Using $tan^{-1}$ Show that
$begingroup$
$$ pi = 2 sqrt3 sum_{n=0}^infty {(-1)^nover(2n+1)3^n}$$
Really have no idea on this one guys. Its a practice question for my calc 2 final. Please help.
calculus sequences-and-series
edited Dec 8 '18 at 22:17
Cameron Williams
22.4k43680
asked Dec 8 '18 at 22:07
user10753145user10753145
112
$endgroup$
add a comment |
$begingroup$
$$ pi = 2 sqrt3 sum_{n=0}^infty {(-1)^nover(2n+1)3^n}$$
Really have no idea on this one guys. Its a practice question for my calc 2 final. Please help.
calculus sequences-and-series
edited Dec 8 '18 at 22:17
Cameron Williams
22.4k43680
asked Dec 8 '18 at 22:07
user10753145user10753145
112
$endgroup$
add a comment |
$begingroup$
$$ pi = 2 sqrt3 sum_{n=0}^infty {(-1)^nover(2n+1)3^n}$$
Really have no idea on this one guys. Its a practice question for my calc 2 final. Please help.
calculus sequences-and-series
edited Dec 8 '18 at 22:17
Cameron Williams
22.4k43680
asked Dec 8 '18 at 22:07
user10753145user10753145
112
$endgroup$
$$ pi = 2 sqrt3 sum_{n=0}^infty {(-1)^nover(2n+1)3^n}$$
Really have no idea on this one guys. Its a practice question for my calc 2 final. Please help.
calculus sequences-and-series
calculus sequences-and-series
edited Dec 8 '18 at 22:17
Cameron Williams
22.4k43680
asked Dec 8 '18 at 22:07
user10753145user10753145
112
edited Dec 8 '18 at 22:17
Cameron Williams
22.4k43680
asked Dec 8 '18 at 22:07
user10753145user10753145
112
edited Dec 8 '18 at 22:17
Cameron Williams
22.4k43680
edited Dec 8 '18 at 22:17
Cameron Williams
22.4k43680
edited Dec 8 '18 at 22:17
Cameron Williams
22.4k43680
22.4k43680
asked Dec 8 '18 at 22:07
user10753145user10753145
112
asked Dec 8 '18 at 22:07
user10753145user10753145
112
asked Dec 8 '18 at 22:07
user10753145user10753145
112
112
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Recall that for $|t|<1$,
$$frac1t=sum_{kgeq0}(1-t)^k$$
Hence we have that
$$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
Integrating from $0$ to $x$ on the RHS:
$$int_0^x frac{mathrm dt}{1+t^2}=arctan x$$
Doing the same on the LHS:
$$int_0^x sum_{kgeq0}(-1)^kt^{2k}mathrm dt=sum_{kgeq0}(-1)^kint_0^x t^{2k}mathrm dt=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Hence we have
$$arctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Now note that
$$arctanbigg(frac1{sqrt{3}}bigg)=sum_{kgeq0}frac{(-1)^k}{2k+1}bigg(frac1{sqrt{3}}bigg)^{2k+1}$$
Hence
$$fracpi6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^{k+frac12}}$$
$$frac{pisqrt{3}}6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
Simplify the radical:
$$fracpi{2sqrt{3}}=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
$$pi=2sqrt{3}sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
QED
answered Dec 8 '18 at 23:18
clathratusclathratus
4,471336
$endgroup$
add a comment |
$begingroup$
HINTS:
First note that
$$arctan(x)=int_0^x frac1{1+t^2},dt$$
Second, represent $frac{1}{1+t^2}$ as a geometric series.
Third, integrate term by term.
