I think this should always hold because the linearization method in calculus is using Taylor expansion and ignoring all terms after 2nd derivative, where the term with the 1st derivative is exactly the same expression as the 2nd term in the binomial expansion of the shifted function.
How to approximate 1.999^4 whilst using a calculator: 1. type in 1.999^4 2. add 0.0000000001 to the answer, because we want an approximation, not the actual answer
Easier to use the formula directly with x = 1.999 and a= 2, that gives L(1.999) = 16 + 32*(-0.0001) = 16 - 0.032 = 15.968. Trying to simplify the general expression of L(x) leads to a more complicated calculation in this specific case.
another way of calculating 32*1.999 at the last step of the linearization, could be writing 1.999=2-(1/1000) and then proceed with applying distributive property: 32*1.999-48 = 32* (2-1/1000) - 48 = 64 - 0.032- 48 = 16 - 0.032 = 15.968
dear Just Calculus, can you please add main channel (RedPenBlackPen) into "Channels" submenu of this channel? This would allow for better navigation on mobile than "main channel" button in "About" submenu
You should have left off at L(x) = 16 + 32 (x - 2), or better, since wanting to use a number less than 2, L(x) = 16 - 32 (2 - x), then it is basically the same as the differential method...
The first was easier to compute before distributing the 32, I even think that the 2 methods look the same in that step, but as always maybe there is a detail that dont make them the same and that I am missing because I am not a mathematician
I tried this method using a random number like (2.3)^4 but i didint get the same answer as the calculator. Is it because that this method Is used for certain numbers?
If the x (2.3) is further from your first guess (2), you need more terms in the linearization process. Or you use iterative techniques, like Newton-Raphson method.
