How to find the value of e^x without using calculator [duplicate]

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1

This question already has an answer here:

• 
  How do pocket calculators calculate exponents?
  
  4 answers
  
  

I want to know how to find the value of $e^x$ where $ x in mathbb R$ without using calculator or using some sorts of function or mathematical algorithm ?

You can assume the value of $x$ yourself ... Eg :- $e^4$ or $e^3.736$

calculus algebra-precalculus logarithms

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edited 1 hour ago

amWhy

191k27222434

asked 1 hour ago

Mark ellon

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Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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marked as duplicate by Michael Lugo, gammatester, Jam, Xander Henderson, Vladhagen 1 min ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So basically you want to calculate the value of $ e^x $ for any $ x in R $ by hand then?
  – Tusky
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Tusky yes it is real number
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do you define the exponential of a set of irrational number?
  – gammatester
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Tusky yes you are right I want to know how to calculate it using hands not calculator
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Even after editing, the question makes no sense. How do you find a mathematical entity without mathematics? What methods are allowed?
  – gammatester
  1 hour ago
  
  

  
  
  
  

 | 
show 4 more comments

up vote
2
down vote

favorite

1

This question already has an answer here:

• 
  How do pocket calculators calculate exponents?
  
  4 answers
  
  

I want to know how to find the value of $e^x$ where $ x in mathbb R$ without using calculator or using some sorts of function or mathematical algorithm ?

You can assume the value of $x$ yourself ... Eg :- $e^4$ or $e^3.736$

calculus algebra-precalculus logarithms

share|cite|improve this question

edited 1 hour ago

amWhy

191k27222434

asked 1 hour ago

Mark ellon

462

New contributor

Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

marked as duplicate by Michael Lugo, gammatester, Jam, Xander Henderson, Vladhagen 1 min ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So basically you want to calculate the value of $ e^x $ for any $ x in R $ by hand then?
  – Tusky
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Tusky yes it is real number
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do you define the exponential of a set of irrational number?
  – gammatester
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Tusky yes you are right I want to know how to calculate it using hands not calculator
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Even after editing, the question makes no sense. How do you find a mathematical entity without mathematics? What methods are allowed?
  – gammatester
  1 hour ago
  
  

  
  
  
  

 | 
show 4 more comments

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

This question already has an answer here:

• 
  How do pocket calculators calculate exponents?
  
  4 answers
  
  

I want to know how to find the value of $e^x$ where $ x in mathbb R$ without using calculator or using some sorts of function or mathematical algorithm ?

You can assume the value of $x$ yourself ... Eg :- $e^4$ or $e^3.736$

calculus algebra-precalculus logarithms

share|cite|improve this question

edited 1 hour ago

amWhy

191k27222434

asked 1 hour ago

Mark ellon

462

New contributor

Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

This question already has an answer here:

• 
  How do pocket calculators calculate exponents?
  
  4 answers
  
  

I want to know how to find the value of $e^x$ where $ x in mathbb R$ without using calculator or using some sorts of function or mathematical algorithm ?

You can assume the value of $x$ yourself ... Eg :- $e^4$ or $e^3.736$

This question already has an answer here:

• 
  How do pocket calculators calculate exponents?
  
  4 answers
  
  

calculus algebra-precalculus logarithms

calculus algebra-precalculus logarithms

share|cite|improve this question

edited 1 hour ago

amWhy

191k27222434

asked 1 hour ago

Mark ellon

462

New contributor

Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 1 hour ago

amWhy

191k27222434

asked 1 hour ago

Mark ellon

462

New contributor

Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

share|cite|improve this question

edited 1 hour ago

amWhy

191k27222434

edited 1 hour ago

amWhy

191k27222434

edited 1 hour ago

amWhy

191k27222434

191k27222434

asked 1 hour ago

Mark ellon

462

New contributor

Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Mark ellon

462

asked 1 hour ago

Mark ellon

462

462

New contributor

Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Mark ellon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

marked as duplicate by Michael Lugo, gammatester, Jam, Xander Henderson, Vladhagen 1 min ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Michael Lugo, gammatester, Jam, Xander Henderson, Vladhagen 1 min ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So basically you want to calculate the value of $ e^x $ for any $ x in R $ by hand then?
  – Tusky
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Tusky yes it is real number
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do you define the exponential of a set of irrational number?
  – gammatester
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Tusky yes you are right I want to know how to calculate it using hands not calculator
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Even after editing, the question makes no sense. How do you find a mathematical entity without mathematics? What methods are allowed?
  – gammatester
  1 hour ago
  
