Jump to content
MWresearch

How do I do that thing where I solve trig equations with imaginary exponents?

Recommended Posts

Sometimes when I have complicated equations involving trig functions I ask wolfram for a solution and it spits out a bunch of imaginary exponents. My guess is it has something to do with the taylor series for e^(i*pi*x) equals cos(x)-isin(x) and e^(-pi*i)=-1, and even in third semester calculus I still haven't seen anything where I solve equations using that identity. But, sometimes I don't want to solve for a number, but rather a formula, and I don't always have wolfram around. So how do I solve complicated trig equations by using Euler's identity?

 

Like let's say I have sqrt(sin(2x)^2+cos(5x)^3)=4. How do I solve that with imaginary/complex exponents?

Share this post


Link to post
Share on other sites

Can you tell me how the sin of something plus the cos of something else will ever add up to 4, whether you take the square root or not?

 

Or are you just extracting the Michael?

Edited by studiot

Share this post


Link to post
Share on other sites

Well if x is a negative imaginary number then cos(-x.i) is equal to cosh(x) and sin(-x.i) will, when squared as above be equal to -sinh^2(x) - these can both give rise to real number answers With magnitude greater than one . So for a given equation such as the above there may well be imaginary or complex values of x that provide solutions .

Share this post


Link to post
Share on other sites

Sometimes when I have complicated equations involving trig functions I ask wolfram for a solution and it spits out a bunch of imaginary exponents.

In Mathematica you can ask it to go from Trig to Exp and vice versa. I assume that wolfram alpha will accept a similar command, but I have never tried it.

Share this post


Link to post
Share on other sites

 

Well if x is a negative imaginary number then cos(-x.i) is equal to cosh(x) and sin(-x.i) will, when squared as above be equal to -sinh^2(x) - these can both give rise to real number answers With magnitude greater than one . So for a given equation such as the above there may well be imaginary or complex values of x that provide solutions .

 

 

x must be real by the OP's own (conventional) notation.

Share this post


Link to post
Share on other sites

I'm fine with complex/imaginary solutions, I just want to know how to calculate them at all. How would I solve the equation I listed for the variable? I can't think of any conventional method that would work, and Wolfram once again gave me complex answers based off of those Euler exponents and things that seem like they were derived from sinh and cosh.

Share this post


Link to post
Share on other sites

There are no solutions as you have chosen to write it.

 

I copy and pasted your equation into Wolfram and it timed out without solution.

 

Please state the question properly.

Share this post


Link to post
Share on other sites

No I actually phrased it just fine, you are the one who assumed I wanted real solutions instead of complex ones, not me. Wolfram gives an answer using those identities, so there must be a way. I really can't think of a reason for you to be so obtuse over an inaccurate assumption.

Share this post


Link to post
Share on other sites

 

No I actually phrased it just fine

 

If you know the answer why ask the question?

 

If x is imaginary, is ix real or imaginary?

 

If you wish to work with a complex number you should use z, but should if you wish to use x instead (and that is being obtuse) then you should not also use ix.

Share this post


Link to post
Share on other sites

To be honest I don't think you actually read the first post. My question isn't anything to do with whether or not a solution is complex or imaginary, it's how to algebraically derive any solution in general using the identity e^(ix)=cos(x)+isin(x) and sinh and cosh.

Share this post


Link to post
Share on other sites

 

MWresearch

To be honest I don't think you actually read the first post. My question isn't anything to do with whether or not a solution is complex or imaginary, it's how to algebraically derive any solution in general using the identity e^(ix)=cos(x)+isin(x) and sinh and cosh.

 

 

That's twice now you have substituted personal attack instead of mathematical argument.

 

That apart, you have now contradicted yourself

 

 

MWresearch

Sometimes when I have complicated equations involving trig functions I ask wolfram for a solution and it spits out a bunch of imaginary exponents. My guess is it has something to do with the taylor series for e^(i*pi*x) equals cos(x)-isin(x) and e^(-pi*i)=-1, and even in third semester calculus I still haven't seen anything where I solve equations using that identity. But, sometimes I don't want to solve for a number, but rather a formula, and I don't always have wolfram around. So how do I solve complicated trig equations by using Euler's identity?

 

Like let's say I have sqrt(sin(2x)^2+cos(5x)^3)=4. How do I solve that with imaginary/complex exponents?

