Two Power Series questions that need to be solved ly

In summary, the conversation discussed two power series questions that needed to be solved urgently. The first question involved finding coefficients c2 through c6 of the power series representation for the function f(x) = 2x (ln(1+x)). The correct power series was determined to be 2x^2 + 2x^3 - 2x^4 + 2x^5 - 2x^6 + ... with coefficients of 0, 0, 2, 2, -2, 2, -2, ... The second question involved finding a partial sum for the power series representation of f(x) = 1/(1+9x^2) and determining the radius of convergence. The
  • #1
reichiru
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Two Power Series questions that need to be solved urgently

Homework Statement


Question 1: The function f(x) = 2x (ln(1+x)) is represented as a power series. Find coefficients c2 through c6 of the power series.
Question 2: Write a partial sum for the power series which represents the function f(x) = 1/(1+(3^2)(x^2)) consisting of the first five non-zero terms. Also find the radius of convergence.

Homework Equations


The Attempt at a Solution


Question 1: Okay so the power series I came up with is (2(-1)n)/n x (xn), using the fact that ln(1+x) = 1/1+x = 1/1-(-x) and finding the partial sum 1-x+x^2-x^3... etc. So the coefficients I got were 1, 2/3, 1/2, 2/5, and 1/3. Unfortunately, the online thing says it's wrong, and I have no idea why...

Question 2: This one completely stumped me, the best I could come up with is to somehow take out the 3^2 so that you're left with 1+x^2 in the denominator, which is much easier to calculate a partial sum out of.

EDIT: Question 2 solved... just Question 1 now...
EDIT2: Never mind now, solved Question 1 as well.
 
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  • #2


Hello there,

For Question 1, it seems like you have the right idea but your power series is not quite accurate. The correct power series representation for f(x) = 2x (ln(1+x)) is 2x^2 + 2x^3 - 2x^4 + 2x^5 - 2x^6 + ... This can be obtained by using the power series representation for ln(1+x) which is x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - x^6/6 + ... and then multiplying it by 2x. So the coefficients you should get are 0, 0, 2, 2, -2, 2, -2, ...

For Question 2, you can first write f(x) as 1/(1+9x^2) and then use the geometric series formula to find the power series representation. The first five non-zero terms are 1 - 9x^2 + 81x^4 - 729x^6 + 6561x^8. The radius of convergence can be found by using the ratio test, which in this case will give you a radius of convergence of 1/3.

Hope this helps! Let me know if you have any further questions. Good luck!
 

FAQ: Two Power Series questions that need to be solved ly

What is a power series and how is it different from a regular series?

A power series is a series where each term is a constant multiplied by a variable raised to a power. This is different from a regular series where each term is a constant multiplied by an integer. Power series are often used in calculus and mathematical analysis to represent functions as an infinite sum of terms.

How do you determine if a power series converges or diverges?

There are several tests that can be used to determine the convergence or divergence of a power series, such as the ratio test and the root test. These tests involve taking the limit of the ratio or root of consecutive terms in the series. If the limit is less than 1, the series converges; if it is greater than 1, the series diverges; and if it is equal to 1, the test is inconclusive.

Can you use a power series to approximate a function?

Yes, power series can be used to approximate functions within a certain radius of convergence. This is known as a Taylor series, where the function is represented as an infinite sum of terms using the derivatives of the function at a specific point. The more terms that are included in the series, the closer the approximation will be to the actual function.

How do you find the radius of convergence for a power series?

The radius of convergence for a power series can be found by using the ratio test or the root test. These tests will give a value for the radius of convergence, which represents the distance from the center point where the series will converge. It is important to note that the series may also converge at the endpoints of the interval, so these should also be checked separately.

Can you use a power series to solve differential equations?

Yes, power series can be used to solve certain types of differential equations. This is known as the method of Frobenius, where a power series is substituted into the differential equation and solved for the coefficients. This method is particularly useful for solving differential equations with singular points or when other methods, such as separation of variables, are not applicable.

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