## Math tricks for everyone

$${z=}\frac{1}{2i}{ln}\frac{x-i}{x+i}$$

or,
$${z=}\frac{1}{2i}{ln}\frac{{-}{(}{1}{-}{x}^{2}{)}{-}{2xi}}{{x}^{2}{+}{1}}$$

We substitute,
x=tan[theta]

Now we have,
$${z}{=}{-}{\frac{1}{2i}}{ln}{[}{cos}{2}{\theta}{+}{i}{sin}{(}{2}{\theta}{)}{]}$$

$${e}^{-2iz}{=}{cos}{2}{\theta}{+}{i}{sin}{2}{\theta}$$
$${cos}{(}{-2z}{)}{+}{i}{sin}{(}{-2z}{)}{=}{[}{cos}{(}{2}{\theta}{)}{+}{i}{sin}{(}{2}{\theta}{)}{]}$$

Therefore,
$${-2z}{=}{2}{\theta}$$
z=-theta=-arctanx
The minus sign is causing trouble
Attached Images
 math_cal.bmp (219.8 KB, 24 views)

 Quote by Anamitra $${z=}\frac{1}{2i}{ln}\frac{x-i}{x+i}$$ or, $${z=}\frac{1}{2i}{ln}\frac{{-}{(}{1}{-}{x}^{2}{)}{-}{2xi}}{{x}^{2}{+}{1}}$$ We substitute, x=tan[theta] Now we have, $${z}{=}{-}{\frac{1}{2i}}{ln}{[}{cos}{2}{\theta}{+}{i}{sin}{(}{2}{\theta}[)]{]}$$ $${e}^{-2iz}{=}{cos}{2}{\theta}{+}{i}{sin}{2}{\theta}$$ $${cos}{-2z}{+}{i}{sin}{-2z}{=}{[}{cos}{(}{2}{\theta}{)}{+}{i}{sin}{(}{2}{\theta}{)}$$ Therefore, $${-2z}{=}{2}{\theta}$$ z=-theta=-arctanx The minus sign is causing trouble
THANX FOR THE POST!

Too bad my browser does not decode Tex, can you post a picture of the derivation?
 Lets examine the following: $${z=}\frac{1}{2i}{ln}\frac{{-}{(}{1}{-}{x}^{2}{)}{-}{2xi}}{{x}^{2}{+}{1}}$$ When ever we use ln(x) in calculus we mean ln[abs(x)] Using the above information[rather by extending it to the area of complex numbers:we are actually utilizing the liberty of changing the sign] we may remove the negative sign from the above relation and obtain what we expect. $$\int\frac{1}{x}{dx}{=}{ln}\mid{x}\mid$$ [The above is of course a standard result] Just think of integrating 1/x from -500 to -36 Evaluation of ln x from -500 to -36 should first read as ln[-36] - ln[-500] before we can simplify to cancel the minus sign.This will happen if the absolute value is not considered. Our formula is: $$\int {f(x)}{dx}{=}{f(b)}{-}{f(a)}$$ Integration on the LHS is from a to b.

 Quote by Anamitra $${z=}\frac{1}{2i}{ln}\frac{x-i}{x+i}$$ or, $${z=}\frac{1}{2i}{ln}\frac{{-}{(}{1}{-}{x}^{2}{)}{-}{2xi}}{{x}^{2}{+}{1}}$$ We substitute, x=tan[theta] Now we have, $${z}{=}{-}{\frac{1}{2i}}{ln}{[}{cos}{2}{\theta}{+}{i}{sin}{(}{2}{\theta}{)}{]}$$ $${e}^{-2iz}{=}{cos}{2}{\theta}{+}{i}{sin}{2}{\theta}$$ $${cos}{(}{-2z}{)}{+}{i}{sin}{(}{-2z}{)}{=}{[}{cos}{(}{2}{\theta}{)}{+}{i}{sin}{(}{2}{\theta}{)}{]}$$ Therefore, $${-2z}{=}{2}{\theta}$$ z=-theta=-arctanx The minus sign is causing trouble
Thanx for the picture. I follow you up to x = tan[theta]

I don't see how you made the substitution in line 7 of the picture. Can you please explain?

Also, z had imaginary values associated with it in line 1, where did it become a real number?

Let me point out as well that Cos(-2z) + iSin(-2z) = Cos(2z) - iSin(2z) so if z is real your equation on line 9 becomes Cos(2z) - iSin(2z) = Cos(28) + iSin(28) and so it is DANGEROUS to say 2z = 28

I have used 8 for theta because of keyboard limitations.

