Wikipedia:Reference desk/Archives/Mathematics/2008 December 26

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December 26

inequality involving a sequence

Resolved

 – Shahab (talk) 04:51, 28 December 2008 (UTC)

Hello. Consider the following sequence of natural numbers given by ordering the elements in the following set: $B=\{b\not \equiv 1{\pmod {c}}|b\in \mathbb {N} \}$. Here c is some fixed natural number. Then prove that for any two elements $b_{h}$ and $b_{h+g}$ of B, $b_{h+g}-b_{h}\leq g+{\Bigg \lceil }{\frac {g}{c-1}}{\Bigg \rceil }$. I want to prove this mathematically. I'll be grateful for any help--Shahab (talk) 09:05, 26 December 2008 (UTC)

Suppose

$d={\Bigg \lceil }{\frac {g}{c-1}}{\Bigg \rceil }$

$\Rightarrow g=d(c-1)-e\quad ,\quad 0\leq e<c-1$

$\Rightarrow g+{\Bigg \lceil }{\frac {g}{c-1}}{\Bigg \rceil }=g+d=dc-e\quad ,\quad 0\leq e<c-1$

Now show that

$b_{h+c-1}=b_{h}+c$

and so

$b_{h+d(c-1)}=b_{h}+dc$

and then

$b_{h+d(c-1)-e}=b_{h+d(c-1)}-e{\text{ or }}b_{h+d(c-1)}-e-1$

so

$b_{h+d(c-1)-e}\leq b_{h+d(c-1)}-e$

I'll let you take it from there. Gandalf61 (talk) 11:03, 26 December 2008 (UTC)

Thanks for the help--Shahab (talk) 04:48, 28 December 2008 (UTC)

Relation between trigonometric functions and exponential / multiplicative reciprocal functions

Hi. A while ago, I stumbled upon an interesting problem. Consider a circle of radius r, and a chord c = 1 unit of length. Let a be the (smaller) arc marked by the chord. The goal of the problem is to develop a function f(n)=r, so that it calculates the minimum length of r, for which a - c <= 10^-n. In other words, find a function that tells you the smallest radius for which the arc and the chord get so close to each other in length, that within an error margin of 10^-n, they are practically equal.

Using the Law of Cosines for an isoceles triangle and the base-10 logarithm, I came to the equation:

$n=-log_{10}(r\times arccos({\frac {2r^{2}-1}{2r^{2}}})-1)$

with the angle expressed in radians. The only problem now is solving it for r, but since r is both inside and outside of the arccosine function, it's tricky. Not even a CAS will solve it algebraically. I noticed, though, that when graphed, $arccos({\frac {2r^{2}-1}{2r^{2}}})$ is similar to a reciprocal function with an even exponent (like ${\frac {1}{x^{2k}}}$) (axis symmetry, two asymptotes, etc...). Can an arccosine function with that type of fraction as an argument be re-written as a reciprocal function? Also, the whole formula above yields, when graphed, a square-root-function-like shape (exponential with negative exponent). Or it could be logarithmic... Any clues as to solve the above equation for r algebraically would be greatly appreciated. (I've tried regression, and came up with weird numbers that I'd like to mean something...) Thanks, sfaefaol 12:30, 26 December 2008 (UTC)

The chord is $\scriptstyle c=2r\sin({\frac {a}{2r}})=1.$ The condition is $\scriptstyle a-c=10^{-n}$ or $\scriptstyle a=1+10^{-n}.$ Substituting gives $\scriptstyle 2r\sin({\frac {1+10^{-n}}{2r}})=1$ where $\scriptstyle r$ is the unknown. Define a new variable $\scriptstyle x$ by $\scriptstyle 2r={\frac {1+10^{-n}}{x}}$ and get the equation $\scriptstyle {\frac {\sin({x})}{x}}={\frac {1}{1+10^{-n}}}.$ Use the Taylor series approximation $\scriptstyle {\frac {\sin({x})}{x}}\approx 1-{\frac {x^{2}}{6}}$ to get the quadratic equation $\scriptstyle 1-{\frac {x^{2}}{6}}={\frac {1}{1+10^{-n}}}.$ Solve: $\scriptstyle x={\sqrt {\frac {6}{10^{n}+1}}}.$ So your result is $r\approx 0.204\times 10^{n/2}.$ Bo Jacoby (talk) 15:55, 26 December 2008 (UTC).

acromegly

What are the chances that 2 brother's in law that did not grow up in same city or live in same city deveolop acromegly/pituitay tumors? —Preceding unsigned comment added by CAElick (talk • contribs) 18:44, 26 December 2008 (UTC)

Maybe you better redirect the question to the Reference desk/Science for more precise information. Anyway, since there is no kinship between them and they come from different places, with no other information I would say the probability are independent. That is, the information that one of them develop a tumor should not affect the probability that the other one will, which remains the same as for any other person.--PMajer (talk) 21:21, 26 December 2008 (UTC)

It sounds like they are independent events, unless there is some commonality which wasn't mentioned, like being exposed to the same chemical while visiting each other. If they are independent events, then just multiply the probabilities of each event to find the probability of both happening simultaneously. However, if you consider the probability that the two would both develop the same disease from the rather large list of rare diseases, not just this specific one, then the chances are much higher. StuRat (talk) 21:29, 26 December 2008 (UTC)

maths quiz

Determine all 3 digit numbers N which are divisible by 11 and where N/11 is equal to the sum of the squares of the digits of NDon deepan (talk) 19:04, 26 December 2008 (UTC)

OK, I've done that. FYI, there are exactly two solutions. Algebraist 19:19, 26 December 2008 (UTC)

This only requires a couple of lines in Mathematica 122.107.203.230 (talk) 22:18, 28 December 2008 (UTC)

For[n=110,n<1000,n+=11,
  a=Mod[Quotient[n,100],10];
  b=Mod[Quotient[n,10],10];
  c=Mod[Quotient[n,1],10];
  If[n/11==a^2+b^2+c^2,Print[n]]
  ]

maths quiz

Given positive real numbers a, b, and c such that a + b + c = 1, show that aabbcc + abbcca + acbacb <=1Don deepan (talk) 19:06, 26 December 2008 (UTC)

AM-GM inequality. Algebraist 19:08, 26 December 2008 (UTC)

have you attempted to solve this problem yet? ♫Deathgleaner 00:37, 27 December 2008 (UTC)

Please define what you mean by aabbcc etc. If you mean a*a*b*b*c*c then all 3 of your terms are identical, so you are asking if aabbcc <= 1/3. If it means something else what? -- SGBailey (talk) 00:20, 28 December 2008 (UTC)