Consider $a,b,c\in \mathbb{R}$ such that $|a{x}^{2}+bx+c|\le 1\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}x\in [0,1]$. Prove that $|a|\le 8\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}},|b|\le 8$ and $|c|\le 1$.

crossoverman9b
2022-06-26
Answered

Consider $a,b,c\in \mathbb{R}$ such that $|a{x}^{2}+bx+c|\le 1\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}x\in [0,1]$. Prove that $|a|\le 8\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}},|b|\le 8$ and $|c|\le 1$.

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Ryan Newman

Answered 2022-06-27
Author has **26** answers

Let $f(x)=a{x}^{2}+bx+c$.

Hence, $f(0)=c$, $f(1)=a+b+c$ and $f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$

After solving of this system we obtain:

$|a|=|2f(1)-4f\left(\frac{1}{2}\right)+2f(0)|\le 2|f(1)|+4|f\left(\frac{1}{2}\right)|+2|f(0)|\le 8$

$|b|=|-f(1)+4f\left(\frac{1}{2}\right)-3f(0)|\le |f(1)|+4|f\left(\frac{1}{2}\right)|+3|f(0)|\le 8$

$|c|=|f(0)|\le 1$

Hence, $f(0)=c$, $f(1)=a+b+c$ and $f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$

After solving of this system we obtain:

$|a|=|2f(1)-4f\left(\frac{1}{2}\right)+2f(0)|\le 2|f(1)|+4|f\left(\frac{1}{2}\right)|+2|f(0)|\le 8$

$|b|=|-f(1)+4f\left(\frac{1}{2}\right)-3f(0)|\le |f(1)|+4|f\left(\frac{1}{2}\right)|+3|f(0)|\le 8$

$|c|=|f(0)|\le 1$

asked 2022-05-19

Determine the value of $b$ for which the system

$\begin{array}{rl}{x}_{1}+4{x}_{2}-3{x}_{3}+2{x}_{4}& =2\\ 2{x}_{1}+7{x}_{2}-4{x}_{3}+4{x}_{4}& =3\\ -{x}_{1}-5{x}_{2}+5{x}_{3}-2{x}_{4}& =b\\ 3{x}_{1}+10{x}_{2}-5{x}_{3}+6{x}_{4}& =4\end{array}$

is soluble, and determine the solution set.

$\begin{array}{rl}{x}_{1}+4{x}_{2}-3{x}_{3}+2{x}_{4}& =2\\ 2{x}_{1}+7{x}_{2}-4{x}_{3}+4{x}_{4}& =3\\ -{x}_{1}-5{x}_{2}+5{x}_{3}-2{x}_{4}& =b\\ 3{x}_{1}+10{x}_{2}-5{x}_{3}+6{x}_{4}& =4\end{array}$

is soluble, and determine the solution set.

asked 2022-09-20

Find homogeneous system of linear equations whose solution space is: V = span((1,-2,4,3),(1,-1,6,4),(3,-8,8,3)).

First I found vectors were linearly dependent, so I discarded the third vector to form a new base. Next I figured system will have 2 free variables and if we would use gauss-jordan elimination on matrix of a system we would get this:

$$\left[\begin{array}{cccc}1& 0& \ast & \ast \\ 0& 1& \ast & \ast \end{array}\right]$$

First I found vectors were linearly dependent, so I discarded the third vector to form a new base. Next I figured system will have 2 free variables and if we would use gauss-jordan elimination on matrix of a system we would get this:

$$\left[\begin{array}{cccc}1& 0& \ast & \ast \\ 0& 1& \ast & \ast \end{array}\right]$$

asked 2021-02-11

solve using substitution

asked 2022-05-19

What are the steps required to solve this system of equations?

The system that I need to solve is

$\begin{array}{rlr}& {i}_{1}+{i}_{2}+{i}_{3}& =0\\ & {i}_{1}+{i}_{4}+{i}_{6}& =0\\ & {i}_{5}+{i}_{6}& ={i}_{2}\\ & -{v}_{s1}+{i}_{1}{r}_{1}+{i}_{3}{r}_{3}-{i}_{4}{r}_{4}+{v}_{s4}& =0\\ & -{i}_{2}{r}_{2}+{v}_{s2}-{i}_{5}{r}_{5}-{i}_{3}{r}_{3}& =0\\ & -{v}_{s4}+{i}_{4}{r}_{4}+{i}_{5}{r}_{5}-{i}_{6}{r}_{6}-{v}_{s6}& =0\text{}\end{array}$

for the variables ${i}_{k}$.

