General

I think i found my long lost brother

@Oyster: @ulti25: @bloodIsShed:

Yeah, I think I've got it.
Thanks a ton

[url=http://i547.photobucket.com/albums/hh471/someStriker/de_sum.jpg]sorry for the horrible handwriting[/url]
hopefully this makes it easier to understand.... maybe

[quote=SunsetDews]Ah, okay, I think I see what you're saying, but would you really have to move the term up by 1?
I guess it'd seem kind of pointless to have the first term equal to 0, but it's not like it would mess anything up would it?[/quote]
A series normally starts with a constant, then with the powers of x increasing.
We move up the n term so we don't have zero. In your case, it probably won't matter.
However in solving differential equations using the power series, [b]yes it is important and it does matter[/b].

[quote=SunsetDews]Ah yeah I think I got it. I took your suggestion and just wrote out the terms for a bit to see what you were saying.
Don't quite see what you're saying near the end but I'll try it out myself and see what I can figure out
Thanks a lot
Ah, okay, I think I see what you're saying, but would you really have to move the term up by 1?
I guess it'd seem kind of pointless to have the first term equal to 0, but it's not like it would mess anything up would it?[/quote]

When you get around to actually differentiating/integrating the functions yourself, just write out a couple of terms and it should be clear why convergence at endpoints might change with differentiating/integrating.

As for what you asked Oyster, no, it probably wouldn't really matter at the moment, but your teacher will throw a fit if you don't move up the lower limit by 1. When I went through it, I was taught the same thing and thought what you just asked too. It's just an acknowledgement that the derivative of a constant is 0 and is pointless to include, so might as well bring up the lower limit.

omg this math is so hard

[quote=ulti25]The key observation is pretty much that you're going to be differentiating/integrating with respect to x and that only a part of the function is really dependent on x.

If you were to write out the terms up until some finite n, you'd notice that the n will be replaced by some number and will simplify to yield a constant for that value of n.

The (-1)^(n + 1) and (n + 1) don't depend on x and can be treated as a constant when differentiating/integrating. That leaves you with the (x - 1)^(n + 1) which is the only part of the function that's dependent on x, which you just hit with the power rule (or whatever it's called)

If differentiating:
(x - 1)^(n+1) = (n + 1)(x - 1)^n

If integrating:
(x - 1)^(n+1) = [(x - 1)^(n + 2)]/(n+2) + C

So, for example, the derivative of f(x) would be (((-1)^(n+1))(n + 1)(x - 1)^n)/(n + 1)

Just remember to add the + C for integration outside of the summation.

As far as convergence goes, the radius of convergence will be the same (know what that is?) for f(x) or any of its derivatives/integrals, but convergence at endpoints might change. Essentially, since you're adding a factor to the numerator when you differentiate, an endpoint that previously converged might now diverge. On the other hand, when you integrate, you're adding a factor to the denominator. An endpoint that previously diverged might now converge due to that additional factor.[/quote]

Ah yeah I think I got it. I took your suggestion and just wrote out the terms for a bit to see what you were saying.
Don't quite see what you're saying near the end but I'll try it out myself and see what I can figure out
Thanks a lot

[quote=Oyster]My bad, scratch that. You probably don't need to worry about that.
~~~
You know how in series, you have summation of n, with n starting at 0 and it approaches infinity at the upper limits,
well depending on how it is written,
ie: f(x) = Sum[x^n], {n,0,infinity}
__f'(x) = Sum[n*x^(n-1)], {n,1,infinity} you would have to move the n term up by 1, b/c if n=0 the first term in the series is 0.[/quote]

Ah, okay, I think I see what you're saying, but would you really have to move the term up by 1?
I guess it'd seem kind of pointless to have the first term equal to 0, but it's not like it would mess anything up would it?

Power series...

If in the form 1/n^p, and P is greater than or equal to one, then the series converges to (1/P-1). If P is between 0 and 1, then the series diverges.

[quote=SunsetDews]What do you mean by {n, 0, infinity} and {n, 1, infinity} mean?
Sorry I'm just unfamiliar with the way it's written[/quote]
I know it's confusing the way it's written b/c it's written on the bottom and top of the summation sign.
n=0 -> lower bound
n=infinity -> upper bound

You know how in series, you have summation of n, with n=0 at the bottom of the summation sign and infinity at the top of the summation sign.
well depending on how it is written,
ie: f(x) = Sum[x^n], {n,0,infinity}
__f'(x) = Sum[n*x^(n-1)], {n,1,infinity} you would have to move the n term up by 1, b/c if n=0 the first term in the series is 0.

[quote=Oyster]You would differentiate/integrate how you would normally do it, with respect to x:

f(x) = Sum{(-1)^(n+1) * [(x-1)^(n+1)] / (n+1)}, with {n, 0, infinity}
f'(x) = Sum{(-1)^(n+1) * [(x-1)^(n)]}, with {n,1,infinity}

However, for series if f(x) is the summation of n, with [b]n=0 going to infinity[/b]
_________________f'(x) is the summation of n, with [b]n=1 going to infinity.[/b][/quote]

What do you mean by {n, 0, infinity} and {n, 1, infinity} mean?

Sorry I'm just unfamiliar with the way it's written

The key observation is pretty much that you're going to be differentiating/integrating with respect to x and that only a part of the function is really dependent on x.

If you were to write out the terms up until some finite n, you'd notice that the n will be replaced by some number and will simplify to yield a constant for that value of n.

The (-1)^(n + 1) and (n + 1) don't depend on x and can be treated as a constant when differentiating/integrating. That leaves you with the (x - 1)^(n + 1) which is the only part of the function that's dependent on x, which you just hit with the power rule (or whatever it's called)

If differentiating:
(x - 1)^(n+1) = (n + 1)(x - 1)^n

If integrating:
(x - 1)^(n+1) = [(x - 1)^(n + 2)]/(n+2) + C

So, for example, the derivative of f(x) would be (((-1)^(n+1))(n + 1)(x - 1)^n)/(n + 1)

Just remember to add the + C for integration outside of the summation.

As far as convergence goes, the radius of convergence will be the same (know what that is?) for f(x) or any of its derivatives/integrals, but convergence at endpoints might change. Essentially, since you're adding a factor to the numerator when you differentiate, an endpoint that previously converged might now diverge. On the other hand, when you integrate, you're adding a factor to the denominator. An endpoint that previously diverged might now converge due to that additional factor.

search up Patrickjmt in utube, this guy's channel has like everything on calculus.

You would differentiate/integrate how you would normally do it, with respect to x:

f(x) = Sum{[(-1)^(n+1)/(n+1)]*[(x-1)^(n+1)]}
f'(x) = Sum{(-1)^(n+1)*[(x-1)^(n)]}

write clearer notes