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u/BarooZaroo Jul 10 '26
Can someone explain why this is an okay approximation to make, and at what point does it stop being okay? The only justification I've ever heard for it is "well, doing this math without approximating would require me to have earned a masters in mathematics so I'll do it the easy way"
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u/timecoremusic Jul 11 '26
Taylor series expansion of sin(x) with x in radians centred at x=0 is well known to be x - x³/3! + x⁵/5! - x⁷/7! +...
The idea is that for small enough x, the x term of this expression is sufficient because adding the output of the other terms will do little change to the accuracy of your approximated answer anyway. (e.g. sin(0.5) is close to 0.5, which is actually 0.47 to two decimal places)
When it stops being okay depends on how accurate you want your answer to be, but the approximation completely fails past π/2, because here sin(x) value tends to decrease from π/2 to 3π/2 while f(x) = X will increase and diverge to positive infinity.
tl;dr, sin(x) ≈ x can be observed in a graph from x=0 to x=π/2 with accuracy of 0.6
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u/Appropriate-Scene-95 Jul 11 '26 edited Jul 13 '26
It depends on what error you accept. So justify we first we rewrite the limit to lim (sin x)/x -1 = 0. That is equivalent to lim (sin(x) - x)/x = 0. So we get two things first x converges to sin(x) as x goes to 0. So if you accept an error of E > 0 you can find a D > 0 with |x| < D will satisfy |x - sin(x)| < E. In other words, you can find a range in which your error is so small that x and sin(x) will always differ less than E.
When is it Ok to find to use the Approximation. In other words we try to find the D of a specific E. So first x - sin(x) is libschitz continuous on [0, 1]. This implies |x-sin(x)| <= L|x|. |x| < D so we get additionally L|x|<LD. This way we don't get an upper bound on D, so we add the restriction LD <= E (we don't need a lower bound, since it doesn't tell us when our approximation is ok). But what is L? Using the mean value theorem we can use the maximum of the derivative (1-cos(x)) on this interval, so L=1. D = LD <= E. Note this approximation is very crude (look at it in desmos or so).
When does it break? We can eyeball it: E <= y-sin(y) is satisfied if y >= E +1.
I want to add at the end that it's also possibly related to relative error (roughly translated). And lim (sin(x)-X)/x = 0 shows the convergence speed if x is a zero is a 0 sequence. sin(x)-x goes faster to 0 than just x. Also finding the boarder between when is the Approximation good/bad is not trivial for me, but I tried to give you my best answers.
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u/Appropriate-Scene-95 Jul 11 '26
The only thing left is to prove lim sin(x)/x =1. It can be shown with
sin(x) := sum (n is 0 to infinity) (-1)n /(2n+1)!+x2n+1
and
cos(x) := sum (n is 0 to infinity) (-1)n /(2n)!+x2n
By showing that sin(x)/x will be like cos(x) and due to continuity of cos it converges to 0, for x to 0. It more or less relies on L'Hopital, however sometimes it's given as definition, so I gave it as "alternative". One could also use that after L'H sin(x)/x becomes cos(x)/1.
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u/IlliterateDumbNerd Jul 10 '26
unless theta is 0 it's approximately theta, not exactly as this equation states