Closed form expression for f(x) = sigma (n = 1 to infinity) for x^n / [n(n+1)]

In summary, the conversation discusses finding a closed-form expression for the power series sigma (n=1 to infinity) x^n / [n(n+1)] using the hint to let g(x) = x * f(x) and compute g''(x), then integrate it twice. The interval of convergence was calculated to be -1 <= x <= 1. The solution process involved finding g(x), g'(x), and g''(x), and recognizing that g''(x) is the power series for 1 / (1-x) at x = 0. Integrating g''(x) twice resulted in g'(x) = -ln(1-x). The conversation ends with a question about how to continue
  • #1
potentenum
1
0

Homework Statement



Consider the power series.

sigma (n=1 to infinity) x^n / [n(n+1)]

if f(x) = sigma x^n / [n(n+1)], then compute a closed-form expression for f(x).

It says: "Hint: let g(x) = x * f(x) and compute g''(x). Integrate this twice to get back to g(x) and hence derive f(x)".

Homework Equations



None? Well, see number 3.

The Attempt at a Solution



OK so i calculated the interval of convergence to be

-1 <= x <=1

and I'm following the hint so.

g(x) = x f( x)
and f(x) = x/2 + x²/6 + x^3 / 12 + x^4 / 20 ...

and x f(x) = x²/2 + x^3 /6 + x^4 / 12 ...

and g(x) = above.

so g'(x) = x + x²/2 + x^3 / 3 + x^4 / 4...

and g''(x) = 1 + x + x² + x^3 + x^4

and I recognized g''(x) to be the power series at x = 0 for

1 / (1 - x)

so i set g'' (x) = 1 / (1-x)

and then the hint said to integrate g''(x) twice, so

g'(x) = - ln (1-x).

now I'm stuck. How can I integrate -ln (1-x) again? or am i supposed to use what i know (i.e. f(1) = 1 and f(-1) = -1 - ln2?) (that is, i calculated the sum of f(x) if x = 1 and x = -1.

THANK YOU!
 
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  • #2
What is the derivative of x ln(x) - x?
 

1. What is a closed form expression?

A closed form expression is a mathematical expression that can be written in a finite number of mathematical operations, such as addition, subtraction, multiplication, division, and exponentiation. It does not involve infinite operations or functions, making it easier to evaluate and manipulate.

2. How is the closed form expression for f(x) = sigma (n = 1 to infinity) for x^n / [n(n+1)] derived?

The closed form expression for this function is derived using techniques from calculus, such as integration and differentiation. By manipulating the function and using mathematical identities, we can simplify it into a closed form expression that can be easily evaluated.

3. Can the closed form expression for f(x) = sigma (n = 1 to infinity) for x^n / [n(n+1)] be used for any value of x?

Yes, the closed form expression can be used for any value of x as long as it satisfies the conditions of convergence. In other words, the value of x must be such that the infinite series converges, which can be determined using convergence tests such as the ratio test or the root test.

4. How is the closed form expression for f(x) = sigma (n = 1 to infinity) for x^n / [n(n+1)] useful?

The closed form expression allows us to quickly and accurately calculate the sum of the infinite series for a given value of x. This can be useful in various applications, such as in physics, engineering, and finance, where infinite series are commonly used to model real-world phenomena.

5. Are there any limitations to using the closed form expression for f(x) = sigma (n = 1 to infinity) for x^n / [n(n+1)]?

Yes, there are limitations to using the closed form expression. It can only be used for a specific type of infinite series, known as a telescoping series, where the terms "cancel out" and the remaining terms can be simplified. It may not be applicable to other types of infinite series, such as geometric or harmonic series.

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