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So basically you want to calculate the value of $ e^x $ for any $ x in R $ by hand then?
  – Tusky
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Tusky yes it is real number
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How do you define the exponential of a set of irrational number?
  – gammatester
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Tusky yes you are right I want to know how to calculate it using hands not calculator
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Even after editing, the question makes no sense. How do you find a mathematical entity without mathematics? What methods are allowed?
  – gammatester
  1 hour ago
  
  

  
  
  
  

So basically you want to calculate the value of $ e^x $ for any $ x in R $ by hand then?
– Tusky
1 hour ago

So basically you want to calculate the value of $ e^x $ for any $ x in R $ by hand then?
– Tusky
1 hour ago

@Tusky yes it is real number
– Mark ellon
1 hour ago

@Tusky yes it is real number
– Mark ellon
1 hour ago

How do you define the exponential of a set of irrational number?
– gammatester
1 hour ago

How do you define the exponential of a set of irrational number?
– gammatester
1 hour ago

@Tusky yes you are right I want to know how to calculate it using hands not calculator
– Mark ellon
1 hour ago

@Tusky yes you are right I want to know how to calculate it using hands not calculator
– Mark ellon
1 hour ago

1

1

Even after editing, the question makes no sense. How do you find a mathematical entity without mathematics? What methods are allowed?
– gammatester
1 hour ago

Even after editing, the question makes no sense. How do you find a mathematical entity without mathematics? What methods are allowed?
– gammatester
1 hour ago

 | 
show 4 more comments

5 Answers
5

active

oldest

votes

up vote
3
down vote

Use Taylor series $e^x = 1 + x + fracx^22! + fracx^33! +...$

share|cite|improve this answer

answered 1 hour ago

Vasya

2,8231514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This works well for small $x$. For larger $x$ it is a lot of work.
  – Ross Millikan
  39 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And it it is very inaccurate for larger negative $x$
  – gammatester
  37 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

I will add to Vasya's answer. We know that for every $x$ real number, whatever,
$$1 + x + fracx^22! + fracx^33! +...$$

converges, and the limit is equal to $e^x$ .

Let's say you want a estimate of $e^7$ within an error of $10^-5$, what you will do is find out the smallest $n$ for which
$$ frac7^nn! < 10^-5 .$$
Given the faster growth of factorials than powers this will happen soon! Let's say, the smallest such $n$ is $10$, then the approximate value of $e^7$ will be
$$ e^x simeq 1 + 7 + frac7^22! + cdots + frac7^1010! cdot $$

I do not know if this is computationally the most efficient, but it sure works in theory.

share|cite|improve this answer

answered 49 mins ago

Behnam Esmayli

1,894315

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how do your actual numbers look like for e.g. $e^-20?$
  – gammatester
  35 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your question? You're adding up those fractions, that's it.
  – Behnam Esmayli
  26 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Behnam Good answer. But I think this isn't the most computationally efficient method as Taylor series are $O(M(n)n^1/2)$ but this can be reduced to $O(M(n)ln n)$ with AGM splitting (source). $M(n)$ represents the complexity of the multiplication algorithm. Also, CORDIC may be comparable to division (i.e. $O(M(n))$), according to slide 7 of (this source).
  – Jam
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Therefore I asked for the actual numbers. The largest term is $43099804.12$ and $e^-20 = 0.20611536224cdot 10^-8.$ The series is convergent for all $x,$ but that does not mean it can be used for actual computing (without arbitrary precision arithmetic).
  – gammatester
  15 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is why I included the disclaimer at the end: I am not sure if this is the most efficient way for machines doing it.
  – Behnam Esmayli
  13 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

HINT:

$$e^x=lim_ntoinfty(1+frac xn)^n$$

share|cite|improve this answer

edited 1 hour ago

Jakobian

2,114519

answered 1 hour ago

Rhys Hughes

4,1461227

• 
  
  
  

  
  

  
  
  
  
  
  

  
  my apologies im stupid. thanks to jakob for the correction
  – Rhys Hughes
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops sorry .....
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is not a practical method of computing $e^x$.
  – TonyK
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also not obvious to me how to get explicit error bounds for a given value of $n$.
  – Jack M
  9 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

For small values of the exponent, you can use the Taylor series.
$$e^a=sum_i=0^infty frac a^ii!$$
In theory you can use this for any $a$, but for $a$ at all large you need lots of terms.

