Determine, with the development point $\,x_0=0\,$, Taylor’s limit formula of the second degree for the function
$$f(x)=2\cos(x)-2\sin(2x)\,,\,\,x\in \reel\,.$$
hint
Remember that the limit formula includes an epsilon function, unlike the approximations.
answer
$$f(x)=2-4x-x^2+x^2\cdot \varepsilon(x)$$
B
A smooth function $f$ of one variable fulfills that $\,f(2)=1\,$ and $\,f’(2)=1\,.$ Its approximating second-degree polynomial with development point $\,x_0=2\,$ fulfills $\,P_2(1)=1\,.$ Determine $\,P_2(x)\,$.
hint
Set up the formula for the approximating polynomial and consider what is known and what you are still missing.
hint
Utilize the given information about the polynomial to set up an equation which you can solve for the last unknown component.
answer
$$P_2(x)=-1+x+(x-2)^2$$
Exercise 2: Taylor’s Formulas and Approximation. By Hand
State Taylor’s limit formula of the second degree for $\,f\,$ with the development point $\,(x_0,y_0)=(0,0)\,$.
hint
See Theorem 21.7.
hint
The single and double partial derivatives may seem overwhelming to compute by hand, but remember that the exponential function is particularly simple in differentiation. You are given a composite function (“a function within a function”), so use the chain rule.
We now wish to compute the value $\,f(\frac 34, \frac12)\,$. This can be hard to do without software aid (and we could establish advanced functions that even computer software would struggle with calculating values for). So, let’s instead find an approximate value of $\,f(\frac 34, \frac12)\,$ via an approximating second-degree polynomial. We can extract such an approximating polynomial from the above established limit formula with development point $\,(0,0)$. On the other hand, the point $\,(\frac 34, \frac 12)\,$ is a little closer to $\,(1,1)$, a point from which it is also relatively easy to develop. So, perhaps we should rather use $\,(1,1\,)$ as the development point? Would that make a significant difference?
D
Provide the approximating polynomials of the second degree, $\,P_2(x,y)\,$ and $\,Q_2(x,y)\,$, for $\,f\,$ with the development points $\,(0,0)\,$ and $\,(1,1)\,$, respectively.
Compute approximating values with $P_2$ and $Q_2$ at the point $\,(\frac 34, \frac 12)\,$. Compare these with a computer-calculated value of $\,f(\frac 34, \frac 12)\,.$
answer
At this point, $P_2$ deviates from $f$ with $\,2.04\%\,$, while $\,Q_2\,$ deviates with $\,0.719\%\,.$ So, even though it takes a little more effort to set up $Q_2$, it might be worth it for better precision.
Exercise 3: Application of an Approximating Polynomial
A function $\,f:\reel^2 \rightarrow\reel\,$ is given by
$$f(x,y)=\sqrt{x^2+y^2}\,.$$
A
Determine the approximating polynomial $\,P_2(x,y)\,$ of the second degree of $\,f\,$ with development point $\,(x_0,y_0)=(3,4)\,.$
The following question will illustrate and analyse the error that we incur by using the second-degree polynomial instead of the exact value.
B
Determine, using the approximation from the previous question, the diagonal length of a rectangle with side lengths 2.9 and 4.2. You may use Maple for the arithmetic calculations.
hint
Imagine the rectangle drawn in an $\,(x,y)$ coordinate system with its corners at the points $\,(0,0)\,$ and $\,(2.9,4.2)\,.$ The Pythagorean Theorem will be useful to find the diagonal.
hint
The distance from the origin to an arbitrary point $\,(x,y)\,$ is a function of two variables. And that function is precisely $\,f(x,y)=\sqrt{x^2+y^2}\,.$
answer
With plenty of decimals, the result is
$$\,P_{2}(2.9,4.2)=5.10400\,.$$
C
Compare this with a Maple-calculated value of the accurate diagonal length.
hint
This means, use Maple to solve $f(2.9,4.2)$ and compare this with the approximate value. Is the difference significant?
answer
The precise diagonal length is $5.10392$, so the approximation is not that bad.
The difference is approximately $8\cdot10^{-5}$, which is not much relative to the diagonal length. Whether this difference in significant or not - whether the approximation can be used for a certain application - depends on the purpose and need for accuracy.
Exercise 4: Diagonalization and Reduction of a Quadratic Form
Structure eigenvectors as columns in a $\mathbf V$ matrix, and eigenvalues as diagonal elements in a $\mathbf{\Lambda}$ matrix. Take care that their orders match!
hint
The $\mathbf Q$ matrix we want must be orthogonal, since it in the given matrix equation is implied that $\mathbf Q^{-1}=\mathbf Q^{\transp}$, meaning its columns must be mutually orthogonal and each of them normalized (of unit length). Note that $\mA$ is symmetric.
hint
$\mA$ has two eigenspaces (you have computed three eigenvalues of which two are equal). A symmetric matrix has orthogonal eigenspaces, so a vector from each will be orthogonal. Now we just need a third.
hint
It is an option to use the GramSchmidt algorithm on the two (linearly independent) eigenvectors that come from the same eigenspace. This might be overkill, though. Since the three span the domain $\reel^3$, then we can just compute the cross product of an eigenvectors from each eigenspace.
answer
There are several possibilities. A possible answer is:
Doing so will result in only square terms with eigenvalues in front and no mixed terms. The new $\widetilde x$ and $\widetilde y$ are then written in the orthonormal eigenbasis consisting of the columns from $\mathbf Q$ as basis vectors.
