Chronological Thread 

< Chronological >      < Thread >

---

I am trying to print a long document with many linked images.
In the document the images are correclty displayed.
When I print or export to pdf I get only blank space instead of images.
Any suggestion?
I include the files in question
PS
I am using Texmacs 1.0.5.10 on Debian unstable.

Attachment: fit_dec0.985_0_6.pdf
Description: Adobe PDF document

Attachment: fit_dec0.985_1_6.pdf
Description: Adobe PDF document

<TeXmacs|1.0.5.10>

<style|generic>

<\body>
  <section|Primi fit>

  Inanzitutto sappiamo che il correlatore si comporta come:

  <\equation*>
    D(t)\<sim\>\<mu\>+<frac|e<rsup|-M*t>|t<rsup|<frac|3|2>>>+<frac|e<rsup|-M(L-*t)>|(L-t)<rsup|<frac|3|2>>>
  </equation*>

  Il caso piu' semplice consiste nel tralasciare il secondo termine e
  concentrarsi sui punti in cui\ 

  <\equation*>
    e<rsup|-M(L-*t)>\<sim\>0
  </equation*>

  In questo caso facendo la derivata del logaritmo si ottiene:

  <\equation*>
    <frac|\<partial\>|\<partial\>t>ln D(t)=<frac|1|\<mu\>+<frac|e<rsup|-M*t>|t<rsup|<frac|3|2>>>><left|(><frac|-M*e<rsup|-M*t>|t<rsup|<frac|3|2>>>-<frac|3|2><frac|*e<rsup|-M*t>|t<rsup|<frac|5|2>>><right|)>
  </equation*>

  Supponiamo inoltre di essere in una situazione in cui
  <with|mode|math|\<mu\>\<gg\>1> otteniamo allora:

  <\equation*>
    <frac|\<partial\>|\<partial\>t>ln D(t)=<frac|1|\<mu\>><left|(>1-<frac|e<rsup|-M*t>|\<mu\>*t<rsup|<frac|3|2>>>+\<ldots\>.<right|)><left|(><frac|-M*e<rsup|-M*t>|t<rsup|<frac|3|2>>>-<frac|3|2><frac|*e<rsup|-M*t>|t<rsup|<frac|5|2>>><right|)>
  </equation*>

  Questo a prim ordine in <with|mode|math|e<rsup|<rsub|-M*t>>> diventa:

  <\equation*>
    <frac|\<partial\>|\<partial\>t>ln D(t)=<frac|1|\<mu\>><frac|e<rsup|-M*t>|t<rsup|<frac|3|2>>><left|(>-M-<frac|3|2><frac|*|t<rsup|>><left|)>
  </equation*>

  La forma adeguata per provare il fit e':

  <\equation*>
    <frac|\<partial\>|\<partial\>t>ln D(t)=A*<frac|e<rsup|-M*t>|t<rsup|<frac|3|2>>>+B*<frac|e<rsup|-M*t>|t<rsup|<frac|5|2>>>
  </equation*>

  Non siamo in grado di distinguere i due termini.

  Una prima approssimazione e' cercare eliminare la correzione infrarossa e
  fare il fit con solo il decadimento esponenziale.

  <\equation*>
    <frac|\<partial\>|\<partial\>t>ln D(t)=C+A*e<rsup|-M*t>*
  </equation*>

  Questo fit da un risultato plausibile ma e' poco giustificato.

  Cerchiamo di riprendere la espressione precedente e lavorare su qualche
  approssimazione

  <\equation*>
    D(t)\<sim\>\<mu\>+<frac|e<rsup|-M*t>|t<rsup|<frac|3|2>>>+<frac|e<rsup|-M(L-*t)>|(L-t)<rsup|<frac|3|2>>>
  </equation*>

  Supponiamo che la derivata la facciamo numerica come
  <with|mode|math|f<rprime|'>(t)=f(t)-f(t-1)>.

