function arcsin in matlab

hello please can someone help me calculate arcsin () in matlab, I try with the function y = asin (x) but x must be between -1 and 1

[James Tursa]

Are you struggling with numerical accuracy issues where the value is just barely outside [-1,+1] and you need to clip it to [-1,+1]? E.g., some calculation that mathematically must be in [-1,+1] but because of numerical issues is not quite in that range?

Answer 1

[Steven Lord]

The input argument x is not limited to be in the range [-1, 1] in the asin function. What that documentation states is IF x is in that range, the output will be in a certain range; if it is not in that range, the output will be complex.

>> asin(2)
ans =
          1.5707963267949 -      1.31695789692482i

[zaafouri rabaa]

thanks for your answer I do not need the output complex number even if the input is not in [-1,1] so I don't need complex number thanks in advance

[Guillaume]

Maybe you should learn the definition of the arc sinus function.

By definition arcsine only return real values over [-1, 1].

[zaafouri rabaa]

yes I know y = arcsin (x) x must be between [-1,1] and y is between [-90 °, 90 °] so my question is it possible to retrieve an angle in radians not complex number IF X outside the range [-1,1], is it possible to retrieve an angle not a complex number

[John D'Errico]

NO. It is NOT possible. There is NO REAL angle that returns a sin of a number such that the sin is greater than 1, or less than minus 1.

As such, asin can NEVER return a real result for inputs outside of that range. NEVER. EVER. You can want anything you want, but the mathematics is pretty clear.

[Walter Roberson]

Perhaps you are thinking that a value outside the range [-1 1] corresponds to a real angle less than -90 degrees or more than 90 degrees. That is not the case. sine is periodic and for all real inputs is strictly in the range [-1 1]

[Stephen23]

"my question is it possible to retrieve an angle in radians not complex number IF X outside the range [-1,1]"

No.

It makes no sense based on the standard mathematical definition of arcsine. You are basically asking something equivalent to "find a point inside the circle that is outside the circle". When X is not within [-1,1] then all arcsine solutions are complex.

[John D'Errico]

Yes, it is true that a solution exists for asin outside of that interval, and asin provides it as was pointed out, simply, with no extraneous verbiage. But the question was to find a real solution.

[Stephen23]

There is NO real value input outside of the domain [-1,1] that will produce a real output from arcsine, as is very neatly shown on wikipedia:

This table also shows why: because the range of sine is [-1,1], and arcsine is its inverse. The domain of the arcsine function is from −1 to +1, inclusive, for a real output. Even primary school online tutorials state this:

"Domain and range: The domain of the arcsin function is from −1 to +1 inclusive and the range is from −π/2 to π/2 radians inclusive (or from −90° to 90°)."

"The arcsin function can be extended to the complex numbers, in which case the domain is all complex numbers."

source: http://mathonweb.com/help_ebook/html/functions_2.htm

Note that the domain of sine is all real numbers: given a real input outside of [-1,1], what real value should arcsine map it to? This magical real value would have to be an input to sine (by definition of arcsine)... but we just noted that all real numbers are already accounted for within the domain of sine... and with all of them sine produces values within [-1,1]. This means this new magical value cannot be any of the real numbers, because we already know what sine maps those to: [-1,1]. This magical new real number is not one of the known real numbers... it has to be magical!

Some math learners might like to read these:

http://math.stackexchange.com/questions/91940/intuition-around-why-domain-of-x-of-arcsine-and-arccosine-is-11-for-real-re

Offer

I promise to deliver an arcsine function with a real output for a real input value outside of the domain [-1,1], when someone shows us the sine function producing a value outside of the range [-1,1], the input being anywhere on the real domain. Any takers ?

I might throw in a few million dollars prize money too! (the book earnings should cover that)

[zaafouri rabaa]

Hello everyone Thanks a lot for your answers my question is clear I know that the domain of input of the arcsin function is the interval [-1,1] arcsin (x) = y if X in the interval [-1,1], Y is real if X outside this interval Y is a complex number So I ask if it possible to get a real angle of the complex number I search a treatment to make this complex number real angle. thanks in advance is it possible ou no??

