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anonymous

  • 5 years ago

What is a Taylor Series for f(x)=ln(2+2x) ?

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  1. anonymous
    • 5 years ago
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    ummm I might be able to do this... we'll need to collaborate though

  2. anonymous
    • 5 years ago
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    they gave the first part: ln(2)+\[\sum_{n=1}^{\infty}\]

  3. anonymous
    • 5 years ago
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    so whats in the sigma, thats important

  4. anonymous
    • 5 years ago
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    thats what we have to try to find lol

  5. anonymous
    • 5 years ago
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    does the question say exactly that? What is a taylor series for...

  6. anonymous
    • 5 years ago
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    take the derivative of ln(x) like 4 or 5 times see if you can see a pattern

  7. anonymous
    • 5 years ago
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    yes, What is a Taylor Series for f(x)=ln(2+2x)

  8. anonymous
    • 5 years ago
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    of and at x=1/2 we know that ln(1) = 0

  9. anonymous
    • 5 years ago
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    \[\sum_{?}^{?} (f^{(n)}(1/2))/n! * (2x-(1/2))^{n}\]

  10. anonymous
    • 5 years ago
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    the general formula would be F^(n) (x) = (-1)^(n+1) (n-1) ! --------------- x^(n)

  11. anonymous
    • 5 years ago
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    thats my best guess but I think I'm taking it too fat by kinda approxximating

  12. anonymous
    • 5 years ago
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    yes it would now try writing that in sigma notation which is what I did up there...not sure if this is right to do...

  13. anonymous
    • 5 years ago
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    >.<

  14. anonymous
    • 5 years ago
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    ha, I have no idea

  15. anonymous
    • 5 years ago
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    Go to the general form of the Taylor Series. T(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + ... and plug the derivatives of ln(2+2x) into the places you see fit (I think in this case it's safe to set a=0).

  16. anonymous
    • 5 years ago
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    I dont know where I would stop using the general form...?

  17. anonymous
    • 5 years ago
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    Until you see the general form that they're taking...and then you can put the general form of the series into your sum symbol.

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