How to compute Euler constant $(e^x)$ to its any power.
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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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|
show 3 more comments
$begingroup$
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
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Maybe you could use $e^{x+y}=e^xe^y$ and a method for small powers. For example $e^{50} = prod_{i=1}^{500} e^{0.1}$.
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– humanStampedist
Dec 4 '18 at 15:12
2
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What is the Feymenn method in this context? Can you provide a link?
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– 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.
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– 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
|
show 3 more comments
$begingroup$
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
$endgroup$
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
edited Dec 4 '18 at 15:18
Key Flex
7,77461232
7,77461232
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
|
show 3 more comments
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
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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
$$
e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
= 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
$$
$endgroup$
add a comment |
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1 Answer
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1 Answer
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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
$$
e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
= 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
$$
$endgroup$
add a comment |
$begingroup$
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
$$
e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
= 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
$$
$endgroup$
add a comment |
$begingroup$
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
$$
e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
= 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
$$
$endgroup$
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
$$
e^{,50} approx 10^{,21} cdot 10^{,{{14} over {1000}}50} = 10^{,21} cdot 10^{,{7 over {10}}}
= 10^{,22} cdot 10^{, - {3 over {10}}} approx 10^{,22} {1 over {root 3 of {10} }} approx {1 over 2}10^{,22}
$$
answered Dec 4 '18 at 17:11
G CabG Cab
18.5k31237
18.5k31237
add a comment |
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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