x(x-y)

String theory scientists are saying they need 10 dimensions (9+1) to make their theory mathematically correct.

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.

 

And yet Einstein managed to derive length contraction, time dilation and mass-energy equivalence, so "must" is an over-reach. You need experiments to determine arbitrary coefficients, but not all equations are under-constrained.

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).

You have to set up a handle for each figure, then when you create a new line, you specify to plot it on that same handle.The plot command by default starts anew each timeSee http://stackoverflow.com/questions/12134406/matlab-several-graphs-in-1-loop-each-iteration-adds-a-line-on-every-figure for example.

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

I still think putting plot inside the loop will do the trick.

I tried that and it does not do the trick - I tried both of these:

 

plot(L,I)

plot(L,I(:,T))

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.

In the case of solving mathematical equations, using dimensionless forms of the equations is more reliable.

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.

I am not familiar with the language you are using. However shouldn't "end" be after "plot"? Alternatively should be a set of answers. The code may be using the last value.

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

function = intensity(L,T,c,h,k)

A = (2.*pi.*(c.^2).*h)./(L.^5);
B = exp((h.*c)./(L.*k.*T));

I = A.*((1)./(B - 1));

Main File

% Part A

% Define constants

h = 6.6260E-34; % Planck's Constant
c = 2.9979E8; % Speed of Light
k = 1.3806E-23; % Boltzmann's Constant

% Define array of radiation wavelengths

L = linspace(100E-9,3000E-9,1000);

% For-loop calling the function for intensity

for T = 3000:1000:8000
= intensity(L,T,c,h,k);
end

plot(L,)

xlabel('Wavelength L (m)');
ylabel('Intensity I (W m^{-2})');
title('Black Body Radiation Spectra');
grid on;

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.

 

[latex]\displaystyle \lim_{x \to +0} \frac{1}{x} = +\infty[/latex]

 

[latex]\displaystyle \lim_{x \to -0} \frac{1}{x} = -\infty[/latex]

 

That's why +Inf is unequal to -Inf

"+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 what I hope to find out. "i" like a value, and like any value it represents something, we know what most other values look like at least in some manner, I don't see how it is scientifically impossible to do the same with "i".

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.

Obviously "i" is a value, it's a value of something and that something can be found as represented by patterns extrapolated from nature, it's no different than any other symbol that we call a "number", it represents some kind of value, and what it represents obviously applies to reality in some way.

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]

I can still have a beautiful apple or an elegant apple, I can still picture an honorable apple, how about you just take the traits of what you expect an honorable person to look like and apply them to an apple?

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).

1/0 = +Infinity

1/0 = 1/(-0) = -1/0 = -(1/0) = -Infinity

so Infinity = -Infinity

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.

..."why can't I take the square root of a banana and eat the imaginary part?"...

Haha!