In mathematics, a telescoping series is a series whose general term
tn
tn=an+1-an
(an)
(an)
The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences.
An early statement of the formula for the sum or partial sums of a telescoping series can be found in a 1644 work by Evangelista Torricelli, De dimensione parabolae.[3]
Telescoping sums are finite sums in which pairs of consecutive terms partly cancel each other, leaving only parts of the initial and final terms.[4] Let
an
an
L
Every series is a telescoping series of its own partial sums.[5]
A | |
A | |
i=1 |
2i-1
And then expanding the telescoping sum we have:
a | |
\sum | |
i=1 |
{(2i-1)}=a2-\cancel{(a-1)2}+\cancel{(a-1)2}-\cancel{(a-2)2}+\cancel{(a-2)2}-...+\cancel{1}-\cancel{1}=a2
So for example
a=5
a2=1+3+5+7+9=25
From where the general formula for any n-th power of an integer A:
A | |
A | |
X=1 |
(Xn-(X-1)n)
This open a breach in the forgotten math fact that any parabola (or polynomial) subtend an area from 0 to an integer abscissa A that can be squared via integral or via a finite sum since using X instead of i, and representing what we are doing on the Cartesian plane will be immediately clear that it is possible to apply an exchange of variable
x=X/K
Step=integer
index
scalefactor=1/K
The new Scaling Rule (holding the same physical Area): From Sum of Integers to Sum of Rationals, then to the Limit
More in general, remembering the new definition for the Sum operator as given into the Abstract, we can first use and push the telescoping Sum properties to the limit (then talk in a modular like concept) to show How to refine a Sum, so working to have at the end not just the same numerical value, but the same physical Area (in square meter f.ex.) squareing with a finite number or rectangles called Gnomons, the Area Below the 1st Derivative of a Parabola
Y=Xn
A\inN+ |
K
Remembering that:
A2*K2=
A ⋅ K | |
\sum | |
X=1 |
(2X-1)
More in general:
An*Kn=
A ⋅ K | |
\sum | |
X=1 |
(Xn-(X-1)n)
Thanks to the known distributive property for the Sum we can left unchanged the value of the Sum if we multiply all the sum by an unitary (in this case quadratic) factor
1=
K2 | |
K2 |
A2=
A | |
\left(\sum | |
X=1 |
2X-1\right)
K2 | |
K2 |
=
A*K | |
\sum | |
X=1 |
2X-1 | |
K2 |
then I'll show how to apply the exchange of variable
X=x*K
A2=
A*K | |
\sum | |
X=1 |
2X-1 | |
K2 |
=
| ||||
\sum | ||||
x=1/K |
{\cancel{K}}}
2x ⋅ \cancel{K | |
And I hope is now clear why it is used
X
i
IF and only IF (IFF) the Upper Limit
A\inN+ |
A | |
A | |
X=1 |
A | ||
2X-1=\sum | \left( | |
x=1/K |
2x | - | |
K |
1 | |
K2 |
\right)=\limK\toinfty
A | ||
\sum | \left( | |
x=1/K |
2x | - | |
K |
1 | |
K2 |
\right)=
A | |
\int | |
0 |
2xdx=A2
That shows what I will call: the Distribution for Power Terms Law, that works for the n-th power in this way:
An=
A | |
\sum | |
X=1 |
Mn=
A | |
\sum | |
x=1/K |
Mn,K
where
Mn
Mn,K
Mn,K={n\choose1}
xn-1 | |
K |
-{n\choose2}
xn-2 | |
K2 |
+{n\choose3}
xn-3 | |
K3 |
-...+/-
1 | |
Kn |
The first
Mn,K
(X-1)n
(Xn-(X-1)n)
M2,K=
2x | |
K |
-
1 | |
K2 |
M3,K=
3x2 | - | |
K |
3x | + | |
K2 |
1 | |
K3 |
M4,K=
4x3 | - | |
K |
6x2 | + | |
K2 |
4x | - | |
K3 |
1 | |
K4 |
M5,K=
5x4 | - | |
K |
10x3 | - | |
K2 |
10x2 | + | |
K3 |
5x | - | |
K4 |
1 | |
K5 |
etc...
Then a list of new manipulation will reveal a new way to solve Power Problems, and conditions let some equality be possible or not, so True or False (as Fermat The Last for
n=2
n<2
A | |
A | |
X=1 |
(Xn-(X-1)
A | |
x=1/K |
({n\choose1}
xn-1 | |
K |
-{n\choose2}
xn-2 | |
K2 |
+{n\choose3}
xn-3 | |
K3 |
-...+/-
1 | |
Kn |
)
This also leads to 2 new sum properties for sum of polynomials can be shownf.ex. into this equalities... it seems no mathematician want to admit...
