1. ## Power series expansion

"By adapting the power series expansion (centred on c = 0) of $\displaystyle h(t) = \frac{1}{1-t}$ , write down a power series expansion (with the indicated centre) for each of the following functions, and indicate an interval over which each series converges.

i. $\displaystyle f(x) = \frac{4}{x^2 - 2x}$ , center c = 1

ii. $\displaystyle g(x) = \frac{2}{(1-x)^3}$ , center c = 0"

I'm not comprehending this question...I think its a mental block. Can anyone rephrase (using different words preferrably in lamen's terms, please) this question? also, the power series expansion is also known as the taylor series expansion?

2. A Power Series is very similar to a Taylor Series, but they're not exactly the same thing. A power series is of the form
Whereas a Taylor Series is of the form
So, you can see that a Taylor Series is a Power Series, but a Power Series is not necessarily a Taylor Series.

Okay, now, it's been a while since I've dealt with a Power Series expansion, but I believe what you're supposed to be doing here is taking your initial series, 1/(1-t), and modifying this series to get the new, listed series.

So, in order to do this, we need to list the steps necessary to change 1/(1-t) into the form 4/(x^2 -2x). List the steps required to change the one series to the other, and then perform these operations on the entire summation. This will be your new Power Series expansion.

3. Hi Math Major,

What you've said makes complete sense to me... finally, maths that I can understand

Ok, so, I've tried to do it to the second function (i.e. g(x)), and I found that it was the second derivative of h(t)... But the first function (i.e. f(x)), its a little difficult, I can see that 2x - x^2 = 1 - (1-x)^2, which might lead me somewhere, but I can't find how its manipulated with regards to h(t)... Can you help here?

Thank you

tsal15

4. Originally Posted by tsal15
Hi Math Major,

What you've said makes complete sense to me... finally, maths that I can understand

Ok, so, I've tried to do it to the second function (i.e. g(x)), and I found that it was the second derivative of h(t)... But the first function (i.e. f(x)), its a little difficult, I can see that 2x - x^2 = 1 - (1-x)^2, which might lead me somewhere, but I can't find how its manipulated with regards to h(t)... Can you help here?

Thank you

tsal15
First, we need to write down the power series expansion for 1/(1-t)

This series is centered at 0, so, for simplicity, we'll begin with the series that is also centered at 0: 2/(1-x)^3

Compare 1/(1-x) to 2/(1-x)^3 Right off the bat, we can see that these two are very similar.

Step 1: Multiply 1/(1-x) by 2.
Result: 2/(1-x)
Step 2: Multiply by 1/(1-x)^2
Result: 2/(1-x)^3

The two equations are now the same. Apply these steps to the above listed series. Multiply the series by 2 and then by 1/(1-x)^2. This will give you a power series expansion for the equation 2/(1-x)^3

The first one is solved in the same method. Alternatively, you could solve by taking the definition of a power series. However, they asked for you to solve it in the above manner, so I suggest doing it that way.

5. Originally Posted by Math Major
First, we need to write down the power series expansion for 1/(1-t)

This series is centered at 0, so, for simplicity, we'll begin with the series that is also centered at 0: 2/(1-x)^3

Compare 1/(1-x) to 2/(1-x)^3 Right off the bat, we can see that these two are very similar.

Step 1: Multiply 1/(1-x) by 2.
Result: 2/(1-x)
Step 2: Multiply by 1/(1-x)^2
Result: 2/(1-x)^3

The two equations are now the same. Apply these steps to the above listed series. Multiply the series by 2 and then by 1/(1-x)^2. This will give you a power series expansion for the equation 2/(1-x)^3

The first one is solved in the same method. Alternatively, you could solve by taking the definition of a power series. However, they asked for you to solve it in the above manner, so I suggest doing it that way.
This is incorrect. Multiplying the power series for $\displaystyle \frac{1}{1-x}$ by $\displaystyle \frac{1}{(1-x)^2}$ does not give the Maclaurin series for $\displaystyle \frac{1}{(1-x)^3}$...it gives a series..but not the correct one. There are two methods....the first I have no idea why anyone would do but knowing that $\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ then $\displaystyle \frac{1}{(1-x)^3}=\left(\sum_{n=0}^{\infty}x^n\right)^3$. Now if you assume that this series converges absolutely (find where it does) then you may use series multiplication. That method is horrible though, instead consider that $\displaystyle \frac{2}{(1-x)^3}=\left(\frac{1}{(1-x)}\right)''$

6. Originally Posted by Mathstud28
This is incorrect. Multiplying the power series for $\displaystyle \frac{1}{1-x}$ by $\displaystyle \frac{1}{(1-x)^2}$ does not give the Maclaurin series for $\displaystyle \frac{1}{(1-x)^3}$...it gives a series..but not the correct one. There are two methods....the first I have no idea why anyone would do but knowing that $\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ then $\displaystyle \frac{1}{(1-x)^3}=\left(\sum_{n=0}^{\infty}x^n\right)^3$. Now if you assume that this series converges absolutely (find where it does) then you may use series multiplication. That method is horrible though, instead consider that $\displaystyle \frac{2}{(1-x)^3}=\left(\frac{1}{(1-x)}\right)''$

so who is right and who is wrong... and what should I be doing?

