# Divergence of the sequence of sine of Fibonacci numbers

Throughout this post, $\{F_n\}$ is the Fibonacci sequence. I’m going to prove that the sequence $\{\sin(F_n)\}$ is divergent, i.e. $\displaystyle \lim_{n\to\infty} \sin(F_n)$ does not exist.

Notation. Let $\mathbb{Q}$ be the set of rational numbers. We denote by $\mathbb{Q}[\sqrt{5}]$ the set $\{p+q\sqrt{5}: \ \ p,q \in \mathbb{Q}\}.$ For example, $\displaystyle \left(\frac{\sqrt{5}+1}{2}\right)^n \in \mathbb{Q}[\sqrt{5}]$ for all integers $n.$

Remark 1. See that $\mathbb{Q} \subset \mathbb{Q}[\sqrt{5}]$ and if $a,b \in \mathbb{Q}[\sqrt{5}],$ then $a \pm b, ab \in \mathbb{Q}[\sqrt{5}].$ Also if $a,b \in \mathbb{Q}[\sqrt{5}]$ and $b \ne 0,$ then $\displaystyle \frac{a}{b} \in \mathbb{Q}[\sqrt{5}].$

Remark 2. It is clear that every element of $\displaystyle \mathbb{Q}[\sqrt{5}]$ is a root of some quadratic polynomial with integer coefficients. We showed here that $\pi \notin \mathbb{Q}$ and we also mentioned، without a proof,  that in fact $\pi$ is not a root of any polynomial with integer  coefficients (see the Remark in that post!). In particular, $\pi \notin \mathbb{Q}[\sqrt{5}].$

Problem. Let $\displaystyle L:=\lim_{n\to\infty} \sin(F_n).$ Show that

i) if $L$ exists, then $L=0$

ii) $L$ does not exist.

Solution. i) Suppose that $L \ne 0.$ Since $F_{n+2}=F_{n+1}+F_n$ and $F_{n-1}=F_{n+1}-F_n,$ we have

$\displaystyle \sin F_{n+2}-\sin F_{n-1}=2\sin F_n \cos F_{n+1}, \ \ \ \ \ \ \ \ (1)$

$\displaystyle \sin F_{n+2}+\sin F_{n-1}=2\sin F_{n+1}\cos F_n. \ \ \ \ \ \ \ \ \ (2)$

Now, since $\displaystyle \lim_{n\to\infty}(\sin F_{n+2}-\sin F_{n-1})=L-L=0,$ we must have $\displaystyle \lim_{n\to\infty}\cos F_n=0,$ by $(1),$ and so, by $\displaystyle (2), \ 2L=\lim_{n\to\infty}(\sin F_{n+2}+\sin F_{n-1})=0,$ contradiction.

ii) Suppose that $L$ exists. Then by i), $L=0.$ For each $n,$ there exists positive integer $k_n$ and $\theta_n \in (-\pi/2, \pi/2)$ such that

$F_n=k_n \pi + \theta_n.$

Since $\displaystyle \lim_{n\to\infty}\sin F_n=0,$ we must have $\displaystyle \lim_{n\to\infty}\sin \theta_n=0$ and hence $\displaystyle \lim_{n\to\infty}\theta_n=0.$ Thus

$\displaystyle \lim_{n\to\infty}(k_{n+1}-k_n-k_{n-1})=\frac{1}{\pi}\lim_{n\to\infty}(\theta_{n-1}+\theta_n-\theta_{n+1})=0.$

So, since $k_n$ are all integers, there exists an integer $m$ such that

$k_{n+1}=k_n+k_{n-1}$

for $n \ge m.$ Thus, by Problem 1 in this post, there exist constants $a,b \in \mathbb{Q}[\sqrt{5}]$ such that

$k_{n+m}=a\varphi^n+b \psi^n, \ \ \ \forall n \ge 0,$

where $\displaystyle \varphi=\frac{\sqrt{5}+1}{2}$ and $\displaystyle \psi=\frac{1-\sqrt{5}}{2}=-\frac{1}{\varphi}.$ Therefore

$F_{n+m}=k_{n+m}\pi + \theta_{n+m}=\pi(a\varphi^n+b\psi^n)+\theta_{n+m}$

and hence

$\displaystyle \frac{F_{n+m}}{\varphi^n}=\pi a + \pi b\frac{\psi^n}{\varphi^n}+\frac{\theta_{n+m}}{\varphi^n},$

which, after taking limit as $n\to\infty,$ gives $\displaystyle \frac{\varphi^m}{\sqrt{5}}=\pi a.$ So $\displaystyle \pi=\frac{\varphi^m}{a\sqrt{5}} \in \mathbb{Q}[\sqrt{5}],$ which is false, by Remark 2. So $L \ne 0$ and thus $L$ doesn’t exist. $\Box$

Exercise. Show that $\displaystyle \lim_{n\to\infty} \cos(F_n)$ does not exist.

