Minimising Without Calculus
Вставка
- Опубліковано 12 чер 2024
- We minimise 5x^2 + y^2 + 2z^2 - 2xy + 4xz + 6z without using calculus. Our approach relies on repeatedly completing the square.
00:00 Intro
00:50 Completing the square
02:43 Terms with x
04:11 Terms with z
Once you said the magic words "complete the square" it was a big enough hint for me. Very cool! Will definitely remember this idea
Simple, yet still possess the magic of math in all its glory!
Thank you for the video!
Waiting for them every Friday which makes weekend better!
Thank you, glad you enjoyed it!
Listening to the video, it finally struck me that you didn't say "zee" but zed! Thanks from Australia!
I have become convinced it becomes a zed when you cross it.
That is great, but I still wonder if he uses the long scale (milliard, not billion)
Like in France
Hello dear Dr Barker (the beautiful mind)
Just on time as always
Thanks
Thank you, good to see that you're still following the channel!
Nice work!
Is there any interesting insight to be gained from thinking in terms of the symmetric matrix of coefficients of the terms? This algorithm you presented certainly seems to work well enough.
I am going to bet a beer the eigenvalues have something to do with whether the function has a min or max or saddle.
1:06 I was so confused for a second because I thought "5x^2 + y ^2 + 2z^2" and "-2xy + 4xz + 6z" were two different expressions. Perhaps you could write such expressions in a single line in the future, or at least make it very clear that's a single expression? Thanks in advance 😅.
How would you solve this with calculus ?
The standard approach would be to take partial derivatives to look for maximum/minimum points. So differentiate everything with respect to x only, keeping y & z fixed, then set ∂f/∂x = 0. We would do the same to set ∂f/∂y = 0 and ∂f/∂z = 0, then solve these 3 simultaneous equations to find values of x, y, z which minimise the function.
Does a similar approach work with lagrange multipliers
Innovative solution! How can we generalise this for any coefficients and if any yz, x, y, or constant terms are present?
Ok I'll minimize: You're a trivial problem. You're nothing.
could this question be solved via the AM-GM inequality as well?
❤
A question- this worked magically well with this set of values. Will it work as well with other sets of values?
of course not in general, differentiate
If you just change the coefficients of the terms that appear in this expression, then the same kind of approach should still work. If you have extra terms, it takes a bit more work to figure out how to complete the squares. It can usually be done, though, if the whole thing is still quadratic.
@@ConManAU зависит от реальной поверхности которая задаётся всякими членами второй степени и ниже с разными коэффициентами,
ну может там гипербола на парабалу?
@@ConManAU if the whole thing is still quadratic
it will, but it might have no minimal value depending on signs +- in front pf the coeficients
I thought it was easier using calculus. Partially differentiate with respect to x, then y, then z, giving three equations in three unknowns. Took about five minutes.
Linear algebra come save us 😭