Ask your own question, for FREE!
MIT 18.01 Single Variable Calculus (OCW) 11 Online
OpenStudy (anonymous):

in the section of Newton's method, professor draw a picture to show us that E2 is approximately equal to the square of E1(video 14, 5:20~6:00). but I don't quite get it, could anybody explains it to me.(better using notations)

OpenStudy (anonymous):

He is making some wild estimations based specifically on the picture that he has drawn; for the specific situation where x1-x is approximately equal to the height of the function at x1. This distance also has to be much less than one; one tenth in this case. He can apparently see some geometric proof in this situation (I'm assuming related to triangles). But, you can at least see what he's talking about if you use a function that satisfies his premise. Let f(x)=0.5x^2-0.5 and use x=1 with x1=1.1 and then run it through Newton's method. His overall point is that in some cases Newton's method will zero in on the answer very quickly, but that is not always the case.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!