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    Multiplication of fourier series

    Check out the Cauchy product, which has to do with multiplying series.. Wikipedia has a good article on it http://en.wikipedia.org/wiki/Cauchy_product" [Broken]
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    Nonlinear ODE by an infinite series expansion

    Okay,when I use a riccati substitution I get \frac{d^2u}{dx^2} - x^2u = 0 But now how to solve this? If I use a series expansion on this I get a solution, but it has a rather ugly recurrence relation. Maybe there is another way to solve this particular ODE, and I am totally brain farting...
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    Nonlinear ODE by an infinite series expansion

    I had done it the way suggested by loveequation, and you do end up with a pattern after you equate for the x^2 coefficient. I spent a couple of hours trying to find another way and was unsuccessful. The recurrence relation we obtain for the coefficients is not nice at all, but would be...
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    DG vs. PDE course

    Well for what it is worth the university I am getting my bachelors at does not even have a differential geometry class and it's the best math school in the state. Granted that state is Nevada...
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    DG vs. PDE course

    mordechai9- To me, and again this is my own opinion, the true practicality of PDE's (and ODE's for that matter) comes from the fact that once we are capable of solving these equations we are able to more accurately model a cornucopia of real life situations. For example in basic mechanics when...
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    DG vs. PDE course

    As far as my experience goes, you are unable to separate PDE's from practicality, as many of the PDE's that we work with (particularly second order) arise from physical problems. But in the end I suppose it is just my opinion. It will largely depend on the instructor how theoretical or...
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    DG vs. PDE course

    I say PDE's, mostly because it will be most immediately useful to you as a physicist. That is not to say that Differential Geometry is not useful, but you will use PDE's before you use DG, at least that is my experience. Also to consider is you own research focus in physics. If you are going...
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    Who can find the solution to this ODE?

    I had not thought of turning it into a Bernoulli equation. Actually though I found out another, more elementary way of doing it using exact differentials: \[ \frac{dy}{dx} = \frac{x^2y^2 - y}{x} \Rightarrow x dy = x^2y^2 dx - y dx \] Divide through by y^2 \[ x\frac{1}{y^2} dy = x^2...
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    Who can find the solution to this ODE?

    Looking for the solution to the following ODE: \[ \frac{dy}{dx} = \frac{x^2y^2 - y}{x} \]
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    Need proof

    First consider the attached diagram, convince yourself of its validity. Please note that since the small right triangle at the top has angle A, then the length of its sides must be a scalar multiple of the right triangle with hypotenuse 1. Clearly, by the Pythagorean theorem the hypotenuse...
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    Maximum velocity of a vehicle moving downhill

    What I am trying to show is that given a hill with a 2% grade, a scooter with a passenger (a combined weight of roughly 500 lbs) can exceed 40mph, where the scooter has a 2 horsepower engine, and has a flatland top speed of 30mph. Now I believe that there may be no max velocity given an...
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    Maximum velocity of a vehicle moving downhill

    The reason I put the max velocity on flat ground was because to my way of thinking, it gives some information about what it's max speed downhill might be. I appreciate the complexity of the issue, and have garnered some insight to the problem through your comments. If I find a reasonable...
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    Maximum velocity of a vehicle moving downhill

    Suppose some vehicle with mass m is moving down a hill with angle \theta . Say that the vehicle is exerting a force parallel to the ground of magnitude f . Suppose that the maximum velocity of the vehicle on flat ground is v_{max} . What is the max velocity of the vehicle down the...
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    How To Find Sides Of Right Triangle With Known Angles And Area?

    Let a be the adjacent, let b be the opposite, and let c be the hypotenuse and A the area. Then \[ c = \sqrt{\frac{2A}{\cos \theta \sin \theta}} \] \[ a= \sqrt{2A \cot \theta} \] \[ b = \sqrt{2A \tan \theta} \]
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