1. ## 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.

2. ##  Originally Posted by WartonMorton 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) ]

3. sumation from k=0 to inf of (-1)^k * k! * (b+x)^[(-1)(k+1)] * (x-a)

valid from (0,1/(b+a)]

4. Originally Posted by WartonMorton 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 $\displaystyle 1/(b+x)$

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 $\displaystyle 1/(1+y)$ ? In this problem,what do you have in place of y ?

Later we can discuss the validity of convergence.

5. sorry don't know how to apply the hint.

6. Originally Posted by WartonMorton 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?

7. Originally Posted by ques 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?
$\displaystyle 1/(b-x)=1/(b+a+(x-a))$

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