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[hops]

Even then, I still don't know how to solve for N.

[hops]

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

[hops]

I'm going to fail this year... I guess it's understandable everyone gave up on me. I can't even solve a single question which is apparently incredibly easy to everyone else that it doesn't even need explanation...

[hops]

Please please please please help I just want to know what to do I'm racking my brain all day all night just trying to even get through one damn question I know I'm dumb please HELP ME

[hops]

Therefore any N such that |a_N| = |s_N+1 - s_N| < 10^-2 will do the trick.

How would I even work out N for a)?

I know I've been super dumb so far but I just want to know this one thing. All I want is just to be able to survive this semester.

[Strife26]

Don't panic Cinder!

I'm pretty sure that you're going to need to take the natural log of both sides, but I'm not familiar enough with ln to know how to proceed from there

[TheDarkStar]

Therefore any N such that |a_N| = |s_N+1 - s_N| < 10^-2 will do the trick.

How would I even work out N for a)?

I know I've been super dumb so far but I just want to know this one thing. All I want is just to be able to survive this semester.

Symbolically find an expression for |s_N+1 - s_N| for a generic s_N| and then solve |s_N+1 - s_N| < 10^-2. Basically you're trying to find an N for which the difference between s_N+1 and s_N is less than 1/100.

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

This is more difficult. Actually, I'm not sure if it's even algebraically solvable. Are you sure you need to solve it? What's the context for this equation?

[hops]

I talked to the teacher and she said that since N is a natural number, I can just solve for it numerically...

[Reelya]

Yeah, that type of problem cannot normally be solved algebraically. For example x = 10^x can only be approximated with computation.

Consider substituting M = N + 1 however, and you get this relationship:

50M <= 2^(M-1)

Which might make things easier to get a feeling for. e.g. when "M" = 10, then 50M is 500 and 2^9 = 512, so the answer is pretty close to 10 (N = 9), and you can iterate approximations for as many decimal places as you like.

[MagmaMcFry]

Guys, you don't actually need to find the smallest N for which this holds. All you need to do is find a number N such that this is true. Pick a number, see if it works. If it doesn't, pick another number.

In problem (a), a₁₀ = 10/1024 < 1/100, and since a_10 is the difference between s_10 and s_9, we can choose N=10 and we're done.

So, okay, the series limit lies between s_n and s_n+1, but so what?

The series value, as well as all partial sums s_k for k>n, all lie between s_N and s_N+1. This is an interval of width less than 1/100. Any two values inside this interval must have a difference of less than 1/100.

[hops]

How do I find the partial sum of the Basel Problem?

[MagmaMcFry]

That is an advanced wizardry problem, not something which you can really work towards, so you have two options. Option 1: You become a math wizard and stumble upon a solution, generally by accident while messing around with something else (half of the wizard part is noticing what you just found). Option 2: you google "partial sums of the Basel problem" to find out if anyone else has done Option 1 before for your particular problem.

[hops]

http://imgur.com/x4q8YAW

This is physics but, does anyone know how they got 5.10 from 5.9? Is this a property of integrals?

[Helgoland]

Yeah. Differentiation is (roughly) inverse to integration.

[hops]

Yeah. Differentiation is (roughly) inverse to integration.

Ehm, what? I don't see how that applies. Why does the integral suddenly become q(s) when divided by delta-s as it approaches 0? And what happened to the rest of the Taylor series expansion?

Actually, I can probably figure out how they got q(s), I just don't really get how they got rid of the Taylor series expansion, since they didn't say this is an approximation I'm assuming they magicked it out somehow.

EDIT: OH