Wikipedia:Reference desk/Archives/Mathematics/2007 February 14

= February 14 =

Clearing up a few things
I want to make sure that this:

$$ \lim_{a^+ \to \infty} \Bigg(\sum_{a=1}^\infty \tfrac{1}{a^n}\Bigg) $$

Reads: "The limit of a as it approaches infinity approximates (with increasing accuracy) the closed form of the bracketed series". Is this correct? Is there a better way of saying this?

164.11.204.51 02:35, 14 February 2007 (UTC)
 * It doesn't just approxmiate the right-hand side, it equals the right-hand side. --Trovatore 02:37, 14 February 2007 (UTC)
 * Whoops, I just noticed that you don't in fact have a right-hand side. So I'm afraid your translation is completely wrong.  It shouldn't be a sentence, but a noun phrase. --Trovatore 02:38, 14 February 2007 (UTC)
 * Why don't I have a right hand side? You're meant to explain these things. I assume you're referring to the limit part which denotes approaching infinity from both sides. I've changed that now, does it make sense?
 * I would say the limit of a as it approaches infinity from the right for the function [blah blah blah] . Wait, is it supposed to be $${a^+ \to \infty}$$ or $${a \to \infty^+}$$? --Wirbelwind ヴィルヴェルヴィント  (talk) 06:32, 14 February 2007 (UTC)
 * It would have to be the latter. How can you approach positive infinity from the left? —The preceding unsigned comment was added by 203.49.220.190 (talk) 09:26, 14 February 2007 (UTC).

The "right-hand side" means part of the equation after the equals sign. You don't have an equals sign (except for the range of the summation, which doesn't count), so you don't have an equation; you have an expression. The translation of an expression into English is not a sentence, but a noun phrase. So the verb "approximates" doesn't belong; you shouldn't have a verb at all. (Well, you can have them in subordinate clauses, but not as a predicate.) --Trovatore 07:59, 14 February 2007 (UTC)


 * Actually, looking at the expression more closely, there's another confusion. This part:
 * $$\Bigg(\sum_{a=1}^\infty \tfrac{1}{a^n}\Bigg)$$
 * does not depend on a at all. Did you mean to sum from n equals 1 to &infin;, rather than from a equals 1 to &infin;?
 * (Not that you can't make sense of it the way you wrote it; it's fine to take the limit as a goes to infinity of something that doesn't depend on a. It's not wrong, just kind of trivial.) --Trovatore 08:05, 14 February 2007 (UTC)


 * I don't know whether you're trying to find a sentence to match the formula, or a forumla to match the sentence. Either runs into problems, because neither makes sense. a can't approach positive infinity from the positive side, because it can't start out higher than infinity. It can only approach it from the negative side, and that's assumed so you don't have to write it. You can't assign a values both under the summation symbol and under the limit symbol, so you probably meant n for the summation. Also, it doesn't mean much to say that as a approaches infinity it also approaches something else, so that's probably not what you're going for with the sentence. Here's something that would make sense, and might be what you're going for: $$ \lim_{a \to \infty} \Bigg(\sum_{n=1}^\infty \tfrac{1}{a^n}\Bigg) = 0 $$, which means "The limit of a geometric series with common ratio 1/a, as a approaches infinity, is zero." This is true. Its pieces are $$ \lim_{a \to \infty} \Bigg(\sum_{n=1}^\infty F(a,n)\Bigg) = result $$ where a and n are separate numbers, F(a,n) is some formula using a and n, and result is what the summation approaches as a approaches infinity. The symbol F takes a and n and does something to them, the summation takes n as input and acts on the formula next to it, the lim symbol takes a as input and acts on the summation next to it, and that must be equal to whatever's on the "right hand side" of the equals sign.Black Carrot 06:36, 24 February 2007 (UTC)

Intergral of Cosine squared X
Hi! I did the intergral of Cosine squared X or Cos2X and well... a peer of mine got me double guessing myself.. I worked it out (if correctly!) to come out to be (1/4)(sin [2x] + x) + C i was wondering if that is correct or is it just [(Sin[2x])/4] + X + C. I know I know it sounds like a silly question--Agester 03:34, 14 February 2007 (UTC)s with just the order of operations but help on this would greatly help me progress without thinking i'm a total idiot... P.S. the difference is just the if the whole segment is over 4 or just the sine part Thanks for input! --Agester 03:00, 14 February 2007 (UTC)


