Finding the $n^{text{th}}$ term
$begingroup$
I have the following sequence:
$$1, 2 cos(x),~ 1+2 cos(2x),~ 2 cos(x)+2 cos(3x)….$$
I need to find the $n^{text{th}}$ term, considering the $1^{st}$ term as $1$, $2^{nd}$ as $2 cos(x)$ and so on.
We can represent it by a recurrence function as:
$$f(n) = f(n-2) + 2cos((n-1) x), f(1) = 1, f(2) = 2cos(x)$$
So what is the non-recursive version of the function, or how to approach to find one?
sequences-and-series trigonometry recurrence-relations
edited Jan 9 at 11:15
Larry
2,55031131
asked Jan 9 at 10:51
ash patelash patel
83
$endgroup$
add a comment |
$begingroup$
I have the following sequence:
$$1, 2 cos(x),~ 1+2 cos(2x),~ 2 cos(x)+2 cos(3x)….$$
I need to find the $n^{text{th}}$ term, considering the $1^{st}$ term as $1$, $2^{nd}$ as $2 cos(x)$ and so on.
We can represent it by a recurrence function as:
$$f(n) = f(n-2) + 2cos((n-1) x), f(1) = 1, f(2) = 2cos(x)$$
So what is the non-recursive version of the function, or how to approach to find one?
sequences-and-series trigonometry recurrence-relations
edited Jan 9 at 11:15
Larry
2,55031131
asked Jan 9 at 10:51
ash patelash patel
83
$endgroup$
add a comment |
$begingroup$
I have the following sequence:
$$1, 2 cos(x),~ 1+2 cos(2x),~ 2 cos(x)+2 cos(3x)….$$
I need to find the $n^{text{th}}$ term, considering the $1^{st}$ term as $1$, $2^{nd}$ as $2 cos(x)$ and so on.
We can represent it by a recurrence function as:
$$f(n) = f(n-2) + 2cos((n-1) x), f(1) = 1, f(2) = 2cos(x)$$
So what is the non-recursive version of the function, or how to approach to find one?
sequences-and-series trigonometry recurrence-relations
edited Jan 9 at 11:15
Larry
2,55031131
asked Jan 9 at 10:51
ash patelash patel
83
$endgroup$
I have the following sequence:
$$1, 2 cos(x),~ 1+2 cos(2x),~ 2 cos(x)+2 cos(3x)….$$
I need to find the $n^{text{th}}$ term, considering the $1^{st}$ term as $1$, $2^{nd}$ as $2 cos(x)$ and so on.
We can represent it by a recurrence function as:
$$f(n) = f(n-2) + 2cos((n-1) x), f(1) = 1, f(2) = 2cos(x)$$
So what is the non-recursive version of the function, or how to approach to find one?
sequences-and-series trigonometry recurrence-relations
sequences-and-series trigonometry recurrence-relations
edited Jan 9 at 11:15
Larry
2,55031131
asked Jan 9 at 10:51
ash patelash patel
83
edited Jan 9 at 11:15
Larry
2,55031131
asked Jan 9 at 10:51
ash patelash patel
83
edited Jan 9 at 11:15
Larry
2,55031131
edited Jan 9 at 11:15
Larry
2,55031131
edited Jan 9 at 11:15
Larry
2,55031131
2,55031131
asked Jan 9 at 10:51
ash patelash patel
83
asked Jan 9 at 10:51
ash patelash patel
83
asked Jan 9 at 10:51
ash patelash patel
83
83
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For odd $n=2k+1$, we have
$$
f_n = f_1+sum_{j=1}^{k}left( f_{2j+1} - f_{2j-1}right)=1+2sum_{j=1}^k cos (2jx).
$$ For even $n=2k$, we have
$$
f_n = f_2+sum_{j=2}^{k}left( f_{2j} - f_{2j-2}right)=2sum_{j=1}^k cos left((2j-1)xright).
$$
answered Jan 9 at 11:11
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Hint.
Try to solve
$$
Z_n = Z_{n-2}+2e^{i(n-1)x}
$$
here $Z_n = Z_n^h+Z_n^p$ and
$$
Z_n^h-Z_{n-2}^h=0
$$
gives
$$
Z_n^h = C_1+C_2(-1)^n
$$
and
$$
Z_n^p-Z_{n-2}^p = 2lambda^n
$$
gives
$$
Z_n^p = frac{lambda ^{n+2} left((lambda -1) (-1)^{2 n+1}+lambda +1right)}{lambda ^2-1}
$$
After that
$$
f_n = Re(Z_n)
$$
answered Jan 9 at 11:07
CesareoCesareo
9,9413518
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For odd $n=2k+1$, we have
$$
f_n = f_1+sum_{j=1}^{k}left( f_{2j+1} - f_{2j-1}right)=1+2sum_{j=1}^k cos (2jx).
$$ For even $n=2k$, we have
$$
f_n = f_2+sum_{j=2}^{k}left( f_{2j} - f_{2j-2}right)=2sum_{j=1}^k cos left((2j-1)xright).
