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\rhead{Name: Hemanta K. Maji} %%% <-- YOUR NAME HERE

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\title{Homework 0\\(0 Points. Practice Problems only.)}

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%%% Problem 1 
\item {\bfseries Approximating Exponentiation.} 
  The goal of this problem is to approximate the function $\exp(-\eps)$, where $0\leq \eps\leq 1$, using polynomials.  
  We shall use the Lagrange form of the Taylor remainder theorem to perform this estimate. 
  
  Let us define the function $f(x) = \exp(-x)$. 
  Verify that $f\p i(x) = (-1)^i\exp(-x)$, where $i\geq0$ and $f\p i$ is the short-had for the $i$-th derivative of the function $f$. 
  Let us set $a=0$. 
  
  Recall that the Taylor's series of the function $f$ is defined as follows. 
  $$f(\eps) = \sum_{i\geq 0} f\p i(0) \frac{\eps^i}{i!}$$
  For $f(x) = \exp(-x)$, we have
  $$\exp(-\eps) = 1 - \eps + \frac{\eps^2}{2!} - \frac{\eps^3}{3!} + \dotsi$$
  
  Recall that the Lagrange form of the Taylor's remainder theorem is defined as follows. 
  For every $\eps$ and $k\geq 0$, there exists $\theta\in(0,1)$ such that
  $$f(\eps) = \left(\sum_{i\geq 0} f\p i(0) \frac{\eps^i}{i!}\right) + f\p{k+1}(\theta\eps) \frac{\eps^{k+1}}{(k+1)!}$$
  Therefore, for our choice of $f(x) = \exp(-x)$, we have
  $$\exp(-\eps) = \overbrace{\left(1 - \eps + \frac{\eps^2}{2!} - \frac{\eps^3}{3!} + \dotsi + (-1)^k\frac{\eps^k}{k!}\right)}^{\text{Approximation: } p_k(\eps)} + \overbrace{(-1)^{k+1}\exp(-\theta\eps)\frac{\eps^{k+1}}{(k+1)!}}^{\text{Remainder: }R_{\eps,k}}$$
  \begin{enumerate}
  \item Prove that for even $k$, the remainder $R_{\eps,k}$ is negative. 
    Therefore, you have proved that $\exp(-\eps) \leq p_{k}(\eps)$. 
  \item Prove that for odd $k$, the remainder $R_{\eps,k}$ is positive. 
    Therefore, you have proved that $\exp(-\eps) \geq p_k(\eps)$. 
  \item Prove that the magnitude of the remainder $\abs{R_{\eps,k}} \leq \frac{\eps^{k+1}}{(k+1)!}$. 
    This proves a bound on the quality of the approximation of $\exp(-\eps)$ by the polynomial $p_k(\eps)$. 
  \end{enumerate} 
  
  ({\footnotesize {\bfseries Remark:}
    Students are encouraged to plot $\exp(-x)$  and the polynomials $p_k(x)$ to understand the bounds proved in this problem.% 
  })


  
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%%% Problem 2 
\item {\bfseries Approximating Logarithm.}
  The goal of this problem is to approximate the function \mbox{$\ln(1-\eps)$}, where $0\leq \eps<1$, using polynomials. 
  
  Let us define $f(x) = \ln(1-x)$. 
  Verify that $f\p i(x) = - \frac{(i-1)!}{(1-x)^i}$, for $i\geq 1$.  
   
  Verify that, for $f(x)=\ln(1-x)$, the Taylor series gives us
  $$\ln(1-\eps) = - {\eps} - \frac{\eps^2}2 - \frac{\eps^3}3 - \dotsi$$
  
  Verify that, for $f(x)=\ln(1-x)$, the Lagrange form of the Taylor's remainder theorem gives us the following. 
  For every $\eps, k$, there exists $\theta\in(0,1)$ such that
  $$\ln(1-\eps) = \overbrace{\left(- {\eps} - \frac{\eps^2}2 - \frac{\eps^3}3 - \dotsi - \frac{\eps^k}k\right)}^{\text{Approximation: }p_k(\eps)} -\frac{1}{(1-\theta\eps)^{k+1}}\frac{\eps^{k+1}}{(k+1)}$$
  \begin{enumerate}
  \item Prove that $\ln(1-\eps) \leq p_k(\eps)$, for all $k\geq 0$. 
  \item How large is the magnitude of the remainder as a function of $k$ and $\eps$? 
  \item Prove that $\ln(1-\eps) \geq p_k(\eps) - \frac{\eps^k}{k}$, for all $0\leq \eps\leq 1/2$. 
  