For large values of the exponent I would use $e^a =10^frac alog (10)$ and the laws of exponents. I know $log(10) approx 2.30$ and I know a number of base $10$ logs. For example if asked for $e^10$ I would do
$$e^10=10^frac 10log 10approx 10^frac 102.30approx 10^4.35approx 2cdot 10^4cdot 10^0.05approx 2cdot 10^4cdot (1+0.05cdot 2.30)approx 22300$$
This was done without a calculator except for the $frac 102.3$ which is not a difficult division. I know $log_102approx 0.30103$. The correct value is about $22026$ so I am not far off.

share|cite|improve this answer

answered 40 mins ago

Ross Millikan

283k23191359

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Won't you need better and better estimates for $log10$ for larger values of $a$?
  – Jack M
  11 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

The Taylor expansion is one way, but if $|x|$ is large, you have to compute quite a few terms before they get small enough to ignore. In particular, if $x$ is large and negative, you end up alternately adding and subtracting large numbers to end up with a small number, which is very bad for accuracy.

A better method is to express $x$ as $n+r$, where $n$ is an integer and $rin [-0.5, 0.5)$. Then the series expansion of $e^r$ converges rapidly, because $|x^n|le 1/2^n$. Now you can calculate $e^x$ as $e^ne^r$, where $e^n$ is just a sequence of multiplications by $e=2.718281828ldots$

(Just for completeness: there is a better way to calculate $e^n$ than multiplying $n$ times, namely exponentiation by squaring. If you are going to program your computation, you will want to look into this $-$ it's not difficult.)

share|cite|improve this answer

answered 30 mins ago

TonyK

39.1k349128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But for larger and larger values of $n$, you need better and better accuracy on the value of $e$. And how do you obtain that? Maybe taking more terms in the series expansion for $e^1$. I'm guessing if you work out the details, your method ends up requiring just as much computational work as just computing a Taylor polynomial for $e^x$.
  – Jack M
  12 mins ago
  

  
  
  
  

add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Use Taylor series $e^x = 1 + x + fracx^22! + fracx^33! +...$

share|cite|improve this answer

answered 1 hour ago

Vasya

2,8231514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This works well for small $x$. For larger $x$ it is a lot of work.
  – Ross Millikan
  39 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And it it is very inaccurate for larger negative $x$
  – gammatester
  37 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

Use Taylor series $e^x = 1 + x + fracx^22! + fracx^33! +...$

share|cite|improve this answer

answered 1 hour ago

Vasya

2,8231514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This works well for small $x$. For larger $x$ it is a lot of work.
  – Ross Millikan
  39 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And it it is very inaccurate for larger negative $x$
  – gammatester
  37 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Use Taylor series $e^x = 1 + x + fracx^22! + fracx^33! +...$

share|cite|improve this answer

answered 1 hour ago

Vasya

2,8231514

Use Taylor series $e^x = 1 + x + fracx^22! + fracx^33! +...$

share|cite|improve this answer

answered 1 hour ago

Vasya

2,8231514

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Vasya

2,8231514

answered 1 hour ago

Vasya

2,8231514

answered 1 hour ago

Vasya

2,8231514

2,8231514

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This works well for small $x$. For larger $x$ it is a lot of work.
  – Ross Millikan
  39 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And it it is very inaccurate for larger negative $x$
  – gammatester
  37 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This works well for small $x$. For larger $x$ it is a lot of work.
  – Ross Millikan
  39 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And it it is very inaccurate for larger negative $x$
  – gammatester
  37 mins ago
  
  

  
  
  
  

This works well for small $x$. For larger $x$ it is a lot of work.
– Ross Millikan
39 mins ago

This works well for small $x$. For larger $x$ it is a lot of work.
– Ross Millikan
39 mins ago

And it it is very inaccurate for larger negative $x$
– gammatester
37 mins ago

And it it is very inaccurate for larger negative $x$
– gammatester
37 mins ago

add a comment | 

up vote
3
down vote

I will add to Vasya's answer. We know that for every $x$ real number, whatever,
$$1 + x + fracx^22! + fracx^33! +...$$

converges, and the limit is equal to $e^x$ .

