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Problem 6_ The polynomials Ho (x) =1 Hi(x) = 2x Hz (x) = 4x2 _ 2 Hz(x) = 8x3 12x are called the Hermite polynomials. Show H {Ho (x), Hi (x), Hz(x), Hz(x)} is a basi...

Question

Problem 6_ The polynomials Ho (x) =1 Hi(x) = 2x Hz (x) = 4x2 _ 2 Hz(x) = 8x3 12x are called the Hermite polynomials. Show H {Ho (x), Hi (x), Hz(x), Hz(x)} is a basis for Qs'Problem 7. Using Problem 6, find each of the following: [1]#, [x]#, [x2]u, and [x?]x.Problem 8. Using your answer in Problem 7, create a 4 X 4 matrix A where each column is one of the vectors from your solution. Find A-1 (use a computer!) Compare this to the matrix you used in Problem 6. Briefly explain what you see, and

Problem 6_ The polynomials Ho (x) =1 Hi(x) = 2x Hz (x) = 4x2 _ 2 Hz(x) = 8x3 12x are called the Hermite polynomials. Show H {Ho (x), Hi (x), Hz(x), Hz(x)} is a basis for Qs' Problem 7. Using Problem 6, find each of the following: [1]#, [x]#, [x2]u, and [x?]x. Problem 8. Using your answer in Problem 7, create a 4 X 4 matrix A where each column is one of the vectors from your solution. Find A-1 (use a computer!) Compare this to the matrix you used in Problem 6. Briefly explain what you see, and why you think it happened:



Answers

Exercises $9-14$ require techniques from Section $3.1 .$ Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for $3 \times 3$ determinants described prior to Exercises $15-18$ in Section $3.1 .$ INote: Finding the characteristic polynomial of a $3 \times 3$ matrix is not easy to do with just row operations, because the variable $\lambda$ is involved.
$$
\left[\begin{array}{rrr}{5} & {-2} & {3} \\ {0} & {1} & {0} \\ {6} & {7} & {-2}\end{array}\right]
$$

So big. I matrix six negative, too. Zero negative too. Nine, zero, five, eight and three. Like that. So once again, we're going to subtract lambda from all of the diagonal diagonal entries and use the equation that we used previously to try and find the characteristic polynomial. So the determinants of this matrix is equal to six minus lambda times the determinant. Ah, nine minutes, Lambda zero eight three minutes. Lambda minus native to times the determinant of negative too. Zero five three minus lander. And then plus zero times the determinant. Uh, negative, too. Nine minutes, Landau. Five and eight. Now we can cancel out the last term because with the zero and the coefficient is gonna end up being zero. Anyway, it's and we're left with these and we're gonna expand them out like this. Six minutes. Land a times, part disease, nine minutes, lamb that times three minus lander minus zero times eight. I know that's gonna cancel out into zero, but let's leave that, as is for now, just to keep the form and then plus two times currencies negative. Two times three months, Lambda minus zero times mine like that. And then we mark the zeros outs like that, and then we're left with six months. Lander times nine minus Lambda Times three minus lander plus negative four times three minus lander. So now we're going to expand these out so that we can get the polynomial. If we were trying to get the roots, we would be trying to fact arises by, for example, group into three months. Land us together and then expanding the polynomial created by six minutes. Linda Times Nyman's Lambda minus four. But this is asking for the polynomial. Instead of the Eigen values, we're going to expand them. Little left term becomes negative Land a cube plus e team Lambda squared minus minus 99 lander and then finally plus 162. And then the expression on the right expands to for Orlando minus 12 making this in total minus land. Acute plus 18 land a squared minus 95 Lambda plus 1 50. And we're done

We're given the Matrix as follows. One zero Negative one to three. Negative. One zero six and zero. Recalling the definition of the characteristic equation which is the determinant of a minus land. I is equal to zero. We get that the determinants of the following matrix should be equal to zero. So one minus lander zero negative one to three minus lander Negative. One zero six in negative. Lambda is equal to zero. Now we're going. Teoh, use the formula that were given from Chapter three, Section one as recommended by the problem. And get that the determinants of this is equal to one minus lander times the determinant Uh, three months Lambda Negative 16 minus lander minus zero times the determinant of to negative 10 Negative lander plus negative one times that a determinant of to three minutes, Lambda zero and six. Basically a word over doing is we take the valley on the first row and then find the determinants of everything else. And then during the same for the other values on the first row like that. And of course, finally like that. So now we're going to expand this out. We get that This is equal to one month's Lambda Times. And it did terminate about two by two. Matrix is simple or three minutes. Lander times negative. Landau minus negative one times six. And the middle term cancels out because the coefficient is just gonna be zero. And then we have minus because the minus one and the men and the determinants of this becomes to times six minus three minutes. Lambda Times zero The two times six is just gonna be 12. And since the three months land that is being multiplied by zero, it doesn't matter. So the term of the end is just gonna be minus 12. Meanwhile, the term in the front is gonna be one month's Lambda Times. Landis squared minus three lander and then minus negative one makes positive and then multiply by 60 plus six minus 12. And that expands as negative land. Acute plus lambda squared plus the real and the squared minus three. Lander minus six. Lander plus six minus 12. Simplifying this a little bit. We get negative land acute plus four lander squared minus nine. Lander minus six. So that's gonna be our characteristic polynomial. Yeah, we're done

We had the matrix four zero zero five three to negative too. Zero to Now we're going to find a characteristic polynomial by subtracting Landau from each of the diagonal entries like this and we're going to do the same thing that we tried for Problem nine, where we used the formula given by section 3.1 and say that the determinants of this is equal to for Moniz Lander times the determinants of three minutes Lander 20 to negative Lander minus zero ties the determinant of five to negative two two minutes. Lambda plus zero times the determinant of five free minus lambda Negative too. And zero Now, because the last two terms have a coefficient of zero, we can cancel them out. So we're left with just four minus lambda times this determinant and that is found by three minus lander times two minutes, Lambda minus two times zero. Which is, of course, zero. So it ends a simplify too. Three months, Miss Lambda Times. Two minutes, Lambda. Now, if you are finding the Eigen values, we can just set this equal to zero and easily see that they're equal to 43 and two. But That's not where that's not what we after in this question. It's finding the characters that polynomial. So we're going to expand this out, and that's going to turn out to be negative. Land a Cube plus nine land a squared minus 26 Lander plus 24 and we're done.

Okay, so we have a minus the eye, which is one minus lambda. 1234 Minus Lambda 1002 Money's Lambda. Now let's take a determinate of that. All right, That's difficult to to minus Lambda everyone one in London for money slammed it. Excuse me on the Cube. Negative barbecue with post its lander squared minus to London Markazi.


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