I think I broke Wolfram?

[GreenPrint]

I think I broke Wolfram??

Homework Statement

I was trying rewrite properly the tangent and arctangent function and was trying to figure out what values of the argument in the natural log was < 0 so that way I could simplify my formulas further because

i ln(-1) = -pi/2
and stuff of the sort

(-x i)/SQRT(x^2 + 1) + SQRT(x^2 +1)/(x^2 + 1) < 0
basically looking for the first x value that would make that a negative number

if you have no idea what I'm talking about don't worry about just tell me why I don't get a solution

http://www.wolframalpha.com/input/?i=solve(-x+i)/SQRT(x^2+++1)+++SQRT(x^2++1)/(x^2+++1)+<+0

go to that site

as you can see it didn't give me a solution
wolfram is suppose to say "no solution exists" when there's not one so what does this mean when it doesn't tell me anything? Did I break it? Is there a solution?

Homework Equations

The Attempt at a Solution

 

[CompuChip]

I ran your expression through Mathematica, here is a plot of its real and imaginary parts.
It looks like Wolfram Alpha is right, and the real part only tends to zero asymptotically.

 

[GreenPrint]

ok is there an answer for what is the first negative number that makes the equation negative becasue i need to know that

(-x i)/SQRT(x^2 + 1) + SQRT(x^2 +1)/(x^2 + 1) < 0

how do i do this then

like I'm looking for the x value that will make that expression negative... what's the first value that will do this? Is there a way to figuer this out

 

[CompuChip]

That is my point... I don't know if you can see my attachment yet, but it clearly shows that
a) the imaginary part is non-zero whenever x is non-zero, so the comparison "Expression < 0" is useless anyway, you can at most speak about "Real part of expression < 0"
b) the real part is always positive, and only tends to zero as |x| goes to infinity (if you can't see my picture, you can ask Wolfram Alpha)

[edit]Actually that last link shows you that the real part of your expression is 1/sqrt(x² + 1) which is non-zero for all real and even complex x.

 

[GreenPrint]

hmmm so that expression can never be negative? hmm i think you uploaded the wrong picture lol as in the picture...

anways you ok it can never be negative but what about
+/- because sqare roots have two solutions
so i would get a negative hmmmm

 

[GreenPrint]

http://www.wolframalpha.com/input/?i=solve(-x+i)/-SQRT(x^2+++1)+-SQRT(x^2++1)/(x^2+++1)+<0

gave me nothing but if i put in like -1 i get an answer

http://www.wolframalpha.com/input/?i=solve(-x+i)/-SQRT(x^2+++1)+-SQRT(x^2++1)/(x^2+++1)+=-1

 

[╔(σ_σ)╝]

When you put in an expression that doesn't make sense you get useless results. Your first problem was trying to order a complex number. What exactly does < mean in the complex plane?

 

[GreenPrint]

When you put in an expression that doesn't make sense you get useless results. Your first problem was trying to order a complex number. What exactly does < mean in the complex plane?

Guess you have a point there ok but I still need to figure out the smallest x value that would make that expression negative so that way I can simplify further and state the domain of my new equation that it applies to

 

[╔(σ_σ)╝]

Your original expression is always a complex number for real x. So it is never negative.
The expression here : http://m.wolframalpha.com/input/?i=solve%28-x+i%29%2F-SQRT%28x%5E2+%2B+1%29+-SQRT%28x%5E2+%2B1%29%2F%28x%5E2+%2B+1%29+%3D-1 is different from what you first posted; there is a negative sign on the second term.

 

[GreenPrint]

Ya because square roots have two solutions one positive and one negative? I just left the positive and negative symbol +/- out or was I just not suppose to put into begin with I thought I was suppose to but left it out in the first expression... ? If you want I can provide some background on where that expression came from let me know

 

[╔(σ_σ)╝]

Square roots have positive and negative sign when you are trying to solve equations such as $$a^{2}= b$$. When a equation involving square roots is defined it is not up to you to include a plus or minus. You only include them if they are a result of an equation involving even roots. Anyway the expressions in the square roots in your question are always positive. In your first expression the term to the right is always positive and greater than zero. The expression to the left is a complex number whose imaginary part is < = 0.