7. Originally Posted by tsal15
so who is right and who is wrong... and what should I be doing?
If I may be so bold to interject. Math Major is wrong and Mathstud right. Please let me explain.

Starting with the power series

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$

then multiplying by
$\displaystyle \frac{2}{(1-x)^2}$
gives
$\displaystyle \frac{2}{(1-x)^3} = \frac{2}{(1-x)^2} \frac{1}{1-x} = \frac{2}{(1-x)^2} \left( 1 + x + x^2 + x^3 + \cdots \right)$

$\displaystyle = \frac{2}{(1-x)^2} + \frac{2x}{(1-x)^2} + \frac{2x^2}{(1-x)^2} + \frac{2x^3}{(1-x)^2} + \cdots$

which is a series but not a power series! I think the idea here (and MathStud said) is to using the standard series

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$

and manipulate this to get your series. Multiply by numbers or a power of x. Adding and substracting and possibly substituting. Allow me to illustrate with the first example. First, we decompose the function by noting that

$\displaystyle \frac{4}{x^2-2x} = \frac{-4}{x(2 - x)} =- 2\left(\frac{1}{x} + \frac{1}{2-x} \right)$

Now we will come up with a power series for each. Note the following:

$\displaystyle \frac{1}{x} = \frac{1}{ (x-1+1)} = \frac{1}{1 + (x-1)}$

and similary

$\displaystyle \frac{1}{2-x} = \frac{1}{2 - (x-1+1)} = \frac{1}{1 - (x-1)}$

So a power series for $\displaystyle \frac{1}{2-x}$ is to use the series
$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$
and substitute $\displaystyle x$ with $\displaystyle (x-1)$. Similarly for $\displaystyle \frac{1}{x}$, substitute for $\displaystyle x$ with $\displaystyle -(x-1)$

So

$\displaystyle \frac{1}{2-x} = 1 +(x-1) + (x-1)^2 + (x-1)^3 + \cdots$
$\displaystyle \frac{1}{x} = 1 - (x-1) + (x-1)^2 - (x-1)^3 + \cdots$

Adding and multiply by $\displaystyle -2$ gives
$\displaystyle \frac{4}{x^2-2x} = - 4 - 4(x-1)^2 - 4(x-1)^4 - \cdots$

Of course the interval of convergence is required but I'll leave that for a later discussion.

8. Originally Posted by danny arrigo
If I may be so bold to interject. Math Major is wrong and Mathstud right. Please let me explain.

Starting with the power series

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$

then multiplying by
$\displaystyle \frac{2}{(1-x)^2}$
gives
$\displaystyle \frac{2}{(1-x)^3} = \frac{2}{(1-x)^2} \frac{1}{1-x} = \frac{2}{(1-x)^2} \left( 1 + x + x^2 + x^3 + \cdots \right)$

$\displaystyle = \frac{2}{(1-x)^2} + \frac{2x}{(1-x)^2} + \frac{2x^2}{(1-x)^2} + \frac{2x^3}{(1-x)^2} + \cdots$

which is a series but not a power series! I think the idea here (and MathStud said) is to using the standard series

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$

and manipulate this to get your series. Multiply by numbers or a power of x. Adding and substracting and possibly substituting. Allow me to illustrate with the first example. First, we decompose the function by noting that

$\displaystyle \frac{4}{x^2-2x} = \frac{-4}{x(2 - x)} =- 2\left(\frac{1}{x} + \frac{1}{2-x} \right)$

Now we will come up with a power series for each. Note the following:

$\displaystyle \frac{1}{x} = \frac{1}{ (x-1+1)} = \frac{1}{1 + (x-1)}$

and similary

$\displaystyle \frac{1}{2-x} = \frac{1}{2 - (x-1+1)} = \frac{1}{1 - (x-1)}$

So a power series for $\displaystyle \frac{1}{2-x}$ is to use the series
$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$
and substitute $\displaystyle x$ with $\displaystyle (x-1)$. Similarly for $\displaystyle \frac{1}{x}$, substitute for $\displaystyle x$ with $\displaystyle -(x-1)$

So

$\displaystyle \frac{1}{2-x} = 1 +(x-1) + (x-1)^2 + (x-1)^3 + \cdots$
$\displaystyle \frac{1}{x} = 1 - (x-1) + (x-1)^2 - (x-1)^3 + \cdots$

Adding and multiply by $\displaystyle -2$ gives
$\displaystyle \frac{4}{x^2-2x} = - 4 - 4(x-1)^2 - 4(x-1)^4 - \cdots$

Of course the interval of convergence is required but I'll leave that for a later discussion.

