Taylor and Maclaurin Series

I'm Professor Dave, let's get to know the 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 al 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 to infinity, but we may want to know something about all of these coefficients. Now that we understand the differentiation, we are ready to expand our understanding of these types of series, so let's talk about this now. Let's say we're looking at this power expansion, the series C sub N multiplied by the quantity X minus A, raised to the power N.

We can describe this with a function, but we might want to solve for these coefficients. In reality, it won't be as complicated 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 the first coefficient here, C zero. So F of A is equal to C zero. With the first coefficient reduced, how do we obtain the next one? This is where differentiation comes into play. Let's say we take the derivative of this function.

C zero will disappear, since the derivative of any constant is zero. Then for the second term, X minus A will disappear, leaving C one. To be clear, if we distributed 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 disappear and C one X will become C one. For the other terms we will use the chain rule. Starting with this one, the two goes down here, leaving an exponent of one, and then we multiply by the derivative of what's inside, which will be just one, so we're left with two C twice the amount X minus A.

The The Second Part The chain rule will always be the same for the rest of these terms, so we can just do the part with the exponent. Three goes down here and the exponent becomes two. Four goes down here and the exponent becomes three. Now, with this expression, which represents the first derivative of the function, if we replace A, all of these terms will disappear, just like when we replace A in the original function. The only thing left is C one. Then F prime of A, or the derivative of F evaluated at A, is equal to C one.

Let's go back to the first derivative expression and take the derivative again. C one disappears and this term becomes two C two. So we have two times three, or six C, three times the quantity X minus A, the next coefficient becomes twelve, and so on. Again, replacing A with At this point we could see a pattern. Taking the third derivative and replacing 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 a three factorial. Plugging A into the fourth derivative of F will give us twenty-four C four, which means that C four is equal to the fourth derivative of F over four factorial.

Therefore, we can set up an expression to find the nth coefficient of the series. C sub N is equal to the nth derivative of F evaluated at A, divided by factorial N. This works for all terms, even the first, if we understand that the zero derivative of a function is simply the function itself, and that the zero factorial is equal to one. To summarize, for the function given by the sum of C sub N times the quantity X minus A raised to the power N, where the absolute value of X minus A is less than the radius of convergence of the series, the coefficients of the function are given by this formula.

If we then generate a new function in the same way, but replace this formula with C sub N, so that the terms of the series look like this, we have just generated the Taylor series of the function F in A, which can also be described like F around A, or F centered on A. So a Taylor series is a type of power series but with the coefficients defined like this. 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 is equal to zero, this binomial simply becomes X to the power of N, and everywhere we replace A, we simply replace 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 on. This is a special type of Taylor series called the Maclaurin series. This is a Taylor series where the expansion takes place around the point X equal to zero. So we definitely need to keep these definitions in mind for a Taylor series and a Maclaurin series, because we're going to apply these expansions to specific functions. Take for example E to the power of X. Let's find the Maclaurin series that represents the function F of X equal to E to the power of N.

Well, what do we know about E to X? The derivative of E to X is E to X, so no matter what derivative we're looking at, we only have E to That means that any derivative of this function evaluated at zero will give us one, so this entire term can disappear, leaving us with to the factorial zero over zero, or simply one, plus X to the factorial one over one, plus X squared over two factorials, and so on. This should sound very familiar to you, since we just derived a form from E to X that we used earlier in this calculus series.

That shape was actually the Maclaurin series for function. Let's make sure we can find the radius of convergence of this function. That will require the proportion test. We set that up and flip the denominator to get X to the (N plus one) factorial over (N plus one), multiplied by N factorial over value of X over N plus one, and as N approaches infinity, it reaches 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 content ourselves with some simple definitions and return to these concepts later.

And with this, we have concluded our calculus study, so it is time to move on to other topics. Before we go, let's check understanding.