Message from JC17-OR in MacGuyver - Skills & Academics #homework-help

2018-02-07 04:32:50 UTC  
2018-02-07 05:10:43 UTC  

@Deleted User My pleasure

2018-02-14 03:09:04 UTC  

I was wondering if someone could help me with a problem for my financial mathematics class?

2018-02-14 03:09:35 UTC  


2018-02-14 03:32:28 UTC  

@Tycho Brahe @Zyzz you guys too may have some insight for @Deleted User ?

2018-02-14 03:34:38 UTC  

I'll go ahead and post a screenshot of the problem so people could see if they would be able to help.

2018-02-14 03:34:57 UTC  

@Deleted User What's the question? If you drop it here then when someone who can help comes on they can work on it

2018-02-14 03:35:13 UTC  

yeah perfect

2018-02-14 03:35:22 UTC

2018-02-14 03:38:56 UTC  

Here are my notes. The example is supposed to align with how to do the problem, I just can’t seem to get the right answer.

2018-02-14 03:38:59 UTC

2018-02-14 03:39:44 UTC  

Let me know if my notes can't be read.

2018-02-14 04:02:15 UTC  

I'm going to get some rest for tonight, but if anyone would still want to look at it and give me some guidance, I would greatly appreciate it. 😃

2018-02-14 04:18:00 UTC  

First, find future value:
FV = Principle\*(1 + rate\*time)
FV = 875*(1 + .1025 * 2)
FV = 1054.375

Then in one year there will be one year on the note left, and the bank is going to discount that at a 17.5% rate, so the discounted value the bank pays satisfies:
FV = DV * (1 + .175 * 1)

Plugging in the value for FV from before:
1054.375 = DV * (1.175)

Dividing both sides by 1.175 we get:

2018-02-14 04:18:38 UTC  

DV = 897.3404255319 = 897.34

2018-02-14 04:20:24 UTC  

So the holder of the note went from 875 to 897.34 in one year, so he has his own future value calculation he can plug values into:
897.34 = 875\*(1 + rate_for_holder \* 1)

dividing both sides by 875 you get:

2018-02-14 04:21:31 UTC  

1.0255314286 = 1 + rate_for_holder
so rate_for_holder = 0.0255314286... ~~ 0.026
so the rate for the holder is 2.6%

2018-02-14 04:22:33 UTC  
2018-02-14 14:52:16 UTC  

Thank you! @ThisIsChris I think I understand it much better now.

2018-02-14 19:15:49 UTC  

My pleasure! @Deleted User

2018-02-19 17:11:50 UTC  

Hey everyone I need some math help. I'm trying to do the final problem and I can't figure it out.

The best I've come up with is that the 2S-S argument only works for 2S -S, if you go above 2S-S to say 3S-2S, then it diverges to infinity. Also the r-value of 2 cannot be used in the geometric series sum equation as it is not between -1&1.

Any help is super appreciated.

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?