  ({\footnotesize{\bfseries Hint:} Use the fact that $\ln(1-\eps) = -\eps -\frac{\eps^2}2 - \frac{\eps^3}3 - \dotsi$, for $0\leq \eps<1$.})
  \end{enumerate} 
  
  ({\footnotesize {\bfseries Remark:}
    Again, students are encouraged to plot $\ln(1-x)$  and the polynomials $p_k(x)$ and the polynomials $p_k(x)-\frac{x^k}{k}$ to understand the bounds proved in this problem.% 
  })
  
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%%% Problem 3
\item {\bfseries AM-GM Inequality.}
  The goal of this problem is to prove the AM-GM inequality using the Jensen's inequality. 
  Let us recall the Jensen's inequality. 
  A function $f$ is convex in the range $[a,b]$ if $f\p 2$ is positive in the range $[a,b]$. 
  Jensen's inequality states that if $f$ is convex in the range $[a,b]$, then
  $$\frac{f(a) + f(b)}2 \geq f\left(\frac{a+b}2\right)$$
  Equality holds if and only if $a=b$. 
  
  A function $f$ is concave in the range $[a,b]$ if $f\p 2$ is negative in the range $[a,b]$. 
  Jensen's inequality states that if $f$ is concave in the range $[a,b]$, then 
  $$\frac{f(a)+f(b)}2 \leq f\left(\frac{a+b}2\right)$$
  Equality holds if and only if $a=b$. 
  
  Let us recall the AM-GM inequality. 
  For positive $a,b$, we have 
  $$\frac{a + b}2 \geq \sqrt{ab}$$
  Equality holds if and only if $a=b$. 
  
  Prove the AM-GM inequality using Jensen's inequality using $f(x) = \ln(x)$ (recall that $\ln(x)$ is concave in $(0,\infty)$). 
  
   
  
\nextoddpage 
%%% Problem 4
\item {\bfseries Cauchy-Schwarz Inequality.}  
  The goal of this problem is to prove the Cauchy-Schwarz inequality using the Jensen's inequality. 
  Let us recall the Cauchy-Schwarz inequality. 
  For positive $a_1,a_2,b_1,b_2$, we have
  $$\left(a_1b_1 + a_2b_2\right) \leq \left(a_1^2 + a_2^2\right)^{1/2}\left(b_1^2 + b_2^2\right)^{1/2}$$
  Equality holds if and only if $\frac{a_1}{b_1} = \frac{a_2}{b_2}$. 
  
  \begin{enumerate}
  \item Let us consider an intermediate inequality. 
    For positive $A,B$, we have
    $$\left(1 + \sqrt{AB}\right) \leq \left(1 + A\right)^{1/2}\left(1+ B\right)^{1/2}$$
    Equality holds if and only if $A=B$. 
    
    Use this intermediate inequality to prove the Cauchy-Schwarz inequality. 
  \item Prove that the function $f(x) = \ln\left(1 + \exp(x)\right)$ is a convex function. 
  \item Prove the intermediate inequality using the Jensen's inequality on the function $f(x) = \ln\left(1 + \exp(x)\right)$. 
  \end{enumerate} 
  
   

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%%% Problem 5 
\item {\bfseries Young's Inequality.}
  The goal of this problem is to prove the Young's inequality using the (general form) of the Jensen's inequality. 
  One can interpret Young's inequality as a generalization of the AM-GM inequality. 
  
  Let us recall the general form of the Jensen's inequality. 
  Suppose $f$ is a convex function in the range $[a,b]$. 
  For any positive $\alpha,\beta$ such that $\alpha+\beta=1$, we have
  $$\alpha f(a) + \beta f(b) \geq f\left(\alpha a + \beta b\right) $$
  Equality holds if and only if $a=b$. 
  
  Note that if we choose $\alpha=\beta= \frac12$, we get the particular form of Jensen's inequality as used in the previous two problems.  
  
  Let us now recall the Young's inequality. 
  Let $p,q$ be H\"older conjugates, \ie, positive reals numbers such that $\frac1p + \frac1q=1$. 
  For positive $a,b$, we have 
  $$ab \leq \frac{a^p}p + \frac{b^q}q$$
  Equality holds if and only if $a^p = b^q$. 
  
  Prove Young's inequality using the general form of the Jensen's inequality on the function $f(x) = \ln(x)$.   
  
  {\footnotesize
  ({\bfseries Remark.} 
    Note that for $p=q=2$, Young's inequality is identical to the AM-GM inequality.% 
  )}
  
  

 
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%%% Problem 6 
\item {\bfseries H\"older's Inequality.} 
  The goal of the problem is to prove the H\"older's inequality using the (general form) of the Jensen's inequality. 
  
  Let us recall H\"older's inequality. 
  Suppose $p,q$ are H\"older conjugates. 
  For positive $a_1,a_2,b_1,b_2$, we have 
  $$\left(a_1b_1 + a_2b_2\right) \leq \left(a_1^p + a_2^p\right)^{1/p}\left(b_1^q + b_2^q\right)^{1/q}$$
  
  \begin{enumerate}
  \item Consider the following intermediate inequality.
    For positive $A,B$, we have
    $$\left(1 + A^{1/p}B^{1/q}\right) \leq \left(1 + A\right)^{1/p}\left(1+B\right)^{1/q}$$
    Prove the H\"older's inequality using the intermediate inequality. 
  \item Prove the intermediate inequality using the general form of the Jensen's inequality on the function $f(x) = \ln(1 + \exp(x))$. 
  \item What is the characterization of achieving equality in the H\"older's inequality? 
  \end{enumerate}









 
 



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