Wikipedia:Reference desk/Archives/Mathematics/2011 November 23

= November 23 =

is this quadratic?
Hi I'm just wondering if you can help me find how to get the answers for this formula.

0=-4000+2000/(1+r)+4000/(1+r)^2

The following is not homework questions specifically but old notes which already say that the answers are:
 * -1/4(square root of 17)-3/4
 * 1/4(square root of 17)-3/4

My limited knowledge tells me the following equation is quadratic, but I'm at a loss how to do this at all. SQRT(number) means square root the number:


 * 0=ax^2+bx+c
 * 0=4000/(1+r)^2+2000/(1+r)-4000

Quadratic formula
 * 0=-b{+/-}SQRT((b^2-4ac)/2a)
 * 0=-2000{+/-}SQRT((2000^2-(4*4000*(-4000)/2*4000)
 * 0=-2000{+/-}SQRT((4000000+32000000)/8000))
 * 0=(x-1932.92)(x-2067.08)

Aaaand that's how far I got. I'm sorry I'm pretty bad at this, but how do I make 1/(1+r)=x to solve for r? --Thebackofmymind (talk) 00:32, 23 November 2011 (UTC)
 * You have misstated the formula. If $$ax^2+bx+c=0$$ then $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$. Widener (talk) 01:02, 23 November 2011 (UTC)

If you could tell me what I need to fix that would be very helpful thanks. --Thebackofmymind (talk) 02:16, 23 November 2011 (UTC)
 * Use Widener's formula, not $$x = -b \pm \sqrt{\frac{b^2-4ac}{2a}}$$. 98.248.42.252 (talk) 06:32, 23 November 2011 (UTC)

Q: how do I make 1/(1+r)=x to solve for r? A: x(1+r)=1, 1+r=1/x, r=1/x&minus;1. Bo Jacoby (talk) 10:01, 23 November 2011 (UTC).

Yes it's quadratic


0 = -4000 + 2000 +     4000               -            1 + r  (1 + r)(1 + r)

To get all the r's above the line, multiply by (1 + r)(1 + r)

= -4000 (1 + r)(1 + r) + 2000 (1 + r) + 4000

= -4000 - 2.4000r - 4000 r.r + 2000 + 2000r + 4000

= -4000 r.r - 6000r + 2000

We have a x.x + b x + c = 0 where a = -4000 b = -6000 c = 2000

then x = -b +/- SQRT (b.b - 4ac) ---                 2a

= 6000 +/- SQRT ( 36 000 000 + 4.4000.2000 ) -                      -8000

= 6000 +/- SQRT ( 68 000 000 ) ---                  -8000

= 6000 +/- 8246..        ---             -8000

= -1.781 and 0.281 to 3 dec. pl.  Cuddlyable3 (talk) 10:33, 23 November 2011 (UTC)


 * Indeed. Alternativerly,
 * $$-4000+\frac{2000}{1+r}+\frac{4000}{(1+r)^2}=0$$


 * $$\Rightarrow -4000(1+r)^2+2000(1+r)+4000=0$$


 * $$\Rightarrow 2(1+r)^2-(1+r)-2=0$$


 * $$\Rightarrow 1+r=\frac{1 \pm \sqrt{1^2+4 \times 2 \times 2}}{4}$$


 * $$\Rightarrow r=-\frac{3}{4} \pm \frac{\sqrt{17}}{4}$$
 * Gandalf61 (talk) 13:27, 23 November 2011 (UTC)

Thank you so much everyone! Laying it out in formulas really help show me where I was wrong! And Cuddlyable3, the r.r means r squared, right? --Thebackofmymind (talk) 23:31, 23 November 2011 (UTC)
 * Yes. Cuddlyable3 (talk) 14:02, 24 November 2011 (UTC)

Differentiable at a single point
Am I correct in asserting that the following real-valued function is differentiable at but a single point, $$x = 0$$?

$$f(x) := \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ x^2      & \text{if } x \notin \mathbb{Q} \end{cases}$$

I'm aware of functions differentiable at but one point over the complex numbers, but I didn't believe it was possible over the reals until I came up with this one. --Leon (talk) 11:24, 23 November 2011 (UTC)
 * That's correct. You can take any nowhere differentiable, continuous at 0 function (such as Weierstrass) and multiply it by x to get a function differentiable only at 0. -- Meni Rosenfeld (talk) 12:01, 23 November 2011 (UTC)


 * It is Baire function of class 2. I'm surprised there's so little about differentiating Baire class 1 functions. Dmcq (talk) 13:35, 23 November 2011 (UTC)

Could someone explain this in more detail. As far as I can see, ƒ is nowhere continuous. I always thought that differentiability was a stronger condition than continuity. — Fly by Night  ( talk )  22:13, 23 November 2011 (UTC)
 * You're right, differentiability is stronger than continuity. What you're missing is that f is continuous at 0. --Trovatore (talk) 22:32, 23 November 2011 (UTC)

Cosine sums of increasing difficulty
1) Given an integer N, what is

$$y_{pk} = \max \left[ \sum_{n=1}^{N} \cos(nx) \right]$$

2) Given an integer N, what is

$$y_{pk} = \max \left[ \sum_{n=1}^{N} \cos(nx + k_n \frac{ \pi}{2}) \right] $$ where $$k_n \in \{0,1,2,3\}$$ at random

3) Given an integer N, how do you find values for $$M_n$$ that minimise

$$y_{pk} = \max\left[ \sum_{n=1}^{N} \cos(nx+k_n\frac{\pi}{2}+M_n)\right] $$

Cuddlyable3 (talk) 14:01, 23 November 2011 (UTC)


 * I redid this so that the equations were done in LaTex rather than ascii art. In future, take a look at this page so you can learn how to do it yourself - it looks much better this way, and is easier to read and understand. The first one I think is merely N (which is trivial to see). The following questions are not so easy, and I won't give the answers to you because you should do your own homework, but you should note that the answer to question 2 must include the values of kn in some way - notice how for N=2 and k1=0, ypk is different for different values of k2 --130.216.55.200 (talk) 22:39, 24 November 2011 (UTC)
 * Thank you for setting my equations in LaTex. They relate not to homework but to a real problem for multi-carrier radio broadcasting. Yes the first is N. Case 2 is also N because it reduces to Case 1 when all kn are zero. The real question is how do you minimise ypk in Case 3 by free choice of the constants Mn. Cuddlyable3 (talk) 03:38, 27 November 2011 (UTC)