We know that: e^x=sum_(n=0)^(oo) (x^n)/(n!) log(x)=sum_(n=1)^(oo) ((-1)^(n+1)(x−1)^n)/(n) Where second series converges when |x−1|<1.It is possible to prove that: e^(log(x))=x for |x−1|<1 using only series representation?

sondestiny120g

sondestiny120g

Answered question

2022-08-11

We know that:
e x = n = 0 x n n !
log ( x ) = n = 1 ( 1 ) n + 1 ( x 1 ) n n
Where second series converges when |x−1|<1.It is possible to prove that:
e log ( x ) = x
for | x 1 | < 1 using only series representation?

Answer & Explanation

neglegir86

neglegir86

Beginner2022-08-12Added 12 answers

log ( x ) = n = 1 ( 1 ) n + 1 ( x 1 ) n n ( x 1 ) ( x 1 ) 2 2 + O ( x 3 ) e log ( x ) = n = 0 log n ( x ) n ! 1 + log ( x ) + log 2 ( x ) 2 + O ( log 3 ( x ) ) 1 + [ ( x 1 ) ( x 1 ) 2 2 ] + [ ( x 1 ) 2 2 + ( x 1 ) 4 2 ( x 1 ) 3 ] x ( x 1 ) 3 + ( x 1 ) 4 2
As | x 1 | < 1, the leading term is the one with the smallest exponent; therefore:
e log ( x ) x

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