ( 2 n + 1 )! ( − 1 ) n x 2 n + 1
HOW for cos x \cos x cos x f ( 0 ) = 1 , f ′ ( 0 ) = 0 , f ′ ′ ( 0 ) = − 1 , … f(0)=1,\;f'(0)=0,\;f''(0)=-1,\dots f ( 0 ) = 1 , f ′ ( 0 ) = 0 , f ′′ ( 0 ) = − 1 , … only even powers :
cos x = 1 − x 2 2 ! + x 4 4 ! − ⋯ = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! \boxed{\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots=\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}} cos x = 1 − 2 ! x 2 + 4 ! x 4 − ⋯ = n = 0 ∑ ∞ ( 2 n )! ( − 1 ) n x 2 n
Intuition Sanity check — paritysin \sin sin odd ⇒ only odd powers. cos \cos cos even ⇒ only even powers. Differentiating sin \sin sin cos \cos cos d d x ( x − x 3 6 ) = 1 − x 2 2 \frac{d}{dx}\big(x-\frac{x^3}{6}\big)=1-\frac{x^2}{2} d x d ( x − 6 x 3 ) = 1 − 2 x 2
Intuition WHY not just differentiateYou can differentiate ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) d d x ln ( 1 + x ) = 1 1 + x \dfrac{d}{dx}\ln(1+x)=\dfrac{1}{1+x} d x d ln ( 1 + x ) = 1 + x 1 geometric series .
HOW. The geometric series (from ∑ r n = 1 1 − r \sum r^n=\frac{1}{1-r} ∑ r n = 1 − r 1 r = − x r=-x r = − x 1 1 + x = 1 − x + x 2 − x 3 + ⋯ ( ∣ x ∣ < 1 ) \frac{1}{1+x}=1-x+x^2-x^3+\cdots\qquad(|x|<1) 1 + x 1 = 1 − x + x 2 − x 3 + ⋯ ( ∣ x ∣ < 1 ) ln ( 1 + 0 ) = 0 \ln(1+0)=0 ln ( 1 + 0 ) = 0 ln ( 1 + x ) = x − x 2 2 + x 3 3 − ⋯ = ∑ n = 1 ∞ ( − 1 ) n − 1 x n n ( − 1 < x ≤ 1 ) \boxed{\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n}}\quad(-1<x\le 1) ln ( 1 + x ) = x − 2 x 2 + 3 x 3 − ⋯ = n = 1 ∑ ∞ n ( − 1 ) n − 1 x n ( − 1 < x ≤ 1 )
Worked example Why does it blow up atx = − 1 x=-1 x = − 1 x = − 1 x=-1 x = − 1 ln ( 0 ) = − ∞ \ln(0)=-\infty ln ( 0 ) = − ∞ − 1 − 1 2 − 1 3 − ⋯ -1-\tfrac12-\tfrac13-\cdots − 1 − 2 1 − 3 1 − ⋯ Why this step? The interval of convergence matches the function's domain trouble.
Intuition WHY it generalises Pascal's triangleFor positive integer n n n n n n infinite series — but built the same way: from derivatives at 0 0 0
HOW. f ( x ) = ( 1 + x ) n f(x)=(1+x)^n f ( x ) = ( 1 + x ) n f ′ ( x ) = n ( 1 + x ) n − 1 f'(x)=n(1+x)^{n-1} f ′ ( x ) = n ( 1 + x ) n − 1 f ′ ′ ( x ) = n ( n − 1 ) ( 1 + x ) n − 2 f''(x)=n(n-1)(1+x)^{n-2} f ′′ ( x ) = n ( n − 1 ) ( 1 + x ) n − 2 f ( k ) ( x ) = n ( n − 1 ) ⋯ ( n − k + 1 ) ( 1 + x ) n − k f^{(k)}(x)=n(n-1)\cdots(n-k+1)(1+x)^{n-k} f ( k ) ( x ) = n ( n − 1 ) ⋯ ( n − k + 1 ) ( 1 + x ) n − k x = 0 x=0 x = 0 f ( k ) ( 0 ) = n ( n − 1 ) ⋯ ( n − k + 1 ) f^{(k)}(0)=n(n-1)\cdots(n-k+1) f ( k ) ( 0 ) = n ( n − 1 ) ⋯ ( n − k + 1 ) k ! k! k ! ( 1 + x ) n = ∑ k = 0 ∞ ( n k ) x k = 1 + n x + n ( n − 1 ) 2 ! x 2 + ⋯ ( ∣ x ∣ < 1 if n ∉ Z ≥ 0 ) \boxed{(1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k=1+nx+\frac{n(n-1)}{2!