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

[dodo]

An old trick about differentiation from Feynman's book "Tips on Physics":

If f(x) is a function whose rule has the form f = k . u^a . v^b . w^c . ...
where u,v,w,... are also functions of x (you can think of u,v,w,... as "sub-expressions" in the rule for f),
and k,a,b,c,... are constants, then

f' = f . (a u'/u + b v'/v + c w'/w + ...)

It can be proven by induction on the number of factors.

Note that the sum within parenthesis has as many terms as "sub-expression" factors in the original, and a sum of zero terms should be regarded as 0.

 

[Samuelb88]

I always admired the fact that $\int_{-1}^1 \frac{\mathrm{d}x}{x} = i\pi$. By a symmetry argument, intuition would suggest the integral is zero since there is infinite negative area from -1 to 0 and infinite positive area from 0 to 1.

 

[Guffel]

I love the fact that the number of partitions into odd parts is the same as the number of partitions in distinct parts. Far from intuitive and finding a bijective proof is difficult (easy to prove using generating functions though). Not really a trick, but the discussion also seems to include curiousities. Sorry if OT.

 

[zetafunction]

I always admired the fact that $\int_{-1}^1 \frac{\mathrm{d}x}{x} = i\pi$. By a symmetry argument, intuition would suggest the integral is zero since there is infinite negative area from -1 to 0 and infinite positive area from 0 to 1.

this can be obtained by SCHOKSTKY'S FORMULA (i do not how to spell the name)

$$(x+ie)^{-1}= -i\pi \delta (x)+P(1/x)$$

since P(1/x) is odd the integral over it vanishes and only the delta function contributes.

 

[agentredlum]

You can use whatever symbol you want, but you can only use that symbol.

O-K Suppose I choose to use 0. Then 1 + 1 has no solution in base 1?

Also note that you get a big problem in the left-most position of any base 1 representation (when using 0) because that asks for 'how many powers of 0^0

the answer of course is 0*0^0 so are you saying that since i don't have any powers of 0 then i don't have to compute 0^0?

Are you saying we make 0 = 1 so even if i pick the SYMBOL 0 we pretend its the NUMBER 1?

Thank you for the post, PLEASE POST MORE!

 

[agentredlum]

I love the fact that the number of partitions into odd parts is the same as the number of partitions in distinct parts. Far from intuitive and finding a bijective proof is difficult (easy to prove using generating functions though). Not really a trick, but the discussion also seems to include curiousities. Sorry if OT.

Thanx for the post, no need to apologize, it is interesting from Number Theory isn't it?

I remember when i was struggling with Number Theory almost all problems at the end of any chapter required a bit of genius.

You never know when someone might look at your post, then the ones above and below and come up with a great idea.

Is this what you are posting about?

http://demonstrations.wolfram.com/TheNumberOfPartitionsIntoOddPartsEqualsTheNumberOfPartitions/

Heres an explanation of partitions. Personally I find partitions fascinating

http://en.wikipedia.org/wiki/Partition_(number_theory )

 

[Guffel]

Is this what you are posting about?

http://demonstrations.wolfram.com/TheNumberOfPartitionsIntoOddPartsEqualsTheNumberOfPartitions/

Heres an explanation of partitions. Personally I find partitions fascinating

http://en.wikipedia.org/wiki/Partition_(number_theory )

That's it. And I agree, partitions are fascinating!

 

[agentredlum]

An old trick about differentiation from Feynman's book "Tips on Physics":

If f(x) is a function whose rule has the form f = k . u^a . v^b . w^c . ...
where u,v,w,... are also functions of x (you can think of u,v,w,... as "sub-expressions" in the rule for f),
and k,a,b,c,... are constants, then

f' = f . (a u'/u + b v'/v + c w'/w + ...)

It can be proven by induction on the number of factors.

Note that the sum within parenthesis has as many terms as "sub-expression" factors in the original, and a sum of zero terms should be regarded as 0.

Can you post a slmple example? How would you do f(x) = (4x^5)*(3x^2) using this trick. I know how to get the right answer the usual way f'(x) = 84x^6 but i couldn't get the trick to work. Hopefully a worked out example will help. THANX FOR THE POST! POST MORE!

