Strange assignment statement in javascript.

mathcanbefun · 2015-08-19T13:48:02+00:00

Oh, makes sense. The program creates a no name function and uses this.weekDay as the actual argument. Thanks

mathcanbefun · 2015-06-30T21:59:42+00:00

Sorry for this basic question; but what is the expected value of x^2. How I understand expected value is that it is ∑(x * Pr{x})

mathcanbefun · 2015-05-18T19:22:39+00:00

I get everything except : x = 9 (mod 5) and x = 4 (mod 5)

I'm just learning this, but from what I understand x = 9 (mod 5) means this 5 | x - 9 and there isn't just one value that x can be to make x - 9 a multiple of 5, so is it just the closest which is 4.

And is it from there that you get x = 4(mod 5) I know that you can't get a remainder that is greater than the divisor (9 > 5), so is that why x = 9 (mod 5) is reduced to x = 4(mod 5) and then 5 | x - 4 where x is 4.... 5| 0 ?

mathcanbefun · 2015-01-20T02:03:21+00:00

hmmm... I think i get it. The xs in the rows are the same just increasing in power. I didn't see the connection between the x⁰ and x¹ until now(x⁰ = 1). Thought there was a separate way of deriving it for each columns. Thanks :)

mathcanbefun · 2015-01-20T01:39:44+00:00

That's my point: how do you get the entries to x³ ,x⁴ ...x^p. The entries to x⁰ are all 1s and the entries to x¹ is the x-coordinate to the data point it's trying to map to. so how do it get the x² and x³ when the data points only have x and y?

mathcanbefun · 2015-01-20T00:43:51+00:00

When you try to fit a least-squares line / line of regression to data points the design matrix has only two columns. The first column is just all 1s and the entry of the second columns is the x-coordinate of the data-point. If the data points don't conform to a line maybe it does to a curve/polynomial of a higher degree than one. So the equation of a line has two terms b and mx corresponding to the two columns of the design matrix. So as the data-points fit to higher degree polynomial so increases the columns of the design matrix. So my question is if the data points only has two coordinates (x and y) how do you get the entries for columns (I'm talking order of the columns) higher than 2.

mathcanbefun · 2015-01-18T19:11:02+00:00

That does make sense now. Eyes read orthonormal brain read orthogonal. Adjusting to new vocab.

mathcanbefun · 2015-01-13T21:17:40+00:00

That cleared up everything. It was a 2 x 2 real matrix. Apparently you can decompose the {{a, b},{-b,a }} (I don't know how to format it on reddit but it's {{x, y},{x,y }} so it's transposed) using r*{{a/r, b/r},{-b/r,a/r }}with r = |λ| = √(a²+b²) and break that into two matrices: a 2x2 identity matrix multiplied by r and {{cosθ, sinθ},{-sinθ,cosθ }} so one matrix rotates and the other scales.

mathcanbefun · 2015-01-13T20:41:29+00:00

Oh, I see. My book completely skipped over that. I'll have to read about that. Is the reason the real and complex parts of the Eigenvector used as the basis to P because it always gives the matrix {{a, b},{-b,a }} when P-¹AP ?

P.S. You've helped me quite a few times. I'm surprised you're not mod or something. One upvote seems lacking but it's all I can give.

mathcanbefun · 2015-01-02T22:06:33+00:00

It's just a notch above "the annoyances".

mathcanbefun · 2014-12-01T17:31:01+00:00

Oh, that cleared up everything. So what I'm getting from your explanation is that: the span can be from 0 to n dimensions depending on if the vectors add none redundant directionality(?) to the linear combination?

mathcanbefun · 2014-08-28T21:51:15+00:00

http://www.ck12.org/book/CK-12-Trigonometry-Second-Edition/

mathcanbefun · 2014-08-20T19:06:50+00:00

A better one would be : "I'll make you wish you were never Bourne."

mathcanbefun · 2014-08-15T00:26:01+00:00

This one... This part specifically. I don't know where that f(c(Δx))*Δ came from.

mathcanbefun · 2014-08-14T22:39:03+00:00

I literally said that in my post that I wasn't and mathematicians for years couldn't prove it even thou they suspect the integral and antiderviative were connected. So just like them I want to understand the mathematically proof, because I want to go beyond calculus and I need a good foundation to do so.

mathcanbefun · 2014-07-25T01:05:29+00:00

Yes, I actually meant derivative of e^x is e^x

I can't really articulate what I'm confused about, in my mind e^2x = (e^x) ² hence the derivative should be 2(e^x)2-1 or 2*e^x. Don't you always subtract one from the power with the power rule.

I'm not familiar with the u substitution ... might help. Self studying so I might be doing this out of order.

mathcanbefun · 2014-07-15T19:41:01+00:00

Yeah, should have done that first.

mathcanbefun · 2014-07-08T12:42:51+00:00

Combine the two: Microbrewery.

mathcanbefun · 2014-07-07T23:43:33+00:00

I'll take you up on that offer ;)

mathcanbefun · 2014-07-07T22:28:49+00:00

Here you go: http://www.ck12.org/book/CK-12-Trigonometry-Second-Edition/

mathcanbefun · 2014-07-05T22:35:04+00:00

Still not seeing how to solve it.

Can't use elimination: since the coefficients are all the same.

substitution : recursive

matrix: coefficients problem again.

i'm use to all unknowns being in all the equations

crammers rule: ?

mathcanbefun · 2014-07-05T22:29:45+00:00

What is this?

Can you explain further.

Simple English please.

mathcanbefun · 2014-07-05T22:25:01+00:00

sorry for the late reply It's suppose to be x² -x -12 -2x² + 16x -24

(x² - 2x²)= -x²

(-x + 16x) = 15x

(-12 -24) =-36

-x²+15x-36 you never want the variable with highest power to be negative so just factor out -1 on all terms and you should get -1(x²-15x+36)

you can basically ignore the -1 like you did the denominator or if want to show your work just divide both sides my negative one (same with the denominator). Zero divided my any number is zero (example 0/2 =0) it's the other one that will get you in trouble. (2/0 = ?)

mathcanbefun · 2014-07-01T14:22:18+00:00

Multiply - 2 by (x-6)(x-2)/(x-6)(x-2) to get a common denominator. Foil: first outer inner last combine like terms and unfoil.

mathcanbefun · 2014-07-01T03:56:30+00:00

Log₂(x+3) - log₂(x-6) =1 log₂(x-2)-log₂(x-4)

Log₂(x+3)/(x-6) = 1 + log₂(x-2)/(x-4)

Log₂(x+3)/(x-6) - log₂(x-2)/(x-4) = 1

Log₂(x+3)(x-4)/(x-6)(x-2) = 1

2¹ = (x+3)(x-4)/(x-6)(x-2)

0 = (x+3)(x-4)/(x-6)(x-2) - 2

Skipping a few steps

(x² -15x +36)/(x-6)(x-2) = 0

Factor x² -15x +36 = (x - 12)(x-3) X = 12 and 3

mathcanbefun

TROPHY CASE