Is there a simple function that can make sin into a line? [$sin(f(x)) = {2xoverpi}$ from $0$ to $piover2$]
$begingroup$
I'm currently working on a synth, and I'd like to find an equation that can go from a triangle wave to sine smoothly. In order to do this, I'm trying to find a function which can be put inside sine $(sin(f(x))$ that will make the output into a straight line from $(0,0)$ to $({piover2}, 1)$.
Unfortunately, my math is a bit rusty at them moment. I could have sworn there was a way to put arcsin, or arccos into sin and have it kind of cancel out the curve, but I can't remember it (or whether or not it was that simple).
Is there any such simple function $f(x)$ that can be put into sin() to result in a flat line from $(0,0)$ to $({piover2}, 1)$? $f(x)$ such that $sin(f(x)) = {2xover pi}$ from $0$ to $piover2$
trigonometry
edited Dec 14 '18 at 6:30
user587054
51911
asked Dec 14 '18 at 5:45
Seph ReedSeph Reed
1187
$endgroup$
add a comment |
$begingroup$
I'm currently working on a synth, and I'd like to find an equation that can go from a triangle wave to sine smoothly. In order to do this, I'm trying to find a function which can be put inside sine $(sin(f(x))$ that will make the output into a straight line from $(0,0)$ to $({piover2}, 1)$.
Unfortunately, my math is a bit rusty at them moment. I could have sworn there was a way to put arcsin, or arccos into sin and have it kind of cancel out the curve, but I can't remember it (or whether or not it was that simple).
Is there any such simple function $f(x)$ that can be put into sin() to result in a flat line from $(0,0)$ to $({piover2}, 1)$? $f(x)$ such that $sin(f(x)) = {2xover pi}$ from $0$ to $piover2$
trigonometry
edited Dec 14 '18 at 6:30
user587054
51911
asked Dec 14 '18 at 5:45
Seph ReedSeph Reed
1187
$endgroup$
add a comment |
$begingroup$
I'm currently working on a synth, and I'd like to find an equation that can go from a triangle wave to sine smoothly. In order to do this, I'm trying to find a function which can be put inside sine $(sin(f(x))$ that will make the output into a straight line from $(0,0)$ to $({piover2}, 1)$.
Unfortunately, my math is a bit rusty at them moment. I could have sworn there was a way to put arcsin, or arccos into sin and have it kind of cancel out the curve, but I can't remember it (or whether or not it was that simple).
Is there any such simple function $f(x)$ that can be put into sin() to result in a flat line from $(0,0)$ to $({piover2}, 1)$? $f(x)$ such that $sin(f(x)) = {2xover pi}$ from $0$ to $piover2$
trigonometry
edited Dec 14 '18 at 6:30
user587054
51911
asked Dec 14 '18 at 5:45
Seph ReedSeph Reed
1187
$endgroup$
I'm currently working on a synth, and I'd like to find an equation that can go from a triangle wave to sine smoothly. In order to do this, I'm trying to find a function which can be put inside sine $(sin(f(x))$ that will make the output into a straight line from $(0,0)$ to $({piover2}, 1)$.
Unfortunately, my math is a bit rusty at them moment. I could have sworn there was a way to put arcsin, or arccos into sin and have it kind of cancel out the curve, but I can't remember it (or whether or not it was that simple).
Is there any such simple function $f(x)$ that can be put into sin() to result in a flat line from $(0,0)$ to $({piover2}, 1)$? $f(x)$ such that $sin(f(x)) = {2xover pi}$ from $0$ to $piover2$
trigonometry
trigonometry
edited Dec 14 '18 at 6:30
user587054
51911
asked Dec 14 '18 at 5:45
Seph ReedSeph Reed
1187
edited Dec 14 '18 at 6:30
user587054
51911
asked Dec 14 '18 at 5:45
Seph ReedSeph Reed
1187
edited Dec 14 '18 at 6:30
user587054
51911
edited Dec 14 '18 at 6:30
user587054
51911
edited Dec 14 '18 at 6:30
user587054
51911
51911
asked Dec 14 '18 at 5:45
Seph ReedSeph Reed
1187
asked Dec 14 '18 at 5:45
Seph ReedSeph Reed
1187
asked Dec 14 '18 at 5:45
Seph ReedSeph Reed
1187
1187
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The sin function has an inverse $sin^{-1}(x)$ defined on $-1<x<1$.
$$sin(sin^{-1}(x)) = x.$$ I'm not sure if this is the answer you were looking for, but you can just scale the inverse sine like $sin((2/pi)x)$ so that at $x = pi/2$, you are still in the range of the inverse sine.
answered Dec 14 '18 at 5:52
D.B.D.B.
