Math Challenge - February 2021

In summary, the conversation covers various topics in mathematics such as Calculus, Measure Theory, Convergence, Infinite Series, Topology, Functional Analysis, Real Numbers, Algebras, Complex Analysis, and Group Theory. The first question discusses a differentiable function with no zeros in a given interval, while the second question involves a measure space and an integrable function. The third question presents multiple statements to prove or find counterexamples for. The fourth question deals with the convergence of a series and the fifth question talks about continuous functions on a Hausdorff space. Moving on, the sixth question explores Banach spaces and Cauchy sequences. The seventh question discusses sets in the topological space of real numbers. The eighth question involves representations
  • #1

fresh_42

Mentor
Insights Author
2023 Award
18,994
23,916
Summary: Calculus, Measure Theory, Convergence, Infinite Series, Topology, Functional Analysis, Real Numbers, Algebras, Complex Analysis, Group Theory1. (solved by @Office_Shredder ) Let ##f## be a real, differentiable function such that there is no ##x\in \mathbb{R}## with ##f(x)=0=f'(x)##.
Show that ##f## has at most finitely many zeros in the interval ##[0,1]##.

2. (solved by @wrobel ) Let ##(X,\Omega,\omega)## be a measure space and ##f## be a ##\omega-##integrable function.
Show that for every ##\varepsilon>0 ## there is a set ##W\in \Omega## such that ##\omega(W)< \infty ## and ##\int_{X-W}|f|\,d\omega <\varepsilon .##

3. (solved by @suremarc ) Prove or find a counterexample to:
(a) ##L^2## convergence implies pointwise convergence.
(b) ##\displaystyle{\lim_{n \to \infty}\int_{0}^{\infty}\dfrac{\sin x^n}{x^n}\,dx}=1##
(c) Let ##(f_n)## be a sequence of measurable functions which converge uniformly to zero on ##[0,\infty).## Then
$$
\lim_{n \to \infty}\int_{[0,\infty)} f_n(x)\,dx=\int_{[0,\infty)} \lim_{n \to \infty} f_n(x)\,dx\,.
$$

4. (solved by @julian ) Let ##(a_n)## be a sequence of positive real numbers such that the series ##\displaystyle{\sum_{n=1}^\infty}a_n=:C<\infty## converges. Show that ##\displaystyle{\sum_{n=1}^\infty\left(\prod_{k=1}^n a_k\right)^{1/n}}\leq e\cdot C.##

5. Let ##(E,\mathcal{T})## be a normal Hausdorff space, and ##U_1,\ldots,U_n## a finite open covering of ##E##. Then there are continuous functions ##g_1,\ldots,g_n\, : \,(E,\mathcal{T})\longrightarrow [0,1]## such that ##g_1+\ldots+g_n=1## on ##E## and ##g_j(E-U_j)=\{0\}## for all ##1\leq j\leq n.##

6. Let ##(X,\Omega,\omega)## be a measure space and ##1\leq p<\infty.## Show that
(a) ##\tilde{L}^p:=L^p(X,\Omega,\omega)## is a Banach space with respect to ##\|\cdot\|_p\,.##
(b) The sequence ##(\|f_n\|_p)\subseteq \mathbb{R}## is bounded for every Cauchy sequence ##(f_n)\subseteq L^p(X,\Omega,\omega).##

7. (solved by @nuuskur ) We know that there are only two sets in ##(\mathbb{R},|\cdot|),## which are open and closed, the empty set and the entire topological space. Prove it.

8. Let ##(V,\alpha)## and ##(W,\beta)## be irreducible representations of an associative, complex algebra ##\mathcal{A}##. Assume that ##V## and ##W## are complex and of countable dimension. Then
$$
\dim \operatorname{Hom}_{\mathcal{A}}(V,W) = \begin{cases}1&\text{ if }(V,\alpha)\cong(W,\beta)\\0&\text{ otherwise }\end{cases}
$$

9. (solved by @StoneTemplePython , @nuuskur ) Let ##U\subseteq\mathbb{C}## be an open connected neighborhood and ##f:U\longrightarrow\mathbb{C}## a holomorphic function. If ##|f|## has a local maximum in ##z_0\in U,## i.e. there is an open neighborhood ##z_0\in U_0\subseteq U## with ##|f(z_0)|\geq |f(z)|## for all ##z\in U_0,## then ##f## is constant.

