How can you simplify the quadratic formula using completing the square?

[Anamitra]

\int_0^x darctanx=\int_0^x \frac{dx}{x^2+1}=\int_0^x \frac{i}{2(x-i)} - \frac{i}{2(x+i)} dx = \frac{i}{2} ln \left( \frac{x-i}{x+i} \right) \Rightarrow \forall x \in \Re, \: arctanx = \frac{i}{2}ln \left( \frac{x-i}{x+i} \right)

Suppose we consider the integral\int\frac{1}{x-i}{dx}

The path of integration has not been specified
[Interestingly the integrand does not satisfy the CR equations[Cauchy-Riemann equations] and hence it is not an analytical function. The derivative cannot be defined uniquely at any particular point.But given a path/route we should be able to define the derivative uniquely for points on the given path[by taking the tangential direction] rendering the integral suitable for the process of integration.Could anybody confirm or de-confirm the last statement?]

Sorry for the mistake.One may replace x by z=x+iy and specify the path along the x-axis.The CR equations do hold.

 

[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

 

[agentredlum]

{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?

 

[Anamitra]

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.

 

[agentredlum]

{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.

I would love to read your thoughts on this.

 

[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.

 

[agentredlum]

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?

 

[Anamitra]

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}{+}<br /> {{(}{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 ]

 

[Anamitra]

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}{)}

Link: http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

 

[Anamitra]

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

 

[agentredlum]

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.

 

[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}

{-}{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.]

 

[agentredlum]

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

 

[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}
{=&gt;}{i}^{2}{=}\pm{1}
Are you ready to believe that?

 

[Guffel]

http://en.wikipedia.org/wiki/Brun%27s_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) &lt; \mathcal{M}\left(twin primes\right) &lt; \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 http://en.wikipedia.org/wiki/Small_set_%28combinatorics%29" 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. :)

 

[Char. Limit]

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}
{=&gt;}{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.

 

[micromass]

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.

 

[Anamitra]

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.]

 

[Anamitra]

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

 

[pwsnafu]

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

Every single line is wrong. The nth-root is multivalued on the complex numbers. You are not allowed to just throw three quarters of your solution set away. This is like saying
(-1)^2 = 1^2 and therefore -1=1.

 

[Anamitra]

x^4-i^4=0
has four solutions.

One of them is i. The others are not "i"

 

[Anamitra]

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.

 

[pwsnafu]

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.

What does "positive value of the fourth root" mean?

You haven't responded to the criticism. The RHS equals the set {1,i,-1,-i}, the LHS equals {i}. They are not the same.

Even if you are talking about principle roots, you aren't keeping track of the branch cut, so the http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities" That's from line 4 to line 5.

 

[agentredlum]

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.

I will give you the benefit of the doubt for line 2 since (-1)(-1)(-1)(-1) = 1

How does line 3 follow in post #69?

Please post picture.

Oh wait...i got it, you are defining i^2 = 1 so with this definition i = i^3

However have you forgotten your own definition in line 6?

Or is it that you want i^2 = 1 sometimes and i^2 = -1 at other times?

 

[agentredlum]

http://en.wikipedia.org/wiki/Brun%27s_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) &lt; \mathcal{M}\left(twin primes\right) &lt; \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 http://en.wikipedia.org/wiki/Small_set_%28combinatorics%29" 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. :)

Thanx for the pos! Please post more.

I find it very interesting that the sum of the reciprocals of the squares converges but the sum of the reciprocals of the primes diverges. In some 'sense' this implies the primes are more numerous than the squares. This is counterintuitive because consecutive primes can have arbitrary numerical difference between them.

 

[agentredlum]

I have a question. For LARGE N, given N consecutive composite numbers, will you find a square among them?

 

[Anamitra]

We have, considering its multiple valued nature,
{(}{i}^{4}{)}^{1/4}{=}{[}{1,-1,i,-i}{]}-------------- (1)

Now i^8=1. Therefore,

{(}{1}*{(}{i}^{4}{)}{)}^{1/4}{=}{[}{1,-1,i,-i}{]}

{(}{i}^{8}*{i}^{4}{)}^{1/4}{=}{[}{1,-1,i,-i}{]}

{(}{i}^{12}{)}^{1/4}{=}{[}{1,-1,i,-1}{]}
Or,

{i}^{3}{=}{[}1,-1,i,-i{]} ----------------- (2)But,

{i}^{3}{=}{[}-i{]} ---------------- (3)

Relations (2) and (3) don't seem to hang together.The possible values of {i}^{3} may have different interpretationsCan we explain this ambuigty in the light of the calculations shown in the Wikipedia link in Post #73 or otherwise?