Let's say you want a estimate of $e^7$ within an error of $10^-5$, what you will do is find out the smallest $n$ for which
$$ frac7^nn! < 10^-5 .$$
Given the faster growth of factorials than powers this will happen soon! Let's say, the smallest such $n$ is $10$, then the approximate value of $e^7$ will be
$$ e^x simeq 1 + 7 + frac7^22! + cdots + frac7^1010! cdot $$

I do not know if this is computationally the most efficient, but it sure works in theory.

share|cite|improve this answer

answered 49 mins ago

Behnam Esmayli

1,894315

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how do your actual numbers look like for e.g. $e^-20?$
  – gammatester
  35 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your question? You're adding up those fractions, that's it.
  – Behnam Esmayli
  26 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Behnam Good answer. But I think this isn't the most computationally efficient method as Taylor series are $O(M(n)n^1/2)$ but this can be reduced to $O(M(n)ln n)$ with AGM splitting (source). $M(n)$ represents the complexity of the multiplication algorithm. Also, CORDIC may be comparable to division (i.e. $O(M(n))$), according to slide 7 of (this source).
  – Jam
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Therefore I asked for the actual numbers. The largest term is $43099804.12$ and $e^-20 = 0.20611536224cdot 10^-8.$ The series is convergent for all $x,$ but that does not mean it can be used for actual computing (without arbitrary precision arithmetic).
  – gammatester
  15 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is why I included the disclaimer at the end: I am not sure if this is the most efficient way for machines doing it.
  – Behnam Esmayli
  13 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

I will add to Vasya's answer. We know that for every $x$ real number, whatever,
$$1 + x + fracx^22! + fracx^33! +...$$

converges, and the limit is equal to $e^x$ .

Let's say you want a estimate of $e^7$ within an error of $10^-5$, what you will do is find out the smallest $n$ for which
$$ frac7^nn! < 10^-5 .$$
Given the faster growth of factorials than powers this will happen soon! Let's say, the smallest such $n$ is $10$, then the approximate value of $e^7$ will be
$$ e^x simeq 1 + 7 + frac7^22! + cdots + frac7^1010! cdot $$

I do not know if this is computationally the most efficient, but it sure works in theory.

share|cite|improve this answer

answered 49 mins ago

Behnam Esmayli

1,894315

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how do your actual numbers look like for e.g. $e^-20?$
  – gammatester
  35 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your question? You're adding up those fractions, that's it.
  – Behnam Esmayli
  26 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Behnam Good answer. But I think this isn't the most computationally efficient method as Taylor series are $O(M(n)n^1/2)$ but this can be reduced to $O(M(n)ln n)$ with AGM splitting (source). $M(n)$ represents the complexity of the multiplication algorithm. Also, CORDIC may be comparable to division (i.e. $O(M(n))$), according to slide 7 of (this source).
  – Jam
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Therefore I asked for the actual numbers. The largest term is $43099804.12$ and $e^-20 = 0.20611536224cdot 10^-8.$ The series is convergent for all $x,$ but that does not mean it can be used for actual computing (without arbitrary precision arithmetic).
  – gammatester
  15 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is why I included the disclaimer at the end: I am not sure if this is the most efficient way for machines doing it.
  – Behnam Esmayli
  13 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

I will add to Vasya's answer. We know that for every $x$ real number, whatever,
$$1 + x + fracx^22! + fracx^33! +...$$

converges, and the limit is equal to $e^x$ .

Let's say you want a estimate of $e^7$ within an error of $10^-5$, what you will do is find out the smallest $n$ for which
$$ frac7^nn! < 10^-5 .$$
Given the faster growth of factorials than powers this will happen soon! Let's say, the smallest such $n$ is $10$, then the approximate value of $e^7$ will be
$$ e^x simeq 1 + 7 + frac7^22! + cdots + frac7^1010! cdot $$

I do not know if this is computationally the most efficient, but it sure works in theory.

share|cite|improve this answer

answered 49 mins ago

Behnam Esmayli

1,894315

I will add to Vasya's answer. We know that for every $x$ real number, whatever,
$$1 + x + fracx^22! + fracx^33! +...$$

converges, and the limit is equal to $e^x$ .

