A pattern in tanx\tan x

if we take a look at value of tan1100\tan^{-1} 100 , using a calculator we can say it is 89.427.
then if we go to tan11000\tan^{-1} 1000 , we get 89.9427
for tan110000\tan^{-1} 10000 it is 89.99427
tan1100000\tan^{-1} 100000 is 89.999427.
So we have a pattern here using which we can predict the value of tan110n\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.

Note by Mathoholic Oja
3 months, 1 week ago

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This is not hard to explain. First, the function you're trying to compute is actually 180πtan1(x),\frac{180}{\pi} \tan^{-1}(x), where the 180π\frac{180}{\pi} is just the factor that converts from radians to degrees.

First note that for x<1|x| < 1 we have 180πtan1(1/x)=90180πtan1(x)=90180π(xx33+x55)=90180πx+180πx33180πx55+ \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=1/10n.x = 1/10^n. Since x3,x5,x^3, x^5, \ldots are tiny compared to x,x, we can get a very good approximation by cutting off those terms. So we get 180πtan1(10n)=90180π110n+180π13103n90180π110n9057.310n \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: 180πtan1(100)9057.3100=89.427180πtan1(1000)9057.31000=89.9427180πtan1(10000)9057.310000=89.99427 \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.

Patrick Corn - 3 months ago

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Wow man What an observation, I cant even comprehend this

Roman Empire - The Fallen Rises - 3 months, 1 week ago

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Thanks

Mathoholic Oja - 3 months, 1 week ago

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Hi! Just a couple of thoughts:

First, the function tan1(x)\tan^{-1}(x) approaches 9090^{\circ} as we input larger and larger numbers, so it makes sense that your values also get closer and closer to 9090^{\circ} .

I actually tried a similar procedure with an initial input of 200200. I kept increasing this input by factors of 1010 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!

David Stiff - 3 months, 1 week ago

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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.

Mathoholic Oja - 3 months, 1 week ago

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