Question

Let f(x)= 5, g(x)=-3 x - 1, and h(x)=-4 x^2.
Consider the inner product \displaystyle = integral from 0 to 4 p(x)q(x) dx in the vector space C^0[0,4]. Use the Gram-Schmidt process to determine an orthonormal basis for the subspace of C^0[0,4] spanned by the functions f(x), g(x), and h(x).... help??

Answer 1

opps sorry copied the \displaystyle. any ideas of where to start?

Answer 2

have done gram-schmidt before with vectors and only remember that it is a real pain. don't know how one does it with inner product given as integral. sorry

Answer 3

let me look in taylor

Answer 4

using functions makes things a little weird thats what throws me for a loop

Answer 5

so are we thinking of f, g and h as vectors? then first step is to normalize them yes?

Answer 6

\[\int_0^2 f^2dx=\int_0^4 25dx=100\]

Answer 7

so normalized we get
\[v_1(x)=\frac{1}{25}\] hope this is the right track. maybe i am all wrong

Answer 8

so your normalizing by squaring the function and dividing by it. so far seems to be equivalent to a unit vector

Answer 9

so what would my vectors be for f(x) g(x) h(x)? i guess if I have these I could solve similarly to the normal gram-schmidt way.

Answer 10

oh i see f(x) would be (1 0 0) g(x) is (-1 -3 0) ect ect

Answer 11

rather (5 0 0)

Answer 12

ok this is what i have written in taylor:
If A_1, A_2, A_3 we start by defining
\[v_1=\frac{A_1}{|A_1|}\] and let\[ B_2 = A_2 -\text{ the projection of} A_1 \text{on } v_1\]

Answer 13

in our case i guess
\[A_1=5\] so \[v_1=\frac{1}{25}\]

Answer 14

now
\[B_2=A_2-(A_2\circ v_1)v_1\] where circle means the integral.

Answer 15

so it looks like
\[B_2=g(x)-(\int_0^4g(x)\times5)\times \frac{1}{25}\]

Answer 16

if i am reading this thing correctly.

Answer 17

inner product means integral and i am just writing down what the process is. i could be way way off

Answer 18

well lets just check

Answer 19

what would our first vector basis be?

Answer 20

I think it's off. . .

Answer 21

\[\frac{1}{25}\] which is
\[f(x)=5\] normalized

Answer 22

we should get a (v1, v2, v3) such that these are the vectors spanning our basis

Answer 23

1/25 is incorrect

Answer 24

ok

Answer 25

you need more than one answer for gram schmidt

Answer 26

no worries i appreciate the help so far

Answer 27

yes we understand that, but we should get an answer for each step, thus 3 vectors spanning our basis

Answer 28

no just one step at a time.
\[\int_0^4 5^2dx = 100\] yes? i thought the first vector in your normalization process was just the first vector normalized. then you go from there. do you happen to know what the first vector is?

Answer 29

first vector is just a constant and norm is the integral so when you divide you get a vector of length one yes?

Answer 30

in normal gram schmidt you normal the first vector, this is your first unit vector, which is the first of the (v1, v2, ...vn) vectors that are orthonormal

Answer 31

Sorry to but in. . .I have my lin alg book next to me now.

Answer 32

by all means how does it work with functions because i cant figure it out

Answer 33

please help because i have only done this with vectors written as coordinates

Answer 34

Ok. . .

Answer 35

oh i see a mistake on my part. yikes. integral is 100 but norm is square root of that, so 10

Answer 36

I think you were doing the right thing. . . <u1, v1> is the def integral of the first 2 functions multiplied together

Answer 37

so normalized first vector is
\[v_1=\frac{1}{2}\]

Answer 38

thats correct

Answer 39

now let me try to find
\[B_2\] which is
\[B_2=A_2-(A_2\circ v_1)\times v_1\]

Answer 40

\[B_2=g(x)-(\int_0^4g(x)\times \frac{1}{2})dx\times {1}{2}\]

Answer 41

should be times 1/2 not times 12

Answer 42

by the way, where are you getting this from so I can understand the process

Answer 43

i am looking at a copy of Taylor advanced calc

Answer 44

ah ok

Answer 45

the integral is -7 so i get
\[B_2=-3x-8\]

Answer 46

now we need
\[\frac{B_2}{||B_2||}\] and
\[||B_2||=\sqrt{\int_0^4 (-3x-8)^2dx}\]

Answer 47

i get 832 for the integral and
\[8\sqrt{13}\] for the norm

Answer 48

mathgirls still there? did i mess up yet?

