Math, asked by BrainlyProgrammer, 4 hours ago

[Maths]
Question:- Find the value of infinite tatration.
\large{ \orange{{\sqrt[x]{x}}^{\sqrt[x]{x}^{\sqrt[x]{x}^{...}}}}}
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Answers

Answered by Anonymous
11

Pre-requisite Knowledge:

Lambert W function

In order to simplify the given tetration, we will be using a special type of function, know as Lambert W function which is also called productlog function.

Lambert W function is defined as W(x) = f^{-1}(x) where  f(x) = x\cdot e^x.

And in general, if we have any equation of the type  x \cdot e^x = y \implies x = W(y) .

Logarithmic properties

We will use the following logarithmic properties to solve this question.

  •  \ln(x) = \log_e(x)
  •  \log(x\cdot y) = \log(x) + \log(y)
  •  x = y \implies e^x = e^y
  •  \log (a^b) = b \log (a)
  •  b^{[\log_b(a)]} = a

 \rule{280}{1}

Evaluation:

Let the given tetration be equal to some variable (let's say y) and we will solve for y.

y =   { \sqrt[x]{x} }^{ { \sqrt[x]{x} }^{ { \sqrt[x]{x} }^{...} } }

 \implies y =   { \sqrt[x]{x} }^{y}

 \implies y =   {x}^{ \frac{y}{x} }

Taking natural log both sides.

 \implies  \ln(y) =   \ln( {x}^{ \frac{y}{x} } )

 \implies  \ln(y) =   \dfrac{y}{x} \cdot\ln(x)

Multiply both sides with -1.

 \implies   - \ln(y) =    - \dfrac{y}{x} \cdot\ln(x)

 \implies    \ln(y^{ - 1} ) =    - \dfrac{y}{x} \cdot\ln(x)

 \implies    \ln \left( \dfrac{1}{y}  \right) =    - \dfrac{y}{x} \cdot\ln(x)

Exponentiating both sides.

 \implies   e^{ \ln ( \frac{1}{y} )} =e^{    - \frac{y}{x} \cdot\ln(x)}

 \implies    \dfrac{1}{y} =e^{    - \frac{y}{x} \cdot\ln(x)}

Multiply both sides with  \dfrac{-y}{x} \cdot \ln(x) .

{ \implies    \dfrac{1}{ \not y}  \times  \left (\dfrac{- \not y}{x} \cdot \ln(x)\right)=e^{    - \frac{y}{x} \cdot\ln(x)} \times  \left(\dfrac{-y}{x} \cdot \ln(x) \right)}

{ \implies    -  \dfrac{ \ln(x)}{x}  =\left(\dfrac{-y}{x} \cdot \ln(x) \right) \cdot e^{   ( - \frac{y}{x} \cdot\ln(x))}}

Now, we can see that this is forming general form of Lambert W function.

{ \implies   W\left( -  \dfrac{ \ln(x)}{x}  \right)=\dfrac{-y}{x} \cdot \ln(x)}

{ \implies    \dfrac{ - x \times W\left( -  \frac{ \ln(x)}{x}  \right)}{ \ln(x)}=y}

Therefore, our required result is:

 \underline{ \boxed{{ \sqrt[x]{x} }^{ { \sqrt[x]{x} }^{ { \sqrt[x]{x} }^{...} } }  = \dfrac{ - x \times W\left( -  \frac{ \ln(x)}{x}  \right)}{ \ln(x)}}}

 \rule{280}{1}

Bonus question:

Let's solve another interesting question involving Lambert W function.

Question:

Solve for x in the equation  x^x = 2 .

Evaluation:

 {x}^{x}  = 2

Taking natural log both sides.

 \implies \ln({x}^{x} ) =  \ln(2)

 \implies x\ln({x}) =  \ln(2)

 \implies  {e}^{ \ln(x)} \ln({x}) =  \ln(2)

Now this is forming usual definition of Lambert W function, so let's introduce it!

 \implies \ln({x}) =  W(\ln(2))

Exponentiating both sides.

 \implies e^{\ln({x})} =  e^{W(\ln(2))}

 \implies x =  e^{W(\ln(2))}

This is the value of  x satisfying the given equation.

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