7

Homework 7.pdf

Suppose \(f : [0, 1] \to \mathbb{R}\) is a continuous function such that \(f([0, 1]) \subset \mathbb{Q}\). Show that \(f\) is a constant function.

Suppose it is not constant function, then \(\exists x_1,x_2\in[0,1]\) such that \(f(x_1)=q_1\in \mathbb{Q}\) and \(f(x_2)=q_2\in\mathbb{Q}\) where \(q_1\neq q_2\)

Then by intermediate value theorem, \(\forall y_0\in(q_1,q_2),\exists x_0\in\left(x_1,x_2\right):y_0=f\left(x_0\right)\)

Then \(y_0\) can take the irrational number and there is also \(x_0\) corresponding to that \(y_0\)

Then it is contradiction to \(f([0, 1]) \subset \mathbb{Q}\), thus \(f\) is a constant function
Decide, for which \(n \in \mathbb{N}\), if the function \(p(x) = x^n\) is strictly increasing. Justify.

If \(x\in\mathbb{R}\), then for \(n=2k-1,k\in\mathbb{N}\), \(p(x)\) is strictly increasing

By theorem we know a continuous function has to be injective, then it is monotonic increasing or decreasing

Since we know \(x^{2k}\) is not injective because \(x^{2n}=(-x)^{2n}\)

Thus \(n\) has to be odd

If \(x\in [0,+\infty)\) or \(x\) belongs to the subset of \([0,+\infty)\), then for all \(n\in\N\) \(p(x) = x^n\) is strictly increasing since it is injective.
Prove that if \(f\) is strictly increasing, then so is its inverse.

Assume \(f:X\rightarrow Y\)

Since \(f\) is strictly increasing, then \(\forall x_1,x_2\) if \(x_1<x_2\), then \(f(x_1)<f(x_2)\)

Suppose its inverse is not strictly increasing, then \(\exists y_1,y_2\) such that \(y_1<y_2\), but \(f^{-1}\left(y_1\right)\geq f^{-1}\left(y_2\right)\)

Since \(f^{-1}\left(y_1\right),f^{-1}\left(y_2\right)\in X\), then \(f^{-1}\left(y_1\right)\geq f^{-1}\left(y_2\right)\Rightarrow f\left(f^{-1}\left(y_1\right)\right)\geq f\left(f^{-1}\left(y_2\right)\right)\Rightarrow y_1\geq y_2\)

Contradiction

Thus if \(f\) is strictly increasing, then so is its inverse.
Prove that the function \(\sinh x = \frac{e^x - e^{-x}}{2}\) is bijective and compute the inverse.
1. Injectivity
  
  Let \(\sinh x_1=\sinh x_2\), then \(\frac{e^{x_1}-e^{-x_1}}{2}=\frac{e^{x_2}-e^{-x_2}}{2}\Rightarrow e^{x_1}-e^{x_2}=\frac{e^{x_2}-e^{x_1}}{e^{x_1}e^{x_2}}\Rightarrow-e^{x_1}e^{x_2}\left(e^{x_1}-e^{x_2}\right)=\left(e^{x_1}-e^{x_2}\right)\)
  
  Then \(e^{x_1}=e^{x_2}\), then \(x_1=x_2\) since \(-e^{x_1}e^{x_2}=1\) is impossible
  
  Thus it is injective 2. Surjectivity
  
  We need to prove \(\forall y\in \R,\exists x\in\R:y = \frac{e^x - e^{-x}}{2}\)
  
  Consider \(x=\ln\left(y+\sqrt{y^2+1}\right)\), then \(\sinh x=\frac{e^{x}-e^{-x}}{2}=\frac{e^{\ln\left(y+\sqrt{y^2+1}\right)}-\frac{1}{e^{\ln\left(y+\sqrt{y^2+1}\right)}}}{2}=\frac{y+\sqrt{y^2+1}-\frac{1}{y+\sqrt{y^2+1}}}{2}=\frac{y^2+y^2+1+2y\sqrt{y^2+1}-1}{2\left(y+\sqrt{y^2+1}\right)}=\frac{2y^2+2y\sqrt{y^2+1}}{2\left(y+\sqrt{y^2+1}\right)}=y\)
  
  Thus it is surjective
  
  Then we compute the inverse
  
  Let \(\frac{e^x - e^{-x}}{2} = y\), then \(e^x - \frac{1}{e^x} = 2y\)
  
