BigMoosie

So you are taking on a loan, do you plan to calculate exactly enough to only pay off the interest or do you want to reduce the loan by paying off the principle over a certain number of years? Also, so I can make my answer sound more relevant to you, please provide some ballpark figures (loan amound, annual interest + compounding period, time to repay if relevant and time between installments) so I can make my answer relevant to you instead of providing complicated algebra with terms you might not even need.

Suppose a bank provides 12% interest, compounded monthly. In this case 12% is the nominal annual interest rate. The monthly interest rate is 12%/12 = 1%, so over a year this compounds to become 1.01^12-1 = 12.68%. This is called the effective interest rate, because it is how much your money will grow if you leave it for 1 year. Now suppose the bank compounds daily instead of monthly, common sense tells us that you will make more money, but how much? Lets see: Daily interest rate = 12%/365 = 0.03288...% = x. Over a year this compounds to: (1+x)^365 - 1 = 12.74%. So it has risen a little bit, but not substantially more, if you then compound hourly, then every second etc. you will notice that the effective interest rate is approaching a limit (it grows very slowly as you make the time interval smaller but it doesn't go to infinity). It turns out the limit approaches e^i-1, where 'i' is the nominal interest rate. With my example of i=12% this approaches 12.7496852% (not much more than compounding daily). The reason continuous compounding is used in finance is that it is both more fair (arbitrary dates to compound at are just that, arbitrary) and very elegant to work with when calculations become complicated. Note: when using compound interest it is common to state i (the 12% in my example) as the 'force of interest' rather than 'the annual interest rate'.

Bignose, thanks, the site looks good, but the functions I am after seem to only have algorithms for real z there (no Complex algorithms.

d/dx f(x) means 'the derivative of f'. This is calculus. If I know what S and V are then I may be able to provide further help, otherwise all I can do is explain what calculus is. But if I were to provide a *perfect* explaination it would likely not be very different to whatever your textbook provides, so I will only give a brief overfiew. When you find the gradient of a line, such as f(x)=2x (in this case the gradient is 2), what is it exactly that you have? You have a number which indicates how quickly the height of the function changes as you move along the x-axis. In this case, move across 1 unit and up 2 (rise/run). Other ways to express rise/run: (change of rise) / (change of run) (change of y) / (change of x) dy / dx Here 'd' is not a variable but an operator, kind of like +-/* are operators, a variable that means 'very small change'. This can also be expressed as: d f(x) / dx But often the f(x) is moved to the right of the fraction so that it is clearer to see what is happening: d/dx f(x) So now 'd/dx' can stand alone and can be read as 'the derivative of'. Luckily with a straight line the gradient doesn't change, hence: d/dx (2x) = 2 That is why there is a constant, it doesn't matter what 'x' is, it will always be 2. If your function is not a straight line (perhaps a porabola, hyperbola or whatever) then the gradient will be different depending on what part of the function you look at, so for derivatives of these functions the answer will be expressed in terms of 'x'. The interesting thing about derivatives is that if you add two functions together, you can also add their derivatives together. For instance if you add 2x and 3x then the gradient is (2+3). This works for any function, not just lines. That is why I was able to move V(X) to the left hand side of the equation, you can add and subtract them fine. (sorry just notice I made a typo in my first post writing 'D(x)' instead of 'V(x)). There are lots of tricks for remembering how to manipulate different expressions to find their derivatives, and there is also a general method that can be used for all functions but is more tedious. These you will need your textbook for.

It seems hard to find algorithms on the web for computing functions like erf() and zeta() etc. Many sites offer algorithmic solutions at a cost, but my project is non profit so I don't want to spend money on it (it will be released free of charge as a web application). So I am wondering if anybody knows a good algorithm for computing these functions for general complex numbers expressed as (a+bi). The ideal algorithm will be one that is iterative or recursive allowing an arbitrary level of accuracy, and solutions that converge quickly. Right now I need 'erf', 'zeta' and 'gamma' (I have one for gamma but it only calculates real arguments). Other popular special functions that cannot be expressed in terms of these three would also be a bouns. Cheers, - Moosie

