Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Taylor series the definition Let F be a function with derivatives of all orders throughout some interval containing A as an interior point. Then the Taylor series generated by the function F at X equal a is the summation of n = 0 to Infinity of F of KA over K factorial times the quantity X -, A to the K. What this equals is F of a + F prime of AX minus a + F double prime a / 2 factorial X -, a ^2 plus... F to the n / a of a / n factorial times X -, A to the N plus... Remember that F to the K that K is actually being a derivative. However many derivatives out. So first derivative out, second derivative out, third derivative out, NTH derivative out. The Maclaurin series occurs when the a = 0, so it's basically just a Taylor series with our A being 0. The linearization of a differential function F at a point a is the polynomial of degree one given by P of X equaling F of a + F prime a times the quantity X -, a. This is really just the equation for a line where we have y -, y one equal MX minus X1 where M is really just our first derivative at the location a. So M equal F prime A and our X1 Y one point is just AF of a. So if you thought about sticking in A for your X1F of A for your Y1 and your F prime a for your M and getting the Y by itself, the Y is going to act like our P of X in this case for our polynomial polynomial function degree. So Taylor polynomial order N. Let F be a function with derivative of order K for K equal 1/2... up to N and some interval containing a as an interior point. Then for any integer N from zero through capital N, the Taylor polynomial of order N generated by F at X equal a is the polynomial P of NX equal F of A + F prime of a * X -, a + F double prime over 2 factorial X -, a ^2 plus... plus F of N of A / n factorial X -, A to the NTH power. So this polynomial is of order N and it terminates. If we look at some examples, we're going to find the Taylor polynomial of order 012 and three generated at A. So in this case, we're given F of X equals the natural log of 1 + X and we're given we're actually given a equaling 0. But if we know that a = 0, we can find F of 0 to be the natural log of 1 + 0. The natural log of one is 0, so our point is going to be 00. So we're going to start by taking the first 3 derivatives because it's saying find the Taylor polynomial of order 012 and three. So F prime of X is going to be 1 / 1 plus XF double prime of X -1 / 1 + X ^2 F triple prime of X +2 / 1 + X ^3. So now we're going to evaluate at 0 the a = 0. For each of those locations, F prime zero is 1, F double prime zero is -1 F triple prime 0 is 2. So now we're going to put them into the formula. Our P sub not is just going to be 0. Our P sub 1 is going to be 0 plus the one over technically one factorial times X -, 0, or just X our P of two. Our P sub not plus the P sub 1, so 0 + X plus our F double prime -1 / 2 factorial X -, 0 quantity squared. If we simplify that, we'd get X -, 1/2 X squared. Our P sub 3 is our first 3 terms added together, so that zero that X that -1 half X ^2 + 2. Our third derivative over 3 factorial X - 0 ^3 3 factorial 1 * 2 * 3 is Six 2 / 6 is 1/3. So our P sub three, our polynomial of third degree is X - 1 half X ^2 + 1/3 X cubed. So that's our polynomial of order three. Let's look at another one. This time we're going to look at F of X equaling cosine X where a equal π force. So F of π force, cosine of π force is root 2 / 2. We're going to find the first derivative, second derivative and 3rd derivative. So if cosine acts as F derivative, cosine is negative sine, derivative of negative sine is negative cosine derivative of negative cosine is sine. So at π force. We're going to evaluate each of those order degrees of the derivatives. So F prime of π force negative root 2 / 2, F double prime of π force negative root 2 / 2, F triple prime π force root 2 / 2. So our P not is just the root 2 / 2, our P sub 1 is going to be that first term. This root 2 / 2 plus this first derivative times X - π force to the first power. Our piece of 2 is going to equal that piece of not minus. The next term, the root 2 over two X - π force, and now the third the second derivative that negative root 2 / 2 / 2 factorial times X - π four squared. If we simplify that up, we'd get root 2 / 2 minus root 2 over two X - π force minus root 2 / 4 X minus Pi 4 ^2. Now our P sub 3 is really just going to be those terms with one more term at the end, and that last term at the end is going to be the third derivative over three factorial times X - π four cubed 3. Factorial 3 * 2 * 1 is 6/6 times 2 is 12. So it's going to simplify into our Taylor polynomial of order 3 is root 2 / 2 minus root 2 / 2 times the quantity X - π force minus root 2 over four X - π force quantity squared plus root 2 / 12 X -π force cubed. If we have a hyperbolic, we're going to have the directions change a little bit and we're going to find the Taylor series at X = 0. Well, at X = 0, these are really just Maclaurin series. So sometimes it says find the Maclaurin series instead of find the Taylor series at X = 0. If the directions said find the Maclaurin series, they might not give you the fact that we're looking at 0 at all. So we're going to start with the fact that F of 0. If we think about E to the 0 -, e to the -0 all over 2, E to the zero is 1, E to the -0 is 1. So 1 - 1 / 2 is just zero. We're going to start with the function, and we're going to find the first derivative. Think about this as 1/2 * e to the X -, e to the negative X is the original function. And we know the derivative of E to the X is just E to the X. And the derivative of E to the negative X means that we have to take the derivative of the exponent, which is -1 So basically the sign in the middle is going to oscillate between negative, positive, negative positive. So if we look at F of 0, we're going to get 0F prime of 0. We're going to get 1F double prime OF00F triple prime of 0 is 1. So if we look at this one, we're going to get 0 / 0 factorial. 