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[Helgoland]

Heh, I didn't even spot the additional summands

[RedWarrior0]

E: major overhaul of the comment, I misunderstood something initially.
All the terms are divided by delta s. But it's a Taylor series, so most of the extra terms are being multiplied by (delta s) to some higher power, so they go to 0 as delta s goes to 0. The first two terms of the Taylor series are the only ones with less than a (delta s)^2. The first cancels with the preexisting negative term, and the second is the dQ/ds term after dividing by delta s. Then there's the preexisting integral term.

All that remains is changing (1/(delta s) * the integral term) into q(s). The limit as delta s goes to 0 makes this 0/0, which is an indeterminate form. I haven't really put effort into solving it, but you should have seen ways to deal with indeterminate forms.

[hops]

Can somebody explain to me how they got 2 from 1 in this?

http://imgur.com/cVmTw1B

[Shadowlord]

As with everything you've been talking and asking about, none of it makes any sense to me. 

[inteuniso]

Can somebody explain to me how they got 2 from 1 in this?

http://imgur.com/cVmTw1B

(a-b)² may be rewritten as a²-2ab+b².

EDIT: I'm stupid.

[da_nang]

How do I solve for N in 50N + 50 <= 2^N?

Not sure if relevant anymore, but hey, it's solvable but not with elementary functions. First things first, we know that 2^N is convex. Since 50*0 + 50 > 2⁰, we know that 50N + 50 intercepts 2^N exactly twice due to the latter being convex. These intercepts occur for one positive N and one negative N. Call them N₁ and N₂ respectively. Thus the region for which 50N + 50 <= 2^N holds will be N <= N₂ or N >= N₁. The get the actual values, you can either solve them numerically...

Spoiler: Or solve the intercepts symbolically: (click to show/hide)

50N + 50 = 2^N <-> 50N + 50 = e^{N ln 2} <-> (N + 1)e^{-N ln 2} = 1/50 <-> (-N ln 2 - ln 2)e^{-N ln 2} = - (ln 2)/50 <-> (-N ln 2 - ln 2)e^{-N ln 2 - ln 2} = - ((ln 2)/50) e^{- ln 2}

<-> -N ln 2 - ln 2 = W_k(- ((ln 2)/50) e^{- ln 2}) <-> N= -W_k(- ((ln 2)/50) e^{- ln 2})/(ln 2) - 1 <-> N₁ = -W_-1(- ((ln 2)/50)e^{- ln 2})/(ln 2) - 1 , N₂ = -W₀(- ((ln 2)/50) e^{- ln 2})/(ln 2) - 1

Where W_k(x) is the kth branch of the Lambert W function. Only W₀(x) and W_-1(x) are real-valued. W₀(x) is defined for x in [-1/e,∞) and is positive for x > 0, where as W_-1(x) is defined for x in [-1/e,0) and is always negative. In fact, W_-1(x) <= W₀(x) for the common domain.

Though it's highly unlikely to for one to solve it symbolically outside of specialist settings.

[hops]

Can somebody explain to me how they got 2 from 1 in this?

http://imgur.com/cVmTw1B

(a-b)² may be rewritten as a²-ab+b².

No, what confuses me is the absolute and the Re(uz₀). How am I supposed to expand an absolute? Triangle inequality and all that.

I mean... like, if I have |a+b|² I can't just rewrite it as |a|² + |b|² - 2|ab| because if I say a = 1 b = 1 then
|1+1|² = |1|² + |1|² - 2|1|
4 = 0

And that's not even what it looks like in the equations.

Sorry if I sound hosstile but I'm kind of annoyed that instead of actually looking at my question you just went to assuming I don't understand how polynomials fucking work.

EDIT: And (a-b)² may be rewritten as a²-2ab+b². So you didn't even get it right in assuming I'm stupid.

[hops]

This is what I did to try to make the numbers work. http://imgur.com/dmVWPTA

I understand polynomials.

[da_nang]

Can somebody explain to me how they got 2 from 1 in this?

http://imgur.com/cVmTw1B

(a-b)² may be rewritten as a²-ab+b².

No, what confuses me is the absolute and the Re(uz₀). How am I supposed to expand an absolute? Triangle inequality and all that.

Simply put, |x|² = x². As for the real part, well, I haven't worked with complex numbers like that before so I'm not sure what's going on there unless there's some context I'm missing.

[MagmaMcFry]

Can somebody explain to me how they got 2 from 1 in this?

http://imgur.com/cVmTw1B

Suppose a and b are complex numbers (represented by vectors in the complex plane) that have the angle C between them. The cosine theorem (in the triangle with corners 0, a, and b) says that |a-b|² = |a|² + |b|²  - 2|a||b|cos C. Now note that |a||b|cos C = Re(āb), so by letting a = ū and b = |u|²z₀ we see that the equations (1) and (2) are equivalent.

[hops]

HOw do I do this? I know how to expand it into geometric series but it wants me to expand it into power series. And why would I even need to use partial fraction decomposition if I'm expanding them into a power series? I just can't see how that would help.

[hops]

How do i find all the values of z as a complex value for sinh(z)^2 = -1/2?

[RedWarrior0]

First one: I'm not certain, but I think the reason your book/instructor say to do partial fraction decomposition is probably with the idea that it's easier to find the power series for 1/(z+2) and 1/(z-2) individually than it is to find the power series for 1/(z^2 -4). Not sure, but that's my thoughts. Not sure I can be of much help on this one, manipulating series hasn't ever been something I've been particularly good at.

Second: Were you given any explanation of the hyperbolic trig functions? The identity sinh(i z) = i sin z can help you here, but I don't know if you're expected to use a different method.

[hops]

The complex logarithm is defined for all complex numbers, right?

[TheDarkStar]

The complex logarithm is defined for all complex numbers, right?

It's defined for all values in C except for real numbers less than or equal to zero.