the tree

If you take a look at the attached image, you should see how the areas under the two graphs are not the same.

Briefly, the difference between [imath]\sum f(x)[/imath] and [imath]\int f(x)[/imath] is that you're not counting the values between the integers. Have you done Riemann sums at all? The Riemann integral is the limit of the Riemann sums of a function as the partitions get finer, but in this case the partitions have a constant width of 1.

Really not sure what you're aiming for here, you're only defining values for a column vector and there's no apparent reason for an algorithm of that shape. from i=1 to N { P[i] := 1/2*i + c mod 256 }; Would be easier. Or, for a single gradient. s := floor(N/256) from i=1 to N { P[i] := i*s } Although it'd help to have any idea whatsoever of what you're trying to do.

Apparently bees can get lots of diseases, who'd have thought Google would know that? Spices are of course used in medicine, from home remedies to providing the active ingredients in proper drugs. In fact they do have antibacterial properties - however randomly throwing spices around without any scientific approach isn't going to help you develop medicines. Ingesting large amounts of something that's usually taken in nutritionally insignificant quantities is not a good idea and could make a sick person a lot more sick. Cooking oil induce vomiting and carelessly mucking about with left over food presents a risk of food poisoning. I'm not sure I even want to comment on how irresponsible it is to experiment on human blood in a home lab.

I think the reputation system is currently a bit rigid. I understand that the ability to give negative reputation would be a bad idea but would it be possible to have a graded system? e.g. give one reputation point for 'heh, that post was kind of amusing' and two or three for 'that post was super duper amazing!'. edit: I definitely think I'd use it more if there was more I could do with it.

Oh ffs. Okay since you had to read that, whateverthehell, I'll be lax on the rules and give a complete answer. You will of course need to fill in the gaps here and there. This isn't an easy integral so bare with me... Assuming you already know that [imath]\frac{\mbox{d}}{\mbox{d}x}\ln( f(x) )=\frac{f'(x)}{f(x)}[/imath] it should be easy enough to work out: [math]\frac{\mbox{d}}{\mbox{d}x} \ln(\sqrt{x+4}+b)=\frac{1}{2\sqrt{x+4}}\cdot\frac{1}{\sqrt{x+4}+b}[/math]. Then you'll need to show that: [math]\frac{1}{\sqrt{x+4}(\sqrt{x+4}+b_1)}-\frac{1}{\sqrt{x+4}(\sqrt{x+4}+b_2)}=\frac{b_2 - b_1}{\sqrt{x+4}(\sqrt{x+4}+b_1)(\sqrt{x+4}+b_2)}[/math] which isn't anything more than basic high school algebra. Thus far you can conclude: [math]\frac{\mbox{d}}{\mbox{d}x} \left( \ln(\sqrt{x+4}+b_1) - \ln(\sqrt{x+4}+b_1) \right)=\frac{1}{2}\frac{b_2 - b_1}{\sqrt{x+4}(\sqrt{x+4}+b_1)(\sqrt{x+4}+b_2)}[/math] Now you should see where I'm going, substitute in [imath]b_1=-2[/imath] and [imath]b_2=2[/imath]. [math]\frac{\mbox{d}}{\mbox{d}x} \left( \ln(\sqrt{x+4}-2) - \ln(\sqrt{x+4}+2) \right)=\frac{1}{2}\frac{4}{x\sqrt{x+4}}[/math] And therefore, with just some scaling to finish up. [math]\int \frac{1}{x\sqrt{x+4}} \mbox \,{d}x = \tfrac{1}{2}\left( \ln (\sqrt{x+4}-2 )-\ln ( \sqrt{x+4}+2 ) \right) + c[/math].

*face* *palm*

What do you mean by this? Do you mean that it'd fail to create sufficient accurate predictive models? That the Lotka–Volterra equations don't work? It is rarely necessary to diagram anything, and particularly in economics there are often too many factors involved for it to be sketched easily. Three vectors? Generally in applied maths you're going to be working with a dense vector field, that is over a bounded space you're talking about uncountably infinite vectors. You'd pick out individual cases for individual initial conditions but even then you'd want a series of results over a time span which would practically mean calculations in the hundreds or thousands. Rn cannot be indexed with R. You cannot do all that much with just lone numbers. Dimensional Analysis? Not really related. You seem awfully opinoinated on something that you don't know what it is. It's not really all that complicated, I mean, it gets complicated once you work with the really advanced stuff but it's all do-able and trained mathematicians are always avaliable. Working on problems that are tradiationally aproached with vector calc, without vectors is far more complicated.

Urm... neither? Those just are things that vector calc is used for.

Just to point it out, applied maths does extend beyond physics. Vector calculus is an important element of quantitative biology, biochemistry and even economics. (not that mathematics needs to be applied in order to be worthy of study)

For the lulz? Why do you think that any mathematics gets done? Represent anything that needs representing both in terms of magnitude and direction?

I think that the one in question, undoubtedly is.

I'd have to side with the quote from Anthony Kennedy. Say a government owned an art gallery, and that art gallery owned a painting that depicted something religious - separation of church and state wouldn't be a good reason for throwing it away.

Welcome to SFN. Hopefully you'll have a good time here. Don't hesitate to start asking and answering questions.

Oh I just thought, it's worth mentioning that the factorial function can be generalised to the Gamma function. [imath]n! = \Gamma(n+1)[/imath] where [imath]\Gamma(z) := \int_0^\infty t^{z-1} e^{-t}\,dt[/imath]. And you should be able to verify that [imath]\Gamma(1)=\Gamma(2)=1[/imath] if you're okay with integration by parts.

Maybe you could start by working out [imath]\frac{\mbox{d}}{\mbox{d}x} \ln(\sqrt{x+a}+b)[/imath].

Intuitively: how many ways do think there are to order zero objects? Slightly less intuitively: [math]n! = n (n-1)![/math] [math]1! = 1 \cdot 0![/math] [math]0! = \tfrac{1!}{1} = 1[/math] Really: that's just the definition.

To be clear with the definitions, a number is rational if it can be expressed as the ratio of two integers and irrational if it is not rational. This is equivalent to having a finite decimal expansion, no points for working out why. Take a rational number that isn't an integer, call it [imath]\frac{p}{q}[/imath] so that [imath]p[/imath] and [imath]q[/imath] have no common factors. [imath]p^2[/imath] and [imath]q^2[/imath] will clearly then have no common factors so [imath]\left( \frac{p}{q} \right)^2[/imath] will be a rational non integer. So no rational non-integer has an integer square, as a corollary no integer can have a rational non-integer square root.

Oh I see, we're talking at cross purposes here. You don't want to be publishing wild speculation in a magazine aimed at the general public. I would go back to ajb's original post, attempt to get something on arXiv. Through the course of your research you will certainly see references to reputable journals, if for instance you find yourself citing a few articles from journals owned by the American Physical Society then your own article may be appropriate for Physical Review Letters.

Unfortunately there's no good explanation unless you're already familiar with definite integration, which if you're actually using Simpson's rule implies that you're not. The is a section on derivation Wikipedia article on Simpon's rule.

If a velocity has horizontal and vertical components vx and vy, then you can think of the object travelling those distances in the x and y directions every unit time, there should be a right angle between them hence the total distance being the hypotenuse of the right angled triangle.

Again, with what area is this in? That sounds, odd. There's little point publishing anything that doesn't contain anything new - not just in an academic context but it any situation.

Ditto that.

Is that a general statement? But really, if you didn't already grasp that an invalid argument can have a true conclusion then you might want to start completely from scratch.

Well (1) is denying the antecedent (2) is affirming the consequent so from inspection they're invalid.