# Integral inequalities (5)

Problem. Suppose that $f:[0,1] \longrightarrow [0,\infty)$ is a continuous function which is not identically zero. Show that

$\displaystyle \int_0^1 \sqrt{f(x)\tan^{-1}x} \ dx < \sqrt{\left(G+\frac{5}{144}+\frac{31\pi-70\ln2}{288}\right)\int_0^1f(x)\ln(1+x)dx},$

where $G$ is the Catalan’s constant.

Solution. By Cauchy-Schwarz inequality

\displaystyle \begin{aligned} \left(\int_0^1 \sqrt{f(x)\tan^{-1}x} \ dx\right)^2=\left(\int_0^1 \sqrt{\frac{\tan^{-1}x}{\ln(1+x)}} \sqrt{f(x)\ln(1+x)} dx\right)^2 \\ \le \int_0^1 \frac{\tan^{-1}x}{\ln(1+x)} \ dx \int_0^1 f(x) \ln(1+x) \ dx.\end{aligned}

So we are done if we show that

$\displaystyle \int_0^1 \frac{\tan^{-1}x}{\ln(1+x)} \ dx < G+\frac{5}{144}+\frac{31\pi-70\ln2}{288}. \ \ \ \ \ \ (*)$

To prove $(*),$ we approximate $\displaystyle \frac{x}{\ln(1+x)}$ with the first four terms of its Maclaurin series expansion.

Claim. For any $x > 0,$ we have

$\displaystyle \frac{x}{\ln(1+x)} < 1+\frac{x}{2}-\frac{x^2}{12}+\frac{x^3}{24}.$

Proof. Let $\displaystyle p(x):=1+\frac{x}{2}-\frac{x^2}{12}+\frac{x^3}{24}$ and $\displaystyle g(x):=\ln(1+x)-\frac{x}{p(x)}, \ x >0.$ To prove the claim, we only need to show that $g$ is increasing for $x \ge 0$ because $g(0)=0.$ So we show that $g'(x) > 0$ for $x > 0.$ Now

$\displaystyle g'(x)=\frac{1}{1+x}-\frac{p(x)-xp'(x)}{(p(x))^2}.$

So we need to show that

$\displaystyle (p(x))^2-(x+1)p(x)+x(1+x)p'(x) > 0$

for $x > 0$ and that’s easy because

$\displaystyle (p(x))^2-(x+1)p(x)+x(1+x)p'(x) = \left(\frac{x^2}{12}-\frac{x^3}{24}\right)^2+\frac{x^4}{8} > 0.$

Now that we have proved the claim, proving $(*)$ is easy. We have

\displaystyle \begin{aligned} \int_0^1 \frac{\tan^{-1}x}{\ln(1+x)} \ dx = \int_0^1 \frac{x}{\ln(1+x)} \cdot \frac{\tan^{-1}x}{x} \ dx < \int_0^1\left(1+\frac{x}{2}-\frac{x^2}{12}+\frac{x^3}{24}\right)\frac{\tan^{-1}x}{x} \ dx \\ =\int_0^1 \frac{\tan^{-1}x}{x} \ dx + \int_0^1\left(\frac{1}{2}-\frac{x}{12}+\frac{x^2}{24}\right)\tan^{-1}x \ dx.\end{aligned}

We have $\displaystyle \int_0^1 \frac{\tan^{-1}x}{x} \ dx=G$ and integration by parts with

$\displaystyle \tan^{-1}x=u, \ \ \ \left(\frac{1}{2}-\frac{x}{12}+\frac{x^2}{24}\right)dx=dv$

gives

$\displaystyle \int_0^1\left(\frac{1}{2}-\frac{x}{12}+\frac{x^2}{24}\right)\tan^{-1}x \ dx=\frac{5}{144}+\frac{31\pi-70\ln2}{288}$

and that completes the solution. $\Box$

# Nice little problems: limit & continuity (3)

Problem. Let $f: [0,1] \longrightarrow \mathbb{R}$ be a continuous function and suppose that there exist $a,b \in (0,1)$ such that

$\displaystyle \int_0^x f(t) \ dt = \int_0^{ax} f(t) \ dt + \int_0^{bx}f(t) \ dt, \ \ \ \forall x \in [0,1].$

Show that if $a+b < 1,$ then $f=0$ and if $a+b=1,$ then $f$ is constant.

