Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Curve fitting a polynomial of degree N is a function of the form F of X equal a sub not plus a sub One X + a sub two X ^2 plus... plus a sub NX sub N where the coefficients a sub zero a sub 1A sub two... A sub N are constants. The data points X1Y1 lies on the curve Y equal F of X provided that F of XI equal Y sub I. The condition that this be true for each I starting at 0 going through N yields the n + 1 equations A sub not plus a sub 1X not plus a sub 2X not squared plus, etcetera. And what this really does is if we think about looking at our a sub not and our a sub one and our a sub two as coefficients that we're trying to solve, then we can come up with the Vandermonde matrix. And the Vandermonde matrix says that basically we're going to take one and then X not and X not squared, X not cubed plus... Till we get to X not to the north. And the reason is we're trying to solve for this a sub not and this a sub one and this a sub two. So if we thought about what's being multiplied by the a sub one, it's the X not or what's multiplied by the a sub two, it's the X not squared. So when we put it into a matrix that we're trying to solve, we're really doing the 1X not X not squared, etcetera. And we're doing that for each of those i's. So for I equal 1, I equal 2, I equal N. Now we have a matrix that we can solve, and if the X coordinates are distinct, then the matrix A is non singular. Thus the system has a unique solution for the coefficients of the mate of the polynomial. So there's a unique polynomial if all of those are unique or if all of those are non singular. Three points determine a circle. Our generic equation of a circle is the quantity X -, H ^2 + y -, K ^2 equal R-squared. If we do some manipulation, we can see X ^2 - 2 H 2 HX plus H ^2 + y ^2 - 2 KY plus K ^2 equal R-squared. Manipulating things around, we can realize that if we do some replacement with generic variables, this could turn into a * X + b * y + C equaling negative X ^2 + y ^2. So if we thought about putting this into a matrix when we're solving for AB and C, because we know our X and our Y, we would end up with a matrix that said XY one, and then whatever that negative X ^2 + y ^2 was, and then XY one for our next point. So that's X ones, X twos. I could have used X knots, etcetera. So, and we'll do an example in a moment. Three points determine a central conic and the central conic is of the form AX squared plus Bxy plus Cy squared equal 1. And remember this B is basically telling us a rotated conic. And so if it's a rotated conic, then we can look at the determinant and figure out whether it's an ellipse, a hyperbola, a parabola, a circle. So looking at a polynomial with the points negative 1115216, we're going to use the equation or the matrices that says one negative one, one and one, then 11515, 12416. This is going to be a cubic because we have three, or a quadratic because we have three points. So what we're really doing is we have Y equal A+, BX plus CX squared. Actually, let's put the Y at the other end. So if we have A + B * -1 + C * -1 ^2 equaling 1 or A + B * 1 + C * 1 ^2 equaling 5A plus B * 2 + C * 2 ^2 equaling 16. So if we're solving for these ABS and CS, we can see that the coefficient on the A is just one. Hence this column of ones. Here the coefficient on the BS is going to be -1 one and two, which is this column. Here the coefficient on the C's -1 ^2, 1 ^2, and 2 ^2. And then our last column is just going to be the Y's or what they were equaling. Now, using our rules and matrices, we're going to come up with the diagonal identity matrix with whatever equals on the end. So if we started with negative 111 and we took row 2, subtract row one and then row three and subtract row one. 1 - 1 zero -1 - 1 row 2O root 2 minus one 1 - -1 two 1 - 1 zero 5 - 1 four. I chose row 2 minus row one instead of row 1 minus row two just because it gave me positive numbers there. Row 3 minus row 103315. So now I'm going to divide by. I'm going to take row two, divide by two. I'm going to take row 3 and divide by three. So one negative 1-110-102-0115 Now if I just take row one and add it to row two, we're going to get 1013. And if I leave that middle row alone for a minute and then we're going to take row 3 minus row two. So 1 - 1 is zero, 1 - 0 one 5 - 2 three. I want to get this one to be gone also. So I'm going to take row 2 Nope row one and subtract row 3. So now we get 100001020013 what that matricy tells us. Because each of the rows is unique, we know that Y is going to equal