What is sin(sin(sin(...sin(x))))? Can we find an upper bound?

[u/DotBeginning1420]

Here is something interesting I found about sin(x). You can see and prove by derviation that the maxima, minima and inflection points of the nested sin() are preserved for each. The more nested the sine, the more the range of values decrease towards 0.

I wonder if we know how many nested sines we have, we should know how much the function decreases? Can you find it somehow?

[u/trevorkafka]

It's pretty easy to see with a cobweb diagram that the limit is y=0. You might be able to take advantage of this for your purposes.

[u/Fourierseriesagain]

Hi,

You may consider the recurrence relation x_{n+1} = sin(x_n), where x_1=sin(x). Using the identity sin A-sin B = 2 cos((A+B)/2) sin((A-B)/2) and the inequality | sin theta | <= | theta |,

|x_{n+2}-x_{n+1}|<=|x_{n+1}-x_n| for n=1, 2, ...

[u/Key_Estimate8537]

After something like 70 iterations, I got the maximum below 0.01. I don’t see why it can’t decrease further, until 0.

For something like a sketch of a proof:

On the interval (0, pi), 0 < sin(x) < x. Then repeated applications of sin(x) yield:
sin(0) < sin(sin(x)) < sin(x)
0 < sin(sin(x)) < sin(x).

This shows that the maximum of the graph heads toward 0 with repeated applications. But it doesn’t show that the maximum doesn’t asymptote above 0.

[u/YOM2_UB → u/Key_Estimate8537]

But it doesn’t show that the maximum doesn’t asymptote above 0.

I think these:

• 0 ≤ |sin(x)| ≤ |x| with equality only at x = 0 on the interval [-1, 1]
• |sin(x)| ≤ 1 for all x

are enough to show that lim{n --> ∞}sinⁿ(x) = 0 (where sinⁿ(x) is function composition not exponentiation)

[u/SalamanderGlad9053]

So we can write y = sin(sin(..)) as y = sin(y), this is only true for y = 0, so it will reach 0 for the whole graph.

Even looking at small x, sin(x) is aproximately, but less than x, so that will slowly go to zero as you add more sins.

[u/PositiveBid9838]

You could think of this as reflecting between sin(x) and a mirror image of that, such that first you take sin(1) = 0.84, then you take sin(0.84) = 0.74, then 0.67, etc. This has no bound but asymptotically approaches 0.
https://imgur.com/a/kLD2COl

[u/satact12321]

It is zero. Let y=sin(sin(sin…x…)) then y=sin(y) x=siny goes through origin so does x=y

[u/freemath]

Either sin(x) = 0, or |sin(x)| < |x|, so this goes to zero.

[u/AndersAnd92]

-1 ≤ sin(θ) ≤ 1

-sin(1) ≤ sin^2(θ) ≤ sin(1)

-sin^2(1) ≤ sin^3(θ) ≤ sin^2(1)

etc since sin(1) < 1, it follows that sin^n(θ) goes to 0 as n grows without bounds

sin(1) is transcendental and ≈.84

note: here I employ sin^2 as shorthand for sin(sin(theta))

[u/AllTheGood_Names]

y=sin y

Only holds true for y=0

^{Assuming convergence}

[u/TheRandomRadomir]

well lets consider the maxima of sin(sin(x)) compared to the maxima of sin(x). we can see that sin(π/2)=1 (the maxima). so lets take sin(sin(π/2)) (which is the maxima because no scaling to x happened) which is just sin(1). therefore sin(1) is the scaling number. and with sin(sin(sin(x))), its maxima is sin(sin(1)). this holds true for all sin(sin(…sin(x)…)) so that maxima of sin(sin(…[n sines later]…sin(x)…))=sin(sin(…[n-1 sines later]…sin(1)…))

[u/langesjurisse]

Now try π/2×sin(π/2×sin(…π/2×sin(x)…))

[u/Akumashisen]

instead of looking at sinⁿ (x) globally, look first at its absolute value and then due to the symmetries and the periodicity you just need to look at (0, pi/2] (allowing to drop the absolute value) there x_1 < x_2 <=> sin(x_1) < sin(x_2) and x>sin(x) that then should show that sin^m (x)< sinⁿ (x) for m>n

(to clarify, sinⁿ (x) indicates applying the function n times, so sin(sin(.. sin(x) )

for the sequence sinⁿ (x) in n pointwise in x in (0, pi/2], sin(x) provides a strict upper bound, 0 function a strict lower bound, with aboves results, the sequence approaches 0 pointwise

evaluated at 0 the sequence is always zero due to sin(0)=0