Find local maxima and minima of impulse train
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I have an impulse train given by
$$sum_j=0^r fraccos bigl(frac2 pi jxr+1 bigr)r+1$$
Plotted for $r=1$ (blue), $5$ (orange) and $8$ (green), this gives:
Ignore the case of $r=1$, as this is a simple $cos$ curve.
Where, in terms of $x$, do the maxima and minima occur (in the given ranges), and what are their values in terms of $x$ and $r$?
Differentiation gives me
$$-sum_j=0^r fracfrac2 pi jr+1sin bigl(frac2 pi jxr+1 bigr)r+1=0$$
but I'm not sure how to proceed from there.
Edit:
As per Yves' suggestion below, I tried to figure out the sum, but I hit a wall even before I get to differentiating:
$$cos x=frace^ix + e^-ix2$$
Therefore
$$sum_j=0^r fraccos bigl(frac2 pi jxr+1 bigr)r+1=frac12(r+1) biggl(sum_j=0^r e^fraci2 pi j xr+1+ sum_j=0^r e^frac-i2 pi j xr+1 biggr)$$
Next,
$$sum_j=0^r-1 a^j=frac1-a^r+11-a$$
and setting $a=e^fraci2 pi xr+1$ (and therefore $a^r+1=e^i2 pi x$), we have
$$frac12(r+1) biggl(sum_j=0^r e^fraci2 pi j xr+1+ sum_j=0^r e^frac-i2 pi j xr+1 biggr)=frac12(r+1) biggl(frac1-e^i2 pi x1-e^fraci2 pi xr+1+frac1-e^-i2 pi x1-e^frac-i2 pi xr+1biggr)$$
$$=frac12(r+1) biggl(fracbigl(1-e^i2 pi xbigr)bigl(1-e^frac-i2 pi xr+1bigr)+bigl(1-e^fraci2 pi xr+1bigr)bigl(1-e^-i2 pi xbigr)bigl(1-e^fraci2 pi xr+1bigr)bigl(1-e^frac-i2 pi xr+1bigr)biggr)$$
...but where do I go from there?
derivatives fourier-analysis fourier-series maxima-minima
asked Sep 5 at 11:33
Richard Burke-Ward
2648
add a comment |Â
up vote
0
down vote
favorite
I have an impulse train given by
$$sum_j=0^r fraccos bigl(frac2 pi jxr+1 bigr)r+1$$
Plotted for $r=1$ (blue), $5$ (orange) and $8$ (green), this gives:
Ignore the case of $r=1$, as this is a simple $cos$ curve.
Where, in terms of $x$, do the maxima and minima occur (in the given ranges), and what are their values in terms of $x$ and $r$?
Differentiation gives me
$$-sum_j=0^r fracfrac2 pi jr+1sin bigl(frac2 pi jxr+1 bigr)r+1=0$$
but I'm not sure how to proceed from there.
Edit:
As per Yves' suggestion below, I tried to figure out the sum, but I hit a wall even before I get to differentiating:
$$cos x=frace^ix + e^-ix2$$
Therefore
$$sum_j=0^r fraccos bigl(frac2 pi jxr+1 bigr)r+1=frac12(r+1) biggl(sum_j=0^r e^fraci2 pi j xr+1+ sum_j=0^r e^frac-i2 pi j xr+1 biggr)$$
Next,
$$sum_j=0^r-1 a^j=frac1-a^r+11-a$$
and setting $a=e^fraci2 pi xr+1$ (and therefore $a^r+1=e^i2 pi x$), we have
$$frac12(r+1) biggl(sum_j=0^r e^fraci2 pi j xr+1+ sum_j=0^r e^frac-i2 pi j xr+1 biggr)=frac12(r+1) biggl(frac1-e^i2 pi x1-e^fraci2 pi xr+1+frac1-e^-i2 pi x1-e^frac-i2 pi xr+1biggr)$$
$$=frac12(r+1) biggl(fracbigl(1-e^i2 pi xbigr)bigl(1-e^frac-i2 pi xr+1bigr)+bigl(1-e^fraci2 pi xr+1bigr)bigl(1-e^-i2 pi xbigr)bigl(1-e^fraci2 pi xr+1bigr)bigl(1-e^frac-i2 pi xr+1bigr)biggr)$$
...but where do I go from there?
derivatives fourier-analysis fourier-series maxima-minima
asked Sep 5 at 11:33
Richard Burke-Ward
2648
Hint: the sum $sum_k cos ktheta$ has a closed-form expression.
– Yves Daoust
Sep 5 at 11:55
Hi Yves. I know that $sum_k cos ktheta$ can be expressed as a telescoping complex geometric series (though sadly, I haven't yet figured out how...). But even if I knew the closed form expression, I'd still be at a loss.
