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[MagmaMcFry]

Here goes: A polynomial of degree n always has n (not necessarily distinct) roots in the complex plane (note the fundamental theorem of algebra).
If a polynomial a₀ + a₁x + a₂x² + ... + a_nxⁿ has the roots b₁, ..., b_n, then it can be written as a_n(x-b₁)(x-b₂)...(x-b_n).
If a+bi is a complex root of a polynomial with only real coefficients, then a-bi is also a root of this polynomial.
To find the roots of a given polynomial, you generally have three options: If the polynomial is quadratic, cubic or quartic, you can always find explicit expressions for the roots (you should be able to memorize or even derive the quadratic formula, but please DO NOT EVER ATTEMPT to manually solve a quartic equation). If you have access to computing power, you can throw your equation into Wolfram Alpha or your graphical calculator. If you don't have either, you are most likely in an exam, and your best bet is to try out some small integers and hope that one of them is a root. With some practice, you can do that just by looking at the equation.
Now that you have a root r, you want to extract the factor (x-r) from your polynomial. You can either do that by long division (just repeatedly subtract (x-r), the process is insanely similar to standard long division), or using Horner's method (if Wikipedia doesn't help, I can explain in more detail). Repeat until there are only linear factors left. That's your factorization.

Now for the polynomial fractions: Given two things a/b and c/d, their sum is (ad+cb)/bd and their product is ac/bd. In your case, you can just use polynomial fractions as things.
To simplify a polynomial fraction, simply factor both the numerator and the denominator polynomial (you should now know how to do that), and do some cancelling out. Finished.

[Trapezohedron]

Having a problem with Mathematics of Investment, specifically the 60-day, 6% method.

So you're supposed to divide the days of the year into 60/360, but what if the days aren't 60?

Like for example, if it's 90/360, you're supposed to divide it into two chunks, one being 60/360 (or I₆₀) and 30/360 (or 1/2 I₆₀).

Now, how do I divide 52 into a fraction of x/y I₆₀ or x/y I₆₀?

For the sake of reference, here's the problem:

Find the ordinary interest using the 60-day, 6% method.

2,340 for 112 days at 6%.

[Karlito]

The 60 day method is a bit silly, unless you don't have access to pen/paper or a calculator. You might divide 112 into 1 chunk of 60 days, 1 chunk of 30 days, 3 chunks of 6 days, and 1 chunk of 4 days.

But again, it's silly unless no one ever taught you long division. As long as you remember that Interest = Principle * Rate * Time, you can calculate the interest for any rate or time period. Assuming a 360 day year, the interest for $2,340 for 112 days at 6% is $2340 * .06 * 112/360 = $43.68.

The 6% method is a sort of mental math trick that takes advantage of the fact that .06 * 60/360 is .01 and 6/360 is .001. Multiplying by a hundredth, a thousandth, and simple factions like 1/2 or 1/3 is easy to do, so you could calculate interest in your head without too much trouble. I'll do your example problem with this method. Interest for a 60 day period is .01 of the principle, $23.40; interest for a 6 day period is .001 of the principle, $2.34. You have 1 60-day, 1 30-day period 3 6-day periods, and 1 4-day period, so add up $23.40 + (1/2)$23.40 + $2.34 + $2.34 + $2.34 + (2/3)$2.34 = $43.68

I'm not sure in what dimension that's a shortcut, but if you were doing this in your head you might round off the extra 4 days, or even the extra 22 days for a quick approximation. Really, this is a perfect example of the unneeded nonsense that gets taught in math classes and causes students to hate the subject.

[Zrk2]

Well this is awkward...

[penguify]

Hey, for this calc problem I need to show that 2sinx-x-xcosx has only one zero across (0, 2pi]. I've tried breaking it into sections like (0, pi/2] and proving that d/dx(2sinx)>=d/dx(x+xcosx), but I always end up with expressions that have an xcosx or xsinx in them, and I don't know what to do about those. Any tips?

(I'm actually trying to maximize the area of a circular segment with a fixed arc length; I know it's best when theta is pi, but I don't know how to prove it)

[Zrk2]

Okay. I just was studying for this type of problem on my midterm.

1. Evaluate at your given points.

2. Use intermediate value theorem to conclude that there must be a 0 on the interval.

3. Take the derivative.

4. Solve for 0.

5. It won't work on the interval.

6. Therefore more than two 0s are impossible.

7. Therefore there is only one 0 on the interval.

[penguify]

3. Take the derivative.

4. Solve for 0.

This is the part I can't get: there's an xcosx in the equation, which differentiates into cosx-xsinx ; and I don't know how to solve equations containing xcosx.


herp derp forgot product rule

[Karlito]

First of all, xcosx differentiates into cosx-xsinx. I don't remember how to solve an equation like this algebraically either (if it's even possible), but you should be able to use a graphing calculator to get some numerical solutions.

EDIT: Ah ha! There's a half-angle identity that can help you. (Well, sort of...)

[penguify]

On obvious differentiation error: durrr, I keep doing that on this problem...

Half-angle IDs? I have to brush up on those I guess...

[Karlito]

It simplifies to x=tan(x/2). I don't think there's anything more you can do to it. Transcendental equations are tricky like that.

[Zrk2]

Basically you should end up unable to solve it. That's how you prove that there is no max/min on the interval.

[Karlito]

Well, zeros do exist, and there is only one on (0,2π], he just has to solve for them numerically, because it can't be done algebraically.

[penguify]

That half-angle works wonderfully (although I'm getting x=2tan(x/2), not 1tan(x/2))! Showing that there are only a few solutions in the interval should be much simpler with a monotonic function like tangent.

[Grek]

First of all, xcosx differentiates into cosx-xsinx. I don't remember how to solve an equation like this algebraically either (if it's even possible), but you should be able to use a graphing calculator to get some numerical solutions.

cos(x) - x*sin(x) = 0 wherever

cos(x) = x*sin(x)

1 = x*sin(x)/cos(x)

1 = x*tan(x)

cot(x) = x

cot(x) - x = 0

at which point you can easily apply newton's method if you can't recall your half-angle identities.

[penguify]

Thanks for the help, Zrk2, Karlito, Grek! I'm using y=2tan(x/2)-x, showing that dy/dx >= 0 for (0,pi)u(pi,2pi], and Mean value theorem to show that there's only one solution for y=0. You guys are awesome!