I need help with a problem involving the nth derivative of arcsin x












1












$begingroup$


I need help with a problem. For context, the section of the textbook the problem is in is about power series. Note that the textbook uses the convention that $f^{(n)}$ represents the $n$th derivative of $f$, and $f^{(0)}(x) = f(x).$ I'll now state the problem exactly as stated in the textbook:



Consider the function $f$ defined by
$f(x) = arcsin x$, for $lvert x rvert leq 1$.
The derivatives of $f(x)$ satisfy the equation
$
(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$, for $n geq 1.
$



The coefficient of $x^n$ in the Maclaurin series for $f(x)$ is denoted by $a_n$. You may assume that the series only contains odd powers of $x$.



$textbf{a.1)}$ Show that, for $n geq 1, (n+1)(n+2)a_{n+2} = n^2 a_n.$



$textbf{a.2})$ Given that $a_1 = 1$, find an expression for $a_n$ in terms of $n$, valid for odd $n geq 3.$



$textbf{b})$ Find the radius of convergence of this Maclaurin series.



$textbf{c})$ Find an approximate value for $pi$ by putting $x = frac{1}{2}$ and summing the first three non-zero terms of this series. Give your answer to $textbf{four}$ significant figures.



I'm stuck on $textbf{a.1}$. The way the question is formulated makes me think you're not supposed to use the actual derivatives of $arcsin$ to solve it, but I can't figure out how to do it. I know that $a_n = frac{f^{(n)}(0)}{n!}$,so I was thinking that if I can find a formula for the nth derivative of $f(x)$, I should be good to go.
I know the derivative of $f(x)$:



$f^prime(x) = frac{d}{dx}arcsin x = frac{1}{sqrt{1- x^2}}$. From here, I can easily also find the second, third, etc. derivatives. However, when I try to come up with a formla for the $textit{nth}$ derivative, I have a problem. I came up with the following formula:



$frac{d^n}{dx^n}arcsin x = (-1)^n prodlimits_{k = 0}^n left(frac{1}{2} - kright)$.



Unfortunately, I have no idea how to proveed from here, as I don't know how to evaluate the product $prodlimits_{k = 0}^n left(frac{1}{2} - kright)$. Anyway, I don't think this is the right approrach, as my textbook hasn't dealt with products yet, only sums. Can anyone help with $textbf{a.1}$?






calculus sequences-and-series derivatives taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you allowed to use the recurrence relation $(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$ ? Then you can simply set $x=0$ to obtain a.1.
    $endgroup$
    – Martin R
    Dec 17 '18 at 13:11












  • $begingroup$
    You made a slight mistake: $a_n=frac{f^{(n)}(0)}{n!}$.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 13:11










  • $begingroup$
    Ah, yes of course. Thanks. I will edit straight away.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:12










  • $begingroup$
    Ah, thanks, I did not think of just letting $x = 0$. Thanks a lot! Seems someone beat you to the actual answer :(
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:16










  • $begingroup$
    Yes – I should write answers, not comments :)
    $endgroup$
    – Martin R
    Dec 17 '18 at 13:20
















1












$begingroup$


I need help with a problem. For context, the section of the textbook the problem is in is about power series. Note that the textbook uses the convention that $f^{(n)}$ represents the $n$th derivative of $f$, and $f^{(0)}(x) = f(x).$ I'll now state the problem exactly as stated in the textbook:



Consider the function $f$ defined by
$f(x) = arcsin x$, for $lvert x rvert leq 1$.
The derivatives of $f(x)$ satisfy the equation
$
(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$, for $n geq 1.
$



The coefficient of $x^n$ in the Maclaurin series for $f(x)$ is denoted by $a_n$. You may assume that the series only contains odd powers of $x$.



$textbf{a.1)}$ Show that, for $n geq 1, (n+1)(n+2)a_{n+2} = n^2 a_n.$



$textbf{a.2})$ Given that $a_1 = 1$, find an expression for $a_n$ in terms of $n$, valid for odd $n geq 3.$



$textbf{b})$ Find the radius of convergence of this Maclaurin series.



$textbf{c})$ Find an approximate value for $pi$ by putting $x = frac{1}{2}$ and summing the first three non-zero terms of this series. Give your answer to $textbf{four}$ significant figures.