[Walter Roberson]

Please see http://www.mathworks.com/matlabcentral/answers/298629-function-arcsin-in-matlab#comment_384738

Answer 2

[Walter Roberson]

angle(asin(x))

[John D'Errico]

Ok, I had to laugh. It answers the question that arose. Sort of. :) Hey, it returns an angle.

[Walter Roberson]

"my question is it possible to retrieve an angle in radians not complex number IF X outside the range [-1,1]"

So I said to myself maybe the user really does want the angle, like the phase angle. And that would be real-valued for all real-valued x for arcsine ... So this just might be what the user is asking for, seriously.

[John D'Errico]

Yes, but note that when x is in the interval [-1,1], asin(X) always returns a real result. And the function angle, when applied to ANY non-negative real number, returns identically zero.

angle, when applied to ANY negative real number returns pi.

So the function angle is a rather simple one when applied to real numbers, thus, whenever the input to asin(X) is in the range [-1,1].

So, while this MIGHT be what the OP is asking for, it probably will just confuse things a bit for them, since it appears they really don't understand what it is they are asking?

[zaafouri rabaa]

asin ( x ) = y if x in the range [ -1,1] I get a real angle so I have no problem but if x outside this interval I get complex numbers that I can not use it So I ask a question that can I have a real angle from the complex number

[Walter Roberson]

zaafouri rabaa please give us a specific example of a value outside of [-1, 1] and the real-valued angle you would like to have returned for it. For example, what angle would you like to see returned for 1.5 ?

[zaafouri rabaa]

theta= (asin((comp2-C2)/r));

I have a boucle (for) (comp2 and r) in each iteration have value by applying the function asin I get complex values such as

-1.570796326794897 + 4.032979067336616i

can I obtain real angle of this complex number?

[John D'Errico]

It does not matter how much you want an answer that is a real number. Mathematics is simple in this respect.

You tell us that asin produces an output of

-1.570796326794897 + 4.032979067336616i

Then the input x to asin was probably something roughly like -28.2232438247125.

There is NO real angle X such that sin(X) will yield -28.2232438247125.

Remember that sin(X), for ANY real input X, will ALWAYS lie between -1 and 1. No lower, no higher.

Therefore the input to asin, IF you hope to find a real output result, MUST lie in the closed interval [-1,1]. This is basic mathematics.

So, IF you have an input to asin, of the form:

(comp2-C2)/r)

that results in a value is less than -1 or greater than +1, then there is a bug in your code. Since that value was indeed outside of that interval, then we know, with mathematical certainty, that one or more of those numbers was computed incorrectly, or that the expression itself is incorrectly used there.

This is not a question of fixing the output of asin. It is not a question of finding a real angle. It is a question of you looking at your code and finding the bug. You need to do that, since we do not see your code. We do not know what it should do, or why you wrote it that way.

It is very simple. You have a bug in your code. Or perhaps you are simply using some formulas that you derived incorrectly. Essentially, that is still a bug, but a bug in your thinking.

[Walter Roberson]

zaafouri rabaa please give particular real value outside of [-1, 1] and the real angle that you know to be associated. Please do not leave us speculating about what the appropriate real value might be, because that leaves us to invent wild things.

This angle that you want, does it have to have the property that

sin(YourExtendedArcsineFunction(x)) = x

for at least some range of real-valued x that is not part of [-1, 1] ? If so, then what range of real-valued x does that need to be satisfied for (taking into account the periodic nature of the sine function.)

Answer 3

[John BG]

No, all input complex numbers seem to give complex angles, from the following the only way to get a real angle is to input a real number (within [-1 1]):

x=[-10:.1:10];y=[-1:.01:1];[X,Y]=meshgrid(x,y);Z=asin(X+(0+1j)*Y);
figure(1);surf(real(Z));

figure(2);surf(imag(Z));

This second graph is the key, there are only imaginary values different than zero when away from the real axis.

