2

I'm trying to find a general formula for the $n$th derivative of $$\dfrac{1}{\sqrt{1 + x^2}}$$ I got up to, \begin{eqnarray*} g^{(0)}(x) &=& g(x) \\ g^{(1)}(x) &=& \dfrac{1}{(1 + x^2)^{1/2}} \\ g^{(2)}(x) &=& \dfrac{-x}{(1 + x^2)^{3/2}} \\ g^{(3)}(x) &=& \dfrac{2x^2 - 1}{(x^2 + 1)^{5/2}} \\ g^{(4)}(x) &=& \dfrac{-6x^3 + 9x}{(x^2 + 1)^{7/2}} \\ g^{(5)}(x) &=& \dfrac{24x^4 - 72x^2 + 9}{(x^2 + 1)^{9/2}} \\ g^{(6)}(x) &=& \dfrac{-120x^5 + 600x^3 - 225x}{(x^2 + 1)^{11/2}} \\ g^{(7)}(x) &=& \dfrac{720x^6 - 5400x^4 + 4050x^2 -225}{(x^2 + 1)^{13/2}} \\ g^{(8)}(x) &=& \dfrac{-5040x^7 + 52920x^5 - 66150x^3 + 11025x}{(x^2 + 1)^{15/2}} \\ \end{eqnarray*}

Except for the first term in the numerator ($n!$), and the power in the denominator, I couldn't find a general pattern for the rest of the coefficients in the numerator. Could anyone shed me some light on this problem? Any idea would be greatly appreciated. Thanks.

roxrook
  • 12,081
  • Is this part of a Taylor Series question, I presume? Also, it seems like your derivatives are off by order $1$. What I mean is, the $0$th order derivative is just your original, $\frac{1}{\sqrt{1+x^2}}$. – Joe Nov 28 '12 at 02:36
  • If this is part of finding the Maclaurin Series for your given function $g(x)$, you can use the well-known fact for the Maclaurin Series for a binomial expansion:

    $$(1+x)^k = \sum_{n=0}^{\infty} {k\choose n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots$$

    In your case,

    $$g(x) = \frac{1}{\sqrt{1+x^2}} = (1+x^2)^{-1/2}$$

    So, $k = -\frac{1}{2}$ and $x = x^2$ per the $(1+x)^k$ part.

     – Joe Nov 28 '12 at 02:45
  • @Joe: Thanks a lot. Yes, it's part of the Taylor series question. – roxrook Nov 28 '12 at 03:51
  • Related: http://math.stackexchange.com/questions/684819/n-th-derivative-of-sinh-1-x/684852#684852 – Tom-Tom May 18 '15 at 07:36

1 Answers1

4

Since this is part of finding the Maclaurin Series for your given function $g(x)$, you can use the well-known fact for the Maclaurin Series for a binomial expansion:

$$(1+x)^k = \sum_{n=0}^{\infty} {k\choose n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots$$

In your case,

$$g(x) = \frac{1}{\sqrt{1+x^2}} = (1+x^2)^{-1/2}$$

So, $k = -\frac{1}{2}$ and $x = x^2$ per the definition of the expansion for $(1+x)^k$

If you need further help writing out the binomial expansion for your function, let me know and I'll fill in some more details.

Note that this only holds when our Taylor Series is about a point $a = 0$. That is, it is a Maclaurin Series. If $a \ne 0$, only then do we need to compute the nth order derivative about $a$. For a given Taylor Series, we can express a function $f$ as:

$$f(x) = \sum_{n=0}^{N} \frac{f^n(a) (x-a)^n}{n!}$$

Joe
  • 4,757
  • 5
  • 35
  • 55
  • Thanks a lot. However, I can't use Maclaurin Series fact. I still believe there must be a general formula for the $n$th derivative. – roxrook Nov 28 '12 at 05:12
  • I'll try working out a formula for the $n$th order derivative then. Give me a few. What point $a$ is your series about? – Joe Nov 28 '12 at 05:22
  • 1
    $a = 0$, I've just figured it out based on your answer. Just plug in $a = 0$, then the remain sequences is : -1, 1, 9, -225, 11025. Once again, thanks a bunch Joe. – roxrook Nov 28 '12 at 05:27
  • 1
    For odd $n$, $g^{(n)}(0)=0$. For even $n=2k$, $g^{(n)}(0)=\frac{(-1)^{k}((2k)!)^2}{4^{k}(k!)^2}$. This is found by expanding $\binom{-1/2}{k}$ and using the fact that $1\cdot3\cdot\cdots\cdot(2k-1)=\frac{(2k)!}{2\cdot4\cdot6\cdot\cdots\cdot(2k)}$. – 2'5 9'2 Nov 28 '12 at 05:45
  • Ah, nice catch, Alex. Good work. – Joe Nov 28 '12 at 05:46