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What's a good approximation to the zeros of the cosine integral?

tough integral involving the Cosine integralCan you think of a good approximation to this integral?An Elliptic Integral - What's the Simplest Answer?Hardy's approximation for the cosineAny good approximation for this integral?What's a good substitution to solve this integral?Why does this approximation of integral of cosine work?Asymptotics of integral - logarithm and cosineIs this an approximation of a cosine?Prove an inequality for the cosine integral

5

2

$begingroup$

I. Logarithmic integral

The logarithmic integral $rm{li}(z)$ has a unique positive zero at $z approx 1.451363$ called the Ramanujan-Soldner constant.

II. Cosine integral

The cosine integral $rm{Ci}(x)$, on the other hand, has infinitely many zeros,

The smallest is nameless and is $x_0 approx 0.61650548$. For the next zeros, I notice they can be approximated by rational multiples of $pi$,

$$begin{array}{|c|c|c|}
hline
n&x_n&npi\
hline
1&3.38&3.14\
2&6.42&6.28\
3&9.52&9.42\
4&12.64&12.56\
hline
end{array}$$

III. Sine integral

The sine integral $rm{Si}(y)$ equated to $frac{pi}2approx 1.57$ also has infinitely many zeros.

The smallest is again nameless and is $y_0 approx 1.92644766$. The next zeros are,

$$begin{array}{|c|c|c|}
hline
n&y_n&(2n+1)tfrac{pi}2\
hline
1&4.89&4.71\
2&7.97&7.85\
3&11.08&10.99\
4&14.20&14.13\
hline
end{array}$$

IV. Questions

1. Do the expressions $x_n - npi$ and $y_n - (2n+1)frac{pi}2$ converge to some non-zero constant?


2. Are there better approximations to the roots $x_n$ and $y_n$?

integration trigonometry logarithms special-functions constants

share|cite|improve this question

edited 15 hours ago

Tito Piezas III

asked 17 hours ago

Tito Piezas IIITito Piezas III

28k369179

$endgroup$

add a comment | 

5

2

$begingroup$

I. Logarithmic integral

The logarithmic integral $rm{li}(z)$ has a unique positive zero at $z approx 1.451363$ called the Ramanujan-Soldner constant.

II. Cosine integral

The cosine integral $rm{Ci}(x)$, on the other hand, has infinitely many zeros,

The smallest is nameless and is $x_0 approx 0.61650548$. For the next zeros, I notice they can be approximated by rational multiples of $pi$,

$$begin{array}{|c|c|c|}
hline
n&x_n&npi\
hline
1&3.38&3.14\
2&6.42&6.28\
3&9.52&9.42\
4&12.64&12.56\
hline
end{array}$$

III. Sine integral

The sine integral $rm{Si}(y)$ equated to $frac{pi}2approx 1.57$ also has infinitely many zeros.

The smallest is again nameless and is $y_0 approx 1.92644766$. The next zeros are,

$$begin{array}{|c|c|c|}
hline
n&y_n&(2n+1)tfrac{pi}2\
hline
1&4.89&4.71\
2&7.97&7.85\
3&11.08&10.99\
4&14.20&14.13\
hline
end{array}$$

IV. Questions

1. Do the expressions $x_n - npi$ and $y_n - (2n+1)frac{pi}2$ converge to some non-zero constant?


2. Are there better approximations to the roots $x_n$ and $y_n$?

integration trigonometry logarithms special-functions constants

share|cite|improve this question

edited 15 hours ago

Tito Piezas III

asked 17 hours ago

Tito Piezas IIITito Piezas III

28k369179

$endgroup$

add a comment | 

$begingroup$

I. Logarithmic integral

The logarithmic integral $rm{li}(z)$ has a unique positive zero at $z approx 1.451363$ called the Ramanujan-Soldner constant.

II. Cosine integral

The cosine integral $rm{Ci}(x)$, on the other hand, has infinitely many zeros,

The smallest is nameless and is $x_0 approx 0.61650548$. For the next zeros, I notice they can be approximated by rational multiples of $pi$,

$$begin{array}{|c|c|c|}
hline
n&x_n&npi\
hline
1&3.38&3.14\
2&6.42&6.28\
3&9.52&9.42\
4&12.64&12.56\
hline
end{array}$$

III. Sine integral

The sine integral $rm{Si}(y)$ equated to $frac{pi}2approx 1.57$ also has infinitely many zeros.

