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

@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

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@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.

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

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. 😃

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:

DV = 897.3404255319 = 897.34

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:

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 14:52:16 UTC

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

My pleasure! @Deleted User

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.

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

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

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

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

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.

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.

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

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.

Do I have that right?

@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

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.

@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

That makes alot of sense.

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

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.

@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

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

Thanks for the help!

@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!) -+...)

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