# Math Challenge - January 2020

• Challenge
• Featured
Mentor

## Summary:

Useful Equations And Inequalities
Calculus
Functional Analysis
Small Groups

## Main Question or Discussion Point

Questions

1.
(solved by @PeroK ) Let $f\, : \,\mathbb{R}\longrightarrow \mathbb{R}$ be a smooth, $2\pi-$periodic function with square integrable derivative, and $\displaystyle{\int_0^{2\pi}}f(x)\,dx = 0\,.$ Prove
$$\int_0^{2\pi} \left[f(x)\right]^2\,dx \leq \int_0^{2\pi} \left[f\,'(x)\right]^2\,dx$$
For which functions does equality hold?

2. (solved by @Antarres ) $M$ be the set of all nonnegative, convex functions $f\, : \,[0,1]\longrightarrow \mathbb{R}$ with $f(0)=0\,.$ Prove
$$\int_0^1 \prod_{k=1}^n f_k(x)\,dx \geq \dfrac{2^n}{n+1}\prod_{k=1}^n \int_0^1 f_k(x)\,dx\quad \forall \;f_1,\ldots,f_n \in M$$
Hint: Define and use $\hat{f}(x)=2x\displaystyle{\int_0^1 f(x)\,dx}\,.$

3. (solved by @Antarres ) Consider the following differential operators on the space of smooth functions $C^\infty(\mathbb{R})$
$$A=2x\cdot \dfrac{d}{dx}\, , \,B=x^2\cdot \dfrac{d}{dx}\, , \,C=-\dfrac{d}{dx}$$
Determine the eigenvectors and a (multiplicative) structure on .

4. (solved by @Antarres ) Prove
$$\sum_{k=0}^n \dfrac{1}{\binom{n}{k}}=\dfrac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\dfrac{2^k}{k}$$

5. (solved by @Antarres ) Calculate
$$\displaystyle{\sum_{k=1}^\infty} \dfrac{1}{\binom{2k}{2}}$$

6. (solved by @Antarres ) Prove for $b>0$
$$\int_{-\infty}^{\infty} f\left(x-\dfrac{b}{x}\right)\,dx=\int_{-\infty}^{\infty}f(x)\,dx$$

7. (solved by @Antarres ) Let be a periodic function. Prove that
$$\int_{-\infty}^{\infty} f(x)\,\dfrac{\sin x}{x}\,dx = \int_{0}^{\pi} f(x)\,dx \text{ and }\int_{-\infty}^{\infty} f(x)\,\dfrac{\tan x}{x}\,dx =\int_{0}^{\pi} f(x)\,dx$$
assuming the integrals exist, i.e. are finite.

8. (a) (solved by @fishturtle1 ) If $\varphi\, : \,G\longrightarrow H$ is a homomorphism of finite groups, then $\operatorname{ord}(\varphi(g))\,|\,\operatorname{ord}(g)$ for all elements $g\in G\,.$
8. (b) (solved by @fishturtle1 ) Determine all group homomorphisms $\varphi\, : \,\mathbb{Z}_4 \longrightarrow \operatorname{Sym}(3)$ and $\psi\, : \,\operatorname{Sym}(3)\longrightarrow \mathbb{Z}_4\,.$

9. Let $\mathbf{a}=(a_1,\ldots,a_n) \in \mathbb{R}_{\geq 0}$ be nonnegative real numbers. The elementary symmetric polynomials are
$$\sigma_k(\mathbf{a}) =\sum_{1\leq j_1<\ldots <j_k\leq n}a_{j_1}a_{j_2}\ldots a_{j_k}$$
and
$$S_k(\mathbf{a}) =\dfrac{1}{\binom{n}{k}}\cdot \sigma_k(\mathbf{a})$$
the corresponding elementary symmetric mean value. Prove
(a) $S_1(\mathbf{a})\geq \sqrt{S_2(\mathbf{a})}\geq \sqrt{S_3(\mathbf{a})}\geq \ldots \geq \sqrt[n]{S_n(\mathbf{a})}$
(b) $S_m(\mathbf{a})^2\geq S_{m+1}(\mathbf{a})\cdot S_{m-1}(\mathbf{a})$ for $m=1,\ldots,n-1$

10. Let $(a_n)_{n\in \mathbb{N}}$ be a sequence of nonnegative real numbers, not all zero. Prove
$$\left(\sum_{n\in \mathbb{N}}a_n\right)^4 < \pi^2 \sum_{n\in \mathbb{N}}a_n^2 \cdot \sum_{n\in \mathbb{N}}n^2a_n^2$$ High Schoolers only

11.
(solved by @etotheipi ) On how many ways can $2020$ be written as a sum of consecutive natural numbers (greater than zero)?

