how do you prove that sin t, sin 2t, sin 3t are linearly independent?

Answer 1

Find three points values of t, t1, t2, and t3, such that the vectors:

(sin t1, sin t2, sin t3)

(sin 2 t1, sin 2 t2, sin 2 t3)

(sin 3 t1, sin 3 t2, sin 3 t3)

are linearly independent.

I'd take t1 = pi/6, t2 = pi/3, t3 = pi/2 so you get vectors:

v1=(1/2, sqrt(3)/2, 1)

v2=(sqrt(3)/2, sqrt(3)/2, 0)

v3=(1, 0, -1)


now, if a* sin t + b*sin 2t + c*sin 3t = 0, then

a*v1 + b*v2 + c*v3 = 0.

But then a = c from the third term.

a * (v1 + v3) = (3a/2, a sqrt(3)/2, 0)

0 = a * (v1 + v3) + c * v2 = (3a/2 + c sqrt(3)/2, (a+c)sqrt(3)/2, 0)

So (a+c) must be zero, or a = -c.

But then the first coordinate is 3a/2 - sqrt(3) a /2

Which can be zero only if a=0, in which case, a=b=c=0.

So the vectors are independent, and hence the functions are independent.

Perhaps a quicker way is to prove first:

sin t and sin 2t are independent.

Then show that:

a sin t + b sin 2t

cannot have period 2 pi/3 without a=b=0. So if

a sin t + b sin 2t = -c sin 3t

then a=b=c=0.

Answer 2

Use the Wronskian. If the Wronskian is not 0 then the functions are linearly independent.

Create a matrix. First row are the functions, second row are the 1st derivatives, and the third row are the 2nd derivatives. The Wronskian is the determinant of this matrix.

| sint .. sin2t ........ sin3t |
| cost 2cos2t .. 3cos3t | = |W|
| -sint -4sint2t . -9sint3t |

|W| = -18sint cos2t sin3t - 3sint sin2t cos3t - 4cost sin2t sin3t
.....+ 2sint cos2t sin3t + 12sint sin2t cos3t + 9cost sin2t sin3t

Which is clearly not 0 so the functions are linearly independent.

To show linear dependence, suppose the functions are 3x and 5x

|3x 5x|
| 3 5 |

|W| = 15x - 15x = 0

Answer 3

The same way you would prove any three vectors are independent: Show that, for any constants c1, c2, c3, the only solution to c1 sin t + c2 sin 2t + c3 sin 3t = 0 is c1=c2=c3=0.

A neat way of showing independence that is tailor-made for functions is to use the Wronskian determinant

W(f1,f2,f3) = det[(f1,f2,f3) (f1',f2',f3') (f1'',f2'',f3'')].

If W is never zero for any t, then the functions f1, f2, and f3 are linearly independent. So compute the determinant

[(sin t, sin 2t, sin 3t)
(cos t, 2 cos 2t, 3 cos 3t)
(-sin t, -4 sin 2t, -9 sin 3t)]

and verify that it never equals zero.

Answer 4

If asint +bsin2t +csin3t = 0 for all t implies a, b, and c =0 then the three are indep.
@ t = pi/2 we get a - c =0 since sinpi/2 =1, sinpi = 0 and
sin3pi/2 = - 1
we also have acost +2bcos2t +3ccos3t =0
@ pi/2 we get -2b =0 since cospi/2 =cos3pi/2 =0 &cospi = -1
so we have b = 0
@pi we get -a -3c =0 since cospi = cos3pi = -1
Using a - c = 0 and this last -a - 3c = 0 we get a = c = 0
So a = b = c =0 which proves indepencence.

Answer 5

the belief is to break up y = a * b * c into y = (a * b) * c so as that y' = (a*b) * c' + (a*b)' * c the place (a*b)' = a' * b + b' * a (observe a' is by-fabricated from a with comprehend to t) -------------------------------- First by-product: enable u = sin(2t)*exp(3t) and v = (a million + 5t + t^2) So y = u*v And enable a = sin(2t), b = exp(3t) Then u = a*b So u' = a*b' + b*a' (Product rule) = sin(2t) * 3exp(3t) + 2cos(2t)*exp(3t) = exp(3t) * (3sin(2t) + 2cos(2t)) additionally v' = 5 + 2t Then y' = v*u' + u*v' = (a million + 5t + t^2) * exp(3t) * (3sin(2t) + 2cos(2t)) + sin(2t) * exp(3t) * (5 + 2t) --------------------------------------... utilising comparable technique, 2nd by-product is: y'' = (3sin(2t) + 2cos(2t)) * [(a million + 5t + t^2) * exp(3t)]' + (a million + 5t + t^2) * exp(3t) * [(3sin(2t) + 2cos(2t))]' + exp(3t) * (5 + 2t) * [sin(2t)]' + sin(2t) * [exp(3t) * (5 + 2t)]' = (3sin(2t) + 2cos(2t)) * exp(3t) * (3 * (a million + 5t + t^2) + (5 + 2t)) + (a million + 5t + t^2) * exp(3t) * (6cos(2t) - 4sin(2t)) + 2 * exp(3t) * (5 + 2t) * cos(2t) + sin(2t) * exp(3t) * (3 * (5+2t) + 2)) and simplify... Very long question, optimistically i did no longer make any stupid errors.