 * I got $$\frac{x}{2} + \frac{\sin(2x)}{4} + C$$

$$\int \cos(x)^2 \, dx = \int \frac{1+ \cos(2x)}{2}\, dx = \frac{x}{2} + \frac{1}{2} \int \cos(2x) \, dx = \frac{x}{2} + \frac{\sin(2x)}{4}$$ You can always verify by differentiation.Dlong 03:12, 14 February 2007 (UTC)

Dlong 03:12, 14 February 2007 (UTC)

hmm... i see.. that does make sense! thank you very much! That saves me from a massive headache! --Agester 03:34, 14 February 2007 (UTC)

Checking some derivatives
I'm trying to figure out whether me or the book is right on this problem:

Let $$f(x) = \cot{ \frac{ \pi x}{10}}$$ and $$g(x) = 5 \sqrt{x}$$

Find $$(f \circ g)' (1)$$

I got does not exist (since one of the factors is $$- \csc^2{ \frac{ \pi}{2}}$$ which evaluates to negative one over zero), but the book has $$- \frac{ \pi}{4}$$. Anyone know what's going on?

Curtmack of the Asylum 19:27, 14 February 2007 (UTC)


 * Yes. $$-\csc^2\left(\frac{ \pi}{2}\right) = -1$$, not 1/0. –King Bee (T • C) 19:31, 14 February 2007 (UTC)
 * Oh duh, I was thinking of secant. *slaps self* Curtmack of the Asylum 21:39, 14 February 2007 (UTC)


 * Don't sweat it. It happens to the best of us. =) –King Bee (T • C) 21:45, 14 February 2007 (UTC)

Homework Help!!! (I know, just hear me out)
So I'm in highschool calculus and my head's about to explode. I'm trying to learn something from the examples in a book so I can do the hw, but I just don't get what happened. There's this equation in which I'm trying to separate out W in terms of t (time). F and k are constants. The original equation is:
 * dW
 * ____ = dt
 * (F-kW)

so the book says to integrate both sides. They put the integral sign on both sides of the equation. Then they create a 1/k to the left of the integral on the left side, and counterbalance it by multiplying the numerator inside the integral by k (so it's just multiplying by a fancy form of 1, since the constant isn't affected by the integral sign). They do this because, "the differential of the denominator, d(F-kW), equals -kdW." so then it becomes integral of (-kdW)/(F-kW). Which they then say, because it's supposedly the "Integral of the reciprocal function," is ln|f-kW| But how the heck is it a reciprocal? it's essentially a derivative of a function over the function, not one over the function. How is that supposed to work? Thanks for any help, Sasha


 * The derivative of that function that you care about (F - kW) is -k. So it is the "derivative of the function over the function", just with a constant missing. –King Bee (T • C) 22:21, 14 February 2007 (UTC)


 * Basically what you did is integration by substitution, except that the substituted variable is not written out explicitly.
 * You have $$\int \frac{1}{F-kW} \,dW$$. You performed the subsitution $$u = F-kW$$; thus $$du = -k\,dW$$ (or $$dW = -\textstyle\frac{1}{k} \,du$$).
 * You then have $$\int \frac{1}{F-kW} \,dW = -\frac{1}{k} \int \frac{1}{u} \,du = -\frac{1}{k} \ln |u| + C = -\frac{1}{k} \ln |F-kW| + C$$
 * --Spoon! 22:40, 14 February 2007 (UTC)

Oh. Ok. So how would one perform similar mathematical magic to a new equation: integral of dM/(100-kM)? Do you multiply it by k over k again? Isn't it sort of deriv over function again? I'm having trouble really getting what the book is saying to do in these cases. And why.

And thanks for the help on that first one.


 * Yep, same deal. W is now M, and F is now 100. Go to town, and good luck! –King Bee (T • C) 22:29, 14 February 2007 (UTC)

So I've hit a new one, if you guys don't mind helping again. The new equation is E = RI + L(dI/dt). E, L, and R are constants. I'm trying to solve for I in terms of t. Apparently, says the answers at the back of the book, this can be rearranged and integrated into


 * I = (E/R)(1-e^(-Rt/L))

I can't figure this one out at all. How does one isolate the I? I tried moving the RI to the left, then multiplying by dt and fiddling around with other things, but I'm just digging myself into a whole. Is the integral of RI + L(dI/dt) equal to the integral of RI plus the integral of L(dI/dt)? Do you think this'll get me anywhere?


 * Try to bring it in the form (something) = dt. --Lambiam Talk  04:49, 15 February 2007 (UTC)

Awesome. That helped a lot. I just never know in what direction to start with this stuff.