Question

approximate the real root of
2x=1/e^x

Answer 1

there are a number of algorithms for this...one simple one is to write it as x = 1/(2e^x) and the algorithm is x_n = 1/2 * 1/e^(x_n-1)

so guess x_0 = 1: x_1 = 1/2

so x_2 = 1/2 * 1/e^(-1/2) = 1/2 * e^(1/2)

so x_3 = 1/2 * 1/e^(1/2*e^(1/2) - 1)

so after the third iteration we get an approximation of x = 0.5960

Answer 2

oh wait...oops, lemme try again:

x_1 = 1/2e

x_2 = 1/2 * 1/e^(1/2e) = 0.4160

x_3 = 1/2 * 1/e^(0.4160) = 0.3298

x_4 = 1/2 * 1/e^(0.3298) = 0.3595

and so on

Answer 3

It's a multiple choice question and these are the possible answers
.352, .335, .352, .355, or 3.521

Answer 4

x_5 = 1/2 * 1/e^(0.3595) = 0.3490

x_6 = 1/2 * 1/e^(0.3490) = 0.3527

x_7 = 1/2 * 1/e^(0.3527) = 0.3514

Answer 5

x_9 = 0.3517

Answer 6

so after 9 iterations, the approximation seems to be roughly 0.3517, which is roughly 0.352

Answer 7

you can also use the newton-rhapsodian algorithm to approximate this

the exact solution is given by W(1/2), where W is the lambert W function

Answer 8

oh ok thanks :)

Answer 9

did you need to use an algorithm or could you just enter this into a calculator? LOL

Answer 10

I think it could've been entered into a calculator, but my teacher wants us to show all our work

Answer 11

using newton-rhapson:

x_n+1 = x_n - f(x_n)/f'(x_n)

so x_n+1 = x_n - (2x_ne^x_n - 1)/2e^x_n(x_n+1)

if we guess x_0 = 1:

x_1 = 0.5920
x_2 = 0.3939
x_3 = 0.3532
x_4 = 0.3517

that's a lot faster

Answer 12

Yeah, that is a lot faster, thanks for the help