10. (solved by @fishturtle1 ) Show that all groups ##G## of order ##pqr## with pairwise distinct primes ##p<q<r## are solvable.
1606835746499-png-png.png


High Schoolers only

11.
(solved by @Not anonymous ) Let ##z## be a natural number with ##1995## decimal digits and ##1\leq n\leq 1994.## Then we note the number, which we get by cutting off the first ##n## digits and append them in the same order at the end of ##z## by ##z^{[n]}.## Show that if ##z## is divisible by ##27,## then all ##z^{[n]}## are divisible by ##27,## too.

12. (solved by @Not anonymous ) Let ##a,b,c,d## be positive real numbers. Prove (in the logically correct order)
$$
\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{d}}\leq \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}
$$

13. (solved by @Not anonymous ) Let ##m\geq 2## be a given natural number. We define a sequence ##(x_0,x_1,x_2,\ldots)## of numbers by ##x_0=0,x_1=1,## and for ##n\geq 0## we set ##x_{n+2}## to be the remainder of ##x_{n+1}+x_n## by division by ##m,## chosen such that ##0\leq x_{n+2} < m.## Decide whether for every ##m\geq 2## there exists a natural number ##k\geq 1,## such that ##x_{k+2}=1\, , \,x_{k+1}=1\, , \,x_k=0.##

14. We define real functions
$$
f_n(x):=x^3+(n+3)\cdot x^2+2n\cdot x -\dfrac{n}{n+1}
$$
for every non-negative integer ##n\geq 0.## Determine all values of ##n,## such that all zeros of ##f_n(x)## are contained in an interval of length ##3.##

15. (a) (solved by @Not anonymous ) Determine the number of all pairs of integers ##(x,y) \in \mathbb{N}_0^2## with ##\sqrt{x}+\sqrt{y}=1993.##
15. (b) Determine for every ##n\in \mathbb{N}## the greatest power of ##2## which divides ##[( 4+\sqrt{18} )^n ].##
 
Last edited:
  • Love
Likes docnet
Physics news on Phys.org
  • #2
Problem #4:

A simple application of the AM-GM leads to the the Harmonic series which is divergent, so let us try to "weight" the terms first

\begin{align*}
(a_1 a_2 \cdots a_n)^{1/n} & = \frac{(c_1 a_1 c_2 a_2 \cdots c_n a_n)^{1/n}}{(c_1 c_2 \cdots c_n)^{1/n}} \\
& \leq \frac{c_1 a_1 + c_2 a_2 + \cdots + c_n a_n}{n (c_1 c_2 \cdots c_n)^{1/n}}
\end{align*}

where we have used the AM-GM inequality. Then

\begin{align*}
\sum_{n=1}^\infty (a_1 a_2 \cdots a_n)^{1/n} & \leq \sum_{n=1}^\infty \frac{1}{n (c_1 c_2 \cdots c_n)^{1/n}} (c_1 a_1 + c_2 a_2 + \cdots c_n a_n) \\
& = \sum_{n=1}^\infty \sum_{k=1}^n \frac{1}{n (c_1 c_2 \cdots c_n)^{1/n}} c_k a_k \\
& = \sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{n (c_1 c_2 \cdots c_n)^{1/n}} c_k a_k
\end{align*}

Let us write ##c_k = \frac{b_k}{b_{k-1}}## with ##c_1 = b_1##, then the above inequality simplifies:

\begin{align*}
\sum_{n=1}^\infty (a_1 a_2 \cdots a_n)^{1/n} & \leq \sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{n (b_n)^{1/n}} \frac{b_k}{b_{k-1}} a_k . \quad (*)
\end{align*}

We have the inequality ##1+x \leq e^x##. Replace ##x## by ##1/k## and then raise both sides to the power of ##k## and it gives the inequality:

\begin{align*}
\left( 1 + \frac{1}{k} \right)^k \leq e
\end{align*}

If we can choose ##b_k## so that:

\begin{align*}
\sum_{n=k}^\infty \frac{1}{n (b_n)^{1/n}} \frac{b_k}{b_{k-1}} = \left( 1 + \frac{1}{k} \right)^k
\end{align*}

then we are done. If we choose ##b_k = (1+k)^k## then

\begin{align*}
\frac{b_k}{b_{k-1}} = \frac{(1+k)^k}{k^{k-1}} = \frac{k (1+k)^k}{k^k}= k \left( 1 + \frac{1}{k} \right)^k
\end{align*}

