Question

Use the approximation e^x is about 1+x+(x^2/2) to estimate e^0.5. The Lagrange remainder will be no greater than what value?

Answer 1

\[1+x+(x^2/2)+ e ^{\theta}x^{3}/6\] this is the aproximate for exponential function with lagrange reminder, where 0< θ < 0.5.
So for 0.5 the aproximate is \[1 + 1/2 + (1/2^{2})/2 = 1 + 1/2 + 1/8\]
The error is less than \[e ^{0.5}1/2^{3}/6 = e ^{0.5}/48\]

Answer 2

ok! thanks so much!