12. (solved by @Not anonymous ) A binary operation on a set is a mapping, which maps a pair from to . E.g. addition is a binary operation on integers. Find two different (i.e. not achievable be renaming the elements) binary operations for which have a neutral element, , and can be inverted: for all there is a with , and are associative: A binary operation on a set $S$ is a mapping, which maps a pair from $S\times S$ to $S$. E.g. addition is a binary operation on integers. Find two different binary operations for $S=\{\,A,B,C,D\,\}$ which have a neutral element, $A\circ X = X$, and can be inverted: for all $X\in S$ there is a $Y\in S$ with $X\circ Y=A$, and are associative: $X\circ (Y\circ Z)=(X\circ Y)\circ Z\,.$

13. (solved by @Not anonymous ) Find all six digit numbers with the following property: If we move the first (highest) digit at the end, we will get three times the original number.

14. (solved by @etotheipi ) The Pell sequence named after the English mathematician John Pell is defined by
$$P(n)= \begin{cases} 0\, , \,n=0\\1 \, , \,n=1\\ P(n-2)+2P(n-1)\, , \,n>1 \end{cases}$$
Calculate the limit $\delta_s:=\lim_{n\to \infty}\dfrac{P(n)}{P(n-1)}\,.$

15. (solved by @Not anonymous ) Consider the graph of $f(x)=1/x$ with $x\geq 1$ and let it rotate around the $x-$axis. This solid of revolution looks like an infinitely long trumpet. Calculate its volume $V$ and its surface $A$.
If we fill it with paint, pour it out again, then we have painted it from inside. Explain this apparent contradiction to the surface you computed.

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• • etotheipi, YoungPhysicist, Greg Bernhardt and 2 others

berkeman
Mentor
Sorry for the dumb question on #1, but...
Would the TRI waveform qualify for the conditions of question? It seems like the ABS of the derivative of the TRI waveform would be larger over that interval than the TRI waveform itself. I'll work out the integral tomorrow if needed. Thanks for the great threads @fresh_42 Mentor
Sorry for the dumb question on #1, but...
Would the TRI waveform qualify for the conditions of question? It seems like the ABS of the derivative of the TRI waveform would be larger over that interval than the TRI waveform itself. I'll work out the integral tomorrow if needed. Thanks for the great threads @fresh_42 What is the TRI waveform? Triangle waveform? Triangles aren't smooth and the first derivative doesn't exist at the peak.

• berkeman
$$\sum_{k=1}^\infty \frac{1}{{2k}\choose{2}} = 2\sum_{k=1}^\infty \frac{1}{2k(2k-1)} = 2\sum_{k=1}^\infty \left(\frac{1}{2k-1} - \frac{1}{2k}\right) = 2\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = 2\ln{2}$$
In the last equality, we used the Taylor series for natural logarithm and plugged $x = 1$ in it:
$$\ln(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$$

I'll take a shot at #8...

