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How Can I calculate the Taylor series of $ln(cos(x))$?
up vote
0
down vote
favorite
I found it in an exercise about limits and i don't know how to solve it.
If it Is possible I would like to understand the steps to calculate It.
Many thanks in advance.
limits trigonometry logarithms
edited 23 hours ago
Tianlalu
2,559632
asked 23 hours ago
Aldo
32
add a comment |
up vote
0
down vote
favorite
I found it in an exercise about limits and i don't know how to solve it.
If it Is possible I would like to understand the steps to calculate It.
Many thanks in advance.
limits trigonometry logarithms
edited 23 hours ago
Tianlalu
2,559632
asked 23 hours ago
Aldo
32
1
The same way you calculate the Taylor series of any function. Do you know how to do that?
– Gerry Myerson
23 hours ago
This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
– hamza boulahia
23 hours ago
Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
– Aldo
23 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I found it in an exercise about limits and i don't know how to solve it.
If it Is possible I would like to understand the steps to calculate It.
Many thanks in advance.
limits trigonometry logarithms
edited 23 hours ago
Tianlalu
2,559632
asked 23 hours ago
Aldo
32
I found it in an exercise about limits and i don't know how to solve it.
If it Is possible I would like to understand the steps to calculate It.
Many thanks in advance.
limits trigonometry logarithms
limits trigonometry logarithms
edited 23 hours ago
Tianlalu
2,559632
asked 23 hours ago
Aldo
32
edited 23 hours ago
Tianlalu
2,559632
asked 23 hours ago
Aldo
32
edited 23 hours ago
Tianlalu
2,559632
edited 23 hours ago
Tianlalu
2,559632
edited 23 hours ago
Tianlalu
2,559632
2,559632
asked 23 hours ago
Aldo
32
asked 23 hours ago
Aldo
32
asked 23 hours ago
Aldo
32
32
1
The same way you calculate the Taylor series of any function. Do you know how to do that?
– Gerry Myerson
23 hours ago
This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
– hamza boulahia
23 hours ago
Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
– Aldo
23 hours ago
add a comment |
1
The same way you calculate the Taylor series of any function. Do you know how to do that?
– Gerry Myerson
23 hours ago
This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
– hamza boulahia
23 hours ago
Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
– Aldo
23 hours ago
1
1
The same way you calculate the Taylor series of any function. Do you know how to do that?
– Gerry Myerson
23 hours ago
The same way you calculate the Taylor series of any function. Do you know how to do that?
– Gerry Myerson
23 hours ago
This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
– hamza boulahia
23 hours ago
This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
– hamza boulahia
23 hours ago
Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
– Aldo
23 hours ago
Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
– Aldo
23 hours ago
add a comment |
5 Answers
5
active
oldest
votes
up vote
1
down vote
accepted
A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:
$$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
You can add more terms if you need to. Now we write:
$$ ln(1+(- frac{1}{2}x^2) ) = dots$$
Do you know the standard Taylor series for this function?
Hint: it is of the form $ln(1+u)$
answered 23 hours ago
WesleyGroupshaveFeelingsToo
816217
add a comment |
up vote
3
down vote
You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.
answered 23 hours ago
Christian Blatter
170k7111324
Or use the taylor series for the tangent, should he happen to know that
– vrugtehagel
21 hours ago
add a comment |
up vote
1
down vote
Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.
answered 21 hours ago
J.G.
18.1k11830
add a comment |
up vote
0
down vote
If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$
answered 21 hours ago
Chase Ryan Taylor
4,35521530
add a comment |
up vote
0
down vote
Since
$$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
we have
$$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
and
$$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$
answered 20 hours ago
Jack D'Aurizio
282k33274653
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:
$$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
You can add more terms if you need to. Now we write:
$$ ln(1+(- frac{1}{2}x^2) ) = dots$$
Do you know the standard Taylor series for this function?
Hint: it is of the form $ln(1+u)$
answered 23 hours ago
WesleyGroupshaveFeelingsToo
816217
add a comment |
up vote
1
down vote
accepted
A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:
$$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
You can add more terms if you need to. Now we write:
$$ ln(1+(- frac{1}{2}x^2) ) = dots$$
Do you know the standard Taylor series for this function?
Hint: it is of the form $ln(1+u)$
answered 23 hours ago
WesleyGroupshaveFeelingsToo
816217
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:
$$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
You can add more terms if you need to. Now we write:
$$ ln(1+(- frac{1}{2}x^2) ) = dots$$
Do you know the standard Taylor series for this function?