– Richard Burke-Ward
Sep 5 at 12:09
Write it down before concluding.
– Yves Daoust
Sep 5 at 12:22
Please see edited version of original question. I still get stuck, just earlier than before!
– Richard Burke-Ward
Sep 5 at 17:17
en.wikipedia.org/wiki/…. This will eventually lead you to a trigonometric equation with no closed-form solution, except for the small values of $r$ (up to four, presumably). You will have to use a numerical method.
– Yves Daoust
Sep 6 at 7:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have an impulse train given by
$$sum_j=0^r fraccos bigl(frac2 pi jxr+1 bigr)r+1$$
Plotted for $r=1$ (blue), $5$ (orange) and $8$ (green), this gives:
Ignore the case of $r=1$, as this is a simple $cos$ curve.
Where, in terms of $x$, do the maxima and minima occur (in the given ranges), and what are their values in terms of $x$ and $r$?
Differentiation gives me
$$-sum_j=0^r fracfrac2 pi jr+1sin bigl(frac2 pi jxr+1 bigr)r+1=0$$
but I'm not sure how to proceed from there.
Edit:
As per Yves' suggestion below, I tried to figure out the sum, but I hit a wall even before I get to differentiating:
$$cos x=frace^ix + e^-ix2$$
Therefore
$$sum_j=0^r fraccos bigl(frac2 pi jxr+1 bigr)r+1=frac12(r+1) biggl(sum_j=0^r e^fraci2 pi j xr+1+ sum_j=0^r e^frac-i2 pi j xr+1 biggr)$$
Next,
$$sum_j=0^r-1 a^j=frac1-a^r+11-a$$
and setting $a=e^fraci2 pi xr+1$ (and therefore $a^r+1=e^i2 pi x$), we have
$$frac12(r+1) biggl(sum_j=0^r e^fraci2 pi j xr+1+ sum_j=0^r e^frac-i2 pi j xr+1 biggr)=frac12(r+1) biggl(frac1-e^i2 pi x1-e^fraci2 pi xr+1+frac1-e^-i2 pi x1-e^frac-i2 pi xr+1biggr)$$
$$=frac12(r+1) biggl(fracbigl(1-e^i2 pi xbigr)bigl(1-e^frac-i2 pi xr+1bigr)+bigl(1-e^fraci2 pi xr+1bigr)bigl(1-e^-i2 pi xbigr)bigl(1-e^fraci2 pi xr+1bigr)bigl(1-e^frac-i2 pi xr+1bigr)biggr)$$
...but where do I go from there?
derivatives fourier-analysis fourier-series maxima-minima
asked Sep 5 at 11:33
Richard Burke-Ward
2648
I have an impulse train given by
$$sum_j=0^r fraccos bigl(frac2 pi jxr+1 bigr)r+1$$
Plotted for $r=1$ (blue), $5$ (orange) and $8$ (green), this gives:
Ignore the case of $r=1$, as this is a simple $cos$ curve.
Where, in terms of $x$, do the maxima and minima occur (in the given ranges), and what are their values in terms of $x$ and $r$?
Differentiation gives me
$$-sum_j=0^r fracfrac2 pi jr+1sin bigl(frac2 pi jxr+1 bigr)r+1=0$$
but I'm not sure how to proceed from there.
Edit:
As per Yves' suggestion below, I tried to figure out the sum, but I hit a wall even before I get to differentiating:
$$cos x=frace^ix + e^-ix2$$
Therefore
$$sum_j=0^r fraccos bigl(frac2 pi jxr+1 bigr)r+1=frac12(r+1) biggl(sum_j=0^r e^fraci2 pi j xr+1+ sum_j=0^r e^frac-i2 pi j xr+1 biggr)$$
Next,
$$sum_j=0^r-1 a^j=frac1-a^r+11-a$$
and setting $a=e^fraci2 pi xr+1$ (and therefore $a^r+1=e^i2 pi x$), we have
$$frac12(r+1) biggl(sum_j=0^r e^fraci2 pi j xr+1+ sum_j=0^r e^frac-i2 pi j xr+1 biggr)=frac12(r+1) biggl(frac1-e^i2 pi x1-e^fraci2 pi xr+1+frac1-e^-i2 pi x1-e^frac-i2 pi xr+1biggr)$$
$$=frac12(r+1) biggl(fracbigl(1-e^i2 pi xbigr)bigl(1-e^frac-i2 pi xr+1bigr)+bigl(1-e^fraci2 pi xr+1bigr)bigl(1-e^-i2 pi xbigr)bigl(1-e^fraci2 pi xr+1bigr)bigl(1-e^frac-i2 pi xr+1bigr)biggr)$$
...but where do I go from there?
derivatives fourier-analysis fourier-series maxima-minima
derivatives fourier-analysis fourier-series maxima-minima
asked Sep 5 at 11:33
Richard Burke-Ward
2648
asked Sep 5 at 11:33
Richard Burke-Ward
2648
edited Sep 5 at 16:45
asked Sep 5 at 11:33
Richard Burke-Ward
2648
asked Sep 5 at 11:33
Richard Burke-Ward
2648
asked Sep 5 at 11:33
Richard Burke-Ward
2648
2648
Hint: the sum $sum_k cos ktheta$ has a closed-form expression.
