## Math(s)

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Saltine
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### Math(s)

Hey here's a maths threads for most of you to safely never visit.
sum yun gai wrote:The solution she was given for the problem below i still think seems like kind of cheating to get a favorable integral, but maybe it's because of the exponential function...
∫ (e^x) / (1 + e^{2x}) dx
Yeah, integrals are like that sometimes, I think. Derivatives are more straightforward, usually, and so verifying an answer is often easier than coming up with it. Which makes the solutions come off as a post-produced trick, like a movie where shards of ceramic jump off the floor and make a vase.

Anyway, my approach was to see that you've got (e^x dx) in there, which is d(e^x). And e^{2x} is (e^x)^2, So replace e^x with u and get
∫ du / (1 + u^2)

which you immediately recognize as being the derivative of the arctangent (1+x^2 in the denominator usually signals some kind of trigonometric nonsense)... that's the magic trick, since just plugging in u = tan v will work, but feels like cheating. If I were to plow through this without knowing about arctangent, I would do this:

∫ du / (1 + iu)(1-iu) ☜ factor the denominator into its (complex!) roots
½ ∫ [ 1/(1+iu) + 1/(1-iu) ] du ☜ partial fraction expansion
½ ln(1+iu) / i + ½ ln(1-iu) / -i ☜ integral of 1/(ax+b) is a natural logarithm
½i [ ln(1-iu) - ln(1+iu) ] ☜ factor out the i
½i ln [(1-iu)/(1+iu)] = v ☜ push the logarithms together and set the answer to v to manipulate it further
ln √[(1+iu)/(1-iu)] = iv ☜ push the ½ into the logarithm and multiply both sides by i
√[(1+iu)/(1-iu)] = e^{iv} ☜ take the exponential of both sides
√[(1+iu)^2/(1+u^2)] = cos v + i sin v ☜ multiply the top and bottom of the fraction by 1+iu on the left, complex exponential on the right
1+iu / √(1+u^2) = cos v + i sin v ☜ take the square root of a square on the left
1/√(1+u^2) = cos v, u/√(1+u^2) = sin v ☜ assume u is real, break into Real and Imaginary parts
u = tan v ☜ divide the right equation by the left equation
v = tan⁻¹ u ☜ whew!

So basically either you cheat or you plow through complex logarithms, exponentials, and a bunch of manipulation, much of which feels like cheating itself. So that means just cheat.

Anyway, the answer is
tan⁻¹ u

and u was just e^x so it's
tan⁻¹ e^x.

Was the solution something like that (without all the complex exponentials)?
--Saltine

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### Re: Math(s)

GAH.
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sum yun gai
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### Re: Math(s)

yeah, that was the exact solution only her class uses "t" instead of "u". so you FAIL! j/k

the part that seems like cheating to me is how you pick e^x as the substitution but find it in both places in the integral. i guess the trick is to just know that one is coming from the du = e^x dx portion and not think too hard about it after that. it's been a few years since i had to do these kinds of problems, and math lends itself to the kind of hand waving and circular logic that kind of pisses me off sometimes
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James
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### Re: Math(s)

Math of s.
SIGNATURE: GONE

PonderThis
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### Re: Math(s)

Saltine make big Ponder feel dumb. Big Ponder sad.

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### Re: Math(s)

James wrote:Math of s.

The erotic awakening of said letter.

I COME BACK FROM MY HIATUS AND FIND A MATHS THREAD IN MY SPAMUSEMENT GODDAMMMMIT
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jvcc
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### Re: Math(s)

Dusk wrote:I COME BACK FROM MY HIATUS AND FIND A MATHS THREAD IN MY SPAMUSEMENT GODDAMMMMIT

SEE WHAT HAPPENS WHEN YOU LEAVE?
ntw3001 wrote:you can't get raped if you always say yes

ntw3001
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### Re: Math(s)

SALTINE HAS TAKEN OVER

HE WAS TOO STRONG

DO YOU SEE WHAT HE HAS DONE

uhh, not meaning to derail the math(s() thread or anything.

fanelian
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### Re: Math(s)

I used to be good at Geometry. Used to enjoy it. I cannot remember any formulas now... except maybe the one for the hypotenuse.

