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.

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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!

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.