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Author Topic: Maths thread.  (Read 142016 times)

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mpsprs

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Re: Maths thread.
« Reply #550 on: December 04, 2015, 05:19:18 pm »
+1

TIL all the binomial coefficients other than (n 0) and (n n) of a prime number are divisible by the prime number, and that this is a sufficient and necessary condition for primality.
Hmm, necessity follows from the expression in terms of factorials, but how do you prove sufficiency? I suppose n choose p isn't divisible by n for prime p dividing n?

Aren't we saying something different?  That p divides (p k) iff p is prime.

You need to tack on a 'for all 0<k<p' condition before the iff. 

But I think Liopoil was thinking about the (proof of the) statement 'if n is not prime, then n does not divide (n k) for some k' by specifically guessing that if p divides n, then n won't divide (n p).  His statement is one direction of yours.  (And the harder to prove)

Witherweaver

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Re: Maths thread.
« Reply #551 on: December 04, 2015, 05:35:18 pm »
0

TIL all the binomial coefficients other than (n 0) and (n n) of a prime number are divisible by the prime number, and that this is a sufficient and necessary condition for primality.
Hmm, necessity follows from the expression in terms of factorials, but how do you prove sufficiency? I suppose n choose p isn't divisible by n for prime p dividing n?

Aren't we saying something different?  That p divides (p k) iff p is prime.

You need to tack on a 'for all 0<k<p' condition before the iff. 

But I think Liopoil was thinking about the (proof of the) statement 'if n is not prime, then n does not divide (n k) for some k' by specifically guessing that if p divides n, then n won't divide (n p).  His statement is one direction of yours.  (And the harder to prove)

Oh, I see, thanks.
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silverspawn

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Re: Maths thread.
« Reply #552 on: January 31, 2016, 11:06:31 am »
0

Can someone explain to me how



?
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liopoil

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Re: Maths thread.
« Reply #553 on: January 31, 2016, 11:14:30 am »
+2

\frac{b}{a + \sqrt{a^2  - b}}

= \frac{b}{a + \sqrt{a^2  - b}} \cdot \frac{a - \sqrt{a^2 - b}}{a - \sqrt{a^2 - b}}

= \frac{b(a - \sqrt{a^2 - b})}{a^2 - (a^2 - b)}

= \frac{b(a - \sqrt{a^2 - b})}{b} = a - \sqrt{a^2 - b}
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silverspawn

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Re: Maths thread.
« Reply #554 on: January 31, 2016, 12:47:33 pm »
+2

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Witherweaver

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Re: Maths thread.
« Reply #555 on: January 31, 2016, 01:17:22 pm »
+2

You should also check the case where you attempt to divide by 0 (i.e., a = sqrt{a^2-b}).
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silverspawn

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Re: Maths thread.
« Reply #556 on: January 31, 2016, 01:52:20 pm »
0

I think it has to be a = -SR(a-b). if that's the case then a = a-b so b = 0 so a = -SR(a) = -a = 0

so a = b = 0 and 0/0 can be anything, so here 0 = 0/0
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faust

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Re: Maths thread.
« Reply #557 on: January 31, 2016, 02:01:21 pm »
0

so a = b = 0 and 0/0 can be anything, so here 0 = 0/0

When I was at high school, I used to think that way. Now I don't like it and I feel like it's much better to say that 0/0 is not defined, as otherwise 0 = 0/0 = 1.
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liopoil

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Re: Maths thread.
« Reply #558 on: January 31, 2016, 02:10:49 pm »
+1

I think it has to be a = -SR(a-b). if that's the case then a = a-b so b = 0 so a = -SR(a) = -a = 0

so a = b = 0 and 0/0 can be anything, so here 0 = 0/0
No, in this case your initial expression doesn't make any sense as faust said. Witherweaver is noting that I claim to be multiplying by "1", but when a = \sqrt{a^2  - b} I am multiplying by 0/0 which can lead to faulty results. However, when this is the case, a^2 = a^2 - b, so b = 0, so the expression is 0 and thus a - \sqrt{a^2 - b} is still correct.
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Kirian