answer
By choosing the columns of $\mathbf Q$ found above as a new orthonormal basis, the reduced $k$ becomes
State an ordinary orthonormal basis for $\,\Bbb R^3\,$ in which the expression of $\,f\,$ does not contain mixed terms. Determine this expression of $f$.
hint
We have already reduced the quadatic terms from $f$. We are missing the linear terms from $f$, which form a first-degree polynomial, as well as the constant. How does the first-degree polynomial look in this new basis?
hint
$\mathbf Q$ is a change-of-basis matrix from the new to the original coordinates. Do a basis change on the coefficients of the linear terms.
$$A = \,\left\{(x,y)\in\reel^2\,|\,-2\leq x \leq2\,,\,\,-2\leq y \leq2\right\}\,.$$
A
$A$ covers a surface in the $(x,y)$ plane. Imagine “lifting (or lowering)” this surface up (or down) to $f$, creating an elevated surface in $(x,y,z)$ space. Let’s denote this elevated surface $M$. In other words, $M$ is the part of $f$ that is located directly above (or below) $A$. Create an illustration of $M$ in Maple.
hint
The illustration can be made by simply limiting the visible part of a plot of $f$ to the range of $A$. Alternatively, parametrize the new surface $M$ and plot it as a coordinate expression with plot3d.
B
What is the largest value that $\,f\,$ attains at the boundary of $\,M\,?$ Do not just read from the graph - a reading can be used to roughly verify afterwards.
hint
Make four restrictions of the function, one to each edge of this rectangular region. They, together with the corners, constitute the boundary. Then find maxima on these restrictions.
hint
The four restrictions are found as $f(-2,y)$, $f(2,y)$, $f(x,-2)$ and $f(x,2)$.
hint
Each restriction is a function of one variable. Finding extrema on those is a highschool task: Differentiate, set equal to zero and solve. After solving for all four, you will have found several $x$ and $y$ values that are candidates for maximum. Find their corresponding function values and compare. Don’t forget to compare with the corner values as well.
answer
The maximum on $f$ at the boundary of $M$ is $-4$, found at the points $(2,0)$ and $(0,2)$.
When looking at the illustration of $M$, it looks like there might be a local maximum at $(0,0)$, and that this maximum has the value $0$. If that is the case, then the graph of $f$ will have a horizontal tangent plane at
$$\,R=(0,0,f(0,0))=(0,0,0)\,.$$
C
Investigate whether this is the case by computing the normal vector for the tangent plane at $\,R\,.$ Justify based on this that the point $\,(x,y)=(0,0)\,$ is a stationary point.
It even looks as if $\,f\,$ has a proper local maximum at this point $\,(0,0)\,$ with the value
$$\,f(0,0)=0\,.$$
D
Discuss with a fellow student the meaning of and difference between the terms maximum, local maximum and proper local maximum.
E
In the same plotting window, plot the approximating second-degree polynomial of $f$ with development point $(0,0)$.
If the observation is correct and $(0,0)$ indeed is a proper local maximum, then $\,f(x,y)\,$ must be negative at points $\,(x,y)\,$ in a close vicinity around $\,(0,0)\,.$ Show that this is true using Taylor’s limit formula of the second degree for $\,f\,$ developed from this point.
hint
You should be able to rewrite the limit formula to
$$f(x,y)=(-3 + \varepsilon(x, y))(x^2 + y^2).$$
What does this tell us about function values at points in the close vicinity of $(0,0)$?
answer
When approaching $(x,y)\to (0,0)$, then
$$f(x,y)=(-3 + \varepsilon(x, y))(x^2 + y^2)$$
will attain negative values in a close vicinity of $(0,0)$, since $\varepsilon(x,y)\to 0$. Thus $(0,0)$ is a proper local maximum.
F
Advanced (skip if time is short):
It also looks as if the point $(2,2)$ is a stationary point and a proper local maximum. Make a similar investigation of $\,f\,$ at and about the point $\,(x,y)=(2,2)\,.$
answer
With Taylor’s limit formula, the function can be rewritten to:
When approaching $(x,y)\to (2,2)$, the epsilon function diminishes and goes towards zero. Also, the squared distance will be small, meaning eventually smaller than $1$. Thus a term smaller than $+3$ is subtracted by $8$, resulting in a negative function value close to $(2,2)$. Thus $(2,2)$ is a proper local maximum.
The above approaches with tangent plane normal vector and limit formula can be cumbersome when the task is to find extrema. Let us below repeat the investigation with the gradient in a much easier fashion.
G
Compute the gradient of $f$, and use it to find all stationary points of $f$.
hint
A stationary point is found where the gradient is the zero vector.
answer
$$\nabla f(x,y)=(3x^2 - 6x,3y^2 - 6y)$$
All stationary points on $f$ are: $(0,0), (0,2), (2,0)$ and $(2,2).$
H
What is the global maximum on $M$?
hint
To show global maximum, consider which of the stationary points that are relevant (which that are located in the interior of $M$). Compare the function value at this point with the functional values of the boundary points found earlier.
answer
We found all stationary points above. The point $(0,0)$ is the only stationary point within the interior of $M$. We already found all other candidates for global maximum from the boundary and corner investigation earlier, and we found all their function values to be negative. Since $f(0,0)=0$, then $(0,0)$ is a global maximum on $M$.
Exercise 6: Teaser Exercise
A function $f\in C^{\infty}(\reel^2)$ fulfills the equations
