Taylor series — derivation from power series: Worked examples

4.3.16 · D3 · Maths › Calculus III — Sequences & Series › Taylor series — derivation from power series

Yeh ek drill page hai. Parent note ne woh ek formula prove kiya jo yahan sab kuch chalata hai:

Notation se pehle, teen reminders seedhi baaton mein:

$f^{(n)} (a)$ ka matlab hai "derivative $n$ baar lo, phir $x$ ko centre $a$ ke barabar set karo" — ek single number.
$n!$ (padho " $n$ factorial") ka matlab hai $n \times (n - 1) \times \dots \times 2 \times 1$ , convention ke saath $0! = 1$ aur $1! = 1$ . Yeh woh number hai jo $n$ differentiations pile up karte hain.
$(x - a)$ centre se displacement hai: aap safe point $a$ se kitna door chal ke aaye hain.

The scenario matrix

Har Taylor problem jo aap miloge in cells mein se ek mein aata hai. Neeche ke examples us cell ke label ke saath hain jo woh cover karte hain, aur milke yeh har row ko hit karte hain.

Cell	Kya cheez ise alag banati hai	Example
A. Centre $a = 0$ , all derivatives equal	derivatives hamesha repeat hote hain	Ex 1 ( $e^{x}$ variant $e^{2 x}$ )
B. Centre $a = 0$ , derivatives signs ke saath cycle karte hain	sirf kuch powers survive karte hain	Ex 2 ( $cos x$ built, sign bookkeeping)
C. Centre $a \neq = 0$	$(x - a)$ ke powers, $x$ ke nahi	Ex 3 ( $x$ about $a = 4$ )
D. Polynomial input (degenerate: series terminate ho jaata hai)	finite, exact, remainder $= 0$	Ex 4
E. Negative / all-sign coefficients	geometric-type, $(x - a)$ ka sign matter karta hai	Ex 5 ( $\frac{1}{1 - x}$ aur $\frac{1}{1 + x}$ )
F. Limiting behaviour / convergence edge	boundary par aur bahar kya hota hai	Ex 6
G. Real-world word problem	build karo, truncate karo, error size estimate karo	Ex 7 (pendulum $sin θ \approx θ$ )
H. Exam twist: ek known series reuse karo	differentiate karne ke bajaye substitute karo	Ex 8 ( $e^{- x^{2}}$ from $e^{x}$ )

Cell A — derivatives jo repeat hote hain

Worked example Ex 1. Maclaurin series of

$e^{2 x}$ (centre $a = 0$ ) Forecast: Aap jaante ho $e^{x} = 1 + x + \frac{x ^{2}}{2} + \dots$ . Guess karo: kya " $2$ " har power ke andar aata hai, ya aage multiply hota hai? Padhne se pehle apna guess likho.

Step 1 — differentiate karo. $f (x) = e^{2 x}$ , toh $f^{'} (x) = 2 e^{2 x}$ , $f^{''} (x) = 4 e^{2 x} = 2^{2} e^{2 x}$ , aur generally $f^{(n)} (x) = 2^{n} e^{2 x}$ . Yeh step kyun? Recipe ko $f^{(n)}$ chahiye. Chain rule har baar ek factor $2$ drop karta hai, toh $n$ derivatives ke baad hum $2^{n}$ carry karte hain.

Step 2 — centre $a = 0$ plug in karo. $f^{(n)} (0) = 2^{n} e^{0} = 2^{n}$ . Yeh step kyun? Coefficients derivative centre par evaluate hone ke baad use karte hain, aur $e^{0} = 1$ .

Step 3 — $n!$ se divide karo. $c_{n} = \frac{2 ^{n}}{n !} \Rightarrow e^{2 x} = \sum_{n = 0}^{\infty} \frac{2 ^{n}}{n !} x^{n} = 1 + 2 x + 2 x^{2} + \frac{4 x ^{3}}{3} + \dots$ Yeh step kyun? Yahi literally $c_{n} = f^{(n)} (a) / n!$ hai.