I'm stuck on $textbf{a.1}$. The way the question is formulated makes me think you're not supposed to use the actual derivatives of $arcsin$ to solve it, but I can't figure out how to do it. I know that $a_n = frac{f^{(n)}(0)}{n!}$,so I was thinking that if I can find a formula for the nth derivative of $f(x)$, I should be good to go.
I know the derivative of $f(x)$:



$f^prime(x) = frac{d}{dx}arcsin x = frac{1}{sqrt{1- x^2}}$. From here, I can easily also find the second, third, etc. derivatives. However, when I try to come up with a formla for the $textit{nth}$ derivative, I have a problem. I came up with the following formula:



$frac{d^n}{dx^n}arcsin x = (-1)^n prodlimits_{k = 0}^n left(frac{1}{2} - kright)$.



Unfortunately, I have no idea how to proveed from here, as I don't know how to evaluate the product $prodlimits_{k = 0}^n left(frac{1}{2} - kright)$. Anyway, I don't think this is the right approrach, as my textbook hasn't dealt with products yet, only sums. Can anyone help with $textbf{a.1}$?






calculus sequences-and-series derivatives taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you allowed to use the recurrence relation $(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$ ? Then you can simply set $x=0$ to obtain a.1.
    $endgroup$
    – Martin R
    Dec 17 '18 at 13:11












  • $begingroup$
    You made a slight mistake: $a_n=frac{f^{(n)}(0)}{n!}$.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 13:11










  • $begingroup$
    Ah, yes of course. Thanks. I will edit straight away.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:12










  • $begingroup$
    Ah, thanks, I did not think of just letting $x = 0$. Thanks a lot! Seems someone beat you to the actual answer :(
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:16










  • $begingroup$
    Yes – I should write answers, not comments :)
    $endgroup$
    – Martin R
    Dec 17 '18 at 13:20














1












1








1





$begingroup$


I need help with a problem. For context, the section of the textbook the problem is in is about power series. Note that the textbook uses the convention that $f^{(n)}$ represents the $n$th derivative of $f$, and $f^{(0)}(x) = f(x).$ I'll now state the problem exactly as stated in the textbook:



Consider the function $f$ defined by
$f(x) = arcsin x$, for $lvert x rvert leq 1$.
The derivatives of $f(x)$ satisfy the equation
$
(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$, for $n geq 1.
$



The coefficient of $x^n$ in the Maclaurin series for $f(x)$ is denoted by $a_n$. You may assume that the series only contains odd powers of $x$.



$textbf{a.1)}$ Show that, for $n geq 1, (n+1)(n+2)a_{n+2} = n^2 a_n.$



$textbf{a.2})$ Given that $a_1 = 1$, find an expression for $a_n$ in terms of $n$, valid for odd $n geq 3.$



$textbf{b})$ Find the radius of convergence of this Maclaurin series.



$textbf{c})$ Find an approximate value for $pi$ by putting $x = frac{1}{2}$ and summing the first three non-zero terms of this series. Give your answer to $textbf{four}$ significant figures.



I'm stuck on $textbf{a.1}$. The way the question is formulated makes me think you're not supposed to use the actual derivatives of $arcsin$ to solve it, but I can't figure out how to do it. I know that $a_n = frac{f^{(n)}(0)}{n!}$,so I was thinking that if I can find a formula for the nth derivative of $f(x)$, I should be good to go.
I know the derivative of $f(x)$:



$f^prime(x) = frac{d}{dx}arcsin x = frac{1}{sqrt{1- x^2}}$. From here, I can easily also find the second, third, etc. derivatives. However, when I try to come up with a formla for the $textit{nth}$ derivative, I have a problem. I came up with the following formula:



$frac{d^n}{dx^n}arcsin x = (-1)^n prodlimits_{k = 0}^n left(frac{1}{2} - kright)$.



