I need help determining the nth derivative of f(x)^k. Ive tried brute forcing it for a pattern but it just gets to large way to fast. Anyone know of a good way of determining it?
#determining the nth derivative of f(x)^k
modern oyster
steep lavaBOT
modern oyster
The main problem is that because of the chain rule and product rule it expands really really fast and becomes really messy.
ocean current
there is no pattern fixed for it
its very question specific
at least till the level of math i have covered
unreal patrol
Try only focusing on the coefficient.
After 1st derivative, it's k
2nd, k(k-1)
3rd, k(k-1)(k-2)
.
.
.
nth, k(k-1)...(k-(n-1))
This is just
$\frac{k!}{(k-(n-2))!}$
crude drumBOT
Galactic Fur
unreal patrol
I hope so. Not really sure if that's the exact one
But somewhat like that should be it
modern oyster
Hmm let me try that one sec.
ocean current
but bro here lets say i had a function (x^2 - 87x + 876)^89
just some random function
wont chain rule affect it
unreal patrol
That will be managed by the derivatives of f(x)
We don't really care about that
Wut we focus here on, are the coefficient terms that are always going to be present
modern oyster
Try only focusing on the coefficient.
After 1st derivative, it's k
2nd, k(k-1)
3...
So if we only look at the first coefficient we get for the first coefficient of the nth derivative:
$\frac{k!*f(x)^{k-n}*f’(x)^n}{(k-n)!}$
But im still abit confused on what to do next?
crude drumBOT
bishop of the flame
unreal patrol
So if we only look at the first coefficient we get for the first coefficient of ...
Yeah. So this will be the coefficient for the first term in the nth derivative
For f(x)
The same will happen with f'(x), but only n-1 times. Cuz it comes after differentiating first time
Then the same with f''(x). But only n-2 times
And it keeps going till nth derivative of f(x). Which will occur only once. Cuz that's the final derivative
modern oyster
Yeah but how to i determine the 2nd third fourth fifth… coefficient. The first and last coefficient are pretty easy to find but its those middle coefficients that are really getting me. Im pretty lost on how to find them.
unreal patrol
idk why I always open such interesting questions when it's time for me to sleep lol. I would've helped you staying here. But @hot karma can help you out while I'm not here. Even @ocean current can, if he has some idea on this.
I'll try this tomorrow morning if you're still in need for help.
ocean current
idk why I always open such interesting questions when it's time for me to sleep ...
mate u always do a great job helping. its all good
modern oyster
Like for instance for the 4th derivative it would be.
$\frac{k!*f(x)^{k-4}*f’(x)^4}{(k-4)!}+…+\frac{k!*f(x)^{k-1}*f^{(4)}(x)}{(k-1)!}$
crude drumBOT
bishop of the flame
modern oyster
Hmm. After some work i can definitely tell there is a pattern. Cant tell what the pattern is but the nth derivative definitely does follow some pattern
modern oyster
Bit of an update. Haven’t made much progress. Just kinda been throwing myself at the wall that is this problem trying to brute force some sort of pattern out of it.
glacial salmon
There's a formula for the nth derivative of f(x)g(x):
$(f g)^{(n)} = \Sigma_{k=0}^n C_n^k f^{(n-k)} g^{(k)}$
crude drumBOT
EQUENOS
glacial salmon
Maybe it will be helpful
hot karma
idk why I always open such interesting questions when it's time for me to sleep ...
i am definitely not allknowing this is also for me difficult
hot karma
i see some pattern but I cant quite write in some compact form
hot karma
kinda thinking if the fourth derivative is worth it in order to see some pattern
@unreal patrol ik you are indian smart guy help😭
se x common
hot karma
Bit of an update. Haven’t made much progress. Just kinda been throwing myself at...
Are you allowed to use this: General Leibniz rule
modern oyster
Yes i can ise that but im not entirely sure how.
hot karma
Yes i can ise that but im not entirely sure how.
Maybe you can do this:
$(f(x))^n = (f(x) \cdot 1)^n = (f(x) \cdot g(x))^{(n)} \Rightarrow \sum_{k=0}^{n} \left(\begin{array}{c} n \ k \end{array}\right) f^{(n-k)}g^{(k)}$
crude drumBOT
adonhs
modern oyster
Thats not entirely what im trying to do because the $(fg)^{(n)}$ the n stands for derivatives while im trying to find the derivative of $f(x)^n$
crude drumBOT
bishop of the flame
modern oyster
I think
unreal patrol
se x common
Lmaoooo. No way you just said that again 😂
And well, your solution seems to be on the right track
It's just that there are a lot of terms
unreal patrol
Each term actually has a pattern 🙂
hot karma
Each term actually has a pattern 🙂
your sentence too
unreal patrol
It's a beautiful sentence
hot karma
there is almost everywhere an a
actually it's terma not term
each terma ctually has a pattern
ngl i am broke i am going
off
unreal patrol
Gg
unreal patrol
You too 😭
glacial salmon
Thats not entirely what im trying to do because the $(fg)^{(n)}$ the n stands fo...
I suppose the idea is to let $g(x) = f(x)^{k-1}$ and use the general Leibniz rule
modern oyster
Ive made alot of progress and i think I’m close. I know for a fact without amplification the number of coefficients for the nth derivative is equal to: $2^n-1$
Also the full nth derivative function will be something along the lines of:
$\sum_{n_1=1}^{n}\frac{kf(x)^{k-n_1}}{(k-n_1)!}\sum_{n_2=1}^{\frac{n!}{n_1!*(n-n_1)!}}(…)$
Also i know for a fact for each individual coefficient if we sum up all derivative values it will add up to n.
crude drumBOT
bishop of the flame
modern oyster
And the coefficients in total are all permutations of n1 numbers that sum up to n, times some constant who’s pattern i have not looked too much into yet
glacial salmon
Another fun fact: the coefficient for $f(x)^{k-n+m}f'(x)^{n-2m}f''(x)^m$ is $(n-m)! \binom{k}{n-m} \binom{n}{2m} \frac{(2m)!}{2^m (m!)^2}$
crude drumBOT
EQUENOS
main umbra
It seems like you're making good progress on understanding the pattern for the nth derivative of ( f(x)^k ). Let me summarize the key points you've mentioned:
- The number of coefficients for the nth derivative is ( 2^n - 1 ).
- The general form of the nth derivative function involves nested summations, with each term having a coefficient determined by ( n_1 ), and the total summation accounts for all permutations of ( n_1 ) numbers that sum up to ( n ).
- Each individual coefficient is a permutation of ( n_1 ) numbers that sum up to ( n ), multiplied by some constant with a pattern you're still exploring.
Additionally, EQUENOS provided a formula for a specific coefficient: the coefficient for ( f(x)^{k-n+m}f'(x)^{n-2m}f''(x)^m ) is given by ( (n-m)! \binom{k}{n-m} \binom{n}{2m} \frac{(2m)!}{2^m (m!)^2} ).
It's great that you're making progress and identifying patterns in the coefficients. If you continue to explore the constants and their patterns, you may uncover a more concise expression for the coefficients. Keep up the good work! If you have specific questions or need further assistance, feel free to ask.
crude drumBOT
notnotnotnoone
glacial salmon
Sounds more epic than it should