Verify karo: Poori series $e^{x}$ ke barabar hai jahan $x$ ko $2 x$ se replace kiya gaya: $\sum (2 x)^{n} / n!$ . Toh " $2$ " har power ke andar $(2 x)^{n}$ ki tarah rehta hai — Ex 8 ke shortcut se match karta hai. Pehle kuch terms: $1 + 2 x + \frac{4 x ^{2}}{2} + \frac{8 x ^{3}}{6} = 1 + 2 x + 2 x^{2} + \frac{4 x ^{3}}{3}$ . ✓

Cell B — cycling derivatives aur sign bookkeeping

Worked example Ex 2. Maclaurin series of

$cos x$ — carefully track karo kaunse terms die hote hain Forecast: $x$ ke kaunse powers appear hote hain — even, odd, ya sab? Aur $x^{4}$ term ka sign kya hoga?

Step 1 — derivative cycle list karo. $cos x \to - sin x \to - cos x \to sin x \to cos x$ (period 4). Yeh step kyun? Trig derivatives cycle karte hain, toh hamesha ke liye compute karne ke bajaye hum repeat spot karte hain.

Step 2 — har ek ko $a = 0$ par evaluate karo. $f (0) = 1, f^{'} (0) = 0, f^{''} (0) = - 1, f^{'''} (0) = 0, f^{(4)} (0) = 1, \dots$ Yeh step kyun? $sin 0 = 0$ har odd-order derivative ko kill karta hai; even ones $+ 1, - 1$ alternate karte hain.

Step 3 — coefficients build karo. Odd $n$ : $c_{n} = 0$ . Even $n = 2 k$ : $c_{2 k} = \frac{( - 1 ) ^{k}}{( 2 k )!}$ . $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \frac{x ^{6}}{6 !} + \dots$ Yeh step kyun? Sirf even derivatives survive karte hain, toh sirf even powers appear hote hain; alternating $\pm 1$ se $(- 1)^{k}$ milta hai.

Verify karo: $x^{4}$ term $+ \frac{1}{4 !} = + \frac{1}{24}$ hai (positive, jaisa $(- 1)^{2} = + 1$ ne predict kiya tha). Term-by-term differentiate karo: $\frac{d}{d x} cos x = - sin x$ , aur indeed $\frac{d}{d x} (1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24}) = - x + \frac{x ^{3}}{6} = - (x - \frac{x ^{3}}{6}) = - sin x$ . ✓ Dekho Maclaurin series — common expansions.

Cell C — ek centre jo zero nahi hai

Yahan picture matter karti hai: hum $x$ ko safe point $x = 4$ ke paas approximate kar rahe hain (jahan answer clean number $2$ hai).

Figure — Taylor series — derivation from power series

Worked example Ex 3. Taylor series of

$x$ about $a = 4$ , pehle teen terms; $4.2$ estimate karo Forecast: Figure dekho. Red curve $x$ hai. Kya parabola (2 terms + curvature) $x = 4$ ke thoda right mein curve ke upar hoga ya neeche? $(x - 4)^{2}$ coefficient ka sign guess karo.

Step 1 — differentiate karo. $f (x) = x^{1/2}$ , $f^{'} (x) = \frac{1}{2} x^{- 1/2}$ , $f^{''} (x) = - \frac{1}{4} x^{- 3/2}$ . Yeh step kyun? Hume pehle teen coefficients ke liye $f, f^{'}, f^{''}$ centre par chahiye.

Step 2 — $a = 4$ plug in karo. $4 = 2$ , $4^{- 1/2} = \frac{1}{2}$ , $4^{- 3/2} = \frac{1}{8}$ . $f (4) = 2, f^{'} (4) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}, f^{''} (4) = - \frac{1}{4} \cdot \frac{1}{8} = - \frac{1}{32}$ Yeh step kyun? Coefficients derivatives centre $a = 4$ par use karte hain, $0$ par nahi.

