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can you please solve this equation with steps

e^(1/x)-x=0

 Jan 20, 2015

Best Answer 

 #3
avatar+99272 
+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!)

 

$$\\\mbox{Consider the red tangent. }\\
$The gradient of this line is f'(x_0)$\\
BUT \mbox{ The tangent is also } \\\\
=rise/run\\\\
=(f(x_0)-0)/(x_0-x_1)\\\\
=f(x_0)/(x_0-x_1)\\\\\\\\
f'(x_0)=f(x_0)/(x_0-x_1)\\\\
f'(x_0)(x_0-x_1)=f(x_0)\\\\
(x_0-x_1)=f(x_0)/f'(x)\\\\
-x_1=[f(x_0)/f'(x_0)]-x_0\\\\
x_1=x_0-[f(x_0)/f'(x)]\\\\$$

.
 Jan 21, 2015
 #1
avatar+21842 
+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:  

$$\small{\text{
$x_0 = 1$
}} \\
\small{\text{
$ x_1 = 1+\frac{e-1}{e+1}= 1.46211715726
$
}} \\
\small{\text{
$
x_2= 1.46211715726 + \frac{ 0.51955244051 }{1.92697260563 }= 1.73173824009
$
}} \\
\small{\text{
$
x_3= 1.73173824009 + \frac{ 0.04975957061}{1.59404698875}= 1.76295411450
$
}} \\
\small{\text{
$
x_4= 1.76295411450+ \frac{ 0.00042115233 }{1.56736524331}= 1.76322281532
$
}} \\
\small{\text{
$
x_5= 1.76322281532+ \frac{ 0.00000002982}{1.56714330612}= 1.76322283435
$
}} \\
\small{\text{
$
x_6= 1.76322283435+ \frac{ 0.000000000000000149544303}{1.56714329041}= 1.76322283435
$
}}$$

$$x\approx 1.76322283435$$

.
 Jan 20, 2015
 #2
avatar+98091 
+8

Thanks, heureka.....

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

 

 Jan 20, 2015
 #3
avatar+99272 
+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
 #4
avatar+98091 
+5

I'm afraid I have to run for the references whenever I have to use this too, Melody.....

It IS a pretty neat how it wotks, though.....and the concept is simple.....!!!

 

 Jan 21, 2015
 #5
avatar+99272 
+5

I do not run to references Chris.

I just work it out myself.  It is easy after you  have done it a few times.

 Jan 21, 2015

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