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Math Help - Power series expansion

  1. #1
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    Post Power series expansion

    "By adapting the power series expansion (centred on c = 0) of 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. f(x) = \frac{4}{x^2 - 2x} , center c = 1

    ii. 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?

    Thank you all in advance
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    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.
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    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
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    Quote Originally Posted by tsal15 View Post
    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.
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  5. #5
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    Quote Originally Posted by Math Major View Post
    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 \frac{1}{1-x} by \frac{1}{(1-x)^2} does not give the Maclaurin series for \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 \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n then \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 \frac{2}{(1-x)^3}=\left(\frac{1}{(1-x)}\right)''
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    Quote Originally Posted by Mathstud28 View Post
    This is incorrect. Multiplying the power series for \frac{1}{1-x} by \frac{1}{(1-x)^2} does not give the Maclaurin series for \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 \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n then \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 \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?
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  7. #7
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    Quote Originally Posted by tsal15 View Post
    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

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

    then multiplying by
    \frac{2}{(1-x)^2}
    gives
     \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)

    = \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

    \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

    \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:

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

    and similary

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

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

    So

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

    Adding and multiply by -2 gives
    \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.
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    Quote Originally Posted by danny arrigo View Post
    If I may be so bold to interject. Math Major is wrong and Mathstud right. Please let me explain.

    Starting with the power series

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

    then multiplying by
    \frac{2}{(1-x)^2}
    gives
     \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)

    = \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

    \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

    \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:

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

    and similary

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

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

    So

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

    Adding and multiply by -2 gives
    \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?
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  9. #9
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    Quote Originally Posted by tsal15 View Post
    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?
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    Quote Originally Posted by danny arrigo View Post
    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
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  11. #11
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    Quote Originally Posted by tsal15 View Post
    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.

    As for the radius of convergence of the series, we start with

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

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

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

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

    which both lead to

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

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

    Quote Originally Posted by tsal15 View Post
    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.

    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+.  ..
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    \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 + ... ]
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    Quote Originally Posted by danny arrigo View Post

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

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

    which both lead to

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

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

    so my mate's method, is not feasible?
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    Quote Originally Posted by Kiwi_Dave View Post
    Surely the easiest way to solve the second problem is just as tsal15 has said.

    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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