You are looking for a thing called fractional derivative and is a part of fractional calculus. In particular it is defined as $$(D^n f) (x)=\frac{1}{\Gamma(n)} \int_0^x (x-t)^{n-1} f(t) dt$$ Note ...

Since the polynomial is quadratic in both $x$ and $y$, you can write it in the form: $$f(x,y) =2 x^2 - x (3 y - 7) - (2 y^2 - 6 y + 4)$$ $$= p^i F_{ij} p^j=(x-x_0,y-y_0)^i F_{ij}(x-x_0,y-y_0)^j$$ ...

Note that the length $S$ of a parametrized curve $s(t)$ is given by $$S=\int_C \sqrt{x'(t)^2+y'(t)^2+z'(t)^2}dt$$ with the integration being along the curve $C$ and $s(t)=(x(t),y(t),z(t))$, given by ...

This is an Emden-Fowler type equation. Your specific case has a particular solution of $$y(x)=\pm\frac{2}{\sqrt{3}} x^{3/2},$$ but even with this, the original ODE doesn't seem to be solvable. As of ...

First set up a function $$f(a,b,c)=\sqrt{a+3} a+\sqrt{b+3} b+\sqrt{c+3} c.$$ Then, set $a=3-b-c$ to eliminate $a$ and get: $$f(b,c)=-\sqrt{-b-c+6} (b+c-3)+\sqrt{b+3} b+c \sqrt{c+3},$$ then find the ...

The absolute value of a number is always positive (or at least zero), ie for both $y_1=2$ and $y_2=-2$ we have $|y_1|=|y_2|=2$. Conversely, $|y|=1$ means $y=\pm 1$ as expected.

Set y=$Y/\sqrt{a}$ and see what happens ;-)

A proof using only the definitions of the derivative and the exponential function as limits, respectively $$y'(x)=\lim_{h\rightarrow0}\left(\frac{y(x+h)-y(x)}{h}\right),~~(1)$$ and $$e^x =\lim_{n\... View answer 0 votes Taking a derivative with respect to x yields$$ -n x^{(-1 + n)} - b_n(x) + b_n(1 + x)=0,$$which is a simple difference equation with solution$$b_n(t)=-n \zeta_{1 - n}(t),$$where \zeta_{1 - n}(t)... View answer 0 votes My guess is that since the covariant volume dV=\sqrt{-g} d^n x, where g=|g_{ij}| is the determinant of the metric, is conserved we have that:$$\sqrt{-g} dV_g=dV_{Euclidian} since the Euclidean ...
Playing around with the series numerically (plotting it etc) it seems it's equal to $\frac12 \sin^2(\frac{x}{2})$. I don't have a proof right now (and I'm too lazy to work one out), but it seems to ...