Taylor Series using Geometric Series and Power Series

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jegues
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Homework Statement


See figure attached.


Homework Equations





The Attempt at a Solution



Okay I think I handled the lnx portion of the function okay(see other figure attached), but I'm having from troubles with the,

[tex]\frac{1}{x^{2}}[/tex]

[tex]\int x^{-2} = \frac{-1}{x} + C[/tex]

How do I deal with the C?

If I can get,

[tex]\frac{-1}{x}[/tex]

I can work with it to get something like the following,

[tex]\frac{\text{first term of geometric series}}{1 - \text{common ratio}}[/tex]

So what do I do about the C? Once I figure this out I can make more of an attempt into shaping,

[tex]\frac{-1}{x}[/tex]

into the form mentioned above.

Any ideas?

Thanks again!
 
Attachments
  • Homework problem 5: Find the Taylor series about 1 for f(x) = 1/x^2 + ln x. Express the answer in sigma notation and determine the open interval of convergence.
    TSQ.jpg
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  • Handwritten derivation on grid paper: working with ln x, d/dx ln x = 1/x written as a geometric series in (x-1), integrated to a power series with constant C found by setting x = 1, giving the series sum over n.
    TSQA1.jpg
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Last edited:
You're going about it backwards. Use

[tex]\frac{1}{x^2} = -\frac{d}{dx}\left(\frac{1}{x}\right)[/tex]
 
vela said:
You're going about it backwards. Use

[tex]\frac{1}{x^2} = -\frac{d}{dx}\left(\frac{1}{x}\right)[/tex]

Alrighty I think I've got a series for,

[tex]\frac{1}{x^{2}}[/tex]

See figure attached. Is this correct?

I can't seem to figure out how to express it sigma notation however.

Any ideas?
 
Attachments
  • Handwritten working on grid paper deriving a series for 1/x^2: 1/x^2 = -d/dx(1/x), 1/x = 1/(1+(x-1)) = 1 - (x-1) + (x-1)^2 - (x-1)^3 + ..., integrated, with c found by setting x = 1 to give c = 2, ending with a question mark.
    TSQA2.jpg
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No, it looks like you integrated the series, but you want to differentiate -1/x to get 1/x2.
 
vela said:
No, it looks like you integrated the series, but you want to differentiate -1/x to get 1/x2.

Whoops!

How does this look? (See figure attached)
 
Attachments
  • Handwritten working on grid paper: Working with 1/x^2, 1/x^2 = -d/dx(1/x), 1/x = 1/(x+1-1) = 1/(1+(x-1)) = 1/(1-(-(x-1))), expanded as 1-(x-1)+(x-1)^2-(x-1)^3+..., then 1/x^2 = 1-2(x-1)+3(x-1)^2-4(x-1)^3+..., and sum from n=0 to infinity of (-1)^n (n+1)(x-1)^n.
    TSQA3.jpg
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