because when we look at the first derivative we multiply it by h, next term (second derivative) cancels and we have next but one term that multiplies by h^3 what means h*h^2 and when we divide it by h, we have h^2
QUESTION: Does having higher Order of ERROR (here O(h^2)) mean that I can diffrentiate a Funtion Upto That order (2) without having any error OR habign the Minimum numer of Order for a Function in order to Have Maximum ACCURACY? can anyone please Explain...thank you
THANKYOU SO MUCH SIR . I AM ABLE TO SCORE COMPLETE 6 MARKS IN MY BSC HONS MATHEMATICS EXAM OF DU ...... GOT 98 OUT OF 100
wow that explanation was soooooo much better than my professor. Thanks!
that has been amazingly helpful, thanks for the effort
Thank you for this useful instruction🌹🌹🌹
how about getting the third and fourth derivative?
I appreciate your work sir
Great explanation sir thank u
great video. Thanks
What about 3rd derivative sir?
what if I wanted a finite difference of a priorly selected Error? say the first derivative of Oh^4 Error?
Okay I might be dumb… but is this man writing backwards or how does this work
Software
He probably mirrored the video in post
Great explanation sir!, but i still dont understand why its equal to O(h^2), please sir or anyone let me know the explanation
because when we look at the first derivative we multiply it by h, next term (second derivative) cancels and we have next but one term that multiplies by h^3 what means h*h^2 and when we divide it by h, we have h^2
QUESTION: Does having higher Order of ERROR (here O(h^2)) mean that I can diffrentiate a Funtion Upto That order (2) without having any error OR habign the Minimum numer of Order for a Function in order to Have Maximum ACCURACY? can anyone please Explain...thank you
Thank you 😊
Thank you
my lord - excellent -which country ? which university ? amarjit india
HKUST, Hong Kong
Thanks proff
Nice
Elegant