How to compute Euler constant $(e^x)$ to its any power.












1












$begingroup$


How to compute $e^x$ ($2.71218...$) to its any power with any shortcut or a method.
I want to know a method to calculate in big powers like $e^{50}$ not small powers, For eg-$0.02$ (using Taylor series or Feymenn method.) If you want to give any alternative method prescribed above for finding small powers, You could give.






exponential-function







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  • 1




    $begingroup$
    Maybe you could use $e^{x+y}=e^xe^y$ and a method for small powers. For example $e^{50} = prod_{i=1}^{500} e^{0.1}$.
    $endgroup$
    – humanStampedist
    Dec 4 '18 at 15:12






  • 2




    $begingroup$
    What is the Feymenn method in this context? Can you provide a link?
    $endgroup$
    – callculus
    Dec 4 '18 at 15:16






  • 2




    $begingroup$
    Maybe calculate $x=e^{50/64}$ by the usual methods; then to calculate $x^2, x^4, x^8, x^{16}, x^{32}, x^{64}=e^{50}$ requires only one additional multiplication each.
    $endgroup$
    – MJD
    Dec 4 '18 at 15:20






  • 4




    $begingroup$
    @IvoTerek: That is a dreadful method in real life. You need so many terms that (1) it takes forever, and (2) rounding errors accumulate unacceptably. (And it's totally useless if $x$ is large and negative.)
    $endgroup$
    – TonyK
    Dec 4 '18 at 15:24








  • 1




    $begingroup$
    Many comments without any reaction of the OP.
    $endgroup$
    – callculus
    Dec 4 '18 at 17:23
















1












$begingroup$


How to compute $e^x$ ($2.71218...$) to its any power with any shortcut or a method.
I want to know a method to calculate in big powers like $e^{50}$ not small powers, For eg-$0.02$ (using Taylor series or Feymenn method.) If you want to give any alternative method prescribed above for finding small powers, You could give.






exponential-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Maybe you could use $e^{x+y}=e^xe^y$ and a method for small powers. For example $e^{50} = prod_{i=1}^{500} e^{0.1}$.
    $endgroup$
    – humanStampedist
    Dec 4 '18 at 15:12






  • 2




    $begingroup$
    What is the Feymenn method in this context? Can you provide a link?
    $endgroup$
    – callculus
    Dec 4 '18 at 15:16






  • 2




    $begingroup$
    Maybe calculate $x=e^{50/64}$ by the usual methods; then to calculate $x^2, x^4, x^8, x^{16}, x^{32}, x^{64}=e^{50}$ requires only one additional multiplication each.
    $endgroup$
    – MJD
    Dec 4 '18 at 15:20






  • 4




    $begingroup$
    @IvoTerek: That is a dreadful method in real life. You need so many terms that (1) it takes forever, and (2) rounding errors accumulate unacceptably. (And it's totally useless if $x$ is large and negative.)
    $endgroup$
    – TonyK
    Dec 4 '18 at 15:24








  • 1




    $begingroup$
    Many comments without any reaction of the OP.
    $endgroup$
    – callculus
    Dec 4 '18 at 17:23














1












1








1





$begingroup$


How to compute $e^x$ ($2.71218...$) to its any power with any shortcut or a method.
I want to know a method to calculate in big powers like $e^{50}$ not small powers, For eg-$0.02$ (using Taylor series or Feymenn method.) If you want to give any alternative method prescribed above for finding small powers, You could give.






exponential-function







share|cite|improve this question











$endgroup$




How to compute $e^x$ ($2.71218...$) to its any power with any shortcut or a method.
I want to know a method to calculate in big powers like $e^{50}$ not small powers, For eg-$0.02$ (using Taylor series or Feymenn method.) If you want to give any alternative method prescribed above for finding small powers, You could give.






exponential-function




exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 15:18









Key Flex

8,28261233




8,28261233










asked Dec 4 '18 at 15:03









Piyush ChoudhuryPiyush Choudhury

63




63








  • 1




    $begingroup$
    Maybe you could use $e^{x+y}=e^xe^y$ and a method for small powers. For example $e^{50} = prod_{i=1}^{500} e^{0.1}$.
    $endgroup$
    – humanStampedist
    Dec 4 '18 at 15:12






  • 2




    $begingroup$
    What is the Feymenn method in this context? Can you provide a link?
    $endgroup$
    – callculus
    Dec 4 '18 at 15:16






  • 2




    $begingroup$
    Maybe calculate $x=e^{50/64}$ by the usual methods; then to calculate $x^2, x^4, x^8, x^{16}, x^{32}, x^{64}=e^{50}$ requires only one additional multiplication each.
    $endgroup$
    – MJD
    Dec 4 '18 at 15:20






  • 4




    $begingroup$
    @IvoTerek: That is a dreadful method in real life. You need so many terms that (1) it takes forever, and (2) rounding errors accumulate unacceptably. (And it's totally useless if $x$ is large and negative.)
    $endgroup$
    – TonyK
    Dec 4 '18 at 15:24








  • 1




    $begingroup$
    Many comments without any reaction of the OP.
    $endgroup$
    – callculus
    Dec 4 '18 at 17:23














  • 1




    $begingroup$
    Maybe you could use $e^{x+y}=e^xe^y$ and a method for small powers. For example $e^{50} = prod_{i=1}^{500} e^{0.1}$.
    $endgroup$
    – humanStampedist
    Dec 4 '18 at 15:12