Unfortunately, I have no idea how to proveed from here, as I don't know how to evaluate the product $prodlimits_{k = 0}^n left(frac{1}{2} - kright)$. Anyway, I don't think this is the right approrach, as my textbook hasn't dealt with products yet, only sums. Can anyone help with $textbf{a.1}$?






calculus sequences-and-series derivatives taylor-expansion







share|cite|improve this question











$endgroup$




I need help with a problem. For context, the section of the textbook the problem is in is about power series. Note that the textbook uses the convention that $f^{(n)}$ represents the $n$th derivative of $f$, and $f^{(0)}(x) = f(x).$ I'll now state the problem exactly as stated in the textbook:



Consider the function $f$ defined by
$f(x) = arcsin x$, for $lvert x rvert leq 1$.
The derivatives of $f(x)$ satisfy the equation
$
(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$, for $n geq 1.
$



The coefficient of $x^n$ in the Maclaurin series for $f(x)$ is denoted by $a_n$. You may assume that the series only contains odd powers of $x$.



$textbf{a.1)}$ Show that, for $n geq 1, (n+1)(n+2)a_{n+2} = n^2 a_n.$



$textbf{a.2})$ Given that $a_1 = 1$, find an expression for $a_n$ in terms of $n$, valid for odd $n geq 3.$



$textbf{b})$ Find the radius of convergence of this Maclaurin series.



$textbf{c})$ Find an approximate value for $pi$ by putting $x = frac{1}{2}$ and summing the first three non-zero terms of this series. Give your answer to $textbf{four}$ significant figures.



I'm stuck on $textbf{a.1}$. The way the question is formulated makes me think you're not supposed to use the actual derivatives of $arcsin$ to solve it, but I can't figure out how to do it. I know that $a_n = frac{f^{(n)}(0)}{n!}$,so I was thinking that if I can find a formula for the nth derivative of $f(x)$, I should be good to go.
I know the derivative of $f(x)$:



$f^prime(x) = frac{d}{dx}arcsin x = frac{1}{sqrt{1- x^2}}$. From here, I can easily also find the second, third, etc. derivatives. However, when I try to come up with a formla for the $textit{nth}$ derivative, I have a problem. I came up with the following formula:



$frac{d^n}{dx^n}arcsin x = (-1)^n prodlimits_{k = 0}^n left(frac{1}{2} - kright)$.



Unfortunately, I have no idea how to proveed from here, as I don't know how to evaluate the product $prodlimits_{k = 0}^n left(frac{1}{2} - kright)$. Anyway, I don't think this is the right approrach, as my textbook hasn't dealt with products yet, only sums. Can anyone help with $textbf{a.1}$?






calculus sequences-and-series derivatives taylor-expansion




calculus sequences-and-series derivatives taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 13:13







Christoffer Corfield Aakre

















asked Dec 17 '18 at 13:07









Christoffer Corfield AakreChristoffer Corfield Aakre

284




284












  • $begingroup$
    Are you allowed to use the recurrence relation $(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$ ? Then you can simply set $x=0$ to obtain a.1.
    $endgroup$
    – Martin R
    Dec 17 '18 at 13:11












  • $begingroup$
    You made a slight mistake: $a_n=frac{f^{(n)}(0)}{n!}$.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 13:11










  • $begingroup$
    Ah, yes of course. Thanks. I will edit straight away.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:12










  • $begingroup$
    Ah, thanks, I did not think of just letting $x = 0$. Thanks a lot! Seems someone beat you to the actual answer :(
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:16










  • $begingroup$
    Yes – I should write answers, not comments :)
    $endgroup$
    – Martin R
    Dec 17 '18 at 13:20


















  • $begingroup$
    Are you allowed to use the recurrence relation $(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$ ? Then you can simply set $x=0$ to obtain a.1.
    $endgroup$
    – Martin R
    Dec 17 '18 at 13:11












  • $begingroup$
    You made a slight mistake: $a_n=frac{f^{(n)}(0)}{n!}$.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 13:11










  • $begingroup$
    Ah, yes of course. Thanks. I will edit straight away.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:12










  • $begingroup$
    Ah, thanks, I did not think of just letting $x = 0$. Thanks a lot! Seems someone beat you to the actual answer :(
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:16










  • $begingroup$
    Yes – I should write answers, not comments :)
    $endgroup$
    – Martin R
    Dec 17 '18 at 13:20
















