# Conversion of a function to a power series

• MathewsMD
In summary, the conversation discusses the representation of the function f(x) = 1/(3-x) as a power series and the different ways to do so. The speaker suggests using the expression for 1/(1-r) = ∑rn and letting r = x - 2 to represent the function as a geometric series. However, there is some confusion over the algebra involved and it is suggested to expand the series around x=0. The conversation also touches on the idea that there are infinite ways to represent the same function as a power series.
MathewsMD
Given the function f(x) = 1/(3-x) it can be represented in a power series by 1/3∑(x/3)n but is there any restriction on saying ∑(x-2)n except for x = 2? In the first case, R = 3 but the second case, R = 1 and on different intervals (i.e. (-3,3) and the other is (1,3). I just simply used the expression for 1/(1-r) = ∑rn and let r = x - 2.

So, any problems? I understand it's a less efficient method to solve this problem but I guess I'm just trying to see different ways to represent the same function.

It's hard to understand what you're asking.

! said:
It's hard to understand what you're asking.

Does f(x) = 1/(3-x) = (1/3)∑(x/3)n for (-3,3)?

MathewsMD said:
Does f(x) = 1/(3-x) = (1/3)∑(x/3)n for (-3,3)?

If the sum is taken for all nonnegative n, then yes. For the infinite series converges to a/(1 - r) where a is the first term and r is the common ratio with |r| < 1.

MathewsMD, apparently I wasn't the only one who was confused by your question. How is this question related to what you asked about ##\sum (x - 2)^n##? For fixed x, that series is also a geometric series that converges if |x - 2| < 0; i.e., if 1 < x < 3. And x certainly can equal 2 here.

Mark44 said:
MathewsMD, apparently I wasn't the only one who was confused by your question. How is this question related to what you asked about ##\sum (x - 2)^n##? For fixed x, that series is also a geometric series that converges if |x - 2| < 0; i.e., if 1 < x < 3. And x certainly can equal 2 here.

Hi, sorry for the ambiguity. I was just in a rush yesterday and will try to clear things up now.

So, I want to express the given function, ## f(x) = \frac {1}{x-2} ## as a power series.

Now, ## \frac {a}{1-r} = ∑_{n=0}^∞ar^n ; -1 < r < 1##

There is one simple method for solving this question, and that is simply dividing f(x) by 2 and then just rearranging. If you need more explanation for this part, just let me know. What I want to do is represent the function in a different way.

Representing the function as ## f(x) = \frac {1}{1 - (x+3)} ; r = x + 3; a = 1, f(x) = ∑_{n=0}^∞ (x+3)^n ; -3 < x < 3##

Now sorry if I my response is still a little convoluted or if my derivation seems trivial and useless. I would just like to know if this is incorrect for any reason. I haven't seen a power series changed in a similar form before, it's always just been dividing out the constant (e.g. 2 in this case) to put the denominator of f(x) in the form of ##1 - ax##. I know this is a much more useful method that will help with more advanced problems, but my main concern is just to understand the concept and figuring out different ways to represent these functions and to see if my understanding is correct. If I'm not mistaken, there are infinite ways to represent the same function, and I am just wondering if mine is correct. Thanks!

MathewsMD said:
Hi, sorry for the ambiguity. I was just in a rush yesterday and will try to clear things up now.

So, I want to express the given function, ## f(x) = \frac {1}{x-2} ## as a power series.

Now, ## \frac {a}{1-r} = ∑_{n=0}^∞ar^n ; -1 < r < 1##

There is one simple method for solving this question, and that is simply dividing f(x) by 2 and then just rearranging. If you need more explanation for this part, just let me know. What I want to do is represent the function in a different way.

Representing the function as ## f(x) = \frac {1}{1 - (x+3)} ; r = x + 3; a = 1, f(x) = ∑_{n=0}^∞ (x+3)^n ; -3 < x < 3##

Now sorry if I my response is still a little convoluted or if my derivation seems trivial and useless. I would just like to know if this is incorrect for any reason. I haven't seen a power series changed in a similar form before, it's always just been dividing out the constant (e.g. 2 in this case) to put the denominator of f(x) in the form of ##1 - ax##. I know this is a much more useful method that will help with more advanced problems, but my main concern is just to understand the concept and figuring out different ways to represent these functions and to see if my understanding is correct. If I'm not mistaken, there are infinite ways to represent the same function, and I am just wondering if mine is correct. Thanks!

