Unraveling the Mysteries of Taylor's Theorem

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Taylor's theorem allows the approximation of a function f(x) at a point (x + δ) using its derivatives, expressed as an infinite series. This series is particularly useful for small values of δ, providing accurate approximations by using just the first few terms. The discussion focuses on finding the Taylor series for ln(1 + δ) and emphasizes the importance of correctly substituting values and calculating derivatives. Participants explore the pattern of terms in the series, noting the need for alternating signs and the role of factorials in simplification. Ultimately, the goal is to derive a general formula for the Taylor series of ln(x + δ) that captures the relationship between the terms and their coefficients.
  • #51
Oblio said:
The n's in the fraction were increasing too though?

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

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

Ok we start with this:

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

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

\frac{1}{x}\delta, -\frac{1}{2x^2}\delta^2, \frac{1}{3x^3}\delta^3, -\frac{1}{4x^4}\delta^4

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

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

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

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

\frac{1}{x}\delta, -\frac{1}{2x^2}\delta^2, \frac{1}{3x^3}\delta^3, -\frac{1}{4x^4}\delta^4

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

In general what is the nth term?

lol well I know that it would be your answer!
How come the -1 requires the ^{n+1} but the others are just ^{n}?
 
  • #55
Oblio said:
lol well I know that it would be your answer!
How come the -1 requires the ^{n+1} but the others are just ^{n}?

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
Right, ok.
So, are n's basically placed where things count up?
 
  • #57
Oblio said:
Right, ok.
So, are n's basically placed where things count up?

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