You guys are too advanced for me.
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Saltine
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### Re: Math(s)

sum yun gai wrote:the part that seems like cheating to me is how you pick e^x as the substitution but find it in both places in the integral.
Oh, sure; that too.
Dusk wrote:I COME BACK FROM MY HIATUS AND FIND A MATHS THREAD IN MY SPAMUSEMENT GODDAMMMMIT
jvcc wrote:SEE WHAT HAPPENS WHEN YOU LEAVE?
ntw3001 wrote:DO YOU SEE WHAT HE HAS DONE
HEY WHY ARE WE ALL YELLING? IS MY AIR CONDITIONER TOO LOUD? I'LL TURN IT OFF BUT IT MIGHT GET TOO HOT IN HERE. Anyway, this thread is now muggy and still full of math so it's even less desirable to nigh all y'all.

fanelian wrote:You guys are too advanced for me.
You never dealt with calculus? It's totally not too advanced for you, I'm sure. Besides, remembering formulas is the Internet's job, not yours.

I'll post more mathematic here when I get a chance, unless someone else has something to ask/discuss.
--Saltine

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### Re: Math(s)

I actually enjoyed trying to figure out what you were talking about based on my experience in AP Calculus many moons ago in high school.
I love this post so much I'm going to take it behind the middle school and get it pregnant!

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### Re: Math(s)

I was advanced in math since 3rd grade AND I really love it. I used to do problems just to do them, like Saltine here. I'm also the person everyone recommends to talk to when anyone need help with math(s) for classes, and I am always willing to help.
However, it's been a long time since I've persued any problems on my own. Therefore, I may participate in more of the math(s) thread, maybe not.
I do support it though. Maybe I'll let WTF Is Juice? know. He's a math(s) dude.

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Saltine
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### Re: Math(s)

So, here's a neat integral thing that I may have mentioned before:
Solve ∫₀¹ [ x⁴ (1-x)⁴ / (1+x²) ] dx.
Note that this result must be positive (since the integrand is non-negative everywhere and only equals zero at the endpoints).
--Saltine

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### Re: Math(s)

Seven.
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### Re: Math(s)

Saltine wrote:So, here's a neat integral thing that I may have mentioned before:
Solve ∫₀¹ [ x⁴ (1-x)⁴ / (1+x²) ] dx.
Note that this result must be positive (since the integrand is non-negative everywhere and only equals zero at the endpoints).
I don't like you pointing out how poor my attempts are at approximating others' baked goods.
Saltine wrote:This is all logically consistent, but the artist does not go on to explain that you love Hitler. See, this is why logicians don't write popular music.

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### Re: Math(s)

Saltine wrote:Solve ∫₀¹ [ x⁴ (1-x)⁴ / (1+x²) ] dx

So, the answer is interesting, but the method of solution that I found was... unpleasant.

Starting off, the denominator suggests that this becomes a trig substitution (also, the answer suggests that this becomes a trig substitution, but I digress). Plug in the appropriate substitution, simplify, and you are left with $\int_0^\frac{\pi}{4} \tan^4{\theta} \left( 1 - \tan{\theta} \right)^4 \, d\theta$. At this point, I'm hoping that there's a trick that I missed, because the obvious way is to expand the term in the parentheses, distribute through, and then weep quietly in the corner while rocking back and forth.

Anyway, some uninteresting horribleness later, we get $\frac{22}{7} - \pi$, or the error term for the standard (low quality) rational approximation of $\pi$.