The system that I need to solve is

$\begin{array}{rlr}& {i}_{1}+{i}_{2}+{i}_{3}& =0\\ & {i}_{1}+{i}_{4}+{i}_{6}& =0\\ & {i}_{5}+{i}_{6}& ={i}_{2}\\ & -{v}_{s1}+{i}_{1}{r}_{1}+{i}_{3}{r}_{3}-{i}_{4}{r}_{4}+{v}_{s4}& =0\\ & -{i}_{2}{r}_{2}+{v}_{s2}-{i}_{5}{r}_{5}-{i}_{3}{r}_{3}& =0\\ & -{v}_{s4}+{i}_{4}{r}_{4}+{i}_{5}{r}_{5}-{i}_{6}{r}_{6}-{v}_{s6}& =0\text{}\end{array}$

for the variables ${i}_{k}$.

asked 2022-05-15

System of equations with multiplication

I want to find all $x,y,z\in \mathbb{R}$ such that

$(x+1)yz=12,(y+1)zx=4,(z+1)xy=4$.

I can multiply all three equations to get

$(x+1)(y+1)(z+1){x}^{2}{y}^{2}{z}^{2}=192$

I can divide the first equation by the second to get

$\frac{(x+1)y}{(y+1)x}}=4$, which simplifies to $3xy+4x-y=0$

None of these seems to help. How can I solve the equations?

I want to find all $x,y,z\in \mathbb{R}$ such that

$(x+1)yz=12,(y+1)zx=4,(z+1)xy=4$.

I can multiply all three equations to get

$(x+1)(y+1)(z+1){x}^{2}{y}^{2}{z}^{2}=192$

I can divide the first equation by the second to get

$\frac{(x+1)y}{(y+1)x}}=4$, which simplifies to $3xy+4x-y=0$

None of these seems to help. How can I solve the equations?

asked 2022-06-07

For what values of k does this system of equations have a unique solution?

$\{\begin{array}{l}y+2kz=0\\ x+2y+6z=2\\ kx+2z=1\end{array}$

I have $\left[\begin{array}{cccc}1& 2& 6& 2\\ 0& 1& 2k& 0\\ k& 0& 2& 1\end{array}\right]$

When I row reduce, I get:

$\left[\begin{array}{cccc}1& 0& 6-4k& 2\\ 0& 1& 2k& 0\\ 0& 0& 2-6k-4{k}^{2}& 1-2k\end{array}\right]$

$\{\begin{array}{l}y+2kz=0\\ x+2y+6z=2\\ kx+2z=1\end{array}$

I have $\left[\begin{array}{cccc}1& 2& 6& 2\\ 0& 1& 2k& 0\\ k& 0& 2& 1\end{array}\right]$

When I row reduce, I get:

$\left[\begin{array}{cccc}1& 0& 6-4k& 2\\ 0& 1& 2k& 0\\ 0& 0& 2-6k-4{k}^{2}& 1-2k\end{array}\right]$

asked 2022-06-24

stability in the periodic orbit and in the singular point

$\dot{x}=-y+\lambda x(36-9{x}^{2}-{y}^{2})\phantom{\rule{0ex}{0ex}}\dot{y}=9x+\lambda y(36-9{x}^{2}-{y}^{2})\phantom{\rule{0ex}{0ex}}\dot{z}=-6z-{\lambda}^{2}{x}^{2}{y}^{2}{z}^{3}$

I want to analyze the stability in the periodic orbit and in the singular point, so for the singular point I take the derived matrix of the linear part, and I got the eigenvalues, wich are ${\lambda}_{1}=-6$, ${\lambda}_{2}=3i$ , $\lambda =-3i$ . I wanted to use Andronov-Vitt but I have two eigenvalues with no real part so I can´t, does anybody can help me with this?

$\dot{x}=-y+\lambda x(36-9{x}^{2}-{y}^{2})\phantom{\rule{0ex}{0ex}}\dot{y}=9x+\lambda y(36-9{x}^{2}-{y}^{2})\phantom{\rule{0ex}{0ex}}\dot{z}=-6z-{\lambda}^{2}{x}^{2}{y}^{2}{z}^{3}$

I want to analyze the stability in the periodic orbit and in the singular point, so for the singular point I take the derived matrix of the linear part, and I got the eigenvalues, wich are ${\lambda}_{1}=-6$, ${\lambda}_{2}=3i$ , $\lambda =-3i$ . I wanted to use Andronov-Vitt but I have two eigenvalues with no real part so I can´t, does anybody can help me with this?