Logo

AskSia

Plus

4. (a) Write the Maclaurin series (including the general term) for $f(x)=\ln (e+...
Jan 27, 2024
4. (a) Write the Maclaurin series (including the general term) for f(x)=ln⁡(e+x)f(x)=\ln (e+x). (b) What is the radius of convergence? (c) Use the first three terms of that series to write an expression that estimates the value of ∫01ln⁡(e+x2)dx\int_{0}^{1} \ln \left(e+x^{2}\right) d x
Solution by Steps
step 1
To write the Maclaurin series for f(x)=ln⁡(e+x) f(x) = \ln(e + x) , we use the general formula for the Maclaurin series expansion of a function f(x) f(x) around x=0 x = 0
step 2
The Maclaurin series for ln⁡(e+x) \ln(e + x) is given by ln⁡(e+x)=∑k=1∞(−1)k(−1+e+x)kk \ln(e + x) = \sum_{k=1}^{\infty} \frac{(-1)^k (-1 + e + x)^k}{k} for | -1 + e + x | < 1
step 3
Simplifying the series, we get ln⁡(e+x)=x−x22e+x33e2−x44e3+x55e4−… \ln(e + x) = x - \frac{x^2}{2e} + \frac{x^3}{3e^2} - \frac{x^4}{4e^3} + \frac{x^5}{5e^4} - \ldots . This is the series expansion including the general term
Answer
ln⁡(e+x)=x−x22e+x33e2−x44e3+x55e4−… \ln(e + x) = x - \frac{x^2}{2e} + \frac{x^3}{3e^2} - \frac{x^4}{4e^3} + \frac{x^5}{5e^4} - \ldots
Key Concept
Maclaurin Series Expansion
Explanation
The Maclaurin series is a Taylor series expansion of a function about 0. It is used to approximate functions using a sum of polynomial terms.
step 1
To find the radius of convergence for the series, we use the ratio test
step 2
The radius of convergence for the series ln⁡(e+x) \ln(e + x) is R=e R = e , since the series converges when |x| < e
Answer
The radius of convergence is e e .
Key Concept
Radius of Convergence
Explanation
The radius of convergence is the distance within which a power series converges. It is found using the ratio test or other convergence tests.
step 1
To estimate the integral ∫01ln⁡(e+x2)dx \int_{0}^{1} \ln(e + x^2) dx using the first three terms of the Maclaurin series, we substitute x2 x^2 for x x in the series
step 2
The first three terms of the series for ln⁡(e+x2) \ln(e + x^2) are x2−x42e+x63e2 x^2 - \frac{x^4}{2e} + \frac{x^6}{3e^2}
step 3
We integrate each term from 0 to 1 to estimate the integral: ∫01x2dx−∫01x42edx+∫01x63e2dx \int_{0}^{1} x^2 dx - \int_{0}^{1} \frac{x^4}{2e} dx + \int_{0}^{1} \frac{x^6}{3e^2} dx
step 4
The integrals are x33∣01−x510e∣01+x721e2∣01 \frac{x^3}{3} \Big|_0^1 - \frac{x^5}{10e} \Big|_0^1 + \frac{x^7}{21e^2} \Big|_0^1
step 5
Evaluating the integrals, we get 13−110e+121e2 \frac{1}{3} - \frac{1}{10e} + \frac{1}{21e^2}
Answer
13−110e+121e2 \frac{1}{3} - \frac{1}{10e} + \frac{1}{21e^2}
Key Concept
Estimating Integrals with Series
Explanation
To estimate the value of an integral using a series, we can integrate the series term by term within the interval of integration.
© 2023 AskSia.AI all rights reserved