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Algebra Level 5

Evaluate the real part of the number $\Huge {i^{i^{i^{.^{.^{.}}}}}}$ .

Note: $i$ is the imaginary number $\sqrt{-1}$

The answer is 0.438.

2 solutions

Digvijay Singh
May 3, 2015

Duplicate question .

Pi Han Goh - 6 years, 1 month ago

whats lambert function......... plz reply..

rajat kharbanda - 6 years, 1 month ago

Lambert W function ... basically it's a special function: to find the inverse of $We^W = f(w)$ . That is, if you have an equation $x e^x =12345$ , then the value of $W(12345)$ is simply the root of the previous equation. Simplest way to solve it is by numerical / approximation method.

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – I saw the solution to the above posted duplicate question.. I had made the equations , but he writes using numerical methods we get value of a and b........ plz sorry to disturb you, thanks.......... could you explain

rajat kharbanda - 6 years, 1 month ago

@Rajat Kharbanda – See equation 12 here , set $x = -\frac 12 \pi i$ .

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – gone over my head...... but thanks anyway

rajat kharbanda - 6 years, 1 month ago

@Pi Han Goh – nice explanation sir, but would you please show me a proof that the infinite tetration converge only when $e^{-e} \le x \le e^{\frac{1}{e}}$

Digvijay Singh - 6 years, 1 month ago

@Digvijay Singh – Hint: what's the extreme values of $x^{\frac 1x}$ ?

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – its $e^{1/e}$ , but how is this connected to the convergence?

Digvijay Singh - 6 years, 1 month ago

@Digvijay Singh – See this .

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – thanks , now i get it...

Digvijay Singh - 6 years, 1 month ago

shit...! i did'nt knw... sorry... should i remove this now (the question)?

Digvijay Singh - 6 years, 1 month ago

It's up to you. I would have removed it if I were you.

Pi Han Goh - 6 years, 1 month ago

Let $i^{i^{i^{.^{.^{.}}}}}=x$

$\therefore x=i^x$

Differentiating both sides with respect to $x$ :

$1=i^x \cdot log_e i$

$\implies 1=x \cdot log_ei$

$\implies x=\dfrac{1}{log_ei}=\dfrac{2}{i.\pi}$

So, real part of $x$ is $0$ .

What's wrong with it ? @Digvijay Singh

Sandeep Bhardwaj - 6 years, 1 month ago

Chew-Seong Cheong
May 4, 2015

$i^{i^{i^{...}}} = A+ Bi = i^{A+Bi}\quad \Rightarrow e^{\frac{\pi i}{2}(A+Bi)} = e^{-\frac{\pi B}{2}} e^{\frac{\pi Ai}{2}} = \sqrt{A^2+B^2} e^{\tan^{-1} {\frac {B}{A}}i}$

$\begin{cases} e^{-\frac{\pi B}{2}} = \sqrt{A^2+B^2} & \Rightarrow A^2+B^2 = e^{-\pi B} & & ...(1) \\ \frac{\pi A}{2} = \tan^{-1} {\frac {B}{A}} & \Rightarrow \tan {\frac{\pi A}{2}} = \dfrac {B}{A} & \Rightarrow B = A \tan {\frac{\pi A}{2}} & ...(2) \end{cases}$

Substituting Eqn. 2 in Eqn. 1:

$A^2+A^2 \tan^2 {\frac{\pi A}{2}}= e^{-\pi A \tan {\frac{\pi A}{2}}}$

Using numerical method, we find that $A=\boxed{0.438282937}$

What do you mean by numerical method ?

Pranjal Jain - 6 years, 1 month ago

I used Excel spreadsheet to calculate $f(A) = A^2+A^2 \tan^2 {\frac{\pi A}{2}} - e^{-\pi A \tan {\frac{\pi A}{2}}}$ . I first estimated a value for $A$ , $A_1$ then I checked for $f(A_1^*) \approx 0$ , then estimated a more accurate value for $A$ , $A_2$ and continued to about $A_{10}$ .

Chew-Seong Cheong - 6 years, 1 month ago

I plotted the graph. Just wondering if there's a method without using calculator/computer.

Pranjal Jain - 6 years, 1 month ago

@Pranjal Jain – I plotted the graph too to check the $f(x) \approx 0$ but with a spreadsheet. If the equation can be easily differentiated, we can use Newton's Method $x_{n+1} = a_n - \dfrac{f(x_n)}{f'(x_n)}$ . Numerical method usually needs calculator/computer.

Chew-Seong Cheong - 6 years, 1 month ago

@Chew-Seong Cheong – Sir, while using newtons meathod, What to take $x_{o}$ initially? How to know what to take?

Md Junaid - 3 years, 10 months ago

@Md Junaid – Usually any near estimate will work.

Chew-Seong Cheong - 3 years, 10 months ago

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The answer is 0.438.

2 solutions

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