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• October 15th, 2017, 04:16
Unsolved Challenge
Evaluate ∠APR in Challenge Questions and Puzzles
As soon as you know that $x = 1/\sqrt2$, it follows (by looking at the triangle $PNA$) that $\overline{AP} = 1$. So $Q,P,R$ all lie on a circle...
3 replies | 113 view(s)
• October 14th, 2017, 16:09
Unsolved Challenge
Evaluate ∠APR in Challenge Questions and Puzzles
\coordinate (Q) at (-5,0) ; \coordinate (P) at (-3.5,3.5) ; \coordinate (R) at (5,0) ; \coordinate (A) at (0,0) ; \coordinate (N) at...
3 replies | 113 view(s)
• October 14th, 2017, 06:05
Unsolved Challenge
Prove ∠A≤60°. in Challenge Questions and Puzzles
Start with a bit of calculus. If $0<b\leqslant c$ and $f(x) = x(b+c-x)$, then $f(x) \geqslant bc$ whenever $b\leqslant x\leqslant c.$ The reason for...
2 replies | 116 view(s)
• October 13th, 2017, 07:04
If $z\in A\cap U^*$ (which is a nonempty set) then $z\notin A \cap U^*\cap V^*$ (because $A \cap U^*\cap V^*$ is empty). In particular, $z\notin... 2 replies | 52 view(s) • October 13th, 2017, 04:04 Starting from the fact that$e^t \to \infty$as$t\to\infty$, it follows that$\lim_{t\to\infty}e^{-t} = \lim_{t\to\infty}\dfrac1{e^t} =...
2 replies | 38 view(s)
• October 11th, 2017, 12:06
You are correct on both counts. The definition of $\text{Arcsin}$ specifies that its domain is $$and its range is$$. If $x$ (in that domain) is...
2 replies | 92 view(s)
• October 8th, 2017, 11:58
Opalg replied to a thread Level surface in Number Theory
Hi Elize, and welcome to MHB! You want to find integer solutions to the equation $x^2 - 6xy +y^2 + 8y = 1$. I would start by completing the square...
1 replies | 90 view(s)
• October 7th, 2017, 03:43
Yes, that is correct. The closure of a set $A$ always includes the whole of $A$.
3 replies | 90 view(s)
• October 6th, 2017, 04:39
No, Theorem 1.8 applies to an arbitrary subset $A$ of the complex plane. I think that maybe you are misled by thinking that an isolated point $x$...
3 replies | 90 view(s)
• October 5th, 2017, 03:07
Sudharaka posted a visitor message on Ackbach's profile
Yes, it's a log time but nice to be back. :)
• October 5th, 2017, 03:03
$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3+\sqrt{3^{2017}}}+\frac{1}{3^2+\sqrt{3^{2017}}}+\cdots+\frac{1}... 2 replies | 113 view(s) • October 4th, 2017, 05:13 That looks like a simple misprint. It should say "as Y is closed". The proof then works smoothly. 3 replies | 4344 view(s) • October 2nd, 2017, 16:35 If (x^2+ax+b)^{-1} = \sum_{k=0}^{\infty}c_kx^k then 1 = (x^2+ax+b)\sum_{k=0}^{\infty}c_kx^k. Compare the coefficients of x^{k+2} and x^{k+3} on... 2 replies | 132 view(s) • October 2nd, 2017, 04:59 Use \lim instead of \text{lim}. The subscript will then automatically appear underneath the limit (in display mode, but not in inline mode:... 2 replies | 82 view(s) • October 2nd, 2017, 04:53 Yes. :) 2 replies | 100 view(s) • October 2nd, 2017, 04:49 Yet another typo! Yes, it should obviously be \lim_{h\to0+}. Robert Bartle was a well-known mathematician, who specialised in writing... 2 replies | 68 view(s) • October 1st, 2017, 07:30 It looks to me as though you have identified another misprint in Bartle and Sherbert. The only way I can make sense of it is that the above sentence... 1 replies | 79 view(s) • September 30th, 2017, 12:06 I am sure that you are right, and that x^* need not be in V. The explanation seems to be that there is a mistake (or at least a misprint) in the... 5 replies | 135 view(s) • September 30th, 2017, 04:28 Let x be a number in$$. One idea would be to use induction to construct a sequence $(s_n)$ of rational numbers such that $|x-s_n| <1/n$. Such a...
2 replies | 61 view(s)
• September 29th, 2017, 14:45
It looks as though the answer should be $I = 4\pi\cos2\alpha\cos\alpha$ (because the constant is integrated over an interval of length $\pi$). (Cool)
3 replies | 130 view(s)
• September 29th, 2017, 14:37
It doesn't really matter, because the extra term when $r=n$ is $\sin(n\pi/n) = \sin\pi = 0$. (Bigsmile)
6 replies | 143 view(s)
• September 28th, 2017, 10:45
(a). Find $S_n$. With the vertices at the points $e^{2\pi ij/n}$ on the unit circle, the distance between $e^{2\pi ij/n}$ and $e^{2\pi ik/n}$ is...
6 replies | 143 view(s)
• September 24th, 2017, 15:13
The way I look at it is this. The proof of 6.3.3 takes place in a background environment consisting of functions that are continuous on  and...
3 replies | 122 view(s)
• September 24th, 2017, 03:36
Oops, yes, of course it should be $(n+3)(n^2+3) = n^3 + 3n^2 + 3n + 9 = (n+1)^3 + 8$. But the only cubes that differ by $8$ are $0$ and $\pm8$. Since...
5 replies | 179 view(s)
• September 23rd, 2017, 11:45
So if $n+3$ and $n^2+3$ are cubes then so is $(n+3)(n^2+3) = n^3 + 3n^2 + 3n = (n+1)^3 - 1$. But two consecutive numbers cannot both be cubes unless...
5 replies | 179 view(s)
• September 23rd, 2017, 09:33
With $x=y=1$, $f(1) + f(1) = f(1)$, from which $f(1) = 0$. Fix $y$, and differentiate the equation $f(x) + f(y) = f(xy)$ with respect to $x$:...
4 replies | 146 view(s)
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