Euler's summation formula : Number Theory #Euler's #summation #formula #: #Number #Theory
Euler's summation formula :
Statement :
If f(x) has
continuous derivative f’ on [a,b] where 0 < a< b , then
𝜮a<n≤b f(n) = ∫
f(t)dt + ∫ (t – [t] ) dt +f(b)( [b] – b) -f(a) ( [a] -a ).
Proof :
Suppose f(x) has
continuous derivative and f’(x) is in the closed interval [a,b] where 0 < a <
b.
Let [a] =m and [b] = k.
Then 𝜮a<n≤b f(n) = 𝜮m<n≤k f(n) = 𝜮 f(n) …(i)
| where in third sigma n is from m+1 to k |
Suppose (n-1) and n
are two integers in [a,b] and t lies between (n-1) and n.
Then ∫n-1n
[t] f’(t) dt = ∫ (n-1) f’(t) dt
= (n-1) ∫ f’(t) dt
= (n-1)
[f(t)] n-1n
= (n-1) ( f(n) – f(n-1)
= n f(n) –
f(n) – (n-1) f(n-1) … (ii)
Taking summation with n = m+1, m+2, … , k on both sides of
(ii) , we get
𝜮 ∫n-1n [t] f’(t) dt = 𝜮 (n f(n) – f(n)
– (n-1) f(n-1) )
⇒ 𝜮 ∫n-1n
[t] f’(t) dt = k f(k) – m f(m) – 𝜮 f(n)
⇒
𝜮 f(n) = k f(k) – m f(m) – 𝜮 ∫n-1n
[t] f’(t) dt
= k f(k) – m f(m) – ∫ab
[t] f’(t) dt
= k f(b) – m f(a) – ∫ab
[t] f’(t) dt … (iii)
Now ∫ab
f(t) dt = ∫ab 1. f(t) dt = b f(b) -a f(a) - ∫0a
t f’(t) dt
∫ab
f(t) dt - b f(b) + a f(a) + ∫0a
t f’(t) dt = 0 … (iv)
From ( iii ) , 𝜮 f(n) = k f(b) – m f(a) – ∫ab [t] f’(t) dt + 0
𝜮 f(n) = k f(b) – m f(a) – ∫ab [t] f’(t) dt + ∫ab
f(t) dt - b f(b) + a f(a) + ∫0a
t f’(t) dt … ( v )
| since replace o by L.H.S. of (iv) |
Now 𝜮
f(n) = ∫ab f(t)dt
+ ∫ab (t – [t] ) dt +f(b)( k – b) -f() ( m -a )
𝜮 f(n) = ∫ab f(t)dt + ∫ab
(t – [t] ) dt +f(b)( [b] – b) -f(a) ( [a] -a ).
*** Hence the
Proof ***
* Fundamental Theorem of Arithmetic
* Historical Introduction to Number Theory
* The Mobius Function 𝝻 ( n ) .
#Euler's #summation #formula #: #Number #Theory

Comments
Post a Comment