 

[elegysix]

sorry for the late reply, here's an image of my post on fractional calculus you asked for.
[PLAIN]http://img835.imageshack.us/img835/5461/fraccalculus.png

 

[dodo]

(Sorry, elegysix, I was replying to the post before yours. :)

Well, maybe your example (4x^5)*(3x^2) = 12 x^7 is a bit too simple to use this.

But try to differentiate instead something like

f(x) = (3x+1)*sqrt(x^2-2)*(x+3)^3

which would be very lenghty if you have to apply the product rule repeatedly, instead you can write in a moment,

f'(x) = (3x+1)*sqrt(x^2-2)*(x+3)^3 . ( 3/(3x+1) + (1/2)*2x/(x^2-2) + 3*(1/(x+3)) )

and then simplify to your taste. Although for me it's more the elegance of the formula.

 

[agentredlum]

(Sorry, elegysix, I was replying to the post before yours. :)

Well, maybe your example (4x^5)*(3x^2) = 12 x^7 is a bit too simple to use this.

But try to differentiate instead something like

f(x) = (3x+1)*sqrt(x^2-2)*(x+3)^3

which would be very lenghty if you have to apply the product rule repeatedly, instead you can write in a moment,

f'(x) = (3x+1)*sqrt(x^2-2)*(x+3)^3 . ( 3/(3x+1) + (1/2)*2x/(x^2-2) + 3*(1/(x+3)) )

and then simplify to your taste. Although for me it's more the elegance of the formula.

WOW! This is a great trick! I did your example above both ways, the long way and the 'trick' way 'trick' way was much better. Then i did my example using trick way and it worked.

So as i understand it in words 'To find the derivative of a product of many different funtions, find the derivative of each, divide by the function, add them all up and multiply by the original function.

Words are crude so in formula.

f = (u)(v)(w)

f'= (u)(v)(w)[u'/u + v'/v + w'/w]

And this can be used for the product of any number of functions.

This is fantastic! And quite elegant i agree. Thanx for the post! PLEASE POST MORE!

 

[agentredlum]

sorry for the late reply, here's an image of my post on fractional calculus you asked for.
[PLAIN]http://img835.imageshack.us/img835/5461/fraccalculus.png[/QUOTE]

Thank you so much for the picture. I really appreciate it. PLEASE POST MORE!

 

[agentredlum]

If you use the common denominator (u)(v)(w) then

f' = (u)(v)(w)[u'(v)(w) + v'(u)(w) + w'(u)(v)]/[(u)(v)(w)]

so f' = u'(v)(w) + v'(u)(w) + w'(u)(v)

Which is a result I've seen in Calculus texts but it's still a neat trick.

 

[TejasB]

Dude x being the root of a quadratic expression two values are expected in your case you took only + when it is +\-. As an expample consider x^2+x-6=0. By your formula roots become complex when they are 2&-3.

 

[agentredlum]

Dude x being the root of a quadratic expression two values are expected in your case you took only + when it is +\-. As an expample consider x^2+x-6=0. By your formula roots become complex when they are 2&-3.

Look at the post again i have plus minus as +-

Now your example x^2 + x - 6 = 0

Step 1 isolate ax^2 term

x^2 = -x + 6

Step 2 identify a,b,c a = 1, b = -1, c = 6

Step 3 plug them into (b +- sqrt(b^2 + 4ac))/(2a)

(-1 +- sqrt((-1)^2 + 4(1)(6)))/(2(1))

(-1 +- sqrt(1+24))/2

(-1 +- 5)/2

x = -6/2 = -3 or x= 4/2 = 2

In perfect agreement.

Thanx for the post! POST MORE PLEASE!

 

[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

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

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

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

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,
$$\pm{4}{=}\pm{4}{i}^{2}$$
$${=>}{i}^{2}{=}\pm{1}$$
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
$$\mathcal{M}\left(\mathbb{N}^2\right) < \mathcal{M}\left(twin primes\right) < \mathcal{M}\left(\mathbb{P}\right) = \mathcal{M}\left(\mathbb{N}\right) = \infty$$
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. :)

 

[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,
$$\pm{4}{=}\pm{4}{i}^{2}$$
$${=>}{i}^{2}{=}\pm{1}$$
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.