A | |
A | |
1 |
A | |
3X | |
x=1/K |
3x2 | - | |
K |
3x | + | |
K2 |
1 | |
K3 |
B | |
=\sum | |
x=A/B |
3Ax2 | - | |
B |
3xA2 | + | |
B2 |
A3 | |
B3 |
or
B | |
A | |
x=A/B |
3Ax2 | - | |
B |
3xA2 | + | |
B2 |
A3 | |
B3 |
B | |
=\sum | |
X=1 |
3A | \left( | |
B |
A | |
B |
X\right)2-
3A2 | \left( | |
B2 |
A | X\right)+ | |
B |
A3 | |
B3 |
That onto the Right Hand of Fermat the Last let one write:
C3-B3=
C-B | |
\sum | |
x=1/K |
3(x+B)2 | - | |
K |
3(x+B) | + | |
K2 |
1 | |
K3 |
C-B | |||||
=\sum | \left( | ||||
|
3(C-B)(x+B)2 | - | |
A |
3(C-B)2(x+B) | + | |
A2 |
(C-B)3 | |
A3 |
\right)
| |||||
=\sum | |||||
|
| - | |||||
A |
| + | ||||||||||
A2 |
(C-B)3 | |
A3 |
so:
C3-B
A | |
X=1 |
| - | |||||
A |
| + | ||||||||||
A2 |
(C-B)3 | |
A3 |
C3-B3=
A | ||||
\sum | ||||
|
| - | |||||
AK |
| + | ||||||||||
(AK)2 |
(C-B)3 | |
(AK)3 |
=
C3-B3=
A | ||||
\sum | ||||
|
| - | |||||
A2 |
| + | ||||||||||
(A2)2 |
(C-B)3 | |
(A2)3 |
=
C3-B3=
A2 | |
\sum | |
X=1 |
| - | |||||
A2 |
| + | ||||||||||
(A2)2 |
(C-B)3 | |
(A2)3 |
a
r
(1-r)
|r|<1,
|r|<1,
is the series of reciprocals of pronic numbers, and it is recognizable as a telescoping series once rewritten in partial fraction form
infty | |
\begin{align} \sum | |
n=1 |
1 | |
n(n+1) |
&{}=
infty | |
\sum | |
n=1 |
\left(
1 | |
n |
-
1 | |
n+1 |
\right)\\ {}&{}=\limN\toinfty
N | |
\sum | |
n=1 |
\left(
1 | |
n |
-
1 | |
n+1 |
\right)\\ {}&{}=\limN\toinfty\left\lbrack{\left(1-
1 | |
2 |
\right)+\left(
1 | |
2 |
-
1 | |
3 |
\right)+ … +\left(
1 | |
N |
-
1 | |
N+1 |
\right)}\right\rbrack\\ {}&{}=\limN\toinfty\left\lbrack{1+\left(-
1 | |
2 |
+
1 | |
2 |
\right)+\left(-
1 | |
3 |
+
1 | |
3 |
\right)+ … +\left(-
1 | |
N |
+
1 | |
N |
\right)-
1 | |
N+1 |
}\right\rbrack\\ {}&{}=\limN\toinfty\left\lbrack{1-
1 | |
N+1 |
}\right\rbrack=1. \end{align}
where Hk is the kth harmonic number.
≠
!
which does not converge as
In probability theory, a Poisson process is a stochastic process of which the simplest case involves "occurrences" at random times, the waiting time until the next occurrence having a memoryless exponential distribution, and the number of "occurrences" in any time interval having a Poisson distribution whose expected value is proportional to the length of the time interval. Let Xt be the number of "occurrences" before time t, and let Tx be the waiting time until the xth "occurrence". We seek the probability density function of the random variable Tx. We use the probability mass function for the Poisson distribution, which tells us that
\Pr(Xt=x)=
(λt)xe-λ | |
x! |
,
x
t
t
xth
\begin{align} f(t)&{}=
d | |
dt |
\Pr(Tx\let)=
d | |
dt |
\Pr(Xt\gex)=
d | |
dt |
(1-\Pr(Xt\lex-1))\ \\ &{}=
d | |
dt |
\left(1-
x-1 | |
\sum | |
u=0 |
\Pr(Xt=u)\right) =
d | |
dt |
\left(1-
x-1 | |
\sum | |
u=0 |
(λt)ue-λ | |
u! |
\right)\ \\ &{}=λe-λ-e-λ
x-1 | |
\sum | |
u=1 |
\left(
λutu-1 | |
(u-1)! |
-
λu+1tu | |
u! |
\right) \end{align}
f(t)=
λxtx-1e-λ | |
(x-1)! |
.
For other applications, see:
A telescoping product is a finite product (or the partial product of an infinite product) that can be canceled by the method of quotients to be eventually only a finite number of factors.[8] [9] It is the finite products in which consecutive terms cancel denominator with numerator, leaving only the initial and final terms. Let
an
an
For example, the infinite product[8] simplifies as