Complex numbers are being stored as an ordered array [a,b], but that doesn't mean they can't be expressed as mod-arg or any other form by being in combination with functions, such as prod(abs([a,b]), cis(arg([a,b]))). The thing is that if real() is used in conjunction with other functions (outside of the 'real'), then often doing the most direct simplification will dig the expression into a hole where it needs to be expanded in order to be simplified further. I was hoping there would be another way of calculating real() in cases where it gets stuck and it can try an alternative heuristic. An example of possible input into the program: "x = 1+4i" "y = x+pi(e^4-sin(ix))" "simplify(y)" -> sum(x, product(sum(pow([e,0], negate(sin(product([0,1], x))))))) It will then try to produce a representation of the value with the real and imaginary parts separated by manipulating the functions then displaying it in a typical mathematical way, it will never try to produce a decimal representation unless the command to do so is sent. It is easy enough to find formulas on the web for expanding things like tan^2(w+z) in different ways but nobody seems to try to do so with real() or imag(), perhaps that is because there is no way. If no way is clear then I will abandon this approach.

Thanks for the quick responses. @Mr Skeptic: The reason I do not want to break it up into real and imaginary parts is that I am developing a mathematical command line program (a bit like a mini mathematica) and I want to express it this way so that my simplifcation sub-routine can simplify a function that uses it with other functions, it's a bit complicated but my goal has an important purpose, despite it being a bit odd. @ the tree: that involves calculating abs(z) which will help with some simplifications (refering to my above point) but calculating abs(z) also breaks it apart to find sqrt(a*a+b*b), if there is a way of calculating abs using my initial restrictions then I will find it very handy. -- My first attempt was to find a complex number (c+di) which when multiplied by z will result in 'a'. After some simple algebra I got: c = a^2/(a^2+b^2) d = -ab/(a^2+b^2) But I have run into the same problem again, expressing c and d in terms of z (without 'a' and 'b' being separate) with elementary functions.

To the original poster chris_75: Another similar oddity exists with statistics. If you look at a population of humans, and try to calculate the probability that a random human has a height of 'x' metres, then for *any* value of 'x' the probability is 0. If all heights have a probability of 0 then it may appear at first that the height is undefined, but that's not true. This is not just being silly, this is how standard statistical theories define such a random variable. But of course, like your paradox, in reality this tends not to hold true since, for example, measuring devices can only be so precise.

Your wording is a little confusing but this sounds like a simple calculus problem, here is what I assume you are asking: S and V are single-valued functions defined over the reals. For what values of 'x' will the gradient at S(x) be greater than the gradient at V(x)? d/dx F(x) will provide the gradient so this will occur when: d/dx S(x) > d/dx V(x) or: d/dx {S(x) - D(x)} > 0

Given z is a complex number: a+bi, the function real is defined as real(z) = a; For my purpose this is not satisfactory, I cannot break z into its real and imaginary parts. I am wondering if it is possible to express the function in terms of elementary functions: +, -, *, /, ^, cos, sine, log, exp.... Only z or constants must be used as paramaters to these functions, never a or b. Other functions that break z up cannot be used, for example: real(z) = z-imag(z) is not a valid solution. My gut feeling tells me this is impossible, I would greatly appreciate a solution or verification that there is none. Thanks a lot! - Moosie

Hi, I am designing an animation library for the web, and I wanted to create an animation interpolation that is more flexible and elegant than using 'sine' or similar. I have spent countless hours trying to solve this, so I'll be extremely thankful for any help offered. My system I devised is this: A partical is at position 0, with velocity 'v' and accelleration 'a'. I want it to move along a function to land at position 'h' with velocity 0 and accelleration 0. Simply put, I want to get a moving object to land somewhere stationry. 'v' and 'a' are provided to the function as attributes so that if the target moves then the curve can adjust without any abruptness. I figured a polynomial of degree 5 would be the simplest curve to do so: [math]f(x) = Px^5 + Qx^4 + Rx^3 + \frac{a}{2} x^2 + vx[/math] If I let the amount of time for the object to land still be t, and using my constraints from above I can get the following equations: [math]Pt^5 + Qt^4 + Rt^3 + \frac{a}{2} t^2 + vt = h[/math] [math]5Pt^4 + 4Qt^3 + 3Rt^2 + at + v = 0[/math] [math]20pt^3 + 12Qt^2 + 6Rt + a = 0[/math] This is a piece of cake to solve if I know what t is. But I want 't' to be automatic, though of course then I have 4 unknowns with 3 equations. The final constraint I want to impose is this: If: v = 0 a = 0 h = 1 Then t = 1 (see diagram) [math]f(x) = 6x^5 - 15x^4 + 10x^3[/math] And, given that function, if the animation is paused and a new function is calculated using the same current velocity, accelleration and relative displacement from target 'h', then it should still arrive at the same time. That is, if I update the target position on each frame even though the target has not moved, then the total time taken for the animation should not be comprimised. As far as I can see this should be enough to result in a unique solution for 't', that solution is all I need. If there is a more appropriate system, such as some kind of spline, then please mention it. Thankyou, BigMoosie