a) Proof: Let $n = \operatorname{ord}(\varphi(g))$ and $m = \operatorname{ord}(g)$. Note, homomorphism maps identity to identity. So, $(\varphi(g)^m = \varphi(g^m) = \varphi(1) = 1$. Assume by contradiction that $n \not\mid m$. Then $m = nq + r$ for integers $q, r$ where $0 < r < n$. Observe, $\varphi(g)^m = \varphi(g)^{nq+r} = \varphi(g)^{nq}\varphi(g)^r = 1\cdot\varphi(g)^r = \varphi(g)^r = 1$. But $0 < r < n$ which contradicts our assumption that $n = \operatorname{ord}(\varphi(g))$. We can conclude $n \mid m$. []

b) Let $S(3) = \operatorname{Sym}(3)$. First, suppose $\varphi: \mathbb{Z}_4 \rightarrow S(3)$ is a homomorphism. Well, $\langle 1 \rangle = \mathbb{Z}_4$, so wherever $1$ goes will determine the homomorphism. Also, $\operatorname{ord}(1) = 4$. So $1 \mapsto (1), (12), (13),$ or $(23)$. This gives us $4$ possible homomorphisms. Clearly, $1 \mapsto (1)$ gives a homomorphism. Let $\sigma$ be any transposition of $S(3)$ and $x, y \in \mathbb{Z}_4$. Suppose $1 \mapsto \sigma$. Then
$$\varphi(x)\varphi(y) = \varphi(1)^x\varphi(1)^y = \sigma^x\sigma^y = \sigma^{x+y} = \varphi(x+y)$$
We can conclude $\varphi$ is a homomorphism.

Now suppose $\psi: S(3) \rightarrow \mathbb{Z}_4$ is a homomorphism. Well, $\operatorname{ord}((123)) = 3$. By part a), we can conclude $(123) \mapsto 0$. Aswell, $(132) \mapsto 0$. Let $\sigma, \gamma$ be any transpositions of $S(3)$. By part a), $\sigma \mapsto 0$ or $2$. Now, we note $\gamma$ and $\sigma$ share exactly one common element. This implies $\sigma\gamma$ is a $3-$ cycle which means it has order $3$. Therefore, if $\sigma \mapsto 0$ and $\gamma \mapsto 2$, then $\psi(\sigma) + \psi(\gamma) = 2$ while $\psi(\sigma\cdot\gamma) = 0$. It follows that all transpositions map to $0$ or all transpositions map to $2$.

We have two functions to verify are homomorphisms. We have $1, (123), (132) \mapsto 0$ necessarily. Clearly, if all the transpositions map to $0$, then we have the identity function which is a homomorphism. So, suppose all transpositions map to $2$. Let $\sigma, \gamma$ be transpositions and $\alpha, \beta$ be $3-$ cycles. Indeed,
$$\psi(\sigma\cdot\gamma) = 0 = 2 + 2 = \psi(\sigma) + \psi(\gamma)$$
$$\psi(\sigma\cdot\alpha) = 2 = 2 + 0 = \psi(\sigma) + \psi(\alpha)$$
$$\psi(\alpha\cdot\beta) = 0 = 0 + 0 = \psi(\alpha) + \psi(\beta)$$
It follows that $\psi$ is a homomorphism. []
[\SPOILER]

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Mentor
$$\sum_{k=1}^\infty \frac{1}{{2k}\choose{2}} = 2\sum_{k=1}^\infty \frac{1}{2k(2k-1)} = 2\sum_{k=1}^\infty \left(\frac{1}{2k-1} - \frac{1}{2k}\right) = 2\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = 2\ln{2}$$
In the last equality, we used the Taylor series for natural logarithm and plugged $x = 1$ in it:
$$\ln(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$$
Correct. You have been lucky, a typo made this one an easy one.

Mentor
I'll take a shot at #8...

a) Proof: Let $n = \operatorname{ord}(\varphi(g))$ and $m = \operatorname{ord}(g)$. Note, homomorphism maps identity to identity. So, $(\varphi(g)^m = \varphi(g^m) = \varphi(1) = 1$. Assume by contradiction that $n \not\mid m$. Then $m = nq + r$ for integers $q, r$ where $0 < r < n$. Observe, $\varphi(g)^m = \varphi(g)^{nq+r} = \varphi(g)^{nq}\varphi(g)^r = 1\cdot\varphi(g)^r = \varphi(g)^r = 1$. But $0 < r < n$ which contradicts our assumption that $n = \operatorname{ord}(\varphi(g))$. We can conclude $n \mid m$. []

