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Math Help - solutions

  1. #1
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    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
    Attached Thumbnails Attached Thumbnails solutions-scan1.jpg  
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  2. #2
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    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,...
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  3. #3
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    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
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  4. #4
    Super Member Rebesques's Avatar
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    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] \subsetC[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.
    Last edited by Rebesques; July 17th 2006 at 04:19 AM.
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  5. #5
    Super Member Rebesques's Avatar
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    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.
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  6. #6
    Super Member Rebesques's Avatar
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    ...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.
    Last edited by Rebesques; July 16th 2006 at 05:44 AM.
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  7. #7
    Super Member Rebesques's Avatar
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    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) \ell^p, \ 0<p<1 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 c_n)" alt="(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.
    Last edited by Rebesques; July 18th 2006 at 02:39 AM.
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