Wikipedia:Reference desk/Archives/Mathematics/2007 September 23

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September 23

DIFFERENTIAL EQUATION

$f(x)=\sum _{i=0}^{x}f^{(x)}(i)$

where $f^{(x)}$ is the xth derivative of f.

How do I start?--Mostargue 03:35, 23 September 2007 (UTC)

How do you take the derivative of a function that only has values at integers? - Rainwarrior 04:58, 23 September 2007 (UTC)

Is it possible that you mean this:

$f(x)=\sum _{i=1}^{\infty }f^{(i)}(x)?$

At least, that would make sense as a mathematical formula.  --Lambiam 07:26, 23 September 2007 (UTC)

One possibility IF f(x) is a/could be a polynomial would to be write f(x) = a0+ a1x+a2x^2 + a3x^4 ..etc, then calculate all the differentials - take the sum and collect all the terms in x^n... eg you'd have a0 = a0 + a1 + 2a2 +3!a3 etc.. - no idea if that what is what you meant (or how to proceed from there..)87.102.17.252 15:13, 23 September 2007 (UTC)

Now if f(x) was expressable as a polynomial (single valued,continous tpyically) - this gives an infinite number of linear equations with infinite unknowns - if I have n linear equations with n unknowns I can solve this and find each unknown (solutions exist) - therefor by extension I can say that if f(x) is a polynomial in xⁿ then there is one such polynomial that satisfies the above relationship.. what it is I can't currently say.87.102.17.252 15:25, 23 September 2007 (UTC)

Is this supposed to be the Taylor series? — Daniel 23:51, 23 September 2007 (UTC)

YES YES THANK YOU LAMBIAM

$f(x)=\sum _{i=1}^{\infty }f^{(i)}(x)?$

is what I meant. I knew something looked wrong...--Mostargue 01:51, 24 September 2007 (UTC)

To solve this, simplify f(x) − f'(x).  --Lambiam 08:09, 24 September 2007 (UTC)

huh? How do I simplify that? Is that supposed to be an equation equal to zero, in that case it would be the exponential function...--Mostargue 08:17, 24 September 2007 (UTC)

$f'(x)=\sum _{i=2}^{\infty }f^{(i)}(x)=f(x)-f'(x)\ \Rightarrow ...$ Gandalf61 09:55, 24 September 2007 (UTC)

so f(x) = 2f'(x)

... hmm... so.. f(x) = 0.5e^2x, f'(x) = e^2x

Is that right?--Mostargue 10:07, 24 September 2007 (UTC)

Not quite. If f(x) = 0.5e^2x then f'(x) = 2f(x). But you want f'(x) = (1/2)f(x). Gandalf61 10:34, 24 September 2007 (UTC)

f(x) = 2e^0.5x , f'(x) = e^0.5x

yes?--Mostargue 10:43, 24 September 2007 (UTC)

That is one solution. f(x) = 0 is another solution, and in fact there is a family of solutions. You have a separable first-order linear ordinary differential equation; see the first example in Examples of differential equations.  --Lambiam 11:23, 24 September 2007 (UTC)

How many unique families of solutions are there?--Mostargue 14:38, 24 September 2007 (UTC)

One, that is, in solving the differential equation you have one constant you can choose freely. If you fix the value of f(0) to some given value, there is exactly one solution.  --Lambiam 17:50, 24 September 2007 (UTC)

Okay. But how did you get to f(x) - f'(x)?--Mostargue 23:04, 24 September 2007 (UTC)

I think it's one of those things you just need to spot (also known as a 'brainwave'). I couldn't get that either but once you've noticed (or been shown) that using it gives solutions it's a good idea to try to remember for the future - it's not algebraically obvious..87.102.10.190 15:41, 25 September 2007 (UTC)

It's a fairly standard way to collapse a really long sum. It's how you simplify the geometric series, for instance. Black Carrot 23:54, 25 September 2007 (UTC)

Indeed. If Z = a + F(a) + F(F(a)) + ... = Σ_i Fⁱ(a), in which operator F distributes over arbitrary sums, then Z = a + F(Z).  --Lambiam 13:03, 28 September 2007 (UTC)

Mathematical programming

How do we solve a problem given by max z=c'x subject to Ax<b and x is greater then or equal to 0? I believe this is a non linear problem but do not know the reason for this. Cheers--Shahab 07:42, 23 September 2007 (UTC)

Assuming I've correctly interpreted all your terms, this is just a basic linear programming problem (with slack variables), and thus the kind of thing the simplex algorithm was written for. Algebraist 13:18, 23 September 2007 (UTC)

But don't linear programming problems have Ax $\leq$ b, whereas my problem has Ax<b. At least that's what the definition on the wikipedia article says. Moreover the definition in my book also doesn't mention strict inequality. Cheers.--Shahab 16:33, 23 September 2007 (UTC)

If you consider that as a difference, you need to step back and tell us what you really want to do. Are you interested in a mathematical question or do you want to solve an actual problem? The mathematical linear programming problem is defined on the real numbers, and even storing a real number on a computer is impossible in general. If you want to solve an actual problem, you will probably use floating point numbers and the difference between equality and near equality will vanish in the face of other precision problems. If you have a purely mathematical question - I have no idea. —Preceding unsigned comment added by 84.187.32.213 (talk) 17:07, 23 September 2007 (UTC)

It's a purely mathematical question.--Shahab 17:53, 23 September 2007 (UTC)

(question) could you explain what is meant by y=c'x (surely not c' times x?)87.102.17.252 18:53, 23 September 2007 (UTC)

maybe c and x are vectors, and c' is the transpose of c? —Tamfang 00:33, 24 September 2007 (UTC)

In general, this problem has no solution, because "max" will not be defined on this set. Here's a one-dimensional example: find max x subject to 10x < 100 and x ≥ 0. It's clear that the set of x satisfying these inequalities is the half-open interval [0,10) = {x | 0 ≤ x < 10 }. But this set has no maximum. It does, however, have a supremum, which is always defined, and you can talk about sup cx subject to Ax < b -- but in this type of problem, this supremum will just be max cx subject to Ax ≤ b, reducing to the usual form for a linear programming problem. Tesseran 21:21, 23 September 2007 (UTC)

You aren't interested in Integer programming, are you? In that case, there is a big difference between < and ≤. 130.88.79.43 12:43, 25 September 2007 (UTC)