Math thread! Come in and have some pi...

[PrimeraRS]

The way I explain it is:

Imagine if the independent variable (x) is very large. Very, very large. About what would you expect the output to be?

ex: y = 3 + 1/x^3

if x is very large, 1/x^3 is really, really, tiny. So the output is about 3+tiny ~ 3

ex: if y = (x^2+x-1)/(2x^2+5)

if x is very large, in the numerator, x^2 "takes over". the "+x-1" matters very little compared to x^2. Similarly the 2x^2 far out-measures the "+5". so the output should be about (x^2)/(2x^2) = 1/2.

ex: if y = (x^3+x^2-1)/(2x^2+x+10)

from the reasoning of the last example, if x is very large, we have y ~ x^3/(2x^2) = x/2. so its about half of something very large which is very large. So y -> infinity.

To find the "best" approximate of the output, long division is used to find the "slant asymptote"

new avi is looking joocy

[SCAR-H]

Hey guys, wondering if you could help me understand something.

I was doing my hwk for pre-calc. And I don't understand why the Horizontal asymptote for the following problem is y=1



From my understanding, wouldn't that mean that the +1 shifts the graph up by 1? or does it give a H.A.?

or am I just doing it wrong.

(1) As you move farther from 3 in either direction, 2 / (x - 3)^2 approaches ZERO
(2) As you move closer to 3 from either direction, 2 / (x - 3)^2 grows without bound

in (1): 1 + smaller and smaller number
in (2): 1 + larger and larger number

Visual

[Chesssbrah]

Richard Feynman on a probabilistic argument regarding Fermat's Last Theorem (before Andrew Wiles' proof), interesting read: http://www.lbatalha.com/blog/feynman...s-last-theorem

[Slacker23]

Richard Feynman on a probabilistic argument regarding Fermat's Last Theorem (before Andrew Wiles' proof), interesting read: http://www.lbatalha.com/blog/feynman...s-last-theorem

Amusingly enough, a formal version of this was in fact proven and was a fairly major result towards FLT.

A few years before FLT was proven, it was proven that the set of integers for which it holds has uniform density 1.

Uniform density is a bit technical to define, but what it basically means is that - regardless of whether FLT is true or not - if you picked a random n different from 2, the probabiliy that x^n+y^n=z^n have any integer solutions is 0.

[killakms]

Need help bruhs



wut i did: (3(b)-4)-(3(-5)-4)/b+5
Calculated to 3b+15/b+5. Then calculated the same terms and got 6

[EmperorRyker]

Need help bruhs



wut i did: (3(b)-4)-(3(-5)-4)/b+5
Calculated to 3b+15/b+5. Then calculated the same terms and got 6

You made a mistake in the last step. How did you get 6? Factor out 3 in 3b+15, and then see what you get for (3b+15)/(b+5).

[killakms]

You made a mistake in the last step. How did you get 6? Factor out 3 in 3b+15, and then see what you get for (3b+15)/(b+5).

oh ok. Didn't know I had to factor. I just took 3b divided it by 3 and 15 divided it by 5 and result gave me 3+3 which gave me 6. Thanks

[Thiiisguy]

Thank you all for the help in my last problem, I understand it now.

I have another question though.

I'm doing a problem where I have two vertical asymptotes at X= -4, and 0 and A horizontal Asymptote of Y.

How come the graph can intersect the H.A. inbetween both V.A.? but not on the othersides of each VA???

I thought a H.A. was across the whole graph. So shouldn't the graph not intercent H.A =0 in the middle?

[EmperorRyker]

Thank you all for the help in my last problem, I understand it now.

I have another question though.

I'm doing a problem where I have two vertical asymptotes at X= -4, and 0 and A horizontal Asymptote of Y.

How come the graph can intersect the H.A. inbetween both V.A.? but not on the othersides of each VA???

I thought a H.A. was across the whole graph. So shouldn't the graph not intercent H.A =0 in the middle?

Don't think of the horizontal asymptote as a line the graph can't touch. It only tells you how the function behaves when x goes to positive or negative infinity (i.e. when x is very large or very negative). So if you have a horizontal asymptote y = 2, then your function evaluates to approximately (or exactly) 2 when x goes to infinity. But the asymptote tells you nothing about the behavior of the function "in the middle". So it can touch it, cross it or stay far away from it there. But ultimately it's a description of the function's asymptotic behavior, asymptotic in this case referring to infinity.

On the other hand, a function by definition of a vertical asymptote can't touch the latter. It's not defined at the x value of that asymptote and it also blows up or down on either side.

[Chrysippus]

Are you in grad school?

Yeah mane, I have my bachelor's in math and am a ways in for phd. I grace this thread with my godlike presence every so often.

But I have been drinking now so I can't math.

But fuking lmao at alexandrov topologies ****en mirin.

Here's a question I asked/investigated/conjectured/proved independently in undergrad that you guys might find intradesting.

The circumference (perimeter) of a circle, expressed in terms of the radius, is the derivative of the area, expressed in terms of radius, with respect to radius.

Conjecture something about how this relates to all regular polygons, and prove your Conjecture.

Dead ****en serieos

[killakms]

Dont understand where i went wrong here. Pls halp..