a X + b at no A+, BX plus CX squared. But this matricy tells us that my A is 0, my B is 2 and my C is 13. Three. That was separate one, then the three. So my answer for this is going to be Y equal two X + 3 X squared is the polynomial that fits the given three points. So the quadratic Y equal two X + 3 X squared goes through negative 1115 and 2:16. Looking at a circle, we know it's going to be of the form AX plus BY plus C equaling negative the quantity X ^2 + y ^2. Now we want to solve for AB and C So when we make our matrices, our coefficients are going to be the X. The YC has a coefficient of one with it, and then we're going to take these two terms and square them each separately and then take the opposite of it. So 3 + 16, then the opposite of it, so -25 S five, 10125 and 100 negative -912 one 81144, negative 225. Sometimes it helps us to choose a different column to get rid of. So we're going to actually start with our ones and zeros in that third column because we already have a one here. If we look here, we'd have to divide by three or multiply by two and subtract. But this way we can just subtract rows. So if we take row 1 minus row 3 and row 2 minus row three, and we leave that row 3 alone, so -912 one negative 225 and then twelve negative 16, zero 214, negative 20100. Now whenever I can, I try to keep my number small so I can look at this top row and realize everything is divisible by 4. And this next row, everything is divisible by two. So I'm going to take this top row, row one, divide by 4 and row 2 and divide by. Actually, let's divide row two by a -2 because that will put a one in this location, which is what I want. So now we're going to have three negative 4050, negative 710, negative 50, negative 912, one negative 225 S Now let's take row 2 and multiply it by 4 and add to row one and row 2 * -12 and add to row 3. So when I do that, let's see what am I going to get? Row 2 * 4 so -28 and three negative 2500 times 4 negative 2 hundreds -150 we're going to leave row two alone. Row 2 * -12 so 84 and -975 zero one -12 and 50 and OK, 375. So now we can see that this row one, we could divide by -25 and we'd get a one in the right location there. So if we take that one and divide by negative 251006, negative 710, negative 50, 75 zero 1375 S, now we're going to take row one, multiply by 7, add to row two, row one, multiply by -75 and add to row 3. Remember, there are lots of ways to do this. This just happens to be the method I'm seeing. OK, so now what that tells us is we're going to have six X -, 8 Y -75 equaling negative X ^2 + y ^2. If we want to come up with the actual equation of the circle and the center and radius, we're going to need to get the X's and Y's all on one side, and then we're going to have to complete our square. We can take that constant to the other side if we want. So we're going to take half a -6 three and square it nine. If I add 9 to one side, we're going to add 9 to the other. I'm going to have take half of the -8 four and square it. If I add 16 to one side, we're going to add 16 to the other. So this is going to factor into X plus oh, that was an ** plus three quantity squared plus y -, 4 quantity squared equaling 75 and 9 and 16100. So our center is -3 four and our radius is 10. What if we wanted to look at a central ellipse with the points 00430 and 55? Remember, our equation for a central ellipse says AX squared plus bxy plus Cy squared equaling 1. So when we do our matricy here, it's going to be the X ^2 term as the coefficient on the A, the X&Y multiplied together as the coefficient on the B, and then the Y ^2. And they're all going to have one in the last column. So then we get 9001 and 252525 and 1:00. So I can change rows and columns. I'm going to take the 2nd row and move it to the first row. So row two to row one and I'm going to divide by 9 and then I'm going to take row one to row 3 and divide by 16 so we get 1001 ninth. So that row is exactly how I need it. 00 one 116th that rows how I need it. So now I'm going to need to deal with this other row and I might go ahead and leave them as 20 fives for yeah, 20 fives for just a second. So now we're going to take row 1 * -25 and add it to row 2. So if we do that, we're going to get 1001 ninth zero 2525 and -25 ninths plus one ninth. So -25 ninths, no -25 row one. Nope. Row one negative 25 ninths plus one. So one is 9 / 9 negative 16 ninths. So now we're going to take row 3 and multiply by -25 and add to row 2. So we're going to get 1001 ninth zero 25 zero negative 25 sixteenths plus -16 ninths -25 sixteenths -16 ninths. Common denominator 144 -48481 / 144 Oops. Put it in the right spot though -481 / 144 and then this one is 00 one 116th. So then finally we're going to take row 2 and divide by 25. So when we do that we're going to get our matricy that says 1001 ninth 010 dividing by 25 negative 481 over 360000 one 116th. So we know that this is going to be one 9th X ^2 -, 481 three 1006 hundredths XY plus 116th y ^2 equaling one. And just because we usually like integer coefficients, we're going to multiply everything through by 3600 and we're going to get four hundred X ^2 -, 481 XY plus 225 Y squared, equaling 3600. Rotation formula for conics, recall is AX squared plus bxy plus Cy squared plus DX plus Cy plus F = 0, where B does not equal 0. The angle of rotation is cotangent to Theta equaling our coefficients a -, C / b. And if it's a non degenerate, non degenerate conic, then it has the property that the determinant b ^2 -, 4 AC. If it equals zero, it's a parabola, it's less than 0, it's an ellipse or a circle, and if it's greater than 0 it's a hyperbola. So on the example that we were just working on. We can look at our cotangent of two, Theta being our A term 400 minus our C225 all over our B -481. So we can find Theta being 1/2 of the cotangent inverse of -175 / 481. So that's approximately -.003175°. So that's going to tell us that this is going to rotate by a very small angle in the negative direction. So then we can figure out what kind of an angle it is or what kind of a figure it is. So our b ^2 minus our four times our A * C, When we compute that, we can see that that's -100 128,639. So it's going to be an ellipse or a circle. Now, with a lot more math, we could figure out exactly what this ellipse or circle would look like as far as how far out it goes and whatnot. But let's just start over, OK? So if we draw our segment and we put in a line very, very, very, very, very small negative value, and then maybe I move that line a little bit up so it looks like it's going through there and now I've lost the whole thing. So then an ellipse is going to be somewhere along that's going to be sort of perpendicular, not even close to perpendicular. OK, I hadn't originally planned to sketch this, so maybe I should just give up. So the idea would be that we could actually figure out the extremes of the ellipse, how far up and down, etcetera. We do know that it's going to be longer on the X than the Y due to the fact that the coefficient on the X was bigger. So let's look at our last type, which is an application problem, and it talks about wanting to figure out a curve fitting for information that's given for the West from 1970, nineteen, 80 and 1990. And we want to come up with a cubic equation for this. So what we're going to do is we're going to actually write a matrix and we're going to let our time equal 0 for 1970. And this is a technology problem. So we can do these in our calculators. If our time is 0 in 1970, it's going to be 10 in 1980 and 20 in 1990. So when we look at this, we're going to come up with a polynomial, so in this case P of T where it's going to equal A+, BT plus CT squared. So we're needing to find AB and C So if we write our matricy, we would have one for our A coefficient. Our B coefficient for 1970 is going to be 0 because our time is 0. Our C coefficient is going to be 0 and the given information was 34.838. Then if we do 1 coefficient 10 years 10 ^2 or 100 and the given 43.171 and then one 2020 ^2 or 452.786. Now if we use our calculator to put this in reduced row echelon form, we can see that our P of T is going to equal 34.838, which we can actually also see from this top line plus .769 two t + .00641 T squared. Then the problem wants us to go on and add 1960 and instead of having a quadratic, make it into a cubic. And what we're really trying to do is we're trying to determine if having more data makes it a better fitting curve or not. So we're going to add and make this into a cubic equation doing the same steps that we did a moment ago. So our new matricy, we're going to let our T be zero for 1960 and 1020 and 30. So we're going to have 100028 point O 531 ten 10 ^2 10 ^3 34.83812020 ^2 20 ^3 43.17113030 ^2 30 ^3 52.786 grabbing our technology. This case, I'm going to use my calculator. I can get my PT equaling 28.053 + .592233 three t + .00907 T squared minus .000044333 T cubed. If we had wanted to add more information we could have taken it to 1950 and we would have had 010203040 and it would have made us a quad or a fourth degree equation. Thank you and have a wonderful.