 

[GreenPrint]

hmmm ok well thanks...
So I think they do have positive and negative signs because I was dealing with an equation like that let me explain

I started out deriving a better equation for tan(x)

got this

tan(x) = (1- e^(-i 2 x) )/(i + i e ^(-i 2 x) ) + pi n

I checked in my calculator and I think it's right

I then said well for arctangent
what does that equal
tan^-1 (x) = i ln ( (-xi)/SQRT(x^2 + 1) + SQRT(x^2 + 1)/(x^2 + 1) ) + pi n
checked in my calculator I think it's right

third I was like what if I knew tan(x) = -3 or some other number how would I solve for the angle... well then it would have two solutions because you wouldn't obey the restrictions when you sued this form of arctangent(-3) because it's an equation not just simply arctan(-3)

I got the following answers

i ln ( (-xi)/SQRT(x^2 + 1) + SQRT(x^2 + 1)/(x^2 + 1) ) + pi n
and
i ln ( (xi)/SQRT(x^2 + 1) - SQRT(x^2 + 1)/(x^2 + 1) ) + pi n

checked in my calculator believe it's right

so as you can see I was wondering if I could simplify this further because of what I know of complex nubmers and the complex logarithm and i ln(i) can be simplified and iln(-1) can as well so that was why I asked that question

so in this case then wouldn't I need to have the +/- sign in front of the SQRTS I can't remember...

 

[╔(σ_σ)╝]

You realize the equation you wrote down is not for tan(x) and tan^-1(x) ?

They are for tan(z) and tan^-1(z).

EDIT

Where did you even get those equations from ? I'm not sure if they are correct.

When x=0 you get tan(0) = pi n ?

 

[GreenPrint]

You realize the equation you wrote down is not for tan(x) and tan^-1(x) ?

They are for tan(z) and tan^-1(z).

EDIT

Where did you even get those equations from ? I'm not sure if they are correct.

When x=0 you get tan(0) = pi n ?

ya I do realize that... ok so then just take the pi n
off of
tan(x) = (1 - e^(-2ix) )/(i + i e^(-2 i x) )
there?

 

[GreenPrint]

And I actually thought they applied to all reals as well because it works fine as well when you use noraml angle measures right like pi/3 pi/2 etc.?

B.T.W. I started out with sine and cosine and used the fact that tangent equals sine/cosine and simplified that is all

B.T.W. thanks for pointing that out for me I forgot that you don't put + pi n for the normal formulas of tangent, sine, cosine, etc. but for tan^-1 and stuff you do... Sorry I'm tired...

 

[╔(σ_σ)╝]

ya I do realize that... ok so then just take the pi n
off of
tan(x) = (1 - e^(-2ix) )/(i + i e^(-2 i x) )
there?

EDIT

This equation is correct, sorry about that.

 

[GreenPrint]

sin(x) = (e^(ix) - e^(-ix))/(2i)
cos(x) = (e^(ix) + e^(-ix))/2
tan(x) = sin(x)/cos(x)
=( (e^(ix) - e^(-ix))/(2i) ) /( (e^(ix) + e^(-ix))/2 )
= ( 2( (e^(ix) - e^(-ix) ) )/ (2i (e^(ix) + e^(-ix)) )
= (e^(ix) - e^(-ix) )/(i((e^(ix) + e^(-ix)))
= (e^(ix)(1 - e^(-2ix)))/(ie^(-ix)(1 + e^(-2ix) ) )
= (1 - e^(-2ix) )/(i (1 + e^(-2ix) ) )
= (1 - e^(-2ix) )/(i + i e^(-2ix) )
?

this will be easier to read
cis(x) = e^(ix) = cos(x) + i sin(x)
cis(-x) = e^(-ix) = cos(x) - i sin(x) = 1/e^(ix)

sin(x) = (cis(x) - cis(-x) )/(2i)
cos(x) = (cis(x) + cis(-x) )/2
tan(x) = sin(x)/cos(x)
=( (cis(x) - cis(-x) )/(2i) )/( (cis(x) + cis(-x) )/2 )
= ( 2(cis(x) - cis(-x)) )/( 2i(cis(x) + cis(-x)) )
= ( (cis(x) - cis(-x) )/( i(cis(x) + cis(-x)) )
=( cis(x)(1 - cis^2(-x)) )/( i cis(x)(1 + cis^2(-x)) )
=( 1 - cis^2(-x) )/( i(1 + cis^2(-x) )
=( 1 - cis^2(-x) )/( i + i cis^2(-x) )
=( 1 - e^(-i2x) )/(i + i e^(-i2x) )

and if you really wanted
=( 1 - 1/e^(i2x) )/(i + i/e^(2ix)
or expressed this way if you wanted I guess
=(1 - 1/cis^2(x) )/(i + i/cis^2(x)

?

done

I don't see what I've done wrong or were I went wrong if i did =(

 

[╔(σ_σ)╝]

Sorry, you are correct.