b) Let $S(3) = \operatorname{Sym}(3)$. First, suppose $\varphi: \mathbb{Z}_4 \rightarrow S(3)$ is a homomorphism. Well, $\langle 1 \rangle = \mathbb{Z}_4$, so wherever $1$ goes will determine the homomorphism. Also, $\operatorname{ord}(1) = 4$. So $1 \mapsto (1), (12), (13),$ or $(23)$. This gives us $4$ possible homomorphisms. Clearly, $1 \mapsto (1)$ gives a homomorphism. Let $\sigma$ be any transposition of $S(3)$ and $x, y \in \mathbb{Z}_4$. Suppose $1 \mapsto \sigma$. Then
$$\varphi(x)\varphi(y) = \varphi(1)^x\varphi(1)^y = \sigma^x\sigma^y = \sigma^{x+y} = \varphi(x+y)$$
We can conclude $\varphi$ is a homomorphism.

Now suppose $\psi: S(3) \rightarrow \mathbb{Z}_4$ is a homomorphism. Well, $\operatorname{ord}((123)) = 3$. By part a), we can conclude $(123) \mapsto 0$. Aswell, $(132) \mapsto 0$. Let $\sigma, \gamma$ be any transpositions of $S(3)$. By part a), $\sigma \mapsto 0$ or $2$. Now, we note $\gamma$ and $\sigma$ share exactly one common element. This implies $\sigma\gamma$ is a $3-$ cycle which means it has order $3$. Therefore, if $\sigma \mapsto 0$ and $\gamma \mapsto 2$, then $\psi(\sigma) + \psi(\gamma) = 2$ while $\psi(\sigma\cdot\gamma) = 0$. It follows that all transpositions map to $0$ or all transpositions map to $2$.

We have two functions to verify are homomorphisms. We have $1, (123), (132) \mapsto 0$ necessarily. Clearly, if all the transpositions map to $0$, then we have the identity function which is a homomorphism. So, suppose all transpositions map to $2$. Let $\sigma, \gamma$ be transpositions and $\alpha, \beta$ be $3-$ cycles. Indeed,
$$\psi(\sigma\cdot\gamma) = 0 = 2 + 2 = \psi(\sigma) + \psi(\gamma)$$
$$\psi(\sigma\cdot\alpha) = 2 = 2 + 0 = \psi(\sigma) + \psi(\alpha)$$
$$\psi(\alpha\cdot\beta) = 0 = 0 + 0 = \psi(\alpha) + \psi(\beta)$$
It follows that $\psi$ is a homomorphism. []
[\SPOILER]
Correct, except that the trivial homomorphism cannot be called identity homomorphism! $\psi(\pi) \equiv $ means $\psi \equiv 0$ whereas $\psi = 1$ is usually preserved for $\psi = id$ which makes no sense here, and even less to call it identity function.

A short remark: The homomorphisms $\psi \, : \,S(3)\longrightarrow \mathbb{Z}_4$ are all induced by the signature function on $S(3)$.

• fishturtle1
Summary:: Useful Equations And Inequalities
Calculus
Functional Analysis
Small Groups

Questions

1.
Let $f\, : \,\mathbb{R}\longrightarrow \mathbb{R}$ be a smooth, $2\pi-$periodic function with square integrable derivative, and $\displaystyle{\int_0^{2\pi}}f(x)\,dx = 0\,.$ Prove
$$\int_0^{2\pi} \left[f(x)\right]^2\,dx \leq \int_0^{2\pi} \left[f\,'(x)\right]^2\,dx$$
For which functions does equality hold?
Regarding question #1: what do you mean by smooth? if you mean differentiable then $f'$ is trivialy squrare integrable.

Mentor
Regarding question #1: what do you mean by smooth? if you mean differentiable then $f'$ is trivialy squrare integrable.
Yes. But it was faster written than thought about.

• Yonatan N
Correct, except that the trivial homomorphism cannot be called identity homomorphism! $\psi(\pi) \equiv $ means $\psi \equiv 0$ whereas $\psi = 1$ is usually preserved for $\psi = id$ which makes no sense here, and even less to call it identity function.

A short remark: The homomorphisms $\psi \, : \,S(3)\longrightarrow \mathbb{Z}_4$ are all induced by the signature function on $S(3)$.
Thank you for the correction and your insight!

