then if we go to $\tan^{-1} 1000$ , we get 89.9427

for $\tan^{-1} 10000$ it is 89.99427

$\tan^{-1} 100000$ is 89.999427.

So we have a pattern here using which we can predict the value of $\tan^{-1} 10^{n}$

for every zero after 100 , add a 9 after the decimal and let the 427 term remain same.

If anyone has an explanation please do share.

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## Comments

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TopNewestThis is not hard to explain. First, the function you're trying to compute is actually $\frac{180}{\pi} \tan^{-1}(x),$ where the $\frac{180}{\pi}$ is just the factor that converts from radians to degrees.

First note that for $|x| < 1$ we have $\begin{aligned} \frac{180}{\pi} \tan^{-1}(1/x) = 90 -\frac{180}{\pi}\tan^{-1}(x) &= 90 - \frac{180}{\pi} \left(x-\frac{x^3}3 + \frac{x^5}5 - \cdots \right) \\ &= 90 - \frac{180}{\pi} x + \frac{180}{\pi} \frac{x^3}3 - \frac{180}{\pi} \frac{x^5}5 + \cdots \end{aligned}$ and now we're plugging in $x = 1/10^n.$ Since $x^3, x^5, \ldots$ are tiny compared to $x,$ we can get a very good approximation by cutting off those terms. So we get $\begin{aligned} \frac{180}{\pi} \tan^{-1}(10^n) &= 90 - \frac{180}{\pi} \frac1{10^n} + \frac{180}{\pi} \frac1{3 \cdot 10^{3n}} - \cdots \\ &\approx 90 - \frac{180}{\pi} \frac1{10^n} \\ &\approx 90 - \frac{57.3}{10^n} \end{aligned}$ and that's that. So: $\begin{aligned} \frac{180}{\pi} \tan^{-1}(100) &\approx 90 - \frac{57.3}{100} = 89.427\\ \frac{180}{\pi} \tan^{-1}(1000) &\approx 90 - \frac{57.3}{1000} = 89.9427 \\ \frac{180}{\pi} \tan^{-1}(10000) &\approx 90 - \frac{57.3}{10000} = 89.99427 \\ \end{aligned}$ and so on.

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Hi! Just a couple of thoughts:

First, the function $\tan^{-1}(x)$ approaches $90^{\circ}$ as we input larger and larger numbers, so it makes sense that your values also get closer and closer to $90^{\circ}$ .

I actually tried a similar procedure with an initial input of $200$. I kept increasing this input by factors of $10$ and saw I similar pattern: just adding nines at the front of the decimal part. I suppose this would happen with most large numbers.

We can't exactly say that adding nines is all we have to do however, because the numbers you listed have been rounded. The remaining digits do change with each calculation.

I'm actually not quite sure though what causes this "add a nine" behaviour. Not sure if it is an actual mathematical property or just the result of using a calculator. Hopefully someone else knows!

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Hi David, Thanks for your info about the fact that it doesnt need to be factors of 10 to show this pattern and I know that I have taken an approximation but still it is mind boggling that we can begin predicting values of tan x where it starts getting unpredictable.

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Wow man What an observation, I cant even comprehend this

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Thanks

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