Okay, can you show your other steps?

Like how you derived arctan(z) ?

 

[╔(σ_σ)╝]

$$arctan(z)=\frac{1}{2}log\frac{i+z}{i-z}$$

Show me how you got your own formula.

 

[GreenPrint]

Then I was like ok if

tan(x) = ( 1 - e^(-i2x) )/(i + i e^(-i2x) )
then what if I know that tan(x) = #
like for example tan(pi/4) = SQRT(2)/2
I was like well what if I knew this instead and didn't know the angle
tan(x) = SQRT(2)/2
then I would have to do this to solve
tan(SQRT(2)/2)
and what if I didn't know what this was equal to? what if I didn't know it was pi/4 hmmm I could just use a calculator but that is cheating =)
so I said ok then

tan(x) = ( 1 - e^(-i2x) )/(i + i e^(-i2x) ) = #
and solved for x the angle and go the solutions
i ln ( (-xi)/SQRT(x^2 + 1) + SQRT(x^2 + 1)/(x^2 + 1) ) + pi n
and
i ln ( (xi)/SQRT(x^2 + 1) - SQRT(x^2 + 1)/(x^2 + 1) ) + pi n

I think there right
I also figured out that if it was just tan^-1(x) and had to obey the restrictions it would just be this equation only

i ln ( (-xi)/SQRT(x^2 + 1) + SQRT(x^2 + 1)/(x^2 + 1) ) + pi n

 

[GreenPrint]

( 1 - e^(-i2x) )/(i + i e^(-i2x) ) = #
1 - e^(-i2x) = #(i + i e^(-i2x) )
1 - e^(-i2x) = #i + #ie^(-i2x)
0 = -1 + e^(-i2x) + #i + #ie^(-i2x)

let x = e^(-i2x)

0 = -1 + x + #i + #i x
0 = x + #ix + #i -1
0 = x(1 + #i) + #i - 1
1 = x(1 + #i) + #i
1 - #i = x(1 + #i)
(1 - #i)/(1 + #i) = x
(1 - #i)/(1 + #i) = e^(-2ix)
ln( (1 - #i)/(1 + #i) ) = -2ix
ln( (1 - #i)/(1 + #i) )/(-2i) = x
(i/2) ln( (1 - #i)/(1 + #i) ) = x
i ln( ((1 - #i)/(1 + #i))^(1/2) = x
i ln ( SQRT(1 - #i)/SQRT(1 + #i) ) = x

I'm just going to show the argument in the natural log to make it simpler to read for a bit then add the rest once the argument is simple

( SQRT(1 - #i)SQRT(1 + #i) )/( SQRT(1 + #i)SQRT(1 + #i) )
SQRT( (1 - #i)(1 + #i) )/(1 + #i)
SQRT( 1 - #^2 i^2)/(1 + #i)
SQRT( 1 + #^2)/(1 + #i)
( SQRT( 1 + #^2)SQRT( 1 + #^2) ) / ( (1 + #i)SQRT( 1 + #^2) )
(1 + #^2)/( (1 + #i)SQRT( 1 + #^2) )
(1 - ix)/SQRT( 1 + #^2)
1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2)

ok now I'll add on the rest

i ln( 1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2) ) = x
then sense um tangent is periodical with pi
i ln( 1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2) ) + pi n = x

done

 

[╔(σ_σ)╝]

There is a mistake here:

(1 - #i)/(1 + #i) = e^(-2ix)
ln( (1 - #i)/(1 + #i) ) = -2ix

1) You are not talking about the natural logarithm anymore but Log(z) (the prinipal value, i presume)
2) x can be a complex number, correct? So log(e^(-2ix)) =/= -2ix

Log( e^(-2iz)) = Log(e^(-2ix + 2y))) = Log(e^(-2ix) *e^(2y))

Remember |e^z| is not always 1. This is the complex exponential we are talking about.I think there are a lot of subtleties associated with what you are trying to do and you have to be careful.

