MAT102 Tutorial W4B

• 1
  • $f:{\mathbb{R}}^2\to \mathbb{R}$
  • $f(x,y)=\frac{1}{x^2+y^2+1}$
  • a: Injective or surjective
    • Injective
      • If I have two outputs equal, the inputs must be the same.
      • $f(a,b)=f(c,d)$
      • Counterexample:
        • $f(1,0)=\frac{1}{2}$
        • $f(-1,0)=\frac{1}{2}$
    • Surjective
      • Have an arbitrary $n$ where $f(a,b)=n$
      • Counterexample:
        • Top and bottom are always positve.
        • So we don't reach the negative values, $f(x,y)>0$
  • b
    • Find the image of ${\mathbb{R}}^2$ under $f$
    • $f({\mathbb{R}}^2)=(0,1]$
    • Proof
    • $f({\mathbb{R}}^2)\subseteq (0,1]$
      • Pick any $z\in f({\mathbb{R}}^2)$
      • There is $(x,y)\in {\mathbb{R}}^2:z=f(x,y)$
      • Want to show that $z\in (0,1]$
      • $z=f(x,y)=\frac{1}{x^2+y^2+1}$
      • $0\le x^2+y^2$
      • $1\le x^2+y^2+1$
      • $1\ge \frac{1}{x^2+y^2+1}>0$
      • Larger than $0$ because $x^2+y^2+1\ge 1$
      • $1\ge z>0$
      • So $z$ is in there.
    • $f({\mathbb{R}}^2)\supseteq (0,1]$
      • Let $a\in (0,1]$
      • Let $x,y\in {\mathbb{R}}^2$
      • $a=\frac{1}{x^2+y^2+1}$
      • $\frac{1}{a}=x^2+y^2+1$
      • $\frac{1}{a}-1-y^2=x^2$
      • $\pm \sqrt{\frac{1}{a}-1-y^2}=x$
      • $\frac{1}{a}=x^2+y^2+1$
      • $\frac{1}{a}-1-x^2=y^2$
      • $\pm \sqrt{\frac{1}{a}-1-x^2}=y$
      • Solution
      • Want to show $w\in f({\mathbb{R}}^2)$
      • $w=f(x,y)$
      • $w=\frac{1}{x^2+y^2+1}$
      • So we need to fix one value and then we can solve for $x$ or $y$, the above approach doesn't work.
      • Fix $x=0$
      • $w=\frac{1}{y^2+1}$
      • $w(y^2+1)=1$
      • $w{y}^2+w=1$
      • $w{y}^2=1-w$
      • $y^2=\frac{1-w}{w}$
      • $y=\pm \sqrt{\frac{1-w}{w}}$
        • $\frac{1-w}{w}\ge 0$
        • $w>0$
      • $(0,\pm \sqrt{\frac{1-w}{w}})$
      • Now we can say $f(0,\pm \sqrt{\frac{1-w}{w}})=w$
      • $=\frac{1}{0^2+\frac{1-w}{w}+1}$
      • $=\frac{1}{\frac{1-w}{w}+\frac{w}{w}}$
      • $=\frac{1}{\frac{1}{w}}$
      • $=w$
      • So $w\in f({\mathbb{R}}^2)$
• 2
  • $F$ is all functions from $f:[0,1]\to \mathbb{R}$.
  • Define $T:F\to F$ as:
    • $(T(f))(x)=f(x)+2f(1-x)$
  • a
    • For an arbitrary $h\in F$, suppose that $(T(h))(x)=0$ for all $x\in [0,1]$
    • Prove that $h(x)=0$ for all $x\in [0,1]$
    • $(T(h))(x)=0$
    • $h(x)+2h(1-x)=0$
    • I know that $h:[0,1]\to \mathbb{R}$
    • $h(x)+2h(1-x)=0$
    • $h(x)=-2h(1-x)$
    • $h(x)=-2\underset{\begin{array}{c}\text{this will provide}\\ \text{the inverse of }\\ [0,1]\text{ for }x\end{array}}{\underbrace{h(1-x)}}$
    • $h(x)=-2h(1-x)$
    • Solution
      • $[T(h)](x)=h(x)+2h(1-x)=0$
      • $[T(h)](1-x)=h(1-x)+2h(1-(1-x))=0$
      • $[T(h)](1-x)=h(1-x)+2h(x))=0$
      • $h(1-x)+2h(x)=0$
      • $h(1-x)=2h(x)$
      • $\frac{1}{2}h(1-x)=h(x)$
      • $h(x)=-2h(1-x)$
      • $\frac{1}{2}h(1-x)=-2h(1-x)$
      • $h(1-x)=-4h(1-x)$
      • $0=-3h(1-x)$
      • $0=h(1-x)$
      • Then we know that $[T(h)](x)=h(x)+2(0)=0$
      • $[T(h)](x)=h(x)=0$
      • So we've proven that $h(x)=0$ for all $x\in [0,1]$
  • b
    • Injective
    • For any $f,g\in F$, if $[T(f)](x)=[T(g)](x)$ then $f(x)=g(x)$
    • $[T(f)](x)-[T(g)](x)=0$
    • $[T(f-g)](x)=0$
    • Let $h=f-g$
    • So we have $[T(h)](x)=0$ for all $x$
    • We know from part A), that this is true.
    • So then $(f-g)(x)=0,\mathrm{\forall }x$
    • $f(x)-g(x)=0$
    • $f(x)=g(x)$
    • Proven
  • c
    • Surjective
    • for $g\in F$
    • There is $f\in F$ where $T(f)=g$
    • $[T(f)](x)=f(x)+2f(1-x)=g(x)$
    • Find $f(x)$ in terms of $g(x)$
    • $[T(f)](1-x)=f(1-x)+2f(x)=g(1-x)$
    • $f(x)+2f(1-x)=g(x)$
    • $f(1-x)+2f(x)=g(1-x)$
    • $2f(1-x)+4f(x)=2g(1-x)$
    • $\begin{array}{cc}& f(x)+2f(1-x)=g(x)\\ -& 2f(1-x)+4f(x)=2g(1-x)\end{array}$
    • $f(x)-4f(x)=g(x)-2g(1-x)$
    • $-3f(x)=g(x)-2g(1-x)$
    • $f(x)=-\frac{g(x)-2g(1-x)}{3}$
    • If we take $f(x)$ as an input to $[T(f)](x)=\left[T(-\frac{g(x)-2g(1-x)}{3})\right]=g(x)$
    • #tk tutorial activity 3