Infrared
Gold Member
Regarding question #1: what do you mean by smooth? if you mean differentiable then $f'$ is trivialy squrare integrable.
Why? $f'$ existing does not imply that $f'$ is square integrable. I think smooth should mean $C^1$ here.

Why? $f'$ existing does not imply that $f'$ is square integrable. I think smooth should mean $C^1$ here.
you're right etotheipi
Gold Member
2019 Award
Suppose the first number is $n$ and the final number is $m$. We then require that $$\sum_{i=n}^{m} i = \frac{m(m+1)}{2} - \frac{n(n-1)}{2} = 2020$$ By multiplying through by 2 and expanding the brackets, we get $$(m^{2} - n^{2}) + (m+n) = 4040 \implies (m+n)(m-n+1) = 4040$$ Since both $(m+n)$ and $(m-n+1)$ are integers, they must be factor pairs of 4040. Note also that for $n>0$, $(m+n) > (m-n+1)$.

The possible pairs $(m+n, m-n+1)$ are then $(4040,1), (2020, 2), (1010, 4), (808, 5), (505, 8), (404, 10), (202, 20), (101, 40)$.

Furthermore, the sum $(m+n) + (m-n+1) = 2m+1$ is odd, so we also want only the factor pairs which sum to an odd number. We are then left with the pairs $(4040,1), (808, 5), (505, 8), (101, 40)$, each of which could then be solved simultaneously to find the endpoints.

So we have 4 ways, if we also include the boring one which only contains 2020!

• PeroK, jim mcnamara and fresh_42
etotheipi
Gold Member
2019 Award
The recurrence relation $P(n) - 2P(n-1) - P(n-2) = 0$ can be solved using a guess of $P(n) = \lambda^{n}$,

$\lambda^{n} -2\lambda^{n-1} - \lambda^{n-2} = 0 \implies \lambda^{2} -2\lambda -1 = 0$

so we obtain $\lambda = 1\pm \sqrt{2}$ which gives the general solution of

$P(n) = A(1+\sqrt{2})^{n} + B(1-\sqrt{2})^{n}$

We can determine the coefficients as follows,

$P(0) = A + B = 0 \implies A = -B$
$P(1) = (A+B) + \sqrt{2}(A-B) =1 \implies A=\frac{1}{2\sqrt{2}}, B = -\frac{1}{2\sqrt{2}}$

So we end up with the following equation for the $n^{th}$ Pell number

$P(n) = \frac{(1+\sqrt{2})^{n} + (1-\sqrt{2})^{n}}{2\sqrt{2}}$

Finally, since $|1-\sqrt{2}| < 1$, $\lim_{n \to \infty} (1-\sqrt{2})^{n} = 0$, and the limit in the question can be determined

$\delta_{s} = \lim_{n \to \infty} \frac{P(n)}{P(n-1)} = \lim_{n \to \infty} \frac{(1+\sqrt{2})^{n} + (1-\sqrt{2})^{n}}{(1+\sqrt{2})^{n-1} + (1-\sqrt{2})^{n-1}} = \lim_{n \to \infty} \frac{(1+\sqrt{2})^{n}}{(1+\sqrt{2})^{n-1}} = 1+\sqrt{2}$

• fresh_42
Math_QED
Homework Helper
2019 Award
The recurrence relation $P(n) - 2P(n-1) - P(n-2) = 0$ can be solved using a guess of $P(n) = \lambda^{n}$,

$\lambda^{n} -2\lambda^{n-1} - \lambda^{n-2} = 0 \implies \lambda^{2} -2\lambda -1 = 0$

so we obtain $\lambda = 1\pm \sqrt{2}$ which gives the general solution of

$P(n) = A(1+\sqrt{2})^{n} + B(1-\sqrt{2})^{n}$

We can determine the coefficients as follows,

$P(0) = A + B = 0 \implies A = -B$
$P(1) = (A+B) + \sqrt{2}(A-B) =1 \implies A=\frac{1}{2\sqrt{2}}, B = -\frac{1}{2\sqrt{2}}$

