Thread: I noticed a pattern in 7^n, expressed as a summation series.

So 1 / 7 ='s 0.14285714285714285714285714285714...

and I was looking and going yeah... 14, 28, 56, but it becomes 57 instead.

So began to overlap the numbers like this,

.14
.0028
.000056
.00000112
.0000000224
.000000000448
.00000000000896
.0000000000001792
.000000000000003584
.00000000000000007168 +
----------------------------------
.14285714285714285714285714285714...

and that's what I got.

I can now express 7 to any power with a little series I developed from this, which I call the ' Highway to Seven '.

7^(-1) = (14 * 10^(- 2)) + (28 * 10^(- 4)) + (56 * 10^(- 6)) + (112 * 10^(- 8)) + (224 * 10^(- 10)) + (448 * 10^(- 12)) + ...

which is actually

7^(-1) = (2^1 * 7 * 10^(- 2)) + (2^2 * 7 * 10^(- 4)) + (2^3 * 7 * 10^(- 6)) + (2^4 * 7 * 10^(- 8)) + (2^5 * 7 * 10^(- 10)) + (2^6 * 7 * 10^(- 12)) + ...

which is

7^(-1) = (2^1 * 7^(-1+2) * 10^(- 2)) + (2^2 * 7^(-1+2) * 10^(- 4)) + (2^3 * 7^(-1+2) * 10^(- 6)) + (2^4 * 7^(-1+2) * 10^(- 8)) + ...

which becomes

7^(n) = (2^1 * 7^(n+2) * 10^(- 2)) + (2^2 * 7^(n+2) * 10^(- 4)) + (2^3 * 7^(n+2) * 10^(- 6)) + (2^4 * 7^(n+2) * 10^(- 8)) + ...

and since

(2^1 * 10^(- 2)) = 1 / 50 or .02

the series then becomes

7^(n) = (7^(n+2) * .02^1) + (7^(n+2) * .02^2) + (7^(n+2) * .02^3) + (7^(n+2) * .02^4) + ...

or rather

7^(n) = (A * B^1) + (A * B^2) + (A * B^3) + (A * B^4) + ...

Where A = 7^(n+2) , B = .02 or 50^-1

And then, if I factor out the A^2, it becomes

A = A * ( (B^1) + (B^2) + (B^3) + (B^4) + ... )

Which I thought was pretty neat since I found that the equation also works with positive or negative square roots or fractional exponents.

For example:

7^(-3/2) = 0.0539949247156... = (2^1 * 7^(1/2) * 10^(- 2 * 1)) + 2^2 * 7^(1/2) * 10^(- 2 * 2) + 2^3 * 7^(1/2) * 10^(- 2 * 3) + 2^4 * 7^(1/2) * 10^(- 2 * 5) + 2^5 * 7^(1/2) * 10^(- 2 * 6) + 2^6 * 7^(1/2) * 10^(- 2 * 6) + 2^7 * 7^(1/2) * 10^(- 2 * 7) + ...


I'm not quite sure I remember how to put this equation in proper Summation form with the big M that fell on it's side 90 degrees CCW.

I thought there was also a simpler way to express ( (B^1) + (B^2) + (B^3) + (B^4) + ... )

Any help or suggestions would be appreciated.

I wonder if this kind of equation can apply to any other numbers besides seven ?

[

Originally Posted by Jeffrey DreamKing

7^(n) = (A * B^1) + (A * B^2) + (A * B^3) + (A * B^4) + ...

Where A = 7^(n+2) , B = .02 or 50^-1
-
And then, if I factor out the A^2, it becomes

A = A * ( (B^1) + (B^2) + (B^3) + (B^4) + ... )

Which I thought was pretty neat since I found that the equation also works with positive or negative square roots or fractional exponents.

The [tex] [/tex] tags convert language into mathematical symbols.

LaTeX:Symbols - AoPSWiki

So your equation can be written:



However, go back through that section I quoted. You're not factoring out A squared are you? Isn't it 7 squared? And what did you do with it?

Originally Posted by Jeffrey DreamKing

Where A = 7^(n+2) , B = .02 or 50^-1

And then, if I factor out the A^2, it becomes

A = A * ( (B^1) + (B^2) + (B^3) + (B^4) + ... )

Oh yeah, good catch. I messed up the simplification. I meant to say that I factor out A instead of A^2.

Maybe I should just stay away from A and B.