1,25018
$endgroup$
add a comment |
$begingroup$
Here my ideas:
Start with differentiating both sides. $$cos(f(x))*f'(x)=1$$ or $f'(x)=frac{1}{cos(f(x))}$
Integrate. I've used Maple to do it.
de := diff(f(x), x) = 2/(Pi*cos(f(x)));
d 2
de := --- f(x) = ------------
dx Pi cos(f(x))
sol := dsolve(de, f(x));
/2 (x + _C1)
sol := f(x) = arcsin|-----------|
Pi /
Checking for result
sol := dsolve(de, f(x), implicit);
1
sol := x - - Pi sin(f(x)) + _C1 = 0
2
So, for your quest,
_C1 = 0. Then your function is $$f(x)=arcsin(frac{2x}{pi})$$
answered Dec 14 '18 at 6:02
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
$begingroup$
I feel dumb. It was really easy to solve:
sin(arcsin(2x/pi))
Also, in case anyone was wondering, I found the actual equation I was looking for. I'm trying to make it so a knob can change a curve from sin to a line incrementally, and found a good one for it.
1-(1-2x/pi)^C. At C=1.757 you get a curve for which the integral of the absolute difference between it and sine from 0 to pi/2 of pretty low, which is more than close enough for my case. And then at C=1 you get a straight line. Pretty neat!
answered Dec 14 '18 at 5:53
Seph ReedSeph Reed
1187
$endgroup$
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
$begingroup$
The sin function has an inverse $sin^{-1}(x)$ defined on $-1<x<1$.
$$sin(sin^{-1}(x)) = x.$$ I'm not sure if this is the answer you were looking for, but you can just scale the inverse sine like $sin((2/pi)x)$ so that at $x = pi/2$, you are still in the range of the inverse sine.
answered Dec 14 '18 at 5:52
D.B.D.B.
1,25018
$endgroup$
add a comment |
$begingroup$
The sin function has an inverse $sin^{-1}(x)$ defined on $-1<x<1$.
$$sin(sin^{-1}(x)) = x.$$ I'm not sure if this is the answer you were looking for, but you can just scale the inverse sine like $sin((2/pi)x)$ so that at $x = pi/2$, you are still in the range of the inverse sine.
answered Dec 14 '18 at 5:52
D.B.D.B.
1,25018
$endgroup$
add a comment |
$begingroup$
The sin function has an inverse $sin^{-1}(x)$ defined on $-1<x<1$.
$$sin(sin^{-1}(x)) = x.$$ I'm not sure if this is the answer you were looking for, but you can just scale the inverse sine like $sin((2/pi)x)$ so that at $x = pi/2$, you are still in the range of the inverse sine.
answered Dec 14 '18 at 5:52
D.B.D.B.
1,25018
$endgroup$
The sin function has an inverse $sin^{-1}(x)$ defined on $-1<x<1$.
$$sin(sin^{-1}(x)) = x.$$ I'm not sure if this is the answer you were looking for, but you can just scale the inverse sine like $sin((2/pi)x)$ so that at $x = pi/2$, you are still in the range of the inverse sine.
answered Dec 14 '18 at 5:52
D.B.D.B.
1,25018
answered Dec 14 '18 at 5:52
D.B.D.B.
1,25018
answered Dec 14 '18 at 5:52
D.B.D.B.
1,25018
answered Dec 14 '18 at 5:52
D.B.D.B.
1,25018
1,25018
add a comment |
add a comment |
$begingroup$
Here my ideas:
Start with differentiating both sides. $$cos(f(x))*f'(x)=1$$ or $f'(x)=frac{1}{cos(f(x))}$
Integrate. I've used Maple to do it.
de := diff(f(x), x) = 2/(Pi*cos(f(x)));
d 2
de := --- f(x) = ------------
dx Pi cos(f(x))
sol := dsolve(de, f(x));
/2 (x + _C1)
sol := f(x) = arcsin|-----------|
Pi /
Checking for result
sol := dsolve(de, f(x), implicit);
1
sol := x - - Pi sin(f(x)) + _C1 = 0
2
So, for your quest,
_C1 = 0. Then your function is $$f(x)=arcsin(frac{2x}{pi})$$
answered Dec 14 '18 at 6:02
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
$begingroup$
Here my ideas:
Start with differentiating both sides. $$cos(f(x))*f'(x)=1$$ or $f'(x)=frac{1}{cos(f(x))}$
Integrate. I've used Maple to do it.
de := diff(f(x), x) = 2/(Pi*cos(f(x)));
d 2
de := --- f(x) = ------------
dx Pi cos(f(x))
sol := dsolve(de, f(x));
/2 (x + _C1)
sol := f(x) = arcsin|-----------|
Pi /
Checking for result
sol := dsolve(de, f(x), implicit);
1
sol := x - - Pi sin(f(x)) + _C1 = 0
2
So, for your quest,
_C1 = 0. Then your function is $$f(x)=arcsin(frac{2x}{pi})$$
answered Dec 14 '18 at 6:02
Kelly ShepphardKelly Shepphard
2298
$endgroup$
add a comment |
$begingroup$
Here my ideas:
Start with differentiating both sides. $$cos(f(x))*f'(x)=1$$ or $f'(x)=frac{1}{cos(f(x))}$
Integrate. I've used Maple to do it.