So we end up with the following equation for the $n^{th}$ Pell number

$P(n) = \frac{(1+\sqrt{2})^{n} + (1-\sqrt{2})^{n}}{2\sqrt{2}}$

Finally, since $|1-\sqrt{2}| < 1$, $\lim_{n \to \infty} (1-\sqrt{2})^{n} = 0$, and the limit in the question can be determined

$\delta_{s} = \lim_{n \to \infty} \frac{P(n)}{P(n-1)} = \lim_{n \to \infty} \frac{(1+\sqrt{2})^{n} + (1-\sqrt{2})^{n}}{(1+\sqrt{2})^{n-1} + (1-\sqrt{2})^{n-1}} = \lim_{n \to \infty} \frac{(1+\sqrt{2})^{n}}{(1+\sqrt{2})^{n-1}} = 1+\sqrt{2}$
There are probably some theorems about this but do we know something about the uniqueness of this solution? I bet there is a unique solution but this is implicitely used: another solution to the recurence relation could lead to another answer. @fresh_42

• etotheipi
13. Find all six digit numbers with the following property: If we move the first (highest) digit at the end, we will get three times the original number.
Any 6-digit positive integer can be represented as $x = 10^5a + b$ where $a \in \{1, 2, \ldots, 9\}, b \in \{0, 1, \ldots, 99999\}$. The number obtained by moving the first digit to the end is $y = 10b + a$. Since it is said that $y = 3x$, we get the equation $10b + a = 3a10^5 + 3b$, which in turn gives $7b = a(300000-1) = 299999a \Rightarrow b = 42857a$. We try values of $a$ from 1 to 9 and find that only the values 1 and 2 are valid since if $a \ge 3$, then $b$ exceeds its maximum acceptable value, 99999. So there are two 6-digit numbers that meet the mentioned condition: 141257 (corresponding to $a = 1$) and 285714 (corresponding to $a = 2$)

15. Consider the graph of $f(x)=1/x$ with $x \geq 1$ and let it rotate around the x-axis. This solid of revolution looks like an infinitely long trumpet. Calculate its volume $V$ and its surface $A$.
If we fill it with paint, pour it out again, then we have painted it from inside. Explain this apparent contradiction to the surface you computed.
To find the volume and surface area, we consider the solid formed by rotation as being made of a series of infinitesimally small cylinders whose central height (or length?) axes lie on the x-axis. For a cylinder lying between $x$ and $(x+\Delta x)$, the height is $\Delta x$ and radius is $1/x$, so its volume is $\pi (1/x)^2 \Delta x$ and surface area is $2 \pi (1/x) \Delta x$ (consider it to be hollow cylinder for surface area). Using these, we compute the volume and surface area of the full solid (for $x$ ranging from 1 to $\infty$) using integration.

$$V = \int_1^\infty \frac {\pi} {x^2} \,dx = \left. -\frac {\pi} {x} \right|_1^\infty = -\frac {\pi} {\infty} + \frac {\pi} {1} = \pi$$

$$A = \int_1^\infty \frac {2\pi} {x} \,dx = \left. 2 \pi \log x \right|_1^\infty = 2 \pi \infty - 2 \pi = \infty$$

I am not sure if I understood correctly the 2nd part of the question, "the apparent contradiction". Assuming that it refers to how a solid with finite volume has infinite surface area, I don't have a math proof, but I can say that intuitively it is related to the following reasons:
(1) surface area and volume do not measure the same thing and their measures are in different orders of magnitude (e.g. $m^2$ and $m^3$ for area and volume respectively if we use meter as the unit)
(2) The ratio of surface area to volume of a solid such as a cylinder or a sphere tends to infinity as the radius tends to 0. For a hollow cylinder of radius $r$ and height $h$, the ratio of surface area to enclosed volume is $\frac {2 \pi r h} {\pi r^2 h} = \frac {2} {r}$ and as the radius approaches 0, this ratio approaches infinity. In the trumpet-like solid in the question, the small constituent cylinders formed by rotating segments of the curve $f(x) = 1/x$ around x-axis have their radii tending to 0 as $x$ becomes larger and larger (the "mouthpiece" side of the trumpet) and for these, the surface area-to-volume ratio approaches infinity, which explains the apparent contradiction (because as we move towards $\infty$ on x-axis, a smaller volume of paint is able to cover a proportionally larger amount of surface area as compared to the situation for smaller values of $x$).

