Numerical Computation Assignment 1

1. Find the binary representation of the base 10 numbers.

• (a) $64={1000000}_2$
• (b) $\frac{31}{64}=(1+2+4+8+16)\times 2^{-6}={0.011111}_2$
• (c) $55.4$
  • Integer part ${55}_{10}={110111}_2$
  • Fractional part ${0.4}_{10}=0.\overline{011}$
    • $0.4\times 2=0.8\to bit0$
    • $0.8\times 2=1.6\to bit1$
    • $0.6\times 2=1.2\to bit1$
    • $0.2\times 2=0.4$ (cycle repeats: $011$)
  • $55.4=110111.\overline{011}$

2. Convert the following binary numbers to base 10.

• (a) ${1010101}_2=2^0+2^2+2^4+2^6=85$
• (b) ${1011.101}_2=2^0+2^1+2^3+2^{-1}+2^{-3}=11.625$
• (c) $1010.\overline{01}$
  • Integer part $1010=2^1+2^3=10$
  • let $x=0.\overline{01}$, then $2^2\times x=1.\overline{01}$, so $2^{2}x-x=1$,$x=\frac{1}{3}$
  • $1010.{\overline{01}}_2=\frac{31}{3}$

3. Explain how to most accurately compute the two roots of the equation $x^2+bx-{10}^{-12}=0$, where $b$ is a number greater than 100.

The standard quadratic formula gives two roots:

$$x=\frac{-b\pm \sqrt{b^2+4\cdot {10}^{-12}}}{2}$$

Calculate the numerically stable root (the one with the large magnitude) using the standard formula. This involves an addition of like-signed numbers, which is stable.

$$x_2=\frac{-b-\sqrt{b^2+4\cdot {10}^{-12}}}{2}\approx -b$$

Use Vieta's formula for the product of roots

$$x_1\cdot x_2=\frac{c}{a}=-{10}^{-12}\$\$\$\$x_1=\frac{{10}^{-12}}{b}$$

4. Evaluate the quantity $x\sqrt{x^2+2}-x^2$ where $x={10}^{12}$, correct to at least 3 decimal places.

$$(x\sqrt{x^2+2}-x^2)\times \frac{x\sqrt{x^2+2}+x^2}{x\sqrt{x^2+2}+x^2}$$

This simplifies to a numerically stable form:

$$\frac{2x^2}{x\sqrt{x^2+2}+x^2}=\frac{2}{\sqrt{1+2/x^2}+1}$$

Now, substitute $x={10}^{12}$:

$$\frac{2}{\sqrt{1+2/{10}^{24}}+1}\approx 1$$

Correct to at least 3 decimal places, the answer is $1.000$.

5. Find the rates of convergence of the following sequences as $n\to \mathrm{\infty }$.

• (a) $\underset{n\to \mathrm{\infty }}{lim}n\sin \frac{1}{n^2}=0$

For $x_n=n\sin \frac{1}{n^2}$

We use the Taylor series approximation $\sin (u)\approx u$ for small $u$. Letting $u=1/n^2$:

$$x_n=n\sin \left(\frac{1}{n^2}\right)\le n\left(\frac{1}{n^2}\right)=\frac{1}{n}$$

The rate of convergence is $O(\frac{1}{n})$.

• (b) $\underset{n\to \mathrm{\infty }}{lim}[\sqrt{n+3}-\sqrt{n}]=0$

For $x_n=\sqrt{n+3}-\sqrt{n}$

We multiply by the conjugate to simplify the expression:

$$x_n=(\sqrt{n+3}-\sqrt{n})\frac{\sqrt{n+3}+\sqrt{n}}{\sqrt{n+3}+\sqrt{n}}=\frac{3}{\sqrt{n+3}+\sqrt{n}}$$

For large $n$, the denominator $\sqrt{n+3}+\sqrt{n}$ is approximately $2\sqrt{n}$. So,$x_n\approx \frac{3}{2\sqrt{n}}$

The rate of convergence is $O(\frac{1}{\sqrt{n}})$

6. Let $f(x)=x^2-3$, find $c$ satisfying the Mean Value Theorem for $f(x)$ on the interval $[1,3]$.

The Mean Value Theorem states that if a function $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists at least one number $c$ in $(a,b)$ such that:

$$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$$

Here, we are given:

• The function $f(x)=x^2-3$

• The interval $[a,b]=[1,3]$

• $f(a)=f(1)=(1{)}^2-3=-2$

• $f(b)=f(3)=(3{)}^2-3=6$

Now, plug these values into the formula:

$$\frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}=\frac{6-(-2)}{2}=\frac{8}{2}=4$$

• $f^{\prime }(x)=2x$

Now, evaluate the derivative at $x=c$:

• $f^{\prime }(c)=2c$

• $c=2$

The value $c=2$ is indeed between 1 and 3.

