- 23 Aug '06 01:03A function is a mathematical construct. Think of it as a black box that takes in a number and converts it by a set of operations into an output. Simple ones could be f(x) = 2x, a function that just doubles the input. This function is defined for all values of x (both real and complex) however some functions only exist for positive x or integer x or similar.

A slightly more complex function example would be f(x) = 1/x^2. This function will take a large value when x is very small and a trailing value as x increases. It will never equal zero and is undefined when x=0. It is also symmetrical as a negative x will give the same answer a positive x of equal magnitude.

Functions don't have to have easily defined actions. They can do different things to different numbers. For example the Heaviside function, written as H(x). For

H(x) = 0 for x < 0

H(x) = 1 for x > 0

So it's one for positive x and zero for negative x. It is undefined at x=0. This function is used to simulate a switch in an electrical circuit and also in math for various driving functions.

And now an example of a function that only exists for positive integers.

f(x) = x!

This function is a simplified version of the gamma function (which is defined for all x) but that doesn't matter. The factorial operator multiplies a positive integer x by all positive integers less than it. Graphing this function would give a series of dots at integer values of x with nothing joining them.

So let's define a function. It will only be defined when n is a positive integer (like the factorial function above) and will take a range of values depending on the properties of n. We will refer to it as u(n).

If n is a prime number then u(n) = -1.

If n is a composite number (i.e. not a prime) then if its prime decomposition contains a repeated prime then u(n) = 0.

If n is a composite and its prime decomposition contains an even number of different primes then u(n) = 1

If n is a composite and its prime decomposition contains an odd number of different primes then u(n) = -1

That's quite complicated so let's have some examples:

u(10) = 1

as 10= 5*2 so an even number of non-repeated primes

u(96) = 0

as 96 = 3*2*2*2*2*2 so a repeated prime

And the first 10 values of this function will be:

1, -1, -1, 0, -1, 1, -1, 0, 0, 1

So we have defined our function. Now is it useful? Well not really. However what if we define a second function. This new function M(n) will be defined as the sum of all u(n) values up to n.

So the first 10 values will be:

1, 0, -1, -1, -2, -1, -2, -2, -2, -1

So now we come to the poser in this whole mess. Show that either; the absolute value (magnitude) of M(n) will never exceed the square root of n or find a value at which it does. - 23 Aug '06 01:28 / 4 edits

Mertens Conjecture (that no such n exists) was proved to be false in 1985 by Odlyzko and te Riele.*Originally posted by XanthosNZ***Show that either; the absolute value (magnitude) of M(n) will never exceed the square root of n or find a value at which it does.**

This result came as quite a shock to the mathematical community as until then it had been proved to be true for all n up to 500000.

In fact, lM(n)l is known to exceed the square root of n for some n less than 3.21*10^64 (Pintz, 1987)

http://mathworld.wolfram.com/MertensConjecture.html - 23 Aug '06 01:41 / 1 edit

Well done. People should feel free to use this thread to post questions about functions or posers involving functions.*Originally posted by ThudanBlunder***Mertens Conjecture (that no such n exists) was proved to be false in 1985 by Odlyzko and te Riele.**

This result came as quite a shock to the mathematical community as until then it had been proved to be true for all n up to 500000.

In fact, lM(n)l is known to exceed the square root of n for some n less than 3.21*10^64 (Pintz, 1987)

http://mathworld.wolfram.com/MertensConjecture.html - 23 Aug '06 01:53 / 3 editsI think this has been posted before but it might be worth another run:

d(x^2)/dx = x

Proof: If we rewrite x^2 as the sum of x x's, and then take the derivative we get

d/dx[x^2]

= d/dx[ x + x + x + ... (x times) ]

= d/dx[x] + d/dx[x] + d/dx[x] ... (x times)

= 1 + 1 + 1 + ... (x times)

= x

QED