Maclaurin series of eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ — derive all: Worked examples

4.3.17 · D3 · Maths › Calculus III — Sequences & Series › Maclaurin series of eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ — deri

Scenario matrix

Har example neeche us cell ke saath tagged hai jo woh fill karta hai. Goal yeh hai: koi bhi reader kisi aisi situation se na mile jise humne skip kiya ho.

Cell	Kya cheez ise alag banati hai	Example
A · positive $x$ , fast series	$e^{x}$ mein small $x > 0$ plug karo	Ex 1
B · negative $x$ , sign bookkeeping	$e^{- x}$ , alternating signs dhyaan se dekho	Ex 2
C · substitution	kisi jaani-pehchani series mein $x$ ki jagah $x^{2}$ / $2 x$ daalo	Ex 3
D · trig, odd input & parity check	scaled angle ka $sin$	Ex 4
E · edge of convergence	$ln (1 + x)$ at $x = 1$ (allowed) vs $x = - 1$ (forbidden)	Ex 5
F · non-integer power (binomial)	$(1 + x)^{1/3}$ , ek infinite series	Ex 6
G · limit via series (the payoff)	$lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$	Ex 7
H · combine two series	$e^{x} sin x$ multiply karo, terms collect karo	Ex 8
I · real-world word problem	pendulum small-angle approximation	Ex 9
J · exam twist / degenerate	$lim_{x \to 0} \frac{e ^{x} - 1 - x}{x ^{2}}$ cells ko mix karta hua	Ex 10

Ex 1 — Cell A · positive $x$ , fast-shrinking terms

Worked example

$e^{0.2}$ ko $x^{3}$ tak ke terms use karke estimate karo. Forecast: guess karo ki answer $1.2$ se thoda upar hoga — apna guess likh lo.

Steps.

Series likho: $e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots$ . Yeh step kyun? $e^{x}$ ki har derivative $0$ par $1$ ke barabar hai, isliye saare coefficients $\frac{1}{n !}$ hain — yahan koi sign ya parity ki tension nahi.
$x = 0.2$ daalo: $1 + 0.2 + \frac{0.04}{2} + \frac{0.008}{6}$ . Yeh step kyun? Hum sirf substitute kar rahe hain; denominator mein factorials baad ke terms ko tiny bana dete hain, isliye truncate karna safe hai.
Add karo: $1 + 0.2 + 0.02 + 0.0013 \overline{3} = 1.2213 \overline{3}$ . Yeh step kyun? Hum chaar numbers jod rahe hain taaki truncated series ek single estimate mein collapse ho jaaye; running total jaldi stable ho jaata hai kyunki har naya term pichle se roughly $10 \times$ chhota hota hai, isliye chaar terms se hi teen decimals fix ho jaate hain.

Verify: sach mein $e^{0.2} = 1.221402 \dots$ ; humara error roughly $0.00007$ hai — pehle dropped term $\frac{0. 2 ^{4}}{24} \approx 0.0000 \overline{6}$ se chhota. ✓

Ex 2 — Cell B · negative $x$ , sign bookkeeping

Worked example

$e^{- x}$ ki series nikalo aur chaar terms se $e^{- 0.5}$ estimate karo. Forecast: kyunki $x > 0$ ke liye $e^{- x} < 1$ hai, guess karo answer $1$ se neeche, $0.6$ ke paas hoga.

Steps.

$e^{x}$ mein $x$ ki jagah $- x$ daalo: $e^{- x} = 1 + (- x) + \frac{( - x ) ^{2}}{2 !} + \frac{( - x ) ^{3}}{3 !} + \dots$ . Yeh step kyun? Ek valid-for-all- $x$ series mein substitution hamesha legal hai — $e$ ke liye koi convergence question nahi.
Signs simplify karo: $(- x)^{2} = + x^{2}$ , $(- x)^{3} = - x^{3}$ , jisse milta hai $e^{- x} = 1 - x + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots$ (signs alternate karte hain). Yeh step kyun? $- 1$ ki odd powers $- 1$ hoti hain, even powers $+ 1$ — yahi alternation ka poora source hai.
$x = 0.5$ daalo: $1 - 0.5 + \frac{0.25}{2} - \frac{0.125}{6} = 1 - 0.5 + 0.125 - 0.0208 \overline{3} = 0.6041 \overline{6}$ .

Verify: sach mein $e^{- 0.5} = 0.606531 \dots$ ; humara chaar-term value $0.60417$ karib hai, aur thoda neeche hai kyunki agla term $+ \frac{0. 5 ^{4}}{24} \approx + 0.0026$ use upar push karta. ✓

Ex 3 — Cell C · substitution ( $x \to x^{2}$ )

Ex 4 — Cell D · trig, scaled angle & parity check

Worked example

$sin (3 x)$ ki series teen nonzero terms tak; isse $sin (0.3)$ estimate karo (toh $3 x = 0.3$ , $x = 0.1$ ). Forecast: sirf odd powers bachenge (sin odd hai); guess karo $sin (0.3) \approx 0.2955$ .