 

 

I cannot see any reference to the hyperbolic sine or cosine in your first post, perhaps you misunderstand Euler's Identity?

 

http://en.wikipedia.org/wiki/Euler's_identity

 

Your first post did, however, make very clear that you regard x as a real number, as does Wikipedia.

Share this post


Link to post
Share on other sites

That's not really a personal attack, it's an objective assessment of your involvement in this thread. I don't think you read the first post, nothing about that is an attack, just a statement of negligence.

 

Mentioning hyperbolic trig functions also isn't a contradiction because it does not exclude the usage of Euler's identity. Hyperbolic trig functions were mentioned originally by imaatfaal and then later incorporated into my search criteria due to the fact that hyperbolic trig functions and complex exponents can be related to each other to find solutions to complicated trginometric equations.

If you use the same website that you used before, Wikipedia, and search "Hyperbolic Trig Functions," you will find an article which shows the identity between hyperbolic trginometric functions and Euler's identity, such as cosh(ix)=(1/2)(e^(ix)+e^(-ix))=cos(x)

 

Furthermore, I did not use the word "real" in my first post. You made a mistake in assuming that I was only looking for real solutions and continue to make mistakes with other irrational assumptions.

Edited by MWresearch

Share this post


Link to post
Share on other sites

So far you have not addressed a single one of my questions about your lack of rigour, each time avoiding the direct question with a personal attack or self contradiction.

 

I asked a perfectly reasonable question

 

if x is imaginary what is ix?

 

Instead of answering you state (wrongly) that I have not read you first post.

 

You further claim material that was not in it and casitgate me for not finding such non existent material.

Share this post


Link to post
Share on other sites

Once again I find your assumptions irrational. Though it is true I did not answer the first question because it seemed to me like you were being purposely troublesome as your question can not possibly be rooted in an understanding of my post, I did answer the question about the relationship of hyperbolic trig functions and Euler's identity. I'm not quite sure why you'd lie so blatantly when it can easily be seen.

 

If x is imaginary, then ix would yield i^2 as a coefficient of the variable, giving a negative answer.

 

I also stated my situation clearly but perhaps repeating it a different way will allow you to see what I am saying. Imatfaal originally brought up hyperbolic functions and after I saw that, I was reminded of how hyperbolic functions are connected to Euler's identity and thus how I should have included them into the criteria of what I was searching for, which by no means excludes the usage of Euler's identity.

 

I said it before and I will say it again. I did not, and was not looking for solutions limited to real answers. You made the mistake of wrongfully assuming I desired numerical solutions which were also limited to real numbers. The word "real" does not exist in my original post and I made no effort whatsoever to state I was looking for only real numerical solutions. You are simply wrong.

Edited by MWresearch

Share this post


Link to post
Share on other sites

 

MWresearch

To be honest I don't think you actually read the first post. My question isn't anything to do with whether or not a solution is complex or imaginary, it's how to algebraically derive any solution in general using the identity e^(ix)=cos(x)+isin(x) and sinh and cosh.

 

 

 

Your words, not mine nor imatfaals or anyone elses.

 

In response

 

I stated that

 

 

I cannot see any reference to the hyperbolic sine or cosine in your first post, perhaps you misunderstand Euler's Identity?

 

and further gave link to a respectable webpage on the subject referred.

 

You responded by claiming that I am lying, and irrational

 

 

MWresearch

Once again I find your assumptions irrational.

 

I'm not quite sure why you'd lie so blatantly when it can easily be seen.

 

Categorically , your post#1 does not contain any reference to the hyperbolic sine or cosine,

 

So who is lying?

 

Finally you go some way to understanding what I am saying

 

 

If x is imaginary, then ix would yield i^2 as a coefficient of the variable, giving a negative answer.

 

 

So if x is imaginary then ix must be real.

 

And if x is imaginary why would anyone bother to add the i, unless they wanted the result to be real?

 

You cannot have your cake and eat it.

Share this post


Link to post
Share on other sites

The entire discussion could have been simplified if you simply said "my bad, I did not notice you were not asking for a real numerical solution" but instead you for some reason go into denial over basic facts because you seem to have an irrational fear of appearing wrong in front of others. I admitted I made a mistake in a couple other threads, but the world didn't end.