 

[GreenPrint]

ok but this is correct for set of real numbers right?

i ln( 1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2) ) + pi n = x

hmmmmmm...

this is a post from a while back I didn't simplify the answer but the answer was correct I can finish it to show you, I didn't use what you told me in the post above and still got the right answer when I was trying to solve for cos(x)=-2 and got the right answer completely ignoring the complex exponential I think I took that into consideration later... let me show you by finishing this proof and simplifying

e^(ix) = cosx + i sinx = cisx
e^(-ix) = cosx - i sinx = cis(-x)
cis(x) + cis(-x) = cosx + i sin x + cosx - i sinx = 2cosx
from which
cosx = (cis(x) + cis(-x) )/2
from which I set it equal to -2 and began to solve
cosx = (cis(x) + cis(-x) )/2 = -2
Multipled by 2 on both sides
cis(x) + cis(-x) ) = -4
Multipled by cisx on both sides
cis^2(x) + 1 = -4cis(x)
set equal to zero by adding -4cis(x) to both sides
cis^2(x) + 4cis(x) + 1 = 0
solved the quadratic for cis(x)
(-4 +/- sqrt(4^2-4(1)))/2
= -2 +/- sqrt(16 - 4)/2
= -2 +/- sqrt(12)/2
= -2 +/- (2sqrt(3))/2
= -2 +/- sqrt(3)
cis(x) = -2 +/- sqrt(3) = e^(ix)
solved for x took natural log of both sides
ln(-2 +/- sqrt(3)) = ln( e^(ix) )
took out exponents and used ln(e) = 1
ln(-2 +/- sqrt(3)) = ix
divided through by i
ln(-2 +/- sqrt(3)) /i = x
simpified for +/-

for +
ln(-2 + sqrt(3)) /i = x
-i ln(-2 + sqrt(3) )
-2 + SQRT(3) is a negative number so then
-i ln( -1(2-sqrt(3)) )
-i (ln(2-sqrt(3)) + ln(-1) )
= -i( ln(2-sqrt(3)) + i pi )
= -i ln(2 - sqrt(3) + pi + 2 pi n

for -
ln(-2 - sqrt(3)) /i
factored out negative one
ln(-(2 + sqrt(3))/i
used the fact that ln(xy) = lnx + lny
( ln(-1) + ln(2 + sqrt(3)) )/i
used the fact that ln(-1) = i pi
( i pi + ln(2 + sqrt(3)) )/i
canceled out the i
pi + ln((2 + sqrt(3)) )/i = x
pi -ln( 2+sqrt(3) ) + 2 pi n

 

[╔(σ_σ)╝]

ok but this is correct for set of real numbers right?

i ln( 1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2) ) + pi n = x

hmmmmmm...

If you restrict z to be only real numbers then this is still not okay.Log(z) = ln|z| + i Arg(z)

(1 - #i)/(1 + #i) = e^(-2ix)
ln( (1 - #i)/(1 + #i) ) = -2ix

Should be
ln $${\frac{|1-#i|}{|1+#i|}$$ + i$$\theta$$

$$\theta$$ still has to be determined.But then why are you trying to derive a formula for arctan(x) as a real valued function?

 

[╔(σ_σ)╝]

ok but this is correct for set of real numbers right?

i ln( 1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2) ) + pi n = x

hmmmmmm...

this is a post from a while back I didn't simplify the answer but the answer was correct I can finish it to show you, I didn't use what you told me in the post above and still got the right answer when I was trying to solve for cos(x)=-2 and got the right answer completely ignoring the complex exponential I think I took that into consideration later... let me show you by finishing this proof and simplifying

e^(ix) = cosx + i sinx = cisx
e^(-ix) = cosx - i sinx = cis(-x)
cis(x) + cis(-x) = cosx + i sin x + cosx - i sinx = 2cosx
from which
cosx = (cis(x) + cis(-x) )/2
from which I set it equal to -2 and began to solve
cosx = (cis(x) + cis(-x) )/2 = -2
Multipled by 2 on both sides
cis(x) + cis(-x) ) = -4
Multipled by cisx on both sides
cis^2(x) + 1 = -4cis(x)
set equal to zero by adding -4cis(x) to both sides
cis^2(x) + 4cis(x) + 1 = 0
solved the quadratic for cis(x)
(-4 +/- sqrt(4^2-4(1)))/2
= -2 +/- sqrt(16 - 4)/2
= -2 +/- sqrt(12)/2
= -2 +/- (2sqrt(3))/2
= -2 +/- sqrt(3)
cis(x) = -2 +/- sqrt(3) = e^(ix)
solved for x took natural log of both sides
ln(-2 +/- sqrt(3)) = ln( e^(ix) )
took out exponents and used ln(e) = 1
ln(-2 +/- sqrt(3)) = ix
divided through by i
ln(-2 +/- sqrt(3)) /i = x
simpified for +/-

for +
ln(-2 + sqrt(3)) /i = x
-i ln(-2 + sqrt(3) )




for -
ln(-2 - sqrt(3)) /i
factored out negative one
ln(-(2 + sqrt(3))/i
used the fact that ln(xy) = lnx + lny
( ln(-1) + ln(2 + sqrt(3)) )/i
used the fact that ln(-1) = i pi
( i pi + ln(2 + sqrt(3)) )/i
canceled out the i
pi + ln((2 + sqrt(3)) )/i = x

let me show you what I did i completely ignored it and somehow still got the right answer I think I took it into consideration when I used the natural log let me show you

You did it without knowing.

