/sci/ - Science & Math

>>7508288
we're gonna need to use another branch of the logarithm for this one, boys

forgot to say, but log(5x) should be ln(5x). Wolframalpha did that, sorry

>>7508288
is it just me or does this explain nothing

>>7508288
it should be 0 as long as the contour isn't an infinite spiral that encloses the origin. Then the imaginary part remains finite, the real part is just log of the magnitude, and the denominator grows as |x|, which dominates log x.

>>7508315
/thread

>>7508288
your function isnt defined for x <= 0, so you cant look for a limit when x-> -infinity

>>7508841
your high school is showing

>>7508297
lold hard

>>7508313
I don't know what more you expect though

I was hoping one of you smart anons could help me with problems 1.5 and 1.6

I don't usually post stuff like this but im desperate and if a kind anon could point me in the right direction or outright solve it, i'd be eternally grateful

>>7508873
do you know what bijection means ? 1.5 is basically that...
what's the problem with 1.6 ?

You need to make an effort on what you need help on, this is basic shit.

>>7508873
The key fact that you need is that s^n is a group

1.5)Yes. Any bijection f from {1,2,3} to itself has a unique bijective inverse satisfying f^-1(f(x))=x for x = 1,2,3. f^-1 is also a permutation.
1.6) Yes. the composition of two bijections from {1,2,3} to itself is again a bijection from {1,2,3} to itself.

Good luck anon

>>7508877
>>7508874

Thank you kind anons could you please provide a little proof for 1.6 at least? I'm dying over here

>>7508288

It's an indeterminate form of inf/inf, apply L'Hopital's rule once and solve.

You'll get: (1/x) / (sqrt(x^2-1). Rewrite to (sqrt(x^2-1) / x^2 which gives 0. Eliminate -1 and you have x / x^2 which is 1/x. Voila'

>>7508884

You can also use symbolab.com to solve them most times. Also this board is not for homework.

>>7508882

I will sketch the proof for you anon.

We need to prove that if f,g:{1,2,3}->{1,2,3} are bijective then f o g is bijective. That means that it's both 1-1 and onto.

1-1:

If f(g(x))=f(g(y)) then g(x) = g(y) using the fact that f is 1-1.
But g(x) = g(y) implies x = y using the fact that g is 1-1. So f o g is 1-1.

onto:

'f is onto' means that given x in {1,2,3} there exists y in {1,2,3} such that f(y) = x.

Suppose we have x in {1,2,3}. Then since f is onto there exists y in {1,2,3} with f(y) = x. But y is in {1,2,3} and g is onto, so there exists z in {1,2,3} s.t. g(z) = y. Hence x = f(g(z)) = f o g (z). So f o g is onto.

>>7508288
??
(my limits are very rusted by the way)

>>7508887

anon delivers. Thank You. Of course, Id be happy for the input of others and will actively be monitoring this thread