The series is:

7^(n) = (7^(n+2) * .02^1) + (7^(n+2) * .02^2) + (7^(n+2) * .02^3) + (7^(n+2) * .02^4) + ...

which becomes

7^(n) = (7^(n+2)) * (.02^1) + .02^2) + .02^3) + .02^4 + ... )

and if I take (7^(n+2)) out of the right side, it becomes

7^(n) / (7^(n+2)) = (.02^1) + .02^2) + .02^3) + .02^4 + ... )

and

7^(n-(n+2)) = (.02^1) + .02^2) + .02^3) + .02^4 + ... )

is

7^(-2) = (.02^1) + .02^2) + .02^3) + .02^4 + ... )


or

1/49 = (1/50)^1 + (1/50)^2 + (1/50)^3 + ...

or

49^-1 = 50^-1 + 50^-2 + 50^-3 + ...

and the calculator says,

1/49 = 0.0204081632653

And that decimal for 1/49 rePeats after 42 places!

Algebraicly, it's easy to show that no matter what a and r are (except for r=1)



But, in particular, if r is less than 1, then (and ) becomes zero when n goes to infinity, so the formula becomes pretty simple:

49^-1 = 50^-1 + 50^-2 + 50^-3 + ...

For that one, on the right hand side, a = 1/50 and r = 1/50, so it is equal to

(1/50)/(1 - 1/50) = (1/50)/(49/50) = 1/49, as you've found.

Using the formula in the other direction, we can start with 1/7 and see that it equals (1/8)/(1 - 1/8), so

1/7 = 1/8 + (1/8)^2 + (1/8)^3 + (1/8)^4 + ...

which, BTW, repeats after 6 places.

Originally Posted by Jeffrey DreamKing

I can now express 7 to any power with a little series I developed from this, which I call the ' Highway to Seven '.

7^(-1) = (14 * 10^(- 2)) + (28 * 10^(- 4)) + (56 * 10^(- 6)) + (112 * 10^(- 8)) + (224 * 10^(- 10)) + (448 * 10^(- 12)) + ...

which is actually

7^(-1) = (2^1 * 7 * 10^(- 2)) + (2^2 * 7 * 10^(- 4)) + (2^3 * 7 * 10^(- 6)) + (2^4 * 7 * 10^(- 8)) + (2^5 * 7 * 10^(- 10)) + (2^6 * 7 * 10^(- 12)) + ...

which is

7^(-1) = (2^1 * 7^(-1+2) * 10^(- 2)) + (2^2 * 7^(-1+2) * 10^(- 4)) + (2^3 * 7^(-1+2) * 10^(- 6)) + (2^4 * 7^(-1+2) * 10^(- 8)) + ...

For that last equation, a= (2^1 * 7^(-1+2) * 10^(- 2)) and r = 2/10^2, so the formula says the right-hand side would be equal to

Originally Posted by Jeffrey DreamKing

Any help or suggestions would be appreciated.

Awesome ! Thanks Mr. Grapes. You just took this equation way farther than I could have ever gotten myself. I think you took it all the way to the end actually.

Originally Posted by Jeffrey DreamKing

I wonder if this kind of equation can apply to any other numbers besides seven ?

Sweet, it does. So now do I get to ask the question of, "What is this useful for ?" ?

How about working with primes ?

1 divided by prime number equals 1 divided by (prime number plus 1) + (1 divided by (prime number plus 1))^2 + ...

or like this,

1/p = (1/p+1) + (1/p+1)^2 + (1/p+1)^3 + ...

or is this equation useful for relating any number to any other number ?

from the previous example,

where 1/7 = 1/8 + (1/8)^2 + (1/8)^3 + (1/8)^4 + ...

1/7 = 8/56

and

1/8 = 7/56

then I have

8/56 = 7/56 + (7/56)^2 + ...

8/56 - 7/56 = 7/56 - 7/56 + (7/56)^2 + ...

1/56 = 0 + (7/56)^2 + (7/56)^3 ...

which is kind of neat because I have gotten rid of the first term in the summation series,

and since

1/56 = 1/57 + (1/57)^2 + ...

0 + (7/56)^2 + (7/56)^3 ... = 1/57 + (1/57)^2 + ...

and if I take and put 1/57 to one side of the equal sign, I get

1/57 = 0 + ( (7/56)^2 - (1/57)^2 ) + ( (7/56)^3 - (1/57)^3 ) ...

1/57 = 0 + ( 1/64 - 1/3249 ) + ( (7/56)^3 - (1/57)^3 ) ...

1/57 = 0+ ( 3185/207936 ) + ( 34296185105/8990607867641856 ) + ...

and since

1/57 = 1/58 + 1/58^2 + ...

0 + ( (7/56)^2 - (1/57)^2 ) + ( (7/56)^3 - (1/57)^3 ) ... = 1/58 + 1/58^2 + ...

