What Is the Limit of Iterated Sine Functions as n Approaches Infinity?

In summary: Your Name]In summary, we can use the Taylor expansion and the squeeze theorem to show that the limit of lim_{n-> /infty} sinsinsin..._{n times} sin(x) is 0. This is because the sequence a_{n+1}, which is equal to the limit of the original function, is always between -a_n and a_n, and since a_n approaches 0, we can conclude that a_{n+1} also approaches 0. Therefore, the limit of the original function is also 0.
  • #1
talolard
125
0
Hello,

Homework Statement


Calculate the limit of [tex] lim_{n-> /infty} sinsinsin..._{n times } sin(x)[/tex] where x is any real number


My attempt:

Using the Taylor expansion we know that [tex]sin(x)=x- \frac{x^3}{3!} +\circ(x^3)[/tex] and that [tex]1 \geq sin(x) \geq -1 [/tex]
If the expression has a limit then every subsequence aproaches the same limit. specifically we can define a sequence
[tex] a_{n+1}=|a_{n}- \frac {a_{n}^3}{3!}| [/tex] and so the limit of this sequence must be the limit of the above function.
We can restrict our inquiry to [tex] a_1\in[-1,1] [/tex] since tex]1 \geq sin(x) \geq -1 [/tex]

We will show by induction that the sequence is montonic decreasing:
[tex] a_{2}=|a_{1}- \frac {a_{1}^3}{3!}| \geq |a_{1}| - |\frac {a_{1}^3}{3!}| /leq a_{1} [/tex] which holds for any value of [tex] a_1 [/tex]
We will assume for n and prove for n+1
[tex] a_{n+1}=|a_{n}- \frac {a_{n}^3}{3!}| \leq^{?} a_{n} \iff | \frac {3!a_{n} - a_{n}^3}{3!} < a_{n}| \iff |3!a_{n} - a_{n}^3| \leq 3!a_{n} -> |3!a_{n} - a_{n}^3| \geq |3!a_{n}| -| a_{n}^3| \leq 3!a_{n} \iff -| a_{n}^3| \leq 0 [/tex]
Or in words, we used the traingle inequality and the fact that the minimum of an absolut value is zero to show that the sequence decreases.

There is no need to show that it is bounded since an absolute value is lowerly bounded by 0 and we have shown that the sequence is monotonically decreasing.
So we knwo the sequence converges to a finite limit and so the original function does as well. but how do i show that the limit is zero?

Thanks for the help,
Tal
 
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  • #2


Hello Tal,

Your approach is definitely on the right track. To show that the limit is zero, we can use the squeeze theorem. We know that the limit of a_{n+1} is equal to the limit of the original function, so if we can show that a_{n+1} approaches 0, then we can conclude that the original function also approaches 0.

To do this, we can use the fact that 1 \geq sin(x) \geq -1 and the Taylor expansion of sin(x) to show that a_{n+1} is always less than or equal to a_{n}. This is because the term \frac{a_n^3}{3!} will always be positive, so subtracting it from a_n will make a_{n+1} smaller than a_n. This means that a_{n+1} is always between -a_n and a_n, and since we know that a_n approaches 0, we can conclude that a_{n+1} also approaches 0. Therefore, the limit of the original function is also 0.

I hope this helps clarify things. Keep up the good work with your mathematical reasoning!


 

FAQ: What Is the Limit of Iterated Sine Functions as n Approaches Infinity?

What is the concept of "limit of infinite sines"?

The limit of infinite sines is a mathematical concept that deals with the behavior of an infinite sequence of sines as the number of terms approaches infinity. It is used to determine the value that a function approaches as its input approaches a specific value.

How is the limit of infinite sines calculated?

The limit of infinite sines is calculated by taking the limit of the sequence of sines as the number of terms approaches infinity. This can be done using various mathematical techniques such as the squeeze theorem, the ratio test, or the comparison test.

What is the significance of the limit of infinite sines in mathematics?

The limit of infinite sines is significant in mathematics as it helps to determine the convergence or divergence of infinite series, which has applications in various fields such as physics, engineering, and finance. It also plays a crucial role in understanding the behavior of functions and their graphs.

Can the limit of infinite sines be negative?

Yes, the limit of infinite sines can be negative. The sign of the limit depends on the specific sequence of sines being evaluated and the value of the input as it approaches the limit. It can be positive, negative, or zero, depending on the behavior of the sequence.

Is the concept of limit of infinite sines applicable to all functions?

No, the concept of the limit of infinite sines is not applicable to all functions. It is specifically used for functions that can be expressed as an infinite series of sines, such as trigonometric functions. It is also applicable to certain power series, exponential functions, and logarithmic functions.

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