Solution. We may assume, without loss of generality, that $a \le b.$
We differentiate the given identity to get

$f(x)=af(ax)+bf(bx) \ \ \ \ \ \ \ \ \ (1)$

and thus

\begin{aligned} f(x)=a(af(a^2x)+bf(abx))+b(af(abx)+bf(b^2x)) \\ =a^2f(a^2x)+2abf(abx)+b^2f(b^2x).\end{aligned}

An easy induction shows that for any integer $n \ge 1$ we have

$\displaystyle f(x)=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}f(a^kb^{n-k}x). \ \ \ \ \ \ \ \ \ (2)$

Let $\epsilon > 0$ be given. Since $f$ is a continuous function, there exists $\delta > 0$ such that $|f(x)-f(0)| < \epsilon$ whenever $0 \le x < \delta.$
Let $x_0 \in [0,1].$ Since $0 < b < 1,$ we have $\displaystyle \ \lim_{n\to\infty}b^nx_0=0$ and so there exists $N > 0$ such that $0 \le b^nx_0 < \delta$ whenever $n \ge N.$
Thus for $n \ge N$ and $0 \le k \le n$ we have $0 \le a^kb^{n-k}x_0 \le b^nx_0 < \delta$ and hence $|f(a^kb^{n-k}x_0)-f(0)| < \epsilon.$ Therefore if $n \ge N,$ then, by $(2),$

\displaystyle \begin{aligned}|f(x_0)-f(0)|=\left|\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}(f(a^kb^{n-k}x_0)-f(0))+f(0)\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}-f(0)\right| \\ \le \sum_{k=0}^n \binom{n}{k}a^kb^{n-k}|f(a^kb^{n-k}x_0)-f(0)|+|f(0)((a+b)^n-1)| \\ < (a+b)^n\epsilon + |f(0)((a+b)^n-1)|. \end{aligned} \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

If $a+b=1,$ then, by $(3), \ |f(x_0)-f(0)| < \epsilon$ and thus $f(x_0)=f(0),$ i.e. $f$ is constant.
If $a+b < 1,$ then $f(0)=0$ (just put $x=0$ in $(1)$ to get $f(0)=0$). Thus, by $(3),$ we have $|f(x_0)| < (a+b)^n\epsilon < \epsilon$ and so $f(x_0)=0,$ i.e. $f=0. \ \Box$

# Flett’s mean value theorem

Problem (T. M. Flett, 1958). Let $f: [a,b] \longrightarrow \mathbb{R}$ be a differentiable function. Show that if $f'(a)=f'(b),$ then there exists $c \in (a,b)$ such that

$f(c)-f(a)=(c-a)f'(c).$

Solution. We consider two cases.

Case 1 : $f'(a)=f'(b)=0.$ Define the function $g: [a,b] \longrightarrow \mathbb{R}$ as follows

$\displaystyle g(x):=\begin{cases} \frac{f(x)-f(a)}{x-a} & \ \ x \in (a,b] \\ 0 & \ \ x = a.\end{cases}$

Clearly $g$ is continuous on $[a,b]$ and differentiable on $(a,b).$ Also for $x \in (a,b]$ we have

$\displaystyle g'(x)=\frac{(x-a)f'(x)-f(x)+f(a)}{(x-a)^2}. \ \ \ \ \ \ \ \ \ (*)$

So we are done if we show that $g'(c)=0$ for some $c \in (a,b).$ To do that, we consider the three possibilities $g(b)=0, \ g(b) > 0, \ g(b) < 0.$
If $g(b)=0,$ then $g(a)=g(b)=0$ and thus, by Rolle’s theorem, there exists $c \in (a,b)$ such that $g'(c)=0.$
Now suppose that $g(b) > 0.$ Then, by $(*),$

$\displaystyle g'(b)=\frac{f(a)-f(b)}{(b-a)^2}=\frac{-g(b)}{b-a} < 0$

and so we cannot have that $g(x) \le g(b)$ for all $x \in (a,b)$ because then we would have the contradiction $\displaystyle g'(b)=\lim_{x\to{b^-}} \frac{g(x)-g(b)}{x-b} \ge 0.$ Thus there exists $u \in (a,b)$ such that $g(u) > g(b).$ Now consider the function $h(x):=g(x)-g(b), \ \ x \in [a,b],$ which is clearly continuous because $g$ is continuous. We have $h(u) > 0$ and $h(a)=-g(b) < 0.$ Thus, by the intermediate theorem, there exists $v \in (a,u)$ such that $h(v)=0.$ So $g(v)=g(b)$ and hence, by Rolle’s theorem, there exists $c \in (v,b)$ such that $g'(c)=0,$ as desired. The proof for the case $g(b) < 0$ is similar.

Case 2 : The general case. Consider the function $F(x):=f(x)-f'(a)x, \ \ x \in [a,b].$ Then $F'(x)=f'(x)-f'(a)$ and so $F'(a)=F'(b)=0.$ Thus, by Case 1, there exists $c \in (a,b)$ such that $F(c)-F(a)=(c-a)F'(c)$ and so

$f(c)-f'(a)c-f(a)+f'(a)a=(c-a)(f'(c)-f'(a)),$

which simplifies to $f(c)-f(a)=(c-a)f'(c). \ \Box$

That $c$ in Flett’s mean value theorem is not necessarily unique. In fact, there could be many of them, as the following example shows.

