# Complex Variables Derivation of arcsine derivative

## Main Question or Discussion Point

Greetings,

I'm working (playing) on a problem involving approximating the arcsin() function.

I've attmpted to verify the known derivative of the arcsin function

(d(arcsin))/dx = 1/sqrt(1-z^2)

I know I have a mistake in my derivation. I've attached an electronic copy of my work.

Can you help resolve my error?

Thanks
-Sparky

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arildno
Homework Helper
Gold Member
Dearly Missed
You can't split the derivative of the sum to a logarithm like that! (2.line, I think)

dextercioby
Homework Helper
BTW, in the LHS you have a function of "x", while in the RHS a function of "z".

arildno,

Thanks - my attempt was to use the chain rule. I was not splitting the sum of the LN(). The derivative of LN(g(z)) is 1/(g(z)). Then I am attempting to multiply by the derivative of what's inside the LN function.

I've attached some additional detail on my work.

Can you offer some additional thoughts and clarification?

Thanks Again
-Sparky_

#### Attachments

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dextercioby
Homework Helper
arildno,

Thanks - my attempt was to use the chain rule. I was not splitting the sum of the LN(). The derivative of LN(g(z)) is 1/(g(z)). Then I am attempting to multiply by the derivative of what's inside the LN function.

I've attached some additional detail on my work.

Can you offer some additional thoughts and clarification?

Thanks Again
-Sparky_
You mean g'(z)/g(z), right ?

Gib Z
Homework Helper
Chain Rule Sparky_, chain rule :P

Gib Z
Homework Helper
$$\arcsin x=y$$

$$\sin y =x$$

$$\frac{dx}{dy} = \cos y$$

$$\cos y= \sqrt{1-\sin^2 y}$$

$$\sin^2 y=x^2$$

$$\cos y = \sqrt{1-x^2}$$

$$\frac{dx}{dy}=\sqrt{1-x^2}$$

$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$$

$$Q.E.D$$

Yes you're are correct - I did mean (g(z) / g'(z))

Hey Gib Z, Dextercioby and arildno,

My first attempt was to derive the the 1/sqrt(1-z^2) with complex analysis.

Now it bugs me that I'm close but there is an error somewhere in the derivation.

I would like some help finding my mistake.

Gib Z, I do agree (naturally) with you're derivation.

What I'm working on is a way to approximate the arcsine function with the natural log function: -i (LN(iz +/- SQRT(1-z^2)) - This is what I'm working on.

This led me to confirm the derivative of this is 1/SQRT(1-z^2)).

If -i (LN(iz +/- SQRT(1-z^2)) is the arcsine function, then the derivative if this must work out to 1 / SQRT(1-z^2)).

Now it really bugs me that I can't find the error.

I've worked it several times now.

BTW - how do you insert equations like that? I'm attaching my word doc showing my work with equation builder?

Thanks again
-Sparky_
Help?

Gib Z
Homework Helper
Ahh 2nd line of working, the d(w)/dz= ...., Its ment to be 1/g(z), so 1/(iz+/-sqrt(1-x^2), not -i/izsqrt(1-x^2) like you have. Also, the same equation, the 2nd part, after the X sign, put brackets, otherwise its wrong.

Gib Z
Homework Helper
O btw to see how we get the writing up, click on the pictures of our writing :D Itll say what we typed to get that up, and itll have a link for more help.

Gib Z -

Can you take a look at my current work? (I've attached it as a word doc.)

Do you find any mistakes?

If there are no mistakes what should my next step be to continue toward the final answer?

Thanks Again
Sparky_

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Gib Z
Homework Helper
Ahh this isnt technicaly an error...but instead of using the chain rule like you do in the document, could you perhaps use it this other way? i find it clearer.

Let whats inside LN equal u. Find du/dx

and then find dy/du, which is the derivative of -i ln(u) in this case. Multiply the 2 results. dy/du * du/dx = dy/dx :_) Sorri I just find the working easier to work with.

ssd
Gib Z -

Can you take a look at my current work? (I've attached it as a word doc.)

