Finding the Taylor series (complex numbers)












1












$begingroup$


I have
$$
{frac{1}{(z+1)(z-2)}}
$$
I did
$$
{frac{1}{(z+1)(z-2)}} = {frac{A}{z+1}}+{frac{B}{z-2}}
$$



and found
$A=-1/3, B=1/3$



So now I have
$$
-{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} = -{frac{1}{3}} sum(-1)^k z^k + {frac{1}{3}} sum(-1)^k (-{frac{z}{2}})^k
$$



But the answer in the book is different. What am I doing wrong?






complex-analysis complex-numbers taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    What's the answer in the book? See that the second summation can be simplified to $sum (z/2)^k$.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 16:25










  • $begingroup$
    The answer in book is $$ {frac{1}{3}} sum((-1)^{n+1}-2^{-n-1}) z^n $$
    $endgroup$
    – user3132457
    Dec 18 '18 at 16:28
















1












$begingroup$


I have
$$
{frac{1}{(z+1)(z-2)}}
$$
I did
$$
{frac{1}{(z+1)(z-2)}} = {frac{A}{z+1}}+{frac{B}{z-2}}
$$



and found
$A=-1/3, B=1/3$



So now I have
$$
-{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} = -{frac{1}{3}} sum(-1)^k z^k + {frac{1}{3}} sum(-1)^k (-{frac{z}{2}})^k
$$



But the answer in the book is different. What am I doing wrong?






complex-analysis complex-numbers taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    What's the answer in the book? See that the second summation can be simplified to $sum (z/2)^k$.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 16:25










  • $begingroup$
    The answer in book is $$ {frac{1}{3}} sum((-1)^{n+1}-2^{-n-1}) z^n $$
    $endgroup$
    – user3132457
    Dec 18 '18 at 16:28














1












1








1





$begingroup$


I have
$$
{frac{1}{(z+1)(z-2)}}
$$
I did
$$
{frac{1}{(z+1)(z-2)}} = {frac{A}{z+1}}+{frac{B}{z-2}}
$$



and found
$A=-1/3, B=1/3$



So now I have
$$
-{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} = -{frac{1}{3}} sum(-1)^k z^k + {frac{1}{3}} sum(-1)^k (-{frac{z}{2}})^k
$$



But the answer in the book is different. What am I doing wrong?






complex-analysis complex-numbers taylor-expansion







share|cite|improve this question











$endgroup$




I have
$$
{frac{1}{(z+1)(z-2)}}
$$
I did
$$
{frac{1}{(z+1)(z-2)}} = {frac{A}{z+1}}+{frac{B}{z-2}}
$$



and found
$A=-1/3, B=1/3$



So now I have
$$
-{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} = -{frac{1}{3}} sum(-1)^k z^k + {frac{1}{3}} sum(-1)^k (-{frac{z}{2}})^k
$$



But the answer in the book is different. What am I doing wrong?






complex-analysis complex-numbers taylor-expansion




complex-analysis complex-numbers taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 16:30









José Carlos Santos

168k22132236




168k22132236










asked Dec 18 '18 at 16:19









user3132457user3132457

1598




1598












  • $begingroup$
    What's the answer in the book? See that the second summation can be simplified to $sum (z/2)^k$.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 16:25










  • $begingroup$
    The answer in book is $$ {frac{1}{3}} sum((-1)^{n+1}-2^{-n-1}) z^n $$
    $endgroup$
    – user3132457
    Dec 18 '18 at 16:28


















  • $begingroup$
    What's the answer in the book? See that the second summation can be simplified to $sum (z/2)^k$.
    $endgroup$
    – rafa11111
    Dec 18 '18 at 16:25










  • $begingroup$
    The answer in book is $$ {frac{1}{3}} sum((-1)^{n+1}-2^{-n-1}) z^n $$
    $endgroup$
    – user3132457
    Dec 18 '18 at 16:28
















$begingroup$
What's the answer in the book? See that the second summation can be simplified to $sum (z/2)^k$.
$endgroup$
– rafa11111
Dec 18 '18 at 16:25




$begingroup$
What's the answer in the book? See that the second summation can be simplified to $sum (z/2)^k$.
$endgroup$
– rafa11111
Dec 18 '18 at 16:25












$begingroup$
The answer in book is $$ {frac{1}{3}} sum((-1)^{n+1}-2^{-n-1}) z^n $$
$endgroup$
– user3132457
Dec 18 '18 at 16:28




$begingroup$
The answer in book is $$ {frac{1}{3}} sum((-1)^{n+1}-2^{-n-1}) z^n $$
$endgroup$
– user3132457
Dec 18 '18 at 16:28










