My first idea was partial fraction decomposition, but what you have got from quotient rule Why partial fraction decomposition ? It is easy to calculate nth derivative of sum and it is easy to calculate nth derivative of geometric series Lets check what we will get if we follow St Eve's idea nth derivative will be in the form P(x)/Q(x) It is easy to predict that Q(x) = (x^3-4x)^{n+1} but will be some troubles with numerator Maybe we can guess some recurrence relation for it For numerators we can create sequence P_{0}(x) = 6x^2-4x-8 P_{n+1}(x) = P_{n}'(x)(x^3-4x) - (n+1)P_{n}(x)(3x^2-4) If we can get rid of P_{n}'(x) then we could get something useful
Another way is, after doing partial fraction decomposition, to find the taylor series of each fraction at an arbitary point, since the coefficients would be the nth derivative of the function over n!.
It would give the nth derivative evaluated at a specific value; specifically the center of the expansion. As opposed to giving the entire function like Michael did.
It only works for this nicely for non-repeating factors in the denominator. Otherwise you end up with a system of equations for the constants with shared repeated roots.
@@mcwulf25 I learned this method in AP Calculus ~15 years ago, so it definitely it not new. The partial fraction decomposition is constructed to hold for all x, so pick x values that make it simple.
Not necessarily. Consider f(x) = (x^2+1)/x for example. I imagine the repeated derivative only becomes zero if the denominator is actually a factor of the numerator.
@@russellsharpe288 That's true - after cancellation of poles and long division (if necessary), you do partial fraction decomposition. There, every remaining pole *x_0* leads to (at least) one non-vanishing term of the form *A / (x - x_0) ^ k, A != 0, k in N* Those terms will never vanish, regardless how often you differentiate them!
That expression is true for ALL x, so we just pick nice x to make calculation easy. You could have picked x=103, 40678 and 3356.49585473 and would also have found A, B AND C, but much harder.
I question why people feel it's important to prove things like this with induction, when it is already clear from how we got it and induction provides no extra insight or certainty. Who for? There are no rigor police
First day comenting this problem: find the function form Natural to natural that relationates the number of sides of a regular polygon to the numbers of points that are intersections between all the lines that connect each vertice of the polygon to all other vertices
wait .. once you have the partial fraction version, why not write each part as a taylor series, and then just add the derivatives together? (ie look to the x^2022 term to identify the coefficient, which should be (2022nd derivative at x = 0) / 2022! )
3:32 I don't understand why we can take X = 2 since the first line is equivalent to the second provided that X is not equal to 2. Same with X = 0 and X = -2.
👍 If A, B, C satisfy that equation for all values of x except for -2, 0 and 2, you can make x tend to -2 and see that, by continuity, -2 also satisfies it. Another argument is that both sides are polynomials of degree 2. If they have 3 common points they must be equal (just like there is only one straight line passing through 2 points). But we know they have in common all points except -2, 0 and 2.
There's a factor in both the numerator and denominator which cancel out. Can't you just restrict the domain to x != 2 and continue the derivative with the simpler polynomial quotient?
The decomposition holds for all x =/= -2, 0, 2 (because that is where 1/(x-2) + 2/x + 3/(x+2) and (6x^2 - 4x - 8)/(x^3 - 4x) are defined). The determination of A, B, and C did use these "disallowed" values, but that can be shown to be okay using continuity (basically, using the fact that polynomials/rational functions are "nice")
Equation (1), where we "guess" the structure of the partial fractions decomposition, is equivalent to the equation (2), which operates on polynomials, for all x except 2,0,-2. But the polynomials are continuous everywhere, so we find proper coefficients for the polynomials (which ARE defined for x=2,0,-2) and then use them for all the the other x's, where rational function is defined
@@pawemarsza9515 huh, you've just made me realize that we don't even need to care about continuity, since it's only the implication "(6x^2 - 4x - 8)/(x^3 - 4x) = A/(x-2) + B/x + C/(x+2) for all x =/= -2,0,2" ⇐ "6x^2 - 4x - 8 = Ax(x+2) + B(x-2)(x+2) + C(x-2)x for all x" that we care about.
No, it's usually not. But using the current year as a value in a problem is nice because it's not _totally_ arbitrary and because it's big enough that we can't brute-force some algorithm n times if n happens to be in the thousands.
I noticed that you decomposed to A/(x-2) and C/(x+2) but calculated A and C with Ax(x+2) and Cx(x-2). Should it not have been Ax(x-2) and Cx(x+2) to give A=3 and C=1 and therefore 3/(x-2) and 1/(x+2) which gives a final result of the 2022th derivative being 2022! * 3/[(x-2)^2023] * 2/(x^2023) * 1/[(x+2)^2023]?
No. Try and recreate the step at 2:33 i.e. multiplying through the numerator by x(x-2)(x+2). You should see e.g. for the A term that the (x-2) terms in the numerator and denominator cancel and you're just left with Ax(x+2)
let's see... maybe factor the numerator? x^2 - 2 x 2/3 - 4/3 = 0 x = 2/3 ± 4/3 in (-2/3, 2) therefore the numerator factors as 2(3x+2)(x-2). The denominator trivially factors as x(x-2)(x+2). The x-2 parts cancel, giving: 2(3x+2)/(x(x+2)). maybe next we should try to split the numerator into two parts, where each part shares a factor with the denominator: 3x+2 = 2x + (x+2), so the fraction splits nicely into two parts: 2/(x+2) + 2/x. the first derivative introduces a factor of (-1), the next introduces (-2), and so on, which means the nth derivative will have the product of integers from 1 to n, multiplied by (-1)^n. This means the final answer should look like: (-1)^n n! [2/(x+2)^(n+1) + 2/x^(n+1)]. For n=2022, we get: 2022! [2/(x+2)^2023 + 2/x^2023]
hmmm it seems my fractions are off here. i missed a factor of 2: x^2 - 2 x 1/3 - 4/3 = 0 x = 1/3 ± sqrt(13)/3, so this does NOT have a factor of (x-2) in there, which means yeah as he did in the video there must be THREE terms, not two, though the result looks very similar
I'd like to think Mathematica would be "smart" enough to avoid trying to differentiate something so many times, and it would instead come up with this kind of method, but I doubt it.
An exam question would probably have a part (a) to find the partial fractions. Then part (b) would say "hence, or otherwise, find the 2022nd derivative.
No, this is an _excellent_ question _because_ of the trick! That said, the sad reality is that not every student can be running on all cylinders every day. If an exam happens to land on a bad day for you, you're less likely to figure everything out. But unless there are lots and lots of exam days to help find the average of your actual performance, then the exams end up selecting for whether or not you're lucky enough to have had your good days line up with test days, rather than how you actually tend to perform.