Example 1. Consider the function $f(x)=\sin x.$ For any integer $n \ge 1,$ let $\alpha(n)$ be the number of solutions of the equation

$f(x)-f(\pi/2)=(x-\pi/2)f'(x)$

in the interval $(\pi/2, n\pi + \pi/2).$ Show that $\alpha(n)=\begin{cases} n-1 & \text{if n is even} \\ n & \text{if n is odd}\end{cases}.$

Solution. We have $f(x)-f(\pi/2)-(x-\pi/2)f'(x)=\sin x - 1 - (x-\pi/2)\cos x.$ Let $g(x):=\sin x - 1-(x-\pi/2)\cos x.$ We want to find $\alpha(n),$ the number of solutions of $g(x)=0$ in the interval $(\pi/2, n\pi+\pi/2).$ We have $g'(x)=(x-\pi/2)\sin x.$
Since $g(\pi/2)=0$ and $g' > 0$ on $(\pi/2, \pi), \ g$ is increasing on $(\pi/2, \pi)$ and so it has no root in that interval.
Now we count the number of roots of $g$ in those subintervals of $(\pi/2, n\pi+\pi/2)$ which are in the form $I_k:=[k\pi, (k+1)\pi]$ for some integer $k \ge 1.$ First see that if $k$ is even, then $g(k\pi) < 0$ and if $k$ is odd, then $g(k\pi) > 0.$ So, by the intermediate value theorem, $g$ has a root in $I_k.$ Since $g'(x) \ne 0$ for all $x \in (k\pi, (k+1)\pi), \ g$ cannot have more than one root in $I_k.$ So $g$ has exactly one root in $I_k.$
Finally, we look for possible roots of $g$ in the subinterval $(n\pi, n\pi+\pi/2).$ Well, if $n$ is odd, then $g(n\pi) > 0$ and $g(n\pi+\pi/2)=-2 < 0.$ So in this case, $g$ has a root. This root is unique because $g'$ has no root in $(n\pi, n\pi+\pi/2).$ But if $n$ is even, then $g(n\pi+\pi/2)=0$ and $g' > 0$ on $[n\pi, n\pi+\pi/2).$ Thus $g < 0$ on $(n\pi, n\pi+\pi/2)$ and thus $g$ has no root in this case.
So we have shown that $g$ has no root in $(\pi/2, \pi)$ but it has exactly one root in each subinterval $[\pi, 2\pi], \ [2\pi, 3\pi], \cdots , [(n-1)\pi, n\pi].$ That gives $n-1$ roots of $g.$
We have also shown that, in $[n\pi, n\pi+\pi/2), \ g$ has no root if $n$ is even and has exactly one root if $n$ is odd. So $\alpha(n),$ the number of roots of $g$ in $(\pi/2, n\pi+\pi/2),$ is $n-1$ for $n$ even and is $n$ for $n$ odd. $\Box$

Example 2. Let $\displaystyle f: [0,1] \longrightarrow \mathbb{R}$ be a continuous function and suppose that $\displaystyle \int_0^1 f(x) \ dx = \int_0^1 xf(x) \ dx.$ Show that there exist $a,b,c,d \in (0,1)$ such that

\displaystyle \begin{aligned} \int_0^a f(x) \ dx =0, \ \int_0^b f(x) \ dx = f(b). \ \int_0^c xf(x) \ dx = 0, \ \int_0^d xf(x) \ dx = df(d).\end{aligned}

Solution. Consider the function

$\displaystyle g(x):=x\int_0^x f(t) \ dt - \int_0^x tf(t) \ dt, \ \ x \in [0,1].$

See that $\displaystyle g'(x)=\int_0^x f(t) \ dt$ and $g(0)=g(1)=0.$ Thus, by Rolle’s theorem, there exists $a \in (0,1)$ such that $g'(a)=0$ and so $\displaystyle \int_0^a f(t) \ dt =0.$
Now, since $g'(0)=g'(a)=0,$ Flett’s mean value theorem gives a $c \in (0,a)$ that satisfies the condition $g(c)-g(0)=cg'(c),$ i.e.

$\displaystyle c \int_0^c f(t) \ dt - \int_0^c tf(t) \ dt=c\int_0^cf(t) \ dt$

and hence $\displaystyle \int_0^c tf(t) \ dt =0.$
Now let $\displaystyle h_1(x):=e^{-x}\int_0^xf(t) \ dt.$ See that $\displaystyle h_1'(x)=e^{-x}\left(f(x)-\int_0^x f(t) \ dt\right)$ and $h_1(0)=h_1(a)=0.$ Thus, by Rolle’s theorem, there exists $b \in (0,a)$ such that $h_1'(b)=0$ and so $\displaystyle \int_0^b f(t) \ dt = f(b).$
Similarly, let $\displaystyle h_2(x):=e^{-x}\int_0^x tf(t) \ dt.$ Then $h_2(0)=h_2(c)=0$ and so $h_2'(d)=0$ for some $d \in (0,c)$ implying that $\displaystyle \int_0^d tf(t) \ dt = df(d). \ \Box$

Execsrie 1. Geometrically, what does Flett’s mean value theorem mean?

