% Factor[a*x^4 + b*x^3 + c*x^2 + d*x + e]
|mbot||bsmntbombdood: e + d*x + c*x^2 + b*x^3 + a*x^4|
|Templarian||Simple Stats problem i'm stuck on multiple choice: The diameter of ball bearings is known to be normally distrbuted with unknown mean and variance. A random sample of size 25 gave a mean 2.5cm. The 95% confidence interval had length 4cm. Then... (A): The sample variance is 4.86. (B). The sample variance is 26.03. (C).The population variance is 4.84. (D). The population variance is 23.47. (E). The sample variance is 23.47. (T|
to find the nullspace of a matrix, do I just set the matrix equal to zero and find its reduced row echelon form?
|FatalError||more or less, you just solve Ax = 0|
|holst||winux: if you have access to Matlab you can QR factorize for a ortogonal basis of both ran(A) and null(A)|
its pretty much the gram schmidth procedure in action for both parts
|winux||i have to do it by hand on my test tomorrow|
|Safrole||You sound like an overachiever.|
you better slow down.
|winux||i actually did it before, but it sort of flew out of my memory|
|holst||so how do you do it by hand. hmm|
|FatalError||you had the right idea with what you said before|
http://en.wikipedia.org/wiki/Nullspace has an example
|holst||that will get the job done. however i dont like their procedure|
what is that "z=z" nonsense
|FatalError||it means it's a free variable|
|holst||essentially they have two equations on R^3 space; thats the intersection of two non-affine planes; the nullspace is the intersection between them; a line. i would like some geometric side notes on the explanation|
on the / as a
|FatalError||the fact that there is 1free variable should be enough to see that it's a line really|
|godtvisken||To find the power series of sin x/x I can just divide each term of the power series of six x by x, right?|
so sin x is x - x^3/6 + x^5/120 ..
and sin x/x would be 1 - x^2/6 + x^4/120 ?
|godtvisken||Kasadkad`: is there a theorem that says this?|
|Kasadkad`||Sure, the "Just Erase Those Three Dots And It's A Polynomial Theorem"|
It's like how you prove things in multivariable calculus by taking the single variable version and making all the letters bold
It just follows from the fact that multiplying power series works exactly like multiplying polynomials
So you can factor out an x
|billfur||A great example until you come to partial derivatives of pratial derivatives.|