  Let \(t = e^x\), then \(t - \frac{1}{t} = 2y\). Since \(t>0\), then \(t^2 - 2yt - 1 = 0\)
  
  Thus \(t = \frac{2y \pm \sqrt{4y^2 + 4}}{2}\), then \(t = y \pm \sqrt{y^2 + 1}\)
  
  Since \(t>0\), then \(t\) has to be \(y+\sqrt{y^2+1}\)
  
  Thus \(e^{x}=y+\sqrt{y^2+1}\), then take log, we have \(x=\ln\left(y+\sqrt{y^2+1}\right)\)
  
  Finally, \(f^{-1}\left(x\right)=\ln\left(x+\sqrt{x^2+1}\right)\)
Show that \(f(x) = \frac{ax + b}{cx + d}\) is injective if and only if \(ad - bc \neq 0\). Determine \(\text{Im}f\) and find \(f^{-1}\).

\(\Leftarrow\)) Take \(f(x_1)=f(x_2)\), then \(\frac{ax_1+b}{cx_1+d}=\frac{ax_2+b}{cx_2+d}\Rightarrow acx_1x_2+bd+adx_1+bcx_2=acx_1x_2+bd+adx_2+bcx_1\Rightarrow adx_1+bcx_2=adx_2+bcx_1\)

Then we have \(ad(x_1-x_2)=bc(x_1-x_2)\Rightarrow ad=bc=0\) or \(x_1=x_2\)

Since \(ad\neq bc\), then \(x_1=x_2\), then \(f\) is injective

\(\Rightarrow\)) Since \(f\) is injective, then if \(f(x_1)=(x_2)\), then \(x_1=x_2\)

Let compute \(f(x_1)=f(x_2)\), then we get \((ad-bc)(x_1-x_2)=0\)

Suppose \(ad-bc=0\), then \(x_1-x_2\) can be any number which is contradiction to \(x_1-x_2=0\)

Thus \(ad-bc\neq 0\)

Then we find \(\text{Im}f\) and find \(f^{-1}\)

\(f(x)=\frac{ax+b}{cx+d}=\frac{\left(cx+d\right)\frac{a}{c}-\frac{ad}{c}+b}{cx+d}=\frac{a}{c}-\frac{\frac{ad}{c}-b}{cx+d}\)

Since \(x\neq \frac {-d}{c}\), then we know \(cx+d\neq 0\), then \(\frac{1}{cx+d}\in\left(-\infty,0\right)\cup\left(0,+\infty\right)\)

Then \(-\frac{1}{cx+d}\in\left(-\infty,0\right)\cup\left(0,+\infty\right)\), then \(\frac{a}{c}-\frac{\frac{ad}{c}-b}{cx+d}\in\left(-\infty,\frac{a}{c}\right)\cup\left(\frac{a}{c},+\infty\right)\)

Thus \(\text{Im}f=\left(-\infty,\frac{a}{c}\right)\cup\left(\frac{a}{c},+\infty\right)\)

Since \(y=\frac{a}{c}-\frac{\frac{ad}{c}-b}{cx+d}\Rightarrow\frac{a}{c}-y=\frac{\frac{ad}{c}-b}{cx+d}\Rightarrow\frac{\frac{a}{c}-y}{\frac{ad}{c}-b}=\frac{1}{cx+d}\Rightarrow cx+d=\frac{\frac{ad}{c}-b}{\frac{a}{c}-y}\Rightarrow cx=\frac{\frac{ad}{c}-b-\frac{ad}{c}+\mathrm{d}y}{\frac{a}{c}-y}=\frac{\mathrm{d}y-b}{\frac{a}{c}-y}\Rightarrow x=\frac{\mathrm{d}y-b}{a-cy}\)

Finally, \(f^{-1}=\frac{dx-b}{a-cx}\)
Let \(a > 0\), we define the exponential function \(a^x := e^{\ln(a)x}\). Prove using the definition of exponential functions that:
1. \(a^{b+c} = a^b a^c\).
  
  By definition, this is equivalent to \(e^{\ln(a)\left(b+c\right)}=e^{\ln(a)b}e^{\ln(a)c}\)
  
  Let consider \(e^{\ln(a)\left(b+c\right)}=e^{\ln(a)b+\ln(a)c}=e^{\ln(a)b}e^{\ln(a)c}\)
  
  Thus \(a^{b+c} = a^b a^c\). 2. \(\left(a^b\right)^c = a^{bc}\).
  
