Homework Statement
integral of x*e^cos(x) from 0 to 6.
Homework Equations
I tried using integration by parts twice but no luck (couldn't find the integral of e^cos(x)). I was hoping I might use De Moivre's theorem but don't think it's applicable. I thought this looked a little like a La...
Yes, I copied the question down correctly. (I'm looking at the question right now and she can indeed row faster than she can walk.) Do you have any idea as to why my set up would give no real solutions? As Dango said x=6 should be a solution, so when I take the derivative I should get 6 as a...
Homework Statement
Jane is 2 miles offshore in a boat and wishes to reach a coastal village 6 miles down a straight shoreline from the point nearest the boat. She can row her boat at 5 mph and can walk at 3 mph. Where should she land her boat to reach the village in the least amount of time...
shouldn't we conclude that
|(x-a)(x+a)|= |x-a||x+a|< (2a+\delta)|x- a|
?
I still want to clearly define delta as a function of x, epsilon, and a (even though a is a constant)
I see the trick that was used to bound |x^2-a^2| and that was neat, but now I am confused.
If I solved...
then x+a approaches 2a...
I feel like I should just take delta = sqrt(epsilon), and I'm fairly confident any delta less than that will suffice. I do not really understand how to show that though.
Homework Statement
Show x^2 is continuous, on all reals, using a delta/epsilon argument.
Let E>0. I want to find a D s.t. whenever d(x,y)<D d(f(x),f(y))<E.
WLOG let x>y
|x^2-y^2|=x^2-y^2=(x-y)(x+y)<D(x+y)
I am trying to bound x+y, but can't figure out how.
Okay I see how to break it down case wise and find N accordingly. That will work nicely.
However, I was hoping to use the reverse triangle inequality but I run into the double abs. value. It just doesn't look right to say that
for any e>0 there exists and N s.t. for any n >N
|an - a| < e...
For a sequence in the reals
{an} converges to a, show {|an|} converges to |a|.
For any e>0 the exists an N s.t. for any n>N |an-a|<e
I want to use this inequality, but there is something funny going on. I do not know how to justify it.
|an-a|\leq||an|-|a||
d is a metric show d'= d/d+1 is a metric
I know d(x,z)\leqd(x,y)+d(y,z). And have been trying to make it fall out of this.
I have been fooling around with the terms but it has not provided to be useful. Any direction would be helpful.
f:X\rightarrowY, A\subsetX
f(Ac)=[f(A)]c implies f bijective.
Just trying to apply the definitions of injective and bijective. The equivalence makes sense but I am having a hard time showing it.
f(x)=f(y) implies x=y and for every y in Y there exists a x in X s.t. f(x)=y.
I mean all I have is...