Steps.

$sin u = u - \frac{u ^{3}}{3 !} + \frac{u ^{5}}{5 !} - \dots$ . Yeh step kyun? $sin$ ki derivatives $0$ par $0, 1, 0, - 1$ cycle karti hain — har even power ko khatam kar deti hain.
$u = 3 x$ substitute karo: $sin (3 x) = 3 x - \frac{( 3 x ) ^{3}}{6} + \frac{( 3 x ) ^{5}}{120} - \dots = 3 x - \frac{27 x ^{3}}{6} + \frac{243 x ^{5}}{120} - \dots$ . Yeh step kyun? $(3 x)^{3} = 27 x^{3}$ ; scale factor power se amplify ho jaata hai — yeh ek common exam trap hai.
Tidy karo: $sin (3 x) = 3 x - 4.5 x^{3} + 2.025 x^{5} - \dots$ . Yeh step kyun? Factorials divide karne se clean decimal coefficients milte hain: $\frac{27}{6} = 4.5$ aur $\frac{243}{120} = 2.025$ . Abhi simplify karna agले step ke numeric substitution ko fractions-free bana deta hai.
$x = 0.1$ daalo: $0.3 - 4.5 (0.001) + 2.025 (0.00001) = 0.3 - 0.0045 + 0.00002025 = 0.29552$ .

Verify: sach mein $sin (0.3) = 0.295520 \dots$ . ✓ Parity check: sirf $x$ ki odd powers hain, jaisa ek odd function mein hona chahiye. ✓

Neeche ka figure sach mein $sin (3 x)$ (teal) ko hamare odd-power polynomial (dashed orange) se compare karta hai: dono curves $x = 0$ ke paas indistinguishable hain aur sirf $∣ x ∣ \approx 1$ ke paas alag hone lagte hain, jo theek yahi batata hai ki $x^{5}$ par truncate karna small input $x = 0.1$ (plum dot) ke liye safe kyun hai. Yeh parity ko bhi visible banata hai — polynomial origin ke baare mein antisymmetric hai, odd function ko mirror karta hua.

Figure — Maclaurin series of eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ — derive all

Ex 5 — Cell E · convergence ki edge

Worked example

$ln (1 + x)$ series se $ln 2$ evaluate karo ( $x = 1$ set karke). Phir explain karo ki $x = - 1$ kyun forbidden hai. Forecast: guess karo $ln 2 \approx 0.69$ ; guess karo $x = - 1$ blow up karta hai.

Steps.

$ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ , $- 1 < x \leq 1$ ke liye valid. Yeh step kyun? Yeh geometric series $\frac{1}{1 + x} = 1 - x + x^{2} - \dots$ ko integrate karne se aaya, jo sirf $∣ x ∣ < 1$ ke liye converge karta hai; right endpoint $x = 1$ bach jaata hai (alternating series), left wala nahi.
$x = 1$ daalo: $ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ (yeh alternating harmonic series hai). Yeh step kyun? $x = 1$ included hai ( $x \leq 1$ ), aur alternating shrinking terms converge karte hain.
Das terms dete hain $1 - 0.5 + 0.3 \overline{3} - 0.25 + 0.2 - 0.1 \overline{6} + 0.142857 - 0.125 + 0.1 \overline{1} - 0.1 = 0.645635$ . Yeh step kyun? Yeh series dhheere converge karti hai — sirf ~1 digit sahi expect karo, aur error agले term $\frac{1}{11} \approx 0.091$ se neeche.
Left endpoint $x = - 1$ : series ban jaati hai $- 1 - \frac{1}{2} - \frac{1}{3} - \dots$ , jo negative harmonic series hai, jo diverge karti hai — $ln (1 + (- 1)) = ln 0 = - \infty$ se match karti hai. Yeh step kyun? Convergence ka interval function ki apni domain trouble ko mirror karta hai.

Verify: sach mein $ln 2 = 0.693147 \dots$ ; hamara 10-term partial sum $0.645635$ usse ek term ke size ke andar hai (slow convergence, par sahi direction mein). ✓

Ex 6 — Cell F · non-integer power (binomial series)

Ex 7 — Cell G · limit via series (classic payoff)

Worked example

$x \to 0 lim \frac{1 - cos x}{x ^{2}}$ evaluate karo. Forecast: $x = 0$ par yeh $\frac{0}{0}$ hai — indeterminate. Guess karo answer $\frac{1}{2}$ hai.

Steps.

