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>> Recall the definition of a Taylor Series.

Â If you want to compute one of these, well you may have to take a lot of

Â derivatives. But that's not your only choice.

Â There are a few other methods as well, including substitution and combination.

Â Let's begin with an example. The displays a substitution method.

Â Consider the function, 1 over X times sine of x squared.

Â Computing on the Taylor Series for this might be a lot of work.

Â Having to differentiate this is not entirely trivial.

Â However, we do know what sine of X is in terms of its Taylor series?

Â Sine of X is X minus X cubed over 3 factorial, plus X to the fifth over 5

Â factorial, et cetera. If instead of using sine of X, we'll use

Â sign of X squared plugging in X squared into the Taylor series for sin of x.

Â And the multiplying that by 1 over X. Well, we obtain something that is going

Â to work. A little bit of simplification gives X

Â minus X to the fifth over 3 factorial, plus X to the ninth over 5 factorial.

Â Etcetera. Another method for obtaining the same

Â answer would be to take the full summation.

Â As K goes from zero to infinity, of negative one to the K times X squared to

Â the 2K plus 1 over 2K plus 1 quantity factorial.

Â By adding up the coeeficients of the X term properly, we can in fact obtain the

Â full Taylor series for this function. As sum K goes from 0 to infinity,

Â negative 1 to the K, X to the 4 K plus 1. Over quantity 2K plus 1 factorial.

Â That means in effect that we have all of the derivatives of this function

Â evaluated at zero in one simple computation.

Â 2:56

There are other methods as well involving a combination of like terms.

Â Consider the example of cosign squared, of X.

Â We know the the Taylor series for cosign of X.

Â It is the familiar one minus X squared over two factorial, etcetera.

Â We can square that and then evaluate the product.

Â How would we do so. If we consider these as very long

Â polynomials, then we could apply what we know from polynomial multiplication.

Â The lowest order term is equal to 1 that is 1 times 1.

Â The next order term consists of 1 times negative X squared over 2 factorial.

Â Plus negative X squared over 2 factorial times 1.

Â The next order terms are of fourth order and consist of X squared over 2 factorial

Â quantity squared. but there are a few others as well.

Â There is an X to the fourth over 4 factorial term.

Â And another X to the fourth over 4 factorial term.

Â All of the other terms are going to be of higher order.

Â One could continue multiplying although the multiplications would become a bit

Â tedious. And now one must simplify these terms.

Â With a little bit of work, it's easy to get the low order terms and to get 1

Â minus X squared plus X to the fourth divided by 3 minus X to the sixth times,

Â 2/45, etcetera. If you want more terms, then you're going

Â to have to do, a bit more work. The one thing that you can note however

Â is that this must be one minus sin squared of X.

Â And so, from this we obtain the Tailor series for sin squared of X as well.

Â 5:52

There are others as well, such as the hyperbolic tangent or tanh of X.

Â As you might guess, this is really sinh of X over cosh of X.

Â That is E to the X minus E to the minus X over E to the X plus E to the minus X.

Â Now, some of the rules that you're familiar with from trigonometry hold in

Â this hyperbolic setting but with a bit of a twist.

Â For example, cosh squared of X minus sinh squared of X is equal to 1.

Â One can compare that with the familiar formula, cosine squared plus sine squared

Â equals 1. What does this formula mean?

Â well, we know the relationship between cosine, sine and points on a unit circle.

Â Namely cosine is the X coordinate, the sine is the y coordinate.

Â Of a point moving along that unit circle. In the hyperbolic trigonometric setting,

Â the same is true but for a hyperbola instead of, for a circle.

Â 7:09

As you might suspect, there are also hyperbolic secant, co-secant and

Â cotangent functions. Let's investigate these hyperbolic trig

Â functions form the point of view of Taylor series co-secant.

Â For example, if we consider the hyperbolic cosine of X as one half E to

Â the X plus one half E to the minus X. Then it's clear how to compute the

Â Taylor's series. Instead of trying to take derivatives,

Â we'll simply substitute in the known series for e to the X, multiplied by one

Â half and then add to it, 1 1/2 times the series for e to the minus X.

Â Which, of course is the Taylor series for e to the X, with minus X substituted in.

Â Leaving negative signs at the odd powers of X.

Â By combining terms according to degree, we see that all of the odd degree terms

Â cancel. And we are left with only the even degree

Â terms. The same as in the expansion of E to the

Â X 1 plus X squared over 2 factorial plus X to the fourth over 4 factorial,

Â etcetera. If we wish to write this out in summation

Â notation, it would be the sum K goes from 0 to infinity X to the 2 K over 2 K

Â quantity factorial. Notice that just like the Taylor series

Â for cosine of X, COSH of X consists of the even powers but with no alternating

Â sign. That is another relationship between the

Â trigonometric and the hyperbolic trigonometric functions.

