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# can you please solve this equation with steps e^(1/x)-x=0

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558
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can you please solve this equation with steps

e^(1/x)-x=0

Jan 20, 2015

### Best Answer

#3
+101769
+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!)

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

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Jan 21, 2015

### 5+0 Answers

#1
+22358
+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:

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

$$x\approx 1.76322283435$$

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Jan 20, 2015
#2
+101424
+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
+101769
+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
+101424
+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
+101769
+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