-2 + sqrt(3) is it negative or positive ?

It's negative.
ln(-2 + sqrt(3)) = ln((-1)(2 - sqrt(3)) = ln(-1) + ln(2 - sqrt(3))


If you had done what I told you, you would have
Log ( -2 + sqrt(3) ) = ln |-2+sqrt(3)| + iarg(-2 + sqrt(3))

Arg(-2 + sqrt(3)) = pi
ln |-2+sqrt(3)| = ln(2- sqrt(3))

My point is that in your calculations you completely ignored the argument of the number you take Log of.

 

[GreenPrint]

see and it was right

http://www.wolframalpha.com/input/?i=(e^(ix)+%2B+e^(-ix))/2+%3D+-2

or my derived formula where i ignored the complex exponential

if cos(y) = x
give

x > 1 U x < -1

then y =

- i ln( -x + SQRT(x^2 - 1) ) + pi + 2 pi n
-i ln(-x - SQRT(x^2 -1) ) + pi + 2 pi n

 

[GreenPrint]

hmmm ok let me look at this stuff

i think I just used the fact that ln(-1) was i pi then I divided by i...

 

[╔(σ_σ)╝]

see and it was right

http://www.wolframalpha.com/input/?i=(e^(ix)+%2B+e^(-ix))/2+%3D+-2

or my derived formula ignoring the complex exponential

if cos(y) = x
give

x > 1 U x < -1

then y =

- i ln( -x + SQRT(x^2 - 1) ) + pi + 2 pi n
-i ln(-x - SQRT(x^2 -1) ) + pi + 2 pi n

You didn't ignore the Log or complex exponential you just simplified things by using only real numbers. If x was complex you would be in trouble.

I will take a closer look at your formula, there is a possibility it works for all real values.
What now ?
Edit Btw you function depends on x to produce x?

i ln( 1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2) ) + pi n = x

 

[╔(σ_σ)╝]

hmmm ok let me look at this stuff

i think I just used the fact that ln(-1) was i pi then I divided by i...

Yes this is true for x as a real number.

 

[GreenPrint]

You didn't ignore the Log or complex exponential you just simplified things by using only real numbers. If x was complex you would be in trouble.

I will take a closer look at your formula, there is a possibility it works for all real values.



What now ?



Edit


Btw you function depends on x to produce x?

i ln( 1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2) ) + pi n = x

lol sorry and thanks for your help!

 

[╔(σ_σ)╝]

lol sorry and thanks for your help!

What did you do here ?

( SQRT( 1 + #^2)SQRT( 1 + #^2) ) / ( (1 + #i)SQRT( 1 + #^2) )
(1 + #^2)/( (1 + #i)SQRT( 1 + #^2) )
(1 - ix)/SQRT( 1 + #^2)- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>?
1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2)

How did you go from step 2 to 3 ?

 

[jackmell]

Homework Statement

I was trying rewrite properly the tangent and arctangent function and was trying to figure out what values of the argument in the natural log was < 0 so that way I could simplify my formulas further because

i ln(-1) = -pi/2
and stuff of the sort

(-x i)/SQRT(x^2 + 1) + SQRT(x^2 +1)/(x^2 + 1) < 0
basically looking for the first x value that would make that a negative number

I'd say x=0 is the only answer that makes the expression a negative number (zero imaginary part) assuming you don't use the principal square root values. That is, use the function:

$$f(x)=-\frac{x i}{\sqrt{x^2 + 1}} - \frac{\sqrt{x^2 + 1}}{x^2 + 1}$$

or:

$$f(x)=\frac{x i}{\sqrt{x^2 + 1}} - \frac{\sqrt{x^2 + 1}}{x^2 + 1}$$

where the root symbol means "principal value".

Also, Wolfram Alpha and Mathematica only use the principal value by default when evaluating a multi-valued function.