Let's say you want a estimate of $e^7$ within an error of $10^-5$, what you will do is find out the smallest $n$ for which
$$ frac7^nn! < 10^-5 .$$
Given the faster growth of factorials than powers this will happen soon! Let's say, the smallest such $n$ is $10$, then the approximate value of $e^7$ will be
$$ e^x simeq 1 + 7 + frac7^22! + cdots + frac7^1010! cdot $$

I do not know if this is computationally the most efficient, but it sure works in theory.

share|cite|improve this answer

answered 49 mins ago

Behnam Esmayli

1,894315

share|cite|improve this answer

share|cite|improve this answer

answered 49 mins ago

Behnam Esmayli

1,894315

answered 49 mins ago

Behnam Esmayli

1,894315

answered 49 mins ago

Behnam Esmayli

1,894315

1,894315

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how do your actual numbers look like for e.g. $e^-20?$
  – gammatester
  35 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your question? You're adding up those fractions, that's it.
  – Behnam Esmayli
  26 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Behnam Good answer. But I think this isn't the most computationally efficient method as Taylor series are $O(M(n)n^1/2)$ but this can be reduced to $O(M(n)ln n)$ with AGM splitting (source). $M(n)$ represents the complexity of the multiplication algorithm. Also, CORDIC may be comparable to division (i.e. $O(M(n))$), according to slide 7 of (this source).
  – Jam
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Therefore I asked for the actual numbers. The largest term is $43099804.12$ and $e^-20 = 0.20611536224cdot 10^-8.$ The series is convergent for all $x,$ but that does not mean it can be used for actual computing (without arbitrary precision arithmetic).
  – gammatester
  15 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is why I included the disclaimer at the end: I am not sure if this is the most efficient way for machines doing it.
  – Behnam Esmayli
  13 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how do your actual numbers look like for e.g. $e^-20?$
  – gammatester
  35 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't get your question? You're adding up those fractions, that's it.
  – Behnam Esmayli
  26 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Behnam Good answer. But I think this isn't the most computationally efficient method as Taylor series are $O(M(n)n^1/2)$ but this can be reduced to $O(M(n)ln n)$ with AGM splitting (source). $M(n)$ represents the complexity of the multiplication algorithm. Also, CORDIC may be comparable to division (i.e. $O(M(n))$), according to slide 7 of (this source).
  – Jam
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Therefore I asked for the actual numbers. The largest term is $43099804.12$ and $e^-20 = 0.20611536224cdot 10^-8.$ The series is convergent for all $x,$ but that does not mean it can be used for actual computing (without arbitrary precision arithmetic).
  – gammatester
  15 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is why I included the disclaimer at the end: I am not sure if this is the most efficient way for machines doing it.
  – Behnam Esmayli
  13 mins ago
  

  
  
  
  

And how do your actual numbers look like for e.g. $e^-20?$
– gammatester
35 mins ago

And how do your actual numbers look like for e.g. $e^-20?$
– gammatester
35 mins ago

I don't get your question? You're adding up those fractions, that's it.
– Behnam Esmayli
26 mins ago

I don't get your question? You're adding up those fractions, that's it.
– Behnam Esmayli
26 mins ago

@Behnam Good answer. But I think this isn't the most computationally efficient method as Taylor series are $O(M(n)n^1/2)$ but this can be reduced to $O(M(n)ln n)$ with AGM splitting (source). $M(n)$ represents the complexity of the multiplication algorithm. Also, CORDIC may be comparable to division (i.e. $O(M(n))$), according to slide 7 of (this source).
– Jam
21 mins ago

@Behnam Good answer. But I think this isn't the most computationally efficient method as Taylor series are $O(M(n)n^1/2)$ but this can be reduced to $O(M(n)ln n)$ with AGM splitting (source). $M(n)$ represents the complexity of the multiplication algorithm. Also, CORDIC may be comparable to division (i.e. $O(M(n))$), according to slide 7 of (this source).
– Jam
21 mins ago

Therefore I asked for the actual numbers. The largest term is $43099804.12$ and $e^-20 = 0.20611536224cdot 10^-8.$ The series is convergent for all $x,$ but that does not mean it can be used for actual computing (without arbitrary precision arithmetic).
– gammatester
15 mins ago

Therefore I asked for the actual numbers. The largest term is $43099804.12$ and $e^-20 = 0.20611536224cdot 10^-8.$ The series is convergent for all $x,$ but that does not mean it can be used for actual computing (without arbitrary precision arithmetic).
– gammatester
15 mins ago

That is why I included the disclaimer at the end: I am not sure if this is the most efficient way for machines doing it.
– Behnam Esmayli
13 mins ago

That is why I included the disclaimer at the end: I am not sure if this is the most efficient way for machines doing it.
– Behnam Esmayli
13 mins ago

add a comment | 

up vote
1
down vote

HINT:

$$e^x=lim_ntoinfty(1+frac xn)^n$$

share|cite|improve this answer

edited 1 hour ago

Jakobian

2,114519

answered 1 hour ago

Rhys Hughes

4,1461227

• 
  
  
  