P.S. Please excuse my LaTeX. Perhaps someone should install jsMath.
Last edited by efnar on Mon Sep 13, 2010 1:28 am, edited 1 time in total.
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efnar
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### Re: Math(s)

As an aside, there is a fairly nice elementary recurrence relationship for $\pi$. It's fairly simple (in that it only uses trigonometry), but it converges very slowly (and is thus useless in practice). If you end up reading through the paper, please read past the technical hand waving regarding convergence of sequences, and other such analysis minutia.
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Saltine
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### Re: Math(s)

Hey, I lured efnar out of hiding with mathematches.
loofah wrote:I don't like you pointing out how poor my attempts are at approximating others' baked goods.
See, I always assumed you threw in the extra 0.00126ish so that nobody could accuse you of shortchanging them.
efnar wrote:The denominator suggests that this becomes a trig substitution (also, the answer suggests that this becomes a trig substitution, but I digress). Plug in the appropriate substitution, simplify, and you are left with $\int_0^\frac{\pi}{4} \tan^4{\theta} \left( 1 - \tan{\theta} \right)^4 \, d\theta$.
oh whoa, that's pretty hardcore. The less punch-yourself-in-the-face way to do it is:
x⁴ (1-x)⁴ = x⁸-4x⁷+6x⁶-4x⁵+x⁴ ☜ binomial expansion and stuff
x⁴ (1-x)⁴ / (1+x²) = x⁶-4x⁵+5x⁴-4x²+4 - 4/(1+x²) ☜ polynomial long division
∫₀¹ [ x⁴ (1-x)⁴ / (1+x²) ] dx = ∫₀¹ [ x⁶-4x⁵+5x⁴-4x²+4 ] dx - ∫₀¹ [ 4 / (1+x²) ] dx .
The left integral is just a polynomial and evaluates to ²²/₇; the right one is the (now easier) trig substitution and evaluates to π, so you end up with
²²/₇ - π > 0
or
²²/₇ > π
which loofah hinted at and you presented anyway.
efnar wrote:P.S. Please excuse my LaTeX. Perhaps someone should install jsMath.
Yeah, but this is just spamusers.com, so I just make do with Unicode.
efnar wrote:As an aside, there is a fairly nice elementary recurrence relationship for $\pi$
Neat; I plan to try it out when I am not in a rush which I am right now bye
--Saltine

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### Re: Math(s)

Ah crap, I now realise that I've spent the summer letting my brain melt into some sort of cheese substitute, meaning that I've forgotten everything except the very, very, very basics of calculus. Oh well, it was fun trying to follow all this, especially around 7 minutes after I've gotten out of bed.

efnar
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### Re: Math(s)

Saltine wrote:The less punch-yourself-in-the-face way to do it is

[snip]

Oh, that's much better, indeed.
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sum yun gai
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### Re: Math(s)

the instructions say: "Use partial fractions to find the integral"

i've got the partial fraction piece, i'm just having trouble finding a way to integrate it. here's the original problem:

ʃ (6x)/(x³ - 8 ) dx

i factored the denominator into (x-2)(x² + 2x + 4) and solved the partial fractions to get 1/(x-2) and (2-x)/(x² + 2x + 4) but i can't integrate the second fraction. help?

edit: got another one i can't factor:

ʃ (x² - 4x + 7)/(x³ - x² + x + 3) dx
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### Re: Math(s)

sum yun gai wrote:i factored the denominator into (x-2)(x² + 2x + 4) and solved the partial fractions to get 1/(x-2) and (2-x)/(x² + 2x + 4) but i can't integrate the second fraction. help?

Don't forget that the sum of the integral is the same as the integral of the sum. You can break down that fraction into ʃx/(x² + 2x + 4) - ʃ2/(x² + 2x + 4)
THen you should be able to find the form from integral tables here:http://www.integral-table.com/

EDIT: efnar suggests that this was likely not meant to be looked up in an integral table, for purposes of an exam (unless your prof is really nice). You have to add some terms in the numerator to force a u-substitution. You'll get something like 1/2 ʃ [(2x-4+6)/(x² + 2x + 4) - 6/(x² + 2x + 4)].
Then you have to complete the square in the denominator of the second expression to force another u-substitution.
You'll get something like ʃ dx/[{(x+1)²/3} +1]. You'll end up with an arctan. Blech. It was quite messy.

