Proof that $exp(x+y) = exp(x)*exp(y)$ using limit definition of $exp(x)$












4












$begingroup$


I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$



I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$



Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$



From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.






real-analysis limits exponential-function







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$endgroup$








  • 9




    $begingroup$
    These should be the limit as $nrightarrowinfty$?
    $endgroup$
    – gd1035
    Nov 27 '18 at 18:59










  • $begingroup$
    Thank you, yes. Fixed.
    $endgroup$
    – TomRatTUM
    Nov 27 '18 at 19:16










  • $begingroup$
    See this answer math.stackexchange.com/a/3000717/72031
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:50
















4












$begingroup$


I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$



I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$



Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$



From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.






real-analysis limits exponential-function







share|cite|improve this question











$endgroup$








  • 9




    $begingroup$
    These should be the limit as $nrightarrowinfty$?
    $endgroup$
    – gd1035
    Nov 27 '18 at 18:59










  • $begingroup$
    Thank you, yes. Fixed.
    $endgroup$
    – TomRatTUM
    Nov 27 '18 at 19:16










  • $begingroup$
    See this answer math.stackexchange.com/a/3000717/72031
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:50














4












4








4





$begingroup$


I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$



I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$



Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$



From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.






real-analysis limits exponential-function







share|cite|improve this question











$endgroup$




I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$



I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$



Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$



From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.






real-analysis limits exponential-function




real-analysis limits exponential-function






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 19:15







TomRatTUM

















asked Nov 27 '18 at 18:57









TomRatTUMTomRatTUM

234




234








  • 9




    $begingroup$
    These should be the limit as $nrightarrowinfty$?
    $endgroup$
    – gd1035
    Nov 27 '18 at 18:59










  • $begingroup$
    Thank you, yes. Fixed.
    $endgroup$
    – TomRatTUM
    Nov 27 '18 at 19:16










  • $begingroup$
    See this answer math.stackexchange.com/a/3000717/72031
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:50














  • 9




    $begingroup$
    These should be the limit as $nrightarrowinfty$?
    $endgroup$
    – gd1035
    Nov 27 '18 at 18:59










  • $begingroup$
    Thank you, yes. Fixed.
    $endgroup$
    – TomRatTUM
    Nov 27 '18 at 19:16










  • $begingroup$
    See this answer math.stackexchange.com/a/3000717/72031
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:50








9




9




$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59




$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59












$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16




$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16












$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50




$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50










2 Answers
2






active

oldest

votes


















1












$begingroup$

$$
\dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
\lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
\lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
$$
Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $x+n$ is not necessarily a positive integer.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:52



















2












$begingroup$

It suffices to show that
$$
lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
$$
But
$$
frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
$$
and as, for suitably large $n$, say $nge n_0$, we have that
$$
frac{1}{2}<1+frac{x+y}{n}<2,
$$
then
$$
left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
$$
Now, for all $zinmathbb R$, we have
$$
left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
$$
since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



Thus
$$
lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
$$
and hence $(1)$ holds.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $$
    \dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
    \lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
    \lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
    $$
    Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      $x+n$ is not necessarily a positive integer.
      $endgroup$
      – Paramanand Singh
      Nov 28 '18 at 1:52
















    1












    $begingroup$

    $$
    \dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
    \lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
    \lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
    $$
    Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      $x+n$ is not necessarily a positive integer.
      $endgroup$
      – Paramanand Singh
      Nov 28 '18 at 1:52














    1












    1








    1





    $begingroup$

    $$
    \dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
    \lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
    \lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
    $$
    Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$






    share|cite|improve this answer









    $endgroup$



    $$
    \dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
    \lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
    \lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
    $$
    Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 19:06









    Samvel SafaryanSamvel Safaryan

    493111




    493111












    • $begingroup$
      $x+n$ is not necessarily a positive integer.
      $endgroup$
      – Paramanand Singh
      Nov 28 '18 at 1:52


















    • $begingroup$
      $x+n$ is not necessarily a positive integer.
      $endgroup$
      – Paramanand Singh
      Nov 28 '18 at 1:52
















    $begingroup$
    $x+n$ is not necessarily a positive integer.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:52




    $begingroup$
    $x+n$ is not necessarily a positive integer.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:52











    2












    $begingroup$

    It suffices to show that
    $$
    lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
    $$
    But
    $$
    frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
    left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
    $$
    and as, for suitably large $n$, say $nge n_0$, we have that
    $$
    frac{1}{2}<1+frac{x+y}{n}<2,
    $$
    then
    $$
    left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
    left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
    $$
    Now, for all $zinmathbb R$, we have
    $$
    left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
    $$
    since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



    Thus
    $$
    lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
    $$
    and hence $(1)$ holds.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      It suffices to show that
      $$
      lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
      $$
      But
      $$
      frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
      left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
      $$
      and as, for suitably large $n$, say $nge n_0$, we have that
      $$
      frac{1}{2}<1+frac{x+y}{n}<2,
      $$
      then
      $$
      left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
      left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
      $$
      Now, for all $zinmathbb R$, we have
      $$
      left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
      $$
      since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



      Thus
      $$
      lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
      $$
      and hence $(1)$ holds.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It suffices to show that
        $$
        lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
        $$
        But
        $$
        frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
        left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
        $$
        and as, for suitably large $n$, say $nge n_0$, we have that
        $$
        frac{1}{2}<1+frac{x+y}{n}<2,
        $$
        then
        $$
        left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
        left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
        $$
        Now, for all $zinmathbb R$, we have
        $$
        left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
        $$
        since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



        Thus
        $$
        lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
        $$
        and hence $(1)$ holds.






        share|cite|improve this answer









        $endgroup$



        It suffices to show that
        $$
        lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
        $$
        But
        $$
        frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
        left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
        $$
        and as, for suitably large $n$, say $nge n_0$, we have that
        $$
        frac{1}{2}<1+frac{x+y}{n}<2,
        $$
        then
        $$
        left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
        left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
        $$
        Now, for all $zinmathbb R$, we have
        $$
        left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
        $$
        since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



        Thus
        $$
        lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
        $$
        and hence $(1)$ holds.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 19:10









        Yiorgos S. SmyrlisYiorgos S. Smyrlis

        63.2k1384163




        63.2k1384163






























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            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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            Can I use Tabulator js library in my java Spring + Thymeleaf project?

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            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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