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 Miscellaneous math
 Posted 2016-07-19, 04:00 PM $\reverse \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!} = \frac{\cos(x) + \cosh(x)}{2}$ $\reverse {n\choose k} = {n\choose n\cdot\sum\limits_{p=0}^{r-1} (-1)^p + k\cdot\prod\limits_{q=0}^{r-1} (-1)},\,r \in \mathbb{Z}$ "Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
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Chruser

 Posted 2016-07-20, 12:56 PM in reply to Chruser's post "Miscellaneous math" Surely, there is a simpler method to select between 1 and 0 depending on odd/even than your summation?
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 WetWired

 Posted 2016-07-20, 06:02 PM in reply to WetWired's post starting "Surely, there is a simpler method to..." The answer is 7
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-Spector-

 Posted 2016-07-20, 08:49 PM in reply to Chruser's post "Miscellaneous math" Wait! What an obtuse way to say that n choose k = n choose k or n choose n-k! Why not just $\reverse {n\choose k} = {n\choose n\cdot{(-1)^r+1\over2} + k\cdot(-1)^{r+1}},\,r \in \mathbb{Z}$ or even $\reverse {n\choose k} = {n\choose n-k}$ Last edited by WetWired; 2016-07-20 at 08:58 PM.
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 WetWired

Posted 2016-07-25, 06:25 PM in reply to WetWired's post starting "Wait! What an obtuse way to say that n..."
WetWired said: [Goto]
 Wait! What an obtuse way to say that n choose k = n choose k or n choose n-k! Why not just $\reverse {n\choose k} = {n\choose n\cdot{(-1)^r+1\over2} + k\cdot(-1)^{r+1}},\,r \in \mathbb{Z}$ or even $\reverse {n\choose k} = {n\choose n-k}$
cuz u a bitch
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-Spector-

Posted 2016-07-31, 07:48 PM in reply to WetWired's post starting "Wait! What an obtuse way to say that n..."
WetWired said: [Goto]
 Wait! What an obtuse way to say that n choose k = n choose k or n choose n-k! Why not just $\reverse {n\choose k} = {n\choose n\cdot{(-1)^r+1\over2} + k\cdot(-1)^{r+1}},\,r \in \mathbb{Z}$ or even $\reverse {n\choose k} = {n\choose n-k}$

Yes, my "formula" is just the result of using the identity $\reverse {n\choose k} = {n\choose n-k}$ repeatedly, followed by a little algebra:

$\reverse {n\choose k} = {n\choose n-k} = {n\choose n-(n-k)} = {n\choose n-(n-(n-k))} = \dots$

I just thought the symmetry between the sum and product (and their upper limits) was fun.
"Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
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Chruser

 Posted 2016-08-01, 06:25 PM in reply to Chruser's post starting "Yes, my "formula" is just the result of..." Since I'm just sitting here shit posting, what's your favorite "fun fact about math"? Mine is: X% of Y is equal to Y% of X
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-Spector-

 Posted 2016-08-02, 12:00 PM in reply to -Spector-'s post starting "Since I'm just sitting here shit..." $\reverse {x\over 100} \times y = {y\over 100} \times x$ $\reverse {x \times y \over 100} = {y \times x \over 100}$ $\reverse {x \times y} = {y \times x}$ $\reverse {x \times y} = {x \times y}$ $\reverse {x} = {x}$ $\reverse {1} = {1}$
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 WetWired

 Posted 2016-08-03, 08:29 AM in reply to -Spector-'s post starting "Since I'm just sitting here shit..." If you had a wire as long as the circumference of the Earth and added just ten meters to it it would float 1.6 meters off the ground everywhere. If you could fold a piece of paper 42 times you would reach the moon and beyond. If you have 23 people in a room there's a 50% chance that two of them share a birthday. If you have 50 people in a room there's a 97% chance that two of them share a birthday. (1/2)+(1/4)+(1/8)+(1/16)+(1/32)+...=1 An infinite surface area can enclose a finite volume. Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" It is to your advantage to switch your choice. 10! seconds is exactly six weeks. If you properly shuffle a deck of cards, chances are the order in that deck has never been seen before. Given a solid ball in 3dimensional space, there exists a decomposition of the ball into a finite number of disjoint subsets, which can then be put back together in a different way to yield two identical copies of the original ball. This is known as the Banach-Tarski paradox.
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Demosthenes

Posted 2016-08-03, 11:44 AM in reply to Demosthenes's post starting "If you had a wire as long as the..."
Demosthenes said: [Goto]
 Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" It is to your advantage to switch your choice.
False. This is only true if he has to show you a door with a goat behind it regardless of your original choice.
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 WetWired

 Posted 2016-08-12, 05:57 PM in reply to WetWired's post starting "False. This is only true if he has to..." WHat in the actual fuck... I can't even remember how to do fractions.
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Kazilla

Posted 2016-08-12, 06:04 PM in reply to Kazilla's post starting "WHat in the actual fuck... I can't even..."
Kazilla said: [Goto]
 WHat in the actual fuck... I can't even remember how to do fractions.

1/2 = .5 = 50%

now you know
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-Spector-

 Posted 2016-08-12, 08:35 PM in reply to -Spector-'s post starting "1/2 = .5 = 50% now you know" Yea but, 3/4 * 5/7 = X I don't even know where to begin. I think you have to find common denominator right? so 28. 21/28 * 20/28 right? Then you are supposed to flip one upside down? 21/28 * 28/20 then multiply them and then some division and you are good right? 100% not trolling btw, I legit forgot how to do that shit.
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Kazilla

 Posted 2016-08-13, 06:47 AM in reply to Kazilla's post starting "Yea but, 3/4 * 5/7 = X I don't even..." Er, no. when you multiply fractions you just multiply numerator * numerator and denominator * denominator so.. 3/4 * 5/7 = 15/28
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-Spector-

 Posted 2016-08-14, 09:08 PM in reply to Kazilla's post starting "Yea but, 3/4 * 5/7 = X I don't even..." You're thinking of addition.
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 WetWired

 Posted 2016-08-15, 05:05 AM in reply to WetWired's post starting "You're thinking of addition." $\reverse \int_0^\infty \frac{\sin(x)}{x} \, \mathrm{d}x=\frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3} \, \mathrm{d}x =\frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13}\cdot\frac{\sin(x/15)}{x/15} \, \mathrm{d}x = \frac{467807924713440738696537864469}{935615849440 640907310521750000}~\pi$ "Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
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Chruser

Posted 2016-09-04, 06:37 PM in reply to Chruser's post starting "\int_0^\infty \frac{\sin(x)}{x} \,..."
Chruser said: [Goto]
 $\reverse \int_0^\infty \frac{\sin(x)}{x} \, \mathrm{d}x=\frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3} \, \mathrm{d}x =\frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13}\cdot\frac{\sin(x/15)}{x/15} \, \mathrm{d}x = \frac{467807924713440738696537864469}{935615849440 640907310521750000}~\pi$
Is there a source, of proof of this? I have no idea how to do those integrals.
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Demosthenes

 Posted 2016-09-05, 08:05 AM in reply to Demosthenes's post starting "Is there a source, of proof of this? I..." He is the source and the proof, accept him.
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-Spector-

Posted 2016-09-06, 03:58 PM in reply to Demosthenes's post starting "Is there a source, of proof of this? I..."
Demosthenes said: [Goto]
 Is there a source, of proof of this? I have no idea how to do those integrals.

Borwein integral.
"Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
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Chruser

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