Can you proceed now?
answered Dec 8 '18 at 22:14
Mark ViolaMark Viola
132k1275173
$endgroup$
add a comment |
$begingroup$
Hint:
$;dfracpi6=arctan biggl(dfrac1{sqrt 3}biggr)$. Also remember that for $|x|<1$,
$$arctan x=sum_{n=0}^infty(-1)^n frac{x^{2n+1}}{2n+1}.$$
answered Dec 8 '18 at 22:17
BernardBernard
121k740116
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Recall that for $|t|<1$,
$$frac1t=sum_{kgeq0}(1-t)^k$$
Hence we have that
$$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
Integrating from $0$ to $x$ on the RHS:
$$int_0^x frac{mathrm dt}{1+t^2}=arctan x$$
Doing the same on the LHS:
$$int_0^x sum_{kgeq0}(-1)^kt^{2k}mathrm dt=sum_{kgeq0}(-1)^kint_0^x t^{2k}mathrm dt=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Hence we have
$$arctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Now note that
$$arctanbigg(frac1{sqrt{3}}bigg)=sum_{kgeq0}frac{(-1)^k}{2k+1}bigg(frac1{sqrt{3}}bigg)^{2k+1}$$
Hence
$$fracpi6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^{k+frac12}}$$
$$frac{pisqrt{3}}6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
Simplify the radical:
$$fracpi{2sqrt{3}}=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
$$pi=2sqrt{3}sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
QED
answered Dec 8 '18 at 23:18
clathratusclathratus
4,471336
$endgroup$
add a comment |
$begingroup$
Recall that for $|t|<1$,
$$frac1t=sum_{kgeq0}(1-t)^k$$
Hence we have that
$$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
Integrating from $0$ to $x$ on the RHS:
$$int_0^x frac{mathrm dt}{1+t^2}=arctan x$$
Doing the same on the LHS:
$$int_0^x sum_{kgeq0}(-1)^kt^{2k}mathrm dt=sum_{kgeq0}(-1)^kint_0^x t^{2k}mathrm dt=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Hence we have
$$arctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Now note that
$$arctanbigg(frac1{sqrt{3}}bigg)=sum_{kgeq0}frac{(-1)^k}{2k+1}bigg(frac1{sqrt{3}}bigg)^{2k+1}$$
Hence
$$fracpi6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^{k+frac12}}$$
$$frac{pisqrt{3}}6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
Simplify the radical:
$$fracpi{2sqrt{3}}=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
$$pi=2sqrt{3}sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
QED
answered Dec 8 '18 at 23:18
clathratusclathratus
4,471336
$endgroup$
add a comment |
$begingroup$
Recall that for $|t|<1$,
$$frac1t=sum_{kgeq0}(1-t)^k$$
Hence we have that
$$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
Integrating from $0$ to $x$ on the RHS:
$$int_0^x frac{mathrm dt}{1+t^2}=arctan x$$
Doing the same on the LHS:
$$int_0^x sum_{kgeq0}(-1)^kt^{2k}mathrm dt=sum_{kgeq0}(-1)^kint_0^x t^{2k}mathrm dt=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Hence we have
$$arctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Now note that
$$arctanbigg(frac1{sqrt{3}}bigg)=sum_{kgeq0}frac{(-1)^k}{2k+1}bigg(frac1{sqrt{3}}bigg)^{2k+1}$$
Hence
$$fracpi6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^{k+frac12}}$$
$$frac{pisqrt{3}}6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
Simplify the radical:
$$fracpi{2sqrt{3}}=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
$$pi=2sqrt{3}sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
QED
answered Dec 8 '18 at 23:18
clathratusclathratus
4,471336
$endgroup$
Recall that for $|t|<1$,
$$frac1t=sum_{kgeq0}(1-t)^k$$
Hence we have that
$$frac1{1+t^2}=sum_{kgeq0}(-1)^kt^{2k}$$
Integrating from $0$ to $x$ on the RHS:
$$int_0^x frac{mathrm dt}{1+t^2}=arctan x$$
Doing the same on the LHS:
$$int_0^x sum_{kgeq0}(-1)^kt^{2k}mathrm dt=sum_{kgeq0}(-1)^kint_0^x t^{2k}mathrm dt=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Hence we have
$$arctan x=sum_{kgeq0}(-1)^kfrac{x^{2k+1}}{2k+1}$$
Now note that
$$arctanbigg(frac1{sqrt{3}}bigg)=sum_{kgeq0}frac{(-1)^k}{2k+1}bigg(frac1{sqrt{3}}bigg)^{2k+1}$$
Hence
$$fracpi6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^{k+frac12}}$$
$$frac{pisqrt{3}}6=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
Simplify the radical:
$$fracpi{2sqrt{3}}=sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
$$pi=2sqrt{3}sum_{kgeq0}frac{(-1)^k}{(2k+1)3^k}$$
QED
answered Dec 8 '18 at 23:18
clathratusclathratus
4,471336
answered Dec 8 '18 at 23:18
clathratusclathratus
4,471336
answered Dec 8 '18 at 23:18
clathratusclathratus
4,471336
answered Dec 8 '18 at 23:18
clathratusclathratus
4,471336
4,471336
add a comment |
add a comment |
$begingroup$
HINTS:
First note that
$$arctan(x)=int_0^x frac1{1+t^2},dt$$
Second, represent $frac{1}{1+t^2}$ as a geometric series.
