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Math Help - Derivative of complex exponential with high powers

  1. #1
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    Exclamation Derivative of complex exponential with high powers

    evaluate:

    d ^15/dx ^15 exp [(1 + sqrt(3)i) x]

    i dont know how to approach this problem with the drivative to the 15th power.

    firstly 1 + sqrt(3)i must be converted to polar form.. then im stuck

    pls help me.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Uhh...

    Quote Originally Posted by tasukete View Post
    evaluate:

    d ^15/dx ^15 exp [(1 + sqrt(3)i) x]

    i dont know how to approach this problem with the drivative to the 15th power.

    firstly 1 + sqrt(3)i must be converted to polar form.. then im stuck

    pls help me.
    Maybe I am being stupid but wouldnt it be f^{(15)}(x)=(\sqrt{3}i)^{15}e^{1+\sqrt{3}ix}=-2187\sqrt{3}ie^{1+\sqrt{3}ix}?
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    Exclamation

    Quote Originally Posted by Mathstud28 View Post
    Maybe I am being stupid but wouldnt it be f^{(15)}(x)=(\sqrt{3}i)^{15}e^{1+\sqrt{3}ix}=-2187\sqrt{3}ie^{1+\sqrt{3}ix}?

    eh? im having trouble understanding as to how u got that. i thought the answer would be in the form of trig and expo functions??? am i wrong?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by tasukete View Post
    eh? im having trouble understanding as to how u got that. i thought the answer would be in the form of trig and expo functions??? am i wrong?
    if you think about it \frac{D[e^{u(x)}]}{dx}=u'(x)\cdot{e^u(x)} right? Now if u'(x) is a constant then we dont need to use chain rule we jsut have keep multiplying it so for example...remember u(x) is a constant then we would have f(x)=e^{u(x)}...then f'(x)=u'(x)\cdot{e^{u(x)}}...now since it is a constant we just yank it out so f''(x)=u'(x)\cdot{\frac{D[e^{u(x)}]}{dx}}...so using this same concept of yanking out the constant we get f^{(n)}(x)=u'(x)^{n}\cdot{e^{u(x)}}...and in this case u'(x) is a consant...so yeah
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Maybe I am being stupid but wouldnt it be f^{(15)}(x)=(\sqrt{3}i)^{15}e^{1+\sqrt{3}ix}=-2187\sqrt{3}ie^{1+\sqrt{3}ix}?

    after obtaining an expression for the derivative...i got a different answer to ures... since the constant would be of high power.. (1 + sqrt(3) i ) ^ 15... expanding it would be silly.

    so wouldnt u need to convert it into polar form and use De Moivre's theorem to simplify it?

    if that is so i got something different to ures... i had -2^15 ????

    hrmm... i just noticed u only included the imaginary part in the constant....y is that?
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  6. #6
    Moo
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    Hello,

    Quote Originally Posted by tasukete View Post
    after obtaining an expression for the derivative...i got a different answer to ures... since the constant would be of high power.. (1 + sqrt(3) i ) ^ 15... expanding it would be silly.

    so wouldnt u need to convert it into polar form and use De Moivre's theorem to simplify it?

    if that is so i got something different to ures... i had -2^15 ????

    hrmm... i just noticed u only included the imaginary part in the constant....y is that?
    Indeed, mathstud got wrong because he forgot 1.

    (1 + sqrt(3) i ) ^ 15
    This is correct.
    But don't bother with De Moivre's theorem

    (1+\sqrt{3} i)^2=(1-3+2\sqrt{3} i)=2(-1+\sqrt{3} i)

    So (1+\sqrt{3} i)^3=(1+\sqrt{3} i)^2(1+\sqrt{3} i)=2\underbrace{(-1+\sqrt{3} i)(1+\sqrt{3} i)}_{(a-b)(a+b)=a^2-b^2}=2(-3-1)=-2^3

    So (1+\sqrt{3} i)^{15}=((1+\sqrt{3} i)^3)^5=(-2^3)^5=-2^{15}
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Wait

    Quote Originally Posted by Moo View Post
    Hello,



    Indeed, mathstud got wrong because he forgot 1.



    This is correct.
    But don't bother with De Moivre's theorem

    (1+\sqrt{3} i)^2=(1-3+2\sqrt{3} i)=2(-1+\sqrt{3} i)

    So (1+\sqrt{3} i)^3=(1+\sqrt{3} i)^2(1+\sqrt{3} i)=2\underbrace{(-1+\sqrt{3} i)(1+\sqrt{3} i)}_{(a-b)(a+b)=a^2-b^2}=2(-3-1)=-2^3

    So (1+\sqrt{3} i)^{15}=((1+\sqrt{3} i)^3)^5=(-2^3)^5=-2^{15}
    I dont think you are right Moo...I just put the answer in my calculator and it says I am right
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  8. #8
    Moo
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    Oh yeah ?

    You forgot the brackets...

    It's (1+sqrt(3) i)x, not 1+sqrt(3) ix


    Calculators are evil especially when the one who uses it got wrong himself
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    Quote Originally Posted by Moo View Post
    Oh yeah ?

    You forgot the brackets...

    It's (1+sqrt(3) i)x, not 1+sqrt(3) ix


    Calculators are evil especially when the one who uses it got wrong himself

    hrmmm...was wondering why i couldnt get stud's answer..
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  10. #10
    Moo
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    You should've asked before... I could have answered you...
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    OHhhh

    Quote Originally Posted by Moo View Post
    Oh yeah ?

    You forgot the brackets...

    It's (1+sqrt(3) i)x, not 1+sqrt(3) ix


    Calculators are evil especially when the one who uses it got wrong himself
    I see why I got it wrong....there needs to be a mandatory LaTeX requirement for clarity in questions
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  12. #12
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    I see why I got it wrong.......there needs to be a mandatory LaTeX requirement for clarity in questions
    His question was clear...all brackets were here...
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