  • 2




    $begingroup$
    What is the Feymenn method in this context? Can you provide a link?
    $endgroup$
    – callculus
    Dec 4 '18 at 15:16






  • 2




    $begingroup$
    Maybe calculate $x=e^{50/64}$ by the usual methods; then to calculate $x^2, x^4, x^8, x^{16}, x^{32}, x^{64}=e^{50}$ requires only one additional multiplication each.
    $endgroup$
    – MJD
    Dec 4 '18 at 15:20






  • 4




    $begingroup$
    @IvoTerek: That is a dreadful method in real life. You need so many terms that (1) it takes forever, and (2) rounding errors accumulate unacceptably. (And it's totally useless if $x$ is large and negative.)
    $endgroup$
    – TonyK
    Dec 4 '18 at 15:24








  • 1




    $begingroup$
    Many comments without any reaction of the OP.
    $endgroup$
    – callculus
    Dec 4 '18 at 17:23








1




1




$begingroup$
Maybe you could use $e^{x+y}=e^xe^y$ and a method for small powers. For example $e^{50} = prod_{i=1}^{500} e^{0.1}$.
$endgroup$
– humanStampedist
Dec 4 '18 at 15:12




$begingroup$
Maybe you could use $e^{x+y}=e^xe^y$ and a method for small powers. For example $e^{50} = prod_{i=1}^{500} e^{0.1}$.
$endgroup$
– humanStampedist
Dec 4 '18 at 15:12




2




2




$begingroup$
What is the Feymenn method in this context? Can you provide a link?
$endgroup$
– callculus
Dec 4 '18 at 15:16




$begingroup$
What is the Feymenn method in this context? Can you provide a link?
$endgroup$
– callculus
Dec 4 '18 at 15:16




2




2




$begingroup$
Maybe calculate $x=e^{50/64}$ by the usual methods; then to calculate $x^2, x^4, x^8, x^{16}, x^{32}, x^{64}=e^{50}$ requires only one additional multiplication each.
$endgroup$
– MJD
Dec 4 '18 at 15:20




$begingroup$
Maybe calculate $x=e^{50/64}$ by the usual methods; then to calculate $x^2, x^4, x^8, x^{16}, x^{32}, x^{64}=e^{50}$ requires only one additional multiplication each.
$endgroup$
– MJD
Dec 4 '18 at 15:20




4




4




$begingroup$
@IvoTerek: That is a dreadful method in real life. You need so many terms that (1) it takes forever, and (2) rounding errors accumulate unacceptably. (And it's totally useless if $x$ is large and negative.)
$endgroup$
– TonyK
Dec 4 '18 at 15:24






$begingroup$
@IvoTerek: That is a dreadful method in real life. You need so many terms that (1) it takes forever, and (2) rounding errors accumulate unacceptably. (And it's totally useless if $x$ is large and negative.)
$endgroup$
– TonyK
Dec 4 '18 at 15:24






1




1




$begingroup$
Many comments without any reaction of the OP.
$endgroup$
– callculus
Dec 4 '18 at 17:23




$begingroup$
Many comments without any reaction of the OP.
$endgroup$
– callculus
Dec 4 '18 at 17:23










1 Answer
1






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$begingroup$

Since
$$
cdots ;{4 over {10}} < {{21} over {50}} < log _{10} e = 0.43429 cdots < {{22} over {50}} < {4 over 9} < {5 over {10}}; cdots
$$
and you can find many other better bounds, depending on the precision that you need.



Then for instance you can get
$$
10^{,21} < e^{,50} < 10^{,22}
$$
or better
$$
e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
= 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
$$






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    1 Answer
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    1 Answer
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    active

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    0












    $begingroup$

    Since
    $$
    cdots ;{4 over {10}} < {{21} over {50}} < log _{10} e = 0.43429 cdots < {{22} over {50}} < {4 over 9} < {5 over {10}}; cdots
    $$
    and you can find many other better bounds, depending on the precision that you need.



    Then for instance you can get
    $$
    10^{,21} < e^{,50} < 10^{,22}
    $$
    or better
    $$
    e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
    = 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since
      $$
      cdots ;{4 over {10}} < {{21} over {50}} < log _{10} e = 0.43429 cdots < {{22} over {50}} < {4 over 9} < {5 over {10}}; cdots
      $$
      and you can find many other better bounds, depending on the precision that you need.



      Then for instance you can get
      $$
      10^{,21} < e^{,50} < 10^{,22}
      $$
      or better
      $$
      e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
      = 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since
        $$
        cdots ;{4 over {10}} < {{21} over {50}} < log _{10} e = 0.43429 cdots < {{22} over {50}} < {4 over 9} < {5 over {10}}; cdots
        $$
        and you can find many other better bounds, depending on the precision that you need.



        Then for instance you can get
        $$
        10^{,21} < e^{,50} < 10^{,22}
        $$
        or better
        $$
        e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
        = 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
        $$






        share|cite|improve this answer









        $endgroup$



        Since
        $$
        cdots ;{4 over {10}} < {{21} over {50}} < log _{10} e = 0.43429 cdots < {{22} over {50}} < {4 over 9} < {5 over {10}}; cdots
        $$
        and you can find many other better bounds, depending on the precision that you need.



        Then for instance you can get
        $$
        10^{,21} < e^{,50} < 10^{,22}
        $$
        or better
        $$
        e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
        = 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 17:11









        G CabG Cab

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        19.7k31339






























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