Power Series Problem: Solve f(3x) = 1/(1 - 3x)

If you want to determine the sum of an infinite series, you need to use convergence tests or, in this case, use the geometric series formula.
  • #1
JProgrammer
20
0
So here is the problem I am trying to solve:

You can combine two (or more) convergent power series on the same interval I. Using the properties of the geometric series, find the power series of the function below.

Series:
f(x) = 1/(1 - x) = sigma k = 0, infinity = 1+ x + x^2 + x^3

Function:
f(3x) = 1/(1 - 3x)

I plug in: expand sum (1/(1 - 3x) to order infinity and it returns with just understanding the 1/1 - 3x part and nothing else, producing a wrong answer.

If someone could tell me what I am doing wrong, I would appreciate it.

Thank you,
 
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  • #2
JProgrammer said:
. . . it returns . . .

Hi JProgrammer, what is "it"?

I believe the correct power series would be

$$1+3x+9x^2+27x^3+81x^4+\cdots$$

convergent for $|x|<\dfrac13$.
 
  • #3
Let u= 3x. Then [tex]f(3x)= f(u)= 1+ u+ u^2+ u^3+ \cdot\cdot\cdot= 1+ 3x+ (3x)^2+ (3x)^3+ \cdot\cdot\cdot= 1+ 3x+ 9x^2+ 27x^3+ \cdot\cdot\cdot[/tex].

"I plug in: expand sum (1/(1 - 3x) to order infinity and it returns with just understanding the 1/1 - 3x part and nothing else, producing a wrong answer."

"Plug in" to what? Don't assume that some program can replace thinking!
 

FAQ: Power Series Problem: Solve f(3x) = 1/(1 - 3x)

What is a power series?

A power series is an infinite series of the form ∑n=0^∞ cn(x-a)n, where cn are constants and a is a fixed number. It is a function that can be represented by an infinite polynomial.

How can I solve f(3x) = 1/(1-3x)?

To solve this power series problem, we can use the substitution method. Let u = 3x, then the equation becomes f(u) = 1/(1-u). We can then use the geometric series formula, ∑n=0^∞ rn = 1/(1-r), where r is the common ratio, to rewrite the equation as f(u) = ∑n=0^∞ (3x)n. This is now in the form of a power series, which we can then manipulate to solve for x.

What is the radius of convergence for this power series?

The radius of convergence for this power series is 1/3. This means that the series will converge when |3x| < 1, or when |x| < 1/3. Beyond this radius, the series will diverge and the power series representation will not be valid.

Can I use this power series to approximate values of f(3x) for values of x beyond the radius of convergence?

No, the power series representation is only valid within its radius of convergence. Beyond this radius, the series will diverge and will not be an accurate approximation of the function f(3x).

How can I determine the interval of convergence for this power series?

The interval of convergence for this power series is the set of all values of x for which the series converges. In this case, the interval of convergence is (-1/3, 1/3). We can determine this interval by using the ratio test, which involves taking the limit as n approaches infinity of |cn+1|/|cn| and setting it equal to the radius of convergence.

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