Power series representation

  • #1
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Homework Statement


write a power series representation of the following:
[tex] \frac{x}{15x^2 +1} [/tex]


Homework Equations


the formula
[tex] \frac{1}{1-x} = 1 + x + x^2 + ... = \sum_{n=0}^{∞} x^n [/tex]


The Attempt at a Solution


we can rewrite the summnd like
[tex] \frac{x}{15} \left( \frac{1}{1+\frac{x^2}{15}} \right)[/tex]
we can write the denominator from the above term as:

[tex] 1 - \left( - \left( \frac{x}{\sqrt{15}} \right)^2 \right) [/tex]

so using the above term we can write the series like:
[tex] \frac{x}{15} \sum_{n=0}^∞ (-1)^n \frac{x^{2n}}{15^{n/2}}[/tex] /known data[/b]

and this simplifies to:

[tex] \sum_{n=0}^{∞} (-1)^n \frac{x^{2n+1}}{15^{n/2 + 1}} [/tex]

is that correct? This is the basis for the second part which asks for the interval of convergence
I cant write absolute value, but here goes:
[tex] \frac{x^2}{\sqrt{15}} < 1[/tex]

[tex] x < \sqrt{\sqrt{15}} [/tex]

This means that the interval is

[tex] \left( -15^{1/4} , 15^{1/4} \right) [/tex]
Unfortunately I m getting the answer wrong as per the computer... can you please take a look and see if this is correct or not?

Thank you for your help. It is greatly appreciated!
 

Answers and Replies

  • #2
vela
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Homework Statement


write a power series representation of the following:
[tex] \frac{x}{15x^2 +1} [/tex]


Homework Equations


the formula
[tex] \frac{1}{1-x} = 1 + x + x^2 + ... = \sum_{n=0}^{∞} x^n [/tex]


The Attempt at a Solution


we can rewrite the summnd like
[tex] \frac{x}{15} \left( \frac{1}{1+\frac{x^2}{15}} \right)[/tex]
This isn't correct. Try multiplying out the denominator to see this.
we can write the denominator from the above term as:

[tex] 1 - \left( - \left( \frac{x}{\sqrt{15}} \right)^2 \right) [/tex]

so using the above term we can write the series like:
[tex] \frac{x}{15} \sum_{n=0}^∞ (-1)^n \frac{x^{2n}}{15^{n/2}}[/tex] /known data[/b]

and this simplifies to:

[tex] \sum_{n=0}^{∞} (-1)^n \frac{x^{2n+1}}{15^{n/2 + 1}} [/tex]

is that correct? This is the basis for the second part which asks for the interval of convergence
I cant write absolute value, but here goes:
[tex] \frac{x^2}{\sqrt{15}} < 1[/tex]

[tex] x < \sqrt{\sqrt{15}} [/tex]

This means that the interval is

[tex] \left( -15^{1/4} , 15^{1/4} \right) [/tex]
Unfortunately I m getting the answer wrong as per the computer... can you please take a look and see if this is correct or not?

Thank you for your help. It is greatly appreciated!
 
  • #3
1,444
2
This isn't correct. Try multiplying out the denominator to see this.
You're right... my bad

the term should go

[tex] x \frac{1}{1-(-15x^2)} [/tex]
which is
[tex] x \sum (-15x^2)^n [/tex]
[tex] \sum (-1)^n 15^n x^{2n+1} [/tex]

is that correct?
And it follows that:
[tex] \left| -15x^2 \right| < 1 [/tex]
and solving this we get
[tex] \left| x^2 \right| < \frac{1}{\sqrt{15}} [/tex]

Is this correct?
 
  • #4
1,444
2
Hey can you let me know if this is corrct what I did?

Thank you for your help
 
  • #5
vela
Staff Emeritus
Science Advisor
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Education Advisor
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Looks good except for the typo in your last line. It should be |x| not |x2|.
 

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