# Numerical method,finding root in iteration method.

• Mar 6th 2010, 02:52 AM
avengerevenge
Numerical method,finding root in iteration method.
hello i have problem with solution. please clear my doubt.

my doubt is how did the writer get x0 = 3.8
how did he get x0 value.
• Mar 6th 2010, 03:21 AM
Quote:

Originally Posted by avengerevenge
hello i have problem with solution. please clear my doubt.

my doubt is how did the writer get x0 = 3.8
how did he get x0 value.

Hi avengerevenge,

Maybe from graphing

$f(x)=2x-log_{10}x-7=0$

Or..

$10^{2x-log_{10}x}=10^7$

$\frac{10^{2x}}{10^{log_{10}x}}>10^7$

$\frac{10^{2x}}{x}>10^7$

$10^{2x}>x10^7$

$x=1,\ 10^2<10^7$

$x=2,\ 10^4<(2)10^7$

$x=3,\ 10^6<(3)10^7$

$x=4,\ 10^8>(4)10^7$

$3
• Mar 6th 2010, 03:37 AM
HallsofIvy
"Find real root of $2x- log_{10}(x)= 7$ by iteration method

The answer is $x= \frac{1}{2}(log_{10}(x)+ 7)$

$x_0= 3.8$
$x_1= \frac{1}{2}(log_{10}(x)+ 7)= 3.7893$
$x_2= \frac{1}{2}(log_{10}(x)+ 7)= 3.7893$

How did he get $x_0= 3.8$? Pretty much an "educated" guess. Under the right conditions, if $x_0$ is any number reasonably close to the answer to begin with, this sequence will converge to the solution to the equation. I suspect that this author already knew the answer and chose 3.8 because it was already correct to the first decimal place.

If I were coming to this equation without knowing the answer I might argue "log(x) is always smaller than x so what happens if I ignore it? The equation becomes 2x= 7 and x= 7/2= 3.5. I think I will try $x_0= 3.5$." Then I would get:

$x_0= 3.5$

$x_1= \frac{1}{2}(log_{10}(3.5)+ 7)= 3.7720$

$x_2= \frac{1}{2}(log_{10}(3.7720)+ 7)= 3.7883$

$x_3= \frac{1}{2}(log_{10}(3.7883)+ 7)= 3.7892$

$x_4= \frac{1}{2}(log_{10}(3.7892)+ 7)= 3.7893$

$x_5= \frac{1}{2}(log_{10}(3.7893)+ 7)= 3.7893$

Since those last two iterations are the same to the four decimal places I am keeping, I can stop here, confident that my answer is correct at least to three decimal places. It took a couple more iterations because my first value was not as close to the correct answer as 3.8 but I still get the same answer.

Notice that if I had started with $x_0= 1$, my first step would have been
$x_1= \frac{1}{2}(log_{10}(1)+ 7)= 7/2= 3.5$

And then the iteration would be the same as above.

If I had started with something really bad, say $x_0= 100$, then I would have

$x_1= \frac{1}{2}(log_{10}(100)+ 7)= 27/2= 13.5$
$x_2= \frac{1}{2}(log_{10}(13.5)+ 7)= 4.0652$
$x_3= \frac{1}{2}(log_{10}(4.0652)+ 7)= 3.8045$
$x_4= \frac{1}{2}(log_{10}(3.8045)+ 7)= 3.7901$

and I am heading right back to 3.7893.
• Mar 6th 2010, 03:40 AM
HallsofIvy
Quote:

Hi avengerevenge,

Maybe from graphing

$f(x)=2x-log_{10}x-7=0$

Or..

$10^{2x}-x=10^7$

Hopefully, not! $10^{2x- log_{10}x}= \frac{10^{2x}}{x}$, not $10^{2x}- x$.

Quote:

$10^{2x}>10^7$

$2x>7\ \Rightarrow\ x>3.5$
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