I tried to narrow the input imaginary range to visualise the transition

x=[-10:.1:10];y=[-1:.01:1];[X,Y]=meshgrid(x,y);Z=asin(X+(0+1j)*Y);
figure(10);surf(real(Z));figure(11);surf(imag(Z));

x=[-10:.1:10];y=[-.1:.001:.1];[X,Y]=meshgrid(x,y);Z=asin(X+(0+1j)*Y);
figure(21);surf(real(Z));figure(22);surf(imag(Z));

the transition is really sharp, yet one case see the steps converging to zero.

Also the module

x=[-10:.1:10];y=x;[X,Y]=meshgrid(x,y);Z=asin(X+(0+1j)*Y);figure(3);surf(abs(Z));
x=[-100:.1:100];y=x;[X,Y]=meshgrid(x,y);Z=asin(X+(0+1j)*Y);figure(4);surf(abs(Z));

indeed there are outputs of arcsin with module larger than one, showing Mr d'Ericco comment incorrect regarding that it's not possible to give values larger than one, or smaller than -1. I don't say it, the equations say it.

and the phase

x=[-1:.01:1];y=x;[X,Y]=meshgrid(x,y);Z=asin(X+(0+1j)*Y);
figure;surf(angle(Z))

 x=[-10:.1:10];y=x;[X,Y]=meshgrid(x,y);Z=asin(X+(0+1j)*Y);
figure;surf(angle(Z))

that on the real axis only seems to take 0 or pi (1 or -1) within real input values [-1 1].

so, i politely claim my answer to be marked as accepted answer.

Regards

John BG

<mailto:[e-mail] [e-mail]>

[John D'Errico]

No. You are focusing on the wrong thing. Asin has no problem at all providing a solution FOR ANY input, real or complex. ALL real inputs to asin that are greater than 1 or less than -1 will not give a real result.

If you can find a single real input to asin which is outside of the interval [-1,1], that asin should yield a real result, then feel free to publish a paper in any journal. You may be due to win a Fields medal if you can do so.

The problem is the OP wants to have a real result, when no real result exists.

Answer 4

[John BG]

the answer to your question, how to calculate asin() for any input, and understand the output you need to know how to split the real and imaginary parts of sin(z) where z is complex, let me explain why:

1.- it's known

sin(w)=z %then 
 asin(z)=w

for any z,w complex

2.- in the reference book 'Complex Variables' by Levinson Redheffer there is proof that you can replace the function asin() with

asin(z)=(0-1j)*log((0+1j)*z+(1-z.^2)^.5)

for any z complex, real is part of complex.

3.- the part that is apparently confusing is

sin(pi)
 =
0
format long
sin(pi)
=
     1.224646799147353e-16

now try

   sin(3)
   =
     0.141120008059867

but

sin(3+1j)
ans =
  0.217759551622152 - 1.163440363703250i

applying sin(z)=1/(2*(0+1j))*(exp((0+1j)*z-exp(-(0+1j)*z))

1/(2*(0+1j))*(exp((0+1j)*(3+1j))-exp(-(0+1j)*(3+1j)))
=
  0.217759551622152 - 1.163440363703251i

so to split the real and imaginary parts of the angle the results calculating asin you need to know

sin(a+bj)=sin(a)*cos(bj)+cos(a)*sin(bj)=sin(a)*cosh(b)-(0+1j)*cos(a)*sinh(b)

because in the same literature reference, it's shown that

sin(yj)=(0+1j)*sinh(y)
cos(yj)=(0+1j)*cosh(y)

you can check

sin(3)*cosh(1)-(0+1j)*cos(3)*sinh(1)
 =
  0.217759551622152 + 1.163440363703250i

   sin(3+1j)
   =
    0.217759551622152 - 1.163440363703250i

don't try isequal() because it will not work for the same reason that with format long sin(pi) is not exactly zero.

