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Math  /  Calculus

QuestionAssignment 4: Problem 13 (1 point)
Find the Taylor polynomial with 4 nonzero terms about x=0x=0 for the function f(x)=∫ex3dx,f(0)=0f(x)=\int e^{x^{3}} d x, f(0)=0.

Studdy Solution

STEP 1

1. We are given the function f(x)=∫ex3 dx f(x) = \int e^{x^3} \, dx .
2. We need to find the Taylor polynomial with 4 nonzero terms about x=0 x = 0 .
3. The Taylor polynomial is centered at x=0 x = 0 , which is also known as the Maclaurin series.
4. We are given that f(0)=0 f(0) = 0 .

STEP 2

1. Find the Taylor series expansion of ex3 e^{x^3} about x=0 x = 0 .
2. Integrate the series term-by-term to find the Taylor series for f(x) f(x) .
3. Identify the first four nonzero terms of the Taylor series for f(x) f(x) .
4. Write the Taylor polynomial using these terms.

STEP 3

Start with the Taylor series expansion for eu e^u at u=0 u = 0 :
eu=1+u+u22!+u33!+⋯ e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots
Substitute u=x3 u = x^3 to find the series for ex3 e^{x^3} :
ex3=1+x3+(x3)22!+(x3)33!+⋯ e^{x^3} = 1 + x^3 + \frac{(x^3)^2}{2!} + \frac{(x^3)^3}{3!} + \cdots
ex3=1+x3+x62+x96+⋯ e^{x^3} = 1 + x^3 + \frac{x^6}{2} + \frac{x^9}{6} + \cdots

STEP 4

Integrate the series term-by-term to find f(x)=∫ex3 dx f(x) = \int e^{x^3} \, dx :
f(x)=∫(1+x3+x62+x96+⋯ ) dx f(x) = \int \left( 1 + x^3 + \frac{x^6}{2} + \frac{x^9}{6} + \cdots \right) \, dx
f(x)=x+x44+x714+x1060+⋯ f(x) = x + \frac{x^4}{4} + \frac{x^7}{14} + \frac{x^{10}}{60} + \cdots

STEP 5

Identify the first four nonzero terms of the Taylor series for f(x) f(x) :
1. x x
2. x44 \frac{x^4}{4}
3. x714 \frac{x^7}{14}
4. x1060 \frac{x^{10}}{60}

STEP 6

Write the Taylor polynomial using these terms:
The Taylor polynomial with 4 nonzero terms is:
P(x)=x+x44+x714+x1060 P(x) = x + \frac{x^4}{4} + \frac{x^7}{14} + \frac{x^{10}}{60}
The Taylor polynomial with 4 nonzero terms for f(x) f(x) about x=0 x = 0 is:
x+x44+x714+x1060 \boxed{x + \frac{x^4}{4} + \frac{x^7}{14} + \frac{x^{10}}{60}}

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