# solutions

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• Jun 17th 2006, 05:14 AM
praveen
solutions
Dear sir/Madam,

I am praveen here,
please find SCANNED Attachments.These contains some problems.
i need solution for these problems.
I request to send solutions.

thanking you

Regards,
Praveen
• Jun 17th 2006, 07:25 PM
ThePerfectHacker
3b)
$x\Gamma(x)=\lim_{n\to\infty}\frac{n^xn!}{(1+x)(2+x )....(n+x)}$
3c)
Therefore the poles are where the denominator is zero,
$(1+x)(2+x)....=0$
Thus,
$x=-1,-2,-3,...$
• Jun 17th 2006, 07:31 PM
ThePerfectHacker
1a)Maybe? Since this infinite series can be diffrenciated it must be countinous. Since diffrenciablility implies couninutitity.
1b)
The solution(s) to
$x^2+\alpha x+\beta=0$
are,
$x=\frac{-\alpha\pm \sqrt{\alpha-4\beta}}{2}$
Since there are pure imaginaries,
$-\alpha =0$ thus, $\alpha =0$
Then you are looking at,
$x^2+\beta=0$
Thus,
$x^2=-\beta$
Has pure imaginaries solutions only when,
$-\beta<0$
Thus,
$\beta>0$
---
Solution,
$\alpha=0,\beta>0$
• Jul 4th 2006, 08:22 AM
Rebesques
1a) If the space is finite, then it is certainly compact.

Conversely, suppose the space X is compact. Consider the (open) covering $\cup_{x\in X} B(x,1/2)$. Since X is compact, there exists a finite subcovering $X\subset B(x_1,1/2)\cup\ldots\cup B(x_n,1/2).$ As our metric is the discreet metric, $B(x_1,1/2)=\{x_1\},\ldots, B(x_n,1/2)=\{x_n\}.$ So $X\subset \{x_1,\ldots, x_n\}\Rightarrow$ 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] $\subset$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 $||T||={\rm sup}_{x\neq 0}\frac{||Tx||_Y}{||x||_X}$,
so
$||T||={\rm sup}_{x\neq 0}\frac{||Tx||_Y}{||x||_X}\geq \frac{||Tx||_Y}{||x||_X}, \ \forall x\in X$
so again
$||T||\geq \frac{||Tx||_Y}{||x||_X}, \ \forall x\in X.$
• Jul 5th 2006, 11:51 AM
Rebesques
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.
• Jul 15th 2006, 11:28 AM
Rebesques
...And just thought of an answer to 2)a).

There is a set of unit vectors $S=\{e_1,\ldots,e_n,\dots\}$ to form a basis (*) for X. Consider any linear mapping $T:X\rightarrow Y$ not to vanish on S (we can have this as Y is nonempty) and define $\overline{T}(e_n)=\frac{n}{||T(e_n)||}T(e_n), \ \forall n\in \mathbb{N}.$ This mapping is well defined as S is a basis.

We then have

$||\overline{T}||={\rm sup}_{||x||=1}\frac{||\overline{T}(x)||}{||x||} \geq {\rm sup}_{n} \frac{||\overline{T}(e_n)||}{||e_n||}=+\infty$

by the way $\overline{T}$ 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. :) :o
• Jul 17th 2006, 06:33 AM
Rebesques
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. :o

i) $\ell^p, \ 0 is reflexive. Just show that for every $f\in \ell^p'$ there exists a sequence $(c_n)\in\ell^q, \ (1/q)+(1/p)=1$ with $||f||=||(c_n)||_{\ell^q}$.

For this, consider a (continuous linear) functional $f$ on $\ell^p$, and let $e_{n}=(0,0,\ldots,0,1,0,\ldots),$ the unit occupying the n-th place. This sequence is a basis for $\ell^p$, as they are linearly independent and $(a_n)=\sum a_n(e_n)$. We exploit linearity to obtain

$f((a_n))=\sum a_n f((e_n))$,

so all functionals determine (and are completely determined by) the sequence $(f((e_n)))=:(c_n)$. To show this belongs to $\ell^q$, take $(a_n)=(|c_1|^{q-1},\ldots,|c_k|^{q-1},0,\ldots)$ for any natural k, and using continuity

$|f((a_n))|\leq ||f|| \cdot||(a_n)||_{\ell^p}$ or

$\bigg\{\sum_{i=1}^k |c_i|^q\bigg\}^{1/q}\leq ||f||$

and since k was arbitrary, $||f||\geq ||(c_n)||_{\ell^q}$. On the other hand, Holder's inequality gives

$|f((b_n))|=|\sum b_n c_n|\leq ||b_n||_{\ell^p}||c_n||_{\ell^q}, \ \forall (b_n)\in \ell^p$

so also $||f||\leq ||(c_n)||_{\ell^q}$, so these are equal.

ii) Direct application of the Open Mapping Theorem: There is
$\epsilon>0$ such that corresponding balls satisfy

$B_Y(0,\epsilon)\subset T(B_X(0,1))$,

so for $y\in Y, \ ||y||_Y=1$ we have $||T^{-1}(\epsilon y)||_X\leq 1$ or $||T^{-1}(y)||_X\leq 1/\epsilon$. So $T^{-1}$ is continuous.