Exercise 2. Let $f: [a,b] \longrightarrow \mathbb{R}$ be a differentiable function. Show that if $f'(a)=f'(b),$ then there exists $c \in (a,b)$ such that $f(b)-f(c)=(b-c)f'(c).$

Exercise 3. Think about this: is there a differentiable function $f: [a,b] \longrightarrow \mathbb{R}$ that satisfies the conditions i) $f'(a)=f'(b)$ and ii) there are infinitely many $c \in (a,b)$ for which $f(c)-f(a)=(c-a)f'(c) ?$ If there is, give one!

# Limit of integrals (17)

Problem. Show that $\displaystyle \lim_{n\to\infty} \int_0^n \frac{dx} {1+n^2 \cos^2 x}=1.$

Solution. Let $\displaystyle I_n:= \int_0^n \frac{dx} {1+n^2 \cos^2 x}=1.$ For $n > 1,$ let $\displaystyle k=\left \lfloor \frac{n}{\pi}-\frac{1}{2}\right \rfloor,$ i.e. $k$ is the integer that satisfies the condition

$\displaystyle \frac{(2k+1)\pi}{2} < n < \frac{(2k+3)\pi}{2}.$

Now

$\displaystyle I_n:=\int_0^{\frac{\pi}{2}} \frac{dx}{1+n^2\cos^2x} + \sum_{i=1}^k \int_{\frac{(2i-1)\pi}{2}}^{\frac{(2i+1)\pi}{2}} \frac{dx}{1+n^2\cos^2x} + \int_{\frac{(2k+1)\pi}{2}}^n \frac{dx}{1+n^2\cos^2x}.$

Put $\tan x = t$ to get

\displaystyle \begin{aligned}I_n=\int_0^{\infty} \frac{dt}{t^2+n^2+1} + \sum_{i=1}^k \int_{-\infty}^{\infty} \frac{dt}{t^2+n^2+1} + \int_{-\infty}^{\tan n} \frac{dt}{t^2+n^2+1} \\ =\frac{(k+1)\pi}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{\tan n}{\sqrt{n^2+1}}\right).\end{aligned}

So, since $\tan^{-1}$ is bounded and $\displaystyle n-\frac{\pi}{2} < (k+1)\pi < n + \frac{\pi}{2},$ we have

$\displaystyle \lim_{n\to\infty} I_n=\lim_{n\to\infty} \frac{(k+1)\pi}{\sqrt{n^2+1}}=1. \ \Box$

# Find the function! (2)

Problem. Let $f:(0,\infty)\rightarrow \mathbb{R}$ be an increasing function such that $\displaystyle f(x)>-\frac{1}{x}$ and $\displaystyle f(x) f \left(f(x)+\frac{1}{x}\right)=1$ for all $x > 0.$ Find $f.$

Solution. I show that $\displaystyle f(x)=\frac{1-\sqrt{5}}{2x}$ for all $x > 0.$ Let $\displaystyle y:=f(x)+\frac{1}{x} > 0.$ Then $\displaystyle f(y)f\left(f(y)+\frac{1}{y}\right)=1$ and that, since $f(x)f(y)=1,$ gives $\displaystyle f\left(f(y)+\frac{1}{y}\right)=f(x).$
So, since $f$ is injective, $\displaystyle f(y)+\frac{1}{y}=x$ and therefore $\displaystyle 1+\frac{f(x)}{y}=xf(x)$ because $f(x)f(y)=1.$ So $y+f(x)=xyf(x)$ and thus, since $\displaystyle y=f(x)+\frac{1}{x},$

$\displaystyle x(f(x))^2-f(x)-\frac{1}{x}=0,$

which gives two solutions

$\displaystyle f_1(x)=\frac{1 - \sqrt{5}}{2x}, \ \ \ f_2(x)=\frac{1+\sqrt{5}}{2x}.$

Now since, for all $x > 0, \ f_1$ is negative and increasing and $f_2$ is positive and decreasing and we want $f$ to be increasing, we must have $f(x) =f_1(x)$ for all $x > 0.$
The only thing left is to show that $\displaystyle f(x)= \frac{1-\sqrt{5}}{2x}$ satisfies the conditions given in the problem and that is easy. $\Box$

# Nice little problems: limit & continuity (2)

Problem 1. Let $\{a_n\}$ be a sequence with $|a_0| < 1$ and $\displaystyle a_n = \sqrt{\frac{1+a_{n-1}}{2}}, \ \ n \ge 1.$
Show that $\displaystyle \lim_{n\to \infty}4^n(1-a_n)=\frac{1}{2}\left(\cos^{-1}a_0\right)^2.$

