Recent content by Mumba

oh man yes sure thanks again ^^

hmm, ok but why ist L(x) = x+1? --> p(x) = x if its a constant function and p(x)=x, shouldn't be P(x+1)=x too?

Should it be like that: If p(x) =1 then, q(x) = p(x+1) = p(x)+p(1) = 1 + 1 = 2, so L[1]= 2!?

Hey i just tried to calcuculate the eigenvectors. But i can't get a sol. The result is always = 0. ?? Edit: Forget it, using the way with a and b, i was able to solve it correct ^^ (without any eigenvectors)

Cool :) Thx jbunniii

yeah its good thx mate! ;)

:D:D cool thanks alooooot

ahh ok so if u change this fpr L(x^2) --> 1 2 1 matrix: 1 1 1 0 1 2 0 0 1 correct? ^^

and then for L(1) --> 1 0 0 L(x) --> 1 1 0 L(x^2) --> 1 0 1 So my matrix would be 1 1 1 0 1 0 0 0 1 ??

So i get then L(1) = 1 L(x) = x+1 L(x^2)=x^2+1 ?

Coool thanks a lot, easy this way. :D

L(p)=q(p)=p(x+1)? wie kommst du da auf L(x)=1+x? Was hast du denn dann für L(1)?

But so i get for L(x) = x^2 + x, L(x^2)=x^3 + x^2 So should i choose a new basis, for example {1+x,x^2,x^3} to get the repr matrix?

the i get as matrix: 4/sqrt(2) 3/sqrt(2) 2/sqrt(2) 1/sqrt(2) correct?

Is [text]K_{S,B1}[/tex] not the same as B1? And the same for B2? So i calculate the inverste and multply them and then I am finished?