Expressing series in terms of a Power Series

In summary, this individual is struggling to express a function in terms of another function. They are trying to express f(x) as a sum of two terms, one for each index j. They have found that the sum becomes 1 when j = 2. They need to integrate these series to find the relationship between the two functions.
  • #1
pkmpad
7
1
Hello and thank you for trying to help.

In spite of the fact that this seems a very simple problem, I do not find myself able to get a solution. Here it goes:

Let $$f(x)=\displaystyle \sum_{k=3}^\infty a_k \frac{x^k}{k(k-1)(k-2)}$$ and $$g(x)=\displaystyle \sum_{k=0}^\infty a_k x^k$$. Express f(x) in terms of g(x).

My attempt has been to express f(x) as
$$\displaystyle\sum_{k=3}^{\infty}a_k\frac{x^k}{k(k-1)(k-2)}=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k}-\sum_{k=3}^{\infty}\frac{a_k x^k}{k-1}+\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k-2}$$

And then

$$f(x)=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k}-x\sum_{k=2}^{\infty}\frac{a_k x^k}{k}+\frac{x^2}{2}\sum_{k=1}^{\infty}\frac{a_k x^k}{k}$$

What should I do next? I suppose that I have to integrate those series, but I have already tried that and I have not been able to get any relationship with g(x).

Any help will be appreciated.
 
Last edited:
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  • #2
Your LaTex is not displaying correctly. I have tried 2 different browsers. It looks like you have f(x) = Sum of (k=3 to infinity) (ak xk) / (k(k-1)(k-2)),
then g(x) = sum of (k=0 to infinity) (ak xk), then I get confused trying to follow your workings, after that.
 
  • #3
scottdave said:
Your LaTex is not displaying correctly. I have tried 2 different browsers. It looks like you have f(x) = Sum of (k=3 to infinity) (ak xk) / (k(k-1)(k-2)),
then g(x) = sum of (k=0 to infinity) (ak xk), then I get confused trying to follow your workings, after that.

Corrected, thank you
 
  • Like
Likes scottdave
  • #4
pkmpad said:
Hello and thank you for trying to help.

In spite of the fact that this seems a very simple problem, I do not find myself able to get a solution. Here it goes:

Let $$f(x)=\displaystyle \sum_{k=3}^\infty a_k \frac{x^k}{k(k-1)(k-2)}$$ and $$g(x)=\displaystyle \sum_{k=0}^\infty a_k x^k$$. Express f(x) in terms of g(x).

My attempt has been to express f(x) as
$$\displaystyle\sum_{k=3}^{\infty}a_k\frac{x^k}{k(k-1)(k-2)}=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k}-\sum_{k=3}^{\infty}\frac{a_k x^k}{k-1}+\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k-2}$$

And then

$$f(x)=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k}-x\sum_{k=2}^{\infty}\frac{a_k x^k}{k}+\frac{x^2}{2}\sum_{k=1}^{\infty}\frac{a_k x^k}{k}$$

What should I do next? I suppose that I have to integrate those series, but I have already tried that and I have not been able to get any relationship with g(x).

Any help will be appreciated.

Using summation index ##j = k-2## your sum becomes
$$f(x) = \sum_{j=1}^{\infty} a_{j+2} \frac{x^{j+2}}{j(j+1)(j+2)}$$
 

1. What is a power series?

A power series is a representation of a function as an infinite sum of terms, where each term is a constant multiplied by a variable raised to a non-negative integer power. It is a way of expressing a function as a polynomial of infinite degree.

2. Why is it useful to express a series in terms of a power series?

Expressing a series in terms of a power series allows us to approximate a function with a polynomial, making it easier to evaluate and manipulate. It also allows us to extend the domain of a function and to find the behavior of a function near a specific point.

3. How do you determine the coefficients of a power series?

The coefficients of a power series can be determined using the Taylor series expansion, which uses derivatives of the function at a specific point. Alternatively, there are formulas for finding the coefficients of well-known functions such as exponential, trigonometric, and logarithmic functions.

4. Can all functions be expressed as a power series?

No, not all functions can be expressed as a power series. The function must be infinitely differentiable at the point where the power series is centered, and the power series must converge within a specific radius of convergence. Some functions, such as step functions and discontinuous functions, cannot be expressed as a power series.

5. How do you determine the radius of convergence for a power series?

The radius of convergence for a power series can be determined using the ratio test or the root test, which involve taking the limit as n approaches infinity of the absolute value of the ratio between consecutive terms or the nth root of the absolute value of each term, respectively. The radius of convergence is the value of x for which the series converges.

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