2
The trick here is to separate places in a function (i.e., which argument is which) and the names of variables occupying those places. Suppose, for instance, that we define
$$
f(x, y) = x^3y
$$
and then ask "What's the derivative of $f(y, x)$ with respect to $x$?" Using the chain rule, one might be tempted to say that it's
$$
\frac{\partial f} {\...
2
The calculus argument is that you have $\frac{dx}{dk}\approx -\frac{x^2}{4}$ suggesting $\int -\frac1{x^2}\, dx \approx \int \frac14\, dk$ i.e. $\frac{1}{x}\approx \frac{k}{4}+B$ for some constant $B$, i.e. $x\approx\frac{1}{\frac{k}{4}+B}$. But here you have a discrete recursion so may need some adjustment.
When $0 < P_0=c \le 1$ you have the bounds $$\...
1
$$\dfrac{dp}{dr}r+p=0$$
Or more generally
$$\dfrac{dp}{dr}F(r)+G(p)=0$$
Is a linear first order homogeneous ODE.
As suggested in the comments can be solved by separation of variables or integrating factor.
1
\begin{gather*}
\frac{dy}{dx} =\frac{y^{3}}{2x^{3}} +\frac{3y}{2x}\\
\\
Let\ y=tx\\
\frac{dy}{dx} =t+x\frac{dt}{dx}\\
t+x\frac{dt}{dx} =\frac{t^{3}}{2} +\frac{3t}{2}\\
x\frac{dt}{dx} =\frac{t^{3} +t}{2}\\
\\
\int \frac{2dt}{t^{3} +t} =\int \frac{dx}{x}\\
\\
Now,\ \int \frac{2dt}{t^{3} +t} =\int \frac{2dt}{t\left( t^{2} +1\right)} =\int \frac{2tdt}{t^{2}\left(...
1
The line
print( np.dot( D, yl ).round( decimals=9 ) )
calculates the integrals of the Fourier transform at set of frequencies using the rectangle integration method:
\begin{equation}
F(k) = \int\limits^{\infty}_{-\infty}e^{-ikx}\left(\partial_x + x\right)e^{-\frac{x^2}{2}}dx=\left.e^{-ikx}e^{-\frac{x^2}{2}}\right\vert^{\infty}_{-\infty} +\int^{\infty}_{-\...
1
Such recursions that look like Bernoulli equations can often be solved using the "Bernoulli trick". That is, consider the corresponding negative power of $P$,
$$
P_k^{-1}=P_{k-1}^{-1}\left(1-\tfrac14P_{k-1}\right)^{-1}
=P_{k-1}^{-1}+\tfrac14+\tfrac1{16}P_{k-1}+\tfrac1{64}P_{k-1}^2+...\ge P_{k-1}^{-1}+\tfrac14
$$
so that
$$
P_k\ge \frac{k}4+P_0^{-1}\...
1
Compare
$$(x^2+6y^2)dx-4xydy=0~~~~(1)$$
with $$Mdx+Ndy=0 \implies \frac{\partial M}{\partial y}=12y,~~ \frac{\partial N}{dx}=-4y $$ It's integrating factor is
$$I=\exp \left(\int \frac{12y +4y}{-4xy}\right)=x^{-4}$$
Multiplying (1) with $x^{-4}$,(1) becomes an exact ODE:
$$(x^{-2}+6x^{-4} y^2) dx- 4x^{-3} ydy==0$$
It's solution is
$$\int (x^{-2}+6x^{-4}y^2) ...
1
We can check if the ODE is exact:
$$\Delta=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=\frac{\partial}{\partial y}(x^2+6y^2)-\frac{\partial}{\partial x}(-4xy)=16y\ne0.$$
So it's not exact, but as
$$\frac{\Delta}{N}=\frac{16y}{-4xy}=-\frac{4}{x}=f(x),$$
we can get an integrating factor with the form
$$\mu(x)=\exp\left(\int f(x)\mathrm{d}x\...
1
Let us compute $f'(u)$ for $u > 0$:
$$ f'(u) = -e^{\alpha u}\log(u)-\alpha u e^{\alpha u}\log(u) - e^{\alpha u} = -e^{\alpha u}(\log(u)+\alpha u \log(u) +1) $$
This is continuous on $(0,+\infty)$, so it is locally bounded on $(0,+\infty)$. Hence, $f$ is locally Lipschitz on $(0,+\infty)$.
However,
$$ \lim_{u \to 0⁺} f'(u) = +\infty, $$
hence $f$ is not ...
Only top voted, non community-wiki answers of a minimum length are eligible
