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Author Topic: Mathematics Help Thread  (Read 226926 times)

Arx

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Re: Mathematics Help Thread
« Reply #2295 on: December 06, 2016, 09:08:25 am »

Since the IMV theorem says that functions can't teleport, if something is bigger than the local max or smaller than the local min, it's not in the interval.

The EVT predicts the existence of a maximum and/or minimum given that the function is continuous on [a,b] and differentiable on (a,b).

Fermat's theorem says that at an extreme value f(c), f'(c) = 0.

That should give you all the tools you need to dismantle and solve this problem, along with a little basic calculus.
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Re: Mathematics Help Thread
« Reply #2296 on: December 06, 2016, 09:17:13 am »

goddammit, I'm not going to wrap my head around this fast enough for the homework deadline

how would i know what the local maximum or minimum is without differentiating?

Any continuous function defined on a compact set has a minimum. This is a theorem you should have heard about by now.
why?
« Last Edit: December 06, 2016, 09:18:52 am by Cinder »
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Arx

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Re: Mathematics Help Thread
« Reply #2297 on: December 06, 2016, 09:23:37 am »

Oh, right! Well, there are three things that can be minima: the endpoints and the EVs. You have the endpoints, so if you find those I have a suspicion one of them will have f(a) = 4, and the other will be larger. That implies that for a point to be a minimum, it must be less than or equal to that. Do you follow?
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Re: Mathematics Help Thread
« Reply #2298 on: December 06, 2016, 09:27:29 am »

...Sorry if I'm being obtuse, but how would I find the extreme value... without differentiating? Or I don't know. I'm just trying to prove a fucntion go below a value. But I don't know what the minimum is. And I can't differentiate.
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Re: Mathematics Help Thread
« Reply #2299 on: December 06, 2016, 09:32:05 am »

I'm really sorry about this, guys, this is why I avoided asking auestions and tried to learn on my own, but there are just so many things I don't understand that I have completely lost track of where to even start.
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Arx

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Re: Mathematics Help Thread
« Reply #2300 on: December 06, 2016, 09:33:44 am »

The point here is that you don't need to find the extreme value - just demonstrate that some c exists so f(c) is a minimum (this is already proved by the EVT), and that it must be <= 4.

Step one to finding a extreme on an interval [a,b] is always to check the endpoints, in this case -2 and 2. Doing so reveals that f(a) and f(b) are ~60some and 4.

This implies that if there's a minimum value in the interval (if it's like a hanging chain shape, increasing towards 2) it must be less than 4. Does that make sense? Try drawing a few sketch graphs if it's still not seeming right.
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Re: Mathematics Help Thread
« Reply #2301 on: December 06, 2016, 09:34:30 am »

Uh, I'm asking abot question ii if it wasn't clear.
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da_nang

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Re: Mathematics Help Thread
« Reply #2302 on: December 06, 2016, 09:35:48 am »

You don't need to find the extrema, just the properties of the minimum.

Spoiler: hint for i (click to show/hide)

Spoiler: hint for ii (click to show/hide)
« Last Edit: December 06, 2016, 09:41:06 am by da_nang »
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Re: Mathematics Help Thread
« Reply #2303 on: December 12, 2016, 04:05:03 pm »

Hi, it's me again. I understand the concept of summing up algebraic series. I understand the concept of summing up alternating series. I understand the concept of summing up geometric series. I.... have no idea what to do to sum up S or Sn in a). It's not like I can split them apart...



For that matter, can I have some tips on understand mathematics? I understand the theory, but... gah. My homework seems to exclusively involve something that push the limit of the theorems' application.

EDIT: Oh, it was a power series... I still would like an explanation though.
« Last Edit: December 12, 2016, 04:50:46 pm by Cinder »
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Re: Mathematics Help Thread
« Reply #2304 on: December 12, 2016, 06:28:57 pm »

This problem does not require you to calculate the limit of the series. It just wants you to find out whether such a limit exists, in the regular and absolute sense.

For convergence, notice that in both cases the individual summands have alternating signs and their absolute values decrease monotonically to zero, so by Leibniz's rule, these series both converge.

For absolute convergence, by definition you simply need to check whether the sum of all an is finite. You can prove this by approximating an from above with the summands of a series you know to be bounded, or disprove it by approximating an from below with the summands of an series you know to be unbounded.

