Approximation of tan(f(n))
$begingroup$
I have the function $f(n)= pi (n+frac{1}{2})+epsilon frac{2pi(n+1/2)-a}{2pi(n+1/2)}$ when $epsilon<<1$, I can't understand what identity I need to use to prove that:
$tan(f(n))approx frac{2pi(n+1/2)}{epsilon(2pi(n+1/2)-a)}$
I tried with taylor but the first element in $f(x)$ taking tan function to $infty$
approximation
edited Jan 11 at 13:50
Jam
4,97921431
asked Jan 11 at 13:14
Daniel VainshteinDaniel Vainshtein
19011
$endgroup$
add a comment |
$begingroup$
I have the function $f(n)= pi (n+frac{1}{2})+epsilon frac{2pi(n+1/2)-a}{2pi(n+1/2)}$ when $epsilon<<1$, I can't understand what identity I need to use to prove that:
$tan(f(n))approx frac{2pi(n+1/2)}{epsilon(2pi(n+1/2)-a)}$
I tried with taylor but the first element in $f(x)$ taking tan function to $infty$
approximation
edited Jan 11 at 13:50
Jam
4,97921431
asked Jan 11 at 13:14
Daniel VainshteinDaniel Vainshtein
19011
$endgroup$
1
$begingroup$
Your function is $f(x)$, but it is written in terms of $n$. Should it be $f(n)$?
$endgroup$
– Calvin Godfrey
Jan 11 at 13:28
$begingroup$
What if $aapprox2pi(n+1/2)$?
$endgroup$
– Barry Cipra
Jan 11 at 13:37
add a comment |
$begingroup$
I have the function $f(n)= pi (n+frac{1}{2})+epsilon frac{2pi(n+1/2)-a}{2pi(n+1/2)}$ when $epsilon<<1$, I can't understand what identity I need to use to prove that:
$tan(f(n))approx frac{2pi(n+1/2)}{epsilon(2pi(n+1/2)-a)}$
I tried with taylor but the first element in $f(x)$ taking tan function to $infty$
approximation
edited Jan 11 at 13:50
Jam
4,97921431
asked Jan 11 at 13:14
Daniel VainshteinDaniel Vainshtein
19011
$endgroup$
I have the function $f(n)= pi (n+frac{1}{2})+epsilon frac{2pi(n+1/2)-a}{2pi(n+1/2)}$ when $epsilon<<1$, I can't understand what identity I need to use to prove that:
$tan(f(n))approx frac{2pi(n+1/2)}{epsilon(2pi(n+1/2)-a)}$
I tried with taylor but the first element in $f(x)$ taking tan function to $infty$
approximation
approximation
edited Jan 11 at 13:50
Jam
4,97921431
asked Jan 11 at 13:14
Daniel VainshteinDaniel Vainshtein
19011
edited Jan 11 at 13:50
Jam
4,97921431
asked Jan 11 at 13:14
Daniel VainshteinDaniel Vainshtein
19011
edited Jan 11 at 13:50
Jam
4,97921431
edited Jan 11 at 13:50
Jam
4,97921431
edited Jan 11 at 13:50
Jam
4,97921431
4,97921431
asked Jan 11 at 13:14
Daniel VainshteinDaniel Vainshtein
19011
asked Jan 11 at 13:14
Daniel VainshteinDaniel Vainshtein
19011
asked Jan 11 at 13:14
Daniel VainshteinDaniel Vainshtein
19011
19011
1
$begingroup$
Your function is $f(x)$, but it is written in terms of $n$. Should it be $f(n)$?
$endgroup$
– Calvin Godfrey
Jan 11 at 13:28
$begingroup$
What if $aapprox2pi(n+1/2)$?
$endgroup$
– Barry Cipra
Jan 11 at 13:37
add a comment |
1
$begingroup$
Your function is $f(x)$, but it is written in terms of $n$. Should it be $f(n)$?
$endgroup$
– Calvin Godfrey
Jan 11 at 13:28
$begingroup$
What if $aapprox2pi(n+1/2)$?
$endgroup$
– Barry Cipra
Jan 11 at 13:37
1
1
$begingroup$
Your function is $f(x)$, but it is written in terms of $n$. Should it be $f(n)$?
$endgroup$
– Calvin Godfrey
Jan 11 at 13:28
$begingroup$
Your function is $f(x)$, but it is written in terms of $n$. Should it be $f(n)$?
$endgroup$
– Calvin Godfrey
Jan 11 at 13:28
$begingroup$
What if $aapprox2pi(n+1/2)$?
$endgroup$
– Barry Cipra
Jan 11 at 13:37
$begingroup$
What if $aapprox2pi(n+1/2)$?
$endgroup$
– Barry Cipra
Jan 11 at 13:37
add a comment |
1 Answer
1
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oldest
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$begingroup$
The first step is to use
$$ tan(x +npi +pi/2) = -frac{cos x}{sin x}$$
The following is easy
answered Jan 11 at 13:33
DamienDamien
60714
$endgroup$
add a comment |
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$begingroup$
The first step is to use
$$ tan(x +npi +pi/2) = -frac{cos x}{sin x}$$
The following is easy
answered Jan 11 at 13:33
DamienDamien
60714
$endgroup$
add a comment |
$begingroup$
The first step is to use
$$ tan(x +npi +pi/2) = -frac{cos x}{sin x}$$
The following is easy
answered Jan 11 at 13:33
DamienDamien
60714
$endgroup$
add a comment |
$begingroup$
The first step is to use
$$ tan(x +npi +pi/2) = -frac{cos x}{sin x}$$
The following is easy
answered Jan 11 at 13:33
DamienDamien
60714
$endgroup$
The first step is to use
$$ tan(x +npi +pi/2) = -frac{cos x}{sin x}$$
The following is easy
answered Jan 11 at 13:33
DamienDamien
60714
answered Jan 11 at 13:33
DamienDamien
60714
answered Jan 11 at 13:33
DamienDamien
60714
answered Jan 11 at 13:33
DamienDamien
60714
60714
add a comment |
add a comment |
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1
$begingroup$
Your function is $f(x)$, but it is written in terms of $n$. Should it be $f(n)$?
$endgroup$
– Calvin Godfrey
Jan 11 at 13:28
$begingroup$
What if $aapprox2pi(n+1/2)$?
$endgroup$
– Barry Cipra
Jan 11 at 13:37