  
  

  
  
  
  
  
  

  
  my apologies im stupid. thanks to jakob for the correction
  – Rhys Hughes
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops sorry .....
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is not a practical method of computing $e^x$.
  – TonyK
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also not obvious to me how to get explicit error bounds for a given value of $n$.
  – Jack M
  9 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

HINT:

$$e^x=lim_ntoinfty(1+frac xn)^n$$

share|cite|improve this answer

edited 1 hour ago

Jakobian

2,114519

answered 1 hour ago

Rhys Hughes

4,1461227

• 
  
  
  

  
  

  
  
  
  
  
  

  
  my apologies im stupid. thanks to jakob for the correction
  – Rhys Hughes
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops sorry .....
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is not a practical method of computing $e^x$.
  – TonyK
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also not obvious to me how to get explicit error bounds for a given value of $n$.
  – Jack M
  9 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

HINT:

$$e^x=lim_ntoinfty(1+frac xn)^n$$

share|cite|improve this answer

edited 1 hour ago

Jakobian

2,114519

answered 1 hour ago

Rhys Hughes

4,1461227

HINT:

$$e^x=lim_ntoinfty(1+frac xn)^n$$

share|cite|improve this answer

edited 1 hour ago

Jakobian

2,114519

answered 1 hour ago

Rhys Hughes

4,1461227

share|cite|improve this answer

share|cite|improve this answer

edited 1 hour ago

Jakobian

2,114519

edited 1 hour ago

Jakobian

2,114519

edited 1 hour ago

Jakobian

2,114519

2,114519

answered 1 hour ago

Rhys Hughes

4,1461227

answered 1 hour ago

Rhys Hughes

4,1461227

answered 1 hour ago

Rhys Hughes

4,1461227

4,1461227

• 
  
  
  

  
  

  
  
  
  
  
  

  
  my apologies im stupid. thanks to jakob for the correction
  – Rhys Hughes
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops sorry .....
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is not a practical method of computing $e^x$.
  – TonyK
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also not obvious to me how to get explicit error bounds for a given value of $n$.
  – Jack M
  9 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  my apologies im stupid. thanks to jakob for the correction
  – Rhys Hughes
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops sorry .....
  – Mark ellon
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is not a practical method of computing $e^x$.
  – TonyK
  47 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's also not obvious to me how to get explicit error bounds for a given value of $n$.
  – Jack M
  9 mins ago
  

  
  
  
  

my apologies im stupid. thanks to jakob for the correction
– Rhys Hughes
1 hour ago

my apologies im stupid. thanks to jakob for the correction
– Rhys Hughes
1 hour ago

Oops sorry .....
– Mark ellon
1 hour ago

Oops sorry .....
– Mark ellon
1 hour ago

1

1

This is not a practical method of computing $e^x$.
– TonyK
47 mins ago

This is not a practical method of computing $e^x$.
– TonyK
47 mins ago

It's also not obvious to me how to get explicit error bounds for a given value of $n$.
– Jack M
9 mins ago

It's also not obvious to me how to get explicit error bounds for a given value of $n$.
– Jack M
9 mins ago

add a comment | 

up vote
0
down vote

For small values of the exponent, you can use the Taylor series.
$$e^a=sum_i=0^infty frac a^ii!$$
In theory you can use this for any $a$, but for $a$ at all large you need lots of terms.

For large values of the exponent I would use $e^a =10^frac alog (10)$ and the laws of exponents. I know $log(10) approx 2.30$ and I know a number of base $10$ logs. For example if asked for $e^10$ I would do
$$e^10=10^frac 10log 10approx 10^frac 102.30approx 10^4.35approx 2cdot 10^4cdot 10^0.05approx 2cdot 10^4cdot (1+0.05cdot 2.30)approx 22300$$
This was done without a calculator except for the $frac 102.3$ which is not a difficult division. I know $log_102approx 0.30103$. The correct value is about $22026$ so I am not far off.

share|cite|improve this answer

answered 40 mins ago

Ross Millikan

283k23191359

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Won't you need better and better estimates for $log10$ for larger values of $a$?
  – Jack M
  11 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

For small values of the exponent, you can use the Taylor series.
$$e^a=sum_i=0^infty frac a^ii!$$
In theory you can use this for any $a$, but for $a$ at all large you need lots of terms.