[ I have used the third brackets for set notation instead of braces]

 

[agentredlum]

We have, considering its multiple valued nature,
{(}{i}^{4}{)}^{1/4}{=}{[}{1,-1,i,-i}{]}-------------- (1)

Now i^8=1. Therefore,

{(}{1}*{(}{i}^{4}{)}{)}^{1/4}{=}{[}{1,-1,i,-i}{]}

{(}{i}^{8}*{i}^{4}{)}^{1/4}{=}{[}{1,-1,i,-i}{]}

{(}{i}^{12}{)}^{1/4}{=}{[}{1,-1,i,-1}{]}
Or,

{i}^{3}{=}{[}1,-1,i,-i{]} ----------------- (2)But,

{i}^{3}{=}{[}-i{]} ---------------- (3)

Relations (2) and (3) don't seem to hang together.The possible values of {i}^{3} may have different interpretationsCan we explain this ambuigty in the light of the calculations shown in the Wikipedia link in Post #73 or otherwise?

[ I have used the third brackets for set notation instead of braces]

x^3 - i^3 = 0 has 3 solutions. NONE of them are 1, -1, -i

the solutions are

i

-sqrt(3)/2 - (1/2)i

sqrt(3)/2 - (1/2)i

so your set of solutions in relation 2 is not correct.

 

[Anamitra]

The following relation should hold
{i}^{3}{=}{[}{1}{,}{-1}{i}{,}{-i}{]},

if {{(}{i}^{12}{)}}^{1/4}{=}{i}^{3}

What you have said is
{{(}{i}^{3}{)}}^{1/3}{=}{[}{i},{\omega}{i},{\omega}^{2}{i}{]}

It does not contradict my assertions or stand against them in any manner.

I have simply tried to show a contradiction in the existing ideas/formulations by deriving two results for
{i}^{3} in Post #77

1. {i}^{3}{=}{[}{1},{-1},{i},{-i}{]}
2. {i}^{3}{=}{[}{-i}{]}

The second one is the one that we use conventionally.

[ I have used the third bracket for set notation instead of braces]

 

[Anamitra]

There are two different options:

1. To formulate "Exception Handling" rules to take care of the inconsistencies. The list may be very large and one might need to expand the list quite often.

2. To allow the imaginary "i" to misbehave[in a restricted manner] by reformulating its definition and the associated rules.

The second option could have far-reaching effects in several areas of physics for example in General Relativity.

 

[agentredlum]

Here is a neat little trick on how to prove all even perfect numbers are also triangular numbers.

Euclid showed that all even perfect numbers are of the form 2^(n - 1)(2^n - 1)

Multiply the expression above by 2 and divide it by 2

2[2^(n - 1)(2^n - 1)]/2

now combine 2 and 2^(n - 1) = 2^n under multiplication, you now get...

2^n(2^n - 1)/2

Now notice that if you let 2^n - 1 = S you get...

(S + 1)S/2

This MUST be a triangular number

All you had to do was multiply and divide by 2 and re-arrange a few things...

 

[Anamitra]

{x}^{4}{=}{1}

Includes two distinct assignments in a combined state: {x}^{2}{=}{+1}

and {x}^{2}{=}{-1}
By the convention suggested in the link below we consider the first one --the positive root when the square-root operation is applied on the first equation.

https://www.physicsforums.com/showpost.php?p=3400199&postcount=67Now let us consider the expression/relation: {i}^{4}{=}{1}

The two inseparable assignments involved in the above equation are: {i}^{2}{=}{+1}
and

{i}^{2}{=}{-1}

There would be a big trouble if we follow the convention given in the above link--as suggested by steadfast conservative thinking.
[We need an exception handling statement/rule here: Better still we reformulate the whole thing]The set theoretical formulation is better disposed in relation to this problem.
{i}^{4}{=}{1}

implies,
{{(}{i}^{4}{)}}^{1/2}{=}{[}{+1,-1}{]}
But the +1 is invariably, a part of the solution set!

