Find out that sets of functions are linearly dependent, Mathematics

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Find out if the following sets of functions are linearly dependent or independent.

 (a) f x ) = 9 cos ( 2x )    g x ) = 2 cos2 ( x ) -  2 sin 2 ( x )

(b) f (t ) = 2t 2                                g (t ) = t 4

Solution:

(a) f x ) = 9 cos ( 2x )     g x ) = 2 cos2 ( x ) -  2 sin 2 ( x )

We'll start via writing down (1) for these two functions.

c(9cos(2x)) + k(2cos2(x)) - 2sin2(x)) = 0

We need to find out if we can find non-zero constants c and k which will make this true for all x or if c = 0 and k = 0 are the only constants which will make this true for all x. it is frequently a fairly not easy process. The process can be simplified along with a good intuition for this type of thing, but that's tough to come by, particularly if you haven't done many of these types of problems.

Under this case the problem can be simplified through recalling

cos2(x) - sin2(x) = cos(2x)

Using this fact our equation turns into,

9c cos ( 2x ) + 2k cos ( 2 x ) = 0

(9c + 2k ) cos ( 2x ) = 0

With such simplification we can notice that this will be zero for any pair of constants c and k which satisfies.

9c + 2k = 0

In between the possible pairs on constants which we could use are the following pairs.

c= 1                 k= -(9/2)

c= 2/9              k = -1

c= -2                k = 9

c= -(7/6)                      k = 21/4

Well I'm sure you can notice there are literally thousands of possible pairs and they can be created as "simple" or as "complicated" as you need them to be.

Thus, we've managed to get a pair of non-zero constants which will make the equation true for all x and hence the two functions are linearly dependent.

(b) f (t ) = 2t 2                                g (t ) = t 4

As from the last part, we'll start through writing down (1) for these functions.

2ct2 + kt4 = 0

In this case there isn't any rapid and simple formula to write one of the functions in terms of another as we did in the first part. Therefore, we're just going to have to notice if we can find constants. We'll begin by noticing that if the original equation is true, so if we differentiate everything we find a new equation which must also be true. Conversely, we've got the subsequent system of two equations in two unknowns.

2ct 2 + kt 4  = 0

4ct + 4kt 3  = 0

 We can solve this system for c and k and notice what we find.  We'll start through solving the second equation for c.

c = -kt2

Then, plug this in the first equation.

2(-kt2)t2 + kt4 = 0

-kt4 = 0

So recall that we are after constants which will make it true for all t. The only manner that it will ever be zero for all t is if k = 0! Therefore, if k = 0 we should also have c = 0.

Thus, we've shown that the only way as,

2ct 2 + kt 4  = 0

It will be true for all t is to need that c = 0 and k = 0. The two functions thus, are linearly independent.

Since we saw in the previous illustrations determining whether two functions are linearly independent or dependent can be a fairly included process. It is where the Wronskian can assist.


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