What is the Taylor expansion for ln(1+z)?

[Wishbone]

the problem reads develop expansion of ln(1+z)

of course I just tried throwing it into the formula for taylor expansions, however I do not know what F(a) is, the problem doesn't specify, so how can I use a taylor series?

 

[Physics Monkey]

What exactly did you find difficult about using the formula? You just take derivatives of your function and evaluate them at z = 0 provided you're expanding about zero. The terms in the series are then \frac{1}{n!} f^{(n)}(0) z^n where in your case f(z) = \ln{(1+z)}.

 

[Wishbone]

z = 0 provided you're expanding about zero.

the problem doesn't say that, see that's what my question is... Am I supposed to assume that?

 

[d_leet]

the problem reads develop expansion of ln(1+z)

of course I just tried throwing it into the formula for taylor expansions, however I do not know what F(a) is, the problem doesn't specify, so how can I use a taylor series?

Well you can derive a taylor expansion for it by using the geometric series and a bit of integration.

 

[Wishbone]

Well you can derive a taylor expansion for it by using the geometric series and a bit of integration.

well it says develop the taylor expansion, I think it means use the taylor expansion.. I don't know tho, its a really stupid, poorly worded question.

 

[d_leet]

well it says develop the taylor expansion, I think it means use the taylor expansion.. I don't know tho, its a really stupid, poorly worded question.

Well you would end up with the same answer, or at least you should, so it may just be easier to derive it from the sum of an infinite geometric series.

 

[Wishbone]

Well you would end up with the same answer, or at least you should, so it may just be easier to derive it from the sum of an infinite geometric series.

ya but what do i use as a? I end up with the same problem...

 

[d_leet]

ya but what do i use as a? I end up with the same problem...

Well if you do it the way I suggested you could just use a, leave it general and just make sure to note the rquirements on a.

 

[Wishbone]

What exactly did you find difficult about using the formula? You just take derivatives of your function and evaluate them at z = 0 provided you're expanding about zero. The terms in the series are then \frac{1}{n!} f^{(n)}(0) z^n where in your case f(z) = \ln{(1+z)}.

ya I must have to do it the other way becausethis wouldn't even work. If I take z=0, then df(0)/dz= undefined

 

[Physics Monkey]

You must have made a mistake, Wishbone. \frac{d}{dz} \ln{(1+z)} = \frac{1}{1+z} which is perfectly well defined at z = 0.

Also, you can expand about whatever point you want, you just have to worry about the radius of convergence. I suggested z = 0 because that is what is usually done in practice with this function.

 

[Wishbone]

You must have made a mistake, Wishbone. \frac{d}{dz} \ln{(1+z)} = \frac{1}{1+z} which is perfectly well defined at z = 0.

Also, you can expand about whatever point you want, you just have to worry about the radius of convergence. I suggested z = 0 because that is what is usually done in practice with this function.

oh geez, you I was taking the derivate of ln z. whoops.

 

[Wishbone]

wait won't the nth derivative on ln (1+z) where z=0 always equal 1?

 

[d_leet]

wait won't the nth derivative on ln (1+z) where z=0 always equal 1?

No... because ypu end up with a number raised to a negative power and so it will osscilate between positive and negative values.

 

[Wishbone]

No... because ypu end up with a number raised to a negative power and so it will osscilate between positive and negative values.

yes but isn't it 1 to a negative power?

 

[d_leet]

yes but isn't it 1 to a negative power?

What is the derivative of something to a negative power though?

 

[Wishbone]

What is the derivative of something to a negative power though?

oh duh! man I do not know what is wrong with me tonight

 

[Wishbone]

hmmmm I did it out, and I didnt get the answer in the book, it says it should look like:

(-1)^{n-1}*z^n/n

 

[d_leet]

hmmmm I did it out, and I didnt get the answer in the book, it says it should look like:

(-1)^{n-1}*z^n/n

And what did you get?

 

[Wishbone]

\sum 0 + (1)z + \frac{\frac{z^2}{z(1+z)^3}}{2} + \frac{\frac{z^3}{-6(1+z)}{3*2}

 

[Wishbone]

\sum 0 + (1)z + \frac{(z^2)}{(2(1+z)^3)*2} + \frac{(z^3)}{(-6(1+z)^3)*3*2 }

 

[d_leet]

\sum 0 + (1)z + frac{1}{2(1+z)^3}(z^2)/2 + frac{1}{-6(1+z)^3}(z^3)/3*2

Is this what you meant? The formatting is horrible but it isn't right that much I can tell, how did you get that?

 

[Wishbone]

ug, you sorry about that is it clearer now?

 

[d_leet]

It's ugly as hell, how did you get that?

 

[Wishbone]

ok i took

F'(z)= 1/(1+z)^2
F''(z) = 1/2(1+z)^3
F'''(z) = 1/-6(1+z)^4

so I do F(0) + F'(z)*(z-0) + F''(z)*(z-0)^2/2! + F'''(z)*(z-0)^3/3!

 

[d_leet]

ok i took

F'(z)= 1/(1+z)^2
F''(z) = 1/2(1+z)^3
F'''(z) = 1/-6(1+z)^4

so I do F(0) + F'(z)*(z-0) + F''(z)*(z-0)^2/2! + F'''(z)*(z-0)^3/3!

