Chapter 14

A First Course in Real Analysis · 26 exercises

Problem 1

Find the solution by the Lagrange multiplier rule. Find the minimum value of \(x_{1}^{2}+3 x_{2}^{2}+2 x_{3}^{2}\) subject to the condition \(2 x_{1}+3 x_{2}+\) \(4 x_{3}-15=0\)

4 step solution

Problem 1

Show that the relation \(F(x, y)=0\) yields \(y\) as a function of \(x\) in an interval \(I\) about \(x_{0}\) where \(F\left(x_{0}, y_{0}\right)=0\). Denote the function by \(f\) and compute \(f^{\prime}\). \(F(x, y) \equiv y^{3}+y-x^{2}=0 ;\left(x_{0}, y_{0}\right)=(0,0)\)

3 step solution

Problem 2

Find the solution by the Lagrange multiplier rule. Find the minimum value of \(2 x_{1}^{2}+x_{2}^{2}+2 x_{3}^{2}\), subject to the condition \(2 x_{1}+3 x_{2}-\) \(2 x_{3}-13=0\)

6 step solution

Problem 2

Show that the relation \(F(x, y)=0\) yields \(y\) as a function of \(x\) in an interval \(I\) about \(x_{0}\) where \(F\left(x_{0}, y_{0}\right)=0\). Denote the function by \(f\) and compute \(f^{\prime}\). \(F(x, y) \equiv x^{2 / 3}+y^{2 / 3}-4=0 ;\left(x_{0}, y_{0}\right)=(1,3 \sqrt{3})\)

4 step solution

Problem 3

Find the solution by the Lagrange multiplier rule. Find the minimum value of \(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\) subject to the conditions \(2 x_{1}+2 x_{2}+\) \(x_{3}+9=0\) and \(2 x_{1}-x_{2}-2 x_{3}-18=0\)

5 step solution

Problem 3

Show that the relation \(F(x, y)=0\) yields \(y\) as a function of \(x\) in an interval \(I\) about \(x_{0}\) where \(F\left(x_{0}, y_{0}\right)=0\). Denote the function by \(f\) and compute \(f^{\prime}\). \(F(x, y) \equiv x y+2 \ln x+3 \ln y-1=0 ;\left(x_{0}, y_{0}\right)=(1,1)\)

5 step solution

Problem 4

Find the solution by the Lagrange multiplier rule. Find the minimum value of \(4 x_{1}^{2}+2 x_{2}^{2}+3 x_{3}^{2}\) subject to the conditions \(x_{1}+2 x_{2}+\) \(3 x_{3}-9=0\) and \(4 x_{1}-2 x_{2}+x_{3}+19=0\)

4 step solution

Problem 4

Evaluate \(\int_{F} K\left(x_{1}, x_{2}\right) d V_{2}(x)\), where \(F\) is bounded by the curves whose equations are given. Perform the integration by introducing variables \(u_{1}, u_{2}\) as indicated. Draw a graph of \(F\) and the corresponding region in the \(u_{1}, u_{2}\)-plane. Find the inverse of each transformation. \(K\left(x_{1}, x_{2}\right)=\left(x_{1}^{2}+x_{2}^{2}\right)^{-3} \cdot F\) is bounded by \(x_{1}^{2}+x_{2}^{2}=2 x_{1}, x_{1}^{2}+x_{2}^{2}=4 x_{1}, x_{1}^{2}+\) \(x_{2}^{2}=2 x_{2}, x_{1}^{2}+x_{2}^{2}=6 x_{2} .\) Mapping: \(x_{1}=u_{1} /\left(u_{1}^{2}+u_{2}^{2}\right), x_{2}=u_{2} /\left(u_{1}^{2}+u_{2}^{2}\right)\)

7 step solution

Problem 4

Show that the relation \(F(x, y)=0\) yields \(y\) as a function of \(x\) in an interval \(I\) about \(x_{0}\) where \(F\left(x_{0}, y_{0}\right)=0\). Denote the function by \(f\) and compute \(f^{\prime}\). \(F(x, y) \equiv \sin x+2 \cos y-\frac{1}{2}=0 ;\left(x_{0}, y_{0}\right)=(\pi / 6,3 \pi / 2)\)

4 step solution

Problem 5

Find the solution by the Lagrange multiplier rule. Find the minimum value of \(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}\) subject to the condition \(2 x_{1}+\) \(x_{2}-x_{3}-2 x_{4}-5=0\)

6 step solution

Problem 5

Suppose that \(x=\left(x_{1}, x_{2}\right), y=\left(y_{1}, y_{2}\right)\) and \(F: \mathbb{R}^{4} \rightarrow \mathbb{R}^{2}\) are such that \(F(x, y)=0\) and the Implicit function theorem is applicable for all \((x, y)\). Denoting the implicitly defined function by \(f\), find a formula in terms of the first partial derivatives of \(F\) for \(\partial f^{1} / \partial x_{1}, \partial f^{1} / \partial x_{2}, \partial f^{2} / \partial x_{1}, \partial f^{2} / \partial x_{2}\).

6 step solution

Problem 5

Give an example of a relation \(F(x, y)=0\) such that \(F\left(x_{0}, y_{0}\right)=0\) and \(F_{2}\left(x_{0}, y_{0}\right)=0\) at a point \(O=\left(x_{0}, y_{0}\right)\), and yet \(y\) is expressible as a function of \(x\) in an interval about \(x_{0}\).

