Unraveling the Mysteries of Taylor's Theorem

[learningphysics]

Also meaning that the first two terms cannot be simplified yet?

So, taking $$\frac{1}{4!}$$ $$\frac{-6}{x^{4}}$$$$\delta^{4}$$

becomes:
$$\frac{-6}{4*3*2*1*x^{4}}$$$$\delta$$$$^{4}$$

= $$\frac{-1}{4x^{4}}$$$$\delta$$$$^{4}$$ ?

Yes, that looks right... so show the first 5 terms as a sum...

 

[Oblio]

leaving me with

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

copied it out too quick...

 

[learningphysics]

leaving me with

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

? What's with the last term?

 

[Oblio]

^^ oops. edited it.

 

[learningphysics]

leaving me with

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

copied it out too quick...

cool... now except for the first term... do you see the pattern of the next 4 terms... can you write the series formula?

 

[Oblio]

yeah i do se it: delta, the denominator and the exponent are all going up by one very clearly.

 

[Oblio]

is a series formula numerous equations again?

 

[learningphysics]

is a series formula numerous equations again?

it's a formula like:

$$e^x = \Sigma_0^{\infty}{\frac{x^n}{n!}}$$

write a fomula like that for this series... you can either substitute in x = 1 now, or after you have the series formula... up to you.

 

[Oblio]

it's a formula like:

$$e^x = \Sigma_0^{\infty}{\frac{x^n}{n!}}$$

write a fomula like that for this series... you can either substitute in x = 1 now, or after you have the series formula... up to you.

$$e^x = \Sigma_0^{\infty}{\frac{x^n}{1+n!^{n+1}}}$$ I'm not sure if I need the exponent in the numerator and whether I must include delta...

 

[learningphysics]

$$e^x = \Sigma_0^{\infty}{\frac{x^n}{1+n!^{n+1}}}$$ I'm not sure if I need the exponent in the numerator and whether I must include delta...

I just used the e^x as an example... don't try to make it look exactly like that. Yes, you need delta...

when you write out the sum of the formula you gave above, do you get back the series you calculated?

 

[Oblio]

$$ln (x+\delta) = \Sigma_0^{\infty}{\frac{x^n}{1+n!^{n+1}}}$$$$\delta$$ $$^{n+1}$$

I'm pretty sure this is still wrong but I'm looking at it check things out..

 

[learningphysics]

I'm pretty sure this is still wrong but I'm looking at it check things out..

But the x's have to be in the denominator. also the 1+n!^(n+1) makes no sense...

 

[Oblio]

I'm not sure how else to do it. (I've handed this assignment in now.. such is deadlines!)

 

[learningphysics]

I'm not sure how else to do it. (I've handed this assignment in now.. such is deadlines!)

Sorry about that. no big deal. My idea was something more along the lines of:

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

Then substituting x = 1. But you did most of the important stuff... calculating the terms... so this is just the last bit...

 

[Oblio]

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

 

[learningphysics]

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

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

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

 

[Oblio]

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

 

[learningphysics]

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?

 

[Oblio]

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 $$^{n}$$?

 

[learningphysics]

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...

 

[Oblio]

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

 

[learningphysics]

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

Yeah.

 

[Oblio]

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