Step 3 — $n!$ se divide karo aur $(x - 4)$ ke powers use karo. $x \approx 2 + \frac{1}{4} (x - 4) - \frac{1}{32 \cdot 2 !} (x - 4)^{2} = 2 + \frac{1}{4} (x - 4) - \frac{1}{64} (x - 4)^{2}$ Yeh step kyun? Basis $(x - a)^{n} = (x - 4)^{n}$ hai — $x$ ke powers nahi (parent note mein is galti ke baare mein warn kiya gaya hai).

Step 4 — $4.2$ estimate karo. Yahan $x - 4 = 0.2$ hai: $2 + \frac{1}{4} (0.2) - \frac{1}{64} (0.2)^{2} = 2 + 0.05 - 0.000625 = 2.049375$ Yeh step kyun? Displacement ko truncated polynomial mein plug karna hi linear-plus-curvature approximation hai (Linear approximation & differentials sirf pehle do terms hain).

Verify karo: True value $4.2 = 2.049390 \dots$ ; hamara estimate $2.049375$ roughly $\approx 1.5 \times 1 0^{- 5}$ off hai — centre ke itne paas excellent hai. $(x - 4)^{2}$ coefficient negative hai, toh parabola curve ke neeche bend karta hai — figure mein concave-down red curve se match karta hai. ✓

Cell D — degenerate: ek polynomial (series ruk jaata hai)

Worked example Ex 4. Taylor series of

$p (x) = x^{3} - 2 x + 1$ about $a = 1$ Forecast: $p$ already ek polynomial hai. Iske Taylor series mein kitne terms honge — finite ya infinite?

Step 1 — zero tak differentiate karo. $p^{'} (x) = 3 x^{2} - 2$ , $p^{''} (x) = 6 x$ , $p^{'''} (x) = 6$ , $p^{(4)} (x) = 0$ . Yeh step kyun? Degree-3 polynomial ka 4th derivative zero hota hai — yeh degenerate case hai jahan series terminate honi chahiye.

Step 2 — $a = 1$ par evaluate karo. $p (1) = 0$ , $p^{'} (1) = 1$ , $p^{''} (1) = 6$ , $p^{'''} (1) = 6$ . Yeh step kyun? Same recipe, centre $a = 1$ .

Step 3 — assemble karo. $p (x) = 0 + 1 (x - 1) + \frac{6}{2 !} (x - 1)^{2} + \frac{6}{3 !} (x - 1)^{3} = (x - 1) + 3 (x - 1)^{2} + (x - 1)^{3}$ Yeh step kyun? Har $p^{(n)} (1)$ ko $n!$ se divide karo; $n \geq 4$ ke liye har term $0$ hai.

Verify karo: Expand karo: $(x - 1) + 3 (x - 1)^{2} + (x - 1)^{3}$ . $u = x - 1$ lo: $u + 3 u^{2} + u^{3}$ ; substitute back karke collect karo — $x^{3} - 2 x + 1$ exactly reproduce hona chahiye (remainder yahan genuinely $0$ hai). Sympy neeche confirm karta hai. Yeh woh ek case hai jahan "series $= f$ everywhere" literally sach hai. ✓

Cell E — negative / all-sign coefficients (geometric flavour)

Worked example Ex 5. Maclaurin series of

$\frac{1}{1 - x}$ aur $\frac{1}{1 + x}$ Forecast: Geometric series kehta hai $\frac{1}{1 - x} = 1 + x + x^{2} + \dots$ . $\frac{1}{1 + x}$ ke liye series guess karo — kya badlega?

Step 1 — $f (x) = (1 - x)^{- 1}$ differentiate karo. $f^{'} (x) = (1 - x)^{- 2}$ , $f^{''} (x) = 2 (1 - x)^{- 3}$ , aur generally $f^{(n)} (x) = n! (1 - x)^{- (n + 1)}$ . Yeh step kyun? Har derivative exponent ko positive factor ki tarah drop karta hai (chain rule $- (1 - x)$ se $+ 1$ deta hai), $n!$ build karta hai.