 

[pwsnafu]

By the convention suggested in the link below we consider the first one --the positive root when the square-root operation is applied on the first equation.

https://www.physicsforums.com/showpost.php?p=3400199&postcount=67

No, you have the convention wrong. I'll write it out in detail. Consider real numbers. The equation

x^2 = b

has two solutions if b is not zero and positive. Let's give them names, say, x_1 and x_2. Now due the property of the reals we know:

1. One of the solutions is greater than the other.
2. If we have one solution, we can obtain the other by multiplying by -1

This means that there exists a real number, y, such that x_1=y and x_2 = -y. The problem is we don't know if y itself is positive or negative. The convention is "we choose y to be positive and write y = \sqrt{b}".

In other words: the convention refers to the square root symbol, not the solution set of the equation x^2 = b. It also only applies to real analysis, not complex analysis.

PS, the j^2 = 1 is called the http://en.wikipedia.org/wiki/Split-complex_number" . Unlike complex numbers, they do not form a field, and so the applications are very limited.

 

[Anamitra]

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.

Possibly you did not notice the words--no exceptions

 

[micromass]

Possibly you did not notice the words--no exceptions

Yes, and that is exactly what pwsnafu tries to explain. The square root symbol implies that you take the positive root. But the equation x^2=b still has multiple values, while \sqrt{b} is only one value.

 

[Anamitra]

\sqrt{{x}^{2}} is defined to be a positive value----exceptions are not allowed as per your statement.

\sqrt{{x}^{4}} is defined to be a positive value--- no exceptions.

Your "exception-less" definition denies the multiple valued nature of {16}^{1/2}

 

[micromass]

Your "exception-less" definition denies the multiple valued nature of {16}^{1/2}

Indeed, that's exactly what it was for. We want to eliminate the multiple-valued functions, and we thus want to give on single value for the square root. If you want the negative value, then you just need to write -\sqrt{16}.

 

[agentredlum]

\sqrt{{x}^{2}} is defined to be a positive value----exceptions are not allowed as per your statement.

\sqrt{{x}^{4}} is defined to be a positive value--- no exceptions.

Your "exception-less" definition denies the multiple valued nature of {16}^{1/2}

sqrt(x^2) = |x| this allows you to use all real numbers x. Does this hold if x is complex?

Example sqrt(i^2) = |i|

The left hand side of the example is sqrt(-1) = i

What is the right hand side of the example? In many contexts I am aware of |i| = 1

So the equation does not hold when moving from Reals to Complex.

 

[micromass]

sqrt(x^2) = |x| this allows you to use all real numbers x. Does this hold if x is complex?

Example sqrt(i^2) = |i|

The left hand side of the example is sqrt(-1) = i

What is the right hand side of the example? In many contexts I am aware of |i| = 1

So the equation does not hold when moving from Reals to Complex.

The square root symbol is only defined for positive real numbers. So writing \sqrt{-1} isn't defined. (Yes, I know that there are math books out there that do use the notation i=\sqrt{-1}, but I still don't consider that notation to be standard).

 

[Anamitra]

{i}^{4}{=}{1}

{{(}{i}^{4}{)}}^{1/2}{=}{1}^{1/2}

{i}^{2}{=}{+}{1}

The "no exceptions" clause disallows any prescription/rule[ an exception handling clause to] handle this situation

 

[Anamitra]

Post #77 has been overlooked by micromass and Pwsnafu in a host of other commitments.
Link:

\https://www.physicsforums.com/showpost.php?p=3401056&postcount=77

 

[micromass]

{i}^{4}{=}{1}

{{(}{i}^{4}{)}}^{1/2}{=}{1}^{1/2}

{i}^{2}{=}{+}{1}

The "no exceptions" clause disallows any prescription/rule[ an exception handling clause to] handle this situation

Well, that just illustrates that (a^b)^c=a^{bc} isn't necessarily true for complex numbers

 

[Anamitra]

{(}{{i}^{4}{)}}^{1/2}{=}{(}{\sqrt{i}{)}}^{4}

You agree to this[Your rule does not suggest this unless you incorporate some exception handling clause]?

 

[micromass]

{i}^{4} is not a complex number

It's not a complex number, so... ?

 

[agentredlum]

The square root symbol is only defined for positive real numbers. So writing \sqrt{-1} isn't defined. (Yes, I know that there are math books out there that do use the notation i=\sqrt{-1}, but I still don't consider that notation to be standard).

You have to define sqrt(-1) = i otherwise you lose Gauss Fundamental Theorem of Algebra...and nobody wants to lose THAT.

 

[micromass]

You have to define sqrt(-1) = i otherwise you lose Gauss Fundamental Theorem of Algebra...and nobody wants to lose THAT.

