Finding laurent series expansion at Infinity












0















Find the laurent series expansion at $infty$ of the follwoing function:



$dfrac{1}{z^2-8z+25}$




Consider $dfrac{1}{z^2-8z+25}$,Replace $z=frac{1}{w}$,we get



$dfrac{1}{z^2-8z+25}=dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



$(w^2-frac{8}{25}w+frac{1}{25})=(w-frac{4}{25})^2+(frac{3}{25})^2$



Hence $dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



$=dfrac{w^2}{25((frac{3}{25})^2{1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}}})} $



The next step involves just the expansion of ${1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}})}^{-1}$



I dont understand if its going in the right direction or not



Can someone please help if I should proceed further






complex-analysis laurent-series







share|cite|improve this question



























    0















    Find the laurent series expansion at $infty$ of the follwoing function:



    $dfrac{1}{z^2-8z+25}$




    Consider $dfrac{1}{z^2-8z+25}$,Replace $z=frac{1}{w}$,we get



    $dfrac{1}{z^2-8z+25}=dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



    $(w^2-frac{8}{25}w+frac{1}{25})=(w-frac{4}{25})^2+(frac{3}{25})^2$



    Hence $dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



    $=dfrac{w^2}{25((frac{3}{25})^2{1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}}})} $



    The next step involves just the expansion of ${1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}})}^{-1}$



    I dont understand if its going in the right direction or not



    Can someone please help if I should proceed further






    complex-analysis laurent-series







    share|cite|improve this question

























      0












      0








      0








      Find the laurent series expansion at $infty$ of the follwoing function:



      $dfrac{1}{z^2-8z+25}$




      Consider $dfrac{1}{z^2-8z+25}$,Replace $z=frac{1}{w}$,we get



      $dfrac{1}{z^2-8z+25}=dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



      $(w^2-frac{8}{25}w+frac{1}{25})=(w-frac{4}{25})^2+(frac{3}{25})^2$



      Hence $dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



      $=dfrac{w^2}{25((frac{3}{25})^2{1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}}})} $



      The next step involves just the expansion of ${1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}})}^{-1}$



      I dont understand if its going in the right direction or not



      Can someone please help if I should proceed further






      complex-analysis laurent-series







      share|cite|improve this question














      Find the laurent series expansion at $infty$ of the follwoing function:



      $dfrac{1}{z^2-8z+25}$




      Consider $dfrac{1}{z^2-8z+25}$,Replace $z=frac{1}{w}$,we get



      $dfrac{1}{z^2-8z+25}=dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



      $(w^2-frac{8}{25}w+frac{1}{25})=(w-frac{4}{25})^2+(frac{3}{25})^2$



      Hence $dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



      $=dfrac{w^2}{25((frac{3}{25})^2{1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}}})} $



      The next step involves just the expansion of ${1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}})}^{-1}$



      I dont understand if its going in the right direction or not



      Can someone please help if I should proceed further






      complex-analysis laurent-series




      complex-analysis laurent-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 28 at 11:37









      Join_PhD

      1467




      1467






















          3 Answers
          3






          active

          oldest

          votes


















          2














          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above






          share|cite|improve this answer























          • But what is the problem with my steps
            – Join_PhD
            Nov 28 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 at 12:03



















          1














          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.






          share|cite|improve this answer























          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 at 11:56



















          1














          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$






          share|cite|improve this answer





















          • But why is my direction not correct?
            – Join_PhD
            Nov 28 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 at 12:01













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017046%2ffinding-laurent-series-expansion-at-infinity%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above






          share|cite|improve this answer























          • But what is the problem with my steps
            – Join_PhD
            Nov 28 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 at 12:03
















          2














          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above






          share|cite|improve this answer























          • But what is the problem with my steps
            – Join_PhD
            Nov 28 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 at 12:03














          2












          2








          2






          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above






          share|cite|improve this answer














          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 at 12:02

























          answered Nov 28 at 11:48









          Shubham

          1,5921519




          1,5921519












          • But what is the problem with my steps
            – Join_PhD
            Nov 28 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 at 12:03


















          • But what is the problem with my steps
            – Join_PhD
            Nov 28 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 at 12:03
















          But what is the problem with my steps
          – Join_PhD
          Nov 28 at 11:54




          But what is the problem with my steps
          – Join_PhD
          Nov 28 at 11:54












          In this manner, you can approach in less number of steps because you know expansion
          – Shubham
          Nov 28 at 11:55




          In this manner, you can approach in less number of steps because you know expansion
          – Shubham
          Nov 28 at 11:55












          @Shubham: Edit $dfrac{1}{1-a}$.
          – Yadati Kiran
          Nov 28 at 12:00




          @Shubham: Edit $dfrac{1}{1-a}$.
          – Yadati Kiran
          Nov 28 at 12:00












          Thanks. Edited @YadatiKiran
          – Shubham
          Nov 28 at 12:03




          Thanks. Edited @YadatiKiran
          – Shubham
          Nov 28 at 12:03











          1














          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.






          share|cite|improve this answer























          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 at 11:56
















          1














          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.






          share|cite|improve this answer























          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 at 11:56














          1












          1








          1






          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.






          share|cite|improve this answer














          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 at 11:55

























          answered Nov 28 at 11:54









          MisterRiemann

          5,7291624




          5,7291624












          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 at 11:56


















          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 at 11:56
















          the way what i have done is it correct?
          – Join_PhD
          Nov 28 at 11:55




          the way what i have done is it correct?
          – Join_PhD
          Nov 28 at 11:55












          Your steps seem correct. They also don't seem to lead to anything fruitful.
          – MisterRiemann
          Nov 28 at 11:56




          Your steps seem correct. They also don't seem to lead to anything fruitful.
          – MisterRiemann
          Nov 28 at 11:56











          1














          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$






          share|cite|improve this answer





















          • But why is my direction not correct?
            – Join_PhD
            Nov 28 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 at 12:01


















          1














          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$






          share|cite|improve this answer





















          • But why is my direction not correct?
            – Join_PhD
            Nov 28 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 at 12:01
















          1












          1








          1






          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$






          share|cite|improve this answer












          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 at 11:58









          José Carlos Santos

          148k22117218




          148k22117218












          • But why is my direction not correct?
            – Join_PhD
            Nov 28 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 at 12:01




















          • But why is my direction not correct?
            – Join_PhD
            Nov 28 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 at 12:01


















          But why is my direction not correct?
          – Join_PhD
          Nov 28 at 11:59




          But why is my direction not correct?
          – Join_PhD
          Nov 28 at 11:59












          I just don't see how you will get the correct answer from that.
          – José Carlos Santos
          Nov 28 at 12:01






          I just don't see how you will get the correct answer from that.
          – José Carlos Santos
          Nov 28 at 12:01




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017046%2ffinding-laurent-series-expansion-at-infinity%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Wiesbaden

          Image
          Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
          Read more

          Marschland

          Image
          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
          Read more

          Dieringhausen

          Image
          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
          Read more