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

[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
$$\mathcal{M}\left(\mathbb{N}^2\right) < \mathcal{M}\left(twin primes\right) < \mathcal{M}\left(\mathbb{P}\right) = \mathcal{M}\left(\mathbb{N}\right) = \infty$$
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 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.

 

[agentredlum]

See below.
x^5 - x + 1 = 0

No idea what you mean by "close" or "extraction of roots".
What? Why?
In real analysis? Hell no!
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.
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.
And again solve x^5 - x + 1 = 0.

Look at this link, what are all those radicals doing there?

http://mathworld.wolfram.com/QuinticEquation.html

What do you know about closure?

Are you saying we don't need radicals to approximate solutions to polynomial equations?

Why are the Complex numbers closed under addition subtraction multiplication division AND EXTRACTION OF ROOTS?

 

[agentredlum]

Why stop at n-1 why not keep going?

You get the same solutions with different numbers inside the cosine and sine.

You 'sound' like you know what you are talking about, throwing around fancy terms...Real Analysis...big deal, but you only see it your way.

I GAURANTEE you that your calculator does not use Real Analysis and cosets to approximate roots of polynomial equations.

1669Newton introduces his iterative method for the numerical approximation of roots.

 

[agentredlum]

Where did you come from? micromass does not need you to defend him/her, he/she is perfectly capable of holding a rational, intelligent stimulating conversation so just chill.

 

[agentredlum]

How does solving x^5 = x - 1 relate to what i said to micromass about the middle term?

Can you factor x^4 + x^2 + 1

??

 

[pwsnafu]

Look at this link, what are all those radicals doing there?
http://mathworld.wolfram.com/QuinticEquation.html

How does solving x^5 = x - 1 relate to what i said to micromass about the middle term?

Can you factor x^4 + x^2 + 1

??

I'm putting these two together. You made the claim that "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." Which is fine. But the http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem" says you can't do this for a general quintic. Specifically the example I gave you cannot be solved using radicals. It's got nothing to do with middle term claim you made or approximations.

What do you know about closure?

That fields are closed under their operations? That the complex numbers are algebraically closed? What do you want to know?

Are you saying we don't need radicals to approximate solutions to polynomial equations?

Well, if you only care about Newton's method, then that method doesn't use nth roots. Only subtraction, division and derivatives.

Why are the real numbers closed under addition subtraction multiplication division AND EXTRACTION OF ROOTS?

Well add/sub/mult/div I'm not going into. There are plenty of books on that. I'm interpreting "extraction of roots" to mean real exponentiation. It is is closed in the sense that a^b is always a real number. But it is only defined when

1. a is non-negative and b is real,
2. b is an odd integer and a is real.

So it's not a binary operation over all the reals. It is closed in the sense that its codomain is still the reals.

Why stop at n-1 why not keep going?

You get the same solutions with different numbers inside the cosine and sine.

Correct that is why you stop. You don't need the rest, they just cycle around. It's where the term "cyclotomic" comes from.

You 'sound' like you know what you are talking about, throwing around fancy terms...Real Analysis...big deal, but you only see it your way.

Borderline ad hominem. Fancy or not, they are the mathematical terms. And this is a mathematical forum.

I GAURANTEE you that your calculator does not use Real Analysis and cosets to approximate roots of polynomial equations.

1669Newton introduces his iterative method for the numerical approximation of roots.

Considering Newton's method uses the derivative, actually it does.
Sometimes Newton's Method results in cycles, and it's useful to know when that happens. Although most of the advanced calculators such as Mathematica have probably moved over to the superior http://en.wikipedia.org/wiki/Jenkins-Traub_method" . I don't know any root finding algorithms that don't use the derivative at all.

Where did you come from? micromass does not need you to defend him/her, he/she is perfectly capable of holding a rational, intelligent stimulating conversation so just chill.

Post number 15.