The ordinary binomial theorem ( a + b ) n (a+b)^n ( a + b ) n n n n positive integer — then it terminates after n + 1 n+1 n + 1 1.02 = ( 1 + 0.02 ) 1 / 2 \sqrt{1.02}=(1+0.02)^{1/2} 1.02 = ( 1 + 0.02 ) 1/2 1 1.03 = ( 1 + 0.03 ) − 1 \frac{1}{1.03}=(1+0.03)^{-1} 1.03 1 = ( 1 + 0.03 ) − 1 n n n fraction or negative number .
The magic: for such n n n ( 1 + x ) n (1+x)^n ( 1 + x ) n infinite series that only makes sense (converges) when = = ∣ x ∣ < 1 = = ==|x|<1== == ∣ x ∣ < 1 == x x x later terms shrink fast , so keeping just 2–3 terms gives an excellent approximation. That's the whole game: turn an ugly root/reciprocal into easy arithmetic.
For integer n n n ( n r ) = n ! r ! ( n − r ) ! \binom{n}{r}=\dfrac{n!}{r!(n-r)!} ( r n ) = r ! ( n − r )! n ! n ≥ r n\ge r n ≥ r ( n n + 1 ) = 0 \binom{n}{n+1}=0 ( n + 1 n ) = 0
For rational/negative n n n product form of the coefficient still does:
( n r ) = n ( n − 1 ) ( n − 2 ) ⋯ ( n − r + 1 ) r ! \binom{n}{r}=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!} ( r n ) = r ! n ( n − 1 ) ( n − 2 ) ⋯ ( n − r + 1 )
Nothing forces this to become zero, so the series never terminates .
Definition General Binomial SeriesFor any real (rational) n n n = = ∣ x ∣ < 1 = = ==|x|<1== == ∣ x ∣ < 1 == ( 1 + x ) n = 1 + n x + n ( n − 1 ) 2 ! x 2 + n ( n − 1 ) ( n − 2 ) 3 ! x 3 + ⋯ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots ( 1 + x ) n = 1 + n x + 2 ! n ( n − 1 ) x 2 + 3 ! n ( n − 1 ) ( n − 2 ) x 3 + ⋯
Let f ( x ) = ( 1 + x ) n f(x)=(1+x)^n f ( x ) = ( 1 + x ) n f ( x ) = ∑ a r x r f(x)=\sum a_r x^r f ( x ) = ∑ a r x r a r = f ( r ) ( 0 ) r ! a_r=\dfrac{f^{(r)}(0)}{r!} a r = r ! f ( r ) ( 0 )
Differentiate repeatedly:
f ( x ) = ( 1 + x ) n ⇒ f ( 0 ) = 1 f(x)=(1+x)^n \Rightarrow f(0)=1 f ( x ) = ( 1 + x ) n ⇒ f ( 0 ) = 1 f ′ ( x ) = n ( 1 + x ) n − 1 ⇒ f ′ ( 0 ) = n f'(x)=n(1+x)^{n-1}\Rightarrow f'(0)=n f ′ ( x ) = n ( 1 + x ) n − 1 ⇒ f ′ ( 0 ) = n f ′ ′ ( x ) = n ( n − 1 ) ( 1 + x ) n − 2 ⇒ f ′ ′ ( 0 ) = n ( n − 1 ) f''(x)=n(n-1)(1+x)^{n-2}\Rightarrow f''(0)=n(n-1) f ′′ ( x ) = n ( n − 1 ) ( 1 + x ) n − 2 ⇒ f ′′ ( 0 ) = n ( n − 1 ) f ′ ′ ′ ( x ) = n ( n − 1 ) ( n − 2 ) ( 1 + x ) n − 3 ⇒ f ′ ′ ′ ( 0 ) = n ( n − 1 ) ( n − 2 ) f'''(x)=n(n-1)(n-2)(1+x)^{n-3}\Rightarrow f'''(0)=n(n-1)(n-2) f ′′′ ( x ) = n ( n − 1 ) ( n − 2 ) ( 1 + x ) n − 3 ⇒ f ′′′ ( 0 ) = n ( n − 1 ) ( n − 2 )
Why this step? Each derivative pulls down one more factor and drops the power by 1, so at x = 0 x=0 x = 0
So the coefficient of x r x^r x r a r = f ( r ) ( 0 ) r ! = n ( n − 1 ) ⋯ ( n − r + 1 ) r ! . a_r=\frac{f^{(r)}(0)}{r!}=\frac{n(n-1)\cdots(n-r+1)}{r!}. a r = r ! f ( r ) ( 0 ) = r ! n ( n − 1 ) ⋯ ( n − r + 1 ) .