Thus, the value of $c$ that satisfies the Mean Value Theorem for $f(x)=x^2-3$ on the interval $[1,3]$ is $2$.

7. Let $f(x)=x$, $g(x)=e^x$, find $c$ satisfying the Mean Value Theorem for integrals with $f(x)$, $g(x)$ in the interval $[0,1]$.

To find $c$ satisfying the Mean Value Theorem for integrals with $f(x)=x$ and $g(x)=e^x$ on the interval $[0,1]$, we use the Weighted Mean Value Theorem for Integrals. This theorem states that if $f$ and $g$ are continuous on $[a,b]$ and $g(x)$ does not change sign on $[a,b]$, there exists a $c\in [a,b]$ such that:

$${\int }_a^{b}f(x)g(x)dx=f(c){\int }_a^{b}g(x)dx$$

For $f(x)=x$, $g(x)=e^x$, and interval $[0,1]$, the theorem gives:

$${\int }_0^{1}x{e}^{x}dx=c{\int }_0^1{e}^{x}dx$$

The integral of $e^x$ is $e^x$. Evaluating from 0 to 1:

$${\int }_0^1{e}^{x}dx=e^x{|}_0^1=e^1-e^0=e-1$$

Use integration by parts with $u=x$, $dv=e^{x}dx$. Then $du=dx$, $v=e^x$. By integration by parts:

$$\int x{e}^{x}dx=uv-\int vdu=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C$$

Evaluating from 0 to 1:

$${\int }_0^{1}x{e}^{x}dx=(x{e}^x-e^x){|}_0^1=((1\cdot e^1-e^1)-(0\cdot e^0-e^0))=(0-(-1))=1$$

Substitute the integrals into the theorem:

$$1=c(e-1)$$

Solving for $c$:

$$c=\frac{1}{e-1}$$

8. Find the Taylor series generated by $f(x)=\frac{1}{x}$ about $x=2$.

Of course. We can find the Taylor series for $f(x)=\frac{1}{x}$ about $x=2$ using the definition of a Taylor series.

The formula for a Taylor series generated by a function $f(x)$ about $x=a$ is:

$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(a)}{n!}(x-a{)}^n$

In our case, $f(x)=\frac{1}{x}$ and the center is $a=2$.

We need to find a pattern for the nth derivative, $f^{(n)}(x)$.

• $f(x)=x^{-1}$

• $f^{\prime }(x)=-1\cdot x^{-2}$

• $f^{″}(x)=(-1)(-2)\cdot x^{-3}=2!\cdot x^{-3}$

• $f^{‴}(x)=(2!)(-3)\cdot x^{-4}=-3!\cdot x^{-4}$

• $f^{(4)}(x)=(-3!)(-4)\cdot x^{-5}=4!\cdot x^{-5}$

Observing the pattern, the nth derivative is:

$f^{(n)}(x)=(-1{)}^{n}n!\cdot x^{-(n+1)}=\frac{(-1{)}^{n}n!}{x^{n+1}}$

Substitute $x=2$ into the general formula for the nth derivative.

• $f(2)=\frac{1}{2}$

• $f^{\prime }(2)=-\frac{1}{2^2}=-\frac{1}{4}$

• $f^{″}(2)=\frac{2!}{2^3}=\frac{2}{8}=\frac{1}{4}$

• $f^{‴}(2)=-\frac{3!}{2^4}=-\frac{6}{16}=-\frac{3}{8}$

The general term for the nth derivative evaluated at $a=2$ is:

$$f^{(n)}(2)=\frac{(-1{)}^{n}n!}{2^{n+1}}$$

So the series is:

$\frac{1}{x}=\frac{1}{2}-\frac{1}{4}(x-2)+\frac{1}{8}(x-2{)}^2-\frac{1}{16}(x-2{)}^3+\dots$

9. Find the degree 4 Taylor polynomial $P_4(x)$ for $f(x)=\sin x$ centered at the point $x=0$. Estimate the maximum possible error when using $P_4(x)$ to estimate $\sin x$ for $|x|\le 0.0001$.

