[math-fun] Exponential function by limits approach:
Hello, A formula of this type is the following known formula: e^(z) = (1+z/n)^n ; n ---- infinity; I found the following formulas: 1) e^(z)=(2-sqrt(1-2*z/n))^n; 2) e^(z)=(3/2+(1/n)*(sqrt(13*n^6-30*z*n^3+36*z^2)/(4)+(12*z-5*n^3)/(8))^(1/3)-3*n/(4*((sqrt(13*n^6-30*z*n^3+36*z^2)/(4)+(12*z-5*n^3)/(8))^(1/3))))^(n^3); 3) e^(z)=(3/2+(1/n)*(sqrt(13*n^14-30*z*n^7+36*z^2)/(4*n^4)+(12*z-5*n^7)/(8*n^4))^(1/3)-3*n/(4*((sqrt(13*n^14-30*z*n^7+36*z^2)/(4*n^4)+(12*z-5*n^7)/(8*n^4))^(1/3))))^(n^7); 4) e^(z) = ((sqrt(36*z^2-30*n^17*z+13*n^34)/(4*n^14)-(5*n^17-12*z)/(8*n^14))^(1/3)/n-(3*n)/(4*(sqrt(36*z^2-30*n^17*z+13*n^34)/(4*n^14)-(5*n^17-12*z)/(8*n^14))^(1/3))+3/2)^n^17 ; and finally 5) e^(k)=((-sqrt((160*n^3*(h-r)^(1/3)-sqrt((9*(h-r)^(2/3)-8*n^2*(h-r)^(1/3)-12*n^4+48*k)/(h-r)^(1/3))*(9*(h-r)^(2/3)+16*n^2*(h-r)^(1/3)-12*n^4+48*k))/(sqrt((9*(h-r)^(2/3)-8*n^2*(h-r)^(1/3)-12*n^4+48*k)/(h-r)^(1/3))*(h-r)^(1/3)))/6+sqrt((9*(h-r)^(2/3)-8*n^2*(h-r)^(1/3)-12*n^4+48*k)/(h-r)^(1/3))/6+n/3)/n+1)^n^4 ; with: r : (192*k*n^2-176*n^6)/27 ; h : (8*sqrt(511*n^12-1380*k*n^8+1872*k^2*n^4-1728*k^3))/27 ; n ---- infinity; FME...
participants (1)
-
françois mendzina essomba2