Math Is Fun Forum

i^2=-1
j^2=1
so k^2=0
e^kx=cosk x+ksink x
e^kx=1+kx+0x/2+0k^2/6...=1+kx
cosk x=1
sink x=x
so tank x=x asink x=x Acosk x=1 acosk x=anything atank x=x
e^ix is on the x^2+y^2=1
e^jx is on the x^2-y^2=1
so e^kx is on the x^2+0y^2=1 or x=+-1
e^ix x is half the area of a sector
e^jx x is half the area of a hyperbola sector thingy
then e^kx x is half the area of a triangle(does work with cosk and sink)
conclusion:the zero factor crumples stuff up a lot

Trying to solve a diff equation and spent a day on it and still it did not work?
Don't panic!I've got a way to quickly solve it.
First,integrate both sides with dx.
Then,take the natural logarithm of both sides.
Next,differentiate both sides with d/dx
Last,raise everything to the power e.
You've just wasted 5 minuites on nonsense integrating!PAAANIC!

so i was making a chemistry simulator
i figured out that doing the interatomic force simulation would be way easier than doing bonds
but question is what are the forces?(cllearly not lorentz forces in between the valence electrons)

∑(-1)^x->1-1+1-1+1.......->1/1-(-1)->1/2
When one takes one
plus one from one plus one
and on and on but ends
anon then starts again,
then some sum to one,
zero sum other ones.
One wonders who'd have won
had stopped not been done;
had he summed every bit
untill the infinite

Lest you should think that such
less well known sums are much
ado about nonsense
I do give these two cents:
The universe has got
an answer which is not
what most would first surmise,
it is a compromise,
and though it gives a laugh
the universe gives it "half".

(actual computation shows 0.5000000000000001,which is done by adding the floatation unit (0.1+0.2) and multipling it by 5/3)