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.
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?
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! )
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
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.
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
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.
I was about to use the quotient rule 😆
Thanks for the partial fractions haha
Hey, I know you from somewhere.
all of us was about to do that
@@ilikepudding09 black pen
red pen
YEEAAAA!!!
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
You should do more problems like these, I really enjoyed doing this and watching the solution in which we did the same btw
Love it! Partial fractions was the way to go, I should have recognized it immediately.
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.
@@JM-us3fr That's why I said the taylor series evaluated at an arbitrary point. That would give you the general formula for the nth derivative.
@@whonyx6680 "evaluated at" is different than "centered at"
@@radadadadee True, my mistake. But you still understood what I meant.
That was fun.
Thank you, professor.
13:18
kind of saw how to do it from the start, but i wouldve totally messed up the partial fractions part (i suck at those)
your 8 gives me anxiety
In my 51 years of calculus, I have never seen that very nice procedure for calculating the partial fraction expansion. Fun. Thanks.
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.
@@brandongroth4569 Thanks.
Black Pen Red Pen seems to have caused that method to go viral!
@@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.
You may wish to look up the Heaviside Cover-up method.
Thank you for this channel!
Nicely explained.
Great lecture, with a perfect explanation of how to derive partial fractions. Michael Penn's a star!
Note that the denominator is a higher power than the numerator. If it was the other way round then the umpteenth derivative would likely be zero.
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!
I would have differentiated that function 2022 times!
But you'd end up finishing the work in 2023 and have to do one more iteration 😀
@@R0M4ur0 It's about 6 differentitation a day, not too bad!
I feel like the quotient rule is just redundant, since it's basically just product rule.
Could you please explain why that substitution of x=2, x=0, x=-2 to find A, B, C works?
you could have almost any other substitutions but with this one a lot of things turn out to be 0 which simplifies calculations
They isolate one term at a time by making the others zero.
Search "A trick I've ignored for long enough"
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.
@@Noam_.Menashe thank you, I saw that video a long time ago but I totally forgot about it. I guess "I've ignored it for long enough"
These sorts of problems are always interesting, once you realize that you need to do partial fraction decomposition it just breaks down quite nicely.
*@Michael Penn* -- That should read "Find the 2,022 *nd* derivative of ..."
Perhaps next time we should do one with quotient rule instead of partial fraction decomposition as overkill or when the structure make that easier.
I hate the quotient rule. I hate it so much that I would rather invert the dominator and use the product rule.
@@robertlunderwood my man really hates the quotient rule so instead uses the quotient rule... legend
@@brendanmiralles3415 Well, I don't have to worry about getting the subtraction wrong or putting the wrong item in the denominator.
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?
are you sure? plugging x=±2 into the numerator doesn't seem to give 0.
Oops, I mistfactored big time
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! )
But I wanted to the the monstrous value of 2022! at the end!
Any chance x is a complex number?
Yes. The functions are all holomorphic, except for the values at the denominators
Now we just need to deal with 2022!
At 11:41 you wrote (x+1) and it should have been (x-2) in the denominator.
looks like a Tylor series
I feel like inducting on 1/(x + r) derivatives would've made you have to write less.
The more I watch your videos, the more I love your videos
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
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.
@@pedroteran5885 So, the argument of extension by continuity is missing.
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
I wonder if this would be possible to solve using Taylor series.
thanks youtube for recommending this, who knew i could actually learn something while having a shit
Great !!
I wonder how the decomposition is valid, since x = ( -2, 0, 2) each cause the function to be dividing by 0?
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.
2022-th or whatever else is almost never the point.
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.
Nice
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)
@@mulletronuk Oh yes, of course. Sorry. My mistake. Forgot about that.
* 2,022nd derivative
That's easy B square
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
faa di bruno formula 😏
Could Mathematica brute-force this?
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.
idk what you're doing or why you're choosing to do specifically that, but its cool and interesting nontheless
Random comment for the youtube algorithm
This would make a terrible exam question. Exam questions shouldn't rely on tricks.
partial fraction decomposition isn't a "trick"; it is taught in all calculus courses
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.
Virtually all of math in practice consists of "tricks"...
I think it would be a good bonus question, the sort of thing that is at the end of a test to relax students rather than stress them.
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.