Solution. A quick induction shows that $0 < a_n < 1$ for $n \ge 1.$ So we can write $a_n=\cos x_n,$ where $x_0 \in (0,\pi)$ and $x_n \in (0, \pi/2)$ for $n \ge 1.$
Now the relation $\displaystyle a_n=\sqrt{\frac{1+a_{n-1}}{2}}$ becomes $\displaystyle \cos x_n = \cos\left(\frac{x_{n-1}}{2}\right)$ and hence $\displaystyle x_n=\frac{x_{n-1}}{2},$ for $n \ge 1$ (don’t forget that $x_0/2 \in (0,\pi/2)$ and so $\cos(x_0/2) > 0$).
Therefore $\displaystyle x_n=\frac{x_0}{2^n}$ and, as a result, $\displaystyle \lim_{n\to\infty} x_n =0.$ Thus

$\displaystyle \lim_{n\to\infty}4^n(1-a_n)=\lim_{n\to\infty}2^{2n+1}\sin^2\left(\frac{x_n}{2}\right)=\lim_{n\to\infty}2^{2n+1}\left(\frac{x_n}{2}\right)^2$
$\displaystyle =\lim_{n\to\infty}2^{2n+1}\left(\frac{x_0}{2^{n+1}}\right)^2 =\frac{x_0^2}{2}=\frac{1}{2}\left(\cos^{-1}a_0\right)^2. \Box$

Problem 2. Let $a, b$ be two distinct real numbers and let $\{x_n\}$ be a sequence that satisfies the following condition

$x_nx_m +ab \le ax_n +bx_m, \ \ \forall \ m,n.$

Show that $\displaystyle \lim_{n\to\infty}\frac{(-1)^{n}n!x_n(x_n+1)}{n^n}=0.$

Solution. The limit is $0$ because the condition forces the sequence $\{x_n\}$ to be bounded (each $x_n$ has to be between $a,b$) and so, since $\displaystyle \lim_{n\to\infty}\frac{n!}{n^n}=0,$ the limit has to be zero too.
To see why each $x_n$ is between $a,b,$ put $m=n$ in the given condition. Then

$\displaystyle 0 \ge x_n^2-(a+b)x_n+ab=(x_n-a)(x_n-b)$

and that can’t happen unless $x_n$ is between $a,b. \ \Box$

Problem 3. Let $f:[0,1] \longrightarrow [0,\infty)$ be a continuous function. Show that

$\displaystyle \lim_{n\to\infty}\left(-n+\sum_{i=1}^ne^{\frac{1}{n}\cdot f\left(\frac{i}{n}\right)}\right)=\int_0^1 f(x) \ dx.$

Solution. Since $f$ is continuous, it’s bounded. So $0 \le f(x) \le M$ for some $M$ and all $x \in [0,1].$ Now

\displaystyle \begin{aligned} \sum_{i=1}^n \frac{f(i/n)}{n} \le \sum_{i=1}^n (e^{f(i/n)/n}-1)=\sum_{i=1}^n \sum_{k=1}^{\infty}\frac{(f(i/n))^k}{n^kk!}=\sum_{i=1}^n \frac{f(i/n)}{n} + \sum_{k=2}^{\infty} \sum_{i=1}^n\frac{(f(i/n))^k}{n^kk!} \end{aligned}

\displaystyle \begin{aligned}\le \sum_{i=1}^n \frac{f(i/n)}{n}+\sum_{k=2}^{\infty}\sum_{i=1}^n \frac{M^k}{n^kk!}=\sum_{i=1}^n \frac{f(i/n)}{n}+\sum_{k=2}^{\infty}\frac{M^k}{n^{k-1}k!} < \sum_{i=1}^n \frac{f(i/n)}{n}+\frac{e^M}{n}.\end{aligned}

So

\displaystyle \begin{aligned} \lim_{n\to\infty}\left(-n+\sum_{i=1}^ne^{\frac{1}{n}\cdot f\left(\frac{i}{n}\right)}\right)=\lim_{n\to\infty}\sum_{i=1}^n (e^{f(i/n)/n}-1)=\lim_{n\to\infty} \sum_{i=1}^n \frac{f(i/n)}{n} =\int_0^1 f(x) \ dx. \ \Box \end{aligned}

# Integral inequalities (4)

Problem. Let $\displaystyle f: [0,1] \longrightarrow \mathbb{R}$ be a differentiable function such that $|f'(x)| \le 1$ for all $x \in [0,1].$ For an integer $n \ge 0,$ let $\displaystyle I_n:=\int_0^1 x^nf(x) \ dx.$
Show that for any integer $n \ge 0,$

$\displaystyle |I_{n+1}| \le \frac{1}{(n+1)(n+2)}\sum_{k=0}^n\left((k+1)|I_k|+\frac{1}{k+2}\right)-\frac{1}{(n+2)(n+3)}.$

(We assume that $f'$ is integrable).