Do you find any mistakes?

If there are no mistakes what should my next step be to continue toward the final answer?

Thanks Again
Sparky_
The last two lines of your work do not appear to follow from the previous line.

I think I'm closer but once again I would appreciate your help.

I've attached my work with some corrections.

I've noticed that there is 2 or more solutions because of the several +/- combinations.

There is obviously one correct solution (maybe obviously is the wrong word) because the derivative (with respect to z) of arcsin(z) = 1/(sqrt(1-z^2))

I can see I have some solutions that are close to this.

first, I do not see a combination that yields 1/(SQRT(1-Z^2)). Can you see how to get there.

How do I mathmatically correctly trash the other solutions?

Thanks again
-Sparky_

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Gib Z
Homework Helper
You should probably try doing it one at a time, derive it once where its always plus, another time for minus.

3rd Line of working, dw/dz, 2nd part, -iz over some stuff, whys it negative iz? should be positive. And in your 2nd line where you times both sides by sqrt(1-z^2), then after that line you multiply by that things conjugate. Rather, multiply both things by their conjugates already.
When multiplying by conjugates, even though its plus minus, state that which ever sign the reader chose to work with, you multiplyed it by the other sign.
Working when starting should look like this:

$$\frac{dw}{dx} = (\frac{iz +/- \sqrt{1-z^2}}{iz +/- \sqrt{1-z^2}} \cdot \frac{1}{iz +/- \sqrt{1-z^2}}) +/- ( \frac{-iz}{iz\sqrt{1-z^2} +/- (1-z^2)} \cdot \frac{iz\sqrt{1-z^2} +/- (1-z^2)}{iz\sqrt{1-z^2} +/- (1-z^2)})$$

Good luck

Gib Z (and others)
Greetings again,

I have some more progress to show and discuss:

I have worked (or attempted) the 4 cases of the various +/- combinations.

The good news - the numerator is always "1". - Which is what the tables have for the correct answer.

More good news - one of the 4 cases (case 3 in my doc) is the answer.

point of uncertainty: I have 3 other cases.

I suspect this has to do with the various quadrants. I have attached my work in a word doc.

I used Excel to attempt to tabulate the results for the 4 cases for the "standard radian values" (0, 30, 45, 60, 90 and so on in radians) just to get some more insight.

I do see there are points where some of the cases are not defined. I'll grant I did not handle the imaginary "i" term correctly in Excel.

1) Am I correct that the cases are for where the function is valid and defined?

2) are my other solutions correct?

3) do you see anything interesting in this derivation of the derivative of arcsine using complex analysis?

it appears I cannont attach my Excel file. - ????
Thanks so much

#### Attachments

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Gib Z
Homework Helper
I can't see anything wrong with the derivation of any of them.

Something I do see is the 2nd case equates to the negative of the actual derivative, and is equal to the derivative of arcCos!! arcsin= -arccos.

As to mathematically ruling them out...if your feeling brave, you could try showing the 2nd derivative of each of these 4 cases. The one which's 2nd derivative is the same as arcsins 2nd derivative is the one.

Those 2nd derivatives should be much easier to calculate than the 1st ones. And this time you don't have to deal with the plus and minuses.

Thanks,

I'll attempt that next step - it will probably be a couple of days at the least - real busy the next two days.

It may be a sign of a sick mind but I'm enjoying this stuff.

I'll be in touch.

Thanks for the advice and help.
-Sparky_

Regarding the arcsine function and the complex variables approach:

z = sin(w)

w = -i*ln(iz +/- SQRT(1-z^2))

I'm trying to plug in numbers for z and "turn the crank" and get back the angle.

Meaning, if z = 0.5, the equation should yield 0.5236 (30 degrees) and if z = 0.707, I should get 0.7854 (45 degrees).

I've tried to do this in Excel. - no luck

I was assuming the "i" terms shifted things by PI/2 - no combination has worked.

How does this equation yield arcsine values? How do I turn the crank and handle the "i" terms.

Thanks
Sparky_