Math Problem Statement

f(x)=2xe^(e^(3x)^2), a=0; taylor series

Solution

To find the Taylor series of f(x)=2xee(3x)2f(x) = 2x e^{e^{(3x)^2}} around a=0a = 0, we need to compute the derivatives of f(x)f(x) at x=0x = 0 and use the Taylor series formula:

f(x)=f(0)+f(0)x+f(0)2!x2+f(0)3!x3+f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots

Let's proceed by evaluating the function and its derivatives at x=0x = 0.

Step 1: Evaluate f(0)f(0)

f(0)=20ee(30)2=0f(0) = 2 \cdot 0 \cdot e^{e^{(3 \cdot 0)^2}} = 0

Since f(0)=0f(0) = 0, the constant term in the Taylor series is zero.

Step 2: Compute the First Derivative, f(x)f'(x)

Using the product rule for derivatives: f(x)=2xee9x2f(x) = 2x \cdot e^{e^{9x^2}} f(x)=2ee9x2+2xddx(ee9x2)f'(x) = 2 \cdot e^{e^{9x^2}} + 2x \cdot \frac{d}{dx}\left(e^{e^{9x^2}}\right)

To differentiate ee9x2e^{e^{9x^2}} with respect to xx, we use the chain rule: ddx(ee9x2)=ee9x2ddx(e9x2)=ee9x2e9x2(18x)\frac{d}{dx}\left(e^{e^{9x^2}}\right) = e^{e^{9x^2}} \cdot \frac{d}{dx}(e^{9x^2}) = e^{e^{9x^2}} \cdot e^{9x^2} \cdot (18x)

Therefore, f(x)=2ee9x2+2xee9x2e9x218x=2ee9x2+36x2ee9x2e9x2f'(x) = 2 e^{e^{9x^2}} + 2x \cdot e^{e^{9x^2}} \cdot e^{9x^2} \cdot 18x = 2 e^{e^{9x^2}} + 36x^2 e^{e^{9x^2}} e^{9x^2}

Now, evaluate f(0)f'(0): f(0)=2ee0=2ef'(0) = 2 e^{e^0} = 2e

So, the coefficient of xx in the Taylor series is 2e2e.

Step 3: Compute the Second Derivative, f(x)f''(x)

This calculation involves differentiating f(x)f'(x) and becomes quite complex. However, for practical purposes, you might only need the first few terms.

After computing higher derivatives, you

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Math Problem Analysis

Mathematical Concepts

Taylor Series
Derivatives
Exponential Functions
Product Rule
Chain Rule

Formulas

Taylor series formula: f(x) = f(0) + f'(0)x + f''(0)/2!x^2 + ...
Product Rule: (fg)' = f'g + fg'
Chain Rule: d/dx(e^g(x)) = e^g(x) * g'(x)

Theorems

Taylor Series Expansion

Suitable Grade Level

Undergraduate (Calculus)