Mixing
Hacking the Sum Index to work with Rational Index [on hold]
Clash Royale CLAN TAG#URR8PPP
up vote
-2
down vote
favorite
I hope you will find this interesting, and pls answer me if any error or some more comment must be added to let this thing be published on ArXiV
Hacking the Sum Index to work with Rational Index:
First of all: instead using the "mute" index "i" we need to use the new "talking" index "X" that is exactly the Ordinate on a Cartesian Plane,
Knowing that you can write any n-th Power of an Integer A as:
$$
A^n= sum_X=1^A (X^n-(X-1)^n)
$$
where we can call: $$M_n=(X^n-(X-1)^n)$$ the Complicate Modulus, or the First Integer Derivate (since on the Cartesian Plane you can see that such Columns are capable to square the area of the First Derivate).
Here how the telescoping sum property is used to square the derivate y'=3x^2 with rectangular columns (called gnomons) having y'_N =3x^2-3x+1 as height and 1 as base. M_n= y'_N can be for so called the Integer First Derivate
Here is the property allow the telecoping sum works: Missing Area is equal to the Exceeding one
Now for the known property of the Sum we can also write:
$$
a^n = (A/K)^n = sum_X=1^A (X^n-(X-1)^n)/K^n
$$
with $ainmathbbQ$
Now taking as example the case n=2,
$$
a^2 = (A/K)^2 = sum_X=1^A 2X/K^2 + 1/K^2
$$
We prove we can Sum Terms resulting by Rational Index Just, since we can apply this variable exchange:
$$ x=X/K$$
Remembering that new Lower limit becomes: $LL= 1/K$
that new Upper limit becomes: $UL= a = A/K$
The new index of the Sum are now: 1/K, 2/K,3/K..... a
And the RATIONAL STEP SUM becomes:
$$
a^2 = (A/K)^2 = sum_X=1^A 2X/K^2 + 1/K^2 = sum_x=1/k^a 2x/K + 1/K^2
$$
Here is why we are allowed to sum Rational Index
Now we can pull $$Ktoinfty$$ having:
$$
a^2 = lim_Ktoinfty sum_x=1/k^a 2x/K + 1/K^2 = int_0^a 2x dx
$$
Where we are now able to rise also an Upper limit that is $$ainmathbbR-Q$$
More in general using the Proper Rational Complicate Modulus $$M_n,K$$ (or what can be called Y'_Q so the Rational First Derivate).
$$
M_n,K = n choose 1x^n-1/K - n choose 2x^n-2/K^2 + n choose 3x^n-3/K^3 -... +/- frac1K^n
$$
For example for $n=3$ is:
$$
M_3,K= 3x^2/K - 3x/K^2 + 1/K^3
$$
So we can write that:
$$
a^n= sum_x=1/k^a M_n,K = lim_Ktoinfty sum_x=1/k^a M_n,K = int_0^a n x^n-1 dx
$$
It was a long trip for me to set-up this stuff that, I hope, will be usefull to better explain several "Power Problems"
Here is how to behave this trick from N to R
proof-verification
asked Aug 3 at 7:18
Stefano Maruelli
254
put on hold as unclear what you're asking by user133281, John Ma, Somos, Adrian Keister, Shailesh Aug 4 at 1:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |Â
up vote
-2
down vote
favorite
I hope you will find this interesting, and pls answer me if any error or some more comment must be added to let this thing be published on ArXiV
Hacking the Sum Index to work with Rational Index:
First of all: instead using the "mute" index "i" we need to use the new "talking" index "X" that is exactly the Ordinate on a Cartesian Plane,
Knowing that you can write any n-th Power of an Integer A as:
$$
A^n= sum_X=1^A (X^n-(X-1)^n)
$$
where we can call: $$M_n=(X^n-(X-1)^n)$$ the Complicate Modulus, or the First Integer Derivate (since on the Cartesian Plane you can see that such Columns are capable to square the area of the First Derivate).
