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

  • Thread starter Lucy Yeats
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  • #1
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Homework Statement



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!)+...

Homework Equations



Perhaps Taylor series?

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.)
 

Answers and Replies

  • #2
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We're studying complex numbers at the moment, so I'm guessing this is to do with real and imaginary parts, if that helps.
 
  • #3
I like Serena
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Hi Lucy Yeats! :smile:


Perhaps Taylor series?
These series are Taylor series! :wink:


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

You only need one more observation.
What is the derivative of S2?
 
  • #4
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The derivative of S2 is S1.
The derivative of S3 is S2.
The derivative of S1 is S3.
 
  • #5
I have a problem solving stellar numbers 1,3,6,10,15 thanks for your help and clarification
for creating the general statement
 
  • #6
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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?
 
  • #7
I like Serena
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The derivative of S2 is S1.
The derivative of S3 is S2.
The derivative of S1 is S3.
Yes... so you can set up a differential equation.
The solutions should correspond to S1, S2, and S3.


I have a problem solving stellar numbers 1,3,6,10,15 thanks for your help and clarification
for creating the general statement
Huh? :confused:
 
  • #8
I like Serena
Homework Helper
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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?
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:)
 
  • #9
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(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!! :-)
 
  • #10
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I get B=0 and A=2/3.
Does that sound right?
 
  • #11
I like Serena
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(1/3)(e^x)+(e^-0.5)(Acos(3x/2)+Bsin(3x/2))

Is that right for sine and cosine?
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]


Thanks for being so helpful!! :-)
:smile:
 
  • #12
117
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I'll check again later- I must have made an error in the algebra.

Thanks! Have a great day!
 

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