My work:

i forgot to add all over b-0

I added common denominators and canceled

[Chrysippus]

^ simplify and do all the steps. 1/(0-4) = -1/4. you're making big mistakes across a single equal sign (the first one). 1/(0-4)*(b-4) is not -4 (though i cant see all your work so i dunno). and (b-4) isn't the LCD of 1/(b-4) and 1/4. you;re also canceling "0-4" with "b-4"

I was drunk in my last post, but consider that problem I posted. It's pretty and not talked about (sort of a recreational problem). If you're unaware.

For example. For an equilateral triangle with side a, A=(sqrt (3)/4)a^2 and P=3a.

Now let a = nr, write A and P in terms of nr.

Set dA/dr equal to P.

Now solve for r in terms of a.


You're gonna smile.

[EmperorRyker]

I was drunk in my last post, but consider that problem I posted. It's pretty and not talked about (sort of a recreational problem). If you're unaware.

For example. For an equilateral triangle with side a, A=(sqrt (3)/4)a^2 and P=3a.

Now let a = nr, write A and P in terms of nr.

Set dA/dr equal to P.

Now solve for r in terms of a.


You're gonna smile.

It seems like an elementary Calc I thing. Even Stewart hints at it hard.

[Chrysippus]

It seems like an elementary Calc I thing. Even Stewart hints at it hard.

It's elementary, but a cute result.

Spoiler below:

Spoiler!

For any regular polygon, the derivative of area is always the perimeter, provided we express area and perimeter as functions of the radius of the inscribed circle. This follows in higher dimensions also.

[EmperorRyker]

It's elementary, but a cute result.

Spoiler below:

Spoiler!

For any regular polygon, the derivative of area is always the perimeter, provided we express area and perimeter as functions of the radius of the inscribed circle. This follows in higher dimensions also.

Yeah, it's definitely a cute result, and it seems to follow directly from noting that if you were integrating for the area, you'd be adding things that look like the perimeter and are infinitesimally thin.

[killakms]

wth mane...



My Work...

[EmperorRyker]

@killakms, where did you get that h in the denominator? Other than that the second to last step looks correct, just rewrite it neatly after all the cancellations.

[killakms]

@killakms, where did you get that h in the denominator? Other than that the second to last step looks correct, just rewrite it neatly after all the cancellations.

idk not sure thought rate of change formula goes over h
Should the h turn into -1?

[EmperorRyker]

idk not sure thought rate of change formula goes over h
Should the h turn into -1?

Oh, right, it does. You didn't have it in your first line, that's why I was confused. On that point, write all lines fully rigorously and don't do things like this, it's confusing and if you mess it up, you won't go where you went wrong. You also won't get as much partial credit on tests.

Anyway, yeah, -h and h at the bottom cancel out, so the answer should be -1/(your original answer).

[killakms]

Oh, right, it does. You didn't have it in your first line, that's why I was confused. On that point, write all lines fully rigorously and don't do things like this, it's confusing and if you mess it up, you won't go where you went wrong. You also won't get as much partial credit on tests.

Anyway, yeah, -h and h at the bottom cancel out, so the answer should be -1/(your original answer).

thanx brah. Got it.
edit:nvm minor mistake

[Chrysippus]

yes. and being regular isnt necessary (though it makes for pretty formulas). i think, though, having something convex matters, so that there is no self-intersection. this same analogy should carry over to n-dimensional volume. dipping an apple in caramel makes the volume it encompasses change by dV = A(x)*dx.

I think we're having a breakdown in communication if you think that being regular isn't necessary.

[hidingwithmusic]

I still haven't gotten my last exam back! Im so nervous, but I think I earned a C+/B-. Im praying that I made a B on that exam! The chapters we are working on now is pretty much only graphing and its easy (polynomials function of higher degree, polynomial division, zero of polynomial, rational graphing). Im only having trouble with these 3. Our homework is from the textbook and my teacher doesn't cover everything... I need help with these three problems.

[GaryOatz]

Handed in my Bachelors Thesis today, feelsgoodman

[Chrysippus]

The way I understood your post, should hold for any simple, closed, piecewise continuous convex curve.

The result for regular polygons is that the relation holds whenever area and perimeter are functions of the radius of the inscribed circle of the polygon. This doesn't work for non-regular polygons because if they aren't regular (or have similar restriction), then the radius of the 'inscribed circle' doesn't determine the area or perimeter of the shape. Given some other properties about the shape, the radius of the inscribed circle can determine area and perimeter, but that won't be the appropriate input variable anymore.

I think we can generalize, but that result definitely doesn't hold for non-regular polygons.

The problem was given to students, and a neat paper was put together.

http://www.math.byu.edu/~mdorff/docs/DorffPaper07.pdf

[killakms]

Am i missing something?

[Chrysippus]

Am i missing something?

Firstly, (-3.3) isn't an interval, and I don't think -00 will be recognized as - infinity.

[killakms]

Firstly, (-3.3) isn't an interval, and I don't think -00 will be recognized as - infinity.

hmm meant 3,3 correct?...also is where i input the 54 correct?

[EmperorRyker]

hmm meant 3,3 correct?...also is where i input the 54 correct?

It is actually, yeah. Same with the local minimum.

[killakms]

wuts wrong with my red box?

[Chrysippus]

wuts wrong with my red box?

You're not wrong but they probably want you to simplify that..