For large values of the exponent I would use $e^a =10^frac alog (10)$ and the laws of exponents. I know $log(10) approx 2.30$ and I know a number of base $10$ logs. For example if asked for $e^10$ I would do
$$e^10=10^frac 10log 10approx 10^frac 102.30approx 10^4.35approx 2cdot 10^4cdot 10^0.05approx 2cdot 10^4cdot (1+0.05cdot 2.30)approx 22300$$
This was done without a calculator except for the $frac 102.3$ which is not a difficult division. I know $log_102approx 0.30103$. The correct value is about $22026$ so I am not far off.

share|cite|improve this answer

answered 40 mins ago

Ross Millikan

283k23191359

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Won't you need better and better estimates for $log10$ for larger values of $a$?
  – Jack M
  11 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

For small values of the exponent, you can use the Taylor series.
$$e^a=sum_i=0^infty frac a^ii!$$
In theory you can use this for any $a$, but for $a$ at all large you need lots of terms.

For large values of the exponent I would use $e^a =10^frac alog (10)$ and the laws of exponents. I know $log(10) approx 2.30$ and I know a number of base $10$ logs. For example if asked for $e^10$ I would do
$$e^10=10^frac 10log 10approx 10^frac 102.30approx 10^4.35approx 2cdot 10^4cdot 10^0.05approx 2cdot 10^4cdot (1+0.05cdot 2.30)approx 22300$$
This was done without a calculator except for the $frac 102.3$ which is not a difficult division. I know $log_102approx 0.30103$. The correct value is about $22026$ so I am not far off.

share|cite|improve this answer

answered 40 mins ago

Ross Millikan

283k23191359

For small values of the exponent, you can use the Taylor series.
$$e^a=sum_i=0^infty frac a^ii!$$
In theory you can use this for any $a$, but for $a$ at all large you need lots of terms.

For large values of the exponent I would use $e^a =10^frac alog (10)$ and the laws of exponents. I know $log(10) approx 2.30$ and I know a number of base $10$ logs. For example if asked for $e^10$ I would do
$$e^10=10^frac 10log 10approx 10^frac 102.30approx 10^4.35approx 2cdot 10^4cdot 10^0.05approx 2cdot 10^4cdot (1+0.05cdot 2.30)approx 22300$$
This was done without a calculator except for the $frac 102.3$ which is not a difficult division. I know $log_102approx 0.30103$. The correct value is about $22026$ so I am not far off.

share|cite|improve this answer

answered 40 mins ago

Ross Millikan

283k23191359

share|cite|improve this answer

share|cite|improve this answer

answered 40 mins ago

Ross Millikan

283k23191359

answered 40 mins ago

Ross Millikan

283k23191359

answered 40 mins ago

Ross Millikan

283k23191359

283k23191359

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Won't you need better and better estimates for $log10$ for larger values of $a$?
  – Jack M
  11 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Won't you need better and better estimates for $log10$ for larger values of $a$?
  – Jack M
  11 mins ago
  
  

  
  
  
  

Won't you need better and better estimates for $log10$ for larger values of $a$?
– Jack M
11 mins ago

Won't you need better and better estimates for $log10$ for larger values of $a$?
– Jack M
11 mins ago

add a comment | 

up vote
0
down vote

The Taylor expansion is one way, but if $|x|$ is large, you have to compute quite a few terms before they get small enough to ignore. In particular, if $x$ is large and negative, you end up alternately adding and subtracting large numbers to end up with a small number, which is very bad for accuracy.

A better method is to express $x$ as $n+r$, where $n$ is an integer and $rin [-0.5, 0.5)$. Then the series expansion of $e^r$ converges rapidly, because $|x^n|le 1/2^n$. Now you can calculate $e^x$ as $e^ne^r$, where $e^n$ is just a sequence of multiplications by $e=2.718281828ldots$

(Just for completeness: there is a better way to calculate $e^n$ than multiplying $n$ times, namely exponentiation by squaring. If you are going to program your computation, you will want to look into this $-$ it's not difficult.)

share|cite|improve this answer

answered 30 mins ago

TonyK

39.1k349128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But for larger and larger values of $n$, you need better and better accuracy on the value of $e$. And how do you obtain that? Maybe taking more terms in the series expansion for $e^1$. I'm guessing if you work out the details, your method ends up requiring just as much computational work as just computing a Taylor polynomial for $e^x$.
  – Jack M
  12 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

The Taylor expansion is one way, but if $|x|$ is large, you have to compute quite a few terms before they get small enough to ignore. In particular, if $x$ is large and negative, you end up alternately adding and subtracting large numbers to end up with a small number, which is very bad for accuracy.

