Will $y=x$ and $y=\sin^n \left(\frac{\pi x}{2} \right)$ meet in the interval $(0,1)$ for all $n \ge 2$?

Question

I am currently learning the basics of real analysis, and still haven't learned the concepts and properties of pointwise convergence and uniform convergence of functions.

Still, I was thinking of a function $f_n : (0,1) \to \mathbb{R}$, $$ f_n(x) = \sin^n \left(\frac{\pi x}{2} \right) $$ and if we took the $n$ to infinity, it would give $$ \lim_{n \to \infty} f_n(x) = 0 $$

From that, I thought that there might exist some $f_n$ that $y=f_n(x)$ does not meet $y=x$.

I thought it this way since for every $x \in (0,1)$, $\lim_{n \to \infty}f_n (x) = 0$, I could use $\epsilon - N$. To be specific, for given $x \in (0,1)$, I have taken $\epsilon = x$, then exists a natural number $N$ such that the following holds: $$ n \ge N \Rightarrow |f_n(x)| < x $$

I thought that this means that for that point $x$, $(x,f_n(x))$ is below $(x,x)$, and hence will not meet $y=x$.

Since I can take such $N$ for every point in $(0,1)$, I thought that $y=f_n(x)$ and $y=x$ will not meet for some large $N$.

But my instructor told me it will always meet since the function is pointwise convergent, but not uniformly convergent. I do not have enough knowledge of these concepts, so I am asking help for them.

My own guess to the failure of my conjecture(?) is that taking the large $N$ is impossible. To take $N$, it has to be bigger or equal to all the other $N$ taken from each $x \in (0,1)$. In other words, $$ N := \sup\{N_x : x \in (0,1)\}$$ where $N_x$ is the $N$ for each $x$. I thought that the set is not upper-bounded hence unable to take $N$.

Thank you in advance for your considerate help.

Answer 1

If $x_n=1-\frac1{2n}$, then
$$f_n(x_n)=\cos^n\left(\frac\pi{4n}\right)\\ =\left(1-\frac{\pi^2}{32n^2}+O(n^{-4})\right)^n\\ =1-\frac{\pi^2}{32n}+O(n^{-2})\gt x_n$$

Answer 2

In this answer, I will provide a lower level (at least to a calculus level) proof that the answer to OP's post is yes. Further, I hope that the way I present the proof can serve as inspiration to anyone else that has a problem that requires chaining lots of inequalities together.

First, define the function

$$g_n(x)=\sin^n\left(\frac{\pi x}{2}\right)-x$$

Then note that for $n\geq 2$ (and $x=\frac{1}{3}$) we have that

$$\sin^n\left(\frac{\pi}{2}\cdot\frac{1}{3}\right)=\sin^n\left(\frac{\pi}{6}\right)=\frac{1}{2^n}<\frac{1}{3}\Rightarrow g_n\left(\frac{1}{3}\right)<0$$

Now, consider the sequence $x_n=1-\frac{1}{f(n)}$ where $f(n):\mathbb{N}\to\mathbb{R}$ is some increasing function such that $f(2)\geq 2$. We want to show that there exists such a function $f(n)$ such that $g_n(x_n)>0$. We choose not to fully spell out $f(n)$ at this point for two reasons: it is easier to write the following computations using simply $f(n)$ and using $f(n)$ provides a good case study on how one might go about proving similar problems. We have that

$$g_n\left(x_n\right)=\sin^n\left(\frac{\pi }{2}\cdot\left(1-\frac{1}{f(n)}\right)\right)-\left(1-\frac{1}{f(n)}\right)$$

But the trig term in this is just

$$\sin^n\left(\frac{\pi }{2}\cdot\left(1-\frac{1}{f(n)}\right)\right)=\sin^n\left(\frac{\pi}{2}-\frac{\pi}{2f(n)}\right)=\cos^n\left(\frac{\pi}{2f(n)}\right)$$

Using the Taylor Series approximation for $\cos(x)$, we have that

$$\cos^n\left(\frac{\pi}{2f(n)}\right)>\left(1-\frac{1}{2}\left(\frac{\pi}{2f(n)}\right)^2\right)^n=\left(1-\frac{\pi^2}{8f(n)^2}\right)^n$$

We may then use the binomial theorem to get that

$$\cos^n\left(\frac{\pi}{2f(n)}\right)>\sum_{k=0}^n\binom{n}{k}\left(-\frac{\pi^2}{8f(n)^2}\right)^k=1+\sum_{k=1}^n\binom{n}{k}\left(-\frac{\pi^2}{8f(n)^2}\right)^k$$

