Binomial theorem for rational indices (approximate values)

3.3.14 · Maths › Sequences & Series

WHY — rational $n$ ke liye formula kyun change hota hai?

Integer $n$ ke liye, $(r n) = \frac{n !}{r ! ( n - r )!}$ ko $n \geq r$ chahiye, aur $(n + 1 n) = 0$ tail ko khatam kar deta hai — series ruk jaati hai.

Rational/negative $n$ ke liye, fractions ke factorials exist nahi karte, lekin coefficient ki product form abhi bhi exist karti hai:

$(r n) = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - r + 1 )}{r !}$

Kuch bhi ise zero nahi banata, isliye series kabhi terminate nahi hoti.

HOW — ise first principles se kaise derive karein (Taylor/Maclaurin route)

Maano $f (x) = (1 + x)^{n}$ . Hum ek power series $f (x) = \sum a_{r} x^{r}$ banate hain jahan $a_{r} = \frac{f ^{(r)} ( 0 )}{r !}$ .

Baar baar differentiate karo:

$f (x) = (1 + x)^{n} \Rightarrow f (0) = 1$
$f^{'} (x) = n (1 + x)^{n - 1} \Rightarrow f^{'} (0) = n$
$f^{''} (x) = n (n - 1) (1 + x)^{n - 2} \Rightarrow f^{''} (0) = n (n - 1)$
$f^{'''} (x) = n (n - 1) (n - 2) (1 + x)^{n - 3} \Rightarrow f^{'''} (0) = n (n - 1) (n - 2)$

Ye step kyun? Har derivative ek aur factor neeche kheenchti hai aur power ko 1 se ghataati hai, isliye $x = 0$ par hume exactly numerator ka pattern milta hai.

Toh $x^{r}$ ka coefficient hai $a_{r} = \frac{f ^{(r)} ( 0 )}{r !} = \frac{n ( n - 1 ) \dots ( n - r + 1 )}{r !} .$

Assemble karo: $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots ■$

Approximation shortcut (80/20 wala)

Figure — Binomial theorem for rational indices (approximate values)

Common Mistakes (Steel-man + Fix)

Flashcards

Rational-index binomial series kis

x

ke liye converge karti hai?

∣ x ∣ < 1

ke liye.

Rational

n

ke liye

(1 + x)^{n}

mein

x^{r}

ka coefficient kya hai?

\frac{n ( n - 1 ) \dots ( n - r + 1 )}{r !}

Positive integer

n

ke liye series terminate kyun hoti hai lekin rational

n

ke liye nahi?

Integer

n

ke liye ek factor

(n - n) = 0

aata hai; rational

n

ke liye koi bhi factor kabhi zero nahi hota, isliye ye infinite hai.

(1 + x)^{- 1}

expand karo.

1 - x + x^{2} - x^{3} + \dots

(1 + x)^{1/2}

ko 3 terms tak expand karo.

1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots

(1 + x)^{n}

ka first-order approximation kya hai?

1 + n x

(998)^{1/3}

approximate karne ki trick kya hai?

Nearest cube factor out karo:

10 (1 - 0.002)^{1/3}

, phir expand karo.

Binomial use karke

\frac{1}{1.03}

ki value 4 dp tak kya hai?

0.9709

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek number jo 1 ke bahut paas hai, jaise $1.02$ . Iska square root lena ya isse 1 divide karna haath se annoying hai. Trick ye hai: jab bhi tum 1 ke paas ki cheezein multiply karte ho, "extra bit" ( $0.02$ ) zyaadatar kaam karta hai, aur extra bit ka extra bit itna tiny hota hai ki tum use ignore kar sakte ho. Toh $1.02$ ≈ 1 plus $0.02$ ka aadha = $1.01$ . Binomial series ek recipe hai jo kehti hai: 1 lo, $n$ times chhota bit add karo, phir thoda correction, aur ruk jao. Jab tak chhota bit 1 se kam hai, corrections shrink hote rehte hain aur tum jaldi ho jaate ho.

Connections

Binomial Theorem for Positive Integral Index — ye terminate hone wala parent case hai.
Maclaurin & Taylor Series — ye note literally $(1 + x)^{n}$ ki Maclaurin series hai.
Infinite Geometric Series — $(1 - x)^{- 1} = 1 + x + x^{2} + \dots$ exactly GP sum $\frac{1}{1 - x}$ hai.
Convergence of Series — $∣ x ∣ < 1$ condition.
Error & Approximation in Calculus — truncation error $\approx$ pehla dropped term.

Binomial theorem for rational indices (approximate values)

WHY — rational $n$ ke liye formula kyun change hota hai?

HOW — ise first principles se kaise derive karein (Taylor/Maclaurin route)

Approximation shortcut (80/20 wala)

Worked Examples

Common Mistakes (Steel-man + Fix)

Forecast-then-Verify

Flashcards

Connections

Concept Map