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What is a meaning of homotopic in Cauchy's theorem for the sets?

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Homework Statement



three quistions':
1) What is a meaning of (homotopic) in Cauchy's theorem for the sets'?
and in sets what is t,s where H(t,s)

2)
(-1)^1/2*(-1)^1/2 = (-1*-1)^1/2=1^1/2=1=/=i^2




3)
In Integral functions' by Reiman Sum
is delta x
It handles any inclination at the point
Concluded that by diminishing its value to zero

Homework Equations





The Attempt at a Solution



2) (-1)^1/2*(-1)^1/2 = i(1)^1/2*i(1)^1/2=i*i=-1=/=1
But the former right way
The result is wrong
logic Sports saysThe right does not lead to an error
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement



three quistions':
1) What is a meaning of (homotopic) in Cauchy's theorem for the sets'?
Two paths in a topological space (or the functions defining those paths) are said to be "homotopic" if one can be "continuously deformed" into the other. More specifically, if f:R-> X is a function from the real numbers to a space X (so that f defines a one-dimensional path in X) and g:R->X is another, then f and g are homotopic if and only if there exist a function [itex]\Phi(t, u)[/itex] where [itex]0\le t\le 1[/itex], u is real number, continuous in both t and u, and such that [itex]\Phi(0, u)= f(u)[/itex], [itex]\Phi(1, u)= g(u)[/itex].

and in sets what is t,s where H(t,s)
I don't recognize this. What is H? What area of "sets" do you mean? Since you mention "homotopic" could this be a homotopy group?


2)
(-1)^1/2*(-1)^1/2 = (-1*-1)^1/2=1^1/2=1=/=i^2
No, axbx= (ab)x does not, in general, hold for complex numbers.



3)
In Integral functions' by Reiman Sum
is delta x
It handles any inclination at the point
"Inclination"? I think that's a mis-translation. delta x measures the slight change in x on which you are basing your Riemann Sum.

Concluded that by diminishing its value to zero
It's a bit more complicated than that. You must let delta x go to 0 while, at the same time letting the number of terms in the sum go to infinity. In any case, I see no question here.

Homework Equations





The Attempt at a Solution



2) (-1)^1/2*(-1)^1/2 = i(1)^1/2*i(1)^1/2=i*i=-1=/=1
But the former right way
The result is wrong
logic Sports saysThe right does not lead to an error
"logic" doesn't work if you start from incorrect premises. As I said above, axbx is not generally true for complex numbers. It is the "former" that is wrong. The definition of "1/2 power" or square root, is that (a1/2)(a1/2)= (a1/2)2= a for any complex number a. (-1)1/2(-1)1/2= -1 is correct.
 
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  • #3
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Thanks you the abundant thanks
:rolleyes:
 

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