x(x-y)

Depends upon which string theory you're referring to - I believe that taking into account all different versions of string theories, the number of dimensions they predict range from 7 to 26 or something like that.

In reply to both comments above: Indeed, I understand that there are a select few cases where physically correct equations can be obtained purely from theoretical analysis and thus perhaps my wording of "must" was too extreme given these cases; however, my overall point (absent the "must" wording) stands. I appreciate the corrections, thank you.

Yes, thought experiments are all very well - but until they can be confirmed by actual experiments, the "results" are largely meaningless (that is not to say that thought experiments are not useful however).

Ah, I see, that was very helpful - thanks.

I tried that and it does not do the trick - I tried both of these: inside the for loop and they both yielded the same result - only one curve was plotted; i.e. all the other temperatures were overwritten except for 8000K. Thus, somehow I need to extract an array for each temperature result for the intensity from the single for loop and then plot a graph of these. Instead, for now, I have just written multiple for loops for each temperature which has given me the graph I wanted - I will keep thinking of a way to do this as described above.

Agreed, but my main point was that obtaining the physically correct equation is not done using mathematics - one must experiment to achieve this.

Also, it's important to note that one generally obtains the correct equation (both dimensionally sound and physically correct) through means of experimentation (i.e. empirically) not simply by using mathematics - of course, the theoretical derivations can help you to achieve the correct equation for some quantity, however at some point you are going to have to do an experiment(s) in order to obtain a meaningful result which is physically correct. This goes for differential equations too, as you may be able to find the general solution of a differential equation through mathematics - however to obtain the complete solution you will need boundary/initial conditions which are obtained through experiment.

First thing that came to my head was "material deformation", so perhaps an experiment involving Young's Modulus of the tissue paper - measuring the stress [latex]\frac{F}{A}[/latex] and strain [latex]\frac{\Delta L}{L}[/latex] on the tissue paper? I don't know, maybe that's silly but it's the first thing I thought of, I'm sure someone here can come up with something more interesting/exciting.

So, I managed to obtain the right plot in the end by using several for loops for the different temperatures. However, I am sure there is a quicker and more efficient way to achieve this - I am fairly new to MATLAB, so I guess I'll learn these things over time.

Hmmm, I see, yes that could be the problem. Perhaps I need another for loop within the first one where I can define the wavelength array too. I'll try to solve this later after some sleep.

The "end" is to denote the end of the for loop - "plot" can be defined inside the for loop but it would have to be written in a different way than simply defining it outside the loop as I have done. If I print to the command window then it is appearing as an array - however, what concerns me is that a lot of the values are equal to zero for some reason.

I have been trying to plot a number of black body radiator spectra curves in MATLAB but I come across a problem - only one curve is plotted when I run the program. Essentially, I am plotting intensity I against wavelength [latex]\lambda[/latex] for an array of temperatures from 3000K to 8000K in steps of 1000K with a wavelegnth array of 100nm to 3000nm with 1000 intervals, using Planck's radiation formula: [latex]I(\lambda,T) = \frac{2\pi c^2 h}{\lambda^5} \left(\frac{1}{e^\frac{hc}{\lambda kT} -1} \right)[/latex] where T is th absolute temperature, c is speed of light, h is Planck's constant and k is Boltzmann's constant. Here is my code: Function File Main File Any help will be greatly appreciated, thanks! P.S: Just in case it's difficult to read - there are dots in front of the operators which matter in the equation in order for the program to recognise element-by-element computation.

"+0" and "-0" are exactly the same quantity - thus rendering your statement as incorrect.

[latex]\displaystyle \int \frac{dx}{a\sin x + b\cos x + c\tan x} \hspace{10mm}a,b,c \in \mathbb{R} \ne 0[/latex] [latex]let \; t = \tan \left(\frac{x}{2}\right) \Rightarrow dt = \frac{1}{2} \sec^2 \left(\frac{x}{2}\right) dx[/latex] [latex]\Rightarrow dx = \frac{2\, dt}{\sec^2\left(\frac{x}{2}\right)} = \frac{2\, dt}{1+ t^2}[/latex] [latex]\sin \left(\frac{x}{2}\right) = \frac{t}{\sqrt{1 + t^2}}[/latex] [latex]\cos \left(\frac{x}{2}\right) = \frac{1}{\sqrt{1 + t^2}}[/latex] [latex]\Rightarrow \sin x = 2\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) = \frac{2t}{1 + t^2}[/latex] [latex]\Rightarrow \cos x = \cos^2 \left(\frac{x}{2}\right) - \sin^2 \left(\frac{x}{2}\right) = \frac{1-t^2}{1+t^2}[/latex] [latex]\Rightarrow \tan x = \frac{2\tan \left(\frac{x}{2}\right)}{1-\tan^2 \left(\frac{x}{2}\right)} = \frac{2t}{1-t^2}[/latex] [latex]\Rightarrow \displaystyle \int \frac{\frac{2\, dt}{1 + t^2}}{\frac{2at}{1 +t^2} + \frac{b(1-t^2)}{1 + t^2} + \frac{2ct}{1-t^2}}[/latex] [latex]\Rightarrow \displaystyle \int \frac{2\, dt}{2at + b(1-t^2) + \frac{2ct(1+t^2)}{1-t^2}}[/latex] Where does one go from here? I multiplied numerator and denominator by 1-t^2 but that just left me with a quadratic over a quartic - I suppose I could try partial fractions but the quartic would be difficult to factorise. P.S. This is not a homework problem, just me "doodling".