de := diff(f(x), x) = 2/(Pi*cos(f(x)));
d 2
de := --- f(x) = ------------
dx Pi cos(f(x))
sol := dsolve(de, f(x));
/2 (x + _C1)
sol := f(x) = arcsin|-----------|
Pi /
Checking for result
sol := dsolve(de, f(x), implicit);
1
sol := x - - Pi sin(f(x)) + _C1 = 0
2
So, for your quest,
_C1 = 0. Then your function is $$f(x)=arcsin(frac{2x}{pi})$$
answered Dec 14 '18 at 6:02
Kelly ShepphardKelly Shepphard
2298
$endgroup$
Here my ideas:
Start with differentiating both sides. $$cos(f(x))*f'(x)=1$$ or $f'(x)=frac{1}{cos(f(x))}$
Integrate. I've used Maple to do it.
de := diff(f(x), x) = 2/(Pi*cos(f(x)));
d 2
de := --- f(x) = ------------
dx Pi cos(f(x))
sol := dsolve(de, f(x));
/2 (x + _C1)
sol := f(x) = arcsin|-----------|
Pi /
Checking for result
sol := dsolve(de, f(x), implicit);
1
sol := x - - Pi sin(f(x)) + _C1 = 0
2
So, for your quest,
_C1 = 0. Then your function is $$f(x)=arcsin(frac{2x}{pi})$$
answered Dec 14 '18 at 6:02
Kelly ShepphardKelly Shepphard
2298
answered Dec 14 '18 at 6:02
Kelly ShepphardKelly Shepphard
2298
answered Dec 14 '18 at 6:02
Kelly ShepphardKelly Shepphard
2298
answered Dec 14 '18 at 6:02
Kelly ShepphardKelly Shepphard
2298
2298
add a comment |
add a comment |
$begingroup$
I feel dumb. It was really easy to solve:
sin(arcsin(2x/pi))
Also, in case anyone was wondering, I found the actual equation I was looking for. I'm trying to make it so a knob can change a curve from sin to a line incrementally, and found a good one for it.
1-(1-2x/pi)^C. At C=1.757 you get a curve for which the integral of the absolute difference between it and sine from 0 to pi/2 of pretty low, which is more than close enough for my case. And then at C=1 you get a straight line. Pretty neat!
answered Dec 14 '18 at 5:53
Seph ReedSeph Reed
1187
$endgroup$
add a comment |
$begingroup$
I feel dumb. It was really easy to solve:
sin(arcsin(2x/pi))
Also, in case anyone was wondering, I found the actual equation I was looking for. I'm trying to make it so a knob can change a curve from sin to a line incrementally, and found a good one for it.
1-(1-2x/pi)^C. At C=1.757 you get a curve for which the integral of the absolute difference between it and sine from 0 to pi/2 of pretty low, which is more than close enough for my case. And then at C=1 you get a straight line. Pretty neat!
answered Dec 14 '18 at 5:53
Seph ReedSeph Reed
1187
$endgroup$
add a comment |
$begingroup$
I feel dumb. It was really easy to solve:
sin(arcsin(2x/pi))
Also, in case anyone was wondering, I found the actual equation I was looking for. I'm trying to make it so a knob can change a curve from sin to a line incrementally, and found a good one for it.
1-(1-2x/pi)^C. At C=1.757 you get a curve for which the integral of the absolute difference between it and sine from 0 to pi/2 of pretty low, which is more than close enough for my case. And then at C=1 you get a straight line. Pretty neat!
answered Dec 14 '18 at 5:53
Seph ReedSeph Reed
1187
$endgroup$
I feel dumb. It was really easy to solve:
sin(arcsin(2x/pi))
Also, in case anyone was wondering, I found the actual equation I was looking for. I'm trying to make it so a knob can change a curve from sin to a line incrementally, and found a good one for it.
1-(1-2x/pi)^C. At C=1.757 you get a curve for which the integral of the absolute difference between it and sine from 0 to pi/2 of pretty low, which is more than close enough for my case. And then at C=1 you get a straight line. Pretty neat!
answered Dec 14 '18 at 5:53
Seph ReedSeph Reed
1187
edited Dec 14 '18 at 15:03
answered Dec 14 '18 at 5:53
Seph ReedSeph Reed
1187
answered Dec 14 '18 at 5:53
Seph ReedSeph Reed
1187
answered Dec 14 '18 at 5:53
Seph ReedSeph Reed
1187
1187
add a comment |
add a comment |
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