3b)
3c)
Therefore the poles are where the denominator is zero,
Thus,
1a)Maybe? Since this infinite series can be diffrenciated it must be countinous. Since diffrenciablility implies couninutitity.
1b)
The solution(s) to
are,
Since there are pure imaginaries,
thus,
Then you are looking at,
Thus,
Has pure imaginaries solutions only when,
Thus,
---
Solution,
1a) If the space is finite, then it is certainly compact.
Conversely, suppose the space X is compact. Consider the (open) covering . Since X is compact, there exists a finite subcovering As our metric is the discreet metric, So X is finite.
Now for C[a,b]. We should show that there exists some numerable and dense subset of C[a,b].
Surely, this cannot be done off the top of our heads. Weierstrass' Approximation Theorem, sais that the set of polynomials P[a,b] C[a,b] is dense.
If we could create a numerable and dense subset of P[a,b], we would be done.
Mumble, mumble... What about polynomials with rational coefficients?
Worth a try.
1b) Remember ,
so
so again
Bored to do any work (again), so lets try 3a.
-Note that sinz=-z+z^3/3!-... so sinz/z^7=(1/z^6)*(-1+...)
This means z=0 is a pole of order 6.
-Now e^{1/z}=1+1/z+1/(2!z^2)+... and so there are infinite terms in the series to diverge at z=0; this means z=0 is an essential singularity.
-Also (1-cosz)/z=(-z^2/2!-z^3/3!-...)/z=-z/2!-z^2/3!-...
so z=0 is a simple root.
...And just thought of an answer to 2)a).
There is a set of unit vectors to form a basis (*) for X. Consider any linear mapping not to vanish on S (we can have this as Y is nonempty) and define This mapping is well defined as S is a basis.
We then have
by the way is defined.
(*) a correction: we can extract such a set S from the actual basis set B -which I admit, need not be countable. Define the operator to be zero on B-S and everything works fine.
Lunch time...
No lunch for those not paid, so let's try 2b), which actually I could have figured out much sooner if I was not wasting braincells on myspace.
i) is reflexive. Just show that for every there exists a sequence with .
For this, consider a (continuous linear) functional on , and let the unit occupying the n-th place. This sequence is a basis for , as they are linearly independent and . We exploit linearity to obtain
,
so all functionals determine (and are completely determined by) the sequence c_n)" alt="(f((e_n)))=c_n)" />. To show this belongs to , take for any natural k, and using continuity
or
and since k was arbitrary, . On the other hand, Holder's inequality gives
so also , so these are equal.
ii) Direct application of the Open Mapping Theorem: There is
such that corresponding balls satisfy
,
so for we have or . So is continuous.