How can we define a constraint saying:
if b is larger than zero then y has to equal 1
if b is equal to zero then y has to equal 0
?
It concerns the following problem:
b[i][k] = amount of vegetable i growing on patch k
y[i][k] = 1 if vegetable i is grown on patch k
= 0 otherwise
So...
I have to find: \int_{0}^{2\pi}\sqrt{t^2+2} dt
I found that \int \sqrt{t^2+2} dt = \frac{t\sqrt{t^2+2}}{2} - arcsin(\frac{t}{\sqrt{2}}) + c
But when I fill in 2\pi I get: \frac{2\pi \sqrt{4\pi ^2+2}}{2}- arcsin(\frac{2\pi }{\sqrt{2}})
but arcsin(\frac{2\pi }{\sqrt{2}}) doesn't exist..
Have I...
I have to find: \int_{0}^{2\pi}\sqrt{t^2+2} dt
I found that \int \sqrt{t^2+2} dt = \frac{t\sqrt{t^2+2}}{2} - arcsin(\frac{t}{\sqrt{2}}) + c
But when I fill in 2\pi I get: \frac{2\pi \sqrt{4\pi ^2+2}}{2}- arcsin(\frac{2\pi }{\sqrt{2}})
but arcsin(\frac{2\pi }{\sqrt{2}}) doesn't exist...
How can the function
f: ℝ² → ℝ : (x,y) |--> {{x^3-y^3}\over{x-y}} if x ≠ y
be defined on the line y=x so that we get a continuous function?Is this correct?: If x=y --> f=0
How can the function
f: ℝ² → ℝ : (x,y) |--> {{x^2+y^2-(x^3y^3)}\over{x^2+y^2}} if (x,y) ≠ (0,0)
be defined in the origin so that we get a continuous function?When I take 'x=y' (so (y,y)) and 'y=x' (so (x,x)) I get:
{{2-y^4}\over{2}}
and
{{2-x^4}\over{2}}
So for the first one I get '1'...
If you simplify them you get:
log(sin(x/2)/(sin(x/2)+cos(x/2))
But if I fill in 0 than I get log(0) but that doesn't exist..
If I use a double angle formulae I also get log(0)