Ah, right ok - so you have knowledge of solving linear equations, quadratic equations and some basic knowledge of complex numbers? Do you have any understanding of calculus? For example, can you do the derivative and integral of [latex]x^2[/latex]? I'm not sure which country you're from, but a good point to start at would be GCSE to AS-Level mathematics; in the latter case, the books written for the OCR syllabus are good (just look up - "Core 1 OCR AS Level" on amazon or in any major bookshop here in the UK if you live here). However, given the fact that you said "Algebra 1 and 2" - I think that's the American system, is it not? Anyway, I'll try to think of some good introductory mathematics books if I have time. I know that "Calculus" which is a part of the Schaum's series of Maths examples books is a good one for practice on calculus questions.

If you square root a number, let's call it x where x > 0, and then square root it again - notice that you will obtain [latex]\sqrt{\sqrt{x}} = \left(x^{\frac{1}{2}}\right)^{\frac{1}{2}} = x^{\frac{1}{4}}[/latex] so, if you perform the square root on this variable x an "infinite amount of times" then you will obtain the term [latex]x^{\frac{1}{\infty}} = x^0 = 1[/latex] Thus, we can say that [latex]\displaystyle \lim_{n \to \infty} x^{\frac{1}{n}} = x^0 = 1[/latex]

This is just getting silly now, this thread no longer belongs in mathematics.

That's because, for all intents and purposes, i doesn't actually exist (it is the imaginary unit after all). Sure the complex number based arithmetic works perfectly, and beautifully as it gives us some of the most elegant equations in Physics and Mathematics, however no matter how hard you try - you cannot perform a square root function on a negative number and obtain a real value, you will always obtain an "imaginary number"; but, as I alluded to above, this is perfectly fine and it clearly works (it's logical).

What level are you at, at the moment? Have you any formal education in mathematics? Or are you starting from the real basics of algebra? For example, if I was to ask if you could solve this linear equation [latex]3x - 5 = 4[/latex] for x, could you do that easily? I'm just asking to see where you are currently, apologies if you are at a higher level than I assume.

Of course, I understand the applications of complex numbers to reality - I have studied, and continue to study, quantum mechanics, electromagnetism, optics and waves, and so I can appreciate the importance (and absolute necessity in the case of QM) of complex numbers in these fields of Physics. However, I maintain the fact that sticking the word apples after i doesn't give any sort of viable, sensible physical quantity and simply does not make sense. What do you actually think "i apples" does mean?

Indeed, as imatfaal stated, one can easily derive an "imaginary expression" for the natural logarithm of a negative number by applying Euler's identity: [latex]Ne^{i\pi} = -N[/latex] where [latex]N \in \mathbb{R}[/latex] [latex]\Rightarrow \ln\left(Ne^{i\pi}\right) = \ln(-N)[/latex] [latex]\Rightarrow \ln N + \ln e^{i\pi} = \ln (-N)[/latex] [latex]\therefore \ln (-N) = i\pi + \ln N[/latex]

Because even if you can do it tenuously through the use of pretentious language and silly poetical nonsense, you cannot do it in mathematics - it just does not make any sense to slap the word apples in front of the imaginary unit i and expect to know what this so-called "quantity" (in your opinion it's a quantity) is. The closest you can get is probably just by saying you have some purely imaginary complex number z of the form z = ai - call a "apples" if you like (it still wouldn't make any sense to, and I don't understand what your motivation behind doing so would be) , but it is just a real coefficient of the imaginary unit (i.e. the imaginary part of z).

It is true that [latex]\displaystyle \lim_{x \to 0} \frac{1}{x} = \infty[/latex] however [latex]\frac{1}{0} \ne \infty[/latex] If 1/0 = n, where n is anything but undefined, then there would have to exist a variable n which when multiplied by zero equals 1. There is no such number (of any type), thus the correct way to represent 1/0 is by simply stating that it is undefined.

Yes, infinity is used in mathematics but the same cannot be said of the arbitrary object set of "apples". Why do you even think i apples should make some sense? It just doesn't, as Bignose said you can't just put words together (or words and the imaginary unit in this case) and expect them to make sense.

Haha!