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Calculus III - 3-Dimensional Space: Equations of Lines

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...

Differential Equations - Fourier Series: Periodic & Orthogonal Functions - ii


Example 2 Show that {sin(nπxL)}n=1 is mutually orthogonal on LxL and on 0xL.

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 n=1 because if we used n=0 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.
LLsin(nπxL)sin(mπxL)dxIn 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,
LLsin(nπxL)sin(mπxL)dx=20Lsin(nπxL)sin(mπxL)dxWe only have two cases to do for the integral here.
n=m_
Not much to this integral. It’s pretty similar to the previous examples second case.

LLsin2(nπxL)dx=20Lsin2(nπxL)dx=0L1cos(2nπxL)dx=(xL2nπsin(2nπxL))|0L=LL2nπsin(2nπ)=LSummarizing up we get,
LLsin2(nπxL)dx=20Lsin2(nπxL)dx=Lnm_
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.

LLsin(nπxL)sin(mπxL)dx=20Lsin(nπxL)sin(mπxL)dx=0Lcos((nm)πxL)cos((n+m)πxL)dx=[L(nm)πsin((nm)πxL)L(n+m)πsin((n+m)πxL)]0L=L(nm)πsin((nm)π)L(n+m)πsin((n+m)π)As with the previous example we know that n and m are both integers a and so both of the sines above must be zero and all together we get,
LLsin(nπxL)sin(mπxL)dx=20Lsin(nπxL)sin(mπxL)dx=0So, we’ve shown that if nm the integral is zero and if n=m the value of the integral is a positive constant and so the set is mutually orthogonal.
We’ve now shown that {sin(nπxL)}n=1 is mutually orthogonal on LxL and on 0xL.
We need to work one more example in this section.

Example 3 Show that {sin(nπxL)}n=1 and {cos(nπxL)}n=0 are mutually orthogonal on LxL.

This example is a little different from the previous two examples. Here we want to show that together both sets are mutually orthogonal on LxL. 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 0xL. 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.
LLsin(nπxL)cos(mπxL)dxThe 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. LxL, and so by Fact 3 above we know the integral must be zero or,
LLsin(nπxL)cos(mπxL)dx=0So, in previous examples we’ve shown that on the interval LxL 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.

  1. {cos(nπxL)}n=0 and {sin(nπxL)}n=1 are mutually orthogonal on LxL as individual sets and as a combined set.
  2. {cos(nπxL)}n=0 is mutually orthogonal on 0xL.
  3. {sin(nπxL)}n=1 is mutually orthogonal on 0xL.
We will also be needing the results of the integrals themselves, both on LxL and on 0xL so let’s also summarize those up here as well so we can refer to them when we need to.

  1. LLcos(nπxL)cos(mπxL)dx={2Lif n=m=0Lif n=m00if nm
  2. 0Lcos(nπxL)cos(mπxL)dx={Lif n=m=0L2if n=m00if nm
  3. LLsin(nπxL)sin(mπxL)dx={Lif n=m0if nm
  4. 0Lsin(nπxL)sin(mπxL)dx={L2if n=m0if nm
  5. LLsin(nπxL)cos(mπxL)dx=0

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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