It is generally accepted that the Amazon basin once drained into the Pacific, and that when the Andes formed it begin flowing in the opposite direction. I am wondering what the process would have been like, I see alternatives: * Would it have changed direction immediately once a certain threshold was met? * Would there have been two separate basins flowing in separate directions and the Eastern one slowly grew in size? * bWould a large part of South America be flooded into a giant dam, with outlets on both sides? I cannot clearly see what would have happened.

Yes, polynomials are always defined for all values of x (hence the infinite domain), and if the highest power is odd then it will tend to infinity in one x-direction and -infinity in the other. Graphing a function makes this clearer, for instance this particular function graphs as: From this it is pretty clear that all y-values are defined, but alternatively the function: f(x) = x^4 -2x^3 - 5x^2 (highest power is 4, even): Clearly this has a minimum which will need to be investigated.

Unless the highest power is even, in which case it will have a global minimum or maximum.

Let f(x) = y then, then to get the range of y =(x+5)^3+2 we need to know what (x+5)^3 + 2 is restricted to, since (x+5)^3 can equal any real number, then so can that +2, therefore y is all real y, then to get the range: y = (x+5)^3+2 Make x the subject x = (y-2)^(1/3) - 5 Just looking at this, you know that (y-2) is all real numbers ®, and R^(1/3) is all real numbers and so is R - 5 is all real numbers therefore x is all real numbers. Look for occasions where exponents or roots are even, that is where you will often encounter restrictions.

yeah i've been running it for several hours and everything is going perfect, ive tried other installs and been so confused, this one is great, im actually finally doing image manipulations in pHp, yay!

Thankyou both, Xampp sounds perfect, downloading it. Will come crawling back for more help if I fail miserably installing with this as well! Cheers.

I want to install apache, mysql, php and GD on my computer. In the past I have wasted ridiculous amounts of time trying to get them all running on old computers and there was always problems with version numbers and I never got the GD installed. I am hoping that there is an easy to install package of all these, with all the most usual settings and options enabled. A gui would be best but is not necessary if there are better distributions. Can you please point me to a project or offer some advice? Thanks alot -moosie

Problem © appears to be the easiest, especially if you think about it in reverse. Consider the equivalent problem, a person takes of from the ground at 4.4 m per s for 10 seconds. Call this height reached A, and this final velocity B. Calculate the length of time taken to fall under gravity before velocity B is acheived, then calculate the height fallen as C. The final answer is A + C. (A) and (B) if I am not mistaken are actually incredibly tedious, I tried to calculate a function for a general case of this problem for perfect accelleration and deceleration (for a game I was making) and never came across a neat solution.

This is much more complicated than I was expecting, thanks so much guys, your help is as always very valuable to me.

I have recently been studying the mathematics behind projectile motion through air resistance. We are given problems where the resistance force is proportional to v2 or v or something else (simple) containing v. I know that shape has alot to do with it but what is normally the function used for approximating air resistance for say, a spherical object. Or are such functions so complex that it would be impossible to solve without experimental data?

I think the main issue here is being open minded, and it comes in all degrees. There are narrow minded people everywhere including those who will completely disregard these "crackpot" theries and those who will stand by them to the day they die. Which is the crackpot? Both. Open mindedness is a personality trait, coupled with arrogance and education (or lack of) and you can bake yourself your very own crackpot.

It is correct.

I don't know about that particular equation you speak about but in general there is no 100% certainty that an equation is correct, it has just managed to stand up against experimental data better than any contenders. And it is not always a case that such equations cannot be derived from simpler premises. I found it interesting that newton's law of cooling and equations for motion etc can all be derived from very simple differential equations.

Light travels slower through materials with different refractive indexes, I can't remember the reason, but I'm pretty sure that temperature has nothing to do with it.