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I know that there are a few questions on how to solve this particular equation, but I'm not very proficient in maths and would like to follow a method which I had a little bit more of a knowledge of and it did seem simpler and best way to put my working for. I have not really studied it in length mathematically but have seen it applied in computer science.

my working

So above is my working for this method, choosing an initial value of 60,000 as that put me quite close to 0 when I used it as a value of n. My problem is is that I genuinely thought it will definitely lead to me an answer by what I read, yet I can't even get x3 without getting a maths error in my calculator.

What does this mean? Does it mean that I can't use this method to find the solution to this question from maths that I know of and I have to go into teaching myself Lambert W to solve this? Did I start with the wrong x1..? I haven't got the practice in Newton's method before so this is quite new to me.

Lukali
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  • Your title has $\log x$, but your text has $\log_2 x$. Can you resolve this please? – TonyK Jun 19 '21 at 23:00
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    As for your calculator error: you are trying to compute $\log_2 C$ when $C$ is negative. But it looks to me like you have a double negative that you have forgotten to convert to a positive ($60,000-\frac{-47,639}{\cdots}$). – TonyK Jun 19 '21 at 23:02
  • @TonyK Thank you I see it was a simple mistake now earlier than the bit I tried to make sure I input correctly into the calculator many times. – Lukali Jun 19 '21 at 23:10

2 Answers2

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Hint

As already said, you have a mistake in your calculations.

You are tryind to find the zero of function $$f(x)=\frac{x \log (x)}{\log (2)}-10^6$$ $$f'(x)=\frac{\log (x)}{\log (2)}+\frac{1}{\log (2)}$$ $$x_{n+1}=x_n-\frac{x_n \log (x_n)-10^6 \log (2)}{\log (x_n)+1}$$ Simplify this last expression $$x_{n+1}=\frac{x_n+10^6 \log (2)}{\log (x_n)+1}$$ As written, less sources for mistakes.

Claude Leibovici
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$$ f= \dfrac{x \log x }{10^6 \log 2}-1$$

Find derivative and apply the procedure you have given.

Narasimham
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