 We start with: Integral= $${z}{=}\frac{1}{2i}{ln}{\mid}\frac{{-}{(}{1}{-}{x}^{2}{)}{-}{2xi}}{{x}^{2}{+}{1}}{\mid}$$ $${=}\frac{1}{2i}{ln}{\mid}{-}\frac {{1}{-}{x}^{2}}{{1}{+}{x}^{2}}{-}\frac{2xi}{{1}{+}{x}^{2}}{\mid}$$ $${=}\frac{1}{2i}{ln}{[}\frac {{1}{-}{x}^{2}}{{1}{+}{x}^{2}}{+}\frac{2xi}{{1}{+}{x}^{2}}{]}$$ Now, $$\frac{{1}{-}{tan}^{2}{\theta}}{{1}{+}{tan}^{2}{\theta}}{=}{cos}{2}{\theta}$$ And, $$\frac{{2}{tan}{\theta}}{{1}{+}{tan}^{2}{\theta}}{=}{sin}{2}{\theta}$$ Substituting x=tan[theta] we may proceed.

 Quote by Anamitra We start with: Integral= $${z}{=}\frac{1}{2i}{ln}{\mid}\frac{{-}{(}{1}{-}{x}^{2}{)}{-}{2xi}}{{x}^{2}{+}{1}}{\mid}$$ $${=}\frac{1}{2i}{ln}{\mid}{-}\frac {{1}{-}{x}^{2}}{{1}{+}{x}^{2}}{-}\frac{2xi}{{1}{+}{x}^{2}}{\mid}$$ $${=}\frac{1}{2i}{ln}{[}\frac {{1}{-}{x}^{2}}{{1}{+}{x}^{2}}{+}\frac{2xi}{{1}{+}{x}^{2}}{]}$$ Now, $$\frac{{1}{-}{tan}^{2}{\theta}}{{1}{+}{tan}^{2}{\theta}}{=}{cos}{2}{\theta}$$ And, $$\frac{{2}{tan}{\theta}}{{1}{+}{tan}^{2}{\theta}}{=}{sin}{2}{\theta}$$ Substituting x=tan[theta] we may proceed.
unfortunately my browser did not decode. Can you send a picture like last time i asked please?

An Alternative Treatment

$$\frac{1}{{1}{+}{x}^{2}}{=}\frac{1}{2i}{[}\frac{1}{x-i}{-}\frac{1}{x+i}{]}$$

$${=}{-}\frac{1}{2i}{[}\frac{1}{i-x}{+}\frac{1}{i+x}{]}$$

$${=}{-}\frac{1}{2i}{i}^{-1}{[}{{(}{1}{-}{x}{/}{i}{)}}^{-1}{+} {{(}{1}{+}{x}{/}{i}{)}}^{-1}{]}$$

Applying the binomial expansion and after cancellations we have:

Integrand=$${[}{1}{-}{x}^{2}{+}{x}^{4}{-}{x}^{6}............{]}$$

On integration we have,
Integral=$${[}{x}{-}{{x}^{3}}{/}{3}{+}{{x}^{5}}{/}{5}{-}{{x}^{7}}{/}{7} ..........{]}$$
= arctan{x}

[link for the expansion of arc tan(x): http://en.wikipedia.org/wiki/Taylor_series ]
Attached Images
 maths_post.bmp (235.1 KB, 7 views)

Integrand=$${[}{1}{-}{x}^{2}{+}{x}^{4}{-}{x}^{6}............{]}$$

The above series is convergent for abs[x]<1. Integration is allowed for such cases.

When abs[x]>1 we may proceed as follows:
Let y=1/x
Now, abs value of y is less than 1

$$\int\frac{1}{{1}{+}{x}^{2}}{dx}{=}{-}\int\frac{1}{{1}{+}{y}^{2}}{dy}$$

Since y<1 , we may proceed exactly in the same manner and get the same
result preceded by a negative sign as expected.

$${tan}^{-1}{(}{1}{/}{x}{)}{=}{\pi}{/}{2}{-}{tan}^{-1}{(}{x}{)}$$

Attached Images
 Cal_M.bmp (259.0 KB, 4 views)

An attachment[.bmp file] in relation to #56 has been uploaded.
Attached Images
 math.bmp (224.0 KB, 8 views)

 Quote by Anamitra An attachment[.bmp file] in relation to #56 has been uploaded.
Thank you sir. It is a clever trick. I understand now almost all of line #7 of picture in post #52, just one more thing, how did minus sighn appear outside the ln? If this is trivial i beg your forgiveness.

Thanx again.
 You can't take a minus sign outside the log symbol.It is important to note the log of a negative number is undefined. The base has to be positive[but not one or zero]. But the log of a number can be negative. Let $${y}{=}{log}_{b}{A}$$ $${-}{log}_{b}{A}{=}{log}_{b}{A}^{-1}$$ By definition A and b are positive. But y can be negative or positive or zero.In the second relation the negative sign outside the log symbol becomes a power of A and not its coefficient. What I have said so far [in this post] is basically in relation to the logarithm of real numbers. For complex numbers , you may go through the link below: Link: http://en.wikipedia.org/wiki/Complex_logarithm [You cannot take a minus sign out side the log symbol in any situation: real or complex.]

