4.3.19
Intuition Ek saanth mein badi baatEk Taylor series ek daraauni function ko ek infinite polynomial mein badal deti hai. Polynomials hi woh cheez hain jo hum sach mein haath se compute kar sakte hain (add, multiply, divide). Toh jab hum koi value approximate karna chahte hain (sin 0.1 e 0.2 atki hui limit evaluate karni hoti hai (0 0 arithmetic ban jaati hai.
Definition Taylor / Maclaurin seriesAgar f a f ( x ) = ∑ n = 0 ∞ n ! f ( n ) ( a ) ( x − a ) n . a = 0 Maclaurin series kehte hain.
WHY n ! f ( n ) ( a ) Ise derive karo. Maano f ( x ) = ∑ c n ( x − a ) n k x = a
f ( k ) ( x ) = ∑ n ≥ k c n ( n − k )! n ! ( x − a ) n − k .
Har term jisme n > k ( x − a ) x = a n = k f ( k ) ( a ) = c k k ! ⇒ c k = k ! f ( k ) ( a ) .
Coefficients ka yahi ek choice hai jo polynomial ko function ki value aur uske saare slopes/curvatures ko a WHY yahi hai.
Definition Taylor polynomial + remainder (Lagrange form)R_N(x)=\frac{f^{(N+1)}(c)}{(N+1)!}(x-a)^{N+1}$$
kisi $c$ ke liye jo $a$ aur $x$ ke beech hai. Remainder $R_N$ hi **error** hai.
WHY ek bound? Hum c ∣ f ( N + 1 ) ∣ ≤ M ∣ R N ( x ) ∣ ≤ ( N + 1 )! M ∣ x − a ∣ N + 1 .
sin ( 0.1 ) Step 1. sin x ≈ x − 6 x 3 Yeh step kyun? Series alternate karti hai aur fast shrink hoti hai; do terms mein hi error bahut chhoti ho jaati hai.
Step 2. 0.1 − 6 0.001 = 0.1 − 0.0001667 = 0.0998333 Kyun? Polynomial par sirf arithmetic hai.
Step 3. Error bound: agla term 120 x 5 = 120 1 0 − 5 ≈ 8 × 1 0 − 8 Kyun? Alternating series ke liye error ≤ 0.0998334
e 0.2 < 1 0 − 4 Step 1. P N ( 0.2 ) = ∑ n = 0 N n ! 0. 2 n Step 2. Bound: [ 0 , 0.2 ] ∣ f ( N + 1 ) ∣ = e c ≤ e 0.2 < 1.3 ∣ R N ∣ ≤ ( N + 1 )! 1.3 0. 2 N + 1 Kyun? M = 1.3 e c Step 3. N = 3 24 1.3 ( 0.2 ) 4 = 24 1.3 ( 0.0016 ) = 8.7 × 1 0 − 5 < 1 0 − 4 Step 4. 1 + 0.2 + 2 0.04 + 6 0.008 = 1 + 0.2 + 0.02 + 0.001333 = 1.221333 1.221403
Intuition WHY series kabhi kabhi L'Hôpital se behtar hoti hain0 0 sabse chhota power cancel karo. Tum immediately dekh sakte ho kaunsa term dominate kar raha hai — baar baar differentiate karne ki zaroorat nahi, koi jhanjhat nahi.
x → 0 lim x 3 sin x − x Step 1. sin x = x − 6 x 3 + 120 x 5 − ⋯ Kyun? Function ko 0 Step 2. sin x − x = − 6 x 3 + 120 x 5 − ⋯ Kyun? x 0 Step 3. x 3 − 6 1 + 120 x 2 − ⋯ Step 4. x → 0 x = − 6 1
x → 0 lim x 2 e x − 1 − x Step 1. e x = 1 + x + 2 x 2 + 6 x 3 + ⋯ Step 2. e x − 1 − x = 2 x 2 + 6 x 3 + ⋯ Step 3. ÷ x 2 = 2 1 + 6 x + ⋯ → 2 1 L'Hôpital se faster kyun? Yahan L'Hôpital ko 2 rounds chahiye; series ek substitution mein de deti hai.
Worked example Ek thoda mushkil:x → 0 lim x − tan x x − sin x Step 1. x − sin x = 6 x 3 − ⋯ tan x = x + 3 x 3 + ⋯ x − tan x = − 3 x 3 − ⋯ Step 2. Leading terms ka ratio: − x 3 /3 x 3 /6 = − 1/3 1/6 = − 2 1
Common mistake Classic errors ko steel-man karo(1) "Limit ke liye bas x = a Sahi lagta hai kyunki series f Kyun galat: poora point yahi hai ki lowest powers cancel hote hain — tumhe itne terms rakhne chahiye ki sabse chhota bachne wala power denominator ki degree ke barabar ho. Fix: denominator ki order tak kam se kam expand karo (ek safe term aur).
(2) "Do terms hamesha kaafi accurate hote hain." Sahi lagta hai chhote x Kyun galat: accuracy ∣ x − a ∣ N + 1 M a x Fix: hamesha remainder bound check karo.
(3) x = 2 ln ( 1 + x ) Sahi lagta hai — yeh toh formula hai. Kyun galat: yeh sirf ∣ x ∣ < 1 Fix: radius of convergence ka khayal rakho; rewrite ya recenter karo.
(4) x 5 /120 Fix: powers ko ek column mein line up karo; degree ke hisaab se cancel karo.
Recall Feynman: ek 12-saal ke bachche ko samjhaoEk wiggly function compute karna mushkil hai. Lekin Newton ne ek magic trick dhoondhi: ek point ke paas, function almost exactly ek simple polynomial ki tarah behave karta hai — pehle uski height se shuru karo, phir uske slope ka ek tukda jodo, phir ek aur chhhota tukda ki woh kaise jhukta hai, aur aise hi aage. Baad ke tukde bahut tezi se chhhote hote jaate hain (1 , 2 , 6 , 24 , … sin ( 0.1 ) 0 0
"Plug, Cancel, Crawl, Conclude"
Limits ke liye: P lug in series, C ancel lowest powers, C rawl x → 0 x C onclude. Approximation ke liye: "Value & slopes match karo, phir baaki ko bound karo."
Maclaurin coefficient c n c n = n ! f ( n ) ( 0 ) f n 0
Taylor polynomial P N R N ( x ) = ( N + 1 )! f ( N + 1 ) ( c ) ( x − a ) N + 1 c a x
Alternating series ke liye error bound kya hai? ≤
lim x → 0 x 3 s i n x − x − 6 1 sin x − x = − x 3 /6 + ⋯
lim x → 0 x 2 e x − 1 − x 2 1
Limit mein kaafi series terms kyun rakhne chahiye? Lowest powers cancel hote hain; tumhe woh term chahiye jo denominator ki degree se match kare taaki woh bach sake.
cos x 1 − 2 ! x 2 + 4 ! x 4 − ⋯
ln ( 1 + x ) − 1 < x ≤ 1
Agar ∣ f ( N + 1 ) ∣ ≤ M ∣ R N ( x ) ∣ ≤ ( N + 1 )! M ∣ x − a ∣ N + 1
Differentiate k times, set x=a
Error bound M over N+1! x^N+1
