It's not hard to show that the function:
g = \frac{1}{2} (c \times r)
is a "vector potential" function for the constant vector "c". That is, that:
\nabla \times g = c
The calculation is straightforward to carry out in Cartesian coordinates, and I won't reproduce it here.
However...
Is it true in general that:
|\int f(x)dx| < \int |f(x)|dx
Not sure if "Triangle Inequality" is the right word for that, but that seems to be what's involved.
Interesting, thanks.
Yes, that makes sense.
Also, if we don't yet "know" what the symbol "e" is supposed to mean, I suppose one could also proceed by supposing that the solution y is in the form of an (indeterminate) polynomial, in the sense of:
y(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + ...
Yes .. in point of fact, the question arose in my mind while I was reading one of the older calculus texts, specifically, Forsythe on Differential Equations. Forsythe mentions passingly that all the "properties" of the logarithm can be derived from its integral representation/definition, but...
I thought about doing a Taylor expansion about x=1, that is to say:
f(1+x) = f(1) + f'(1)x + \frac{1}{2!}f''(1)x^2 + \frac{1}{3!}f'''(1)x^3 + ...
f(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + ...
but I'm not sure this leads anywhere fruitful.. First, we'd have to get...
If we define a function f(x) such that:
f(x) = \int_{1}^x \frac{dt}{t}
for x>0, so that:
f(y) = \int_{1}^y \frac{dt}{t}
and
f(xy) = \int_{1}^{xy} \frac{dt}{t}
is there a way, using just these "integral" definitions, to prove that:
f(x) + f(y) = f(xy)
Clearly, the function...
I had grown up thinking that the transcendental functions (to wit, e(x), sin(x), cos(x), etc) were (somewhat) arbitrary functions, "defined" simply by the fact that they could not be expressed as polynomials (or some such loose definition).
Indeed, from Wikipedia, we read that:
But now I'm...
So in my first equation above, what I should have written, should have been:
\Delta T \approx \frac{\partial T}{\partial x} \Delta x
+\frac{1}{2}\frac{\partial^2 T}{\partial x^2}\Delta x^2
+ \frac{1}{6} \frac{\partial^3 T}{\partial x^3} \Delta x^3 + ...
Yes?
In physics texts, its customary to write (and even to define the gradient as) the following:
dT = (\nabla T) \cdot dl
Working in Cartesian coordinates, we can expand this as follows:
dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial...
One way to get the same result, w/o introducing line integrals, would be as follows.
You have a divergence-less vector field F, so you know:
\nabla \cdot F = 0
and
F = \nabla \times A
for some vector field A. From Gauss we have that:
\large{ \int_V (\nabla \cdot F) dV =...