Plugging this all back into $g_n(x_n)$, we get that

$$g_n\left(x_n\right)=\sin^n\left(\frac{\pi }{2}\cdot\left(1-\frac{1}{f(n)}\right)\right)-\left(1-\frac{1}{f(n)}\right)$$

$$>1+\sum_{k=1}^n\binom{n}{k}\left(-\frac{\pi^2}{8f(n)^2}\right)^k-\left(1-\frac{1}{f(n)}\right)$$

$$=\sum_{k=1}^n\binom{n}{k}\left(-\frac{\pi^2}{8f(n)^2}\right)^k+\frac{1}{f(n)}$$

Now, this is a complicated expression, but we can bound it by

$$>\frac{1}{f(n)}-\sum_{k=1}^n\binom{n}{k}\left(\frac{\pi^2}{8f(n)^2}\right)^k$$

Further, we have that

$$\sum_{k=0}^n\binom{n}{k}=(1+1)^n=2^n\Rightarrow \binom{n}{k}<2^n$$

Then the full expression can be bounded by

$$>\frac{1}{f(n)}-2^n \sum_{k=1}^n\left(\frac{\pi^2}{8f(n)^2}\right)^k>\frac{1}{(2n)!}-2^n \sum_{k=1}^\infty \left(\frac{\pi^2}{8f(n)^2}\right)^k$$

This is simply a convergent geometric series, which gives us

$$=\frac{1}{f(n)}-\frac{2^n\pi^2}{8f(n)^2-\pi^2}>\frac{1}{f(n)}-\frac{2^n 4^2}{8f(n)^2-4^2}$$

$$=\frac{1}{f(n)}-\frac{2^{n+1}}{f(n)^2-2}=\frac{f(n)^2-2-2^{n+1}f(n)}{f(n)(f(n)^2-2)}>\frac{f(n)^2-2^{2n+2}-2^{n+1}f(n)}{f(n)(f(n)^2-2)}$$

Now, recall that we are trying to show that $g_n(x_n)>0$. Thus, we simply need to pick $f(n)$ such that the numerator of the above expression is positive (since the denominator is always positive). This is easy (you may have noticed that we have simplified things in a particular way) as we may simply take $f(n)=2^{n+2}=2\cdot 2^{n+1}$. Then

$$f(n)^2-2^{2n+2}-2^{n+1}f(n)=4\cdot 2^{2n+2}-2^{2n+2}-2\cdot 2^{2n+2}=2^{2n+2}>0$$

We conclude that

$$g_n\left(1-\frac{1}{2^{n+2}}\right)>0$$

for all $n\geq 2$. Then by the mean value theorem, there exists $x_0\in \left(\frac{1}{3},1-\frac{1}{2^{n+2}}\right)$ such that $g_n(x_0)=0$ and we are done.

Answer 3

When $n \geq 2$ increases, the root moves to a solution which is closer and closer to $x=1^-$ (ignoring the trivial solutions $x=0$ and $x=1$).

To see what happens, using Taylor and the binomial expansion, you face $$\frac{x-\sin ^n\left(\frac{\pi x}{2}\right)}{x-1}=1+\frac{\pi ^2 n}{8} (x-1)+\left(\frac{\pi ^4 n}{192}-\frac{\pi ^4 n^2}{128}\right) (x-1)^3+O\left((x-1)^5\right)$$

So, a first approximation is just $$x_{(1)}=1-\frac{8}{\pi ^2 n}$$ For axample for $n=10$, this would give $x\sim 0.918943$ while the exact solution (given by Newton method) would be $0.915564$.

We could improve using Householder method, the first iterate of which being $$x=1-\frac{24 n}{3 n \left(\pi ^2 n-4\right)+8}$$ which, for $n=10$ gives $0.915756$ which is much better.

The next iterative method (unnamed) would give $$x=1-\frac{8 \left(3 n \left(\pi ^2 n-4\right)+8\right)}{\pi ^2 n \left(3 \pi ^2 n^2-24 n+16\right)}$$ which, for $n=10$ gives $0.915626$ which is still better.

Another thing which coudl be done is to continue the Taylor expansion and use series reversion to get, as approximation $$x=1+t+\frac 1 \pi \sum_{k=1}^p {a_k} (\pi t)^{2k+1}\qquad \text{with}\qquad t=-\frac{8}{\pi ^2 n}$$

the first $a_k$ terms being $$\left\{\frac{3 n-2}{48} ,\frac{105 n^2-120 n+28}{11520},\frac{6615 n^3-10500 n^2+4788 n-592}{3870720}\right\}$$