Mathematical Question

[fireandwater]

Here's a pattern for double digit numbers. I'll represent the larger number as "x" and the smaller number as "y".
So for 63, x=6 and y=3. What I found is...

(x^2-y^2)/(x+y)=x-y

So with 63,

(6^2-3^2)/(6+3)=6-3

Are there any anomalies to this pattern? Is it worth pursuing?

 

[fireandwater]

Thanks to Yamialpha for telling me about it, he was the one who found it.

 

[Yamialpha]

Thanks to Yamialpha for telling me about it, he was the one who found it.

Funny, I don't ever remember me telling you that you could post about it

 

[RoboMastodon]

Here's a pattern for double digit numbers. I'll represent the larger number as "x" and the smaller number as "y".
So for 63, x=6 and y=3. What I found is...

(x^2-y^2)/(x+y)=x-y

So with 63,

(6^2-3^2)/(6+3)=6-3

Are there any anomalies to this pattern? Is it worth pursuing?

It's basic algebra....
(x^2-y^2)=(x+y)(x-y)
therefore...
(x^2-y^2)/(x+y)=(x-y)
It works for any two numbers x and y.
For example: 13 and 11.
13^2-11^2 = 169-121=48
13+11=24
48/24 =2=13-11
The only "anomalies" is when x+y =0, and you can't divide by 0.

 

[MartinM]

The only "anomalies" is when x+y =0, and you can't divide by 0.

Ah, but x+y=0 => x^2-y^2 = 0

 

[RoboMastodon]

This thread is now about 2=1 proofs.
a=b
a^2=ab
2a^2=ab+a^2
2a^2-2ab=ab+a^2-2ab
2(a^2-ab)=a^2-ab
2=1

-1=-1
sqrt(-1/1)=sqrt(1/-1)
sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1)
cross multiply
sqrt(-1)sqrt(-1)=sqrt(1)sqrt(1)
i*i=1
-1=1
-1/2=1/2
1=2

a=sum((-1)^n,n,1,infinity)
a=a
1+-1+1+-1+...+(-1)^n=1+-1+1+-1+...+(-1)^n
(1+-1)+(1+-1+)+...+(1+-1)=1+(-1+1)+(-1+1)+...+(-1+1)
0+0+0+0...=1+0+0+0
0=1
1=2

e^(-pi*i)=-1
ln(e^(pi*i))=ln(-1)
-pi*i=pi*i
-1=1
-1/2=1/2
1=2

 

[Yamialpha]

It's basic algebra....
(x^2-y^2)=(x+y)(x-y)
therefore...
(x^2-y^2)/(x+y)=(x-y)
It works for any two numbers x and y.
For example: 13 and 11.
13^2-11^2 = 169-121=48
13+11=24
48/24 =2=13-11
The only "anomalies" is when x+y =0, and you can't divide by 0.

That's the proof I tried to explain to him, but he didn't think it was correct because of 0. And he actually took calculus in college...

 

[fireandwater]

And he actually took calculus in college...

What's that supposed to mean exactly?

 

[Yamialpha]

What's that supposed to mean exactly?

What that means is that it seems pretty suspicious to me when you claim to have taken college calculus courses but yet you need two 16-year olds to explain to you an extremely obvious answer to a simple algebra question. I told you the same thing RoboMastodon told you, and you wouldn't believe me. I haven't forgotten the fact that you didn't even know how to work with summation notation or the definite integral even after your college classes.

 

[Yamialpha]

Now do you believe that proof showing that no polynomial can be used for primes exclusively?

 

[RoboMastodon]

I'm really bored so I shall now prove the equation in your avatar:

Take the function sin(x)/x and do a taylor expansion on it...

Now factor it into an infinite product over all the zeros. {Sin(x)/x=0|x=+/- pi*n}

Combine each pair into a difference of squares.

Extract the coefficient of x^2. (i.e. the only possible way to get an x^2 term is to multiply the second term in each factor by 1 and then sum them all together). This is what you get, and notice what it equals.

However, recall from the taylor expansion that the x^2 coefficient of Sin(x)/x is -1/6 So...

=)

Now do you believe that proof showing that no polynomial can be used for primes exclusively?

I started proving this but I decided that now I must sleep since I've been up for about 36 hours (I don't like number theory that much anyway, I'm more of a calculus/diff eq's kind of guy).

 

[fireandwater]

Now do you believe that proof showing that no polynomial can be used for primes exclusively?

Not sure, I've already forgotten it, lol.

 

[arunma]

Hey Robo, how do you type equations into your posts? I know it has something to do with a tool on Wikipedia, but can you show me how to do it? Thanks.

 

[Yamialpha]

Hey Robo, how do you type equations into your posts? I know it has something to do with a tool on Wikipedia, but can you show me how to do it? Thanks.

I was wondering the same thing. You're the first person (aside from myself) to solve my avatar.

 

[RoboMastodon]

Hey Robo, how do you type equations into your posts? I know it has something to do with a tool on Wikipedia, but can you show me how to do it? Thanks.

go to www.wikipedia.com search for sandbox, go to wikipedia:sandbox.
and use that.
<math>\int_a^b f(x^2) \frac{\partialD g(x^3,y)}{\partialD x} dx</math>
or you can download and install LaTeX
but I've never gone through the trouble of doing it.

 

[RoboMastodon]

I was wondering the same thing. You're the first person (aside from myself) to solve my avatar.

I believe Euler beat us both by about 200 years.

 

[Yamialpha]

I believe Euler beat us both by about 200 years.

Actually I was just referring to people on CF. Thank you for the tip about Wikipedia, btw.

 

[MartinM]

Alternatively, you can use this handy little applet: http://sciencesoft.at/index.jsp?link=latex&lang=en

There are also numerous programs written for converting LaTeX equations directly to various image formats, though most are for Unix or Linux.

 

[Angel of God]

Here's a pattern for double digit numbers. I'll represent the larger number as "x" and the smaller number as "y".
So for 63, x=6 and y=3. What I found is...

(x^2-y^2)/(x+y)=x-y

So with 63,

(6^2-3^2)/(6+3)=6-3

Are there any anomalies to this pattern? Is it worth pursuing?

All the numbers are a solution.
(x^2-y^2)=(x+y)*(x-y)

therefore:

(x+y)*(x-y)/(x+y)=x-y
x-y=x-y

0=0 all the numbers for x and y are a solution.


It's very simple!

 

[arunma]

Thanks Robo, I'll try that. Maybe I'll prove the Fundamental Theorem of Calculus...just for fun.