Differentiation Marathon!

For college and university level advanced Mathematics
Asif Hossain
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Re: Problem 4

Unread post by Asif Hossain » Tue Apr 06, 2021 12:01 pm

Mehrab4226 wrote:
Tue Apr 06, 2021 9:37 am
Anindya Biswas wrote:
Tue Apr 06, 2021 1:08 am
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?
IG $f^{(2)}(x)=f"(x)$
Hmm..Hammer...Treat everything as nail

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Anindya Biswas
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Re: Problem 4

Unread post by Anindya Biswas » Tue Apr 06, 2021 12:49 pm

Mehrab4226 wrote:
Tue Apr 06, 2021 9:37 am
Anindya Biswas wrote:
Tue Apr 06, 2021 1:08 am
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?
$f''(x)$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: Problem 4

Unread post by Asif Hossain » Tue Apr 13, 2021 11:04 am

Anindya Biswas wrote:
Tue Apr 06, 2021 12:49 pm
Mehrab4226 wrote:
Tue Apr 06, 2021 9:37 am
Anindya Biswas wrote:
Tue Apr 06, 2021 1:08 am
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 :P
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. :P
Hmm..Hammer...Treat everything as nail

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Problem 5

Unread post by Asif Hossain » Thu Apr 22, 2021 7:43 pm

Problem 5:
Find $\frac{d}{dx}(\frac{x^3}{\sqrt{x^2+a}})$
where $a$ is constant.
Hmm..Hammer...Treat everything as nail

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Mehrab4226
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Re: Problem 5

Unread post by Mehrab4226 » Thu Apr 22, 2021 10:21 pm

Asif Hossain wrote:
Thu Apr 22, 2021 7:43 pm
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.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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Mehrab4226
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Location:Dhaka, Bangladesh

Problem: 6

Unread post by Mehrab4226 » Thu Apr 22, 2021 10:25 pm

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?
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

~Aurn0b~
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Re: Problem: 6

Unread post by ~Aurn0b~ » Fri Apr 23, 2021 1:50 am

Mehrab4226 wrote:
Thu Apr 22, 2021 10:25 pm
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?
Let $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$
$\Rightarrow y=\sqrt{x+y} \Rightarrow y^2=x+y$
$\Rightarrow\frac{dy^2}{dx}=\frac{dx}{dx}+\frac{dy}{dx}$
$\Rightarrow\frac{dy^2}{dy}\cdot \frac{dy}{dx}=1+\frac{dy}{dx}$
$\Rightarrow2y\cdot \frac{dy}{dx}-\frac{dy}{dx}=1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2y-1}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}-1}$

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.

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Mehrab4226
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Location:Dhaka, Bangladesh

Re: Differentiation Marathon!

Unread post by Mehrab4226 » Fri Apr 23, 2021 5:17 pm

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
Screenshot 2021-04-23 17.21.48.png (101.04KiB)Viewed 7878 times
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: Differentiation Marathon!

Unread post by ~Aurn0b~ » Fri Apr 23, 2021 11:26 pm

Mehrab4226 wrote:
Fri Apr 23, 2021 5:17 pm
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

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Problem 7

Unread post by Asif Hossain » Sun Apr 25, 2021 10:25 am

Evaluate $f^{(n)}(x)$ for $f(x)=x^{n}e^x$ at $x=0$ where $f^{(n)}(x)$ denotes $n$-th derivative of $f(x)$
Hmm..Hammer...Treat everything as nail

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