It’s Professor Dave, let’s learn about Taylor and Maclaurin series.
In the previous tutorial, we talked about power series.
These were series in the following form, C sub N times X to the N, from zero to infinity.
We mentioned that the sum of a power series can be represented by a function, with terms
involving each power of X all the way to infinity, but we may want to know something about all
of these coefficients.
Now that we understand differentiation, we are ready to expand our understanding of this
kind of series, so let’s talk about this now.
Say we are looking at this power expansion, the series C sub N times the quantity X minus
A, raised to the N power.
We can describe this with a function, but we may want to solve for these coefficients.
It actually won’t be as tricky as you think, given the difference X minus A in each of
these terms.
If we find F of A, all of these binomials become zero, and therefore all of these terms
become equal to zero, except for the first coefficient here, C zero.
So F of A equals C zero.
With the first coefficient down, how do we get the next one?
This is where differentiation comes into play.
Let’s say we take the derivative of this function.
C zero will go away, since the derivative of any constant is zero.
Then for the second term, X minus A will go away, leaving C one.
To be clear, if we were to distribute before taking the derivative, we would get C one
X minus C one A, and since C one A is just a number, that will go away, and C one X will
become C one.
For the other terms we will use the chain rule.
Starting with this one, the two comes down here, leaving an exponent of one, and then
we multiply by the derivative of what’s inside, which will just be one, so we are
left with two C two times the quantity X minus A. The second part of the chain rule will
always be one for the rest of these terms, so we can just do the part with the exponent.
Three comes down here and the exponent becomes two.
Four comes down here and the exponent becomes three.
Now with this expression, which represents the first derivative of the function, if we
plug in A, all of these terms will disappear, just like when we plugged A into the original function.
The only thing left is C one.
So F prime of A, or the derivative of F evaluated at A, equals C one.
Let’s go back to the expression for the first derivative and take the derivative again.
C one goes away, and this term becomes two C two.
Then we have two times three, or six C three times the quantity X minus A, this next coefficient
becomes twelve, and so forth.
Again, plugging in A for X will isolate the two C two term, so F double prime of A equals
two C two, or C two equals F double prime of A over two.
By now we might see a pattern.
Taking the third derivative and plugging in A will give us F triple prime of A equals
six C three, or C three equals F triple prime of A over six, which can also be expressed
as three factorial.
Inputting A into the fourth derivative of F will give us twenty-four C four, meaning
that C four equals the fourth derivative of F over four factorial.
We can therefore set up an expression to find the nth coefficient of the series.
C sub N equals the nth derivative of F evaluated at A, divided by N factorial.
This works for all the terms, even the first, if we understand that the zeroth derivative
of a function is simply the function itself, and that zero factorial equals one.
So to summarize, for the function given by the sum of C sub N times the quantity X minus
A to the N power, where the absolute value of X minus A is less than the radius of convergence
for the series, the coefficients of the function are given by this formula.
If we then generate a new function of the same form, but substitute this formula in
for C sub N, such that the terms in the series look like this, we have just generated the
Taylor series of the function F at A, which can also be described as F about A, or F centered
at A. So a Taylor series is a type of power series but with the coefficients defined like so.
Now that we understand what a Taylor series is, let’s also define a special case.
For the same type of function, but where A equals zero, this binomial becomes simply
X to the N, and everywhere we were plugging in A, we simply plug in zero.
So we get F of zero, plus F prime of zero over one factorial times X, plus F double
prime of zero over two factorial times X squared, and so forth.
This is a special kind of Taylor series called a Maclaurin series.
This is a Taylor series where the expansion is about the point X equals zero.
So we should definitely be aware of these definitions for a Taylor series and Maclaurin
series, because we are going to apply these expansions to specific functions.
Take for example E to the X.
Let’s find the Maclaurin series that represents the function F of X equals E to the X.
That will be the nth derivative of E to the X evaluated at zero, over N factorial, times
X to the N. Well what do we know about E to the X?
The derivative of E to the X is E to the X, so no matter what derivative we are looking
at, we just have E to the X.
Also, we are plugging in zero every time, and anything to the zero power is one.
That means any derivative of this function evaluated at zero will give us one, so this
entire term can just go away, leaving us with X to the N over N factorial.
We can therefore quite easily begin writing out the series, with X to the zero over zero
factorial, or simply one, plus X to the one over one factorial, plus X squared over two
factorial, and so forth.
This should look very familiar, as we have just derived a form for E to the X that we
utilized earlier in this calculus series.
That form was actually the Maclaurin series for the function.
Let’s make sure we can find the radius of convergence for this function.
That will require the ratio test.
We set that up and flip the denominator to get X to the (N plus one) over (N plus one)
factorial, times N factorial over X to the N.
After some simplification, we can cancel some terms, and we get the absolute value of X
over N plus one, and as N approaches infinity, this goes to zero.
Zero is less than one, so the series converges for all values of X and the radius of convergence
is infinite.
Taylor series have many applications, but for now we will be satisfied with some simple
definitions and come back to these concepts later.
And with that, we have wrapped up our survey of calculus, so it’s time to move on to
some other topics.
Before we leave, let’s check comprehension.