Assemble:
( 1 + x ) n = 1 + n x + n ( n − 1 ) 2 ! x 2 + n ( n − 1 ) ( n − 2 ) 3 ! x 3 + ⋯ ■ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3+\cdots \qquad\blacksquare ( 1 + x ) n = 1 + n x + 2 ! n ( n − 1 ) x 2 + 3 ! n ( n − 1 ) ( n − 2 ) x 3 + ⋯ ■
Intuition Why "keep 2 terms" worksIf x x x 0.02 0.02 0.02 x 2 = 0.0004 x^2=0.0004 x 2 = 0.0004 x 3 = 0.000008 x^3=0.000008 x 3 = 0.000008 x x x The first correction n x nx n x
( 1 + x ) n ≈ 1 + n x ( if x very small ) (1+x)^n \approx 1 + nx \quad(\text{if }x\text{ very small}) ( 1 + x ) n ≈ 1 + n x ( if x very small ) ( 1 + x ) n ≈ 1 + n x + n ( n − 1 ) 2 x 2 ( for more accuracy ) (1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 \quad(\text{for more accuracy}) ( 1 + x ) n ≈ 1 + n x + 2 n ( n − 1 ) x 2 ( for more accuracy )
Worked example 1 — Approximate1.02 \sqrt{1.02} 1.02 1.02 = ( 1 + 0.02 ) 1 / 2 \sqrt{1.02}=(1+0.02)^{1/2} 1.02 = ( 1 + 0.02 ) 1/2 x = 0.02 x=0.02 x = 0.02 n = 1 2 n=\tfrac12 n = 2 1 ≈ 1 + 1 2 ( 0.02 ) − 1 8 ( 0.02 ) 2 = 1 + 0.01 − 0.00005 = 1.00995 \approx 1 + \tfrac12(0.02) - \tfrac18(0.02)^2 = 1 + 0.01 - 0.00005 = 1.00995 ≈ 1 + 2 1 ( 0.02 ) − 8 1 ( 0.02 ) 2 = 1 + 0.01 − 0.00005 = 1.00995 Why this step? x = 0.02 x=0.02 x = 0.02 x 2 x^2 x 2 = 1.009950 … =1.009950\ldots = 1.009950 …
Worked example 2 — Approximate1 1.03 \dfrac{1}{1.03} 1.03 1 ( 1 + 0.03 ) − 1 = 1 − 0.03 + 0.03 2 − ⋯ = 1 − 0.03 + 0.0009 = 0.9709 (1+0.03)^{-1}=1-0.03+0.03^2-\cdots = 1-0.03+0.0009 = 0.9709 ( 1 + 0.03 ) − 1 = 1 − 0.03 + 0.0 3 2 − ⋯ = 1 − 0.03 + 0.0009 = 0.9709 Why this step? We used the ( 1 + x ) − 1 (1+x)^{-1} ( 1 + x ) − 1 x = 0.03 x=0.03 x = 0.03 = 0.97087 … =0.97087\ldots = 0.97087 …
Worked example 3 — Force the "1 + x 1+x 1 + x ( 998 ) 1 / 3 (998)^{1/3} ( 998 ) 1/3 998 = 1000 ( 1 − 0.002 ) 998 = 1000(1-0.002) 998 = 1000 ( 1 − 0.002 ) ( 998 ) 1 / 3 = 10 ( 1 − 0.002 ) 1 / 3 (998)^{1/3}=10\,(1-0.002)^{1/3} ( 998 ) 1/3 = 10 ( 1 − 0.002 ) 1/3 ≈ 10 [ 1 + 1 3 ( − 0.002 ) ] = 10 ( 1 − 0.000667 ) = 9.99333 \approx 10\left[1+\tfrac13(-0.002)\right]=10(1-0.000667)=9.99333 ≈ 10 [ 1 + 3 1 ( − 0.002 ) ] = 10 ( 1 − 0.000667 ) = 9.99333 Why this step? We factored out the nearest perfect cube to make ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 ( 1 ± small ) (1\pm\text{small}) ( 1 ± small )
