Calculating Hyperbolic Sin faster than using a standard power seriesIs there a way to get trig functions...
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Calculating Hyperbolic Sin faster than using a standard power series
Is there a way to get trig functions without a calculator?Broken Calculator: only certain unary functions work.How do we define $sin(theta)$ or $cos(theta)$Help with Trigonometric FunctionsHow would Taylor Series work?Understanding power series and their representation of functionsSeries Convergence of $sumlimits_{n=1}^{infty} frac{cos(npi/3)}{n!}$ by Ratio TestOrthogonal Function Dirac Delta SeriesHow do calculators calculate the value of trigonometric functions?How to calculate a “rational” sine/cosine to specific digits of precision?Calculating trig functions faster than with standard power series
$begingroup$
Using $$ sinh x = x + tfrac{x^3}{3!}+ tfrac{x^5}{5!} + tfrac{x^7}{7!}+ cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in Is there a way to get trig functions without a calculator? using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.
Thank you very much.
sequences-and-series trigonometry
edited 3 hours ago
MPW
30.5k12157
asked 4 hours ago
Bill BollingerBill Bollinger
362
New contributor
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Using $$ sinh x = x + tfrac{x^3}{3!}+ tfrac{x^5}{5!} + tfrac{x^7}{7!}+ cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in Is there a way to get trig functions without a calculator? using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.
Thank you very much.
sequences-and-series trigonometry
edited 3 hours ago
MPW
30.5k12157
asked 4 hours ago
Bill BollingerBill Bollinger
362
New contributor
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Using $$ sinh x = x + tfrac{x^3}{3!}+ tfrac{x^5}{5!} + tfrac{x^7}{7!}+ cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in Is there a way to get trig functions without a calculator? using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.
Thank you very much.
sequences-and-series trigonometry
edited 3 hours ago
MPW
30.5k12157
asked 4 hours ago
Bill BollingerBill Bollinger
362
New contributor
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Using $$ sinh x = x + tfrac{x^3}{3!}+ tfrac{x^5}{5!} + tfrac{x^7}{7!}+ cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in Is there a way to get trig functions without a calculator? using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.
Thank you very much.
sequences-and-series trigonometry
sequences-and-series trigonometry
edited 3 hours ago
MPW
30.5k12157
asked 4 hours ago
Bill BollingerBill Bollinger
362
New contributor
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
MPW
30.5k12157
asked 4 hours ago
Bill BollingerBill Bollinger
362
New contributor
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
MPW
30.5k12157
edited 3 hours ago
MPW
30.5k12157
edited 3 hours ago
MPW
30.5k12157
30.5k12157
asked 4 hours ago
Bill BollingerBill Bollinger
362
New contributor
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Bill BollingerBill Bollinger
362
asked 4 hours ago
Bill BollingerBill Bollinger
362
362
New contributor
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Bill Bollinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Note that $$sinh x=frac{e^x-e^{-x}}2$$
So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.
answered 3 hours ago
AndreiAndrei
12.7k21128
$endgroup$
add a comment |
$begingroup$
Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.
The infinite series is
$$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.
But suppose that you write instead
$$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
$$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.
You could use the same trick for most functions expressed as infinite series.
answered 1 hour ago
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Note that $$sinh x=frac{e^x-e^{-x}}2$$
So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.
answered 3 hours ago
AndreiAndrei
12.7k21128
$endgroup$
add a comment |
$begingroup$
Note that $$sinh x=frac{e^x-e^{-x}}2$$
So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.
answered 3 hours ago
AndreiAndrei
12.7k21128
$endgroup$
add a comment |
$begingroup$
Note that $$sinh x=frac{e^x-e^{-x}}2$$
So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.
answered 3 hours ago
AndreiAndrei
12.7k21128
$endgroup$
Note that $$sinh x=frac{e^x-e^{-x}}2$$
So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=lim_{ntoinfty}left(1+frac xnright)^n$$
For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+frac x{2^k}$, then you repeat the $y=ycdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this article.
answered 3 hours ago
AndreiAndrei
12.7k21128
answered 3 hours ago
AndreiAndrei
12.7k21128
answered 3 hours ago
AndreiAndrei
12.7k21128
answered 3 hours ago
AndreiAndrei
12.7k21128
12.7k21128
add a comment |
add a comment |
$begingroup$
Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.
The infinite series is
$$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.
But suppose that you write instead
$$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
$$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.
You could use the same trick for most functions expressed as infinite series.
answered 1 hour ago
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
$begingroup$
Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.
The infinite series is
$$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.
But suppose that you write instead
$$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
$$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.
You could use the same trick for most functions expressed as infinite series.
answered 1 hour ago
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
$begingroup$
Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.
The infinite series is
$$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.
But suppose that you write instead
$$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
$$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.
You could use the same trick for most functions expressed as infinite series.
answered 1 hour ago
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.
The infinite series is
$$sinh(x)=sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.
But suppose that you write instead
$$sinh(x)=sum_{n=0}^infty T_n qquad text{where} qquad T_n=frac{x^{2n+1}}{(2n+1)!}qquad text{and} qquad T_0=x$$ then
$$T_{n+1}= frac {t,, T_n}{(2n+2)(2n+3)}qquad text{where} qquad t=x^2$$ This would be much less expensive in terms of basic operations.
You could use the same trick for most functions expressed as infinite series.
answered 1 hour ago
Claude LeiboviciClaude Leibovici
123k1157135
answered 1 hour ago
Claude LeiboviciClaude Leibovici
123k1157135
answered 1 hour ago
Claude LeiboviciClaude Leibovici
123k1157135
answered 1 hour ago
Claude LeiboviciClaude Leibovici
123k1157135
123k1157135
add a comment |
add a comment |
Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.
Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.
Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.
Bill Bollinger is a new contributor. Be nice, and check out our Code of Conduct.
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