#3**+10 **

This is yet another formula that I have not memorised.

I work it out every time.

(This is not presenting properly as fractions because LaTex is playing up!)

$$\\\mbox{Consider the red tangent. }\\

$The gradient of this line is f'(x_0)$\\

BUT \mbox{ The tangent is also } \\\\

=rise/run\\\\

=(f(x_0)-0)/(x_0-x_1)\\\\

=f(x_0)/(x_0-x_1)\\\\\\\\

f'(x_0)=f(x_0)/(x_0-x_1)\\\\

f'(x_0)(x_0-x_1)=f(x_0)\\\\

(x_0-x_1)=f(x_0)/f'(x)\\\\

-x_1=[f(x_0)/f'(x_0)]-x_0\\\\

x_1=x_0-[f(x_0)/f'(x)]\\\\$$

Melody
Jan 21, 2015

#1**+10 **

**e^(1/x)-x=0**

Root-Finding Algorithm with Newton's method: $$\small{\text{

$

\boxed{ x_{i+1} = x_i - \dfrac{f(x_i)}{f'(x_i)} } \quad f{(x)} = e^{\frac{1}{x}}-x $ and $ f'(x) = -\frac{ e^{\frac{1}{x}}}{x^2}-1 $ we have $ x_{i+1} = x_i + \dfrac{ e^{\dfrac{1}{x_i}}-x_i }

{\frac{ e^{ \left(\dfrac{1}{x_i} \right) } } {x_i^2}+1}

$

}$$

We start with:

$$\small{\text{

$x_0 = 1$

}} \\

\small{\text{

$ x_1 = 1+\frac{e-1}{e+1}= 1.46211715726

$

}} \\

\small{\text{

$

x_2= 1.46211715726 + \frac{ 0.51955244051 }{1.92697260563 }= 1.73173824009

$

}} \\

\small{\text{

$

x_3= 1.73173824009 + \frac{ 0.04975957061}{1.59404698875}= 1.76295411450

$

}} \\

\small{\text{

$

x_4= 1.76295411450+ \frac{ 0.00042115233 }{1.56736524331}= 1.76322281532

$

}} \\

\small{\text{

$

x_5= 1.76322281532+ \frac{ 0.00000002982}{1.56714330612}= 1.76322283435

$

}} \\

\small{\text{

$

x_6= 1.76322283435+ \frac{ 0.000000000000000149544303}{1.56714329041}= 1.76322283435

$

}}$$

$$x\approx 1.76322283435$$

heureka
Jan 20, 2015

#2**+8 **

Thanks, heureka.....

Notice how fast this coverges by using Newton's Method......the 4th and 5th terms agree to seven decimal places ....!!!!

CPhill
Jan 20, 2015

#3**+10 **

Best Answer

This is yet another formula that I have not memorised.

I work it out every time.

(This is not presenting properly as fractions because LaTex is playing up!)

$$\\\mbox{Consider the red tangent. }\\

$The gradient of this line is f'(x_0)$\\

BUT \mbox{ The tangent is also } \\\\

=rise/run\\\\

=(f(x_0)-0)/(x_0-x_1)\\\\

=f(x_0)/(x_0-x_1)\\\\\\\\

f'(x_0)=f(x_0)/(x_0-x_1)\\\\

f'(x_0)(x_0-x_1)=f(x_0)\\\\

(x_0-x_1)=f(x_0)/f'(x)\\\\

-x_1=[f(x_0)/f'(x_0)]-x_0\\\\

x_1=x_0-[f(x_0)/f'(x)]\\\\$$

Melody
Jan 21, 2015