Math Problem Statement

for the numerator: ln ⁑ ( 1 + 4 π‘₯ ) β‰ˆ 4 π‘₯ βˆ’ ( 4 π‘₯ ) 2 2 + 𝑂 ( π‘₯ 3 )

4 π‘₯ βˆ’ 8 π‘₯ 2 + 𝑂 ( π‘₯ 3 ) ln(1+4x)β‰ˆ4xβˆ’ 2 (4x) 2

​ +O(x 3 )=4xβˆ’8x 2 +O(x 3 ) 𝑒 2 π‘₯ βˆ’ 1 β‰ˆ 2 π‘₯ + ( 2 π‘₯ ) 2 2 + 𝑂 ( π‘₯ 3 )

2 π‘₯ + 2 π‘₯ 2 + 𝑂 ( π‘₯ 3 ) e 2x βˆ’1β‰ˆ2x+ 2 (2x) 2

​ +O(x 3 )=2x+2x 2 +O(x 3 ) ( 1 + 2 π‘₯ ) 3 βˆ’ 1

1 + 6 π‘₯ + 12 π‘₯ 2 + 8 π‘₯ 3 βˆ’ 1

6 π‘₯ + 12 π‘₯ 2 + 8 π‘₯ 3 (1+2x) 3 βˆ’1=1+6x+12x 2 +8x 3 βˆ’1=6x+12x 2 +8x 3

Solution

The expressions provided are approximations using Taylor series expansions. Let's break down each approximation:

1. ln⁑(1+4x)\ln(1 + 4x)

Using the Taylor series expansion for ln⁑(1+y)\ln(1 + y) around y=0y = 0, we get: ln⁑(1+y)β‰ˆyβˆ’y22+O(y3)\ln(1 + y) \approx y - \frac{y^2}{2} + O(y^3)

Substituting y=4xy = 4x: ln⁑(1+4x)β‰ˆ4xβˆ’(4x)22+O((4x)3)\ln(1 + 4x) \approx 4x - \frac{(4x)^2}{2} + O((4x)^3) ln⁑(1+4x)β‰ˆ4xβˆ’8x2+O(x3)\ln(1 + 4x) \approx 4x - 8x^2 + O(x^3)

2. e2xβˆ’1e^{2x} - 1

Using the Taylor series expansion for eye^y around y=0y = 0, 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