Determine with the development point $\,x_0=0\,$ Taylor’s limit formula of second degree for the function
$$f(x)=2\cos(x)-2\sin(2x)\,,\,\,x\in \reel\,.$$
answer
$\,f(x)=2-4x-x^2+x^2\cdot \epsilon(x)\,.$
B
Tease-exercise: A smooth function $f$ of one variable fulfill that $\,f(2)=1\,$, $\,f’(2)=1\,$ and $\,P_2(1)=1\,.$ Determine the approximating polynomial of second degree $\,P_2(x)\,$ for $\,f\,$ with the development point $\,x_0=2\,$.
hint
Use the two first informations to state $P_2(x)$. Actually there i only one unkown. Use the last information to determine this.
State Taylor’s limit formula of second degree for $\,f\,$ with the development point $\,(x_0,y_0)=(0,0)\,$ in standard form.
hint
See Theorem 21.7.
B
Determine the gradient $\,\nabla f(0,0)\,$ and the Hessian matrix $\,\mathbf H f(0,0)\,,$ and write in matrix form Taylor’s limit formula of second degree for $\,f\,$ with the development point $\,(x_0,y_0)=(0,0)\,.$
hint
See Theorem 21.8.
Intro to the following question:
We now wish to find and approximate value for $\,f(\frac 34, \frac12)\,$ from an approximating second-degree polynomial for $\,f\,.$ Of course, it is easy to use the approximating second-degree polynomial with the development point $\,(0,0)\,$ that we readily get from the first question. On the other hand $\,(\frac 34, \frac 12)\,$ lies a little closer to $\,(1,1\,)$ from which is also relatively easy to develop. So perhaps on should rather use $\,(1,1\,)$ as the development point? Which difference does this make?
C
Determine the approximating polynomials of second degree, $\,P_2(x,y)\,$ and $\,Q_2(x,y)\,$ for $\,f\,$ with the development points $\,(0,0)\,$ and $\,(1,1)\,$, respectively. Determine their values in the point $\,(\frac 34, \frac 12)\,$ and compare to a computer value of $\,f(\frac 34, \frac 12)\,.$
In the following question we shall illustrate the error that we get if we apply the second-degree polynomial in stead fo the exact value.
B
Determine using the result in the first question the length of diagonal in a rectangle with the side length 2.9 and 4.2 (you may use Maple in the computation).
hint
The distance from the origin to an arbitrary point $\,(x,y)\,$ is a function of 2 variables. Yes, it is exactly $\,f(x,y)=\sqrt{x^2+y^2}\,.$
hint
Imagine that the rectangle is paced in the $\,(x,y)$-coordinatsystem such that the diagonal connects the points $\,(0,0)\,$ and $\,(2.9,4.2)\,.$
answer
The length can be approximated by the use of $\,P_2\,$ that we found above with the development point $\,(3,4)\,.$ In short $\,P_{2}(2.9,4.2)=5.10400\,.$
C
Compare to a Maple–value for the length of the diagonal.
answer
The computer value is 5.10392, so the approximation is not that bad.
D
Is the difference significant?
answer
The difference is approximately $8\cdot10^{-5}$, which is not much in relation to length of the diagonal.
$$A = \,\left\{(x,y)\in\reel^2\,|\,-2\leq x \leq2\,,\,\,-2\leq y \leq2\right\}\,$$
illustrations using Maple where first you see the graph for $\,f\,$ alone, then together with its approximating second-degree polynomial with the development point $\,(0,0)\,.$
B
What is the largest value that $\,f\,$ attains at the boundary of $\,A\,?$
C
It looks as if the graph for $\,f\,$ must have the horizontal tangent plane in the point of tangency
$$\,R=(0,0,f(0,0))=(0,0,0)\,.$$
Show this by determining a normal vector for the tangent plane in $\,R\,.$ And explain that the point $\,(x,y)=(0,0)\,$ is a stationary point.
D
In fact it also looks as if $\,f\,$ has a proper local maximum in the point $\,(0,0)\,$ with the value
$$\,f(0,0)=0\,.$$
This means that $\,f(x,y)\,$ must be negative when $\,(x,y)\,$ is sufficient close to $\,(0,0)\,.$ Show using Taylor’s limit formula of second degree for $\,f\,$ the development point $\,(0,0)\,$ that this is correct.
E
Advanced: Make a similar investigation of $\,f\,$ in and about the point $\,(x,y)=(2,2)\,.$
Exercise 5: Diagonalization and Reduction of a Quadratic Form
Using Eigenvectors you see that $\,\mA\,$ has $-4$ as a single eigenvalue, but $-1$ as a double eigenvalue. Therefore some effort must be done in order to find an orthonormal basis for $\,\reel^3\,$ consisting of eigenvectors for $\,\mA\,.$
hint
To use GramSchmidt i $\,\reel^3\,$ is probably like using a sledgehammer to crack a nut. Recommendation: Choose an eigenvector from the eigenspace $\,E_{-4}\,$ and one from $\,E_{-1}\,.$ Norm these, and find their cross product, then you have three usable vectors for stating $\,\mathbf{Q}\,$.
answer
There are more possibilities, here we have chosen:
Note that $\,f\,$ can be divided into two terms, the first is a quadratic form, let us call this $\,k\,,$ and the second term is a first degree polynomial.
G
Determine the expression $\,k(x,y,z)\,,$ and convert this to a matrix form, and reduce this.
hint
Reduction means: Find an ordinary orthonormal basis for $\,\Bbb R^3\,$ in which the expression for $\,k\,$ is without mixed terms.
hint
You have probably noticed that the matrix form for $\,k(x,y,z)\,,$ is identical with $\,\mA\,,$ so you can just use the basis with corresponding diagonalization, we made above.
answer
If we choose the columns in the $\mathbf Q$ found above as a new orthonormal basis, we get this reduction of $k:$
Find an ordinary orthonormal basis for $\,\Bbb R^3\,$ in which the expression for $\,f\,$ is without mixed terms. Determine the expression.
hint
We use the orthonormal basis found above, then the quadratix form is in place. Now the question is how does the first-degree polynomial that is part of $\,f\,,$ look in this basis.
hint
Use $\,\mathbf Q\,$ found above as a change of base matrix.
answer
If we choose the columns in the matrix $\mathbf Q$ found above as a new orthonormal basis, $\,f\,$ will get this form:
