- 26 Aug '15 13:21 / 7 editsI want the function of x ( f(x) ) from just one mathematical clue in the form of just one rule about the definite integrals of the curve of the graph for f(x).

All variables/constants here are positive real numbers only.

On the graph for f(x), for some arbitrary constant K where K is some specific x value and where K>0, for any x value that is smaller than K i.e. x<K, then f(x) = 0. But I am not interested here in f(x) for x<K but rather f(x) for x≥K.

The only clue we got what that function is this; if we have a=x1 and b=x2 where a≥K and b≥K and a<b, then the integral of the interval of x from x=a to x=b on that graph is given by:

ʃ [a, b] = (K/a) – (K/b)

The only things I have deduce from that clue so far is that the curve for f(x) for x≥K is wholly above the x-axis (thus making all integrals positive ) and that f(x) tends to zero as x tends to +infinity.

But now I am stuck.

So what is the formula for that f(x)?

Any insight will be greatly appreciated.

Not sure but I think I have also deduced, from non-mathematical deduction from its application I have in mind, that:

ʃ [K, +∞] = 1

Is that right? - 26 Aug '15 14:05 / 1 edit

Edit: posted in reply to a post that was later edited to say he understood it...this is a little redundant now.*Originally posted by humy***Although I don't understand that proof ( "-1/x"? cannot see how that relates but I bet I am being completely stupid here ), I greatly appreciate your insight. "K/x^2" somehow intuitively looks about right to me even if I can't see why.**

Identity 1:

d/dx(x^p) = p*x^(p-1), where p is some parameter independent of x.

Identity 2:

ʃ [a, b] dx f(x) = F(a) – F(b) <=> dF(x)/dx = f(x)

F(x) is the antiderivative that Kazet referred to.

We are given:

ʃ [a, b] dx f(x) = (K/a) – (K/b)

so using F(x) = K/x in identity 2 we get d/dx(K/x) = f(x)

The minus sign comes from putting p = -1 into the identity at the start:

d/dx(1/x) = d/dx(x^-1) = -x^-2 = -1/x^2

giving us:

f(x) = -K/x^2 - 02 Sep '15 18:50I normally am good at basic algebra. But, while working on something, I accidentally found that:

( x/(x + 1) ) – ( x/(x + 2) ) = x/( (x + 1)(x + 2) )

and I cannot see why the above is true with the simple rules of algebra I know of.

Can somebody break it down for me with several algebraic intermediate steps so I can see how you can simplify ( x/(x + 1) ) – ( x/(x + 2) ) to x/( (x + 1)(x + 2) ) ? - 02 Sep '15 19:01 / 9 edits

Arr, after not getting it for ages, I suddenly got it just literally only one min after I posted this!*Originally posted by humy***I normally am good at basic algebra. But, while working on something, I accidentally found that:**

( x/(x + 1) ) – ( x/(x + 2) ) = x/( (x + 1)(x + 2) )

and I cannot see why the above is true with the simple rules of algebra I know of.

Can somebody break it down for me with several algebraic intermediate steps so I can see how you can simplify ( x/(x + 1) ) – ( x/(x + 2) ) to x/( (x + 1)(x + 2) ) ?

Sorry about that.

Perhaps this would be a nice bit of maths exercise for one of you if you don't look at my answer below first.

The answer is:

( x/(x + 1) ) – ( x/(x + 2) )

= ( x(x + 2) /((x + 1) (x + 2) ) ) – ( x(x + 1)/((x + 1) (x + 2) ) )

= (x(x + 1) – x(x + 2)) / ((x + 1)(x + 2))

= x( (x + 1) – (x + 2) ) / ((x + 1)(x + 2) ) (factorize the (x(x + 1) – x(x + 2)) part)

= x( x + 1 – x + 2 ) / ((x + 1)(x + 2))

= x( 1 ) / ((x + 1)(x + 2))

= x / ((x + 1)(x + 2)) - 02 Sep '15 19:28

Or more simply just multiply each side of the equation by (x+1)(x+2):*Originally posted by humy***Arr, after not getting it for ages, I suddenly got it just literally only one min after I posted this!**

Sorry about that.

Perhaps this would be a nice bit of maths exercise for one of you if you don't look at my answer below first.

The answer is:

( x/(x + 1) ) – ( x/(x + 2) )

= ( x(x + 2) /((x + 1) (x + 2) ) ) – ( x(x + 1)/((x + 1) (x + 2 ...[text shortened]...

= x( x + 1 – x + 2 ) / ((x + 1)(x + 2))

= x( 1 ) / ((x + 1)(x + 2))

= x / ((x + 1)(x + 2))

LHS * (x+1)(x+2) = x(x+2) - x(x + 1) = x

RHS * (x+1)(x+2) = x

When you have a problem like that where you are trying to prove a known result you don't need to derive it, showing it's true is enough. Incidentally this is a partial fractions problem and you should give the Wikipedia page on partial fractions a look.

https://en.wikipedia.org/wiki/Partial_fraction_decomposition - 04 Sep '15 21:05 / 1 editI am stuck on another problem.

I have produced a very complex mathematical algorithm that seems to output decimals that seem to be one natural number divided by another natural number in a form of a top-heavy fraction but I fail to see the pattern between input x (x>0, x ∈ ℕ) and output y (y ≥ 1).

here are the first 10 inputs and outputs:

input x | output y

1 → 1

2 → 3 / 2

3 → 19 / 12

4 → 29 / 18

5 → 1169 / 720

6 → 5869 / 3600

7 → 41183 / 25200

8 → 288731 / 176400

9 → 128461 / 78400

10 → 10413181 / 6350400

...

Note I don't know how much each output fraction has been inadvertently cancelled down from the unknown general formula, whatever that unknown general formula is, so you must take that into account.

Any ideas? - 05 Sep '15 19:12 / 3 editsWhat I really want to know, far more than the equation for the above pattern (although that would be far better than nothing! ), is the general algebraic formula for:

∑ [X = x, X = +∞] 1/( (X^2)(X + 1) ) where (x>0, x ∈ ℕ)

This is so, to work out that sum (which is for a very important probability model that indirectly helps to solve the problem of induction ), we don't have to use an inefficient iteration to get a good approximation of it but rather use vastly less computation time by deriving the sum directly from a general equation.

So far, I tried in vain to obtain one. But I have worked out that, for arbitrary 'high' x input values, we can derive a reasonable approximation of it with:

∑ [X = x, X = +∞] 1/( (X^2)(X + 1) ) ≈ 1/( 2(x^2) )

where (x>0, x ∈ ℕ)

and the ~1/( 2(x^2) ) approximation is always at least just slightly too low and never too high. And the approximation gets better and better with every increase in x.

Any ideas? - 05 Sep '15 19:21

There doesn't seem to be a simple answer to the sum. Try the following Wolfram Alpha query:*Originally posted by humy***What I really want to know, far more than the equation for the above pattern (although that would be far better than nothing! ), is the general algebraic formula for:**

∑ [X = x, X = +∞] 1/( (X^2)(X + 1) ) where (x>0, x ∈ ℕ)

This is so, to work out that sum (which is for a very important probability model that indirectly helps to solve the problem of in ...[text shortened]... r too high. And the approximation gets better and better with every increase in x.

Any ideas?

Sum[1/( (X^2)(X + 1) ),{X,x,\infty}]