Solution. Integration by parts gives

$\displaystyle I_k=\frac{f(1)}{k+1}-\frac{1}{k+1}\int_0^1x^{k+1}f'(x) \ dx.$

So

\displaystyle \begin{aligned} \frac{1}{(n+1)(n+2)}\sum_{k=0}^n (k+1)I_k=\frac{f(1)}{n+2}-\frac{1}{(n+1)(n+2)}\int_0^1\sum_{k=0}^nx^{k+1}f'(x) \ dx \\ =I_{n+1}+\frac{1}{n+2}\int_0^1x^{n+2}f'(x) \ dx - \frac{1}{(n+1)(n+2)}\int_0^1\sum_{k=0}^nx^{k+1}f'(x) \ dx\end{aligned}

and therefore

$\displaystyle I_{n+1}=\frac{1}{(n+1)(n+2)}\left(\sum_{k=0}^n(k+1)I_k + \int_0^1\left(\sum_{k=0}^nx^{k+1}-(n+1)x^{n+2}\right)f'(x) \ dx \right).$

Thus, since

$\displaystyle \sum_{k=0}^n x^{k+1}-(n+1)x^{n+2}=\sum_{k=0}^n(x^{k+1}-x^{n+2}) \ge 0$

and $|f'(x)| \le 1,$ on $[0,1],$ we have

\displaystyle \begin{aligned} |I_{n+1}| \le \frac{1}{(n+1)(n+2)}\left(\sum_{k=0}^n(k+1)|I_k|+\int_0^1 \left(\sum_{k=0}^nx^{k+1}-(n+1)x^{n+2}\right) dx\right) \\ = \frac{1}{(n+1)(n+2)}\sum_{k=0}^n\left((k+1)|I_k|+\frac{1}{k+2}\right)-\frac{1}{(n+2)(n+3)}. \ \Box \end{aligned}

Example. Let’s check the above result for the function $f(x)=x, \ \ x \in [0,1].$
Then, for any integer $n \ge 0,$ we have $\displaystyle I_n:=\int_0^1 x^nf(x) \ dx = \frac{1}{n+2}$ and so

\displaystyle \begin{aligned} \frac{1}{(n+1)(n+2)}\sum_{k=0}^n\left((k+1)|I_k|+\frac{1}{k+2}\right)-\frac{1}{(n+2)(n+3)} \\ =\frac{1}{(n+1)(n+2)}\sum_{k=0}^n\left(\frac{k+1}{k+2}+\frac{1}{k+2}\right)-\frac{1}{(n+2)(n+3)} \\ =\frac{1}{n+2}-\frac{1}{(n+2)(n+3)}=\frac{1}{n+3}=I_{n+1}.\end{aligned}

Exercise. Show that if $f: [0,1] \longrightarrow \mathbb{R}$ satisfies the conditions given in the above problem and if $\displaystyle \left|\int_0^1 f(x) \ dx\right| \le \frac{1}{2},$ then $\displaystyle \left|\int_0^1x^nf(x) \ dx\right| \le \frac{1}{n+2}$ for all integers $n \ge 0.$
Hint. Induction!

# An application of linear algebra in Calculus

This post is for those who have some knowledge of linear algebra.

Definition. For an integer $n \ge 1,$ the $n\times n$ Hilbert matrix is defined by $H_n=[a_{ij}],$ where

$\displaystyle a_{ij}=\frac{1}{i+j-1}, \ \ 1 \le i,j \le n.$

It is known that $H_n$ is invertible and if $H_n^{-1}=[b_{ij}],$ then $\displaystyle \sum_{i,j}b_{ij}=n^2.$ We are going to use these two properties of Hilbert matrices to solve the following calculus problem.

Problem. Let $n \ge 1$ be an integer and let $f : [0,1] \longrightarrow \mathbb{R}$ be a continuous function. Suppose that $\displaystyle \int_0^1 x^kf(x) \ dx = 1$ for all  $0 \le k \le n-1.$ Show that $\displaystyle \int_0^1 (f(x))^2 dx \ge n^2.$

Solution. Since $H_n,$ the $n\times n$ Hilbert matrix, is invertible, there exist real numbers $p_0, p_1, \cdots , p_{n-1}$ such that

$\displaystyle \sum_{i=1}^n\frac{p_{i-1}}{i+j-1}=1, \ \ \ 1 \le j \le n.$

So the polynomial $\displaystyle p(x)=\sum_{k=0}^{n-1}p_kx^k$ satisfies the conditions

$\displaystyle \int_0^1x^k p(x) \ dx =1, \ \ \ 0 \le k \le n-1.$

Clearly $\displaystyle \sum_{k=0}^{n-1}p_k$ is the sum of all the entries of $H_n^{-1}$ and so $\displaystyle \sum_{k=0}^{n-1}p_k=n^2.$ Now let $f$ be a real-valued continuous function on $[0,1]$ such that

$\displaystyle \int_0^1x^kf(x) \ dx = 1, \ \ \ 0 \le k \le n-1.$

Let $p(x)$ be the above polynomial.Then since

$\displaystyle (f(x))^2-2f(x)p(x)+(p(x))^2 =(f(x)-p(x))^2 \ge 0,$

integrating gives

\displaystyle \begin{aligned} \int_0^1 (f(x))^2dx \ge 2\int_0^1f(x)p(x) \ dx -\int_0^1(p(x))^2dx=2\sum_{k=0}^{n-1}p_k \int_0^1 x^kf(x) \ dx - \\ \sum_{k=0}^{n-1}p_k\int_0^1x^kp(x) \ dx = 2\sum_{k=0}^{n-1}p_k-\sum_{k=0}^{n-1}p_k=\sum_{k=0}^{n-1}p_k =n^2. \ \Box \end{aligned}

# Scary looking integrals (1)

For a general form of the following problem, see Exercise 3.

