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

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- Jun 17th 2006, 05:14 AMpraveensolutions
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 PMThePerfectHacker
3b)

$\displaystyle x\Gamma(x)=\lim_{n\to\infty}\frac{n^xn!}{(1+x)(2+x )....(n+x)}$

3c)

Therefore the poles are where the denominator is zero,

$\displaystyle (1+x)(2+x)....=0$

Thus,

$\displaystyle x=-1,-2,-3,...$ - Jun 17th 2006, 07:31 PMThePerfectHacker
1a)Maybe? Since this infinite series can be diffrenciated it must be countinous. Since diffrenciablility implies couninutitity.

1b)

The solution(s) to

$\displaystyle x^2+\alpha x+\beta=0$

are,

$\displaystyle x=\frac{-\alpha\pm \sqrt{\alpha-4\beta}}{2}$

Since there are pure imaginaries,

$\displaystyle -\alpha =0$ thus, $\displaystyle \alpha =0$

Then you are looking at,

$\displaystyle x^2+\beta=0$

Thus,

$\displaystyle x^2=-\beta$

Has pure imaginaries solutions only when,

$\displaystyle -\beta<0$

Thus,

$\displaystyle \beta>0$

---

Solution,

$\displaystyle \alpha=0,\beta>0$ - Jul 4th 2006, 08:22 AMRebesques
1a) If the space is finite, then it is certainly compact.

Conversely, suppose the space X is compact. Consider the (open) covering $\displaystyle \cup_{x\in X} B(x,1/2)$. Since X is compact, there exists a finite subcovering $\displaystyle X\subset B(x_1,1/2)\cup\ldots\cup B(x_n,1/2).$ As our metric is the discreet metric, $\displaystyle B(x_1,1/2)=\{x_1\},\ldots, B(x_n,1/2)=\{x_n\}.$ So $\displaystyle X\subset \{x_1,\ldots, x_n\}\Rightarrow$ 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]$\displaystyle \subset$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 $\displaystyle ||T||={\rm sup}_{x\neq 0}\frac{||Tx||_Y}{||x||_X}$,

so

$\displaystyle ||T||={\rm sup}_{x\neq 0}\frac{||Tx||_Y}{||x||_X}\geq \frac{||Tx||_Y}{||x||_X}, \ \forall x\in X$

so again

$\displaystyle ||T||\geq \frac{||Tx||_Y}{||x||_X}, \ \forall x\in X.$ - Jul 5th 2006, 11:51 AMRebesques
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 AMRebesques
...And just thought of an answer to 2)a).

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

We then have

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

by the way $\displaystyle \overline{T}$ 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 AMRebesques
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) $\displaystyle \ell^p, \ 0<p<1$ is reflexive. Just show that for every $\displaystyle f\in \ell^p'$ there exists a sequence $\displaystyle (c_n)\in\ell^q, \ (1/q)+(1/p)=1$ with $\displaystyle ||f||=||(c_n)||_{\ell^q}$.

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

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

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

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

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

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

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

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

ii) Direct application of the Open Mapping Theorem: There is

$\displaystyle \epsilon>0$ such that corresponding balls satisfy

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

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