  Quindi la derivata del logaritmo diventa <with|mode|math|<frac|d|dt>log
  (f(t))=<frac|f(t)-f(t-1)|f(t)>=1-<frac|f(t-1)|f(t)>>.

  Nel nostro caso in prima approssimazione
  <with|mode|math|f(t)=\<mu\>+><with|mode|math|e<rsup|-M*t>><with|mode|math|+e<rsup|-M(L-*t)>>\ 

  Quindi possiamo cercare di fare un fit con la seguente espressione:

  <\equation*>
    <frac|\<partial\>|\<partial\>t>ln D(t)=1-<frac|\<mu\>+a*e<rsup|-M*t>+b*e<rsup|-M*(L-t)>|\<mu\>+*e<rsup|-M*t>+e<rsup|-M(L-*t)>>
  </equation*>

  In questo caso <with|mode|math|a=e<rsup|M>, b=e<rsup|-M>>.

  Supponiamo adesso che <with|mode|math|\<mu\>\<gg\>1>.

  Allora possiamo sviluppare il denominatore in potenze di
  <with|mode|math|1/\<mu\>>.

  E otteniamo a prim'ordine:

  <\equation*>
    <frac|\<partial\>|\<partial\>t>ln D(t)=1-<left|(>1+<frac|a*e<rsup|-M*t>+b*e<rsup|-M*(L-t)>|\<mu\>><right|)><left|(>1-*<frac|e<rsup|-M*t>+e<rsup|-M(L-*t)>|\<mu\>><right|)>
  </equation*>

  <\equation*>
    <frac|\<partial\>|\<partial\>t>ln D(t)=<frac|(1-a)*e<rsup|-M*t>+(1-b)*e<rsup|-M*(L-t)>|\<mu\>>
  </equation*>

  <section|Primi risultati numerici: solo decadimento>

  <subsection|6x6x6x12 10000 sweeps e 1500 termalizzazione 1Hb 3 or>

  Nel caso di un reticolo piccolo abbiamo dei primi risultati dove mostriamo
  il fit considerando solo il caso <with|mode|math|<frac|\<partial\>|\<partial\>t>ln
  D(t)> e considerando entrambe i casi <with|mode|math|<frac|\<partial\>|\<partial\>t>ln
  D(t), -<frac|\<partial\>|\<partial\>t>ln D(t)>:

  \ <with|mode|math|\<beta\>=0.985>

  <postscript|fit_dec0.985_0_6.pdf|*4/8|*4/8||||> with
  <with|mode|math|\<chi\><rsup|2>=>

  <postscript|fit_dec0.985_1_6.pdf|*4/8|*4/8||||>with
  <with|mode|math|\<chi\><rsup|2>=1.953>

  \;
</body>

<\initial>
  <\collection>
    <associate|language|american>
  </collection>
</initial>

<\references>
  <\collection>
    <associate|auto-1|<tuple|1|?>>
    <associate|auto-2|<tuple|2|?>>
    <associate|auto-3|<tuple|2.1|?>>
  </collection>
</references>

<\auxiliary>
  <\collection>
    <\associate|toc>
      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|Primi
      fit> <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-1><vspace|0.5fn>

      <vspace*|1fn><with|font-series|<quote|bold>|math-font-series|<quote|bold>|Primi
      risultati numerici: solo decadimento>
      <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-2><vspace|0.5fn>

      <with|par-left|<quote|1.5fn>|6x6x6x12 10000 sweeps e 1500
      termalizzazione 1Hb 3 or <datoms|<macro|x|<repeat|<arg|x>|<with|font-series|medium|<with|font-size|1|<space|0.2fn>.<space|0.2fn>>>>>|<htab|5mm>>
      <no-break><pageref|auto-3>>
    </associate>
  </collection>
</auxiliary>

Attachment: perlista.pdf
Description: Adobe PDF document

---

• No images in pdf or printed output, Luca, 11/28/2005

---

Archive powered by MHonArc 2.6.19.