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Re: Maths thread.
« Reply #559 on: January 31, 2016, 02:16:25 pm »
+1

Isn't it easier just to multiply both sides through by [a + sqrt(a2 + b)] ?
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liopoil

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Re: Maths thread.
« Reply #560 on: January 31, 2016, 02:37:08 pm »
0

Isn't it easier just to multiply both sides through by [a + sqrt(a2 + b)] ?
Uh, yes, it is I guess. Though technically this is assuming the result unless you show it is reversible (which is obviously is, but still).
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Titandrake

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Re: Maths thread.
« Reply #561 on: January 31, 2016, 03:49:34 pm »
0

Isn't it easier just to multiply both sides through by [a + sqrt(a2 + b)] ?
Uh, yes, it is I guess. Though technically this is assuming the result unless you show it is reversible (which is obviously is, but still).

Not reversible if (a + \sqrt{a^2 - b}) = 0 which can happen for a < 0, b = 0. In that case, the identity is actually wrong. I believe that's the only case though.

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silverspawn

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Re: Maths thread.
« Reply #562 on: January 31, 2016, 03:52:35 pm »
0

Okay, what's the deal with everyone using TeX Notation when you don't even turn them into images?  ???
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Titandrake

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Re: Maths thread.
« Reply #563 on: January 31, 2016, 07:07:06 pm »
0

Okay, what's the deal with everyone using TeX Notation when you don't even turn them into images?  ???

Basic TeX notation is readable at a glance, and it's a lot more effort to turn them into images and upload them. If people understand what you mean, the actual format doesn't matter as much. It's just a forum post.
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silverspawn

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Re: Maths thread.
« Reply #564 on: February 01, 2016, 10:38:50 am »
0

I see. Although you don't have to "upload", In case of lio's calculation, all I had to do was to copy the code into a wiki page, add <math> tags, and it complied flawlessly (then you can copy the image codes), which is why I wondered if maybe you use a tool that generates the code.

It seems a bit strange to me, because I don't find TeX sourcecode pretty to look at, and isn't it easier to say SR(4) than \sqrt{4}?
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DStu

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Re: Maths thread.
« Reply #565 on: February 01, 2016, 10:58:27 am »
+3

I see. Although you don't have to "upload", In case of lio's calculation, all I had to do was to copy the code into a wiki page, add <math> tags, and it complied flawlessly (then you can copy the image codes), which is why I wondered if maybe you use a tool that generates the code.

It seems a bit strange to me, because I don't find TeX sourcecode pretty to look at, and isn't it easier to say SR(4) than \sqrt{4}?

probably depends on how often you have written \sqrt{4}.
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Kirian

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Re: Maths thread.
« Reply #566 on: February 01, 2016, 11:20:09 am »
+2

I see. Although you don't have to "upload", In case of lio's calculation, all I had to do was to copy the code into a wiki page, add <math> tags, and it complied flawlessly (then you can copy the image codes), which is why I wondered if maybe you use a tool that generates the code.

It seems a bit strange to me, because I don't find TeX sourcecode pretty to look at, and isn't it easier to say SR(4) than \sqrt{4}?

Even without using Tex, programming languages almost all use sqrt() and not sr().
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Awaclus

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Re: Maths thread.
« Reply #567 on: February 01, 2016, 12:00:46 pm »
0

I would probably have to google what it means if "SR(4)" came up in a math notation somewhere, whereas I instantly know what sqrt means.
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Re: Maths thread.
« Reply #568 on: February 01, 2016, 12:59:29 pm »
0

I see. Although you don't have to "upload", In case of lio's calculation, all I had to do was to copy the code into a wiki page, add <math> tags, and it complied flawlessly (then you can copy the image codes), which is why I wondered if maybe you use a tool that generates the code.

