Summing infinite series $sum_n=0^inftyfraccos^2n+1x2n+1$.
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How should one approach questions of this kind:
What is the sum of
$$sum_n=0^inftyfraccos^2n+1x2n+1$$
For $x=0$ this reduces to $1+1/3+1/5+...$ which diverges by the integral test. In fact, the answer is supposed to be $logcot (x/2)$. How do we arrive at it?
calculus real-analysis sequences-and-series
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up vote
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down vote
favorite
How should one approach questions of this kind:
What is the sum of
$$sum_n=0^inftyfraccos^2n+1x2n+1$$
For $x=0$ this reduces to $1+1/3+1/5+...$ which diverges by the integral test. In fact, the answer is supposed to be $logcot (x/2)$. How do we arrive at it?
calculus real-analysis sequences-and-series
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
How should one approach questions of this kind:
What is the sum of
$$sum_n=0^inftyfraccos^2n+1x2n+1$$
For $x=0$ this reduces to $1+1/3+1/5+...$ which diverges by the integral test. In fact, the answer is supposed to be $logcot (x/2)$. How do we arrive at it?
calculus real-analysis sequences-and-series
How should one approach questions of this kind:
What is the sum of
$$sum_n=0^inftyfraccos^2n+1x2n+1$$
For $x=0$ this reduces to $1+1/3+1/5+...$ which diverges by the integral test. In fact, the answer is supposed to be $logcot (x/2)$. How do we arrive at it?
calculus real-analysis sequences-and-series
edited Aug 19 at 10:04
Nosrati
20.7k41644
20.7k41644
asked Aug 19 at 8:46
Shahab
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4,62612474
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2 Answers
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$$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
for $|t|<1$. Substitute $t=cos x$.
$$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
1-cos x.$$
This can be simplified further using half-angle formulae.
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Where $xneq kpi$ we have $|cos x|<1$ and
$$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
integration gives
$$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
therefore
$$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
$$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
for $|t|<1$. Substitute $t=cos x$.
$$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
1-cos x.$$
This can be simplified further using half-angle formulae.
add a comment |Â
up vote
8
down vote
$$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
for $|t|<1$. Substitute $t=cos x$.
$$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
1-cos x.$$
This can be simplified further using half-angle formulae.
add a comment |Â
up vote
8
down vote
up vote
8
down vote
$$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
for $|t|<1$. Substitute $t=cos x$.
$$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
1-cos x.$$
This can be simplified further using half-angle formulae.
$$sum_n=1^inftyfract^2n+12n+1=tanh^-1t=frac12logfrac1+t1-t.$$
for $|t|<1$. Substitute $t=cos x$.
$$sum_n=1^inftyfraccos^2n+1x2n+1=frac12logfrac1+cos x
1-cos x.$$
This can be simplified further using half-angle formulae.
answered Aug 19 at 8:51
Lord Shark the Unknown
87.5k953114
87.5k953114
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add a comment |Â
up vote
1
down vote
Where $xneq kpi$ we have $|cos x|<1$ and
$$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
integration gives
$$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
therefore
$$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$
add a comment |Â
up vote
1
down vote
Where $xneq kpi$ we have $|cos x|<1$ and
$$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
integration gives
$$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
therefore
$$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Where $xneq kpi$ we have $|cos x|<1$ and
$$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
integration gives
$$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
therefore
$$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$
Where $xneq kpi$ we have $|cos x|<1$ and
$$sum_n=0^infty cos^2nx=dfrac11-cos^2 x$$
integration gives
$$sum_n=0^infty int cos^2nx(-sin x) dx=intdfrac-sin x1-cos^2 x dx$$
therefore
$$sum_n=0^infty dfraccos^2n+1x2n+1=lncotdfracx2$$
answered Aug 19 at 10:14
Nosrati
20.7k41644
20.7k41644
add a comment |Â
add a comment |Â
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