Euler's Number Identity












2












$begingroup$


In pre-calc, we learned that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(1+frac{1}{n}right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = sum_{k=0}^{n} {n choose k} cdot x^{n-k}cdot y^{k}$$
$$x = 1, spacespacespacespacespacespacespace y = frac{1}{n}$$
Plugging that in gives:
$$lim_{ntoinfty}left(1+frac{1}{n}right)^{n} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot frac{1}{n^k}right)$$
That means that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot y^{k}right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
    $endgroup$
    – Doug M
    Dec 12 '18 at 18:59
















2












$begingroup$


In pre-calc, we learned that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(1+frac{1}{n}right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = sum_{k=0}^{n} {n choose k} cdot x^{n-k}cdot y^{k}$$
$$x = 1, spacespacespacespacespacespacespace y = frac{1}{n}$$
Plugging that in gives:
$$lim_{ntoinfty}left(1+frac{1}{n}right)^{n} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot frac{1}{n^k}right)$$
That means that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot y^{k}right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
    $endgroup$
    – Doug M
    Dec 12 '18 at 18:59














2












2








2


1



$begingroup$


In pre-calc, we learned that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(1+frac{1}{n}right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = sum_{k=0}^{n} {n choose k} cdot x^{n-k}cdot y^{k}$$
$$x = 1, spacespacespacespacespacespacespace y = frac{1}{n}$$
Plugging that in gives:
$$lim_{ntoinfty}left(1+frac{1}{n}right)^{n} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot frac{1}{n^k}right)$$
That means that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot y^{k}right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks.






algebra-precalculus







share|cite|improve this question











$endgroup$




In pre-calc, we learned that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(1+frac{1}{n}right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = sum_{k=0}^{n} {n choose k} cdot x^{n-k}cdot y^{k}$$
$$x = 1, spacespacespacespacespacespacespace y = frac{1}{n}$$
Plugging that in gives:
$$lim_{ntoinfty}left(1+frac{1}{n}right)^{n} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot frac{1}{n^k}right)$$
That means that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot y^{k}right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 19:29









saulspatz

15.9k31331




15.9k31331










asked Dec 12 '18 at 18:45









IlikemathIlikemath

164




164








  • 3




    $begingroup$
    I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
    $endgroup$
    – Doug M
    Dec 12 '18 at 18:59














  • 3




    $begingroup$
    I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
    $endgroup$
    – Doug M
    Dec 12 '18 at 18:59








3




3




$begingroup$
I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
$endgroup$
– Doug M
Dec 12 '18 at 18:59




$begingroup$
I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
$endgroup$
– Doug M
Dec 12 '18 at 18:59










4 Answers
4






active

oldest

votes


















1












$begingroup$

$$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
but I believe that you have got confused here. The easiest way to do it is:
$$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
and so we get that:
$$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
$$=sum_{n=0}^inftyfrac{1}{n!}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
      $endgroup$
      – Ilikemath
      Dec 12 '18 at 18:52



















    0












    $begingroup$

    First note
    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
    For the fixed $m<n$,
    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
    Let $ntoinfty$, one has
    $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
    So for any $m$,
    $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
    Let $mtoinfty$, one has
    $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$






      share|cite|improve this answer









      $endgroup$













        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%2f3037073%2feulers-number-identity%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        $$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
        but I believe that you have got confused here. The easiest way to do it is:
        $$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
        and so we get that:
        $$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
        $$=sum_{n=0}^inftyfrac{1}{n!}$$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          $$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
          but I believe that you have got confused here. The easiest way to do it is:
          $$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
          and so we get that:
          $$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
          $$=sum_{n=0}^inftyfrac{1}{n!}$$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            $$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
            but I believe that you have got confused here. The easiest way to do it is:
            $$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
            and so we get that:
            $$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
            $$=sum_{n=0}^inftyfrac{1}{n!}$$






            share|cite|improve this answer









            $endgroup$



            $$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
            but I believe that you have got confused here. The easiest way to do it is:
            $$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
            and so we get that:
            $$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
            $$=sum_{n=0}^inftyfrac{1}{n!}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 12 '18 at 19:12









            Henry LeeHenry Lee

            2,054219




            2,054219























                0












                $begingroup$

                Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                  $endgroup$
                  – Ilikemath
                  Dec 12 '18 at 18:52
















                0












                $begingroup$

                Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                  $endgroup$
                  – Ilikemath
                  Dec 12 '18 at 18:52














                0












                0








                0





                $begingroup$

                Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.






                share|cite|improve this answer











                $endgroup$



                Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 12 '18 at 18:54

























                answered Dec 12 '18 at 18:47









                OmGOmG

                2,502822




                2,502822












                • $begingroup$
                  Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                  $endgroup$
                  – Ilikemath
                  Dec 12 '18 at 18:52


















                • $begingroup$
                  Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                  $endgroup$
                  – Ilikemath
                  Dec 12 '18 at 18:52
















                $begingroup$
                Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                $endgroup$
                – Ilikemath
                Dec 12 '18 at 18:52




                $begingroup$
                Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                $endgroup$
                – Ilikemath
                Dec 12 '18 at 18:52











                0












                $begingroup$

                First note
                $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
                For the fixed $m<n$,
                $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
                Let $ntoinfty$, one has
                $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
                So for any $m$,
                $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
                Let $mtoinfty$, one has
                $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  First note
                  $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
                  For the fixed $m<n$,
                  $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
                  Let $ntoinfty$, one has
                  $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
                  So for any $m$,
                  $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
                  Let $mtoinfty$, one has
                  $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    First note
                    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
                    For the fixed $m<n$,
                    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
                    Let $ntoinfty$, one has
                    $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
                    So for any $m$,
                    $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
                    Let $mtoinfty$, one has
                    $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$






                    share|cite|improve this answer









                    $endgroup$



                    First note
                    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
                    For the fixed $m<n$,
                    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
                    Let $ntoinfty$, one has
                    $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
                    So for any $m$,
                    $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
                    Let $mtoinfty$, one has
                    $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 12 '18 at 20:34









                    xpaulxpaul

                    23k24455




                    23k24455























                        0












                        $begingroup$

                        Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$






                            share|cite|improve this answer









                            $endgroup$



                            Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 12 '18 at 20:54









                            Frank W.Frank W.

                            3,7351321




                            3,7351321






























                                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%2f3037073%2feulers-number-identity%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

                                Plaza Victoria

                                Image
                                Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                                Read more

                                Puebla de Zaragoza

                                Image
                                Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
                                Read more

                                Musa

                                Image
                                Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
                                Read more