Epsilon delta surroundings question

nhrock3
Messages
403
Reaction score
0
i need to prove that for every delta
i call d=delta
e=epsilon
1/|x+1|>e
we can choosr any e we want
so they took e=1/4
but because the innqualitty needs to work for every delta
they took x=min{2,(1+d)/2}
for d>2 it takes x= 1+d /2
for d<2 takes x=2
uppon what logic they found this formula x=min{2,(1+d)/2}
 
Physics news on Phys.org
nhrock3 said:
i need to prove that for every delta
i call d=delta
e=epsilon
1/|x+1|>e
we can choosr any e we want
so they took e=1/4
but because the innqualitty needs to work for every delta
they took x=min{2,(1+d)/2}
for d>2 it takes x= 1+d /2
for d<2 takes x=2
uppon what logic they found this formula x=min{2,(1+d)/2}

I don't understand the question. You say "for every delta", but then the equation that follows doesn't have a delta in it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top