- #1

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Its a similiar to the problem i ve postet before. Maybe if i can solve the first problem i can solve this to. But what i dont understand is that notation. I ve circled it with a red line.

Does anyone know what this means?

Thx

Mumba

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- Thread starter Mumba
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- #1

- 27

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Its a similiar to the problem i ve postet before. Maybe if i can solve the first problem i can solve this to. But what i dont understand is that notation. I ve circled it with a red line.

Does anyone know what this means?

Thx

Mumba

- #2

dx

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- #3

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But what does then Ap=q mean? Or how can i calculate L for 1, x and x^2?

- #4

dx

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Ap = p(x+1) defines the action of L.

So for p = 1, you get L(1) = x + 1 and so on.

So for p = 1, you get L(1) = x + 1 and so on.

- #5

dx

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So, if p = 1 + x, then L(p) = 1 + (x + 1) = 2 + x.

- #6

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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?

- #7

dx

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See my post above.

L(x) will be x + 1 instead of x² + x.

L(x) will be x + 1 instead of x² + x.

- #8

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L(p)=q(p)=p(x+1)?

wie kommst du da auf L(x)=1+x?

Was hast du denn dann für L(1)?

wie kommst du da auf L(x)=1+x?

Was hast du denn dann für L(1)?

- #9

dx

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So, if p(x) = 1 + x, then q(x) = p(x + 1) = 1 + x + 1 = x + 2 (i.e. L[1 + x] = 2 + x)

If p(x) = 1, then q(x) = p(x + 1) = 1. (i.e L[1] = 1 )

- #10

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So i get then

L(1) = 1

L(x) = x+1

L(x^2)=x^2+1

?

L(1) = 1

L(x) = x+1

L(x^2)=x^2+1

?

- #11

dx

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L[x²] = (x+1)² = x² + 2x + 1

- #12

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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

??

- #13

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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? ^^

matrix:

1 1 1

0 1 2

0 0 1

correct? ^^

- #14

dx

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Yes, thats correct.

- #15

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:D:D

cool thx alooooot

cool thx alooooot

- #16

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So, if p(x) = 1 + x, then q(x) = p(x + 1) = 1 + x + 1 = x + 2 (i.e. L[1 + x] = 2 + x)

If p(x) = 1, then q(x) = p(x + 1) = 1. (i.e L[1] = 1 )

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!?

- #17

dx

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No, p(x) is a constant function, i.e. it has the same value for any x. So p(x + 1) would still be 1.

- #18

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but why ist L(x) = x+1?

--> p(x) = x

if its a constant function and p(x)=x, shouldnt be P(x+1)=x too???

- #19

dx

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For a given polynomial p, we get the L[p] by replacing every x with x + 1.

So if p = 1, we do nothing since there is no x, and L[1] = 1

If p = x, we replace x by x + 1, and we get L[x] = x + 1

If p = x

- #20

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oh man

yes sure thx again ^^

yes sure thx again ^^

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