Step 2 — $a = 0$ par: $f^{(n)} (0) = n!$ , toh $c_{n} = \frac{n !}{n !} = 1$ . $\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + \dots$ Yeh step kyun? Sab coefficients $1$ hain — yeh exactly geometric series hai.

Step 3 — $\frac{1}{1 + x}$ ke liye $x \to - x$ replace karo. $\frac{1}{1 + x} = \sum_{n = 0}^{\infty} (- x)^{n} = 1 - x + x^{2} - x^{3} + \dots$ Yeh step kyun? Substitution re-differentiating se behtar hai (Cell H trick); $(x - a)^{n}$ ka sign ab alternate karta hai.

Verify karo: $x = 0.5$ par: $\frac{1}{1 + 0.5} = 0.6667$ . Partial sum $1 - 0.5 + 0.25 - 0.125 + 0.0625 = 0.6875$ ; char aur terms $0.6667$ ki taraf tighten karte hain. Sign pattern $+, -, +, -$ confirmed. ✓

Cell F — boundary par aur us ke bahar limiting behaviour

Worked example Ex 6.

$\frac{1}{1 - x} = \sum x^{n}$ actually kahan hold karta hai? Forecast: Formula ne clean coefficients $c_{n} = 1$ diye. Kya iska matlab hai yeh har $x$ ke liye $\frac{1}{1 - x}$ ke barabar hai? Padhne se pehle guess karo.

Step 1 — andar test karo: $x = 0.5$ . $\sum (0.5)^{n} = \frac{1}{1 - 0.5} = 2$ . ✓ Yeh step kyun? Radius ke andar terms shrink karte hain, sum true value pe converge karta hai.

Step 2 — boundary $x = 1$ test karo. Terms hain $1 + 1 + 1 + \dots$ — kabhi settle nahi hota. Aur $\frac{1}{1 - 1}$ undefined hai. Yeh step kyun? Edge par terms 0 ki taraf shrink nahi karte, toh series diverge karti hai — boundary ek genuine wall hai.

Step 3 — bahar test karo: $x = 2$ . Terms $1, 2, 4, 8, \dots$ blow up karte hain, phir bhi $\frac{1}{1 - 2} = - 1$ ek perfectly finite number hai. Yeh step kyun? Yeh parent note ki teesri galti concrete roop mein hai: derivation coefficients fix karta hai lekin radius of convergence ke bahar equality promise nahi karta.

Verify karo: Radius of convergence $1$ hai. $x = 2$ par true value $- 1$ finite hai lekin series nonsense hai — coefficients "sahi" the aur series phir bhi fail karti hai. Dekho Power series — radius & interval of convergence aur Taylor's theorem with remainder. ✓

Cell G — real-world word problem

Worked example Ex 7. Pendulum: physicists

$sin θ \approx θ$ kyun likhte hain Statement: Pendulum ki motion $\ddot{θ} = - \frac{g}{L} sin θ$ follow karta hai. Simple harmonic motion paane ke liye, engineers $sin θ$ ko $θ$ se replace karte hain. $θ = 1 0^{\circ}$ ke swing ke liye, error kitna bada hai?

Forecast: $1 0^{\circ}$ chhota lagta hai. $θ$ use karne ka $sin θ$ ki jagah percentage error guess karo: $1%$ se kam, kuch $%$ , ya $10%$ se zyada?

Step 1 — series likho (radians mein!). Ex 2 ke sibling se: $sin θ = θ - \frac{θ ^{3}}{6} + \dots$ . Yeh step kyun? First-order truncation (Linear approximation & differentials) exactly $sin θ \approx θ$ hai; agla term $- \frac{θ ^{3}}{6}$ leading error hai.

