Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. The binomial series for -1 less than X less than one. If we have the form of the quantity 1 + X to the M that equals one plus the summation of K equal 1 to Infinity of M choose K of X to the K where M choose one is m / 1 factorial, M choose two is m * m - 1 / 2 factorial. M choose K is m * m - 1 M -2... m -, K + 1 / K factorial for K greater than or equal to three. In this situation, M no longer needs to be an integer, so if we have 1 + X to the -1, we're going to think of that as this one plus the summation K equal 1 to Infinity. It's going to be -1 to the KX to the K. So we have 1 + X to the M is the formula we're going to use. We need to figure out the coefficients. The coefficients -1 choose one, so -1 / 1 factorial would be -1 negative one choose two. So we'd have -1 at times one less than -1 / 2 factorial. That simplifies to one. -1 choose three, we'd have -1 * -2 * -3 all over three factorial, which is going to give us a -1. If we had -1 to the K, we'd have -1 negative two -3... to -1 -, K + 1 all over K factorial. So if we looked at that -1 -, K + 1, we could see that's really negative K. We could factor out -1 from each term, and then we'd have 1 * 2 * 3 * 4... times K, so that's really K factorial. So we'd have -1 times each other K quantity of times over K factorial. The K factorial over K factorial would cancel, which leaves us that -1 to the K, which is what we had right here. So now if we actually want to expand it out, we would have 1 + -1 to the X + 1 to the X ^2 + -1 to the X ^3 + 1 to the X fourth plus... -1 to the KX to the K plus... We look at another example, 1 - 2 / X to the one third. So this is going to be one plus summation K1 to Infinity of 1/3. Choose K of the quantity -2 / X to the K. So when we look at this, we'd have 1 + 1/3 times the -2 / X + 1/3 * 1 less than 1/3 / 2 factorial -2 X squared. Plus 1/3 * 1/3 - 1 * 1/3 - 2 one less again over three factorial -2 / X ^3, etcetera. So what happens is when we simplify it up, we get 1 - 2 / 3 X -4 over nine X ^2 - 40 / 81 X cubed plus... If we look at another one, we'd have 1 + X ^2 ^3, so we'd have one plus bringing the exponent down 3 / 1 factorial times X ^2 to the first. The next term 3 * 1 less than 3 or 3 - 1 / 2 factorial X ^2 ^2 + 3 times. 3 - 1 three -2 / 3 factorial X ^2 ^3 + 3 * 3 - 1 three minus two 3 - 3 over 4 factorial X ^2 to the 4th. Now, because we have this 3 - 3, every term after this is going to end up being zero because we're going to have a 3 - 3 and every term after this. So when we simplify this up, we really get one plus three X ^2 + 3 X to the 4th plus X to the 6th. Now this is actually an interesting one because it was to an integer, it was to a cube, Pascal's triangle. If we start with ones, a triangle of ones, an equilateral triangle, and put ones diagonally going out. Pascal's triangle tells us the coefficients of anything of the quantity A+B to the north. So if we start with our diagonal, our triangle of ones, we put our diagonals of 1 going out. Then what happens is every two numbers we would add together to get a sum underneath it. So one and one is 2/1 and two is 3, two and one is 3. One and three is 4/3 and three is 6, Three and one is 4-1 and four is 5. Four and six is 10, Six and four is 10, four and one is five, etcetera. And what this does is it tells us what the coefficients ought to be. So if I had a + b ^3, I'd look for the row that had a three in the second location. It's always whatever the exponent is. So my coefficients would be 1331. If we come back here, we can see a coefficient of 1/3, 3:00, and 1:00. And then what happens is this one gets taken to the third power. So whatever this term is. So we'd have one a cubed plus three a ^2 b + 3 AB squared. Plus 1B cubed and this particular instance our a is one and our B is X ^2 and this is indeed what we would get when we expanded out looking at an integral zero to -2 of E to the negative X -, 1 over XDX. And what we want to do is we want to approximate an error with. Magnitude less than 10 to the negative 3rd. So we're going to start by understanding we're going to factor out the 1 / X because we have a formula for E to the X. So if we have a formula for E to the X, we're going to use that same formula for E to the negative X, putting in a negative X every time we see our X. So we get zero to .2 of one over XE to the negative X is going to be 1 -, X + X ^2 / 2 four two factorial. Minus X ^3 / 3 factorial plus X to the 4th over 4 factorial, minus X to the fifth over 5 factorial... -1 DX. So this +1 and this -1 are going to cancel. Then we're going to take this 1 / X and multiply it through by everything else that remains. So we're going to end up with -1 + 1 half X - 1 six X ^2 plus 124th X ^3 - 1 twentieth X to the 4th plus... DX. When we do the actual integration here, we're going to get negative X + 1/4 X squared minus 118th X ^3 + 196 X to the 4th -1 / 600 X to the fifth plus... Now remember, we want an error that's. Less than 10 to the negative 3rd. So we really need 4 decimal places to be the same. So when I put in the .2, the first one is just negative .2. When I combine the next the first two terms or the next term with the first one, I get -.19, the first three terms -.1904. And I could actually stop looking at the four because if I need 10 to the negative third precision, I need to go out four places. Then first four terms added together -.19 O 428. Well, these first four terms are the same now in the third term out and the fourth term out. So it's approximately -.190 the three degrees 10 to the negative 3rd that we need with the error. The absolute value of the error less than or equal to that fourth term or when it started repeating. I needed the first three to get that precision, but it's the next one out that the errors got to be less than or equal to. We can also look at the limits, so the limit as Theta goes to 0 sine Theta minus Theta plus Theta cubed over 6 divided by Theta to the fifth. This one's important because what happens is we can have multiple terms that can help get rid of single terms if we need to. So we know a formula for sine Theta that's just Theta minus Theta cubed over three factorial plus three Theta to the fifth over 5 factorial minus Theta the 7th over 7 factorial plus... So this Theta and this negative Theta cancel. This negative Theta cubed over 3 factorial and the positive Theta cubed over six is going to cancel. So then we end up with these terms left over the Theta the 5th. Well, Theta the 5th over Theta the 5th is 1. So 1 / 5 factorial, Theta the 7th over Theta the 5th would give me Theta squared. If we'd gone out to the next one, Theta to the 9th over Theta to the fifth would be Theta to the 4th plus... So when Theta goes to 0, all of these terms down here are going to go to zero, and we really just end up with 1 / 5 factorial or 1 / 120. If we look at another one, X squared minus two, X ^3 + 2 ^2, X ^2 / 2 factorial minus, etcetera, this time what it wants us to do is it wants us to actually come up with the sum of the series. So the first thing to realize is we're not starting with one if we factor out an X ^2. We then are starting with one. So if I pull out an X ^2 I get 1 minus two X + 2 ^2 X ^2 / 2 factorial -2 ^3 X ^3 / 3 factorial, etcetera. Well, this looks very similar to one of our patterns. Our pattern says all the terms had to be positive. And it said that it was something to the end power. So I'm going to put the 2X together, the 2X together, the 2X together. Because I can see in each term the exponents were to the same power. Now to get rid of this negative if I thought about putting the negative inside with the term. I'd get one plus the quantity -2 X plus the quantity -2 X squared. Well, we know a negative squared would give us that positive over two factorial plus the quantity negative two X ^3 / 3 factorial, while a negative cubed is a negative plus the quantity -2 X to the 4th over 4 factorial. A negative to the 4th would be positive. Well, we know that this formula in the parentheses is just E to the X, but in this case, our X is going to be the -2 X, so we get the X ^2. We factored out E to the -2 X as something that's equivalent to the original series. Thank you and have a wonderful day.