Here how the telescoping sum property is used to square the derivate y'=3x^2 with rectangular columns (called gnomons) having y'_N =3x^2-3x+1 as height and 1 as base. M_n= y'_N can be for so called the Integer First Derivate
Here is the property allow the telecoping sum works: Missing Area is equal to the Exceeding one
Now for the known property of the Sum we can also write:
$$
a^n = (A/K)^n = sum_X=1^A (X^n-(X-1)^n)/K^n
$$
with $ainmathbbQ$
Now taking as example the case n=2,
$$
a^2 = (A/K)^2 = sum_X=1^A 2X/K^2 + 1/K^2
$$
We prove we can Sum Terms resulting by Rational Index Just, since we can apply this variable exchange:
$$ x=X/K$$
Remembering that new Lower limit becomes: $LL= 1/K$
that new Upper limit becomes: $UL= a = A/K$
The new index of the Sum are now: 1/K, 2/K,3/K..... a
And the RATIONAL STEP SUM becomes:
$$
a^2 = (A/K)^2 = sum_X=1^A 2X/K^2 + 1/K^2 = sum_x=1/k^a 2x/K + 1/K^2
$$
Here is why we are allowed to sum Rational Index
Now we can pull $$Ktoinfty$$ having:
$$
a^2 = lim_Ktoinfty sum_x=1/k^a 2x/K + 1/K^2 = int_0^a 2x dx
$$
Where we are now able to rise also an Upper limit that is $$ainmathbbR-Q$$
More in general using the Proper Rational Complicate Modulus $$M_n,K$$ (or what can be called Y'_Q so the Rational First Derivate).
$$
M_n,K = n choose 1x^n-1/K - n choose 2x^n-2/K^2 + n choose 3x^n-3/K^3 -... +/- frac1K^n
$$
For example for $n=3$ is:
$$
M_3,K= 3x^2/K - 3x/K^2 + 1/K^3
$$
So we can write that:
$$
a^n= sum_x=1/k^a M_n,K = lim_Ktoinfty sum_x=1/k^a M_n,K = int_0^a n x^n-1 dx
$$
It was a long trip for me to set-up this stuff that, I hope, will be usefull to better explain several "Power Problems"
Here is how to behave this trick from N to R
proof-verification
asked Aug 3 at 7:18
Stefano Maruelli
254
put on hold as unclear what you're asking by user133281, John Ma, Somos, Adrian Keister, Shailesh Aug 4 at 1:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
I already post in the edit title: I received 2 dislike since it was not posted with the right tag: proof verification ? I know the subject will be acid, but I can confirm that in more than 10 years of work on, all seems now in the right place...
– Stefano Maruelli
2 days ago
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I hope you will find this interesting, and pls answer me if any error or some more comment must be added to let this thing be published on ArXiV
Hacking the Sum Index to work with Rational Index:
First of all: instead using the "mute" index "i" we need to use the new "talking" index "X" that is exactly the Ordinate on a Cartesian Plane,
Knowing that you can write any n-th Power of an Integer A as:
$$
A^n= sum_X=1^A (X^n-(X-1)^n)
$$
where we can call: $$M_n=(X^n-(X-1)^n)$$ the Complicate Modulus, or the First Integer Derivate (since on the Cartesian Plane you can see that such Columns are capable to square the area of the First Derivate).
Here how the telescoping sum property is used to square the derivate y'=3x^2 with rectangular columns (called gnomons) having y'_N =3x^2-3x+1 as height and 1 as base. M_n= y'_N can be for so called the Integer First Derivate
Here is the property allow the telecoping sum works: Missing Area is equal to the Exceeding one
Now for the known property of the Sum we can also write:
$$
a^n = (A/K)^n = sum_X=1^A (X^n-(X-1)^n)/K^n
$$
with $ainmathbbQ$
Now taking as example the case n=2,
$$
a^2 = (A/K)^2 = sum_X=1^A 2X/K^2 + 1/K^2
$$
We prove we can Sum Terms resulting by Rational Index Just, since we can apply this variable exchange:
$$ x=X/K$$
Remembering that new Lower limit becomes: $LL= 1/K$
that new Upper limit becomes: $UL= a = A/K$
The new index of the Sum are now: 1/K, 2/K,3/K..... a
And the RATIONAL STEP SUM becomes:
$$
a^2 = (A/K)^2 = sum_X=1^A 2X/K^2 + 1/K^2 = sum_x=1/k^a 2x/K + 1/K^2
$$
Here is why we are allowed to sum Rational Index
Now we can pull $$Ktoinfty$$ having:
$$
a^2 = lim_Ktoinfty sum_x=1/k^a 2x/K + 1/K^2 = int_0^a 2x dx
$$
Where we are now able to rise also an Upper limit that is $$ainmathbbR-Q$$
More in general using the Proper Rational Complicate Modulus $$M_n,K$$ (or what can be called Y'_Q so the Rational First Derivate).