4 step solution

Problem 6

Find the solution by the Lagrange multiplier rule. Find the minimum value of \(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}\) subject to the conditions \(x_{1}-\) \(x_{2}+x_{3}+x_{4}-4=0\) and \(x_{1}+x_{2}-x_{3}+x_{4}+6=0\)

5 step solution

Problem 6

Show that the relation \(F\left(x_{1}, x_{2}, y\right)=0\) yields \(y\) as a function of \(\left(x_{1}, x_{2}\right)\) in a neighborhood of the given point \(P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)\). Denoting this function by \(f\), compute \(f_{, 1}\) and \(f_{, 2}\) at \(P\). \(F\left(x_{1}, x_{2}, y\right) \equiv x_{1}^{3}+x_{2}^{3}+y^{3}-3 x_{1} x_{2} y-4=0 ; P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=(1,1,2)\)

4 step solution

Problem 7

Find the solution by the Lagrange multiplier rule. Find the points on the curve \(4 x_{1}^{2}+4 x_{1} x_{2}+x_{2}^{2}=25\) which are nearest to the origin.

7 step solution

Problem 7

Show that the relation \(F\left(x_{1}, x_{2}, y\right)=0\) yields \(y\) as a function of \(\left(x_{1}, x_{2}\right)\) in a neighborhood of the given point \(P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)\). Denoting this function by \(f\), compute \(f_{, 1}\) and \(f_{, 2}\) at \(P\). \(F\left(x_{1}, x_{2}, y\right) \equiv e^{y}-y^{2}-x_{1}^{2}-x_{2}^{2}=0 ; P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=(1,0,0)\)

4 step solution

Problem 8

Find the solution by the Lagrange multiplier rule. Find the points on the curve \(7 x_{1}^{2}+6 x_{1} x_{2}+2 x_{2}^{2}=25\) which are nearest to the origin.

6 step solution

Problem 8

Show that the relation \(F\left(x_{1}, x_{2}, y\right)=0\) yields \(y\) as a function of \(\left(x_{1}, x_{2}\right)\) in a neighborhood of the given point \(P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)\). Denoting this function by \(f\), compute \(f_{, 1}\) and \(f_{, 2}\) at \(P\). \(F\left(x_{1}, x_{2}, y\right) \equiv x_{1}+x_{2}-y-\cos \left(x_{1} x_{2} y\right)=0 ; P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=(0,0,-1)\)

6 step solution

Problem 9

Evaluate the integral $$ \int_{F} x_{3} d V_{3}(x) $$ by changing to spherical coordinates: \(x_{1}=\rho \cos \varphi \sin \theta, x_{2}=\rho \sin \varphi \sin \theta, x_{3}=\) \(\rho \cos \theta\), where \(F\) is the region determined by the inequalities \(0 \leqslant x_{1}^{2}+x_{2}^{2} \leqslant x_{3}^{2}\), \(0 \leqslant x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \leqslant 1, x_{3} \geqslant 0\)

6 step solution

Problem 9

A vector function \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) is given. Verify that the Inverse function theorem is applicable and find the inverse function \(g\). \(y_{1}=x_{1}, y_{2}=x_{1}^{2}+x_{2}\)

5 step solution

Problem 9

Show that the relation \(F\left(x_{1}, x_{2}, y\right)=0\) yields \(y\) as a function of \(\left(x_{1}, x_{2}\right)\) in a neighborhood of the given point \(P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)\). Denoting this function by \(f\), compute \(f_{, 1}\) and \(f_{, 2}\) at \(P\). \(F\left(x_{1}, x_{2}, y\right) \equiv x_{1}+x_{2}+y-e^{x_{1} x_{2} y}=0 ; P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=\left(0, \frac{1}{2}, \frac{1}{2}\right)\)

5 step solution

Problem 11

(a) Find the maximum of the function \(x_{1}^{2} \cdot x_{2}^{2} \cdots x_{n}^{2}\) subject to the side condition \(\sum_{i=1}^{n} x_{i}^{2}=1\) (b) If \(\sum_{i=1}^{n} x_{i}^{2}=1\), show that \(\left(x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}\right)^{1 / n} \leqslant 1 / n\). (c) If \(a_{1}, a_{2}, \cdots, a_{n}\) are positive numbers, prove that $$ \left(a_{1} \cdot a_{2} \cdots a_{n}\right)^{1 / n} \leqslant \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} $$ [The geometric mean of \(n\) numbers is always less than or equal to the arithmetic mean.]

5 step solution

Problem 12

A vector function \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) is given. Verify that the Inverse function theorem is applicable and find the inverse function \(g\). \(y_{1}=x_{1} \cos \left(\pi x_{2} / 2\right), y_{2}=x_{1} \sin \left(\pi x_{2} / 2\right), x_{1}>0,-1

4 step solution

Problem 12

Suppose that \(F(x, y, z)=0\) is such that the functions \(z=f(x, y), x=g(y, z)\), and \(y=h(z, x)\) all exist by the Implicit function theorem. Show that $$ f_{1} \cdot g_{, 1} \cdot h_{, 1}=-1 $$ This formula is frequently written $$ \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z}=-1 $$

4 step solution

Problem 13

Find an example of a relation \(F\left(x_{1}, x_{2}, y\right)=0\) and a point \(P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)\) such that \(P\) satisfies the relation, and \(F_{, 1}\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=F_{, 2}\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=F_{, 3}\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=0\) yet \(y\) is a function of \(\left(x_{1}, x_{2}\right)\) in a neighborhood of \(P .\)

4 step solution

Problem 14

Suppose that the Implicit function theorem applies to \(F(x, y)=0\) so that \(y=f(x)\). Find a formula for \(f^{\prime \prime}\) in terms of \(F\) and its partial derivatives.

5 step solution

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