##\frac {1}{1 - (x+3)}## isn't equal to ##\frac {1}{x-2}##. There are an infinite number of ways to represent the same function as a geometric series. You can expand around any x value. If they just say "power series" they probably want you to expand it around x=0, i.e. as a series in powers of x. To do that write ##\frac {1}{x-2}=\frac{-1/2}{1-(x/2)}##.

Dick said:
##\frac {1}{1 - (x+3)}## isn't equal to ##\frac {1}{x-2}##. There are an infinite number of ways to represent the same function as a geometric series. You can expand around any x value. If they just say "power series" they probably want you to expand it around x=0, i.e. as a series in powers of x. To do that write ##\frac {1}{x-2}=\frac{-1/2}{1-(x/2)}##.

Sorry, I meant ## f(x) = \frac {1}{2-x} ##

MathewsMD said:
Sorry, I meant ## f(x) = \frac {1}{2-x} ##

##\frac {1}{1 - (x+3)}## isn't equal to ## \frac {1}{2-x} ## either. You've got to start getting the algebra right.

Dick said:
##\frac {1}{1 - (x+3)}## isn't equal to ## \frac {1}{2-x} ## either. You've got to start getting the algebra right.

Representing the function as ## f(x) = \frac {1}{1 - (x - 1)} ; r = x - 1; a = 1, f(x) = ∑_{n=0}^∞ (x-1)^n ; 0 < x < 2##

Third time's the charm. My mistake :)

Anything wrong with this? And yes, I understand what you did and it definitely makes sense solving with that method as well (it's more applicable to more advanced problems). I was just wondering if there's anything mathematically incorrect in my posted work?

MathewsMD said:
Given the function f(x) = 1/(3-x) ...

MathewsMD said:
Hi, sorry for the ambiguity. I was just in a rush yesterday and will try to clear things up now.

So, I want to express the given function, ## f(x) = \frac {1}{x-2} ## as a power series.

MathewsMD said:
Representing the function as ## f(x) = \frac {1}{1 - (x - 1)} ; r = x - 1; a = 1, f(x) = ∑_{n=0}^∞ (x-1)^n ; 0 < x < 2##

Third time's the charm. My mistake :)

Anything wrong with this? And yes, I understand what you did and it definitely makes sense solving with that method as well (it's more applicable to more advanced problems). I was just wondering if there's anything mathematically incorrect in my posted work?
I'm really confused.

You started of with the "given function" as f(x) = 1/(x - 2).

In a later post, you said the "given function" was f(x) = 1/(3 - x).

In your last post, you have f(x) = 1/(1 - (x - 1)), which is not equal to either of the functions above. 1/(1 - (x - 1)) = 1/(2 - x).

So, what exactly is the function you're working with?

1.

## What is the purpose of converting a function to a power series?

The purpose of converting a function to a power series is to approximate the function using a polynomial expression. This allows for easier calculation and evaluation of the function at various points.

2.

## How is a function converted to a power series?

A function can be converted to a power series by using the Taylor series expansion, which involves finding the derivatives of the function at a specific point and plugging them into a formula. The resulting power series will have coefficients that represent the values of the derivatives of the function at that point.

3.

## What are the benefits of using a power series to represent a function?

One benefit is that a power series can be used to approximate a function at any point within its interval of convergence. It also allows for easier calculation of the function and its derivatives at various points. Additionally, a power series can be used to approximate complex functions that may not have a simple algebraic expression.

4.

## What is the interval of convergence in a power series?

The interval of convergence in a power series is the range of values for which the series will converge and accurately represent the original function. This interval can be determined by using techniques such as the ratio test or the root test.

5.

## Can any function be represented by a power series?

No, not all functions can be represented by a power series. The function must be analytic, meaning it can be expressed as an infinite sum of powers of x, and it must have a finite interval of convergence. Functions with discontinuities, such as step functions, cannot be represented by a power series.

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