Problem. Show that

$\displaystyle I:=\int_0^{2\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx=2\pi e^{2/3}.$

Solution. We make the substitution $\displaystyle \tan\left(\frac{x}{2}\right)=2\tan\left(\frac{t}{2}\right).$ Then

$\displaystyle I=2\int_0^{\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx=8e^{5/8}\int_0^{\pi}\exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)\frac{dt}{5-3\cos t}. \ \ \ \ \ (1)$

Now, since $\displaystyle \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)$ is the real part of

$\displaystyle \exp\left(\frac{\cos t}{8}\right)\exp\left(\frac{i\sin t}{8}\right)=\exp\left(\frac{1}{8}e^{it}\right)$

and

$\displaystyle \exp\left(\frac{1}{8}e^{it}\right)=\sum_{n=0}^{\infty}\frac{1}{8^nn!}e^{nit}=\sum_{n=0}^{\infty}\frac{\cos(nt)}{8^nn!}+i\sum_{n=0}^{\infty} \frac{\sin(nt)}{8^nn!},$

we have

$\displaystyle \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)=\sum_{n=0}^{\infty} \frac{\cos(nt)}{8^nn!}.$

Thus, by $(1),$

$\displaystyle I=8e^{5/8}\sum_{n=0}^{\infty}\frac{1}{8^nn!}\int_0^{\pi} \frac{\cos(nt)}{5-3\cos t} \ dt=\frac{4}{3}e^{5/8}\sum_{n=0}^{\infty}\frac{(-1)^n}{8^nn!}\int_0^{2\pi} \frac{\cos(nt)}{\cos t + \frac{5}{3}} \ dt. \ \ \ \ \ \ \ (2)$

But here (Problem 2) I showed that

$\displaystyle \int_0^{2\pi} \frac{\cos(nt)}{\cos t + c} \ dt=\frac{2\pi}{\sqrt{c^2-1}}(-c+\sqrt{c^2-1})^n,$

for integers $n \ge 0$ and real numbers $c > 1.$ Thus

$\displaystyle \int_0^{2\pi} \frac{\cos(nt)}{\cos t + \frac{5}{3}} \ dt=(-1)^n\frac{\pi}{2 \cdot 3^{n-1}}$

and so, by $(2),$

$\displaystyle I=2\pi e^{5/8} \sum_{n=0}^{\infty} \frac{1}{24^nn!}=2\pi e^{5/8}e^{1/24}=2\pi e^{2/3}. \ \Box$

Exercise 1 (for those who are familiar with complex analysis).
Show that $\displaystyle \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right)$ is the real part of $\displaystyle \exp\left(\frac{e^{ix}+2}{e^{ix}+3}\right).$ Now apply Cauchy’s integration formula to give another solution for the above problem.

Exercise 2. Show that

$\displaystyle \int_0^{2\pi} \exp\left(\frac{3+3\cos x}{5+4\cos x}\right) \cos \left( \frac{\sin x}{5+4\cos x} \right) dx=2\pi \sqrt{e}.$

Exercise 3. Given real numbers $a,b$ with $b \ne \pm 1,$ let

$\displaystyle I(a,b):=\int_0^{2\pi}\exp\left(\frac{1+ab+(a+b)\cos x}{b^2+1+2b\cos x}\right) \cos\left(\frac{(b-a)\sin x}{b^2+1+2b\cos x}\right)dx.$

Show that

$\displaystyle I(a,b)=\begin{cases} 2\pi e & \text{if} \ |b| < 1 \\ 2\pi e^{a/b} & \text{if} \ |b| > 1.\end{cases}$

Hint. Make the substitution $\displaystyle \tan\left(\frac{x}{2}\right)=\left|\frac{b+1}{b-1}\right|\tan \left(\frac{t}{2}\right)$ and just do as I did in the solution of the above problem.

Remark. We also have $I(a,\pm 1)=2\pi e^{\min\{1,\pm a\}}.$ That quickly follows if we make the substitution $\displaystyle \tan\left(\frac{\pi}{2}\right)=t$ and use the identity $\displaystyle \int_0^{\infty} \frac{\cos(\alpha x)}{x^2+1} \ dx=\frac{\pi}{2}e^{-|\alpha|},$ which holds for all real numbers $\alpha$ but I have not discussed it in this blog yet.