and

\begin{align*}
\sum_{n=k}^\infty \frac{1}{n (b_n)^{1/n}} \frac{b_k}{b_{k-1}} & = \sum_{n=k}^\infty \frac{1}{n (n+1)} k \left( 1 + \frac{1}{k} \right)^k \\
& = \frac{1}{k} \cdot k \left( 1 + \frac{1}{k} \right)^k \\
& = \left( 1 + \frac{1}{k} \right)^k \quad (**)
\end{align*}

where we have used that

\begin{align*}
\sum_{n=k}^N \frac{1}{n(n+1)} = \sum_{n=k}^N \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{k} - \frac{1}{N+1} .
\end{align*}

To summarise, by substituting ##(**)## into ##(*)## we have:

\begin{align*}
\sum_{n=1}^\infty (a_1 a_2 \cdots a_n)^{1/n} & \leq \sum_{k=1}^\infty \left( 1 + \frac{1}{k} \right)^k a_k \\
& \leq e \cdot \sum_{k=1}^\infty a_k .
\end{align*}
 
  • Like
Likes benorin
  • #3
Suppose for a contradiction [itex]\emptyset\neq A\neq\mathbb R[/itex] is clopen. Take [itex]a\in A[/itex] and [itex]b\notin A[/itex]. We can assume [itex]a<b[/itex]. Put [itex]t := \sup A\cap [a,b][/itex] (it exists, because [itex]b[/itex] is an upper bound). Take [itex]x_n\in A\cap [a,b][/itex] such that [itex]x_n\to t[/itex]. Since [itex]A\cap [a,b][/itex] is closed, we have [itex]t\in A\cap [a,b][/itex]. In particular [itex]t<b\notin A[/itex]. Then [itex](t,b] \subseteq \mathbb R\setminus A[/itex] must hold, but that contradicts openness of [itex]A[/itex].
 
  • #4
item 2:

By definition of the Lebesgue integral for any ##\epsilon>0## there exists a simple function ##\phi,\quad 0\le \phi\le |f|## such that

$$\int_X|f|d\mu-\epsilon\le \int_X\phi d\mu\le \int_X|f|d\mu.$$

Since ##\phi## is a simple function we have

$$\int_X\phi=\int_W\phi,\quad \mu(W)<\infty.$$

It remains to write

$$\int_W|f|d\mu+\int_{X\backslash W}|f|d\mu-\epsilon\le \int_W\phi d\mu\le\int_W|f|d\mu$$
 
  • #5
wrobel said:
item 2:

By definition of the Lebesgue integral for any ##\epsilon>0## there exists a simple function ##\phi,\quad 0\le \phi\le |f|## such that

$$\int_X|f|d\mu-\epsilon\le \int_X\phi d\mu\le \int_X|f|d\mu.$$

Since ##\phi## is a simple function we have

$$\int_X\phi=\int_W\phi,\quad \mu(W)<\infty.$$

It remains to write

$$\int_W|f|d\mu+\int_{X\backslash W}|f|d\mu-\epsilon\le \int_W\phi d\mu\le\int_W|f|d\mu$$
Please define ##\phi ## and ##W## explicitly and prove the properties. Your argument is a little bit like: ... because of the definition of Lebesgue integrals. Sure, that's why the statement is true. And where did you use the properties of a measure?
 
  • #6
fresh_42 said:
Please define and explicitly and prove the properties
There is no need to define ##\phi## explicitly. As I said the existence of such a ##\phi## is a part of definition of the Lebesgue integral. See Folland: Real Analysis: Modern Techniques and Their Applications.
##W## is also clear ##W=\{x\in X\mid\phi(x)>0\}.## Since ##\phi(X)## is a finite set it follows that
$$W=\{x\in X\mid\phi(x)\ge c\},\quad c=\min_W\phi>0.$$
(If ##f=0## then there is nothing to speak about)
If ##\mu(W)=\infty## then ##\int_X\phi d\mu=\int_W\phi d\mu\ge c\mu(W)=\infty.## But it is not so. Thus ##\mu(W)<\infty##
 