 

So, let's go over the facts:

 

Fact #1, apparently the most important fact: Nowhere did I ask for only numerical solution and only in terms of real numbers

 

Fact #2: You repeatedly stated I was asking for a real solution and ignored my multiple statements that I in fact was not looking for a real numerical solution and I specifically stated I wanted generalized solutions and the process by which they were obtained.

 

Fact #3: I never stated hyperbolic trigonometric functions were in the first post, but rather I specifically stated that imaatfaal brought them up first but is still connected to information in the first post, which you are in denial over.

 

Conclusion: Nothing remotely resembling "productive" or "helpful" has resulted from your presence on this thread and it is being prolonged in a negative direction as a result of your irrationally aggressive attitude, therefore the best course of action for the site and thread members is for you to leave the thread.

 

At this point, you do not have to state any fault, you just have to stop posting and everyone can move on.

Edited by MWresearch

Share this post


Link to post
Share on other sites

Mathematica gives me some not particularly nice looking solutions for the equation that you give in the first post using ArcCos and the square root.

 

I can then evaluate this to get approximate numerical values.

Share this post


Link to post
Share on other sites

One of the solutions I got had those imaginary exponents that look like they were transformed by some Euler identity involving imaginary exponents. I don't really know what all those imaginary exponents are used for exactly, but they come up a lot when I plug complicated equations into wolframalpha and I want to know how to use them by hand because I do a lot of algebraic manipulation, and I don't always need to come up with solutions the traditional way. Is solving complicated trig equations even the primary purpose of Euler's identity and imaginary hyperbolic functions?

As far as I can tell, all of those equations involving trigonometric identities on the complex plane are part of something called "complex analysis," and I'm told that kind of stuff will come up in engineering, but I have yet to see how.


I also do some audio engineering and I am loosely familiar with a fourier transformation" being related to trginometric functions over time which use complex exponents in some way.

Share this post


Link to post
Share on other sites

Is solving complicated trig equations even the primary purpose of Euler's identity and imaginary hyperbolic functions?

Maybe not the best choice of words, but yes, using Euler's formula can be a useful way of solving certain equations and simplifying expressions.

Share this post


Link to post
Share on other sites

 

Conclusion: Nothing remotely resembling "productive" or "helpful" has resulted from your presence on this thread and it is being prolonged in a negative direction as a result of your irrationally aggressive attitude, therefore the best course of action for the site and thread members is for you to leave the thread.

 

 

This repeated violation of forum rules that you signed up to is counterproductive.

 

 

 

Your first post is a mixed up jumble that even normally mild ajb comments on

 

 

Maybe not the best choice of words, but yes, using Euler's formula can be a useful way of solving certain equations and simplifying expressions.

 

 

I don't know why you are determined to restrict the discussion to Eulers formulae or why you specified this had to be done in your first post.

I don't know where you are coming from that you need to solve complex analysis problems, but this is particularly common in electrical engineering.

 

Eulers formulae forms the basis of some useful analysis methods, but it is only part of the story and not used directly in much real worl working.

You need more than this, in particular some knowledge of complex analysis to understand the working rather than just take the formulae on trust.

 

I will work through a simpler problem than you have presented in post#1; the problem is simpler but embodies most of what ( I think/guess) you need.

 

Evaluate sin-1(2)

 

This does not converge for any real number, but does converge for some complex number, z = (x+iy); where x and y are real.

 

We proceed as follows

 

2 = sin(x+iy) = sin(x)cos(iy) + cos(x)sin(jy) = sin(x) cosh(y) + icos(x)sinh(y)

 

Equat real and imaginary parts

 

sin(x)cosh(y) = 2

cos(x)sinh(y) = 0

 

From the second equality either cos(x)=0 or sinh(y) = 0

If sin(y) = 0 then y = 0 and cosh(y) = 1

This makes the first equality impossible since x is real

 

Hence cos(x) = 0

Hence [math]\sin \left( x \right) = \pm 1[/math]

Hence from the first equality [math]\cosh \left( y \right) = \pm 2[/math]

 

But since y is real

[math]\cosh \left( y \right) \ge 1[/math]

 

Hence cosh(y) = 2 and sin(x) = 1

 

Hence [math]x = \frac{\pi }{2} + 2n\pi [/math] where n is any integer

 

and

[math]y = {\cosh ^{ - 1}}\left( 2 \right) = \ln \left( {2 \pm \sqrt 3 } \right) = \pm \ln \left( {2 + \sqrt 3 } \right)[/math]

 

Hence

 

[math]{\sin ^{ - 1}}\left( 2 \right) = \frac{\pi }{2} + 2n\pi \pm i\ln \left( {2 + \sqrt 3 } \right) = 1.57 \pm 1.32i + 2n\pi [/math]

Which is complex.

