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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 vector. Note as we

Differential Equations - Series Solutions: Power Series - ii



Example 3 Write the following as a series that starts at n=5 instead of n=3.
n=3n2an1(x+4)n+2

To start the series to start at n=5 all we need to do is notice that this means we will increase the starting point by 2 and so all the other n’s will need to decrease by 2. Doing this for the series in the previous example would give,
n=3n2an1(x+4)n+2=n=5(n2)2an3(x+4)n

Now, as we noted when we started this discussion about index shift the whole point is to get our series into terms of (xx0)n. We can see in the previous example that we did exactly that with an index shift. The original exponent on the (x+4) was n+2. To get this down to an n we needed to decrease the exponent by 2. This can be done with an index that increases the starting point by 2.
Let’s take a look at a couple of more examples of this.


Example 4 Write each of the following as a single series in terms of (xx0)n.
  1. (x+2)2n=3nan(x+2)n4n=1nan(x+2)n+1
  2. xn=0(n5)2bn+1(x3)n+3

a) (x+2)2n=3nan(x+2)n4n=1nan(x+2)n+1 
First, notice that there are two series here and the instructions clearly ask for only a single series. So, we will need to subtract the two series at some point in time. The vast majority of our work will be to get the two series prepared for the subtraction. This means that the two series can’t have any coefficients in front of them (other than one of course…), they will need to start at the same value of n and they will need the same exponent on the xx0.
We’ll almost always want to take care of any coefficients first. So, we have one in front of the first series so let’s multiply that into the first series. Doing this gives,
n=3nan(x+2)n2n=1nan(x+2)n+1Now, the instructions specify that the new series must be in terms of (xx0)n, so that’s the next thing that we’ve got to take care of. We will do this by an index shift on each of the series. The exponent on the first series needs to go up by two so we’ll shift the first series down by 2. On the second series will need to shift up by 1 to get the exponent to move down by 1. Performing the index shifts gives us the following,
n=1(n+2)an+2(x+2)nn=2(n1)an1(x+2)nFinally, in order to subtract the two series we’ll need to get them to start at the same value of n. Depending on the series in the problem we can do this in a variety of ways. In this case let’s notice that since there is an n-1 in the second series we can in fact start the second series at n=1 without changing its value. Also note that in doing so we will get both of the series to start at n=1 and so we can do the subtraction. Our final answer is then,
n=1(n+2)an+2(x+2)nn=1(n1)an1(x+2)n=n=1[(n+2)an+2(n1)an1](x+2)n

b) xn=0(n5)2bn+1(x3)n+3 

In this part the main issue is the fact that we can’t just multiply the coefficient into the series this time since the coefficient doesn’t have the same form as the term inside the series. Therefore, the first thing that we’ll need to do is correct the coefficient so that we can bring it into the series. We do this as follows,
xn=0(n5)2bn+1(x3)n+3=(x3+3)n=0(n5)2bn+1(x3)n+3=(x3)n=0(n5)2bn+1(x3)n+3+3n=0(n5)2bn+1(x3)n+3We can now move the coefficient into the series, but in the process of we managed to pick up a second series. This will happen so get used to it. Moving the coefficients of both series in gives,
n=0(n5)2bn+1(x3)n+4+n=03(n5)2bn+1(x3)n+3We now need to get the exponent in both series to be an n. This will mean shifting the first series up by 4 and the second series up by 3. Doing this gives,
n=4(n9)2bn3(x3)n+n=33(n8)2bn2(x3)nIn this case we can’t just start the first series at n=3 because there is not an n3 sitting in that series to make the n=3 term zero. So, we won’t be able to do this part as we did in the first part of this example.
What we’ll need to do in this part is strip out the n=3 from the second series so they will both start at n=4. We will then be able to add the two series together. Stripping out the n=3 term from the second series gives,
n=4(n9)2bn3(x3)n+3(5)2b1(x3)3+n=43(n8)2bn2(x3)nWe can now add the two series together.
75b1(x3)3+n=4[(n9)2bn3+3(n8)2bn2](x3)nThis is what we’re looking for. We won’t worry about the extra term sitting in front of the series. When we finally get around to finding series solutions to differential equations we will see how to deal with that term there.

There is one final fact that we need take care of before moving on. Before giving this fact for power series let’s notice that the only way for
a+bx+cx2=0
to be zero for all x is to have a=b=c=0.
We’ve got a similar fact for power series.


Fact


If,
n=0an(xx0)n=0for all x then,
an=0,n=0,1,2,

This fact will be key to our work with differential equations so don’t forget it.

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