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emcrazy14

  • one year ago

Urgent help needed: Expand the following up to the term in x^3:

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  1. emcrazy14
    • one year ago
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    \[\frac{ 1+x }{ \sqrt{1-x} }\]

  2. Jibrilpresbury
    • one year ago
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    idk

  3. satellite73
    • one year ago
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    expand in a power series i take it

  4. satellite73
    • one year ago
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    about zero i hope first term is \(f(0)=1\)

  5. satellite73
    • one year ago
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    first derivative is \[f'(x)=\frac{3-x}{2\sqrt{1-x}^3}\] so second term is \[f'(0)x=\frac{3}{2}x\]

  6. satellite73
    • one year ago
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    yikes, now we need the derivative of that beast

  7. satellite73
    • one year ago
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    do i have to do that too?

  8. emcrazy14
    • one year ago
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    Yes, please?

  9. satellite73
    • one year ago
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    grrr

  10. satellite73
    • one year ago
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    lets cheat

  11. satellite73
    • one year ago
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    looks like it is \[f''(x)=\frac{7-x}{4\sqrt{1-x}^5}\]

  12. satellite73
    • one year ago
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    making the next term \[\frac{f''(0)}{2}x^2=\frac{7}{8}x^2\]

  13. satellite73
    • one year ago
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    one more?

  14. emcrazy14
    • one year ago
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    So I find the derivative of that too?

  15. satellite73
    • one year ago
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    \[f'^{(3)}(x)=\frac{33-3x}{8\sqrt{1-x}^7}\]

  16. satellite73
    • one year ago
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    i think

  17. satellite73
    • one year ago
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    if so then \(\frac{f^{(3)}(0)}{3!}x^3=\frac{33}{16}x^3\)

  18. satellite73
    • one year ago
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    hmm let me check if this is right

  19. satellite73
    • one year ago
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    no messed up \(3!=6\) doh

  20. satellite73
    • one year ago
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    third term is \(\frac{11}{16}x^3\)

  21. satellite73
    • one year ago
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    k lets check the whole dam thing and see if it is right

  22. satellite73
    • one year ago
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    http://www.wolframalpha.com/input/?i=%281%2Bx%29%2Fsqrt%281-x%29 yup !!

  23. emcrazy14
    • one year ago
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    Okay, thank you so much! :)

  24. satellite73
    • one year ago
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    yw

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