Solving Problems Using Iterative Process

The equation $x^4-1-x=0$ has a root, $\alpha ,$ between $x=1$ and $x=2.$

Use your iterative formula, with a starting value of $x_1=1.5,$ to find $\alpha$ correct to $2$ decimal places. Give the result of each iteration to $4$ decimal places.

The terms of a sequence, defined by the iterative formula $x_{n+1}=\ln \left(x_n^2+4\right),$ converge to the value $\alpha .$ The first term of the sequence is $2.$

Find the value of $\alpha$ correct to $2$ decimal places. Give each term of the sequence you find to $4$ decimal places.

Define the Fibonacci numbers. Calculate the value of the ${12}^{\mathrm{th}}$ and the ${13}^{\mathrm{th}}$ Fibonacci numbers. The $9^{\mathrm{th}}$ and ${10}^{\mathrm{th}}$ terms in the sequence are $21$ and $34$.

Define the golden ratio with example.

The equation $x^2-3x+1=0$ has a positive root $\alpha$. Show that this equation can be rearranged as $x=3-\frac{1}{x}$

Use the iterative formula $x_{n+1}=3-\frac{1}{x_n}$ with a starting value of $x_1=3$ to find $\alpha$ correct to $3$ decimal places. Give the result of each iteration to $3$ decimal places, where appropriate.

The equation $x^2+x-3=0$ has a negative root $\alpha$. Show that this equation can be rearranged as $x=\frac{3}{x}-1$

Use the iterative formula $x_{n+1}=\frac{3}{x_n}-1$ with a starting value of $x_1=-2.5$ to find $\alpha$ correct to $2$ decimal places. Give the result of each iteration to $4$ decimal places, where appropriate.

The equation $x^2-x-1=0$ has a root positive root $\alpha$. Show that the equation can be rearranged as $x=\frac{1}{x}+1$.

Use the iterative formula $x_{n+1}=\frac{1}{x_n}+1$ with a starting value of $x_1=2$ to find $\alpha$ correct to $3$ decimal places. Give the result of each iteration to $4$ decimal places, where appropriate.

Use the iterative formula $x_{n+1}=\frac{1}{x_n}+1$ with a starting value of $x_1=2$ to find $\alpha$ correct to $3$ decimal places. Give the result of each iteration to $4$ decimal places, where appropriate.

Use the iterative formula $x_{n+1}=\frac{3}{x_n}-1$ with a starting value of $x_1=-2.5$ to find $\alpha$ correct to $2$ decimal places. Give the result of each iteration to $4$ decimal places, where appropriate.

The equation $x^3=8-\frac{x}{2}$ has one real root, denoted by $\alpha .$

Find, by calculation, the pair of consecutive integers between which $\alpha$ lies.

Find the value of $t$ correct to $3$ decimal places using an iterative process based on the equation $t=\sqrt[3]{\frac{3t^2+5}{4}}$. Given that the value of $t$ lies between $1$, and $2$.

The equation $x^3-7x^2+1=0$ has two positive roots, $\alpha$ and $\beta$, which are such that $\alpha$ lies between $0$ and $1$and $\beta$ lies between $6$ and $7 .$

By using an iterative formula $x_{n+1}=\sqrt[3]{7x_n^2+1}$, carry out suitable iterations to find the value of $\beta$ correct to $2$ decimal places.

The equation $x^3-7x^2+1=0$ has two positive roots, $\alpha$ and $\beta$, which are such that $\alpha$ lies between $0$ and $1$and $\beta$ lies between $6$ and $7 .$

By using an iterative formula $x_{n+1}=\sqrt{\frac{x_n^3+1}{7}}$, carry out suitable iterations to find the value of $\alpha$ correct to $2$ decimal places. Give the value of each of your iterations to $4$ decimal places.    [Write the value of $\alpha$ in the answer box].

The equation $\frac{\sin (x-1)}{2x-3}+1=0$ has a root, $\alpha ,$ between $x=1$ and $x=1.4$.

Using an iterative formula based on the equation $x=\frac{3-\sin (x-1)}{2}$ with a suitable starting value, find the value of $\alpha$ correct to $3$ decimal places.

The equation $\frac{\sin (x-1)}{2x-3}+1=0$ has a root, $\alpha ,$ between $x=1$ and $x=1.4$.

The equation $x^4-1-x=0$ has a root, $\alpha ,$ between $x=1$ and $x=2.$ Use your iterative formula, with a starting value of $x_1=1.5,$ to find $\alpha$ correct to $2$ decimal places.

The equation $x^4-1-x=0$ has a root, $\alpha ,$ between $x=1$ and $x=2.$

The equation $x^3+5x-7=0$ has a root $\alpha$ between $1$and $1.2$.

The equation $\ln \left(x+1\right)=-2x+4$ has a root $\alpha$ between $x=1$ and $x=2.$

If the equation $ln\left(x+1\right)+2x-4=0$ has a root $\alpha$ between $x=1$ and $x=2$ then using an iterative formula $x_{n+1}=\frac{4-ln\left(x_n+1\right)}{2}$ with an initial value of $x_1=1.5$, find the value of $\alpha$ correct to$3$ decimal places.