Taylor's Series

  • Thread starter Oblio
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  • #51
learningphysics
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The n's in the fraction were increasing too though?

Ah yes, you're right. I messed up... should be:

[tex]ln(x+\delta) = ln(x) + \Sigma_{n=1}^{n=\infty}(-1)^{n+1}\frac{(\delta)^n}{nx^n}[/tex]
 
  • #52
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Could you do a step by step to getting that from the equation? (please)
 
  • #53
learningphysics
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Could you do a step by step to getting that from the equation? (please)

Ok we start with this:

[tex]ln(x+\delta) = ln(x) + \frac{1}{x}\delta - \frac{1}{2x^2}\delta^2 + \frac{1}{3x^3}\delta^3 - \frac{1}{4x^4}\delta^4[/tex]

instead of a series... let's just look at the sequence...

[tex]\frac{1}{x}\delta, -\frac{1}{2x^2}\delta^2, \frac{1}{3x^3}\delta^3, -\frac{1}{4x^4}\delta^4[/tex]

first term (n=1) is [tex]\frac{1}{x}\delta[/tex]
second term(n=2) is [tex]-\frac{1}{2x^2}\delta^2[/tex]

In general what is the nth term?
 
  • #54
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Ok we start with this:

[tex]ln(x+\delta) = ln(x) + \frac{1}{x}\delta - \frac{1}{2x^2}\delta^2 + \frac{1}{3x^3}\delta^3 - \frac{1}{4x^4}\delta^4[/tex]

instead of a series... let's just look at the sequence...

[tex]\frac{1}{x}\delta, -\frac{1}{2x^2}\delta^2, \frac{1}{3x^3}\delta^3, -\frac{1}{4x^4}\delta^4[/tex]

first term (n=1) is [tex]\frac{1}{x}\delta[/tex]
second term(n=2) is [tex]-\frac{1}{2x^2}\delta^2[/tex]

In general what is the nth term?

lol well I know that it would be your answer!
How come the -1 requires the [tex]^{n+1}[/tex] but the others are just [tex]^{n}[/tex]?
 
  • #55
learningphysics
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lol well I know that it would be your answer!
How come the -1 requires the [tex]^{n+1}[/tex] but the others are just [tex]^{n}[/tex]?

because the n = 1 term needs to have a plus sign...then the signs alternate... if I use n instead of n + 1.... then the signs won't come out right... the n=1 term will come out with a - sign... ie: -(1/x)delta... and all the signs will be the opposite of what they need to be... test it out...
 
  • #56
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Right, ok.
So, are n's basically placed where things count up?
 
  • #57
learningphysics
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Right, ok.
So, are n's basically placed where things count up?

Yeah.
 
  • #58
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lol that's not so hard... I thought I had to make the n's count...
 

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