 Quote by Anamitra You can't take a minus sign outside the log symbol.It is important to note the log of a negative number is undefined. The base has to be positive[but not one or zero]. But the log of a number can be negative. Let $${y}{=}{log}_{b}{A}$$ By definition A and b are positive. But y can be negative or positive or zero. What I have said so far [in this post] is in relation to the logarithm of real numbers. For complex numbers , you may go through the link below: Link: http://en.wikipedia.org/wiki/Complex_logarithm
Thanx, it all makes sense now, so the minus sign was causing trouble because it was a typo.
For myself, I liked that derivation very much because you used clever algebra and sqrt(-1) that is why i tried so hard to understand it.
 A Simple Paradox to Sort Out $$\sqrt{16}{=}\sqrt{-4*-4}$$ $${=}\sqrt{-4}{*}\sqrt{-4}$$ $${=}\pm{2i}{*}\pm{2i}$$ $${=}\pm{4}{i}^{2}$$ Therefore, $$\pm{4}{=}\pm{4}{i}^{2}$$ $${=>}{i}^{2}{=}\pm{1}$$ Are you ready to believe that?
 Brun's theorem states that the sum of the inverses of the twin primes converges, i.e. $$\sum_{p:p+2\in{\mathbb{P}}}\left(\frac{1}{p}+\frac{1}{p+2}\right)=B$$ where B is Brun's constant. So, even if there are infinitely many twin primes (which has not yet been proven), the above sum converges. This is not the truth for the naturals and the primes, i.e. $$\sum_{p\in{\mathbb{P}}}\frac{1}{p}=\infty$$ and $$\sum_{n\in{\mathbb{N}}}\frac{1}{n}=\infty$$ But for the squares (and cubes etc) of the naturals, the sum converges. $$\sum_{n\in{\mathbb{N}}}\frac{1}{n^2}=\frac{\pi^2}{6}$$ Brun's theorem is in my opinion a beautiful result and is worth a post in itself. BUT, I have been thinking that this is somehow related to cardinality. Let's define a measure $$\mathcal{M}\left(A\right)$$which is the sum of the reciprocals of the elements in the set A. Then $$\mathcal{M}\left(\mathbb{N}^2\right) < \mathcal{M}\left(twin primes\right) < \mathcal{M}\left(\mathbb{P}\right) = \mathcal{M}\left(\mathbb{N}\right) = \infty$$ So in the sense of the measure M, the set of all squares is a "less dense" set than the twin primes, which in turn is "less dense" than the set of the primes and so on. This measure can be generalized to any function f(n). $$\mathcal{M}_f\left(A\right)=\sum_{n\in{A}}f\left(n\right)$$ My guess is that this (or something related) has been done before. In that case, where can I find more information? Fake edit: Of course I found what I was looking for about two seconds before I was about to post my ramblings. Well, since I spent an hour trying to figure out how to use LaTeX, I'll post it anyway. :)

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 Quote by Anamitra A Simple Paradox to Sort Out $$\sqrt{16}{=}\sqrt{-4*-4}$$ $${=}\sqrt{-4}{*}\sqrt{-4}$$ $${=}\pm{2i}{*}\pm{2i}$$ $${=}\pm{4}{i}^{2}$$ Therefore, $$\pm{4}{=}\pm{4}{i}^{2}$$ $${=>}{i}^{2}{=}\pm{1}$$ Are you ready to believe that?
The problem with this is the transition to the second line. Fact is, square roots only for-sure have this property:

$$\sqrt{xy} = \sqrt{x}\sqrt{y}$$

When x and y are positive.

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 Quote by Char. Limit The problem with this is the transition to the second line. Fact is, square roots only for-sure have this property: $$\sqrt{xy} = \sqrt{x}\sqrt{y}$$ When x and y are positive.
The problem with the poster is even more fundamental. He says that

$$\sqrt{16}=\pm 4$$

which is simply untrue. The square root is defined to be a positive value, no exceptions.

 Quote by Char. Limit The problem with this is the transition to the second line. Fact is, square roots only for-sure have this property: $$\sqrt{xy} = \sqrt{x}\sqrt{y}$$ When x and y are positive.
This is definitely the correct answer. And Evo has provided a prompt reply.In fact I got the same answer from Ask Dr Math when I sent them this problem a few months back.

Now to make the definition of "i" consistent we have the rule indicated by Evo.It is there in the texts.
Query: If the definition of "i" is re-formulated so that i^2 is =1 in certain exceptional cases it is going to have interesting effects in many areas of physics for example in the area of general relativity. Is it necessary to do that , to keep such a provision?
[This is of course a speculative query. I am not staking any type of claim anywhere in regard of this.]

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