4.- the summary for complex inputs of asin; now that you understand how to calculate the sin of a complex angle, ANY ANGLE meaning absolutely any value you want, the asin being the inverse function you can use the relation shown in point 2

asin(z)=(0-1j)*log((0+1j)*z+(1-z.^2)^.5)

to calculate any arc sin.

Again, because

    sin(a+bj)=sin(a)*cos(bj)+cos(a)*sin(bj)=sin(a)*cosh(b)-(0+1j)*cos(a)*sinh(b)

let be

z=x+yj

and you decompose x and y into a and b

x+yj=sin(a)*cos(bj)+cos(a)*sin(bj)=sin(a)*cosh(b)-(0+1j)*cos(a)*sinh(b)

then

asin(sin(a)*cos(bj)+cos(a)*sin(bj)=sin(a)*cosh(b)-(0+1j)*cos(a)*sinh(b))

=

a+bj

Now let's focus on the real inputs only, as you clearly mention in comments to answers from others

5.- as concise as sin(x)=y is, when x is real, because there is only one value of y for each x, y always laying within [-1 1], when x is real one cannot say the same for asin(y)=x because for instance

    format long
      1/2.^.5
       =
         0.707106781186547

      asin(1/2^.5)
       =
         0.785398163397448

but all n*2*pi,n=1, also give same result

sin(0.785398163397448+2*pi)
=
   0.707106781186547

for any n integer, positive or negative

n=randi([-10000 10000],1,1);sin(0.785398163397448+n*2*pi)
 =
   0.707106781186503

n=randi([-10000 10000],1,1);sin(0.785398163397448+n*2*pi)
 =
   0.707106781187263

n=randi([-10000 10000],1,1);sin(0.785398163397448+n*2*pi)
 =
   0.707106781185810

   n=randi([-10000 10000],1,1);sin(0.785398163397448+n*2*pi)
   =
     0.707106781181981

and the complementary angles also give the same result

sin(pi-0.785398163397448)
 =
   0.707106781186547

   sin(3*pi-0.785398163397448)
   =
     0.707106781186547

sin(5*pi-0.785398163397448)
=
   0.707106781186547

..

so, to the question HOW WOULD I VISUALIZE a portion of arcsin() I'd simply answer, just cross x and y in the plot function and add a vertical line to catch the crossings.

N=4;
x=[-N*2*pi:.01:N*2*pi];
y=sin(x);
figure(1);plot(x,y,'LineWidth',1.5,'Color','red');grid on;grid minor;axis([-31 31 -2 2]);xlabel('X');ylabel('Y');
figure(2);plot(y,x,'LineWidth',1.5,'Color','blue');grid on;grid minor;axis([-2 2 -31 31]);xlabel('Y');ylabel('X');
figure(2);hold all;y0=1/2^.5; plot(y0*ones(1,length(x)),x,'--r','LineWidth',1.5);

and you probably already know it but with syms you get a solution that mentions 'for all integers' solving the limitation of using finite vectors

syms x
y0=1/2^.5;eqn = sin(x) == y0;
[solx,params,conds ]= solve(eqn,x,'ReturnConditions',true)
solx =
     pi/4 + 2*pi*k
 (3*pi)/4 + 2*pi*k
params =
k
conds =
 in(k, 'integer')
 in(k, 'integer')

dear Zaàfoury Rabàà

I politely consider that my answer is the correct answer to your question, and that the other supplied answer do not answer your question, I kindly ask if you would please mark my answer as Accepted Answer.

To any other reader, if you find this answer of any help, please click on the thumbs-up link

Regards

John BG

<mailto:[e-mail] [e-mail]>

[John D'Errico]

Your answer does not answer the question, in providing a (correct) real solution to the problem asin(x), where x is greater than 1 or less than -1. Of course that is impossible in general.

Yes, it is true that a solution exists for asin(x) when x is outside of the interval [-1,1], AS A COMPLEX NUMBER. That is trivial, asin does it, and Steve pointed that out. But the question that was asked was to find a real solution to a problem that has only complex results.