}x^2+\cdots}\quad(|x|<1\text{ if }n\notin\mathbb{Z}_{\ge0}) ( 1 + x ) n = k = 0 ∑ ∞ ( k n ) x k = 1 + n x + 2 ! n ( n − 1 ) x 2 + ⋯ ( ∣ x ∣ < 1 if n ∈ / Z ≥ 0 ) ( n k ) = n ( n − 1 ) ⋯ ( n − k + 1 ) k ! \dbinom{n}{k}=\dfrac{n(n-1)\cdots(n-k+1)}{k!} ( k n ) = k ! n ( n − 1 ) ⋯ ( n − k + 1 )
1 + x \sqrt{1+x} 1 + x n = 1 2 n=\tfrac12 n = 2 1 1 + 1 2 x + 1 2 ( − 1 2 ) 2 x 2 + ⋯ = 1 + 1 2 x − 1 8 x 2 + ⋯ 1+\tfrac12 x+\dfrac{\frac12(-\frac12)}{2}x^2+\cdots=1+\tfrac12 x-\tfrac18 x^2+\cdots 1 + 2 1 x + 2 2 1 ( − 2 1 ) x 2 + ⋯ = 1 + 2 1 x − 8 1 x 2 + ⋯ Why this step? Just feed n = 1 2 n=\tfrac12 n = 2 1 1.21 \sqrt{1.21} 1.21 x = 0.21 x=0.21 x = 0.21 1 + 0.105 − 0.0055 = 1.0995 ≈ 1.1 1+0.105-0.0055=1.0995\approx1.1 1 + 0.105 − 0.0055 = 1.0995 ≈ 1.1
Common mistake Steel-man the common errors(a) Forgetting the n ! n! n ! Wrong idea: "coefficients are just the derivatives." Why it feels right: the value-matching at x = 0 x=0 x = 0 Fix: each d n d x n x n = n ! \frac{d^n}{dx^n}x^n=n! d x n d n x n = n ! n ! n! n ! (b) Using sin x = x − x 3 3 \sin x = x-\frac{x^3}{3} sin x = x − 3 x 3 Feels right because integrating dropped exponents. Fix: the denominator is the factorial 3 ! = 6 3!=6 3 ! = 6 3 3 3 (c) Plugging x = 2 x=2 x = 2 ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) Feels right — it's a formula, right? Fix: convergence only for − 1 < x ≤ 1 -1<x\le1 − 1 < x ≤ 1 ln 3 \ln 3 ln 3 (d) Sign slip in geometric series: 1 1 + x \frac{1}{1+x} 1 + x 1 r = − x r=-x r = − x + + +
Recall Feynman: explain to a 12-year-oldImagine you only know where a runner is right now , plus their speed, plus how fast that speed is changing, and so on. That's enough to predict their whole path near you! A Maclaurin series does this for a curve: it copies the height, the slope, the bend, the bend-of-the-bend at x = 0 x=0 x = 0 1 , x , x 2 , x 3 , … 1, x, x^2, x^3,\dots 1 , x , x 2 , x 3 , …
Mnemonic Remember the five"Every Sane Cat Loves Bananas" → E ˣ (all 1 n ! \frac{1}{n!} n ! 1 S in (odd, alt), C os (even, alt), L n(1+x) (alt, / n /n / n B inomial. And: "factorials downstairs always."