One of my study buddies said the following,

find how f(x) has been altered from looking like h(t) then apply it to the nth sum of h(t) and then we've got to find I, which is the interval of the h(t) i.e. |x|<1 but for f(x) it will be -1<x<1 .....

is this correct?

9. Originally Posted by tsal15
One of my study buddies said the following,

find how f(x) has been altered from looking like h(t) then apply it to the nth sum of h(t) and then we've got to find I, which is the interval of the h(t) i.e. |x|<1 but for f(x) it will be -1<x<1 .....

is this correct?
I'm not sure what your buddy is saying here? Are you trying to find the interval where your series converges?

10. Originally Posted by danny arrigo
I'm not sure what your buddy is saying here? Are you trying to find the interval where your series converges?
so, the sum term in the end (according to my mate) is supposed to look like \sum_{n=0}^{\infty}4x^n (where x = (x-1)^2)...Then to find the Interval of convergence...apparently h(t ) converges when |x|<1, therefore, the new series converges between 0<x<1 (he changed his mind last night from -1<x<1) and he believes that this is all that is required from the question...

I showed him what you said, and he would like to know why you took that approach...he sends his thanks

thank you Danny

11. Originally Posted by tsal15
so, the sum term in the end (according to my mate) is supposed to look like \sum_{n=0}^{\infty}4x^n (where x = (x-1)^2)...Then to find the Interval of convergence...apparently h(t ) converges when |x|<1, therefore, the new series converges between 0<x<1 (he changed his mind last night from -1<x<1) and he believes that this is all that is required from the question...

I showed him what you said, and he would like to know why you took that approach...he sends his thanks

thank you Danny
Why this approach - it's one that can be mastered and works fairly well.

$\displaystyle (1)\;\;\;\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$

this series converges for $\displaystyle -1 < x < 1$ or $\displaystyle |\,x\,| < 1$

Since we substituted
$\displaystyle x = (x-1)$
and
$\displaystyle x = -(x-1)$
into the series, we do the same for interval of convergence

$\displaystyle |\,(x-1)\,| < 1$ and $\displaystyle |\,-(x-1)\,| < 1$

$\displaystyle |\,(x-1)\,| < 1$ or $\displaystyle 0 < x < 2$.

If we had different intervals, we we have to only consider the intersection of the two intervals.

12. Originally Posted by tsal15
Hi Math Major,

What you've said makes complete sense to me... finally, maths that I can understand

Ok, so, I've tried to do it to the second function (i.e. g(x)), and I found that it was the second derivative of h(t)... But the first function (i.e. f(x)), its a little difficult, I can see that 2x - x^2 = 1 - (1-x)^2, which might lead me somewhere, but I can't find how its manipulated with regards to h(t)... Can you help here?

Thank you

tsal15
Surely the easiest way to solve the second problem is just as tsal15 has said.

$\displaystyle g(x)=\frac{d^2}{dx^2}(\frac 1{1-x})=\frac{d^2}{dx^2}(1+x+x^2+x^3+...)=2+6x+12x^2+. ..$

13. $\displaystyle \frac{4}{x^2 - 2x} = \frac{4}{(x^2 - 2x + 1) - 1} = -\frac{4}{1 - (x-1)^2} = -4[1 + (x-1)^2 + (x-1)^4 + (x-1)^6 + ... ]$

14. Originally Posted by danny arrigo

Since we substituted
$\displaystyle x = (x-1)$
and
$\displaystyle x = -(x-1)$
into the series, we do the same for interval of convergence

$\displaystyle |\,(x-1)\,| < 1$ and $\displaystyle |\,-(x-1)\,| < 1$

$\displaystyle |\,(x-1)\,| < 1$ or $\displaystyle 0 < x < 2$.

If we had different intervals, we we have to only consider the intersection of the two intervals.
i thought we substituted $\displaystyle x = (x-1)^2$ instead of $\displaystyle x = (x-1)$

so my mate's method, is not feasible?

15. Originally Posted by Kiwi_Dave
Surely the easiest way to solve the second problem is just as tsal15 has said.

$\displaystyle g(x)=\frac{d^2}{dx^2}(\frac 1{1-x})=\frac{d^2}{dx^2}(1+x+x^2+x^3+...)=2+6x+12x^2+. ..$
Yes thats what I've been thinking he he I feel better that someone is thinking like i am he he

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