If you learned the rational root theorem, that should help with the second one.
Saltine wrote:This is all logically consistent, but the artist does not go on to explain that you love Hitler. See, this is why logicians don't write popular music.

sum yun gai
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### Re: Math(s)

loofah wrote:Blech. It was quite messy.

yes. i already knew that i'm having problems going from "OMFG this is such a mess" to "OMFG that *was* such a mess"

also, the only textbook i have in my possession that has anything to do with the rational root theorem is a book on number theory so it's probably safe to assume i have not seen it presented to me in my studies. and if i haven't seen it, i'm very positive my girlfriend (whose homework this is) has never seen it either although, looking at the actual method i think i have probably known how to do this, i just never heard it called that before... and as usual, i stopped after trying a couple that didn't work.
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### Re: Math(s)

sum yun gai wrote:also, the only textbook i have in my possession that has anything to do with the rational root theorem is a book on number theory so it's probably safe to assume i have not seen it presented to me in my studies.

You don't have to use the rational root theorem, but it helps you find rational roots for factoring cubics (and higher order polynomials). Depending on the problem, you can also "perform a magic substitution", "cleverly group then factor", "just see it" or "guess correctly". The rational roots theorem is commonly presented in a pre-calculus or algebra course.

The whole "finding roots of higher order polynomials" thing is a bit of a duck... (err... "canard"). For second order polynomials, there is the quadratic equation. For third order polynomials there is the (rather unfortunate and inaccurately named) Cardano's method, and Ferrari's method will (eventually) give you roots of fourth order polynomials.

For fifth order polynomials and higher, a surprising thing happens: there can not exist general methods of getting solutions! There are a few ways to see why this is, but all of them (that I know of) require fairly advanced (in the US, this would normally be presented in the fourth year of an undergraduate mathematics program, or the first year of a graduate mathematics program) methods developed by Abel and Galois in the early 19th century. The standard way of explaining this uses group theory: such general solutions can't exist because the group S_n is not a solvable group for n>=5.

My point is that finding roots of arbitrary polynomials is hard, and when problems require that you do such things, they are almost always constructed to have at least some "nice" roots. Most "nice" roots are rational numbers, hence the utility of the rational root theorem in this (artificial) context.
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Saltine
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### Re: Math(s)

So, are we supposed to solve these here or just point you in the right direction?

For (x³-x²+x+3) I just "saw" that -1 should be a root (three terms in x and then then number 3), so you can divide it by (x+1) and then you're left with (x²-2x+3).

ʃ (x²-4x+7)/(x³-x²+x+3) dx
ʃ 2/(x+1) dx + ʃ (1-x)/(x²-2x+3) dx ⇦ partial fractions!
ʃ 2/(x+1) dx + ʃ -1/2u du ⇦ substitute u=x²-2x+3, du=-2(1-x).
2 ʃ 1/(x+1) d(x+1) -½ ʃ 1/u du ⇦ note that dx = d(x+1) and move out coefficients
2 ln (x+1) - ½ ln u ⇦ integrate
2 ln (x+1) - ½ ln (x²-2x+3) ⇦ desubstitute u

The first one, ʃ (6x)/(x³- 8) dx is just god-awful as far as I'm concerned. The tip I'd give is to make sure that once you have a partial fraction piece like

2-x / x²+2x+4

you want to try to get the numerator to be (a multiple of) the derivative of the denominator, so your magic substitution will work: The derivative of the denominator is 2x+2. You can express (2-x) as (3-(x+1)). Which gives you

3 / x²+2x+4 - (x+1)/ x²+2x+4

The left side is a substitution of u=(x+1)/√3, du=1/√3 to get some fun with arctangent. The right side is a substitution for v = x²+2x+4, dv=2x+2 to get a logarithm.

That should get you there, but it's still awful.
--Saltine

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