Answer 49

shes gone

Answer 50

which means we get -3x-8/8sqrt(13)

Answer 51

that is what i get, but really i would not bet my beer money on it

Answer 52

do you have any idea what the answer is?

Answer 53

you seemed to know that the first was 1/2

Answer 54

its a webwork with unlimited attempts so all i can do is try, but i have no idea what the sanwer is nor can i find a decent example to explain the process

Answer 55

can you try that answer? because there is no use in continuing if that is wrong

Answer 56

its not correct

Answer 57

damn

Answer 58

yeah

Answer 59

ill figure it out. no worries i appreciate the help!

Answer 60

sorry. it is
\[g(x)=-3x-1\] yes?

Answer 61

yeah i tried 8sqrt(13)*(-3x-1)

Answer 62

integral is right i checked it

Answer 63

integral is -14 then multiply by
\[v_1\] which gives is 1/2 giving -7 and oh damn!

Answer 64

it says SUBTRACT what a bush league error.

Answer 65

lordamercy it is
\[g(x)-(-7)=g(x)+7=-3x+6\]

Answer 66

ok now lets try it with that. the norm is
\[\sqrt{\int_0^4(-3x+6)^2dx}\]

Answer 67

integral is 48 norm is
\[4\sqrt{3}\]

Answer 68

making
\[v_2=\frac{-3x+6}{4\sqrt{3}}\]

Answer 69

yes nice

Answer 70

what an elementary mistake. subtracting -7 by subtracting 7. try that and see if we hae better luck

Answer 71

yeah its correct

Answer 72

whew. now i feel like somewhat less of an idiot. ok one more to go yes?

Answer 73

yep. ill definitely have to review this process, it is not like the regular gram- schmidt

Answer 74

\[B_3=A_3-(A_3\circ v_1)v_1-(A_3\circ v_2)v_2\]

Answer 75

just so i am understanding the terminology here, b3 is my projection of my third vector onto my second?

Answer 76

\[A_3=h(x)=-4x^2\]
\[(A_3\circ v_1)=\int_0^4(-4x^2)(\frac{1}{2})dx\times \frac{1}{2}\]

Answer 77

\[B_3\] is orthogonal to
\[v_1\] and
\[v_2\]

Answer 78

finally
\[(A_3\circ v_2)v_2=\int(-4x^2)(\frac{-3x+6}{4\sqrt{3}})dx\times (-3x+6)\]

Answer 79

\[(A_3 \circ v_1)=\int_0^4x^2dx=\frac{64}{3}\]

Answer 80

this is a pain just like i remembered

Answer 81

well if i could give you 50 medals i definitely would lol

Answer 82

\[\int_0^4 \frac{3x^3-6x^2}{\sqrt{3}}dx=\frac{64\sqrt{3}}{3}\]\]

Answer 83

multiply by
\[-3x+6\] so get
\[64\sqrt{3}x+128\]

Answer 84

so
\[B_3=-4x^2-\frac{64}{3}-64\sqrt{3}x-128\]

Answer 85

guess we could clean that up.

Answer 86

i get
\[-4x^2-64\sqrt{3}x-\frac{448}{3}\] but by now i could have made a mistake.

Answer 87

and finally... we need
\[\sqrt{\int_0^4 (-4x^2-64\sqrt{3}x-\frac{448}{3})^2dx}\]

Answer 88

i get something really ugly.

Answer 89

lemme try

Answer 90

oh damn damn damn

Answer 91

uh oh a mistake somewhere?

Answer 92

oh no i thought i caught another stupid error but it is ok

Answer 93

i wolframed it and got approx 837

Answer 94

hold on i may have made another stupid error

Answer 95

another goddamn minus sign

Answer 96

\[(A_3\circ v_2)=-64\sqrt{3}x+128\]

Answer 97

so when we SUBTRACT we get
\[B_3=-4x^2+64\sqrt{3}x-128-\frac{64}{3}\]

Answer 98

\[B_3=-4x^2+64\sqrt{3}x-\frac{448}{3}\]