  By definition, this is equivalent to prove \(\left(e^{\ln(a)b}\right)^{c}\left.=\right.e^{\ln(a)bc}\)
  
  Let consider \(\left(e^{\ln(a)b}\right)^{c}=e^{\ln\left(e^{\ln(a)b}\right)\cdot c}=e^{\ln\left(a^{b}\right)\cdot c}=e^{\ln a^{bc}}=a^{bc}\)
  
  Thus \(\left(a^b\right)^c = a^{bc}\). 3. \(a^n = a \cdot a \cdot \ldots \cdot a\) (\(n\) times) for all \(n \in \mathbb{N}\).
  
  By definition, this is equivalent to prove \(\left(e^{\ln a}\right)^{n}=\left(e^{\ln a}\right)\cdot\ldots\cdot\left(e^{\ln a}\right)\)
  
  Consider \(\left(e^{\ln a}\right)^{n}=\left(e^{\ln a}\right)^{\left(1+1+\cdots+1\right)}=\left(e^{\ln a}\right)\cdot\ldots\cdot\left(e^{\ln a}\right)\)
  
  Thus \(a^n = a \cdot a \cdot \ldots \cdot a\) (\(n\) times) for all \(n \in \mathbb{N}\). 4. \(a^{\frac{1}{n}} = \sqrt[n]{a}\) for all \(n \in \mathbb{N}\).
  
  By definition, this is equivalent to prove \(\left(e^{\ln a}\right)^{\frac{1}{n}}=\sqrt[n]{e^{\ln a}}\)
  
  Consider \(\left(e^{\ln a}\right)^{\frac{1}{n}}=\left(e^{\ln a}\right)^{\frac{1}{n}}=e^{\frac{1}{n}\ln a}=e^{\ln a^{\frac{1}{n}}}=e^{\ln\sqrt[n]{a}}=\sqrt[n]{a}\)
  
  Thus \(a^{\frac{1}{n}} = \sqrt[n]{a}\) for all \(n \in \mathbb{N}\) 5. \(a^{\frac{m}{n}} = \sqrt[n]{a^m}\) for all \(n \in \mathbb{N}, m \in \mathbb{Z}\).
  
  By definition, this is equivalent to prove \(\left(e^{\ln a}\right)^{\frac{m}{n}}=\left(\sqrt[n]{e^{\ln a}}\right)^{m}\)
  
  Consider \(\left(e^{\ln a}\right)^{\frac{m}{n}}=\left(\left(e^{\ln a}\right)^{m}\right)^{\frac{1}{n}}=\left(a^{m}\right)^{\frac{1}{n}}\) then use (4) we get \(=\sqrt[n]{a^{m}}\)
  
  Thus \(a^{\frac{m}{n}} = \sqrt[n]{a^m}\) for all \(n \in \mathbb{N}, m \in \mathbb{Z}\) 6. \(a^x\) is strictly increasing for \(a > 1\) and strictly decreasing for \(0 < a < 1\).
  
  Proof
  
  Let take \(x_1>x_2\), then \(\frac{a^{x_1}}{a^{x_2}}=\frac{e^{\left(\ln a\right)x_1}}{e^{\left(\ln a\right)x_2}}=e^{\left(\ln a\right)x_1-\left(\ln a\right)x_2}\)
  
  If it is strictly increasing, then \(e^{\left(\ln a\right)x_1-\left(\ln a\right)x_2}>1\Rightarrow\left(\ln a\right)x_1-\left(\ln a\right)x_2>0\Rightarrow\left(\ln a\right)x_1>\left(\ln a\right)x_2\)
  
  Thus \(\ln a>0\Rightarrow a>1\)
  
  If it is strictly decreasing, then \(0<e^{\ln ax_1-\ln ax_2}<1\Rightarrow\left(\ln a\right)x_1-\left(\ln a\right)x_2<0\Rightarrow\left(\ln a\right)x_1<\left(\ln a\right)x_2\)
  
  Thus \(\ln a<0\Rightarrow0<a<1\)
  
  Thus \(a^x\) is strictly increasing for \(a > 1\) and strictly decreasing for \(0 < a < 1\).

‍