The smallest is again nameless and is $y_0 approx 1.92644766$. The next zeros are,

$$begin{array}{|c|c|c|}
hline
n&y_n&(2n+1)tfrac{pi}2\
hline
1&4.89&4.71\
2&7.97&7.85\
3&11.08&10.99\
4&14.20&14.13\
hline
end{array}$$

IV. Questions

1. Do the expressions $x_n - npi$ and $y_n - (2n+1)frac{pi}2$ converge to some non-zero constant?


2. Are there better approximations to the roots $x_n$ and $y_n$?

integration trigonometry logarithms special-functions constants

share|cite|improve this question

edited 15 hours ago

Tito Piezas III

asked 17 hours ago

Tito Piezas IIITito Piezas III

28k369179

$endgroup$

share|cite|improve this question

edited 15 hours ago

Tito Piezas III

asked 17 hours ago

Tito Piezas IIITito Piezas III

28k369179

share|cite|improve this question

edited 15 hours ago

Tito Piezas III

asked 17 hours ago

Tito Piezas IIITito Piezas III

28k369179

asked 17 hours ago

Tito Piezas IIITito Piezas III

28k369179

asked 17 hours ago

Tito Piezas IIITito Piezas III

28k369179

2 Answers
2

active

oldest

votes

6

$begingroup$

If you look here, you will see that the zeros are given by
$$a+frac{1}{a}-frac{16}{3, a^3}+frac{1673}{15, a^5}-frac{507746}{105
, a^7}+Oleft(frac{1}{a^9}right)$$ where $a=k pi$ for the cosine integral and $a=left(k+frac 12right)pi$ for the sine integral.

If you want a more compact form, you could use
$$a+frac{240 a^2+3739}{a(240 a^2+5019)}$$ I built it from the above.

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

Claude LeiboviciClaude Leibovici

127k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I should have noticed that $a+frac1{a}$ was a closer approximation.
  $endgroup$
  – Tito Piezas III
  15 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

The roots of the Cardinal Sine occur at $npi$, and these correspond to the alternating minima and maxima of the Sine Integral. The roots of the latter will be found asymptotically in the middle and the constant is zero. (For high $n$, $dfrac1{2npi+x}$ tends to a constant over a period.)

More quantitatively, using a linear approximation for $xin[0,2pi]$,

$$frac{sin(2npi+x)}{2npi+x}approxfrac{sin(2npi+x)}{2npi}left(1-frac x{2npi}right)$$

and

$$int_0^xfrac{sin(2npi+t)}{2npi+t}dtapproxfrac{(x-2npi) cos x - sin(x)}{(2npi)^2}.$$

This expression cancels when

$$cos x=frac{xcos(x)-sin(x)}{2npi}=Oleft(frac1nright)$$ so that the bias on the root vanishes like $dfrac1n$. The higher order terms in the development will not change this behavior.

share|cite|improve this answer

edited 14 hours ago

answered 14 hours ago

Yves DaoustYves Daoust

134k676232

$endgroup$

add a comment | 

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6

$begingroup$

If you look here, you will see that the zeros are given by
$$a+frac{1}{a}-frac{16}{3, a^3}+frac{1673}{15, a^5}-frac{507746}{105
, a^7}+Oleft(frac{1}{a^9}right)$$ where $a=k pi$ for the cosine integral and $a=left(k+frac 12right)pi$ for the sine integral.

If you want a more compact form, you could use
$$a+frac{240 a^2+3739}{a(240 a^2+5019)}$$ I built it from the above.

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

Claude LeiboviciClaude Leibovici

127k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I should have noticed that $a+frac1{a}$ was a closer approximation.
  $endgroup$
  – Tito Piezas III
  15 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

If you look here, you will see that the zeros are given by
$$a+frac{1}{a}-frac{16}{3, a^3}+frac{1673}{15, a^5}-frac{507746}{105
, a^7}+Oleft(frac{1}{a^9}right)$$ where $a=k pi$ for the cosine integral and $a=left(k+frac 12right)pi$ for the sine integral.

If you want a more compact form, you could use
$$a+frac{240 a^2+3739}{a(240 a^2+5019)}$$ I built it from the above.

