# math

posted by .

A trigonmetric polynomial of order n is t(x) =
c0 + c1 * cos x + c2 * cos 2x + ... + cn * cos nx
+ d1 * sin x + d2 * sin 2x + ... + dn * sin nx

The output vector space of such a function has the vector basis:
{ 1, cos x, cos 2x, ..., cos nx, sin x, sin 2x, ..., sin nx }

Use the Gram-Schmidt process to find an orthonormal basis using the inner-product:
<f,g> = definite integral over [0,2*pi] of: f(x) * g(x) dx

------------------------------------------------------

The book gives the following answer for the orthonormal basis.

g0 = 1/sqrt(2*pi)
g1 = 1/sqrt(pi) * cos x
gn = 1/sqrt(pi) * cos nx

I've done the first two vectors using Gram-Schmidt and my answers don't match:

Original basis vector u0 = 1
Orthogonal basis vector v0 = 1
Orthonormal basis vector g0 = v0/|v0| = sqrt(2/pi)

Original basis vector u1 = cos x
Orthogonal basis vector v1 = u1 - <u1, g0> * g0
= cos x - sqrt(2/pi) * sqrt(2/pi) = cos x - 2/pi
Orthonormal basis vector g1 = v1/|v1| = (cos x - 2/pi) / sqrt((pi^2 - 8)/(4*pi))
= (cos x - 2/pi) * sqrt((4*pi)/(pi^2 - 8))

What am I doing wrong?

• math -

Original basis vector u0 = 1
Orthogonal basis vector v0 = 1
Orthonormal basis vector g0 = v0/|v0|

|V0|^2 = 2 pi --->

g0 = 1/sqrt(2 pi)

Original basis vector u1 = cos x
Orthogonal basis vector v1 = u1 - <u1, g0> * g0 = cos x because
<cos(x), 1> = 0

• math -

Thanks Count Iblis!

I was mistakenly integrating with pi/2 instead of 2*pi and every time I redid the problem, I just remade the same mistake without noticing it.

Your help pointed out the issue. Thanks so much!

## Similar Questions

1. ### trig

Reduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The quantity …
2. ### tigonometry

expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b) …
3. ### Trig

Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v - u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos …

5. ### pre-cal

Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x -48
6. ### Mathematics - Trigonometric Identities

Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y …
7. ### TRIG!

Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 …
8. ### Calc.

Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= …
9. ### calculus

Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= …
10. ### Trig

Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(-1/5)Sin(3/5) = 0.389418 Sin(s-t) =sin(s)cos(t) - cos(s)sin(t) =sin(-3/5)cos(1/5) - cos(1/5)sin(3/5) …

More Similar Questions