Worked example 4 — First-order only:( 1.001 ) 50 (1.001)^{50} ( 1.001 ) 50 x = 0.001 x=0.001 x = 0.001 n = 50 n=50 n = 50 ≈ 1 + 50 ( 0.001 ) = 1.05 \approx 1+50(0.001)=1.05 ≈ 1 + 50 ( 0.001 ) = 1.05 Why this step? n x = 0.05 nx=0.05 n x = 0.05 50 ⋅ 49 2 ( 0.001 ) 2 ≈ 0.0012 \frac{50\cdot49}{2}(0.001)^2\approx0.0012 2 50 ⋅ 49 ( 0.001 ) 2 ≈ 0.0012
x x x Why it feels right: the integer binomial theorem has no restriction, so students assume the same freedom.
The fix: For rational n n n infinite and only converges when = = ∣ x ∣ < 1 = = ==|x|<1== == ∣ x ∣ < 1 == x = 5 x=5 x = 5 ( 1 ± small ) (1\pm\text{small}) ( 1 ± small )
( a + b ) n (a+b)^n ( a + b ) n n n n Why it feels right: we're used to ( a + b ) n (a+b)^n ( a + b ) n The fix: Factor a a a ( a + b ) n = a n ( 1 + b a ) n (a+b)^n=a^n\left(1+\tfrac{b}{a}\right)^n ( a + b ) n = a n ( 1 + a b ) n ∣ b a ∣ < 1 \left|\tfrac{b}{a}\right|<1 a b < 1 ( 1 + x ) n (1+x)^n ( 1 + x ) n
Common mistake Sign errors in( 1 − x ) n (1-x)^n ( 1 − x ) n Why it feels right: you memorise ( 1 + x ) n (1+x)^n ( 1 + x ) n x → − x x\to -x x → − x The fix: Replace x x x − x -x − x ( 1 − x ) 1 / 2 = 1 − 1 2 x − 1 8 x 2 − ⋯ (1-x)^{1/2}=1-\tfrac12 x-\tfrac18 x^2-\cdots ( 1 − x ) 1/2 = 1 − 2 1 x − 8 1 x 2 − ⋯ x 2 x^2 x 2 ( − x ) 2 = x 2 (-x)^2=x^2 ( − x ) 2 = x 2 stays negative here — expand carefully!).
Recall Predict before you computeQ: Roughly, is 0.96 \sqrt{0.96} 0.96 0.98 0.98 0.98 Forecast: 0.96 = ( 1 − 0.04 ) 1 / 2 ≈ 1 − 0.02 − 1 8 ( 0.0016 ) = 0.9798 \sqrt{0.96}=(1-0.04)^{1/2}\approx1-0.02-\tfrac18(0.0016)=0.9798 0.96 = ( 1 − 0.04 ) 1/2 ≈ 1 − 0.02 − 8 1 ( 0.0016 ) = 0.9798 just below 0.98 0.98 0.98 0.97979 … 0.97979\ldots 0.97979 …
For which x x x Coefficient of x r x^r x r ( 1 + x ) n (1+x)^n ( 1 + x ) n n n n n ( n − 1 ) ⋯ ( n − r + 1 ) r ! \dfrac{n(n-1)\cdots(n-r+1)}{r!} r ! n ( n − 1 ) ⋯ ( n − r + 1 ) .
Why does the series terminate for positive integer n n n n n n For integer
n n n a factor
( n − n ) = 0 (n-n)=0 ( n − n ) = 0 appears; for rational
n n n no factor is ever zero, so it's infinite.
