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Summing a series- Taylor series/ complex no.s?

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the sums of the following series:
    S1=1+(x^3)/(3!)+(x^6)/(6!)+...
    S2=x+(x^4)/(4!)+(x^7)/(7!)+...
    S3=(x^2)/(2!)+(x^5)/(5!)+(x^8)/(8!)+...

    2. Relevant equations

    Perhaps Taylor series?

    3. The attempt at a solution

    I spotted that adding S1+S2+S3=e^x, but I don't know how to proceed.
    Thanks in advance for helping. :-)

    (Our teacher did warn us that this is a tough question.)
     
  2. jcsd
  3. Nov 6, 2011 #2
    We're studying complex numbers at the moment, so I'm guessing this is to do with real and imaginary parts, if that helps.
     
  4. Nov 6, 2011 #3

    I like Serena

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    Hi Lucy Yeats! :smile:


    These series are Taylor series! :wink:


    Sharp!

    You only need one more observation.
    What is the derivative of S2?
     
  5. Nov 6, 2011 #4
    The derivative of S2 is S1.
    The derivative of S3 is S2.
    The derivative of S1 is S3.
     
  6. Nov 6, 2011 #5
    I have a problem solving stellar numbers 1,3,6,10,15 thanks for your help and clarification
    for creating the general statement
     
  7. Nov 6, 2011 #6
    I have another idea:
    If S2=S1" and S3=S1', and S1+S2+S3=e^x,
    can I write a differential equation in S1?
    S1"+S1'+S1=e^x
    If you solve for S1, you get (1/3)(e^x)+(e^-0.5)(Acos3x+Bsin3x)
    This is a general solution, so how would I get rid of the constants A and B?
    Does this method look okay, or am I looking at this the wrong way?
     
  8. Nov 6, 2011 #7

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    Yes... so you can set up a differential equation.
    The solutions should correspond to S1, S2, and S3.


    Huh? :confused:
     
  9. Nov 6, 2011 #8

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    I was just suggesting that! :approve:

    Solve for the boundary conditions that S1(0)=1, S1'(0)=0, and S1''(0)=0.

    But before you do, you should check the period of your cosine and sine.
    (I get a different period. :shy:)
     
  10. Nov 6, 2011 #9
    (1/3)(e^x)+(e^-0.5)(Acos(3x/2)+Bsin(3x/2))

    Is that right for sine and cosine?

    Thanks for being so helpful!! :-)
     
  11. Nov 6, 2011 #10
    I get B=0 and A=2/3.
    Does that sound right?
     
  12. Nov 6, 2011 #11

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    Actually, I get:
    [tex]{e^x \over 3}+ e^{-{x \over 2}}(A \cos({\sqrt 3 \over 2} x)+B \sin({\sqrt 3 \over 2} x))[/tex]


    :smile:
     
  13. Nov 6, 2011 #12
    I'll check again later- I must have made an error in the algebra.

    Thanks! Have a great day!
     
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