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Does this mean that ##I_{max} = 4\pi n I_0## all the time? If so, then that would be all the coincidence needed.

Good point, Lev. The integral that you want to compute is the one around the branch cut. I think you would agree that the larger contour with the notch cut out encloses a region where the function is analytic, and therefore would have an integral equal to zero. Instead of using Cauchy's thm to...

Give ball A some velocity and mass. Ball B and ball C have the same mass but zero velocity. Assume an elastic collision. Conserving kinetic energy gives something like: ##\left(v^2_A m_A + v^2_Bm_B+v^2_Cm_C\right)_{pre}=\left(v^2_A m_A + v^2_Bm_B+v^2_Cm_C\right)_{post} ## Since all the balls are...

From what I've seen, AI is normally a blend of CE and advanced statistics (i.e. mathematics). I think that math would be the best 2nd degree, since it would be able to complement CE well and is a good springboard into the advanced physics needed to understand quantum computing.

The way you have this written is that -1 is greater than x AND x is greater than 1. Is that even possible? Remember when you negate an AND statement, like -1<x<1 which is read -1 is less than x AND x is less than 1, you will get an OR statement.

You should try to do this. At time = 0, you get ## u(x,0)=v(x,0) + h(x) ## Your cosine transform is just in terms of x, right? And for each n, you have an ODE to solve in terms of a function of t, which should be of the form: ## f'(t) = g(t) + c ## where your functions of x are treated as...

I am not sure that there is a standard, since it likely depends on your application. This is simply the Taylor approximation of the cosine function: ## \cos x = 1 - \frac12 x^2 + \frac{1}{24}x^4 - \frac{1}{720}x^6 ...## Therefore, you can cut off the higher order terms whenever you feel they are...

I believe this is usually handled by letting your C2 by a polynomial and solving for the boundary conditions.

That makes sense.

You will probably want to be careful of your divide by zero errors when x = +/- a.

Oh, I see. I read that information as saying that ##f(0,t) = U_0, \quad f(L,t) = U_1##. In this case, you seem to have mixed boundary conditions then. ##\frac{\partial u}{\partial x} (0,t) = 0, u(L,t) = U_1 ## To use the separation of variables after your last step, you will find an appropriate...

Got it. Since you didn't make that explicit in the proof, I was not following your logic. In that case, the only thing missing are the infinite endpoints. If ##\exists m \in \mathbb{E} \, s.t. \forall x \in E, x \geq m## and likewise if ##\exists M \in \mathbb{E} \, s.t. \forall x \in E, x...

To separate the variables, you assume that your function ##F## is a product of two functions, one dependent only on x, ##X(x)## and one dependent only on t ##T(t)##. If you let ## T(0) = 1##, then your initial conditions will fully describe ##X(x)##. In the case that ## \frac{\partial...

In your write-up, you refer to ##f(a_i) ## and ##f(b_i)##. Since ##f## is only defined in ##E##, wouldn't this mean that all your points ##a_i, b_i \in E## as well? I gather that the function you defined is sort of a 'connect the dots' sort of linear interpolation between disjoint points. What...

@Aows, I have not worked much with the FFCT, but it seems like the method is much like that of other transforms. After you have rewritten the derivatives, you should separate the variables, setting F(x,t) = X(x)T(t) and use standard ODE methods to solve for T(t). What is unclear to me it that...

@Aows, The FFCT assumes that ##C(x,t) = \sum_{n = 0}^{N} a_n \cos \frac{n\pi x}{L}## Where ##a_0 (t)= \frac{1}{L} \int_0^L f(x,t) dx \\ a_n(t) = \frac{2}{L} \int_0^L \cos \frac{n \pi x}{L} f(x,t) dx.## Apply the transform to the PDE, as you have done: ##\frac{\partial^2 f}{\partial x^2} =...

I would recommend starting with the definitions of T_1 space, infinite subset, and limit point. Use your definitions as fuel for your proof.

It sounds like you are assuming that there are an infinite number of points, and showing that there are an infinite number of subsets.

@Aows, I second the recommendation that you type out your problem and work. This will make it easier for us to follow your work and provide assistance. In your initial conditions, do you have ## u_t (x, 0) = 0, u(x, 0) = 6\sin ( \pi x ) - 3\sin (4\pi x) ## It looks like maybe it is an ##n\pi x##...

That's exactly what I meant. It has to be true for all the terms up to and including the current one, then you can inductively prove it for the next one ##a_{k+1}##. If the floor function is confusing, sometimes you can just get rid of it. Remember that the floor function is always less than...

I recommend you look at it this way: ##a_k = a_{\left\lfloor \frac{k}{5} \right\rfloor} +a_{\left\lfloor \frac{3k}{5} \right\rfloor} + k## assume ##a_k \leq 20k ## Then show that ##a_{k+1} \leq a_k + 20 \leq 20(k+1).##

You are not being asked for Z, and all your X values are given. What you are missing is the sigma, or standard deviation. In case your google search for "how to calculate standard deviation" didn't return any results, here is a start. Standard deviation is defined as the square root of the...

@Aows, Let's take a look at where these things came from: We write things in dimensionless terms so we can talk about the math in general. Then, based on the physics, apply some assumption, like: Which tells you that the pressure is decreasing as you move away from the well. In your source...

You can quote my username and PF as the source. I am currently an Asst. Prof of Mathematics.

Fitting curves really is a black art. You can apply some standard methods, but when you want to apply some additional finesse, everyone will tell you it depends, so there are few authoritative standards. By definition, the un-weighted model will minimize your un-weighted error. By changing the...