$begingroup$
Are you allowed to use the recurrence relation $(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$ ? Then you can simply set $x=0$ to obtain a.1.
$endgroup$
– Martin R
Dec 17 '18 at 13:11






$begingroup$
Are you allowed to use the recurrence relation $(1 - x^2)f^{(n + 2)}(x) - (2n + 1)xf^{(n + 1)}(x) - n^2 f^{(n)}(x) = 0$ ? Then you can simply set $x=0$ to obtain a.1.
$endgroup$
– Martin R
Dec 17 '18 at 13:11














$begingroup$
You made a slight mistake: $a_n=frac{f^{(n)}(0)}{n!}$.
$endgroup$
– Mindlack
Dec 17 '18 at 13:11




$begingroup$
You made a slight mistake: $a_n=frac{f^{(n)}(0)}{n!}$.
$endgroup$
– Mindlack
Dec 17 '18 at 13:11












$begingroup$
Ah, yes of course. Thanks. I will edit straight away.
$endgroup$
– Christoffer Corfield Aakre
Dec 17 '18 at 13:12




$begingroup$
Ah, yes of course. Thanks. I will edit straight away.
$endgroup$
– Christoffer Corfield Aakre
Dec 17 '18 at 13:12












$begingroup$
Ah, thanks, I did not think of just letting $x = 0$. Thanks a lot! Seems someone beat you to the actual answer :(
$endgroup$
– Christoffer Corfield Aakre
Dec 17 '18 at 13:16




$begingroup$
Ah, thanks, I did not think of just letting $x = 0$. Thanks a lot! Seems someone beat you to the actual answer :(
$endgroup$
– Christoffer Corfield Aakre
Dec 17 '18 at 13:16












$begingroup$
Yes – I should write answers, not comments :)
$endgroup$
– Martin R
Dec 17 '18 at 13:20




$begingroup$
Yes – I should write answers, not comments :)
$endgroup$
– Martin R
Dec 17 '18 at 13:20










1 Answer
1






active

oldest

votes


















1












$begingroup$

The given differential equation shows you that



$$
f^{(n + 2)}(0) - n^2 f^{(n)}(0) = 0.
$$



Then using Taylor,



$$(n+2)!,a_{n+2}-n^2,n!,a_n=0.$$



Simplify.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I will accept the answer in a few minutes when I can.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:17











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The given differential equation shows you that



$$
f^{(n + 2)}(0) - n^2 f^{(n)}(0) = 0.
$$



Then using Taylor,



$$(n+2)!,a_{n+2}-n^2,n!,a_n=0.$$



Simplify.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I will accept the answer in a few minutes when I can.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:17
















1












$begingroup$

The given differential equation shows you that



$$
f^{(n + 2)}(0) - n^2 f^{(n)}(0) = 0.
$$



Then using Taylor,



$$(n+2)!,a_{n+2}-n^2,n!,a_n=0.$$



Simplify.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I will accept the answer in a few minutes when I can.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:17














1












1








1





$begingroup$

The given differential equation shows you that



$$
f^{(n + 2)}(0) - n^2 f^{(n)}(0) = 0.
$$



Then using Taylor,



$$(n+2)!,a_{n+2}-n^2,n!,a_n=0.$$



Simplify.






share|cite|improve this answer









$endgroup$



The given differential equation shows you that



$$
f^{(n + 2)}(0) - n^2 f^{(n)}(0) = 0.
$$



Then using Taylor,



$$(n+2)!,a_{n+2}-n^2,n!,a_n=0.$$



Simplify.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 13:15









Yves DaoustYves Daoust

128k675227




128k675227












  • $begingroup$
    Thanks! I will accept the answer in a few minutes when I can.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:17


















  • $begingroup$
    Thanks! I will accept the answer in a few minutes when I can.
    $endgroup$
    – Christoffer Corfield Aakre
    Dec 17 '18 at 13:17
















$begingroup$
Thanks! I will accept the answer in a few minutes when I can.
$endgroup$
– Christoffer Corfield Aakre
Dec 17 '18 at 13:17




$begingroup$
Thanks! I will accept the answer in a few minutes when I can.
$endgroup$
– Christoffer Corfield Aakre
Dec 17 '18 at 13:17


















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