A better method is to express $x$ as $n+r$, where $n$ is an integer and $rin [-0.5, 0.5)$. Then the series expansion of $e^r$ converges rapidly, because $|x^n|le 1/2^n$. Now you can calculate $e^x$ as $e^ne^r$, where $e^n$ is just a sequence of multiplications by $e=2.718281828ldots$

(Just for completeness: there is a better way to calculate $e^n$ than multiplying $n$ times, namely exponentiation by squaring. If you are going to program your computation, you will want to look into this $-$ it's not difficult.)

share|cite|improve this answer

answered 30 mins ago

TonyK

39.1k349128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But for larger and larger values of $n$, you need better and better accuracy on the value of $e$. And how do you obtain that? Maybe taking more terms in the series expansion for $e^1$. I'm guessing if you work out the details, your method ends up requiring just as much computational work as just computing a Taylor polynomial for $e^x$.
  – Jack M
  12 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

The Taylor expansion is one way, but if $|x|$ is large, you have to compute quite a few terms before they get small enough to ignore. In particular, if $x$ is large and negative, you end up alternately adding and subtracting large numbers to end up with a small number, which is very bad for accuracy.

A better method is to express $x$ as $n+r$, where $n$ is an integer and $rin [-0.5, 0.5)$. Then the series expansion of $e^r$ converges rapidly, because $|x^n|le 1/2^n$. Now you can calculate $e^x$ as $e^ne^r$, where $e^n$ is just a sequence of multiplications by $e=2.718281828ldots$

(Just for completeness: there is a better way to calculate $e^n$ than multiplying $n$ times, namely exponentiation by squaring. If you are going to program your computation, you will want to look into this $-$ it's not difficult.)

share|cite|improve this answer

answered 30 mins ago

TonyK

39.1k349128

The Taylor expansion is one way, but if $|x|$ is large, you have to compute quite a few terms before they get small enough to ignore. In particular, if $x$ is large and negative, you end up alternately adding and subtracting large numbers to end up with a small number, which is very bad for accuracy.

A better method is to express $x$ as $n+r$, where $n$ is an integer and $rin [-0.5, 0.5)$. Then the series expansion of $e^r$ converges rapidly, because $|x^n|le 1/2^n$. Now you can calculate $e^x$ as $e^ne^r$, where $e^n$ is just a sequence of multiplications by $e=2.718281828ldots$

(Just for completeness: there is a better way to calculate $e^n$ than multiplying $n$ times, namely exponentiation by squaring. If you are going to program your computation, you will want to look into this $-$ it's not difficult.)

share|cite|improve this answer

answered 30 mins ago

TonyK

39.1k349128

share|cite|improve this answer

share|cite|improve this answer

answered 30 mins ago

TonyK

39.1k349128

answered 30 mins ago

TonyK

39.1k349128

answered 30 mins ago

TonyK

39.1k349128

39.1k349128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But for larger and larger values of $n$, you need better and better accuracy on the value of $e$. And how do you obtain that? Maybe taking more terms in the series expansion for $e^1$. I'm guessing if you work out the details, your method ends up requiring just as much computational work as just computing a Taylor polynomial for $e^x$.
  – Jack M
  12 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But for larger and larger values of $n$, you need better and better accuracy on the value of $e$. And how do you obtain that? Maybe taking more terms in the series expansion for $e^1$. I'm guessing if you work out the details, your method ends up requiring just as much computational work as just computing a Taylor polynomial for $e^x$.
  – Jack M
  12 mins ago
  

  
  
  
  

But for larger and larger values of $n$, you need better and better accuracy on the value of $e$. And how do you obtain that? Maybe taking more terms in the series expansion for $e^1$. I'm guessing if you work out the details, your method ends up requiring just as much computational work as just computing a Taylor polynomial for $e^x$.
– Jack M
12 mins ago

But for larger and larger values of $n$, you need better and better accuracy on the value of $e$. And how do you obtain that? Maybe taking more terms in the series expansion for $e^1$. I'm guessing if you work out the details, your method ends up requiring just as much computational work as just computing a Taylor polynomial for $e^x$.
– Jack M
12 mins ago

add a comment |