I have not used weights in Matlab before, but I think there is a need to normalize them. Using 1/SE seems to make the most sense, since SE is the standard error of the mean, which is what you are trying to plot. To normalize, I would recommend something like ## w = \frac{1}{1+|\overline{SE} -...

What if you were to let M be the mass of the aircraft? What would you get? Can you work out the acceleration components?

The simplest answer is that your negative sign would have to be included in the definition of ##y_{ch}.## However, if you are trying to describe a different physical environment by switching the sign, (leaving the rest of the values unchanged), you end up with infinitely increasing pressure at...

I am not sure what more to say. Depending on the problem you are working on, the right side could be any number of things. Each would have its own particular solution. If you consider that instead of -1 on the right, you had an arbitrary constant C, would that change things? Only a little. You...

The reason is that it is convenient to do. Generally, the ##p_D## value is decreasing as your radial component increases. The first expression gives you an idea of all the factors that contribute to the specific rate. By lumping them all together, you can focus on only the part of the equation...

That's what I get based on the conversions and approximations I used. It seems high, but I think that is why recently many consumer groups have advocated against ozone-producing air purifiers -- because they replace one type of pollutant with another. Ozone does naturally dissipate, so if you...

ppb is parts per billion, right? I can walk through the process, but I cannot vouch for how accurate the approximations are. According to Wikipedia, air at 20 degrees Celsius is about 1.2 kg per m^3. And molar mass of dry air is 0.0289644 kg/mol. This means that ## \frac{1.2}{0.0289644} ##...

For part a, I agree with Orodruin, there seems to be no such restriction on what t should be, other than non-zero. For part b, your development with the vectors you used was right. I suspect if you get the correct form for part a, you will get the correct solution for b. Try solving for the...

I think the answer is yes...but you have to be careful. If you are able to write a basis vector for one space in terms of basis vectors from another, then clearly, those vectors will be in the intersection. However, your space A is simply the xy-plane. It could have been written with other basis...

I'm not sure if my definition of partial fractions is the same as yours, but ##\int \frac{1}{\sqrt{1 -x^2} }\, dx ## and ##\int \frac{x}{\sqrt{1 -x^2}} \, dx## both have known integrals.

What if you multiplied by 1, i.e. ## 1= \frac{\sqrt{1-x}}{\sqrt{1-x}}## Then you might have something that looks like a known integral using trig functions.

Right. Your set ##\Omega_f## has elements which are periods of ##f##. Say you have one element that you know for sure, ##\omega_0##. Then, if your set ##\Omega_f## is not a discrete set, then you can form a sequence of subsets. Assume there exists a sequence of open neighborhoods, radius =...

They are periods because they are elements of Omega, the set of periods. So ##f(x + \omega_0) = f(x) ##. There is no assumption about the continuity of the function, only the periodicity. You are trying to show that ##\Omega_f## is discrete by proving (by contradiction) that it can only be...

Most of your steps look good. Remember that ##u^2 = f(x,y) ## implies ##u = \pm \sqrt{f(x,y)}.## This is important because you need to be consistent and keep ##\frac vu = -1##.

It looks to me like that came from the unprimed t in your original equation. ## t' = -\frac{R(t')}{c} +t## so ##\nabla t' = -\nabla \frac{R(t')}{c} +\nabla t## Is it reasonable to write ##\nabla t = \frac{-c}{R}\mathbf{R}##?

There is no rule against a sequence being composed of one repeated entry...however, for some fixed ##\omega_1## and ##\omega_2## with ##\omega_1 \neq \omega_2##, there is a fixed distance ##|\omega_1-\omega_2|<0##, therefore there will, at some point be a high enough n, such that ##\frac1n <...

The part where the angles have equal measure is the assumption that this triangle ABC is isosceles. If T1T2 is parallel to AB, then you have all kinds of tools to fill in the measures of the angles. I suppose these assumptions are easy enough to justify based on the definition of the inscribed...

I don't have a complete solution at this point, but I think the fact that p is monic tells you to start with the leading coefficients. You should be able to constrain the possible q's based on that.

Since you are using contradiction, you are assuming that the set is not discrete, right? If the set is discrete, then there is a radius around ##\omega_0## where no members of the neighborhood are in the set ##\Omega_f##. Because this is not true, you should be able to find members in...

I must have lost track of the negative sign somewhere--your new solutions seem to be in the right neighborhood. Look at the coefficient on your ##a_n## integral. It is the same as the one you used on your ##a_0## integral. It should be twice as big.

It sounds like it might be asking about even/odd ... which coefficients will be non-zero in your Fourier expansion. Maybe even the period, like the argument of your sines and cosines, etc.

You were right to use the half-period formula and notice it was an even function. You did not write it like one f(x) = |x|, not f(x) = x. For ##a_0##, it looks like you did 1/2pi, instead of 1/pi for the full period of pi. On the half-period formula, you want to double the result to get the full...

Your work looks solid at first glance...maybe an algebraic error, I'll go back and look at the details soon. Remember, your final form should be something like: ## f(x) = a_0 + \sum_{n=1}^\infty a_n \cos 2n x ##

You seem to be mixing up the formula. Your base for the exponent should be 1 + the percentage growth for each time increment...in this case months. If you were given a yearly growth rate and asked to figure out monthly growth, then n would be 12, but in this case, n is 1, and you can let t...

I see what they did there. Since the wire in infinitely long, B is only dependent on r, the distance from the line. So, if you slice up the equilateral triangle vertically, along lines with the same distance to the wire, you get ##\int\int B \, dA = \int B h(r) dr, ## where ##h(r)## is the...