Use a power series to approximate the value of the integral with an error of less than 0.0001 1 1 e-x8 dx 2 Σ n = 0
Use a power series to approximate the value of the integral with an error of less than 0.0001 1 1 e-x8 dx 2 Σ n = 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Using Power Series for Integral Approximation**
In this example, we aim to approximate the value of the given integral using a power series, ensuring that the error is less than 0.0001. The integral is defined as follows:
\[
\int_{0}^{1} e^{-x^8} \, dx
\]
### Step-by-Step Solution:
Given:
\[
e^{-x^8} = \sum_{n=0}^{\infty} \frac{(-x^8)^n}{n!}
\]
Using this power series expansion and term-by-term integration, the integral becomes:
\[
\int_{0}^{1} e^{-x^8} \, dx \approx \sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-x^8)^n}{n!}\, dx
\]
### Diagram Explanation:
Unfortunately, in the provided image, there are certain boxes indicating missing steps or results, which will need to be filled in as follows:
1. **Power Series Expansion**:
\[
e^{-x^8} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{8n}}{n!}
\]
2. **Integral Transformation Using Power Series**:
\[
\int_{0}^{1} e^{-x^8} \, dx \approx \sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-1)^n x^{8n}}{n!}\, dx
\]
The term-by-term integration gives:
\[
\int_{0}^{1} \frac{(-1)^n x^{8n}}{n!} \, dx = \frac{(-1)^n}{n!} \int_{0}^{1} x^{8n} \, dx
\]
3. **Integration of Each Term**:
\[
\int_{0}^{1} x^{8n} \, dx = \frac{1}{8n+1}
\]
Hence, the integral can be approximated by summing the series:
\[
\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(8n+1)}
\]
To ensure the error is within the desired tolerance of 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F795f20db-de84-4e64-b0fe-ecefbfde6505%2Fc64fe2c9-ceff-4ea3-a41e-ec398a028e25%2Fhr3hiai.jpeg&w=3840&q=75)
Transcribed Image Text:**Using Power Series for Integral Approximation**
In this example, we aim to approximate the value of the given integral using a power series, ensuring that the error is less than 0.0001. The integral is defined as follows:
\[
\int_{0}^{1} e^{-x^8} \, dx
\]
### Step-by-Step Solution:
Given:
\[
e^{-x^8} = \sum_{n=0}^{\infty} \frac{(-x^8)^n}{n!}
\]
Using this power series expansion and term-by-term integration, the integral becomes:
\[
\int_{0}^{1} e^{-x^8} \, dx \approx \sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-x^8)^n}{n!}\, dx
\]
### Diagram Explanation:
Unfortunately, in the provided image, there are certain boxes indicating missing steps or results, which will need to be filled in as follows:
1. **Power Series Expansion**:
\[
e^{-x^8} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{8n}}{n!}
\]
2. **Integral Transformation Using Power Series**:
\[
\int_{0}^{1} e^{-x^8} \, dx \approx \sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-1)^n x^{8n}}{n!}\, dx
\]
The term-by-term integration gives:
\[
\int_{0}^{1} \frac{(-1)^n x^{8n}}{n!} \, dx = \frac{(-1)^n}{n!} \int_{0}^{1} x^{8n} \, dx
\]
3. **Integration of Each Term**:
\[
\int_{0}^{1} x^{8n} \, dx = \frac{1}{8n+1}
\]
Hence, the integral can be approximated by summing the series:
\[
\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(8n+1)}
\]
To ensure the error is within the desired tolerance of 0.
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