To find this N, have a closer look at the proof of Leibniz's rule. If sn are the partial sums of a series satisfying Leibniz's criterion (summands alternate and their absolute values go monotonically to zero), then for any m > n, sm must lie in between sn and sn+1, and of course the series limit must also lie in between. Therefore any N such that |aN| = |sN+1 - sN| < 10-2 will do the trick.
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Strife26

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Re: Mathematics Help Thread
« Reply #2305 on: December 12, 2016, 07:34:02 pm »

Alrighty, I really need to start learning my trig functions and their derivatives. Final is in an hour and a half.

Anyone around who can explain what the heck the 2nd Fundamental means in practice? I get that it's summation is

F(x) = integral from a to X of f(t)dt
then
F'(x) = f(x)

but I can't seem to wrap my head around what this is actually meaning or used for.
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Culise

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Re: Mathematics Help Thread
« Reply #2306 on: December 12, 2016, 08:01:23 pm »

It's been a while for me, but isn't what you just said the first fundamental theorem of calculus?  The second is:
If f is a continuous real-valued function on some closed interval [a, b] with the anti-derivative F defined for all x in that same closed interval:
abf(x)dx = F(b) - F(a)
Basically, the second fundamental is a quick and elegant way to solve an integral on a closed interval by simply solving the anti-derivative for its end points.  For example, if you're looking for the integral of f(x) = 2x between 1 and 5, you just need an antiderivative; for instance, we can use the simplest such antiderivative F(x) = x2.  With that, you shove both ends of the interval in there to get F(5) - F(1), subtract the start from the end, solve each function (25 - 1), and pop out with 24. 

The first fundamental theorem, in case you meant to ask about that, is simply something that tells you that integration and derivation are actually tied together as flip sides of the same thing by way of the anti-derivative, in an analogous sense to multiplication and division or addition and subtraction  Basically, if you've got the integral of some blah (f) with respect to some ugh (x), and you take the derivative of that integral with respect to the same ugh, you pop out the original blah.  As for what you use it for, if you ever want to clean out an integral to get at the equation inside, such as in solving equations, you can do so by taking the derivative.  Or, on the flip side, if you want to get inside of a derivative, take the integral. 

Another thing that the first part tells you is that if you've got a continuous function, there is *some* set of functions out there that are anti-derivatives of it.  It doesn't even have to be continuous over the entire system of real numbers; as long as there's an interval of continuity, that interval has some set of anti-derivatives.  As well, if you ever want to find the area under a line on a chart between two points, that's another use for your anti-derivative.  More generally, calculus is used all over the place, and the fundamental theorem of calculus is at its heart.
« Last Edit: December 12, 2016, 08:07:38 pm by Culise »
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hops

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Re: Mathematics Help Thread
« Reply #2307 on: December 13, 2016, 07:00:09 am »

This problem does not require you to calculate the limit of the series. It just wants you to find out whether such a limit exists, in the regular and absolute sense.

For convergence, notice that in both cases the individual summands have alternating signs and their absolute values decrease monotonically to zero, so by Leibniz's rule, these series both converge.

For absolute convergence, by definition you simply need to check whether the sum of all an is finite. You can prove this by approximating an from above with the summands of a series you know to be bounded, or disprove it by approximating an from below with the summands of an series you know to be unbounded.

To find this N, have a closer look at the proof of Leibniz's rule. If sn are the partial sums of a series satisfying Leibniz's criterion (summands alternate and their absolute values go monotonically to zero), then for any m > n, sm must lie in between sn and sn+1, and of course the series limit must also lie in between. Therefore any N such that |aN| = |sN+1 - sN| < 10-2 will do the trick.
My main issue is with the N, because I'm not sure how I can meaningfully choose an N without being  able to calculate the sum. I understand the Leibniz Criterion but could you explain how you get |sN+1 - sN| < 10-2? I can't really see any meaning for "value of the series" in this context other than the infinite series, and so shouldn't the condition be |sn - s| < 10-2 for n >= N?

EDIT: I think I'm able to parse all the explanation up to the last one... So, okay, the series limit lies between sn and sn+1, but so what?
« Last Edit: December 13, 2016, 07:27:46 am by Cinder »
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Re: Mathematics Help Thread
« Reply #2308 on: December 13, 2016, 09:13:13 am »

I really still don't know how to find N without calculating the sum...
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Re: Mathematics Help Thread
« Reply #2309 on: December 13, 2016, 09:37:24 am »

I just really want to understand this one bit... I'm sorry I always post so much on this thread with stupidly easy problems, but I'm really struggling, guys.
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