Question Number 114802 by aurpeyz last updated on 21/Sep/20

Answered by aurpeyz last updated on 21/Sep/20

is this solution correct?

Commented bymr W last updated on 21/Sep/20

yes

Commented byaurpeyz last updated on 21/Sep/20

what if i do it this way? 1−2(3x)+((6(9x^2 ))/(2!))+... from (1+b)^(−n) =1−nb +((−n(−n−1))/(2!))b^2 is this correct too? and is there any diffrence?

Commented bymr W last updated on 21/Sep/20

expansion at x=0: Σa_k x^k expansion at x=∞: Σb_k (1/x^k ) expansion at x=c: Σc_k (x−c)^k

Commented byaurpeyz last updated on 22/Sep/20

pls can you explain this?

Commented bymr W last updated on 22/Sep/20

how about read a mathematics book by yourself?

Commented byaurpeyz last updated on 22/Sep/20

okay sir

Commented byaurpeyz last updated on 23/Sep/20

sir. with my study. i think the solution is wrong. from the solution. it is only valid if ∣(1/(3x))∣<1 which will yield x>(1/3) or x<((−1)/3) which implies the range of validity of the expansion. if i test the expansion by substituting a value in the region lf validity. x=1 (1+3(1))^(−2) ≠(1/(9(1)))−(2/(27(1)))+... but if expand it in this approach ⇒ (1+3x)^(−2) =1−6x+(((6)9x^2 )/(2!))... which is valid if the ∣3x∣<1 i.e. ∣x∣<(1/3) if x=0 [within the range of validity. ] (1+0)^(−2) =1−6(0)+...=1=1