Well you need to evaluate the f primes, and I still say it would be much eaiser to derive this from the geometric series.

 

[Wishbone]

Well you need to evaluate the f primes, and I still say it would be much eaiser to derive this from the geometric series.

well I don't know how to show natural logs as a geometric series, that's why i couldn't try that

 

[d_leet]

well I don't know how to show natural logs as a geometric series, that's why i couldn't try that

I never said to show a natural log as a geometric series, I said to derive it from one, what happens if you integrate the formula for the sum of an infinite geometric series.

 

[Wishbone]

I never said to show a natural log as a geometric series, I said to derive it from one, what happens if you integrate the formula for the sum of an infinite geometric series.

well it depends on what you start out with right?

 

[d_leet]

well it depends on what you start out with right?

I don't know what you mean.

 

[Wishbone]

I don't know what you mean.

well i guess, no, i don't know how to

 

[Wishbone]

im sorry, these topics were not covered in class.

 

[d_leet]

im sorry, these topics were not covered in class.

Well do you know what the formula for the sum of an infinite geometric series? Maybe try it this way, take the derivative of the ln(1 + z) and then try to find a geometric series for that and then integrate it to find one for the log...

 

[VietDao29]

Well do you know what the formula for the sum of an infinite geometric series? Maybe try it this way, take the derivative of the ln(1 + z) and then try to find a geometric series for that and then integrate it to find one for the log...

d_leet, I think it might be better if you hint him more clearly. AFAIK, he has tried his best to understand the problem. I haven't seen such a 3-page thread that reaches nowhere like this one for a long time.
Wishbone, I think you should re-read your textbook to understand the concept of Taylor's series a little bit more clearer. And try again to see if you can do this problem.
You, however can do it by using geometric series as d_leet has pointed out.
For a common ratio |z| < 1, we have:
\frac{a}{1 - z} = \sum_{k = 0} ^ {\infty} az ^ k, so apply it here, we have:
\frac{1}{1 + z} = \sum_{k = 0} ^ {\infty} (-z) ^ k, \ \forall |z| &lt; 1 (a geometric series with a common ratio -z, and the first term is 1).
Integrate both sides gives:
\int \frac{dz}{1 + z} = \int \left( \sum_{k = 0} ^ {\infty} (-z) ^ k \right) dz
Now, hopefully, you can go from here, right? :)

 

[assyrian_77]

I haven't seen such a 3-page thread that reaches nowhere like this one for a long time.

Then take a look at this thread below: "Urgend Geometric series question".

Sorry for the off-topic.

 

[shmoe]

ok i took

F'(z)= 1/(1+z)^2
F''(z) = 1/2(1+z)^3
F'''(z) = 1/-6(1+z)^4

so I do F(0) + F'(z)*(z-0) + F''(z)*(z-0)^2/2! + F'''(z)*(z-0)^3/3!

These derivatives are incorrect. If F(z)=ln(1+z) then:

F&#039;(z)=(1+z)^{-1}

try finding F'', F''', etc. again. Carefully.

 

[Axiom_137]

For a common ratio |z| < 1, we have:
\frac{a}{1 - z} = \sum_{k = 0} ^ {\infty} az ^ k, so apply it here, we have:
\frac{1}{1 + z} = \sum_{k = 0} ^ {\infty} (-z) ^ k, \ \forall |z| &lt; 1 (a geometric series with a common ratio -z, and the first term is 1).
Integrate both sides gives:
\int \frac{dz}{1 + z} = \int \left( \sum_{k = 0} ^ {\infty} (-z) ^ k \right) dz

This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

\int \left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz

ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}

Plug in z=0 and the sum is zero...My calculator says ln(i) is not zero. ln(i)= i*(pi/2)

So is there seems to be a problem using this method with complex variables.

(Or have I made a mistake with my math?)

 

[d_leet]

This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

\left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz

ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}

Plug in z=0 and the sum is zero...


My calculator says ln(i) is not zero. ln(i)= i*(pi/2)

So is there seems to be a problem using this method with complex variables.

(Or have I made a mistake with my math?)

I think that sum only holds for |z|<1, that is strictly less than one, whereas |i|=1 so one should not expect the equation to hold for z=i since i does not satisfy the necessary inequality.

 

[Axiom_137]

I think that sum only holds for |z|<1, that is strictly less than one, whereas |i|=1 so one should not expect the equation to hold for z=i since i does not satisfy the necessary inequality.

For the geometric series, the inequality for this problem would be :

|-iz|<1 or simply |z|<1

z=0 satisfies |z|<1

 

[d_leet]

For the geometric series, the inequality for this problem would be :

|-iz|<1 or simply |z|<1

z=0 satisfies |z|<1

Yes you're absolutely right, I have no idea what I was thinking, and I cannot think of a reason as to why the sum does not work other than the fact that the series does, at least, correctly calculate the real part of ln(i)

 

[HallsofIvy]

Please, tell us what you think the derivative of 1/(1+ z) is!

 

[Hurkyl]

This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

\int \left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz

ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}

Plug in z=0 and the sum is zero...

You forgot the constant of integration.