Expand ( 1 + x ) − 1 (1+x)^{-1} ( 1 + x ) − 1 1 − x + x 2 − x 3 + ⋯ 1-x+x^2-x^3+\cdots 1 − x + x 2 − x 3 + ⋯ Expand ( 1 + x ) 1 / 2 (1+x)^{1/2} ( 1 + x ) 1/2 1 + 1 2 x − 1 8 x 2 + ⋯ 1+\tfrac12 x-\tfrac18 x^2+\cdots 1 + 2 1 x − 8 1 x 2 + ⋯ First-order approximation of ( 1 + x ) n (1+x)^n ( 1 + x ) n Trick to approximate ( 998 ) 1 / 3 (998)^{1/3} ( 998 ) 1/3 Factor out nearest cube:
10 ( 1 − 0.002 ) 1 / 3 10(1-0.002)^{1/3} 10 ( 1 − 0.002 ) 1/3 , then expand.
Value of 1 1.03 \dfrac{1}{1.03} 1.03 1
Recall Feynman: explain to a 12-year-oldImagine a number very close to 1, like 1.02 1.02 1.02 0.02 0.02 0.02 the extra bit of the extra bit is so tiny you can ignore it. So 1.02 \sqrt{1.02} 1.02 half of 0.02 0.02 0.02 1.01 1.01 1.01 n n n
1, then n x n x n x "
( 1 + x ) n = 1 ⏟ start + n x ⏟ main + n ( n − 1 ) 2 x 2 ⏟ polish + ⋯ (1+x)^n = \underbrace{1}_{\text{start}} + \underbrace{nx}_{\text{main}} + \underbrace{\tfrac{n(n-1)}{2}x^2}_{\text{polish}}+\cdots ( 1 + x ) n = start 1 + main n x + polish 2 n ( n − 1 ) x 2 + ⋯ "One, en-ex, en-en-minus-one over two."
Binomial Theorem for Positive Integral Index — the terminating parent case.Maclaurin & Taylor Series — this note is literally the Maclaurin series of ( 1 + x ) n (1+x)^n ( 1 + x ) n Infinite Geometric Series — ( 1 − x ) − 1 = 1 + x + x 2 + ⋯ (1-x)^{-1}=1+x+x^2+\cdots ( 1 − x ) − 1 = 1 + x + x 2 + ⋯ 1 1 − x \frac{1}{1-x} 1 − x 1 Convergence of Series — the ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 Error & Approximation in Calculus — truncation error ≈ \approx ≈
terminates after n+1 terms
Approximation 1 + nx + ...
Estimate sqrt 1.02, 1/1.03
Intuition Hinglish mein samjhoDekho, jab power ek positive integer ho, tab ( 1 + x ) n (1+x)^n ( 1 + x ) n 1.02 = ( 1 + 0.02 ) 1 / 2 \sqrt{1.02}=(1+0.02)^{1/2} 1.02 = ( 1 + 0.02 ) 1/2 1 1.03 = ( 1 + 0.03 ) − 1 \frac{1}{1.03}=(1+0.03)^{-1} 1.03 1 = ( 1 + 0.03 ) − 1 infinite ban jaati hai. Ye series sirf tab kaam karti hai jab ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 x x x n ( n − 1 ) ⋯ r ! \frac{n(n-1)\cdots}{r!} r ! n ( n − 1 ) ⋯
Faayda kya hai? x x x x 2 , x 3 x^2, x^3 x 2 , x 3 ( 1 + x ) n ≈ 1 + n x + n ( n − 1 ) 2 x 2 (1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 ( 1 + x ) n ≈ 1 + n x + 2 n ( n − 1 ) x 2 n x nx n x
Sabse important trick: agar number ( 1 ± chhota ) (1\pm\text{chhota}) ( 1 ± chhota ) ( 998 ) 1 / 3 = 10 ( 1 − 0.002 ) 1 / 3 (998)^{1/3}=10(1-0.002)^{1/3} ( 998 ) 1/3 = 10 ( 1 − 0.002 ) 1/3 x = − 0.002 x=-0.002 x = − 0.002 9.99333 9.99333 9.99333