The formula for the n-th degree Taylor polynomial of a function $f(x)$ centered at $x=a$ is:

$P_n(x)=f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(n)}(a)}{n!}(x-a{)}^n$

• $f(x)=\sin x⟹f(0)=\sin (0)=0$

• $f^{\prime }(x)=\cos x⟹f^{\prime }(0)=\cos (0)=1$

• $f^{″}(x)=-\sin x⟹f^{″}(0)=-\sin (0)=0$

• $f^{‴}(x)=-\cos x⟹f^{‴}(0)=-\cos (0)=-1$

• $f^{(4)}(x)=\sin x⟹f^{(4)}(0)=\sin (0)=0$

Now, substitute these values into the Taylor polynomial formula:

$$P_4(x)=f(0)+\frac{f^{\prime }(0)}{1!}x+\frac{f^{″}(0)}{2!}{x}^2+\frac{f^{‴}(0)}{3!}{x}^3+\frac{f^{(4)}(0)}{4!}{x}^4$$$$=0+\frac{1}{1!}x+\frac{0}{2!}{x}^2+\frac{-1}{3!}{x}^3+\frac{0}{4!}{x}^4$$$$P_4(x)=x-\frac{1}{6}{x}^3$$

The degree 4 Taylor polynomial for $f(x)=\sin x$ centered at $x=0$ is:

$P_4(x)=x-\frac{x^3}{6}$

The formula for the Lagrange Remainder is:

$R_n(x)=\frac{f^{(n+1)}(\xi (x))}{(n+1)!}(x-x_0{)}^{n+1}$

where $\xi (x)$ is a value between the center $x_0$ and the point $x$.

In our problem, the parameters are:

• Function: $f(x)=\sin x$

• Polynomial degree: $n=4$

• Center: $x_0=0$

• Interval for estimation: $|x|\le 0.0001$

We are finding the remainder for the degree 4 polynomial, $P_4(x)$, which is denoted as $R_4(x)$. We substitute $n=4$ and $x_0=0$ into the formula:

$R_4(x)=\frac{f^{(4+1)}(\xi (x))}{(4+1)!}(x-0{)}^{4+1}=\frac{f^{(5)}(\xi (x))}{5!}{x}^5$

We need to calculate the 5th derivative of $f(x)=\sin x$, which is $f^{(5)}(x)$.

• $f^{\prime }(x)=\cos x$

• $f^{″}(x)=-\sin x$

• $f^{‴}(x)=-\cos x$

• $f^{(4)}(x)=\sin x$

• $f^{(5)}(x)=\cos x$

Now, substitute this derivative back into the expression for the remainder:

$R_4(x)=\frac{\cos (\xi (x))}{5!}{x}^5$

The error is the absolute value of the remainder, $|R_4(x)|$. We need to find its maximum possible value on the interval $|x|\le 0.0001$.

$|R_4(x)|=\left|\frac{\cos (\xi (x))}{5!}{x}^5\right|=\frac{|\cos (\xi (x))|\cdot |x{|}^5}{5!}$

To maximize this expression, we find the maximum value of each component in the numerator:

1. Bound for $|\cos (\xi (x))|$:

By definition, $\xi (x)$ is between $x_0=0$ and $x$. Since $|x|\le 0.0001$, $\xi (x)$ must be within the interval $[-0.0001,0.0001]$. We know that for any real number $z$, the maximum value of $|\cos (z)|$ is 1. Therefore, we can use 1 as an upper bound for this term.

$max|\cos (\xi (x))|\le 1$

2. Bound for $|x{|}^5$:

On the given interval $|x|\le 0.0001$, the maximum value of $|x{|}^5$ occurs at the endpoints, $x=\pm 0.0001$.

$max|x{|}^5=(0.0001{)}^5=({10}^{-4}{)}^5={10}^{-20}$

3. $(n+1)!=5\times 4\times 3\times 2\times 1=120$

Maximum Error $\le \frac{max|\cos (\xi (x))|\cdot max|x{|}^5}{5!}$

Maximum Error $\le \frac{1\cdot {10}^{-20}}{120}$

The maximum possible error when using $P_4(x)$ to estimate $\sin x$ for $|x|\le 0.0001$ is $\frac{{10}^{-20}}{120}$.