Hint: it is of the form $ln(1+u)$
answered 23 hours ago
WesleyGroupshaveFeelingsToo
816217
A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:
$$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
You can add more terms if you need to. Now we write:
$$ ln(1+(- frac{1}{2}x^2) ) = dots$$
Do you know the standard Taylor series for this function?
Hint: it is of the form $ln(1+u)$
answered 23 hours ago
WesleyGroupshaveFeelingsToo
816217
edited 23 hours ago
answered 23 hours ago
WesleyGroupshaveFeelingsToo
816217
answered 23 hours ago
WesleyGroupshaveFeelingsToo
816217
answered 23 hours ago
WesleyGroupshaveFeelingsToo
816217
816217
add a comment |
add a comment |
up vote
3
down vote
You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.
answered 23 hours ago
Christian Blatter
170k7111324
Or use the taylor series for the tangent, should he happen to know that
– vrugtehagel
21 hours ago
add a comment |
up vote
3
down vote
You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.
answered 23 hours ago
Christian Blatter
170k7111324
Or use the taylor series for the tangent, should he happen to know that
– vrugtehagel
21 hours ago
add a comment |
up vote
3
down vote
up vote
3
down vote
You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.
answered 23 hours ago
Christian Blatter
170k7111324
You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.
answered 23 hours ago
Christian Blatter
170k7111324
answered 23 hours ago
Christian Blatter
170k7111324
answered 23 hours ago
Christian Blatter
170k7111324
answered 23 hours ago
Christian Blatter
170k7111324
170k7111324
Or use the taylor series for the tangent, should he happen to know that
– vrugtehagel
21 hours ago
add a comment |
Or use the taylor series for the tangent, should he happen to know that
– vrugtehagel
21 hours ago
Or use the taylor series for the tangent, should he happen to know that
– vrugtehagel
21 hours ago
Or use the taylor series for the tangent, should he happen to know that
– vrugtehagel
21 hours ago
add a comment |
up vote
1
down vote
Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.
answered 21 hours ago
J.G.
18.1k11830
add a comment |
up vote
1
down vote
Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.
answered 21 hours ago
J.G.
18.1k11830
add a comment |
up vote
1
down vote
up vote
1
down vote
Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.
answered 21 hours ago
J.G.
18.1k11830
Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.
answered 21 hours ago
J.G.
18.1k11830
answered 21 hours ago
J.G.
18.1k11830
answered 21 hours ago
J.G.
18.1k11830
answered 21 hours ago
J.G.
18.1k11830
18.1k11830
add a comment |
add a comment |
up vote
0
down vote
If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$
answered 21 hours ago
Chase Ryan Taylor
4,35521530
add a comment |
up vote
0
down vote
If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$
answered 21 hours ago
Chase Ryan Taylor
4,35521530
add a comment |
up vote
0
down vote
up vote
0
down vote
If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$
answered 21 hours ago
Chase Ryan Taylor
4,35521530
If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$
answered 21 hours ago
Chase Ryan Taylor
4,35521530
answered 21 hours ago
Chase Ryan Taylor
4,35521530
answered 21 hours ago
Chase Ryan Taylor
4,35521530
answered 21 hours ago
Chase Ryan Taylor
4,35521530
4,35521530
add a comment |
add a comment |
up vote
0
down vote
Since
$$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
we have
$$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
and
$$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$
answered 20 hours ago
Jack D'Aurizio
282k33274653
add a comment |
up vote
0
down vote
Since
$$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
we have
$$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
and
$$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$
answered 20 hours ago
Jack D'Aurizio
282k33274653
add a comment |
up vote
0
down vote
up vote
0
down vote
Since
$$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
we have
$$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
and
$$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$
answered 20 hours ago
Jack D'Aurizio
282k33274653
Since
$$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
we have
$$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
and
$$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$
answered 20 hours ago
Jack D'Aurizio
282k33274653
answered 20 hours ago
Jack D'Aurizio
282k33274653
answered 20 hours ago
Jack D'Aurizio
282k33274653
answered 20 hours ago
Jack D'Aurizio
282k33274653
282k33274653
add a comment |
add a comment |
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1
The same way you calculate the Taylor series of any function. Do you know how to do that?
– Gerry Myerson
23 hours ago
This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
– hamza boulahia
23 hours ago
Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
– Aldo
23 hours ago