Question

@Miracrown

Answer 1

just part a) how do I apply Gram-Schmidt to this, \[w = Span\left\{ (1+x^2),(x-x^2) \right\}\]

Answer 2

\[p = (1+x^2)~~~\text{and}~~~~q = x-x^2\] so \[<p,q> = -4\] that's not orthogonal but idk how to do this

Answer 3

I think we let \[\vec v_1 = <p,q>\]

Answer 4

wait no

Answer 5

\[\vec v_1 = 1+x^2\]

Answer 6

lolol

Answer 7

\[\vec v_2 = (x-x^2)-\frac{ <(x-x^2) \cdot \vec v_1> }{ <\vec v_1, \vec v_1> } \vec v_1\] actuall idk

Answer 8

we start with one of the vectors itself, then to get the next orthgonal one, we take the next one and subtract it's projection off

Answer 9

yeah

Answer 10

I guess I'm confused because of the inner product

Answer 11

so, we would take v2 - proj_u1(v2)

Answer 12

yeah but what is v1 and v2 l0l

Answer 13

thats my main confusion

Answer 14

v1 and v2 are 1+x^2 and x - x&2

Answer 15

naw gurl

Answer 16

in either order you like xD

Answer 17

wut

Answer 18

i think v1 is 1+x^2 but if we're using gram-schmidt v2 is a bit tricky

Answer 19

i think its that ugly thing i put up there, but what do i do with

Answer 20

\[<p,q>=p(-1)q(-1)+p(0)q(0)+p(1)q(1)\]

Answer 21

plug and chug where

Answer 22

Well, we have to project. So, we would get proj_u1(v2) = <v2,u1> / <u1,u1> * u1

Answer 23

and YES

Answer 24

so, if v1 = 1+x^2 then so does u1

Answer 25

what is u1 now

Answer 26

so v2 = x - x^2 so v2 = x - x^2 so we want to evaluate those at -1, 0 , and 1

Answer 27

u1 = v1

Answer 28

where the v's are the original functions, and the u's are the orthoganol ones cx

Answer 29

ok I see you just made v1 = u1 ahaha, but i still dont see, with the given info, i understand its the projection but

Answer 30

ok, lets go through et

Answer 31

\[\vec v_1 = 1+x^2\]

Answer 32

ok watch me whip

Answer 33

then watch me nei nei

Answer 34

first let's evaluate the functions at -1, 0, and 1
v1(-1) = 2, v1(0) = 1, v1(1) = 2

Answer 35

so, u1(-1) = 2, u1(0) = 1, u1(1) = 2 as well for v2

Answer 36

v2(-1) = -2, v2(0) = 0, v2(1) = 0

Answer 37

so, <v2,u1> = 2*-2+1*0+2*0 = -4

Answer 38

Wait, so lets let \[\vec v_1 = \vec u_1\]
where \[\vec u_1 = 1+x^2\] then we have \[\vec v_2 = \vec u_2 - \frac{ <\vec u_2, \vec v_1 >}{ <\vec v_1, \vec v_1> } \vec v_1\] where \[\vec u_2 = x-x^2\]

Answer 39

and <u1,u1> = -2*-2 + 0*0 + 0*0 = 4

Answer 40

so the projection is
proj_u1(v2) = <v2,u1> / <u1,u1> * u1 = -4/4 * (1+x^2) = -1-x^2

Answer 41

so, we get u2 = v2 - proj_u1(v2)

Answer 42

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

Answer 43

can you use latex idk what you're doing rofl, i mean i think you're missing some info

Answer 44

i don't know how to latex and i don't like et so go with what i have pls

Answer 45

wooow

Answer 46

ok

Answer 47

i think what i said was right

Answer 48

ganeshie didnt even look and bounced

Answer 49

i got a medal frm him tho lmao

Answer 50

wooow

Answer 51

oh was it chu e.e

Answer 52

look at what i said

Answer 53

ill give u one after

Answer 54

\(\color{blue}{\text{Originally Posted by}}\) @Astrophysics
Wait, so lets let \[\vec v_1 = \vec u_1\]
where \[\vec u_1 = 1+x^2\] then we have \[\vec v_2 = \vec u_2 - \frac{ <\vec u_2, \vec v_1 >}{ <\vec v_1, \vec v_1> } \vec v_1\] where \[\vec u_2 = x-x^2\]
\(\color{blue}{\text{End of Quote}}\)

Answer 55

YES THAT WORKS!!!

Answer 56

and then just plug shat in

Answer 57

like

Answer 58

i think thats where i was having the real trouble

Answer 59

i just figured this out

Answer 60

though, you are using v's and u's in the opposite way that I was, but I can adopt your notation

Answer 61

haha yeah its because when i use gram-schmidt the first time i let v1 = x1 etc but that would make it convoluted

Answer 62

cha cha

Answer 63

im gonna keep bumping this thread cause medals

Answer 64

lol we are doomed to a little convolution no matter what we do

Answer 65

ok so plugging in the numbers is where i need your help really

Answer 66

because im a dog and have paws its sort of hard

Answer 67

plug in -1, 0, and 1 into the vectors they give us so that we can take the inner product in the projection

Answer 68

okie so we have

Answer 69

\[p(x) = 1+x^2\]
\[q(x)=x-x^2\]
\[p(-1) = 2\]
\[p(0)=1\]
\[p(1) = 2\]

\[q(-1) = -1-(-1)^2 = -2, ~~~q(0) = 0, ~~~q(1) = 0\] ooooh

Answer 70

lookin' gud

Answer 71

So we have \[< \vec u_2 , \vec v_1> = q(-1)p(-1)+q(0)p(0)+q(1)p(1)\]

Answer 72

YAS

Answer 73

if you are thinking about it in terms of a direct product, then yes, you could do that, but it's not really what they want you to be thinking

Answer 74

yeah I see

Answer 75

So we have

Answer 76

\[< \vec u_2, \vec v_1> = (-2)(2)+0+0 = -4\]

Answer 77

Yerp

Answer 78

\[\vec v_2 = \vec u_2 - \frac{ <\vec u_2, \vec v_1 >}{ <\vec v_1, \vec v_1> } \vec v_1 \]

\[<\vec v_1, \vec v_1> = ||\vec v_1||^2 \] \[||\vec v_1||^2 = q(-1)^2+q(0)^2 +q(1)^2 = (-2)^2 = 4\]

Answer 79

Also yerp :D

Answer 80

Ok yay so far so good, then \[\vec v_2 = (x-x^2)-\frac{ 4 }{ 4 }(1+x^2) \implies (x-x^2)-(1+x^2) \]
\[\implies (x-x^2-1-x^2) = (-2x^2+x-1)\]

Answer 81

don't forget one of those was a -4

Answer 82

Oh crap

Answer 83

Ty, I was thinking about this other theorem for a sec haha ortho. decomposition theorem or w/e ok so we have \[\vec v_2 = (x-x^2)+1(1+x^2) \implies (x-x^2+1+x^2) = (x+1)\]

Answer 84

Merp :P

Answer 85

I do like the sound of that

Answer 86

Haha oh yea

Answer 87

<33333333 I just needed you to watch, since you're such an inspiration

Answer 88

No worries hehehehe and awww thx <3

Answer 89

ty ty the rest should be good :D

Answer 90

All da best man!

Answer 91

:D

Answer 92

just bounces

Answer 93

lel

Answer 94

ganeshie saw your name medalled and bounceddd

Answer 95

hahaha he could sense it rofl