Consider the function f(x) = x- 4 In x
Show that the equation f(x) = 0 has a solution in the interval (1,2)
Show that for this function f the Newton - raphson method formula can be expressed as xn +1 = 4x n (1-4 In x)/4 - x n (n = 0,1,2....).
Consider the function f(x) = x- 4 In x
Show that the equation f(x) = 0 has a solution in the interval (1,2)
Show that for this function f the Newton - raphson method formula can be expressed as xn +1 = 4x n (1-4 In x)/4 - x n (n = 0,1,2....).
Hello, fair_lady0072002!
I'll do the second part . . . it's trickier.
And I believe there's an extra "4" is the answer . . .
Consider the function: $\displaystyle f(x) \:= \:x- 4\ln x$
(b) Show that for this function $\displaystyle f$, the Newton- Raphson method formula
can be expressed as: .$\displaystyle x_{n +1} \;= \;\frac{4x_n(1 - 4\ln x_n)}{4 - x_n}$? for $\displaystyle n = 0,\,1,\,2\,\hdots$
Formula: .$\displaystyle x_{n+1}\;=\;x_n - \frac{f(x_n)}{f'(x_n)} $
I'm going to drop the subscripts (temporarily) . . .
We have: .$\displaystyle f'(x) \:=\:1 - \frac{4}{x}$
Then: .$\displaystyle x_{n+1}\;=\;x - \frac{x - 4\ln x}{1 - \frac{4}{x}} $
Multiply top and bottom of the fraction by $\displaystyle x:\;\;x_{n+1}\;=\;x - \frac{x^2 - 4x\ln x}{x - 4}$
Get a common denominator: .$\displaystyle x_{n+1} \;= \;\frac{x}{1}\cdot\frac{x-4}{x-4} - \frac{x^2 - 4x\ln x}{x - 4}$
$\displaystyle x_{x+1}\;= \;\frac{x^2 - 4x - (x^2 - 4x\ln x)}{x - 4} \;= \;\frac{x^2 - 4x - x^2 + 4x\ln x}{x - 4} \;=$ $\displaystyle \frac{-4x + 4x\ln x}{x - 4} $
Factor: .$\displaystyle x_{n+1}\;=\;\frac{-4x(1 - \ln x)}{-(4 - x)}\quad\Rightarrow\quad \boxed{x_{n+1} \;= \;\frac{4x_n(1 - \ln x_n)}{4 - x_n}} $
Hello again, fair_lady0072002!
I hope this suffices as a solution . . .
Consider the function: $\displaystyle f(x) \:= \:x- 4\ln x$
Show that the equation $\displaystyle f(x) = 0$ has a solution in the interval (1,2)
It can shown that $\displaystyle f(x)$ is continuous for all $\displaystyle x > 0.$
We find that: .$\displaystyle f(1) \;= \;1 - 4\ln1 \;= \;1$
. . . . . . and: .$\displaystyle f(2) \;= \;2 - 4\ln 2 \;\approx \;\text{-}0.773$
Since the continuous function changes from positive to negative over the interval,
. . it must achieve $\displaystyle f(x) = 0$ somewhere on the interval.