# Homework Help: I think I broke Wolfram?

1. Aug 24, 2010

### GreenPrint

I think I broke Wolfram?!?!

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. The attempt at a solution

Last edited: Aug 24, 2010
2. Aug 24, 2010

### CompuChip

Re: I think I broke Wolfram?!?!

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.

#### Attached Files:

• ###### re-im.jpg
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3. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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

4. Aug 24, 2010

### CompuChip

Re: I think I broke Wolfram?!?!

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)

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.

5. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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

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

6. Aug 24, 2010

### GreenPrint

7. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

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?

8. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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

9. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

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.

10. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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 in to 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

11. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

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.

12. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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 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...

Last edited: Aug 24, 2010
13. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

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 ?

Last edited: Aug 24, 2010
14. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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?

15. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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...

16. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

EDIT

This equation is correct, sorry about that.

Last edited: Aug 24, 2010
17. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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 =(

Last edited: Aug 24, 2010
18. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

Sorry, you are correct.

Okay, can you show your other steps?

Like how you derived arctan(z) ?

19. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

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

Show me how you got your own formula.

20. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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

21. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

( 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

Last edited: Aug 24, 2010
22. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

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.

23. Aug 24, 2010

### GreenPrint

Re: I think I broke Wolfram?!?!

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
(-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

Last edited: Aug 24, 2010
24. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

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?

25. Aug 24, 2010

### ╔(σ_σ)╝

Re: I think I broke Wolfram?!?!

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.