Let $f$ be an infinitely differentiable function. Prove that, \[\frac{d^n}{dx^n}\left(e^xf(x)\right)=e^x\sum_{k=0}^{n}{n\choose k}f^{(k)}(x)\] for all nonnegative integer $n$.
For the right hand side,
If k=2, does $f^{(k)}(x)$ mean
$f^{(2)}(x)=f''(x)$ this? or,
$f^{(2)}(x)=f^{2}(x)=f(f(x))$ this?
Let $f$ be an infinitely differentiable function. Prove that, \[\frac{d^n}{dx^n}\left(e^xf(x)\right)=e^x\sum_{k=0}^{n}{n\choose k}f^{(k)}(x)\] for all nonnegative integer $n$.
For the right hand side,
If k=2, does $f^{(k)}(x)$ mean
$f^{(2)}(x)=f''(x)$ this? or,
$f^{(2)}(x)=f^{2}(x)=f(f(x))$ this?
Let $f$ be an infinitely differentiable function. Prove that, \[\frac{d^n}{dx^n}\left(e^xf(x)\right)=e^x\sum_{k=0}^{n}{n\choose k}f^{(k)}(x)\] for all nonnegative integer $n$.
For the right hand side,
If k=2, does $f^{(k)}(x)$ mean
$f^{(2)}(x)=f''(x)$ this? or,
$f^{(2)}(x)=f^{2}(x)=f(f(x))$ this?
$f''(x)$
A lazy solution
Let $P(n)$ be the assertion.
$P(1)$ is easy to see to be true.
Assume $P(1),...,P(n)$ is true so $\frac{d^n}{dx^n}(e^xf(x))=e^x\sum_{k=0}^{n}{n\choose k}f^{(k)}(x)$
Since,${n\choose {k-1}}+{{n-1}\choose {k}}={{n+1}\choose k}$Differentiating again yields the result.
Problem 5
Posted: Thu Apr 22, 2021 7:43 pm
by Asif Hossain
Problem 5:
Find $\frac{d}{dx}(\frac{x^3}{\sqrt{x^2+a}})$
where $a$ is constant.
Problem 5:
Find $\frac{d}{dx}(\frac{x^3}{\sqrt{x^2+a}})$
where $a$ is constant.
If my calculations are correct, the ans should be
\[ \frac{2x^4+3ax^2}{(x^2+a)\sqrt{x^2+a}} \]
The work is too difficult for me to type so I will just mention my procedure.
I first used $\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{g(x)^2}$.
Then find the individual derivatives.
Problem: 6
Posted: Thu Apr 22, 2021 10:25 pm
by Mehrab4226
This is actually a self made problem, just for fun.
Find $\frac{d}{dx}\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$
Note:
There might be 2 derivatives. I don't know 2 derivatives are a thing or not. But just for the fun, we can do it, right?
This could've also been solved by obtaining $\frac{dx}{dy}$ and using the fact that $(\frac{dx}{dy})^{-1}=\frac{dy}{dx}$
I dont have any good derivative problem, so anyone can post the new problem.
Re: Differentiation Marathon!
Posted: Fri Apr 23, 2021 5:17 pm
by Mehrab4226
I have a different approach,
let,
$\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}} = y$
Then
$\sqrt{x+y}=y$
Or,$x+y=y^2$
Or,$y^2-y-x=0$
Or,$y=\frac{-(-1)\pm \sqrt{(-1)^2-4(1)(-x)}}{(2)(1)}$
Or,$y=\frac{1\pm \sqrt{1+4x}}{2}$
Differentiating on both sides we get,
$\frac{d}{dx}y=\pm \frac{1}{\sqrt{1+4x}} $
Which is weird I know, maybe it is true for only the positive values[Because the graph seems to get closer to that],
$\frac{d}{dx}y=\frac{1}{\sqrt{1+4x}}$
I don't know anything now. Please reply and say if it is acceptable.
Graph:
Screenshot 2021-04-23 17.21.48.png (101.04KiB)Viewed 9111 times
I have a different approach,
let,
$\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}} = y$
Then
$\sqrt{x+y}=y$
Or,$x+y=y^2$
Or,$y^2-y-x=0$
Or,$y=\frac{-(-1)\pm \sqrt{(-1)^2-4(1)(-x)}}{(2)(1)}$
Or,$y=\frac{1\pm \sqrt{1+4x}}{2}$
Differentiating on both sides we get,
$\frac{d}{dx}y=\pm \frac{1}{\sqrt{1+4x}} $
Which is weird I know, maybe it is true for only the positive values[Because the graph seems to get closer to that],
$\frac{d}{dx}y=\frac{1}{\sqrt{1+4x}}$
I don't know anything now. Please reply and say if it is acceptable.
Graph:
Screenshot 2021-04-23 17.21.48.png
for non-negative $x$, $4x+1\geq1\Rightarrow \sqrt{4x+1}\geq 1 \Rightarrow 1-\sqrt{4x+1}\leq 0.$
but $y$ cannot have negative value for nonegative x. then again if $4x+1\geq0 \Rightarrow x\geq -\frac{1}{4}$(Which has to be true as y cannot be complex)
I think $x$ wasnt well defined in the problem
Problem 7
Posted: Sun Apr 25, 2021 10:25 am
by Asif Hossain
Evaluate $f^{(n)}(x)$ for $f(x)=x^{n}e^x$ at $x=0$ where $f^{(n)}(x)$ denotes $n$-th derivative of $f(x)$
