LOGARITHM

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IF y=x+x²/2+x³/3+x⁴/4+....... THEN PROVE THAT x=y-y²/2!+y³/3!+y⁴/4!+.......

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kumar krishna agarwal ·2012-06-14 23:24:46

here it should be mentioned that mod(x)<=1

then using series expansion of ln(1-x) we have
-ln(1-x) = x+x2/2+x3/3+x4/4+.......=y
hence,
ln(1-x)=-y
(1-x)=e^(-y)
x= 1-e^-y
using series expansion for e^-y we get the desired
result.

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Shambo Mohanty THANK YOU K.K.A. IT IS THE RIGHT ANSWER
Upvote·0· Reply ·2012-06-15 02:46:42
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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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LOGARITHM

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