# Message from ThisIsChris in MacGuyver - Skills &amp; Academics#homework-help

2018-02-19 22:46:28 UTC

If you talk about 3S you're just bringing more things in that you don't need.

2018-02-19 22:48:32 UTC

when you do 2S-S you should have the same number of elements on both sides, or at least write it in the form of a sum

2018-02-19 22:49:31 UTC

If you subtract these two sums, you should subtract the first from the first, the second from the second, etc.

2018-02-20 01:02:43 UTC

So what you're saying is that the proof relies on ignoring standard rules of subtraction with equal terms.

2018-02-20 01:06:33 UTC

It makes the assumption that you can subtract the 2+4+8+... of the S from the 2S while leaving the 1 untouched which negates the idea of infinity because it is adding terms to S that it is not adding to 2S to leave the 1 untouched.

2018-02-20 01:11:27 UTC

When they subtract terms in the example, they're kind of cherry picking which to subtract first. It's not written in a strict mathematical way, so it flies under the radar. Imagine you said s = 1+1+1+..... You could say s -s = (1+1+...) -(1+....) = 1, yet s-s should be 0.

2018-02-20 01:11:42 UTC

So you can point out that it's not written in a very precise way.

2018-02-20 01:20:04 UTC

That makes sense. The proof isn't valid because it relies on a specific way to subtract the sums which doesn't follow mathematical rules.

2018-02-20 01:20:14 UTC

Do I have that right?

2018-02-20 01:25:12 UTC

@JC17-OR correct, the proof relies on the sum being the same if you are allowed to rearrange terms, but you are not guaranteed to be able to do that if the series is not absolutely convergent

2018-02-20 01:26:01 UTC

Or you could say that they are using a different number of terms in each series in order to get the result that did. Which means that they didn't acttually double s, since s and 2s should have the same number of terms. If both series have the same number of terms, they end up with 2s-s = 2^N - 1, where N is the number of terms.

2018-02-20 01:30:12 UTC

@JC17-OR You can also just prove that the series diverges just by making the terms rigorous:
Define S_n = sum i from 0..n of i^2
Then S = lim n->inf S_n

S_n you can compute explicitly, because it is a finite sum you can do:
2\*S_n - S_n = 2\*(n+1) - 1
i.e. S_n = 2^(n+1) - 1

so S = lim n-> inf 2^(n+1) - 1 = inf

2018-02-20 01:30:47 UTC

That makes alot of sense.

2018-02-20 01:31:32 UTC

Any thoughts on the 4th problem. My teacher hasn't been much help and my classmates are as lost as I am.

2018-02-20 01:36:48 UTC

I figured I could take out the 1/sqrt(2pi) and have the integral e^(-x^2/2), integrate that and do the Taylor series that way, but that didn't work out.

2018-02-20 01:38:21 UTC

@JC17-OR Unforunately I have to eat dinner now, but here is the lead, the derivative of F is f(x), but you know the taylor series for f(x), just take the taylor series for e^x and plug in (-x^2) where x is

2018-02-20 01:39:50 UTC

So the taylor series for F is just the antiderivative of the taylor series for e^x * 1/sqrt(2pi) with (-x^2) plugged into x

2018-02-20 01:43:46 UTC

Thanks for the help!

2018-02-20 01:50:16 UTC

@JC17-OR You're welcome! Here's the full demonstration in case you want it:
F is the antiderivative of f so first find the taylor series of f:
f(x) = (1/sqrt(2pi)) * e^(-x^2)
taylor series for e^y is:
e^y = 1 + y + y^2/2 + y^3/3! + y^4/4! +...
plug in y=-x^2
e^(-x^2) = 1 - x^2 + x^4/2 - x^6/3! + y^8/4! -+...
so f(x) = (1/sqrt(2pi) * (1 - x^2 + x^4/2 - x^6/3! + y^8/4! -+...)
so F(x) = (1/sqrt(2pi)) \* (x - x^3/3 + x^5/10 - x^7/(7\*3!) + x^9/(9\*4!) -+...)

2018-02-20 02:33:10 UTC

Does anybody have an idea for the power series from the given Taylor series?

2018-02-20 05:38:33 UTC

@JC17-OR do you mean the radius of convergence?

2018-02-20 05:39:15 UTC

because the taylor series *is* the power series

2018-02-20 05:42:21 UTC

Anyway I forgot to mention the radius of convergence is infinity, because f(x) is a probability density function, so the integral of f(x) for x from -infinity to infinity is 1.

2018-02-20 15:13:46 UTC

I'm dumb that's what I meant to say. I figured it out, thanks for all your help.

2018-02-20 16:18:30 UTC

@JC17-OR You're welcome!

2018-03-05 22:38:10 UTC

does anyone know anything about lognormal and weibull distributions?

2018-03-05 23:50:38 UTC

<@&387091385075105804> ^^^

2018-03-25 21:04:05 UTC

Need some help with a math problem @here

2018-03-25 21:05:55 UTC

I have to solve a linear inequality problem:
A delivery driver makes \$52 each day that she works and makes approx. \$8 in tips for each delivery. If she wants to make \$220 in one day at least how many deliveries does she need to make?

2018-03-25 21:10:11 UTC

Isn't that just 220 = 52 + 8x

2018-03-25 21:10:16 UTC

Is the \$52 a base pay?

2018-03-25 21:10:34 UTC

If so @Jacob is correct

2018-03-25 21:10:55 UTC

Is is not (220 - 52) \ 8 = x

2018-03-25 21:11:54 UTC

Oh jacob beat me to it. Jacob wrote the more proper expression tbh.

2018-03-25 21:12:23 UTC

Do you solve for x or just leave it as that expression?

2018-03-25 21:12:42 UTC

Solve for x

2018-03-25 21:12:46 UTC

You solve for X but he got you started.

2018-03-25 21:13:40 UTC

Since you get "approximately 8 dollars per tip' the answer will be "about" (whatever x is ) deliveries

2018-03-25 21:14:11 UTC

Okay that makes sense.

2018-03-25 21:14:25 UTC

Thanks. Your goy here can't do maths to save himself

2018-03-25 21:15:43 UTC

Everyone has strengths and weaknesses. This one's pretty basic though. I used Kahn Academy alot in college when I had a foreign teacher who didn't explain things in an articulate manner.