No, you have to define i^2=1 for that. That is not the same as saying \sqrt{-1}=i. The square root operator is only defined for positive real numbers! But just because we didn't define \sqrt{-1} as -1, doesn't mean that i^2=1 isn't true!

 

[agentredlum]

No, you have to define i^2=1 for that. That is not the same as saying \sqrt{-1}=i. The square root operator is only defined for positive real numbers! But just because we didn't define \sqrt{-1} as -1, doesn't mean that i^2=1 isn't true!

Then how would you solve x^2 = -1

By inspection? By factoring?

The extraction of roots must be made legitimate over all finite polynomials with integer co-efficients.

i^2 = -1 does not do this

sqrt(-1) = i does.

 

[micromass]

Then how would you solve x^2 = -1

By inspection? By factoring?

The extraction of roots must be made legitimate over all finite polynomials with integer co-efficients.

i^2 = -1 does not do this

sqrt(-1) = i does.

Well, how would you solve x^2=2+i?? By saying x=\sqrt{2+i}?? How helpful...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of x^n=-1:

\cos(\pi k /n)+i\sin(\pi k /n),~k\in \{0,...,n-1\}

but by your method, you would only find x=\sqrt[n]{x} (even if we would allow the square root on complex numbers). That's not really what we want, is it??

 

[agentredlum]

Well, how would you solve x^2=2+i?? By saying x=\sqrt{2+i}?? How helpful...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of x^n=-1:

\cos(\pi k /n)+i\sin(\pi k /n),~k\in \{0,...,n-1\}

but by your method, you would only find x=\sqrt[n]{x} (even if we would allow the square root on complex numbers). That's not really what we want, is it??

micromass asked...

"Well, how would you solve x^2=2+i?? By saying x=\sqrt{2+i}?? How helpful..."

I would throw caution to the wind and let the chips fall where they may...THEN...

I would put this particular example into the quadratic formula and then sustitute back to confirm i got the right answer. Messere Gauss proved I can do it this way.

If you post a more complicated example which does not have a closed form solution i can still get close to the answer by using the 4 binary operations AND THE EXTRACTION OF ROOTS.

The ability to extract roots of real numbers is necessary, this means you need to extract roots of negative values.

The Real numbers are 1 dimensional, Complex numbers NOT on the vertical axis have 2 distinct dimensions.

How fortunate that i is on the vertical axis and has length 1? So we can use it as a 'cheat' to simplify the algebra.

You are correct as far as the cyclotomic x^n = a + bi is concerned

However using sines and cosines to get the answer of course produces multi values since they are periodic!

This is a neat trick, and i have solved x^n = a + bi in Linear Algebra so i know what you mean and agree with you on most of your points.

micromass also said...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of x^n=-1:

To the best of my knowledge this method you describe works only if you don't have a MIDDLE TERM.

x^2 = (1 + 2i)x + i would not work with this method but easily works using quadratic formula.

 

[pwsnafu]

You have to define sqrt(-1) = i otherwise you lose Gauss Fundamental Theorem of Algebra...and nobody wants to lose THAT.

See below.

If you post a more complicated example which does not have a closed form solution i can still get close to the answer by using the 4 binary operations AND THE EXTRACTION OF ROOTS.

x^5 - x + 1 = 0

No idea what you mean by "close" or "extraction of roots".

The ability to extract roots of real numbers is necessary,

What? Why?

this means you need to extract roots of negative values.

In real analysis? Hell no!

The Real numbers are 1 dimensional, Complex numbers NOT on the vertical axis have 2 distinct dimensions.

How fortunate that i is on the vertical axis and has length 1? So we can use it as a 'cheat' to simplify the algebra.

The complex numbers are defined as R[X]/(X²+1) and i is defined as the cosets of X. That's the definition! What is this talk of "cheat" or "square roots"?

Please answer the following: how much complex analysis have you done? Because you are arguing about stuff they teach you in the first week.

You are correct as far as the cyclotomic x^n = a + bi is concerned

However using sines and cosines to get the answer of course produces multi values since they are periodic!

Here is micromass's result: \cos(\pi k /n)+i\sin(\pi k /n),~k\in \{0,...,n-1\}. Did you notice that k = 0, 1, ..., n-1 and stops? It's a finite number of solutions. Specifically, you get n, agreeing with the fundamental theorem of algebra.

micromass also said...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of x^n=-1:

To the best of my knowledge this method you describe works only if you don't have a MIDDLE TERM.

x^2 = (1 + 2i)x + i would not work with this method but easily works using quadratic formula.

And again solve x^5 - x + 1 = 0.