$$
answered Jan 9 at 11:11
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
For odd $n=2k+1$, we have
$$
f_n = f_1+sum_{j=1}^{k}left( f_{2j+1} - f_{2j-1}right)=1+2sum_{j=1}^k cos (2jx).
$$ For even $n=2k$, we have
$$
f_n = f_2+sum_{j=2}^{k}left( f_{2j} - f_{2j-2}right)=2sum_{j=1}^k cos left((2j-1)xright).
$$
answered Jan 9 at 11:11
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
For odd $n=2k+1$, we have
$$
f_n = f_1+sum_{j=1}^{k}left( f_{2j+1} - f_{2j-1}right)=1+2sum_{j=1}^k cos (2jx).
$$ For even $n=2k$, we have
$$
f_n = f_2+sum_{j=2}^{k}left( f_{2j} - f_{2j-2}right)=2sum_{j=1}^k cos left((2j-1)xright).
$$
answered Jan 9 at 11:11
SongSong
18.6k21651
$endgroup$
For odd $n=2k+1$, we have
$$
f_n = f_1+sum_{j=1}^{k}left( f_{2j+1} - f_{2j-1}right)=1+2sum_{j=1}^k cos (2jx).
$$ For even $n=2k$, we have
$$
f_n = f_2+sum_{j=2}^{k}left( f_{2j} - f_{2j-2}right)=2sum_{j=1}^k cos left((2j-1)xright).
$$
answered Jan 9 at 11:11
SongSong
18.6k21651
answered Jan 9 at 11:11
SongSong
18.6k21651
answered Jan 9 at 11:11
SongSong
18.6k21651
answered Jan 9 at 11:11
SongSong
18.6k21651
18.6k21651
add a comment |
add a comment |
$begingroup$
Hint.
Try to solve
$$
Z_n = Z_{n-2}+2e^{i(n-1)x}
$$
here $Z_n = Z_n^h+Z_n^p$ and
$$
Z_n^h-Z_{n-2}^h=0
$$
gives
$$
Z_n^h = C_1+C_2(-1)^n
$$
and
$$
Z_n^p-Z_{n-2}^p = 2lambda^n
$$
gives
$$
Z_n^p = frac{lambda ^{n+2} left((lambda -1) (-1)^{2 n+1}+lambda +1right)}{lambda ^2-1}
$$
After that
$$
f_n = Re(Z_n)
$$
answered Jan 9 at 11:07
CesareoCesareo
9,9413518
$endgroup$
add a comment |
$begingroup$
Hint.
Try to solve
$$
Z_n = Z_{n-2}+2e^{i(n-1)x}
$$
here $Z_n = Z_n^h+Z_n^p$ and
$$
Z_n^h-Z_{n-2}^h=0
$$
gives
$$
Z_n^h = C_1+C_2(-1)^n
$$
and
$$
Z_n^p-Z_{n-2}^p = 2lambda^n
$$
gives
$$
Z_n^p = frac{lambda ^{n+2} left((lambda -1) (-1)^{2 n+1}+lambda +1right)}{lambda ^2-1}
$$
After that
$$
f_n = Re(Z_n)
$$
answered Jan 9 at 11:07
CesareoCesareo
9,9413518
$endgroup$
add a comment |
$begingroup$
Hint.
Try to solve
$$
Z_n = Z_{n-2}+2e^{i(n-1)x}
$$
here $Z_n = Z_n^h+Z_n^p$ and
$$
Z_n^h-Z_{n-2}^h=0
$$
gives
$$
Z_n^h = C_1+C_2(-1)^n
$$
and
$$
Z_n^p-Z_{n-2}^p = 2lambda^n
$$
gives
$$
Z_n^p = frac{lambda ^{n+2} left((lambda -1) (-1)^{2 n+1}+lambda +1right)}{lambda ^2-1}
$$
After that
$$
f_n = Re(Z_n)
$$
answered Jan 9 at 11:07
CesareoCesareo
9,9413518
$endgroup$
Hint.
Try to solve
$$
Z_n = Z_{n-2}+2e^{i(n-1)x}
$$
here $Z_n = Z_n^h+Z_n^p$ and
$$
Z_n^h-Z_{n-2}^h=0
$$
gives
$$
Z_n^h = C_1+C_2(-1)^n
$$
and
$$
Z_n^p-Z_{n-2}^p = 2lambda^n
$$
gives
$$
Z_n^p = frac{lambda ^{n+2} left((lambda -1) (-1)^{2 n+1}+lambda +1right)}{lambda ^2-1}
$$
After that
$$
f_n = Re(Z_n)
$$
answered Jan 9 at 11:07
CesareoCesareo
9,9413518
edited Jan 9 at 11:30
answered Jan 9 at 11:07
CesareoCesareo
9,9413518
answered Jan 9 at 11:07
CesareoCesareo
9,9413518
answered Jan 9 at 11:07
CesareoCesareo
9,9413518
9,9413518
add a comment |
add a comment |
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