Master Maclaurin coefficient formula a n = f ( n ) ( 0 ) n ! a_n=\dfrac{f^{(n)}(0)}{n!} a n = n ! f ( n ) ( 0 ) , so
f ( x ) = ∑ f ( n ) ( 0 ) n ! x n f(x)=\sum \frac{f^{(n)}(0)}{n!}x^n f ( x ) = ∑ n ! f ( n ) ( 0 ) x n Why the n ! n! n ! d n d x n x n = n ! \frac{d^n}{dx^n}x^n=n! d x n d n x n = n ! , the only surviving term at
x = 0 x=0 x = 0 Series for e x e^x e x ∑ n = 0 ∞ x n n ! = 1 + x + x 2 2 ! + ⋯ \sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\cdots ∑ n = 0 ∞ n ! x n = 1 + x + 2 ! x 2 + ⋯ , valid all
x x x Series for sin x \sin x sin x ∑ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! = x − x 3 6 + x 5 120 − ⋯ \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots ∑ ( 2 n + 1 )! ( − 1 ) n x 2 n + 1 = x − 6 x 3 + 120 x 5 − ⋯ (odd powers)
Series for cos x \cos x cos x ∑ ( − 1 ) n x 2 n ( 2 n ) ! = 1 − x 2 2 + x 4 24 − ⋯ \sum \frac{(-1)^n x^{2n}}{(2n)!}=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots ∑ ( 2 n )! ( − 1 ) n x 2 n = 1 − 2 x 2 + 24 x 4 − ⋯ (even powers)
Series for ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) x − x 2 2 + x 3 3 − ⋯ x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots x − 2 x 2 + 3 x 3 − ⋯ , valid
− 1 < x ≤ 1 -1<x\le1 − 1 < x ≤ 1 How to derive ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) Integrate
1 1 + x = 1 − x + x 2 − ⋯ \frac{1}{1+x}=1-x+x^2-\cdots 1 + x 1 = 1 − x + x 2 − ⋯ term-by-term
Binomial series ( 1 + x ) n (1+x)^n ( 1 + x ) n ∑ k ( n k ) x k = 1 + n x + n ( n − 1 ) 2 x 2 + ⋯ \sum_k \binom{n}{k}x^k=1+nx+\frac{n(n-1)}{2}x^2+\cdots ∑ k ( k n ) x k = 1 + n x + 2 n ( n − 1 ) x 2 + ⋯ ,
∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 if
n ∉ Z ≥ 0 n\notin\mathbb Z_{\ge0} n ∈ / Z ≥ 0 First three terms of 1 + x \sqrt{1+x} 1 + x 1 + x 2 − x 2 8 + ⋯ 1+\frac{x}{2}-\frac{x^2}{8}+\cdots 1 + 2 x − 8 x 2 + ⋯ Why only odd powers in sin \sin sin sin \sin sin is an odd function ⇒ even-power coefficients vanish
Taylor series (general centre a) — Maclaurin is the a = 0 a=0 a = 0 Geometric series — backbone for ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) ( 1 + x ) − 1 (1+x)^{-1} ( 1 + x ) − 1 Radius & interval of convergence — where these series are validEuler's formula — e i x = cos x + i sin x e^{ix}=\cos x+i\sin x e i x = cos x + i sin x L'Hôpital & limits via series — e.g. lim x → 0 sin x x = 1 \lim_{x\to0}\frac{\sin x}{x}=1 lim x → 0 x s i n x = 1 Binomial theorem (integer n) — special terminating case
Match all derivatives at 0
Coefficient rule an equals fn0 over n!
Geometric series 1 over 1+x
Intuition Hinglish mein samjhoMaclaurin series ka idea simple hai: koi bhi smooth function ko hum ek polynomial se replace kar sakte hain, lekin sirf x = 0 x=0 x = 0 x = 0 x=0 x = 0 f ( x ) = ∑ f ( n ) ( 0 ) n ! x n f(x)=\sum \frac{f^{(n)}(0)}{n!}x^n f ( x ) = ∑ n ! f ( n ) ( 0 ) x n n ! n! n ! x n x^n x n n n n n ! n! n !
Paanch important series yaad rakho. e x e^x e x e x e^x e x e 0 = 1 e^0=1 e 0 = 1 1 n ! \frac{1}{n!} n ! 1 sin x \sin x sin x cos x \cos x cos x sin , cos , − sin , − cos \sin,\cos,-\sin,-\cos sin , cos , − sin , − cos sin \sin sin cos \cos cos
ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) 1 1 + x = 1 − x + x 2 − ⋯ \frac{1}{1+x}=1-x+x^2-\cdots 1 + x 1 = 1 − x + x 2 − ⋯ r = − x r=-x r = − x − 1 < x ≤ 1 -1<x\le1 − 1 < x ≤ 1 ( 1 + x ) n (1+x)^n ( 1 + x ) n n n n n n n ( n k ) \binom{n}{k} ( k n )
Ye topic kyun matter karta hai? Kyunki complicated functions ko approximate karna, limits nikalna (jaise sin x x → 1 \frac{\sin x}{x}\to 1 x s i n x → 1 0 0 0 n ! n! n !