  • #7
wrobel said:
There is no need to define ##\phi## explicitly. As I said the existence of such a ##\phi## is a part of definition of the Lebesgue integral. See Folland: Real Analysis: Modern Techniques and Their Applications.
##W## is also clear ##W=\{x\in X\mid\phi(x)>0\}.## Since ##\phi(X)## is a finite set it follows that
$$W=\{x\in X\mid\phi(x)\ge c\},\quad c=\min_W\phi>0.$$
(If ##f=0## then there is nothing to speak about)
If ##\mu(W)=\infty## then ##\int_X\phi d\mu=\int_W\phi d\mu\ge c\mu(W)=\infty.## But it is not so. Thus ##\mu(W)<\infty##
Your idea stands and falls with ##0\leq \phi \leq |f|##. Hence I want to see a prove for the existence and its properties. The same holds for ##W##. Because it is Lebesgue theory is not sufficient.
 
  • #8
fresh_42 said:
Hence I want to see a prove for the existence and its properties.
Screen21-02-02 21-23-51.png


fresh_42 said:
Because it is Lebesgue theory is not sufficient.
do not understand what this means
 
  • #9
Problem 1

Suppose there are infinitely many zeros. Let ##a_n## be an infinite sequence of them. Since [0,1] is compact, ##a_n## must have a convergent subsequence, call it ##b_n## and let the limit be ##b##. Since f is continuous, ##f(b) = \lim_{n\to \infty} f(b_n) = 0##.

f is differentiable at b, and since the limit defining the derivative exists we can evaluate if picking and sequence that converges to zero. In particular, for ##h_n=b_n -b##,
$$f'(b)= \lim_{n\to \infty} \frac{f(b+h_n)-f(b)}{h_n}.$$

But ##f(b)=0## and ##f(b+h_n)=f(b_n)=0##, so we get that ##f'(b)=0##. Thus if f has infinitely many zeros, there must exist some point ##b## such that ##f(b)=f'(b)=0##.
 
  • Like
Likes PeroK
  • #10
Office_Shredder said:
Problem 1

Suppose there are infinitely many zeros. Let ##a_n## be an infinite sequence of them. Since [0,1] is compact, ##a_n## must have a convergent subsequence, call it ##b_n## and let the limit be ##b##. Since f is continuous, ##f(b) = \lim_{n\to \infty} f(b_n) = 0##.

f is differentiable at b, and since the limit defining the derivative exists we can evaluate if picking and sequence that converges to zero. In particular, for ##h_n=b_n -b##,
$$f'(b)= \lim_{n\to \infty} \frac{f(b+h_n)-f(b)}{h_n}.$$

But ##f(b)=0## and ##f(b+h_n)=f(b_n)=0##, so we get that ##f'(b)=0##. Thus if f has infinitely many zeros, there must exist some point ##b## such that ##f(b)=f'(b)=0##.
I liked this one as it runs cross (analytical) country once although the statement itself doesn't look like. I find it beautiful, however, this is only my taste.
 
  • #11
wrobel said:
View attachment 277319do not understand what this means
I haven't assumed a Lebesgue measure, hence some property of a measure has to be used.
 
  • #12
fresh_42 said:
I haven't assumed a Lebesgue measure, hence some property of a measure has to be used.
When he says Lebesgue integral, he means the integral for the given measure, not a specific integral for the Lebesgue measure on the reals.
 
  • Like
Likes wrobel
  • #13
fresh_42 said:
I haven't assumed a Lebesgue measure, hence some property of a measure has to be used.
I have provided the complete proof. If you disagree point the mistake
 
  • #14
wrobel said:
I have provided the complete proof. If you disagree point the mistake
I haven't said it was wrong, I just want to see the path from given conditions to the statement, not just saying it is the definition.
 
  • #15
Once again:
first step:
$$\int|f|d\mu:=\sup\Big\{\int\phi d\mu:\quad 0\le \phi\le |f|,\quad \phi \mbox{ is a simple function}\Big\}$$
second step:
If ##A\subset \mathbb{R}## then by definition
$$\sup A$$ is a number such that for any ##\epsilon>0## there exists $$a\in A:\quad a\ge \sup A-\epsilon$$
what will you have combining step 1 and step 2?
 
  • #16
wrobel said:
Once again:
first step:
$$\int|f|d\mu:=\sup\Big\{\int\phi d\mu:\quad 0\le \phi\le |f|,\quad \phi \mbox{ is a simple function}\Big\}$$
second step:
If ##A\subset \mathbb{R}## then by definition
$$\sup A$$ is a number such that for any ##\epsilon>0## there exists $$a\in A:\quad a\ge \sup A-\epsilon$$
what will you have combining step 1 and step 2?
You are right, and I have my peace. I just wanted to see the steps rather than say by definition! This was lazy, not intuitive, and most of all nobody could learn anything from it. You didn't even explained the used definitions.

