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

[agentredlum]

Also, 1+1+1=111 in base 1.

YOU'RE RIGHT TOO!

Thanks to all posting

I'm not sure the digit 0 would be allowed in base 0.

 

[Char. Limit]

No digits would be allowed in base 0. After all, the base number shows how many different "numbers" you can allow. Base 1 has one number: 1. Base 2 has two numbers: 0 and 1. The amount of numbers in base 0 is... the empty set!

 

[agentredlum]

Basically, Base 1 only allows one digit: 1. Base 2 allows 1 and 0. That's why 1+1+1=111 in base 1... actually, 1+1+1...+1 = 111...1 in base 1.

Oh, so I can't use 0 in base 1

Too bad the symmetry is broken.

 

[agentredlum]

No digits would be allowed in base 0. After all, the base number shows how many different "numbers" you can allow. Base 1 has one number: 1. Base 2 has two numbers: 0 and 1. The amount of numbers in base 0 is... the empty set!

Good point about base 0.

I understand why we would pick 1 as the only allowable number for base 1...for consistency.

so symmetry is exchanged for consistency...i can live with that.

 

[chiro]

Oh, so I can't use 0 in base 1

Too bad the symmetry is broken.

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

 

[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}{+}<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.