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Math Help - series and validity

  1. #1
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    series and validity

    Expand g(x) as indicated and specify the values of x for which the expansion is valid.

    g(x) = (b+x)^(-1) in powers of x-a, a is not equal to -b.
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  2. #2
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    Smile

    Quote Originally Posted by WartonMorton View Post
    Expand g(x) as indicated and specify the values of x for which the expansion is valid.

    g(x) = (b+x)^(-1) in powers of x-a, a is not equal to -b.
    Hint : Use b+ x = b + a + ( x - a ) = (b+a) [ 1 + ( x - a)/(b + a) ]
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  3. #3
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    sumation from k=0 to inf of (-1)^k * k! * (b+x)^[(-1)(k+1)] * (x-a)

    valid from (0,1/(b+a)]
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  4. #4
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    Quote Originally Posted by WartonMorton View Post
    sumation from k=0 to inf of (-1)^k * k! * (b+x)^[(-1)(k+1)] * (x-a)

    valid from (0,1/(b+a)]
    Sorry, that is not right ! From where do you have b+x as well as x-a ? What are the steps that you used to obtain the above step ?

    You have 1/(b+x)

    Step 1 :- What do you have after making the necessary substitutions as explained by my hint? Remember that you want it as powers of x-a.

    Step 2 :- How do you expand 1/(1+y) ? In this problem,what do you have in place of y ?

    Later we can discuss the validity of convergence.
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  5. #5
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    sorry don't know how to apply the hint.
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  6. #6
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    Quote Originally Posted by WartonMorton View Post
    sorry don't know how to apply the hint.

    b-x = b+a + (x-a)

    So, in place of b-x you can substitute b+a + (x-a) as they are equal.
    Can you take this forward?
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  7. #7
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    Quote Originally Posted by ques View Post
    b-x = b+a + (x-a)

    So, in place of b-x you can substitute b+a + (x-a) as they are equal.
    Can you take this forward?
    1/(b-x)=1/(b+a+(x-a))
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