Â Does the same hold for the hyperbolic SINH.

Â Well let's investigate and see. Following the same method as before we

Â will use the Taylor series for E to the X.

Â And then the Taylor series for E to the minus X but now instead of adding these

Â two terms together, we are going to subtract the ladder from the former.

Â This leads to a cancellation of all the even powered terms and distributing the

Â minus sign through and adding, we obtain all of the odd degree terms in the Taylor

Â Series for E to the X. Thus the sum K goes from 0 to infinity.

Â X to the 2K plus 1 over 2K plus 1 quantity factorial.

Â Just like sign. We can continue our exploration of these

Â functions by proceeding as if the Taylor series are like long polynomials.

Â Hence computing things like integrals or derivatives can be done, term by term.

Â Let's consider what the derivative of the hyperbolic sine of X would be.

Â Well, we can differentiate the terms of the Taylor Series.

Â Since the derivative of X to the 2K plus 1 equals 2K plus 1 times X to the 2K.

Â We can see by dividing by 2K plus 1 quantity factorial and summing as K goes

Â from 0 to infinity. That, the derivative is the sum.

Â K goes from 0 to infinity of X to the 2K over 2K, quantity factorial.

Â That is simply the hyperbolic cosine of X.

Â In similarity to what happens with trig functions the derivative of sinh is cosh.

Â Likewise, what is the derivative of cosh of X?

Â If we differentiate term by term we can see that the derivative of X to the 2K is

Â 2K times X to the 2K minus 1. Now we have to perform a shift in the

Â index, here. In order to avoid problems with what

Â happens when K equals 0. Re-indexing properly gives us the sum, K

Â goes from 0 to infinity of X to the 2 K plus 1 over 2 K plus 1 quantity

Â factorial. That is the hyperbolic sine of X.

Â And so we see that the hyperbolic trade functions are very nice.

Â The derivative of sinh is cosh. The derivative of cosh is sinh.

Â And this becomes clear from the Taylor series.

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Now we're used to writing out the Taylor series.

Â Term by term, of course, since there are infinitely many, we can't write them all.

Â We typically use a plus .... or an ellipses to conote what happens in

Â the tail of the series. But, there's another terminology that we

Â will begin using. That of H O T or Higher Order Terms.

Â This is an informal way of saying, etcetera and it can be a bit helpful.

Â For example, you might say that cosine of X is 1 minus X squared over 2 plus Higher

Â Order Terms. We could say that sign of X is X plus

Â higher order terms or X minus X cubed over 3 factorial plus HOT.

Â And we could stop whenever we feel like. That's a convenient thing to do when

Â performing computations on Taylor series. We'll use a more formal bookkeeping

Â mechanism soon. Let's look at this in the context of a

Â simple example. 1 minus 2 X times E to the sine of X

Â squared. We'll begin with the Taylor series for

Â sine of X squared, which is simply X squared minus X to the sixth over 6 plus

Â higher order terms. What happens when we exponentiate this?

Â We use the Taylor series for e to the X, 1 plus our quantity above plus 1 1/2

Â times that quantity squared plus one 1 1/6 times that quantity cubed, etcetera.

Â Notice that I am using plus higher order terms liberally in order to ignore things

Â that aren't going to matter for the first few terms.

Â To simplify this, we go degree by degree. The constant term is one.

Â There's only 1 quadratic term, namely X squared.

Â Are there any fourth order terms? Well, yes there is.

Â 1/2 times X to the fourth. In the sixth degree terms there are two

Â of them. But notice, the coefficients balance each

Â other out. And so the coefficient of the sixth order

Â term is 0. We are left with a Taylor series of 1

Â plus X squared plus X to the fourth over 2 plus higher order terms.

Â Now to get the Taylor series for our original function f.

Â We simply take 1 and subtract 2X times the quantity obtained above.

Â I'll let you show, with a little simplification, that this is 1 minus 2X

Â minus 2X cubed minus X to the fifth plus higher order terms.

Â If you want more terms, you can get them, it will take more work.

Â >> We've now learned not only what Taylor series are but how to compute them

Â quickly and cleanly. Along the way, we've been introduced to

Â two new characters in our story. The hyperbolic sine and cosine.

Â In our next lesson we're going to consider what happens when things don't

Â work out so well. We'll deal with issues of convergence.

Â