– Yves Daoust
Sep 5 at 11:55
Hi Yves. I know that $sum_k cos ktheta$ can be expressed as a telescoping complex geometric series (though sadly, I haven't yet figured out how...). But even if I knew the closed form expression, I'd still be at a loss.
– Richard Burke-Ward
Sep 5 at 12:09
Write it down before concluding.
– Yves Daoust
Sep 5 at 12:22
Please see edited version of original question. I still get stuck, just earlier than before!
– Richard Burke-Ward
Sep 5 at 17:17
en.wikipedia.org/wiki/…. This will eventually lead you to a trigonometric equation with no closed-form solution, except for the small values of $r$ (up to four, presumably). You will have to use a numerical method.
– Yves Daoust
Sep 6 at 7:12
add a comment |Â
Hint: the sum $sum_k cos ktheta$ has a closed-form expression.
– Yves Daoust
Sep 5 at 11:55
Hi Yves. I know that $sum_k cos ktheta$ can be expressed as a telescoping complex geometric series (though sadly, I haven't yet figured out how...). But even if I knew the closed form expression, I'd still be at a loss.
– Richard Burke-Ward
Sep 5 at 12:09
Write it down before concluding.
– Yves Daoust
Sep 5 at 12:22
Please see edited version of original question. I still get stuck, just earlier than before!
– Richard Burke-Ward
Sep 5 at 17:17
en.wikipedia.org/wiki/…. This will eventually lead you to a trigonometric equation with no closed-form solution, except for the small values of $r$ (up to four, presumably). You will have to use a numerical method.
– Yves Daoust
Sep 6 at 7:12
Hint: the sum $sum_k cos ktheta$ has a closed-form expression.
– Yves Daoust
Sep 5 at 11:55
Hint: the sum $sum_k cos ktheta$ has a closed-form expression.
– Yves Daoust
Sep 5 at 11:55
Hi Yves. I know that $sum_k cos ktheta$ can be expressed as a telescoping complex geometric series (though sadly, I haven't yet figured out how...). But even if I knew the closed form expression, I'd still be at a loss.
– Richard Burke-Ward
Sep 5 at 12:09
Hi Yves. I know that $sum_k cos ktheta$ can be expressed as a telescoping complex geometric series (though sadly, I haven't yet figured out how...). But even if I knew the closed form expression, I'd still be at a loss.
– Richard Burke-Ward
Sep 5 at 12:09
Write it down before concluding.
– Yves Daoust
Sep 5 at 12:22
Write it down before concluding.
– Yves Daoust
Sep 5 at 12:22
Please see edited version of original question. I still get stuck, just earlier than before!
– Richard Burke-Ward
Sep 5 at 17:17
Please see edited version of original question. I still get stuck, just earlier than before!
– Richard Burke-Ward
Sep 5 at 17:17
en.wikipedia.org/wiki/…. This will eventually lead you to a trigonometric equation with no closed-form solution, except for the small values of $r$ (up to four, presumably). You will have to use a numerical method.
– Yves Daoust
Sep 6 at 7:12
en.wikipedia.org/wiki/…. This will eventually lead you to a trigonometric equation with no closed-form solution, except for the small values of $r$ (up to four, presumably). You will have to use a numerical method.
– Yves Daoust
Sep 6 at 7:12
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2906155%2ffind-local-maxima-and-minima-of-impulse-train%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Hint: the sum $sum_k cos ktheta$ has a closed-form expression.
– Yves Daoust
Sep 5 at 11:55
Hi Yves. I know that $sum_k cos ktheta$ can be expressed as a telescoping complex geometric series (though sadly, I haven't yet figured out how...). But even if I knew the closed form expression, I'd still be at a loss.
– Richard Burke-Ward
Sep 5 at 12:09
Write it down before concluding.
– Yves Daoust
Sep 5 at 12:22
Please see edited version of original question. I still get stuck, just earlier than before!
– Richard Burke-Ward
Sep 5 at 17:17
en.wikipedia.org/wiki/…. This will eventually lead you to a trigonometric equation with no closed-form solution, except for the small values of $r$ (up to four, presumably). You will have to use a numerical method.
– Yves Daoust
Sep 6 at 7:12