Mentor
There are probably some theorems about this but do we know something about the uniqueness of this solution? I bet there is a unique solution but this is implicitely used: another solution to the recurence relation could lead to another answer. @fresh_42
We have $\delta_s=2+\delta_s^{-1}$ which has a positive and a negative solution, $1\pm \sqrt{2}$. Since $P(n)> P(n-1)$ we are left with the positive one. I'm not sure I understood what you mean. It's basically the analog situation as with the golden ratio and the Fibonacci sequence.

Two quantities are in the silver ratio if the ratio of the sum of the smaller and twice the larger of those quantities, to the larger quantity, is the same as the ratio of the larger one to the smaller one. This leads to
$$\dfrac{2a+b}{a}=\dfrac{a}{b}=\delta_s =[2;2,2,2,\ldots ]$$
where the last representation is the continued fraction of the silver ratio.

Mentor
Any 6-digit positive integer can be represented as $x = 10^5a + b$ where $a \in \{1, 2, \ldots, 9\}, b \in \{0, 1, \ldots, 99999\}$. The number obtained by moving the first digit to the end is $y = 10b + a$. Since it is said that $y = 3x$, we get the equation $10b + a = 3a10^5 + 3b$, which in turn gives $7b = a(300000-1) = 299999a \Rightarrow b = 42857a$. We try values of $a$ from 1 to 9 and find that only the values 1 and 2 are valid since if $a \ge 3$, then $b$ exceeds its maximum acceptable value, 99999. So there are two 6-digit numbers that meet the mentioned condition: 141257 (corresponding to $a = 1$) and 285714 (corresponding to $a = 2$)
The idea is correct, the result is not. $10^5 \cdot 1 + 42,857\cdot 1 = 142,857 \neq 141,257$.
$285,714$ is correct.

Math_QED
Homework Helper
2019 Award
We have $\delta_s=2+\delta_s^{-1}$ which has a positive and a negative solution, $1\pm \sqrt{2}$. Since $P(n)> P(n-1)$ we are left with the positive one. I'm not sure I understood what you mean. It's basically the analog situation as with the golden ratio and the Fibonacci sequence.

Two quantities are in the silver ratio if the ratio of the sum of the smaller and twice the larger of those quantities, to the larger quantity, is the same as the ratio of the larger one to the smaller one. This leads to
$$\dfrac{2a+b}{a}=\dfrac{a}{b}=\delta_s =[2;2,2,2,\ldots ]$$
where the last representation is the continued fraction of the silver ratio.
What I mean: is there a theorem that garantuees that this equation has a unique solution? Because the question solver solves the equation and then uses the solution. What if there is more than one solution (probably there isn't, but what garuantees this)?

Mentor
To find the volume and surface area, we consider the solid formed by rotation as being made of a series of infinitesimally small cylinders whose central height (or length?) axes lie on the x-axis. For a cylinder lying between $x$ and $(x+\Delta x)$, the height is $\Delta x$ and radius is $1/x$, so its volume is $\pi (1/x)^2 \Delta x$ and surface area is $2 \pi (1/x) \Delta x$ (consider it to be hollow cylinder for surface area). Using these, we compute the volume and surface area of the full solid (for $x$ ranging from 1 to $\infty$) using integration.

$$V = \int_1^\infty \frac {\pi} {x^2} \,dx = \left. -\frac {\pi} {x} \right|_1^\infty = -\frac {\pi} {\infty} + \frac {\pi} {1} = \pi$$

$$A = \int_1^\infty \frac {2\pi} {x} \,dx = \left. 2 \pi \log x \right|_1^\infty = 2 \pi \infty - 2 \pi = \infty$$