 

Other examples might be

 

Using

[math]{e^z} = {e^{\left( {x + iy} \right)}} = {e^x} \bullet {e^{iy}} = {e^x}\left( {\cos y + i\sin y} \right)[/math]

 

[math]e\left( {2 + 3i} \right) = {e^2}\left( {\cos 3 + i\sin 3} \right) = \left( { - 7.32 + 1.04i} \right)[/math]

 

There are many more, but we cannot go over them, unless you stop attacking me and start cooperating.

 

Edited by studiot

Share this post


Link to post
Share on other sites

Hmm. So it seems like there's nothing really tricky about it, you just do a direct substitution using those particular complex identities and use logarithms to solve for a numerical answer in the end. It might be useful if there were also identities harnessing the inverse sine and inverse hyperbolic functions. As something else to explore, the gamma function wouldn't happen to be connected to complex analysis because of its identity gamma(x)(gamma(1-x)=pi*csc(pi*x) would it? I am not seeing much about the gamma function on google related to all of this, but I would imagine someone explored it.


I also can't find much on the identities of the inverse functions involving complex arguments in google. Where is a list of identities with things like arcsin(ix) and arccosh(ix)?

Share this post


Link to post
Share on other sites

Do you understand the difference between an identity and an equality?

 

The gamma function finds use in differential equations, including those such as Bessels, which do not have solutions in terms of elementary functions.

Since another word for solving a differntial equation is integration, the gamma function can be used to simplify certain integrals. This is where its connection to trigonometric functions through the reflection formula ( http://en.wikipedia.org/wiki/Reflection_formula )comes in.

 

For instance


[math]\int\limits_0^{\frac{\pi }{2}} {\tan \theta d\theta = } \int\limits_0^{\frac{\pi }{2}} {{{\sin }^{\frac{1}{2}}}\theta {{\cos }^{\frac{{ - 1}}{2}}}\theta d\theta = \frac{1}{2}} \Gamma \left( {\frac{3}{4}} \right)\Gamma \left( {\frac{1}{4}} \right) = \frac{\pi }{{\sqrt 2 }}[/math]

 

 

The issue of inverse functions is more tricky and brings in the subject of convergence quite strongly.

You need to understand the idea of domain and co domain (or range) for functions and inverse functions and its implications which leads to the idea of 'radius of convergence' in complex analysis. This has its counterpart in the 'interval of convergence' in real analysis.

Share this post


Link to post
Share on other sites

That integral seems too specific to be useful, could you expand more on the generalized process by which it is used? Could it be used anytime you have trigonometric functions or any time you have exponents with opposite signs? Would the gamma function ever be useful in equations which aren't restricted to being represented as integrals?

Share this post


Link to post
Share on other sites

 

That integral seems too specific to be useful

 

 

That was just an example. I don't choose difficult ones for the sake of it. And you did ask about the connection to trigonometric functions.

 

How did you get on with the Wiki article?

The idea was to help you find the right search phrases and terms to look up.

 

Just like for trigonometric functions there are some easy to calculate values and the connection to trig I mentioned allow these to be used.

But yes there are tables of gamma function available, or if you could find some other way to calculate them, you could put random values into these formulae.

For instance there is a table in the back of Kreysig 'Advanced Engineering Maths'. You will also find a useful connection to Laplace transforms here.

 

I really am struggling to know what level to put an answer as you don't reply to my questions. which are only designed to help.

I meant what I said about cooperation.

Share this post


Link to post
Share on other sites

The only question I see is "do you understand the difference between an equation and an identity?" After looking up the difference, I do not understand complex analysis or the gamma function any better, so I chose to refrain from commenting on an element of the discussion that would lead it in an unproductive direction. Otherwise, what other questions did you have?


At this point I think I'm just missing is inverse trig identities with complex arguments, the rest is just practice with using those trig identities in equations.


Although I don't completely understand how a Fourier transformation decomposes trigonometric functions using these imaginary/complex trig and Euler identities, but the integral itself looks like a hybrid between the gamma function and Euler's identity.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.