2 Answers
2






active

oldest

votes


















2












$begingroup$

We have if $|r|<1$, then $$frac{1}{1-r}=sum_{i=0}^infty r^i$$



begin{align}
-{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} &=-{frac{1}{3}} sum(-1)^k z^k - frac16 times frac{1}{1-left( frac{z}2right)}
\&= -{frac{1}{3}} sum(-1)^k z^k - {frac{1}{6}} sum ({frac{z}{2}})^k \
&= frac13 left[sum(-1)^{k+1} z^k - {frac{1}{2}} sum ({frac{z}{2}})^k right]\
&= frac13 sumleft[(-1)^{k+1} - {2}^{-k-1} right]z^k\
end{align}






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    It is indeed true that$$frac1{(z+1)(z-2)}=-frac13timesfrac1{z+1}+frac13timesfrac1{z-2}$$and that$$frac1{z+1}=sum_{k=0}^infty(-1)^kz^k.$$However,$$frac1{z-2}=-frac1{2-z}=-sum_{k=0}^inftyfrac{z^k}{2^{k+1}}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      What formula did you use for series of $-{frac{1}{2-z}}$?
      $endgroup$
      – user3132457
      Dec 18 '18 at 16:31










    • $begingroup$
      $$frac1{a-z}=frac1atimesfrac1{1-frac za}=frac1asum_{k=0}^inftyleft(frac zaright)^k=sum_{k=0}^inftyfrac{z^k}{a^{k+1}}.$$
      $endgroup$
      – José Carlos Santos
      Dec 18 '18 at 17:16













    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%2f3045355%2ffinding-the-taylor-series-complex-numbers%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    We have if $|r|<1$, then $$frac{1}{1-r}=sum_{i=0}^infty r^i$$



    begin{align}
    -{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} &=-{frac{1}{3}} sum(-1)^k z^k - frac16 times frac{1}{1-left( frac{z}2right)}
    \&= -{frac{1}{3}} sum(-1)^k z^k - {frac{1}{6}} sum ({frac{z}{2}})^k \
    &= frac13 left[sum(-1)^{k+1} z^k - {frac{1}{2}} sum ({frac{z}{2}})^k right]\
    &= frac13 sumleft[(-1)^{k+1} - {2}^{-k-1} right]z^k\
    end{align}






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      We have if $|r|<1$, then $$frac{1}{1-r}=sum_{i=0}^infty r^i$$



      begin{align}
      -{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} &=-{frac{1}{3}} sum(-1)^k z^k - frac16 times frac{1}{1-left( frac{z}2right)}
      \&= -{frac{1}{3}} sum(-1)^k z^k - {frac{1}{6}} sum ({frac{z}{2}})^k \
      &= frac13 left[sum(-1)^{k+1} z^k - {frac{1}{2}} sum ({frac{z}{2}})^k right]\
      &= frac13 sumleft[(-1)^{k+1} - {2}^{-k-1} right]z^k\
      end{align}






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        We have if $|r|<1$, then $$frac{1}{1-r}=sum_{i=0}^infty r^i$$



        begin{align}
        -{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} &=-{frac{1}{3}} sum(-1)^k z^k - frac16 times frac{1}{1-left( frac{z}2right)}
        \&= -{frac{1}{3}} sum(-1)^k z^k - {frac{1}{6}} sum ({frac{z}{2}})^k \
        &= frac13 left[sum(-1)^{k+1} z^k - {frac{1}{2}} sum ({frac{z}{2}})^k right]\
        &= frac13 sumleft[(-1)^{k+1} - {2}^{-k-1} right]z^k\
        end{align}






        share|cite|improve this answer











        $endgroup$



        We have if $|r|<1$, then $$frac{1}{1-r}=sum_{i=0}^infty r^i$$



        begin{align}
        -{frac{1}{3}}times {frac{1}{z+1}}+{frac{1}{3}}times{frac{1}{z-2}} &=-{frac{1}{3}} sum(-1)^k z^k - frac16 times frac{1}{1-left( frac{z}2right)}
        \&= -{frac{1}{3}} sum(-1)^k z^k - {frac{1}{6}} sum ({frac{z}{2}})^k \
        &= frac13 left[sum(-1)^{k+1} z^k - {frac{1}{2}} sum ({frac{z}{2}})^k right]\
        &= frac13 sumleft[(-1)^{k+1} - {2}^{-k-1} right]z^k\
        end{align}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 18 '18 at 16:32

























        answered Dec 18 '18 at 16:26









        Siong Thye GohSiong Thye Goh

        103k1468119




        103k1468119























            1












            $begingroup$

            It is indeed true that$$frac1{(z+1)(z-2)}=-frac13timesfrac1{z+1}+frac13timesfrac1{z-2}$$and that$$frac1{z+1}=sum_{k=0}^infty(-1)^kz^k.$$However,$$frac1{z-2}=-frac1{2-z}=-sum_{k=0}^inftyfrac{z^k}{2^{k+1}}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What formula did you use for series of $-{frac{1}{2-z}}$?
              $endgroup$
              – user3132457
              Dec 18 '18 at 16:31










