In this section we need to take a look at the equation of a line in R 3 R 3 . As we saw in the previous section the equation y = m x + b y = m x + b does not describe a line in R 3 R 3 , instead it describes a plane. This doesn’t mean however that we can’t write down an equation for a line in 3-D space. We’re just going to need a new way of writing down the equation of a curve. So, before we get into the equations of lines we first need to briefly look at vector functions. We’re going to take a more in depth look at vector functions later. At this point all that we need to worry about is notational issues and how they can be used to give the equation of a curve. The best way to get an idea of what a vector function is and what its graph looks like is to look at an example. So, consider the following vector function. → r ( t ) = ⟨ t , 1 ⟩ r → ( t ) = ⟨ t , 1 ⟩ A vector function is a function that takes one or more variables, one in this case, and returns a...
Example 2 Show that is mutually orthogonal on and on .
First, we’ll acknowledge from the start this time that we’ll be showing orthogonality on both of the intervals. Second, we need to start this set at because if we used the first function would be zero and we don’t want the zero function to show up on our list.
As with the first example all we really need to do is evaluate the following integral.
In this case integrand is the product of two odd functions and so must be even. This means that we can again use Fact 3 above to write the integral as,
We only have two cases to do for the integral here.
Not much to this integral. It’s pretty similar to the previous examples second case.
Summarizing up we get,
As with the previous example this can be a little messier but it is also nearly identical to the third case from the previous example so we’ll not show a lot of the work.
As with the previous example we know that and are both integers a and so both of the sines above must be zero and all together we get,
So, we’ve shown that if the integral is zero and if the value of the integral is a positive constant and so the set is mutually orthogonal.
As with the first example all we really need to do is evaluate the following integral.
In this case integrand is the product of two odd functions and so must be even. This means that we can again use Fact 3 above to write the integral as,
We only have two cases to do for the integral here.
Not much to this integral. It’s pretty similar to the previous examples second case.
Summarizing up we get,
As with the previous example this can be a little messier but it is also nearly identical to the third case from the previous example so we’ll not show a lot of the work.
As with the previous example we know that and are both integers a and so both of the sines above must be zero and all together we get,
So, we’ve shown that if the integral is zero and if the value of the integral is a positive constant and so the set is mutually orthogonal.
We’ve now shown that is mutually orthogonal on and on .
We need to work one more example in this section.
Example 3 Show that and are mutually orthogonal on .
This example is a little different from the previous two examples. Here we want to show that together both sets are mutually orthogonal on . To show this we need to show three things. First (and second actually) we need to show that individually each set is mutually orthogonal and we’ve already done that in the previous two examples. The third (and only) thing we need to show here is that if we take one function from one set and another function from the other set and we integrate them we’ll get zero.
Also, note that this time we really do only want to do the one interval as the two sets, taken together, are not mutually orthogonal on . You might want to do the integral on this interval to verify that it won’t always be zero.
So, let’s take care of the one integral that we need to do here and there isn’t a lot to do. Here is the integral.
The integrand in this case is the product of an odd function (the sine) and an even function (the cosine) and so the integrand is an odd function. Therefore, since the integral is on a symmetric interval, i.e. , and so by Fact 3 above we know the integral must be zero or,
So, in previous examples we’ve shown that on the interval the two sets are mutually orthogonal individually and here we’ve shown that integrating a product of a sine and a cosine gives zero. Therefore, as a combined set they are also mutually orthogonal.
Also, note that this time we really do only want to do the one interval as the two sets, taken together, are not mutually orthogonal on . You might want to do the integral on this interval to verify that it won’t always be zero.
So, let’s take care of the one integral that we need to do here and there isn’t a lot to do. Here is the integral.
The integrand in this case is the product of an odd function (the sine) and an even function (the cosine) and so the integrand is an odd function. Therefore, since the integral is on a symmetric interval, i.e. , and so by Fact 3 above we know the integral must be zero or,
So, in previous examples we’ve shown that on the interval the two sets are mutually orthogonal individually and here we’ve shown that integrating a product of a sine and a cosine gives zero. Therefore, as a combined set they are also mutually orthogonal.
We’ve now worked three examples here dealing with orthogonality and we should note that these were not just pulled out of the air as random examples to work. In the following sections (and following chapter) we’ll need the results from these examples. So, let’s summarize those results up here.
- and are mutually orthogonal on as individual sets and as a combined set.
- is mutually orthogonal on .
- is mutually orthogonal on .
We will also be needing the results of the integrals themselves, both on and on so let’s also summarize those up here as well so we can refer to them when we need to.
With this summary we’ll leave this section and move off into the second major topic of this chapter : Fourier Series.
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