# Taylor's Series

learningphysics
Homework Helper
The n's in the fraction were increasing too though?

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

$$ln(x+\delta) = ln(x) + \Sigma_{n=1}^{n=\infty}(-1)^{n+1}\frac{(\delta)^n}{nx^n}$$

Could you do a step by step to getting that from the equation? (please)

learningphysics
Homework Helper
Could you do a step by step to getting that from the equation? (please)

$$ln(x+\delta) = ln(x) + \frac{1}{x}\delta - \frac{1}{2x^2}\delta^2 + \frac{1}{3x^3}\delta^3 - \frac{1}{4x^4}\delta^4$$

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?

$$ln(x+\delta) = ln(x) + \frac{1}{x}\delta - \frac{1}{2x^2}\delta^2 + \frac{1}{3x^3}\delta^3 - \frac{1}{4x^4}\delta^4$$

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?

How come the -1 requires the $$^{n+1}$$ but the others are just $$^{n}$$?

learningphysics
Homework Helper
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...

Right, ok.
So, are n's basically placed where things count up?

learningphysics
Homework Helper
Right, ok.
So, are n's basically placed where things count up?

Yeah.

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