            • $begingroup$
              $$frac1{a-z}=frac1atimesfrac1{1-frac za}=frac1asum_{k=0}^inftyleft(frac zaright)^k=sum_{k=0}^inftyfrac{z^k}{a^{k+1}}.$$
              $endgroup$
              – José Carlos Santos
              Dec 18 '18 at 17:16


















            1












            $begingroup$

            It is indeed true that$$frac1{(z+1)(z-2)}=-frac13timesfrac1{z+1}+frac13timesfrac1{z-2}$$and that$$frac1{z+1}=sum_{k=0}^infty(-1)^kz^k.$$However,$$frac1{z-2}=-frac1{2-z}=-sum_{k=0}^inftyfrac{z^k}{2^{k+1}}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What formula did you use for series of $-{frac{1}{2-z}}$?
              $endgroup$
              – user3132457
              Dec 18 '18 at 16:31










            • $begingroup$
              $$frac1{a-z}=frac1atimesfrac1{1-frac za}=frac1asum_{k=0}^inftyleft(frac zaright)^k=sum_{k=0}^inftyfrac{z^k}{a^{k+1}}.$$
              $endgroup$
              – José Carlos Santos
              Dec 18 '18 at 17:16
















            1












            1








            1





            $begingroup$

            It is indeed true that$$frac1{(z+1)(z-2)}=-frac13timesfrac1{z+1}+frac13timesfrac1{z-2}$$and that$$frac1{z+1}=sum_{k=0}^infty(-1)^kz^k.$$However,$$frac1{z-2}=-frac1{2-z}=-sum_{k=0}^inftyfrac{z^k}{2^{k+1}}.$$






            share|cite|improve this answer









            $endgroup$



            It is indeed true that$$frac1{(z+1)(z-2)}=-frac13timesfrac1{z+1}+frac13timesfrac1{z-2}$$and that$$frac1{z+1}=sum_{k=0}^infty(-1)^kz^k.$$However,$$frac1{z-2}=-frac1{2-z}=-sum_{k=0}^inftyfrac{z^k}{2^{k+1}}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 16:28









            José Carlos SantosJosé Carlos Santos

            168k22132236




            168k22132236












            • $begingroup$
              What formula did you use for series of $-{frac{1}{2-z}}$?
              $endgroup$
              – user3132457
              Dec 18 '18 at 16:31










            • $begingroup$
              $$frac1{a-z}=frac1atimesfrac1{1-frac za}=frac1asum_{k=0}^inftyleft(frac zaright)^k=sum_{k=0}^inftyfrac{z^k}{a^{k+1}}.$$
              $endgroup$
              – José Carlos Santos
              Dec 18 '18 at 17:16




















            • $begingroup$
              What formula did you use for series of $-{frac{1}{2-z}}$?
              $endgroup$
              – user3132457
              Dec 18 '18 at 16:31










            • $begingroup$
              $$frac1{a-z}=frac1atimesfrac1{1-frac za}=frac1asum_{k=0}^inftyleft(frac zaright)^k=sum_{k=0}^inftyfrac{z^k}{a^{k+1}}.$$
              $endgroup$
              – José Carlos Santos
              Dec 18 '18 at 17:16


















            $begingroup$
            What formula did you use for series of $-{frac{1}{2-z}}$?
            $endgroup$
            – user3132457
            Dec 18 '18 at 16:31




            $begingroup$
            What formula did you use for series of $-{frac{1}{2-z}}$?
            $endgroup$
            – user3132457
            Dec 18 '18 at 16:31












            $begingroup$
            $$frac1{a-z}=frac1atimesfrac1{1-frac za}=frac1asum_{k=0}^inftyleft(frac zaright)^k=sum_{k=0}^inftyfrac{z^k}{a^{k+1}}.$$
            $endgroup$
            – José Carlos Santos
            Dec 18 '18 at 17:16






            $begingroup$
            $$frac1{a-z}=frac1atimesfrac1{1-frac za}=frac1asum_{k=0}^inftyleft(frac zaright)^k=sum_{k=0}^inftyfrac{z^k}{a^{k+1}}.$$
            $endgroup$
            – José Carlos Santos
            Dec 18 '18 at 17:16




















            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045355%2ffinding-the-taylor-series-complex-numbers%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

            Willebadessen

            Image
            Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more

            Residenzschloss Arolsen

            Image
            Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
            Read more