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

Claude LeiboviciClaude Leibovici

127k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I should have noticed that $a+frac1{a}$ was a closer approximation.
  $endgroup$
  – Tito Piezas III
  15 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If you look here, you will see that the zeros are given by
$$a+frac{1}{a}-frac{16}{3, a^3}+frac{1673}{15, a^5}-frac{507746}{105
, a^7}+Oleft(frac{1}{a^9}right)$$ where $a=k pi$ for the cosine integral and $a=left(k+frac 12right)pi$ for the sine integral.

If you want a more compact form, you could use
$$a+frac{240 a^2+3739}{a(240 a^2+5019)}$$ I built it from the above.

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

Claude LeiboviciClaude Leibovici

127k1158135

$endgroup$

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

Claude LeiboviciClaude Leibovici

127k1158135

answered 17 hours ago

Claude LeiboviciClaude Leibovici

127k1158135

answered 17 hours ago

Claude LeiboviciClaude Leibovici

127k1158135

1

$begingroup$

The roots of the Cardinal Sine occur at $npi$, and these correspond to the alternating minima and maxima of the Sine Integral. The roots of the latter will be found asymptotically in the middle and the constant is zero. (For high $n$, $dfrac1{2npi+x}$ tends to a constant over a period.)

More quantitatively, using a linear approximation for $xin[0,2pi]$,

$$frac{sin(2npi+x)}{2npi+x}approxfrac{sin(2npi+x)}{2npi}left(1-frac x{2npi}right)$$

and

$$int_0^xfrac{sin(2npi+t)}{2npi+t}dtapproxfrac{(x-2npi) cos x - sin(x)}{(2npi)^2}.$$

This expression cancels when

$$cos x=frac{xcos(x)-sin(x)}{2npi}=Oleft(frac1nright)$$ so that the bias on the root vanishes like $dfrac1n$. The higher order terms in the development will not change this behavior.

share|cite|improve this answer

edited 14 hours ago

answered 14 hours ago

Yves DaoustYves Daoust

134k676232

$endgroup$

add a comment | 

1

$begingroup$

The roots of the Cardinal Sine occur at $npi$, and these correspond to the alternating minima and maxima of the Sine Integral. The roots of the latter will be found asymptotically in the middle and the constant is zero. (For high $n$, $dfrac1{2npi+x}$ tends to a constant over a period.)

More quantitatively, using a linear approximation for $xin[0,2pi]$,

$$frac{sin(2npi+x)}{2npi+x}approxfrac{sin(2npi+x)}{2npi}left(1-frac x{2npi}right)$$

and

$$int_0^xfrac{sin(2npi+t)}{2npi+t}dtapproxfrac{(x-2npi) cos x - sin(x)}{(2npi)^2}.$$

This expression cancels when

$$cos x=frac{xcos(x)-sin(x)}{2npi}=Oleft(frac1nright)$$ so that the bias on the root vanishes like $dfrac1n$. The higher order terms in the development will not change this behavior.

share|cite|improve this answer

edited 14 hours ago

answered 14 hours ago

Yves DaoustYves Daoust

134k676232

$endgroup$

add a comment | 

$begingroup$

The roots of the Cardinal Sine occur at $npi$, and these correspond to the alternating minima and maxima of the Sine Integral. The roots of the latter will be found asymptotically in the middle and the constant is zero. (For high $n$, $dfrac1{2npi+x}$ tends to a constant over a period.)

More quantitatively, using a linear approximation for $xin[0,2pi]$,

$$frac{sin(2npi+x)}{2npi+x}approxfrac{sin(2npi+x)}{2npi}left(1-frac x{2npi}right)$$

and

$$int_0^xfrac{sin(2npi+t)}{2npi+t}dtapproxfrac{(x-2npi) cos x - sin(x)}{(2npi)^2}.$$

This expression cancels when

$$cos x=frac{xcos(x)-sin(x)}{2npi}=Oleft(frac1nright)$$ so that the bias on the root vanishes like $dfrac1n$. The higher order terms in the development will not change this behavior.

share|cite|improve this answer

edited 14 hours ago

answered 14 hours ago

Yves DaoustYves Daoust

134k676232

$endgroup$

share|cite|improve this answer

edited 14 hours ago

answered 14 hours ago

Yves DaoustYves Daoust

134k676232

answered 14 hours ago

Yves DaoustYves Daoust

134k676232

answered 14 hours ago

Yves DaoustYves Daoust

134k676232