HomeWork ONE

1. 1.3

   a $\epsilon_{\rm machine}\times \beta^{E_{min}}$, where machine epsilon $\epsilon_{\rm machine}=1/beta^{p-1}$, $\beta$ is the base of the system and $E_{min}$ is lowest possible value of exponent.

   b $\epsilon_{\rm machine}\times \beta^{E_{max}}$ with $E_{max}$ being maximum value of exponent,

   c

   $$2^{53-24}-1=536870911.$$

2. 1.8

   a

   $$\sum\limits_{k=0}^{1000}\frac{e^k}{e^k+1} = \sum\limits_{k=0}^{1000}\frac{1}{e^{-k+1}} =1000.03583648423874/$$
   Here we divided numerator and denominator by $e^k$ to avoid overflow.

   b

   $$\sum\limits_{k=0}^{1000}\frac{\cosh k}{1+\sinh k} = \sum\limits_{k=0}^{1000}\frac{1+e^{-2k}}{1+2 e^{-k}-e^{-2 k}} = 1000.38056571994274.$$

   Here we divided numerator and denominator by $e^k$ to avoid overflow.

   c

   $$\sum\limits_{k=0}^{1000}\sqrt{3+e^k} - \sum\limits_{n=0}^{1000}\... ...m\limits_{k=0}^{1000}\frac{2}{\sqrt{3+e^k}+\sqrt{1+e^k}}= 1.92929676762289274.$$

   d

   $$\frac{\sum\limits_{k=0}^{1000}e^k}{\sum\limits_{n=0}^{1000} n e^n... ...00}e^{k-1000}}{\sum\limits_{n=0}^{1000} n e^{n-1000}} = 0.00100058231560098515.$$

   Here we divided numerator and denominator by $e^{1000}$ to avoid overflow.

   e

   Calculate

   $$X=\sum\limits_{k=1}^{1000}k\left[ \sin\left(\pi(k^{10}+1/k)\right)- \sin\left(\pi(k^{10}-1/k)\right) \right].$$

   Since $\sin(\pi+x) = -\sin(x)$, and $\sin[\pi-x]=\sin[x]$, and $\sin[2 \pi+x] = \sin[x]$, and using the fact that even number raised to the power of 10 is even, and odd number raised to teh power of 10 is odd, we obtain

   $$X=2 \sum\limits_{k=1}^{1000}(-1)^k k \sin\left(\frac{\pi}{k}\right) =4.70173884586449512.$$

3. 1.9 Homer states
   $$3987^{12}+4365^{12}=4472^{12}.$$
   Is it true?

   a If Horner right,

   $$X_1= 3987^{12}+4365^{12}-4472^{12}$$
   should be 0, yet Matlab returns $1.2119e+33$

   b If Horner is right,

   $$X_2=\left(3987^{12}+4365^{12}\right)^{\frac{1}{12}}-4472$$
   should be zero, yet Matlab returns $7.0577e-09$.

   c If Horner is right,

   $$X_3=\frac{\left(3987^{12}+4365^{12}\right)}{4472^{12}}$$
   should be $1$, yet Matlab returns $1.000000000018943$

   d If Horner is right,

   $$X_4=\left[\left(3987^{12}+4365^{12}\right) ^{\frac{1}{12}}\right]^{12} - 4472$$
   should be zero, yet Matlab returns $1.211616482290016e+33$

   e The product of two $p$ digit numbers may have up to $2 p$ digits. Therefore $3987^{12}$ may have up to 48 digits, which is longer then Matlab's 16 digits. Therefore we can not use doubles in Matlab to either prove or disprove this conjecture.

   Note that $X_1$ can be explicitly calculated on a computer with inifinite precision (check out vpa command in matlab), because it is an integer. Using the “sym” command, or Mathematica allows us to get the answer:

   $$X_1=1211886809373872630985912112862690.$$

4. 1.11 Here we deal with
   $$f(x)=\frac{e^x-1}{x},$$
   for small values of $x$. Expanding $e^x\simeq 1+x,$ we obtain
   $$f(x)\simeq \frac{(1+x)-1}{x}.$$
   This is very close to the example we studied in class, so all information can be taken directly from lecture notes
5. 1.13 Here we study for small $x$ the function
   $$f(x)=\frac{\ln(1-x)}{x}.$$
   Taking into account that
   $$\ln(1-x)\simeq -x - x^2/2 - x^3/3-x^4/4,$$
   the numerator should be equal to $x$ for small $x$. Yet, the numerator is computed by first $(1-x)$ and then taking the logarithm. Therefore for small values of $x$ the numerator is approximately given by
   $$(1-x)-1$$
   we therefore may use information from lectures to characterize behaviour of this function.

6. 1.17 a

   Use Taylor’s theorem to show that

   $$\vert { \rm sin} x-{\rm sin} x_f \vert \le \vert x-x_f \vert$$ (31)

   Denoting $x=x_f+\delta$ and using
   $$\sin(x=x_f+\delta) \simeq \sin(x)+\delta\cos(x_f),$$
   we obtain
   $$\sin(x)-\sin(x_f)\simeq \delta\cos(x_f).$$
   Since $\cos(x_f)\le 1$, we obtain (32).

   b Why

   $$\vert x-x_f\vert\le \vert\tilde{ x_f} - \tilde{ \tilde{ x_f}}\vert$$ (32)

   This is due to “rounding to nearest”, $x$ is rounded to $x_f$, and $x_f$ is either $\tilde{ x_f}$ or $\tilde{ \tilde{ x_f}}$. In other words, the distance between a number and its floating point representation is less than half a distance between nearest two floaing point numbers.

   c

   Since $2^E<x<2^{E+1}$ and $2^E<x_f<2^{E+1}$, we get $\vert x-x_f\vert<2^E$. More precisely, $x-x_f\le \epsilon e^E$. Use this together with (32) and (33) to get

   $$\vert\sin x-\sin x_f\vert\le \epsilon 2^{E-1}$$ (33)

   d This question implies that
   

   $$x<L\le 2^{E+1}.$$ (34)

   
   If this property is not satisfied, then the result can not be obtained.

   To obtain this result, write

   $$\epsilon 2^{E-1}=\frac{1}{4}\epsilon 2^{E+1},$$
   then use (35).

   Author's note: in (d) the right hand side shoule be $eps*L/2$ and in (e) the right hand side should be $eps*(L+2)/2$.

   e The RHS can be rewritten as

   $$\frac{1}{4}\epsilon(L+4) = \frac{\epsilon L}{4}+\epsilon.$$
   Now use $\vert sin x_f-s_f\vert\le \epsilon$ to obtain the desired result.

   f Using $\epsilon=10^{-16}$ we get $L=4E8$. Picture shows that $10^{-8}$ accuracy is obtained at $k=25$, i.e. $L\simeq 3E7$. In other words the estimate is optimistic.