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December 1[edit]

characteristic function of a pdf[edit]

hello,i have a question .what is the characteristic function of the Gaussion pdf?

From the article normal distribution, the characteristic function is

{\begin{aligned}\chi _{X}(t;\mu ,\sigma )&{}=M_{X}(it)=\mathrm {E} \left[\exp(itX)\right]\\&{}=\int _{-\infty }^{\infty }{\frac {1}{\sigma {\sqrt {2\pi }}}}\exp \left(-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}\right)\exp(itx)\,dx\\&{}=\exp \left(i\mu t-{\frac {\sigma ^{2}t^{2}}{2}}\right).\end{aligned}}

81.153.219.80 10:38, 1 December 2007 (UTC)[reply]

You could always look at characteristic function. HTH, Robinh 20:08, 3 December 2007 (UTC)[reply]

Even better is to have a look at the article Characteristic function (probability theory). But, as already stated above, the answer to the specific question is found at Normal distribution. --Lambiam 21:03, 3 December 2007 (UTC)[reply]

Computing multiplicative order of a base mod composite[edit]

How to do I "correctly" calculate the multiplicative order of a base mod p*q when I already know the multiplicative orders of base mod p and base mod q and the base does not have to be a primitive root of either modulus. I tried multiplying the two smaller orders but when I checked the composite order was actually smaller than the product of the two smaller orders.

Assuming p and q are coprime and a is invertible mod p and mod q, the order of a mod pq is the lowest common multiple of the orders of a mod p and mod q. Algebraist 16:26, 1 December 2007 (UTC)[reply]

A challenge...[edit]

Hopefully this is not out of place on this reference desk, but... I need an expression that ends up in this number: 45093512 It is meant as a joke, and the expression should be conveyable in clean text where possible, though quad(x) can be used. The point is to disguise the number so it can only be found by those who understand the math, so I though this would be a fun challenge for some people. So, reverse-engineer the number, "45093512". Thanks. 213.167.126.215 22:44, 1 December 2007 (UTC)[reply]

Reducing a number to it's prime factors is always a fun way to hide it. in this case, 45093512 = 2 x 2 x 2 x 5636689. I'm sure others can think of more interesting ways to obfuscate your number though :) --Monorail Cat 23:55, 1 December 2007 (UTC)[reply]

Digits 139973392–139973399 in the fractional part of the decimal expansion of

\pi

.[1] Fredrik Johansson 00:05, 2 December 2007 (UTC)[reply]

According to Plouffe's Inverter [2] it is the first 8 decimals of 1/Trott/ln(Salem)/2**(1/3), or of TreeGr2/ln(Artin)^2/Zeta(7). But don't ask me what it means. Maybe 2**(1/3) is 2 raised to 1/3 = cuberoot(2). But the next solution says "^2" and probably means exponentiation there. One notation for exponent < 1 and another for > 1? That would be odd. PrimeHunter 01:19, 2 December 2007 (UTC)[reply]

I don't think it is related to the magnitude of the exponent (see, for example, this where both 2^(1/3) and 2**(1/3) are used), but rather to the table used (all instances I've found of ** come from m414) and the role this exponentiation has in the expression. -- Meni Rosenfeld (talk) 09:21, 2 December 2007 (UTC)[reply]

QUIDK in base 36 220.237.181.140 04:26, 2 December 2007 (UTC) sqrt(5)[reply]

Similarly to Monorail Cat's suggestion, you could use some variation of 2+2*3*3*3*3*3*5*7*11*241. You can group the factors, e.g. 2+66*105*241*27. – b_jonas 09:30, 2 December 2007 (UTC)[reply]

What do you mean by "quad(x)" btw? – b_jonas 09:32, 2 December 2007 (UTC)[reply]

Nicely done, b_jonas - 2,3,5,7,11 and 241 are much more elegant primes to use - I just got lazy and used an online prime factorization utility.. figured it hadn't given me the best ones :) --Monorail Cat 09:38, 2 December 2007 (UTC)[reply]

Sorry, I meant "sqrt(x)", or others like "sin(something)" etc. The Norwegian language got the better of me a second ;P. Factors, though cool, are a bit simple. I was thinking something that wouldn't be obvious how to calculate, even with a calculator. The best think I came up with so far is ((4!-3!-1!)^3 + 1802)^2 + 2287. It includes factorials at least, which wouldn't be an obvious operation to some people. 213.167.126.215 14:34, 2 December 2007 (UTC)[reply]

You might try, making it the result of a difficult definite integral. Find a function and interval to which has some decimal or transidental value, then multiply it by the appropriate quantity to give the desired answer, then work backwards. A math-wiki 10:48, 3 December 2007 (UTC)[reply]

Another good way, is using a different base, 36 was suggested, but I know of a way notate any base (this is just my of doing it, I haven't checked into any whether or not my method here is in regular use or not) Lets convert your number to base 631, you might be wondering how in the world I'm going to find 631 different charachters, well I'm not going to. Instead, I will use base ten numbers to indicate place values, and then separate those with commas. at the end in a subscript I will note the base which is being used so your number in base 631 is..... $113,160,359_{631}$ for example. To work it out for any base you desire, simply pick your base, then take successive powers of that base untill the power n such the $b^{n}$ is larger than 45093512 in this case or your number k in general, then take $b^{n-1}$ and divide k by it, take the integer part as the place value for the place $b^{n-1}$ now you have found the place value $a_{n-1}$ , subtract $a_{n-1}b^{n-1}$ from $k$ and get $k'$ repeat this till you get to $n-(n-1)$ then once you finish that calculation your $k^{(n)}$ will be the last place value. Then add the subscript for the place value and your done. A math-wiki 11:17, 3 December 2007 (UTC)[reply]

Have you considered modular arithmetic? P-adics? Complex numbers? Or, since you allow function calls, how about things like ord(S₄), the order of the fourth symmetric group? (Which, as it happens, is a factorial.) It would help to know who is supposed to decipher this, what the number means, and how long the expression needs to be. Black Carrot (talk) 04:25, 5 December 2007 (UTC)[reply]