It seems a bit strange to me, because I don't find TeX sourcecode pretty to look at, and isn't it easier to say SR(4) than \sqrt{4}?

I would not assume SR(4) means square root without some context.  It looks like the name of some group or something.  But \sqrt{} is pretty unambiguous.
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Witherweaver

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Re: Maths thread.
« Reply #569 on: February 01, 2016, 01:01:21 pm »
0

And if you normally type in TeX/LaTeX (like I do), it's second nature.  Even when writing emails at work, I will use LaTeX notation to talk about equations or mathematical expressions.
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DStu

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Re: Maths thread.
« Reply #570 on: February 01, 2016, 01:25:56 pm »
0

I see. Although you don't have to "upload", In case of lio's calculation, all I had to do was to copy the code into a wiki page, add <math> tags, and it complied flawlessly (then you can copy the image codes), which is why I wondered if maybe you use a tool that generates the code.

It seems a bit strange to me, because I don't find TeX sourcecode pretty to look at, and isn't it easier to say SR(4) than \sqrt{4}?

I would not assume SR(4) means square root without some context.  It looks like the name of some group or something.  But \sqrt{} is pretty unambiguous.


http://www.thesrgroup.com/
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liopoil

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Re: Maths thread.
« Reply #571 on: February 01, 2016, 04:12:04 pm »
+1

I see. Although you don't have to "upload", In case of lio's calculation, all I had to do was to copy the code into a wiki page, add <math> tags, and it complied flawlessly (then you can copy the image codes), which is why I wondered if maybe you use a tool that generates the code.

It seems a bit strange to me, because I don't find TeX sourcecode pretty to look at, and isn't it easier to say SR(4) than \sqrt{4}?
I actually didn't know an easy way to generate the pictures, but I could tell that you knew LaTeX from your post, so that's what seemed easiest to do.
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Re: Maths thread.
« Reply #572 on: February 05, 2016, 07:13:30 pm »
+1

So random idea I thought of today - and actually ended up fully solving while posting since I realised I missed something obvious for working out the answer.

What is the smallest list of n numbers (where n is a positive integer) such that:

The mode is n
The median is n
The mean is n
The range is n

Write down your list of numbers as well.

Harder version: As above, but with a variance of n.

Bonus points in both cases for proving your solutions.

The first version is relatively straight forward. I would say that a talented A level mathematician (16-18 year old level) could probably solve it and maybe prove it. The second version is a lot harder, especially to prove and if you find the whole family of solutions for the lowest n (spoiler: There's more than one solution to the second one).

I do have solutions to all of these, which I'll save for now so people can have a go and because I noticed an improvement to my answer to the second one and want to give a complete answer
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Re: Maths thread.
« Reply #573 on: February 05, 2016, 07:59:32 pm »
0

What do you mean by the smallest list? By definition the list must have n elements, so they will all be the same size.
Unless I am looking for the smallest n that you can do this for?
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WanderingWinder

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Re: Maths thread.
« Reply #574 on: February 05, 2016, 08:11:29 pm »
+2

So random idea I thought of today - and actually ended up fully solving while posting since I realised I missed something obvious for working out the answer.

What is the smallest list of n numbers (where n is a positive integer) such that:

The mode is n
The median is n
The mean is n
The range is n

Write down your list of numbers as well.

Harder version: As above, but with a variance of n.

Bonus points in both cases for proving your solutions.

The first version is relatively straight forward. I would say that a talented A level mathematician (16-18 year old level) could probably solve it and maybe prove it. The second version is a lot harder, especially to prove and if you find the whole family of solutions for the lowest n (spoiler: There's more than one solution to the second one).

I do have solutions to all of these, which I'll save for now so people can have a go and because I noticed an improvement to my answer to the second one and want to give a complete answer

4 (2,4,4,6) and 7 (3.5,3.5,7,7,7,10.5,10.5). 6 almost works for the second part - you can get a variance as close to 6 as you like without ever quite making it there.
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