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swissgirl

  • 2 years ago

What are the first 3 terms of the Taylor expansion for the function \(f(x)=e^{2x}\)

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  1. swissgirl
    • 2 years ago
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    \(f(x)=e^{2x}\) \(f'(x)=2e^{2x}\) \(f''(x)=4e^{2x}\)

  2. perl
    • 2 years ago
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    the taylor series is , does it say about x = 0 ? i assume it is

  3. swissgirl
    • 2 years ago
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    I am assuming that x=0 too

  4. swissgirl
    • 2 years ago
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    I am not sure why we assume x=0 in this case but whtvr

  5. perl
    • 2 years ago
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    sum f^(n) (a) (x-a)^n / n! = f(a) + f'(a) (x-a) + f '' (a) (x-a)^2/2! + ...

  6. perl
    • 2 years ago
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    we dont have to, it just makes it much cleaner

  7. perl
    • 2 years ago
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    and by cleaner i mean simpler , since x - 0 = x , instead of x - 1 or x - 2 , etc

  8. perl
    • 2 years ago
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    so now lets plug in f(0) + f'(0) *x + f ' ' (0) * x^2/2 + ... , since we want just the first three terms

  9. swissgirl
    • 2 years ago
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    So our answer wld be 1+2+2?

  10. perl
    • 2 years ago
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    e^(2*0) + 2*e^(2*0) * x + 4*e^(2*0) *x^2/2

  11. perl
    • 2 years ago
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    , remember a taylor series is approximating a complicated function (here exponential) with a less complicated function (polynomial)

  12. perl
    • 2 years ago
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    1 , 2x , 4x^2/2

  13. perl
    • 2 years ago
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    ok?

  14. swissgirl
    • 2 years ago
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    Yaaaa I think I get it

  15. swissgirl
    • 2 years ago
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    I get it lol just need to know how to enter it into a program. Just gotta fiddle around. THANKS

  16. swissgirl
    • 2 years ago
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    THANKKSSSSS :))))))))

  17. swissgirl
    • 2 years ago
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    Ok figured it out. Thanks for ur help

  18. perl
    • 2 years ago
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    :D

  19. perl
    • 2 years ago
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    where is my medal?

  20. perl
    • 2 years ago
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    thanks

  21. swissgirl
    • 2 years ago
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    U got ur medal even b4 i got my answer :P

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