Anyway, I will take the consequences. If you can do it better, then do it.
 
  • #17
Technically, every theorem follows from relevant definitions. Reminds me of debates with my students about "why need we to prove it, if it is true any way?" :DD
 
  • Like
Likes fresh_42
  • #18
hint for #5: assume you have a compact metric space E, and try it for that case.

hint for #9: prove if your function is not constant then the derivative of some order is non zero at the point where the maximum occurs. Then assume the first derivative is non zero and get a contradiction. (you will need to assume the inverse function theorem.) then assume the nth derivative is non zero, and try to write your function as the nth power of a function whose first derivative is non zero, and again get a contradiction.
 
  • Like
Likes wrobel
  • #19
nuuskur said:
Technically, every theorem follows from relevant definitions. Reminds me of debates with my students about "why need we to prove it, if it is true any way?" :DD
This does not do justice to the answers given by @wrobel.
 
  • Like
Likes nuuskur
  • #20
S.G. Janssens said:
This does not do justice to the answers given by @wrobel.
To be clear: @wrobel's proof was correct. That wasn't the question. I only would have preferred a proof that gives an example of ##W## and how ##\phi## can be found. The definitions of the supremum and simple functions include these steps.
 
  • #21
S.G. Janssens said:
This does not do justice to the answers given by @wrobel.
Agreed, I was just having fun. Apologies to all wounded parties.
 
  • #22
fresh_42

Are you sure that 5 does not require some compactness assumption on ##E##, local compactness or something like that ?

By the way: normal space ##\Longrightarrow## Hausdorff space
 
  • #23
wrobel said:
fresh_42

Are you sure that 5 does not require some compactness assumption on ##E##, local compactness or something like that ?
Yes, I'm sure, at least I cannot see it. We are given a finite covering and the statement is about this covering. Any is only necessary if we need to guarantee its existence.
By the way: normal space ##\Longrightarrow## Hausdorff space
This depends on the author. It is not automatically the case. Listing both avoids confusion. If it comes to the separation axioms, then it is always necessary to look up the definition, esp. for normal and regular.
 
  • Like
Likes wrobel
  • #24
a different approach for 9
(with some details omitted as the big complex analysis theorems are intertwined and I'm not sure what all is allowed for assumption) .

Inside ##U_0##, select a circle around ##z_0## with constant radius ##r##.

Now applying Cauchy's Integral Formula and integrate around the perimeter of this circle
##f(z_0)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-z_0}dz##

now apply modulus function and what amounts to triangle inequality
##\big \vert f(z_0)\big \vert =\big \vert\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-z_0}dz\big \vert \leq \frac{1}{2\pi }\int_\gamma \frac{\vert f(z) \vert }{\vert z-z_0\vert }dz \leq \frac{1}{2\pi }\int_\gamma \frac{\vert f(z_0)\vert }{r}dz\big \vert = \frac{\vert f(z_0)\vert \cdot 2 \pi r}{2\pi r} = \vert f(z_0)\vert ##
checking the equality conditions for triangle inequality, and using the fact that ##f## is holomorphic (really: continuous), this implies ##f## is constant on ##\gamma##.

Hence ##g(z):= f (z)-f(z_0)## has uncountably many zeros in ##U##. And e.g. applying Principle of Isolated Zeros, we see that ##g## is constant ##\implies f## is constant.
 
  • #25
There's really nothing to prove other than the well known theorems :rolleyes:
By Modulus maximum principle [itex]f[/itex] must be constant on [itex]U_0\subseteq U[/itex].
By the Identity theorem, since [itex]U[/itex] is open and connected, [itex]f[/itex] must be constant on [itex]U[/itex].
 
  • #26
nuuskur said:
There's really nothing to prove other than the well known theorems :rolleyes:
By Modulus maximum principle [itex]f[/itex] must be constant on [itex]U_0\subseteq U[/itex].
By the Identity theorem, since [itex]U[/itex] is open and connected, [itex]f[/itex] must be constant on [itex]U[/itex].
That was the idea. Both proofs are easy enough such that they can be figured out, and the statements are important enough to be learnt.

