
#55
Jul1011, 05:28 AM

P: 460

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. I would love to read your thoughts on this. 



#56
Jul1011, 06:36 AM

P: 621

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



#57
Jul1011, 08:19 PM

P: 460





#58
Jul1011, 08:56 PM

P: 621

An Alternative Treatment
[tex]\frac{1}{{1}{+}{x}^{2}}{=}\frac{1}{2i}{[}\frac{1}{xi}{}\frac{1}{x+i}{]}[/tex] [tex]{=}{}\frac{1}{2i}{[}\frac{1}{ix}{+}\frac{1}{i+x}{]}[/tex] [tex]{=}{}\frac{1}{2i}{i}^{1}{[}{{(}{1}{}{x}{/}{i}{)}}^{1}{+} {{(}{1}{+}{x}{/}{i}{)}}^{1}{]}[/tex] Applying the binomial expansion and after cancellations we have: Integrand=[tex]{[}{1}{}{x}^{2}{+}{x}^{4}{}{x}^{6}............{]}[/tex] On integration we have, Integral=[tex]{[}{x}{}{{x}^{3}}{/}{3}{+}{{x}^{5}}{/}{5}{}{{x}^{7}}{/}{7} ..........{]}[/tex] = arctan{x} [link for the expansion of arc tan(x): http://en.wikipedia.org/wiki/Taylor_series ] 



#59
Jul1011, 09:22 PM

P: 621

Integrand=[tex]{[}{1}{}{x}^{2}{+}{x}^{4}{}{x}^{6}............{]}[/tex]
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 [tex]\int\frac{1}{{1}{+}{x}^{2}}{dx}{=}{}\int\frac{1}{{1}{+}{y}^{2}}{dy}[/tex] Since y<1 , we may proceed exactly in the same manner and get the same result preceded by a negative sign as expected. [tex]{tan}^{1}{(}{1}{/}{x}{)}{=}{\pi}{/}{2}{}{tan}^{1}{(}{x}{)}[/tex] Link: http://en.wikipedia.org/wiki/Inverse...tric_functions 



#60
Jul1011, 10:47 PM

P: 621

An attachment[.bmp file] in relation to #56 has been uploaded.




#61
Jul1011, 11:51 PM

P: 460

Thanx again. 



#62
Jul1111, 12:37 AM

P: 621

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 [tex]{y}{=}{log}_{b}{A}[/tex] [tex]{}{log}_{b}{A}{=}{log}_{b}{A}^{1}[/tex] 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.] 



#63
Jul1111, 12:50 AM

P: 460

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. THANK YOU FOR YOUR POSTS! THANK YOU FOR YOUR RESPONSES! PLEASE POST MORE!! 



#64
Jul1211, 11:19 AM

P: 621

A Simple Paradox to Sort Out
[tex]\sqrt{16}{=}\sqrt{4*4}[/tex] [tex]{=}\sqrt{4}{*}\sqrt{4}[/tex] [tex]{=}\pm{2i}{*}\pm{2i}[/tex] [tex]{=}\pm{4}{i}^{2}[/tex] Therefore, [tex]\pm{4}{=}\pm{4}{i}^{2}[/tex] [tex]{=>}{i}^{2}{=}\pm{1}[/tex] Are you ready to believe that? 



#65
Jul1211, 11:19 AM

P: 29

Brun's theorem states that the sum of the inverses of the twin primes converges, i.e.
[tex]\sum_{p:p+2\in{\mathbb{P}}}\left(\frac{1}{p}+\frac{1}{p+2}\right)=B[/tex] 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. [tex]\sum_{p\in{\mathbb{P}}}\frac{1}{p}=\infty[/tex] and [tex]\sum_{n\in{\mathbb{N}}}\frac{1}{n}=\infty[/tex] But for the squares (and cubes etc) of the naturals, the sum converges. [tex]\sum_{n\in{\mathbb{N}}}\frac{1}{n^2}=\frac{\pi^2}{6}[/tex] 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 [tex]\mathcal{M}\left(A\right)[/tex]which is the sum of the reciprocals of the elements in the set A. Then [tex]\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[/tex] 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). [tex]\mathcal{M}_f\left(A\right)=\sum_{n\in{A}}f\left(n\right) [/tex] 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. :) 



#66
Jul1211, 01:20 PM

PF Gold
P: 1,930

[tex]\sqrt{xy} = \sqrt{x}\sqrt{y}[/tex] When x and y are positive. 



#67
Jul1211, 01:23 PM

Mentor
P: 16,565

[tex]\sqrt{16}=\pm 4[/tex] which is simply untrue. The square root is defined to be a positive value, no exceptions. 



#68
Jul1211, 08:30 PM

P: 621

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 reformulated 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.] 



#69
Jul1211, 11:35 PM

P: 621

For Applications of the Exception:
[tex]{i}{=}{{(}{i*i*i*i}{)}}^{1/4}[/tex] [tex]{i}{=}{(}{i*i*i*i}{)}^{1/4}[/tex] [tex]{i}{=}{{(}{i}^{3}*{i}^{3}*{i}^{3}*{i}^{3}{)}}^{1/4}[/tex] [tex]{i}{=}{{(}{i}^{12}{)}}^{1/4}[/tex] [tex]{i}{=}{i}^{3}[/tex] [tex]{i}{=}{i}[/tex] Could any anybody provide me with the full list of exceptions in relation to “i” ? 



#70
Jul1211, 11:51 PM

Sci Advisor
P: 778

[tex](1)^2 = 1^2[/tex] and therefore 1=1. 



#71
Jul1311, 12:11 AM

P: 621

x^4i^4=0
has four solutions. One of them is i. The others are not "i" 



#72
Jul1311, 12:17 AM

P: 621

In the first line of post #69 the positive value of the fourth root on the RHS is matching with the value on the LHS.
This is supposed to continue to the last line unless you create a new rule or some exception. 


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