Find nth derivative of e^m(arcsinx) where x=0

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In summary, you are trying to solve a problem involving finding the nth derivative, but are having difficulty. After doing the first part, you are able to prove the first equation, but are having difficulty in finding the nth derivative. You have found that it is the same as finding the ninth derivative, which is by induction. You have also been able to derive a formula for the nth derivative.
  • #1
Kumail Haider
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Homework Statement
Find nth derivative of e^m(arcsinx) where x=0
Relevant Equations
Leibnitz theorem
(f.b)n=fn.b +nfn-1.b1+.........+f.bn
Here is the problem that I'm trying to solve. I've done the first part that is prove but I'm. facing difficulty in finding nth derivative
IMG_20230421_092939.jpg


I'm attaching pics of my attempt of solving this problem as I've no idea that how to type all these mathematical expressions.
Can anyone please guide me that how to find the nth derivative (it always confuses me )
I will be thankful for your guidance.
IMG_20230421_093330.jpg
IMG_20230421_093444.jpg
IMG_20230421_092953.jpg
 
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  • #2
Kumail Haider said:
I've no idea how to type all these mathematical expressions.
:smile: The ##\LaTeX## Guide link is at the bottom left of the edit window !

But I grant you it's an awful lot of work... pure masochism to typeset all that stuff :wink:

So you start with
$$y^{(0)} = y = e^{m\arcsin x} \ \Rightarrow \ y^{(1)} = {m\over \sqrt{1-x^2}} \; e^{m\arcsin x} = {m\over \sqrt{1-x^2}} \; y$$and so $$
\begin {align*} y^{(2)} &= \left ( {m\over \sqrt{1-x^2}} \right )' \;y +{m\over \sqrt{1-x^2}} \; y' \\ \ \\
&={x\over 1-x^2} \left ( {m\over \sqrt{1-x^2}} \right ) \;y + {m^2\over 1-x^2} \;y \\ \ \\
&= {x\over 1-x^2}\; y^{(1)} + {m^2\over 1-x^2} \;y ^{(0)} \qquad\Rightarrow \\ \ \\
\left (1-x^2 \right )\; y^{(2)} &= x \; y^{(1)} + m^2\;y^{(0)} \tag 1
\end {align*}
$$Leibnitz proof is by induction, so with ##(1)## we have a statement ##P_{n=0}## :
$$\begin {align*} \left (1-x^2 \right )\; y^{(n+2)} &= (2n+1)\;x \; y^{(n+1)} + (n^2+m^2)\;y^{(n)} \tag 2\end {align*} $$ which is true for all ##n## if we can prove ##P_{n}\Rightarrow P_{n+1}##.
To do that we differentiate ##(2)##:
$$\begin {align*} -2x \; y^{(n+2)} + \left (1-x^2 \right ) \; y^{(n+3)} &= (2n+1) \; y^{(n+1)} + (2n+1)\;x \; y^{(n+2)} + (n^2+m^2)\;y^{(n+1)} \\ \ \\
\left (1-x^2 \right ) \; y^{(n+3)} &= (2n+3)\;x \; y^{(n+2)} + (n^2+ 2n+1+m^2)\;y^{(n+1)} \\ \ \\
\left (1-x^2 \right ) \; y^{(n+3)} &=\left (2(n+1)+1\right )\;x \; y^{(n+2)} + \left ( (n+1)^2+ m^2\right )\;y^{(n+1)}
\end {align*} $$which is ##P_{n+1}##.

So there ! But you had already done that, witness the pictures :
Kumail Haider said:
I've done the first part that is prove
Now on to the ninth derivative, one way or the other.
Cheating isn't allowed, I suppose (anyway, the ninth derivative looks horrendous and I don't know if I can trust the result when my cheater substitutes ##x=0## :wink:)

And therefore, all I can come up with is brute force: use ##(2)## four times to ripple down from ##y^{(9)}## to ##y^{(1)}## which is ##m##.

That is from ##n=7## in three steps of 2 to to ##n=1##.

Fortunately, all ##y^{(n+1)}## have a factor ##x## in front, so they drop out at ##x=0##.