$cos x$ ki jagah uski series daalo: $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ . Yeh step kyun? Series ek mysterious $\frac{0}{0}$ ko seedhi algebra mein badal deti hain — yeh limits ka series route hai.
Phir $1 - cos x = \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \dots$ . Yeh step kyun? $1$ cancel ho jaate hain; bachne wala leading term $\frac{x ^{2}}{2}$ hai, jo neeche ke $x^{2}$ se match karta hai.
$x^{2}$ se divide karo: $\frac{1 - cos x}{x ^{2}} = \frac{1}{2} - \frac{x ^{2}}{24} + \dots$ . Yeh step kyun? $x^{2}$ cancel karna $\frac{0}{0}$ ko hata deta hai — yahi toh poora point hai.
$x \to 0$ jaane do: $x$ wale baaki saare terms khatam ho jaate hain, $\frac{1}{2}$ bachta hai. Yeh step kyun? Jab indeterminacy khatam ho jaaye toh expression ek ordinary polynomial hai, isliye limit sirf $0$ par uski value hai — constant term $\frac{1}{2}$ , kyunki $- \frac{x ^{2}}{24} \to 0$ .

Verify: L'Hôpital do baar lagate hain: $lim \frac{s i n x}{2 x} = lim \frac{c o s x}{2} = \frac{1}{2}$ . ✓ Numeric: $x = 0.01$ par, $\frac{1 - c o s 0.01}{0.0001} = 0.499996$ . ✓

Ex 8 — Cell H · do series multiply karna

Ex 9 — Cell I · real-world word problem (pendulum)

Worked example Ek pendulum ka sach wala period

$sin θ$ involve karta hai; "small-angle" model $sin θ$ ki jagah $θ$ use karta hai. $θ = 1 0^{\circ}$ ke liye, us swap ka relative error kya hai? Forecast: $1 0^{\circ}$ small hai — guess karo error percent ka fraction hoga.

Steps.

Convert karo: $1 0^{\circ} = \frac{10 π}{180} = 0.174533$ rad. Yeh step kyun? $sin$ ki series ko angle radians mein chahiye — uske derivative rules radians assume karte hain.
Series: $sin θ = θ - \frac{θ ^{3}}{6} + \dots$ , toh small-angle model $\frac{θ ^{3}}{6}$ drop karta hai. Yeh step kyun? $sin θ$ ki jagah $θ$ rakhna literally $- \frac{θ ^{3}}{6}$ term aur aage wale terms throw away karna hai.
Relative error $\approx \frac{θ - sin θ}{sin θ} \approx \frac{θ ^{3} /6}{θ} = \frac{θ ^{2}}{6}$ . Yeh step kyun? Leading value ke upar leading error ek clean $\frac{θ ^{2}}{6}$ rule of thumb deta hai.
Plug in karo: $\frac{0.17453 3 ^{2}}{6} = \frac{0.030462}{6} = 0.005077$ , yaani roughly $0.51%$ .

Verify: direct: $sin (0.174533) = 0.173648$ ; $\frac{0.174533 - 0.173648}{0.173648} = 0.005097$ , yaani $\approx 0.51%$ . ✓ $\frac{θ ^{2}}{6}$ estimate ekdum sahi baitha — small angle, tiny error.

Ex 10 — Cell J · exam twist / mixed indeterminate

Recall Kaun si cell kaun si thi? (self-test)

Positive small- $x$ estimate ::: Ex 1 (Cell A, $e^{0.2}$ ) $e^{- x}$ se alternating signs ::: Ex 2 (Cell B) cos ke andar $x^{2}$ substitute karna ::: Ex 3 (Cell C) Scaled trig angle & parity ::: Ex 4 (Cell D) Endpoint $x = 1$ kaam karta hai, $x = - 1$ fail ::: Ex 5 (Cell E) Non-integer power infinite series ::: Ex 6 (Cell F) $\frac{1 - c o s x}{x ^{2}} \to \frac{1}{2}$ ::: Ex 7 (Cell G) $e^{x} sin x$ multiply karna ::: Ex 8 (Cell H) Pendulum small-angle error ::: Ex 9 (Cell I) $\frac{e ^{x} - 1 - x}{x ^{2}} \to \frac{1}{2}$ ::: Ex 10 (Cell J)

Connections

Parent topic (Hinglish) — woh derivations jinhe yeh examples exercise karte hain
L'Hôpital & limits via series — Ex 7 & Ex 10 series route hain
Radius & interval of convergence — Ex 5 ka endpoint drama
Binomial theorem (integer n) — Ex 6 non-integer $n$ ke saath
Geometric series — Ex 5 mein $ln (1 + x)$ ke peeche ka engine
Euler's formula — Ex 8 ke series-multiplication idea ka cousin