0 factorial, by definition is 1 * X -, 0 to the zero. Now that's really just one also. So our first term is really just whatever F of the first term is. But I'm trying to help you see a pattern here. I hope so. Plus the first derivative over 1 factorial and one factorial is really just One X - 0 to the first plus the second derivative. So 0 / 2 factorial X - 0 ^2 plus the 3rd derivative 1 / 3 factorial X - 0 ^3 Plus. If we saw the pattern, we'd see that the next term's going to be 0 again. So 0 / 4 factorial X - 0 to the 4th. So 0 times anything is 0. Remember, 0 factorial is just one. So we haven't broken anything, any rules by dividing by zero there. So we get 0 + X + 0 + 1 / 3 factorial X ^3 + 0 + 1 / 5 factorial X to the fifth plus... So what's really happening is this is going to be skipping every other term. So if we look at this, we could think of having A1 factorial and then a three factorial and A5 factorial. So we can see that we're going to have two n + 1 factorial or we're only hitting the odds and then the exponents 135 are also odds SO2N plus one. Now the fact that it's an odd factorial doesn't mean we're only hitting odds on our way down. It means that we're going 3 * 2 * 1 or 5 * 4 * 3 * 2 * 1. To figure out where we start, we need it to be 1. So if we put in zero here, we can see 2 * 0 + 1 is the first power, 2 * 0 + 1 is the one factorial. So our summation n = 0 to Infinity of 1 / 2 N plus one factorial X to the two n + 1. If we look at this next, 1F of X equal E to the X / 2 and once again it's a Maclaurin series, so we know that we want X equal 0. So F of X equal E to the X / 2 F prime X = 1/2 E to the X / 2 F double prime X1 4th E to the X / 2, F triple X equal 1/8 E to the X / 2. If we evaluate all these AT0F of 0 E to the 0 powers, one 1/2 * E to the 0 is going to be 1/2 and then 1/4 and then 1/8 as we can see over here. So we'd have 1 / 0 factorial. Technically we could think of this as an X -, 0 to the zero, but we know that anything to the 0 power is going to be 1. Then we have the 1/2 / 1 factorial X -, 0 to the 1st 1/4 / 2 factorial X - 0 ^2 + 1/8 / 3 factorial X - 0 ^3 plus... So when we're trying to write it as a summation, we can see that this time we have the denominators increasing by powers of 2248. The next one would be 16, so 2 to the north. The factorial is going to be N factorial and we have X to the N power. If we thought about starting at 0 and going to Infinity, this next 1F of X equals sine of X / 2. Once again, it's some chlorine series, so we want to figure out what happens when our a is 0. So our first derivative is going to be 1/2 cosine X / 2, our second derivative negative 1/4 sine X / 2, our 3rd derivative -1 eighth cosine X / 2, our 4th derivative 116th sine X / 2, our 5th derivative 1 / 32, cosine X / 2. Now in a minute, you're going to see hopefully why I did so many of them this time, because I need to see a pattern. So if we evaluate all those at zero, we're going to get zero, a half, zero -1 eighth zero, and 132nd. So in reality, three of those became 0. So I only really have three that I can look for for a pattern. So if we have 0 + 1/2 / 1 factorial X + 0 / 2 factorial X ^2 + -1 eight three factorial X ^3 + 0 / 4 factorial X to the 4th plus 1 / 32 five factorial X to the fifth plus... So the three that we really have to look for, because we're going to ignore the 0 ones, is 1/2 X negative 1/8 / 3 factorial X ^3 + 1 / 32 five factorial X to the fifth plus... So first thing we're going to notice is we're having an oscillating term between positives and negatives. So we're going to have -1 to the N power. If N is zero -1 to the zero is going to give us a positive -1 to the first would give us a negative -1 to the second would give us a positive. Our exponents are going by odds for the XS. So 135 S we're going to have two n + 1 there. The two 832 are just powers OF2SO2 to the two n + 1 and then 2N plus one factorial. Remember, if I start at 0 * 0, zero plus one is 1 and one factorial is just one. So we could think of that with A1 factorial underneath it if we wanted to. So 135, the factorials are going to correspond usually with what we're doing at this point with the exponents. So now the directions are going to change. We're going to find a Taylor series. So we're going to find the Taylor series generated by F at X equal a. So F of X equals two X ^3 + X ^2 + 3 X -8, where A equal 1. If we take enough derivatives, we know eventually we're going to go to 0. So the first derivative is going to be six X ^2 + 2 X +3, the 2nd derivative 12X plus 2, the 3rd derivative 12, The 4th derivative is 0, and every derivative after that is 0. So that's one of the interesting things that can happen with Taylor series. If we start with a polynomial, we know that eventually we're going to have a term that goes to 0, and every derivative passed that'll be 0. So then if we figure out what's happening at one, we get F of one is -2 and F prime of one is 11, and F double prime of one is 14. F triple prime 112 FN of one is 0 for N greater than or equal to four. So just literally putting it in our formula -2 / 0 factorial X -, 1 to the 0 + 11 / 1 factorial X -, 1 to the first plus 14 / 2 factorial X - 1 ^2 + 12 / 3 factorial X - 1 ^3 + 0 / 4 factorial X - 1 to the fourth plus, we can see that all the rest are going to be 0 because all of those derivatives out there are 0. So if we simplify this up, we're going to get a polynomial that says -2 plus 11 times the quantity X -, 1 + 7 times the quantity X - 1 ^2 + 2 times the quantity X - 1 ^3. Thank you and have a wonderful day.