Math Problem Statement
for the numerator: ln β‘ ( 1 + 4 π₯ ) β 4 π₯ β ( 4 π₯ ) 2 2 + π ( π₯ 3 )
4 π₯ β 8 π₯ 2 + π ( π₯ 3 ) ln(1+4x)β4xβ 2 (4x) 2
β +O(x 3 )=4xβ8x 2 +O(x 3 ) π 2 π₯ β 1 β 2 π₯ + ( 2 π₯ ) 2 2 + π ( π₯ 3 )
2 π₯ + 2 π₯ 2 + π ( π₯ 3 ) e 2x β1β2x+ 2 (2x) 2
β +O(x 3 )=2x+2x 2 +O(x 3 ) ( 1 + 2 π₯ ) 3 β 1
1 + 6 π₯ + 12 π₯ 2 + 8 π₯ 3 β 1
6 π₯ + 12 π₯ 2 + 8 π₯ 3 (1+2x) 3 β1=1+6x+12x 2 +8x 3 β1=6x+12x 2 +8x 3
Solution
The expressions provided are approximations using Taylor series expansions. Let's break down each approximation:
1.
Using the Taylor series expansion for around , we get:
Substituting :
2.
Using the Taylor series expansion for around , we get: [ e^y \approx 1 +
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Math Problem Analysis
Mathematical Concepts
Taylor Series
Approximations
Formulas
Taylor series expansion
Theorems
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Suitable Grade Level
Advanced High School