0 + ( (7/56)^2 + -(1/57)^2 + -1/58^2) + ( (7/56)^3 + -(1/57)^3 + -1/58^3)) + ... = 1/58

and since

1/58 = 1/59 +...


0 + ( (7/56)^2 + -(1/57)^2 + -1/58^2) + ( (7/56)^3 + -(1/57)^3 + -1/58^3)) + ... = 1/59 +...

0 + ( (7/56)^2 + -(1/57)^2 + -1/58^2 + -1/59^2) + ( (7/56)^3 + -(1/57)^3 + -1/58^3 + -1/59^3)) + ... = 1/59

but that's more complex than simple.

If I start with 1/1,

1/1 = 1/2 + 1/2^2 + 1/2^3 +...

1/1-1/2 = 1/2-1/2 + 1/2^2 + 1/2^3 +...

1/2 = 0 + 1/2^2 + 1/2^3 +...

(which is the first time I've seen the equation become reflective where 1/n = 0 + 1/n^2 +... or rather, a = 0 + a^2 +... )

and since

1/2 = 1/3 + 1/3^2 + 1/3^3 +...
then

1/2^2 + 1/2^3 +... = 1/3 + 1/3^2 + 1/3^3 +...

and rearranges to become

(1/2^2 + -1/3^2) + (1/2^3 + -1/3^3) +... = 1/3

and it will follow that

(1/2^2 + -1/3^2 + -1/4^2) + (1/2^3 + -1/3^3 + -1/4^3) +... = 1/4

and so on, and so forth, which is boring.

If I back up to when I was dealing with 1/2 and I got the form ( a = 0 + a^2 +... )

and then I take that form and substitute it back into the 1/2 ='s 1/3 +....

where 1/3 ='s a, then

1/2 = 1/3 + 1/3^2 + 1/3^3 +...

is

a = a^2 + a^3 + a^4 +...

where 1/3 = a^2,
1/3^2 = a^3,
1/3^3 = a^4, ...

then

1/2 = (1/3^1/2)^2 + (1/3^2/3)^3 + (1/3^3/4)^4 + ...

or

1/2 = 3^(-1/2)^2 + 3^(-2/3)^3 + 3^(-3/4)^4 + ...

or

2^(-1) = 3^(-1/2)^2 + 3^(-2/3)^3 + 3^(-3/4)^4 + ...

and that's as far as that goes.

Going down the rabbit hole again,

1/2 = 1/2^2 + 1/2^3 +...

is

1/2^1 = 1/2^2 + 1/2^3 +...

1/2^1 - 1/2^2 = 1/2^2 - 1/2^2 + 1/2^3 +...

1/2^2 = 1/2^3 +...

1/2^3 = 1/2^4 +...

which is the same old same old

but if I take and start substituting backwards and branching out, and start with,

1/2 = 1/4 + 1/8 + 1/16 + 1/32 + ...

and since 1/4 = 1/8 + 1/16 + 1/32 + ...

1/2 = 1/8 + 1/16 + 1/32 + ...
+ 1/8 + 1/16 + 1/32 + ...
_____________________
2/8 + 2/16 + 2/32 +...

which is still 1/4 + 1/8 + 1/16 + 1/32 + ...

and no matter how far back I substitute, it will keep giving me that form.

If I go back to

1/1 = 1/2 + 1/2^2 + 1/2^3 +...

and start substituting with 1/2 = 1/3 +...

I can get something funny like,

1/1 = 1/3 + 1/2^2 + 1/3^2 + 1/2^3 + 1/3^2 + ...

or

1/1 = 1/3 + 1/4 + 1/8 + 1/9 + 1/16 + 1/27...

so I can see things start to spread out number-wise.

If I keep substituting, I can get,

1/1 = 1/6 + 1/7 + 1/8 + 1/9 + 1/16 + 1/17 + 1/25 + 1/26 + 1/27 + 1/32,36,37,49,64,65,81,125,126,128,...

so maybe the more I substitute, the more even of a spread there will be and with less gaps between the first numbers of the series.

And I'd like to point out that ( 1/4 + 1/8 + 1/16 + 1/32 + ... ) will not equal 1, but will always be some fraction short of one, and just go on forever.

Sound familiar ? Maybe I should say that ( 1/4 + 1/8 + 1/16 + 1/32 + ... ) is equal to 0.99999999999999999999999999... ?

Clipboard getting fuller, computer getting slower, must sleep...

Originally Posted by Jeffrey DreamKing

Awesome ! Thanks Mr. Grapes.

de nada

Sound familiar ? Maybe I should say that ( 1/4 + 1/8 + 1/16 + 1/32 + ... ) is equal to 0.99999999999999999999999999... ?

Yes, yes, of course it is.