Step 2 — convert aur evaluate karo. $1 0^{\circ} = 0.17453$ rad. True $sin (0.17453) = 0.17365$ . Yeh step kyun? Trig series sirf radians mein $θ$ ke saath valid hain; degrees har power ko corrupt kar denge.

Step 3 — relative error. $\frac{θ - s i n θ}{s i n θ} = \frac{0.17453 - 0.17365}{0.17365} = 0.00509 \approx 0.51%$ Yeh step kyun? Dropped term $\frac{θ ^{3}}{6}$ $θ^{3}$ ki tarah scale karta hai, toh small angles chhota error dete hain — isiliye approximation trusted hai.

Verify karo: Error $\approx 0.51%$ — $1%$ se kam, toh $1 0^{\circ}$ swing safely "simple harmonic" hai. Units: dono quantities dimensionless (radians), ratio pure percent hai. ✓

Cell H — exam twist: reuse karo, re-derive mat karo

Worked example Ex 8. Maclaurin series of

$e^{- x^{2}}$ (koi differentiation nahi!) Forecast: $e^{- x^{2}}$ ko baar baar differentiate karna nightmare hai. Koi shortcut hai? Guess karo kaise.

Step 1 — known series se shuru karo. $e^{u} = \sum_{n = 0}^{\infty} \frac{u ^{n}}{n !} = 1 + u + \frac{u ^{2}}{2} + \frac{u ^{3}}{6} + \dots$ . Yeh step kyun? Recipe legitimate hai lekin slow; ek known series mein substitution same unique coefficients deta hai (uniqueness parent derivation se guaranteed hai).

Step 2 — $u = - x^{2}$ substitute karo. $e^{- x^{2}} = \sum_{n = 0}^{\infty} \frac{( - x ^{2} ) ^{n}}{n !} = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} x ^{2 n}}{n !} = 1 - x^{2} + \frac{x ^{4}}{2} - \frac{x ^{6}}{6} + \dots$ Yeh step kyun? $(- x^{2})^{n} = (- 1)^{n} x^{2 n}$ — sirf even powers, alternating signs, original se factorials.

Step 3 — ek coefficient par sanity check karo. $x^{4}$ ka coefficient $\frac{( - 1 ) ^{2}}{2 !} = \frac{1}{2}$ hai. Yeh step kyun? Ek quick check ki substitution ne factorials intact rakhe.

Verify karo: Direct differentiation $f^{(4)} (0) /4!$ deta hai. $f^{(4)} (0) = 12$ compute karne par, $x^{4}$ coefficient $12/24 = \frac{1}{2}$ hai — substitution se match karta hai. Substitution jeet gaya. ✓

Recall Kaunsa cell kaunsa hai? (self-test)

Ek polynomial ki Taylor series finite hoti hai ::: yeh terminate ho jaati hai (remainder $0$ ), Cell D. $x$ about $a = 4$ ke liye powers ::: $(x - 4)$ ke powers hain, Cell C. Kya $\sum x^{n} = \frac{1}{1 - x}$ at $x = 2$ ::: nahi — radius ke bahar series diverge karti hai chahe value finite ho, Cell F. Fastest route to $e^{- x^{2}}$ ki series ::: $u = - x^{2}$ ko $e^{u}$ mein substitute karo, Cell H.

Connections

Taylor series — derivation from power series (index 4.3.16) (woh recipe jo har example run karta hai)
Maclaurin series — common expansions (Ex 1, 2, 5, 8 sab yahan rehte hain)
Linear approximation & differentials (Ex 3, 7 first-order truncation use karte hain)
Power series — radius & interval of convergence (Ex 6 ki wall at $x = 1$ )
Taylor's theorem with remainder (Ex 3, 7 mein error size)
Geometric series (Ex 5 ke all-ones coefficients)
L'Hôpital's rule (leading Taylor terms aksar isse replace kar dete hain)