Guess you all recognized that I begin to run out of good ideas. Algebra is usually not touched even if the problems are really easy. Geometry is a horror to check or correct without a common picture (plus that I do not know the correct English names of the objects, e.g. angle bisector). Functional analysis is either well-known, used, or a nightmare about the many categories (e.g. different convergences, Sobolev). Most of the interesting problems require an even more advanced framework than algebra does. Calculations like integrals, ODE, or inequalities are either used up or simply boring. And now that you just quote the theorems or corollaries I want to be proven, this source is exhausted, too. Any ideas?
 
  • Like
Likes StoneTemplePython and BvU
  • #27
fresh_42 said:
That was the idea. Both proofs are easy enough such that they can be figured out, and the statements are important enough to be learnt.

Guess you all recognized that I begin to run out of good ideas. Algebra is usually not touched even if the problems are really easy. Geometry is a horror to check or correct without a common picture (plus that I do not know the correct English names of the objects, e.g. angle bisector). Functional analysis is either well-known, used, or a nightmare about the many categories (e.g. different convergences, Sobolev). Most of the interesting problems require an even more advanced framework than algebra does. Calculations like integrals, ODE, or inequalities are either used up or simply boring. And now that you just quote the theorems or corollaries I want to be proven, this source is exhausted, too. Any ideas?
How about a way for people to suggest problems? Of course it has to be in a way that doesn't give them away before the list is posted? Say we send you problems for you to choose from?
 
  • #28
martinbn said:
Say we send you problems for you to choose from?
I have nobody on my ignore list ...

I just need to know one answer to each suggestion:
Will you moderate the answers given, and if not please attach a solution.
 
  • #29
fresh_42 said:
I have nobody on my ignore list ...

I just need to know one answer to each suggestion:
Will you moderate the answers given, and if not please attach a solution.
Does that mean that suggestions are welcomed? Once i did send you a problem, but heard nothing about it. May be it wasn't suitable, but you didn't say anything, so i assumed that you didn't want any.
 
  • #30
martinbn said:
Does that mean that suggestions are welcomed?
Sure. I think it was even mentioned in the early days.
Once i did send you a problem, but heard nothing about it. May be it wasn't suitable, but you didn't say anything, so i assumed that you didn't want any.
Sorry, that I don't remember this, so I can't say what happened. I only need to know who will moderate it: send me a PM if solved by whom and in which post, or give me the solution (for dummies) so that I can moderate it.
 
  • #31
fresh_42 said:
Sure. I think it was even mentioned in the early days.

Sorry, that I don't remember this, so I can't say what happened. I only need to know who will moderate it: send me a PM if solved by whom and in which post, or give me the solution (for dummies) so that I can moderate it.
Yes, the problem should come with a detailed solution, so that you an deside if it fits the chalange.
 
  • #32
There are a bunch of unsolved questions from previous threads, maybe have a challenge month where you just repost them and we work together to try to crack them.
 
  • Like
Likes PhDeezNutz, Keith_McClary, wrobel and 1 other person
  • #34
Office_Shredder said:
There are a bunch of unsolved questions from previous threads, maybe have a challenge month where you just repost them and we work together to try to crack them.

I went back and solved a couple of the previous problems I just didn't post the answers. I'd like another crack at some of the other previous problems myself. I think some of the questions require me learning a bit of new maths, and for that reason I'd like to have a go at them myself.

I thought problems originally were being posted by various members. Has @fresh_42 been setting all the problems of late? Impressive.

Now and then I've been putting together a few problems when I can, which I was going to give @fresh_42 at some point.

I posted one problem last month because the problems got done early. You can have a go at that if you want :smile:. It doesn't require that you learn new maths, you solve the problem by taking some quantity to be an integer and expressing it another way that allows you to make estimations (hint: integrals) and making postulates.
 
Last edited:
  • Like
Likes fresh_42
  • #35
Office_Shredder said:
unsolved questions from previous threads
I spent some time on November 2018 #8 .
 

Suggested for: Math Challenge - February 2021

3
Replies
93
Views
10K
2
Replies
69
Views
3K
2
Replies
42
Views
6K
2
Replies
61
Views
7K
3
Replies
100
Views
6K
3
Replies
80
Views
4K
3
Replies
93
Views
6K
4
Replies
114
Views
6K
3
Replies
102
Views
6K
2
Replies
46
Views
4K
Back
Top