Question

T: P_2 ----> R
T(P(x))= int(from_1 to 1)P(x)dx
find kernel of T
help, please

Answer 1

sorry what do you mean of p_2 ?

Answer 2

it's polynomial degree 2

Answer 3

@terenzreignz you have nothing to do here, it's linear algebra.

Answer 4

sending me away? :)

Answer 5

no, just not want to waste your talent.

Answer 6

your strong is not this stuff, I have many question from cal need your help. but now I stuck with this question, after i have it done, I post my cal , you have no way to escape from me. hehehhe

Answer 7

Who know? right? what if because of my sentence, you become strong in this field, too?

Answer 8

Oh well...

Answer 9

\[\huge T: \mathbb{R}_2[x]\rightarrow \mathbb{R}\]
\[\huge T[f(x)]=\int\limits_{-1}^1f(x)dx\]

Answer 10

hold on, P_2 not R2

Answer 11

P_2 R_2
what's the difference we both mean polynomials of degree 2 :)

Answer 12

of course, they are different, P_2 is a polynomial degree 2 and R2 is a function with 2 variables

Answer 13

for example R2 is something like ax +y =0 and P2 is a0 + a1x +q2x^2

Answer 14

don't mistake \(\large \mathbb{R}^2 \) for \(\large \mathbb{R}_2\)

Answer 15

hey guy, R^2 =R_2 it's my prof terminology

Answer 16

Well, anyway, notation aside, I mean polynomials of degree 2.

Answer 17

dealt

Answer 18

You know
\[\large f(x) = ax^2 + bx + c \ \ \ \ \ a,b,c\in \mathbb{R} \ \ \ a\neq 0\]

Answer 19

yeap

Answer 20

\[\huge \int\limits_{-1}^1f(x)dx =\frac{2a}3+2c\]

Answer 21

hehe...not get.

Answer 22

ok, got

Answer 23

Okay?
Now
\[\large \ker(T) = \left\{ax^2+bx+c \ \ | \ \ \ \frac{2a}{3}+2c = 0\right\}\]

Answer 24

got it

Answer 25

Well....
\[\large \frac{2a}{3}+2c = 0\]\[c=-\frac{a}{3}\]

Answer 26

hold on, T (P(x)) = {int of (Px) | your condition, } not just ax^2 + bx +c, right?

Answer 27

T(P(x)) = integral of P(x)
that's it
ker(T) that's where the conditional set kicks in

Answer 28

and ker(T) when the condition =0 . I got it

Answer 29

yeap we meet there

Answer 30

Okay... so, the kernel of T is just
\[\huge \left\{ ax^2 + bx - \frac{a}3 \ \ \ \ \ | \ \ \ \ \ a,b \in \mathbb{R}\right\}\]

Answer 31

that's it?

Answer 32

Yep :)
Try integrating that from -1 to 1
you'll see it's zero :D

Answer 33

hehehe... you think who have to say thank you to other?

Answer 34

It's Terenze !!! for both black and white meanings !! cheer up

Answer 35

I'm cheering up :)

Answer 36

\[\huge T[f(x)]=\int\limits_{-1}^1f(x)dx\]
Terence, write a triple int and a double int . I need those code, whatever you construct, I just need the codes

Answer 37

to write a double integral simply write \iint
here's the result of writing \iint
\[\huge \iint\]
To write a double integral with a region under, say, R write \iint\limits_{R}
here's the result
\[\Large \iint\limits_{R}\]

To write a triple integral, write \iiint
\[\huge \iiint\]

Answer 38

how about the limit on?

Answer 39

\[\huge \iint\]

Answer 40

\iiint\limits_{}
same concept basically
\[\huge \iiint\limits_{S}\]

Answer 41

hey, friend, I mean specific limits, some thing like int_ from 0 to 1 and then int from 1 to 2 and then int from c to d

Answer 42

well, just write multiple integrals, then.
there's no other way
\int\limits_{a}^{b} \int\limits_{c}^{d}
\[\huge \int\limits_a^b\int\limits_c^d\]

Answer 43

thanks a lot. But I gotta run. will catch up with you later

Answer 44

sure

Answer 45

not this stuff,

Answer 46

let me repost, someone messed it up .and send you the link

Answer 47

@hoa there is a difference between multiple and iterated integrals.