Welcome to our lesson
on linear dependent functions.
In this video, we'll define linear dependent
and linear independent functions
and then show functions are linear dependent.
A set of functions f sub one of x through f sub n of x
is linear dependent on an interval I
if there exists constants c sub one through c sub n
not all zero that satisfy this equation here.
Notice how we have multiples of the functions,
where the sum is equal to zero.
To better understand this concept,
let's consider two functions, f sub one and f sub two
that are linear dependent.
Which means, c sub one times f sub one of x
plus c sub two times f sub two of x
must equal zero.
Let's go ahead and solve this for f sub one of x.
So subtract this quantity here on both sides,
giving us c sub one times f sub one of x
equals negative c sub two times f sub two of x.
Now divide both sides by c sub one,
which would give us f sub one of x
equals negative c sub two divided by c sub one,
which is just another constant times f sub two of x.
So we can say that two functions are linear dependent
if one is a constant multiple of the other.
In this case, the constant would be
negative c sub two divided by c sub one.
Two functions are linear independent
if neither is a constant multiple of the other
on a given interval.
Let's take a look at our first example.
We want to determine the values of c sub one and c sub two
to verify the given functions are linear dependent.
Which means c sub one, some constant,
times e to the power of two x plus c sub two
times e raised to the power of two
times a quantity x minus one
must equal zero.
Let's start by distributing here.
Which would give us c sub one e to the two x
plus c sub two e to the power of,
this would be two x minus two,
which I'm going to write as two x plus negative two,
just to emphasize that we can now apply
this property of exponents here,
meaning because we have a sum here,
we can write this as a product with the same base.
So we can write this as c sub one e to the two x
plus c sub two
e to the two x
times e to the power of negative two equals zero.
With the negative two, we just know they're constant,
we can write this using a positive exponent,
if we move it down to the denominator.
So now let's write this as c sub one e to the two x
plus, since c sub two is also a constant,
let's write this as c sub two divided by e squared
e to the two x equals zero.
Now if we take a look at the exponential parts,
notice how they're both e raised to the power of two x,
and therefore if these are opposites,
this sum would be zero.
So let's go ahead and let c sub two be equal to e squared,
which would make this be positive one e to the two x,
and therefore if we let c sub one be equal to negative one,
this would be negative e to the power of two x.
So just to verify,
if c sub one is negative one,
and you have negative one e to the power of two x,
plus c sub two is e squared,
notice how here we have negative e to the two x,
and here we'd have plus e to the two x,
which is equal to zero,
and therefore we verified the given functions
are linear dependent with this value of c sub one
and this value of c sub two.
Let's take a look at another example.
Again we want to verify these three functions
are linear dependent by determining the values
of c sub one, c sub two, and c sub three,
that would satisfy the given equation.
c sub one times three
plus c sub two times cosine squared x
plus c sub three times sine squared x
equals zero.
Well, having this cosine squared x and sine squared x here
should remind us of this identify here.
So if we let c sub two be equal to one,
and c sub three be equal to one,
this sum here would be equal to positive one,
so if we can make this product here negative one,
we can verify these are linear dependent functions.
So again, this equals one if c sub two equals one
and c sub three equals one.
Well, three is the same as three over one.
So we multiply by the reciprocal or 1/3,
this would be positive one
and since we want negative one,
we'll let c sub one be equal to negative 1/3.
Again, let's go ahead and verify.
We'd have negative 1/3 times three plus again
c sub two and c sub three are equal to one,
so we just have cosine squared x
plus sine squared x
equals zero,
while this product here is negative one,
and this sum here is equal to positive one,
verifying that we do have three linear dependent functions,
verified by these three values of c.
Let's take a look at one more example.
Where here we have four different functions,
so we'll have to find c sub one through c sub four.
So we'd have c sub one times sine squared x
plus c sub two times cosine squared x
plus c sub three times secant squared x
plus c sub four times tangent squared x
must equal zero for some values of c
if these are linear dependent functions.
Well, from our last example,
we should be able to recognize,
that if we let c sub one be equal to one
and c sub two be equal to one,
this sum here would be equal to positive one.
So now if we can determine the value
of c sub three and c sub four,
that would make this sum equal to negative one,
we would satisfy the equation.
So looking at the identify here,
tan squared x plus one equals secant squared x,
notice if we subtracted one on both sides,
and subtracted secant squared on both sides,
we would have tangent squared x
minus secant squared x
equals negative one.
Which means, if we let c sub four be equal to positive one,
and c sub three be equal to negative one,
to get the minus secant squared x,
we would satisfy the equation.
So again, let's go ahead and verify this.
c sub one and c sub two are one,
we have sine squared x plus cosine squared x,
and then c sub three is equal to negative one,
so we have minus secant squared x,
c sub four is one,
so plus tangent squared x equals zero.
Well, this sum is one.
And we just showed that tangent squared x
minus secant squared x
is equal to negative one,
and one minus one is equal to zero.
So the functions are linear dependent
and we now know this because we found the values
of c sub one through c sub four
that satisfy this equation here.
Now when setting differential equations,
we're normally more concerned
about determining linear independent functions,
which will be the focus of the next lesson.
I hope you found this helpful.