$$
M_n,K = n choose 1x^n-1/K - n choose 2x^n-2/K^2 + n choose 3x^n-3/K^3 -... +/- frac1K^n
$$
For example for $n=3$ is:
$$
M_3,K= 3x^2/K - 3x/K^2 + 1/K^3
$$
So we can write that:
$$
a^n= sum_x=1/k^a M_n,K = lim_Ktoinfty sum_x=1/k^a M_n,K = int_0^a n x^n-1 dx
$$
It was a long trip for me to set-up this stuff that, I hope, will be usefull to better explain several "Power Problems"
Here is how to behave this trick from N to R
proof-verification
asked Aug 3 at 7:18
Stefano Maruelli
254
I hope you will find this interesting, and pls answer me if any error or some more comment must be added to let this thing be published on ArXiV
Hacking the Sum Index to work with Rational Index:
First of all: instead using the "mute" index "i" we need to use the new "talking" index "X" that is exactly the Ordinate on a Cartesian Plane,
Knowing that you can write any n-th Power of an Integer A as:
$$
A^n= sum_X=1^A (X^n-(X-1)^n)
$$
where we can call: $$M_n=(X^n-(X-1)^n)$$ the Complicate Modulus, or the First Integer Derivate (since on the Cartesian Plane you can see that such Columns are capable to square the area of the First Derivate).
Here how the telescoping sum property is used to square the derivate y'=3x^2 with rectangular columns (called gnomons) having y'_N =3x^2-3x+1 as height and 1 as base. M_n= y'_N can be for so called the Integer First Derivate
Here is the property allow the telecoping sum works: Missing Area is equal to the Exceeding one
Now for the known property of the Sum we can also write:
$$
a^n = (A/K)^n = sum_X=1^A (X^n-(X-1)^n)/K^n
$$
with $ainmathbbQ$
Now taking as example the case n=2,
$$
a^2 = (A/K)^2 = sum_X=1^A 2X/K^2 + 1/K^2
$$
We prove we can Sum Terms resulting by Rational Index Just, since we can apply this variable exchange:
$$ x=X/K$$
Remembering that new Lower limit becomes: $LL= 1/K$
that new Upper limit becomes: $UL= a = A/K$
The new index of the Sum are now: 1/K, 2/K,3/K..... a
And the RATIONAL STEP SUM becomes:
$$
a^2 = (A/K)^2 = sum_X=1^A 2X/K^2 + 1/K^2 = sum_x=1/k^a 2x/K + 1/K^2
$$
Here is why we are allowed to sum Rational Index
Now we can pull $$Ktoinfty$$ having:
$$
a^2 = lim_Ktoinfty sum_x=1/k^a 2x/K + 1/K^2 = int_0^a 2x dx
$$
Where we are now able to rise also an Upper limit that is $$ainmathbbR-Q$$
More in general using the Proper Rational Complicate Modulus $$M_n,K$$ (or what can be called Y'_Q so the Rational First Derivate).
$$
M_n,K = n choose 1x^n-1/K - n choose 2x^n-2/K^2 + n choose 3x^n-3/K^3 -... +/- frac1K^n
$$
For example for $n=3$ is:
$$
M_3,K= 3x^2/K - 3x/K^2 + 1/K^3
$$
So we can write that:
$$
a^n= sum_x=1/k^a M_n,K = lim_Ktoinfty sum_x=1/k^a M_n,K = int_0^a n x^n-1 dx
$$
It was a long trip for me to set-up this stuff that, I hope, will be usefull to better explain several "Power Problems"
Here is how to behave this trick from N to R
proof-verification
asked Aug 3 at 7:18
Stefano Maruelli
254
edited 2 days ago
asked Aug 3 at 7:18
Stefano Maruelli
254
asked Aug 3 at 7:18
Stefano Maruelli
254
asked Aug 3 at 7:18
Stefano Maruelli
254
254
put on hold as unclear what you're asking by user133281, John Ma, Somos, Adrian Keister, Shailesh Aug 4 at 1:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by user133281, John Ma, Somos, Adrian Keister, Shailesh Aug 4 at 1:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
I already post in the edit title: I received 2 dislike since it was not posted with the right tag: proof verification ? I know the subject will be acid, but I can confirm that in more than 10 years of work on, all seems now in the right place...
– Stefano Maruelli
2 days ago
add a comment |Â
I already post in the edit title: I received 2 dislike since it was not posted with the right tag: proof verification ? I know the subject will be acid, but I can confirm that in more than 10 years of work on, all seems now in the right place...
– Stefano Maruelli
2 days ago
I already post in the edit title: I received 2 dislike since it was not posted with the right tag: proof verification ? I know the subject will be acid, but I can confirm that in more than 10 years of work on, all seems now in the right place...
– Stefano Maruelli
2 days ago
I already post in the edit title: I received 2 dislike since it was not posted with the right tag: proof verification ? I know the subject will be acid, but I can confirm that in more than 10 years of work on, all seems now in the right place...
– Stefano Maruelli
2 days ago
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
I already post in the edit title: I received 2 dislike since it was not posted with the right tag: proof verification ? I know the subject will be acid, but I can confirm that in more than 10 years of work on, all seems now in the right place...
– Stefano Maruelli
2 days ago