I am not sure if I understood correctly the 2nd part of the question, "the apparent contradiction". Assuming that it refers to how a solid with finite volume has infinite surface area, I don't have a math proof, but I can say that intuitively it is related to the following reasons:
(1) surface area and volume do not measure the same thing and their measures are in different orders of magnitude (e.g. $m^2$ and $m^3$ for area and volume respectively if we use meter as the unit)
(2) The ratio of surface area to volume of a solid such as a cylinder or a sphere tends to infinity as the radius tends to 0. For a hollow cylinder of radius $r$ and height $h$, the ratio of surface area to enclosed volume is $\frac {2 \pi r h} {\pi r^2 h} = \frac {2} {r}$ and as the radius approaches 0, this ratio approaches infinity. In the trumpet-like solid in the question, the small constituent cylinders formed by rotating segments of the curve $f(x) = 1/x$ around x-axis have their radii tending to 0 as $x$ becomes larger and larger (the "mouthpiece" side of the trumpet) and for these, the surface area-to-volume ratio approaches infinity, which explains the apparent contradiction (because as we move towards $\infty$ on x-axis, a smaller volume of paint is able to cover a proportionally larger amount of surface area as compared to the situation for smaller values of $x$).
I have to say o.k. but all of my fibers protest. Please, please stop immediately to treat $\infty$ like a number! It isn't and you will sooner or later get into big trouble and produce unnecessary mistakes. The correct way to treat it is:
$$V=\int_1^\infty \dfrac{\pi}{x^2}\,dx = \pi \left[-\dfrac{1}{x}\right]_1^\infty = \pi \left(\left(-\lim_{x\to \infty}\dfrac{1}{x}\right)-(-1)\right)=\pi$$
The paradoxon I meant is the following:

The object is known as Gabriel's horn, and if we filled in paint, poured it out again, and thus have painted an infinitely large inner surface with a finite amount of paint!
The solution is, that in reality paint has a certain thickness, so we didn't need to "fill" the entire horn, only a finite part of it.

Mentor
What I mean: is there a theorem that garantuees that this equation has a unique solution? Because the question solver solves the equation and then uses the solution. What if there is more than one solution (probably there isn't, but what garuantees this)?
\begin{align*}
\delta_s &=\lim_{n \to \infty}\dfrac{P(n-2)+2P(n-1)}{P(n-1)}\\
&=\lim_{n \to \infty}\dfrac{P(n-2)}{P(n-1)}+2\\
&= 2+\delta_s^{-1}
\end{align*}
with two solutions from which only one holds for an increasing sequence starting positive.

Infrared
Gold Member
@fresh_42 I think that argument assumes the limit exists in the first place. Do you have a way to see this?

There are probably some theorems about this but do we know something about the uniqueness of this solution? I bet there is a unique solution but this is implicitely used: another solution to the recurence relation could lead to another answer.
Given the values of $P(0)$ and $P(1)$, the rest of the Pell numbers $P(n), n>1$ are clearly uniquely defined by the recurrence. @etotheipi found a sequence with the right values at $0$ and $1$ that satisfies the recurrence, so it must agree with the Pell sequence.

A little more formally (though this is not necessary), the set of real-valued sequences satisfying the Pell recurrence formula is two-dimensional, so every such sequence is a linear combination of the power solutions @etotheipi found.

• Math_QED
Mentor
@fresh_42 I think that argument assumes the limit exists in the first place. Do you have a way to see this?
You can use a convergence theorem for continued fractions applied to $\delta_s =[2;2,2,2,\ldots ]$ or much easier prove $2\leq \dfrac{P(n)}{P(n-1)} < 3$ per induction.

Math_QED
Homework Helper
2019 Award
@fresh_42 I think that argument assumes the limit exists in the first place. Do you have a way to see this?

Given the values of $P(0)$ and $P(1)$, the rest of the Pell numbers $P(n), n>1$ are clearly uniquely defined by the recurrence. @etotheipi found a sequence with the right values at $0$ and $1$ that satisfies the recurrence, so it must agree with the Pell sequence.

A little more formally (though this is not necessary), the set of real-valued sequences satisfying the Pell recurrence formula is two-dimensional, so every such sequence is a linear combination of the power solutions @etotheipi found.
Thanks this was actually very obvious. I'm not sure why I doubted this. Probably a lack of sleep due to exams!