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
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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
New contributor
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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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
edited 2 hours ago
MPW
30.5k12157
30.5k12157
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asked 3 hours ago
Bill BollingerBill Bollinger
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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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 2 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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 11 mins ago
Claude LeiboviciClaude Leibovici
123k1157135
123k1157135
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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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