Third, integrate term by term.
Can you proceed now?
answered Dec 8 '18 at 22:14
Mark ViolaMark Viola
132k1275173
$endgroup$
add a comment |
$begingroup$
HINTS:
First note that
$$arctan(x)=int_0^x frac1{1+t^2},dt$$
Second, represent $frac{1}{1+t^2}$ as a geometric series.
Third, integrate term by term.
Can you proceed now?
answered Dec 8 '18 at 22:14
Mark ViolaMark Viola
132k1275173
$endgroup$
add a comment |
$begingroup$
HINTS:
First note that
$$arctan(x)=int_0^x frac1{1+t^2},dt$$
Second, represent $frac{1}{1+t^2}$ as a geometric series.
Third, integrate term by term.
Can you proceed now?
answered Dec 8 '18 at 22:14
Mark ViolaMark Viola
132k1275173
$endgroup$
HINTS:
First note that
$$arctan(x)=int_0^x frac1{1+t^2},dt$$
Second, represent $frac{1}{1+t^2}$ as a geometric series.
Third, integrate term by term.
Can you proceed now?
answered Dec 8 '18 at 22:14
Mark ViolaMark Viola
132k1275173
answered Dec 8 '18 at 22:14
Mark ViolaMark Viola
132k1275173
answered Dec 8 '18 at 22:14
Mark ViolaMark Viola
132k1275173
answered Dec 8 '18 at 22:14
Mark ViolaMark Viola
132k1275173
132k1275173
add a comment |
add a comment |
$begingroup$
Hint:
$;dfracpi6=arctan biggl(dfrac1{sqrt 3}biggr)$. Also remember that for $|x|<1$,
$$arctan x=sum_{n=0}^infty(-1)^n frac{x^{2n+1}}{2n+1}.$$
answered Dec 8 '18 at 22:17
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Hint:
$;dfracpi6=arctan biggl(dfrac1{sqrt 3}biggr)$. Also remember that for $|x|<1$,
$$arctan x=sum_{n=0}^infty(-1)^n frac{x^{2n+1}}{2n+1}.$$
answered Dec 8 '18 at 22:17
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Hint:
$;dfracpi6=arctan biggl(dfrac1{sqrt 3}biggr)$. Also remember that for $|x|<1$,
$$arctan x=sum_{n=0}^infty(-1)^n frac{x^{2n+1}}{2n+1}.$$
answered Dec 8 '18 at 22:17
BernardBernard
121k740116
$endgroup$
Hint:
$;dfracpi6=arctan biggl(dfrac1{sqrt 3}biggr)$. Also remember that for $|x|<1$,
$$arctan x=sum_{n=0}^infty(-1)^n frac{x^{2n+1}}{2n+1}.$$
answered Dec 8 '18 at 22:17
BernardBernard
121k740116
answered Dec 8 '18 at 22:17
BernardBernard
121k740116
answered Dec 8 '18 at 22:17
BernardBernard
121k740116
answered Dec 8 '18 at 22:17
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
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