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anonymous

  • 5 years ago

Find the Maclaurin series for f(x) = sin^2(x)

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  1. anonymous
    • 5 years ago
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    use sin^2(x) =1/2 * (1-cos2x) cos 2x - use def cosx from table

  2. anonymous
    • 5 years ago
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    forgot to add that, it says to (Use the Identity 2sinxcosx = sin(2x)

  3. anonymous
    • 5 years ago
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    f(x)=sin^2(0) f'(x)=2*sin(x)*cos(x)

  4. anonymous
    • 5 years ago
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    Ouch! seems easier to use:sin^2(x) =1/2 * (1-cos2x) - no multiplication...

  5. anonymous
    • 5 years ago
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    therefore f'(x)=sin(2x) f''(x)=2cos(2x)

  6. anonymous
    • 5 years ago
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    what is 2sinxcosx has to do with sin^2(x)...? did you mean in initial date: sin2x or sin^2(x)

  7. anonymous
    • 5 years ago
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    f'''(x)=-4sin(2x)=-4*f'(x)

  8. anonymous
    • 5 years ago
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    inik, it's because the first derivative of the function is 2cos(x)sin(x) which is sin(2x), according to the identity given by asker.

  9. anonymous
    • 5 years ago
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    putting it into maclaurin's series im getting 0+0+ 2x^2/2! + 0... am i doing something wrong

  10. anonymous
    • 5 years ago
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    got that... I thought he asked for Maclaurin series representation...I don't use derivatives. there is known presentation of cos & sin - why not to use it? it's simple

  11. anonymous
    • 5 years ago
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    not using derivatives for Maclaurin series? How do you do that?

  12. anonymous
    • 5 years ago
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    James Stuart Essential Calculus ch8... I'm trying to find ref on Internet - let you know

  13. anonymous
    • 5 years ago
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    huykma, i think you are doing it right, when you get to f''''(0) the term would be something times f''(0) and the cycle goes on and on.

  14. anonymous
    • 5 years ago
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    thanks

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