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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)[reply]

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)[reply]

If you could tell me what I need to fix that would be very helpful thanks. --Thebackofmymind (talk) 02:16, 23 November 2011 (UTC)[reply]

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)[reply]

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−1. Bo Jacoby (talk) 10:01, 23 November 2011 (UTC).[reply]

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)[reply]

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)[reply]

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)[reply]

Yes. Cuddlyable3 (talk) 14:02, 24 November 2011 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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 k_n in some way - notice how for N=2 and k₁=0, y_pk is different for different values of k₂ --130.216.55.200 (talk) 22:39, 24 November 2011 (UTC)[reply]

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 k_n are zero. The real question is how do you minimise y_pk in Case 3 by free choice of the constants M_n. Cuddlyable3 (talk) 03:38, 27 November 2011 (UTC)[reply]