That means: at ##x=0## we have ##y^{(9)} |_{x=0} ## ##= (7^2+m^2) \;y^{(7)} |_{x=0}## and so on.

So brute force yields $$y^{(9)} |_{x=0} = (49+m^2)(25+m^2)(9+m^2)(1+m^2)m$$And I am way too lazy to work this out myself. But it turns out the cheater is pretty consistent with ##m(m^8 + 84 m^6 + 1974 m^4 + 12916 m^2 + 11025)## as before. Hurray!

And as a farewell present, usable for the umpteenth derivative as well:$$y^{(2n+1)} |_{x=0} = m\;\prod_{k=1}^n \left( (2k-1)^2+m^2\right ) $$I think ...

[edit] And it's pure fun to do ##y^{(2n)} |_{x=0} ## and find (almost) the same expression ! Am I overdoing...?

(but now I have robbed you of your exercise... violating PF rules. Sorry about that, but I couldn't resist :rolleyes: )

Let me pick up at your

1682076583519.png
##A## is correct, but ##B## seems to be in error. I leave it to you to find the minus sign in your picture ....

:smile:

##\ ##
 
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  • #3
Kumail Haider said:
Can anyone please guide me that how to find the nth derivative (it always confuses me )
It looks like you have the basic idea. Just keep writing terms and look for the pattern (if it isn't apparent from what you've done already).
 
  • #4
BvU said:
Now on to the ninth derivative, one way or the other.
nth ≠ ninth :)
 
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  • #5
BvU said:
Let me pick up at your

View attachment 325221##A## is correct, but ##B## seems to be in error. I leave it to you to find the minus sign in your picture ....

:smile:

##\ ##
Thanks a lot for your guidance😊

So, B = −(1−m^2) (m)

In that(which I've done)
I write equation ##B## down as :
−(1−m^2) = (m^2−1)
It means that I've multipled the whole bracket with this negative sign. Am I picking the right point?
 
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  • #6
vela said:
nth ≠ ninth :)
How on earth did that 9 creep into my brain !
That means that ##y^{(2n)} |_{x=0}## comes into the picture too. Good thing I had fun deriving $$
y^{(2n)} |_{x=0} =\prod_{k=0}^n \left( (2k-1)^2+m^2\right )$$so almost the same expression -- but I am at a loss trying to merge the two into a single one..... help !
Kumail Haider said:
Probably B = −(1−m^2 )(m)
Am I write know?
I think it's going wrong here already:

1682095008918.png


That should be a minus sign, which turns your 3 into my ##(2) ## and then it agrees with

1682095630235.png
 
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  • #7
BvU said:
View attachment 325240

That should be a minus sign, which turns your 3 into my ##(2) ## and then it agrees with
That should be a minus sign, which turns your 3 into my ##(2) ## and then it agrees with

View attachment 325242
Thanks🙂
I think I get the point. I was making mistake in taking derivative of √(1−x ²) it should be −2x /2√(1−x ²)

Is this the mistake that I was making?
 
  • #8
Indeed...
 
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1. What is the general formula for finding the nth derivative of e^m(arcsinx) where x=0?

The general formula for finding the nth derivative of e^m(arcsinx) where x=0 is (-1)^n * m^n * e^(m * arcsinx) / (1-x^2)^n.

2. How do you determine the value of m in the formula for finding the nth derivative?

The value of m can be determined by looking at the exponent of e in the original function. For example, if the original function is e^2(arcsinx), then m=2.

3. Can the nth derivative of e^m(arcsinx) where x=0 be simplified further?

Yes, the nth derivative can be simplified further by factoring out the common terms and using the binomial theorem to expand the (1-x^2)^n term.

4. What is the significance of setting x=0 in the original function?

Setting x=0 in the original function allows us to find the nth derivative at the point where the arcsinx function is undefined. This is known as the Maclaurin series expansion, which is a special case of the Taylor series expansion.

5. Can the formula for finding the nth derivative of e^m(arcsinx) where x=0 be applied to other functions?

Yes, the formula for finding the nth derivative can be applied to other functions that have a similar form, such as e^m(arctanx) or e^m(arccosx). However, the value of m and the range of x may differ for each function.

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