How to find the 10th derivative of f(x)=\sqrt{1+x^2} using binomial series?

[DivGradCurl]

(a) Use the binomial series to find the Maclaurin series of

$$f(x)=\sqrt{1+x^2}$$

$$f(x)=\sqrt{1+x^2} = \left( 1+x^2 \right) ^{1/2} = \sum _{n=0} ^{\infty} \binom{1/2}{n} x^{2n} = \binom{1/2}{0} + \binom{1/2}{1}x^2 + \binom{1/2}{3}x^6 + \binom{1/2}{4}x^8 + \cdots$$

$$f(x)= 1 + \frac{1}{2}x^2 + \frac{\left( \frac{1}{2} \right)\left( -\frac{1}{2} \right)}{2!}x^4 + \frac{\left( \frac{1}{2} \right)\left( -\frac{1}{2} \right) \left( -\frac{3}{2} \right) }{3!}x^6 + \frac{\left( \frac{1}{2} \right)\left( -\frac{1}{2} \right) \left( -\frac{3}{2} \right) \left( -\frac{5}{2} \right) }{4!}x^8 + \cdots$$


$$f(x)= 1 + \frac{1}{2}x^2 + \sum _{n=2} ^{\infty} (-1)^{n-1} \frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-3)}{n!2^n}x^{2n}$$

(a) Use (a) to evaluate $$f^{(10)}(0).$$ (answer: 99,225)

I've had some difficulty to see how this result is obtained, since the $$x^{2n}$$ in the series above seems to give 0 despite the differentiation steps . Can anybody help me clarify this?

Thanks

 

[AKG]

You have a bunch of terms in the form $a_0x^0 + a_2x^2 + a_4x^4 + \dots$. Take the tenth derivative, and you will get:

$$a_0(0) + a_2(0) + \dots + a_{10}\frac{d^{10}}{dx^{10}}x^{10} + a_{12}\frac{d^{10}}{dx^{10}}x^{12} + \dots$$

The first few terms will obviously give zero. The terms with powers higher than 10 will have some x remaining in the tenth derivative, so those will also go to zero. Only the term containing the 10th power will remain in some form. I believe you will get:

$$\frac{(1\cdot 3\cdot 5\cdot 7)(10\cdot 9\cdot 8\cdot 7\cdot 6\cdot)}{32} = 99225$$

 

[DivGradCurl]

Just one thing...

$$\frac{(1\cdot 3\cdot 5\cdot 7)(10\cdot 9\cdot 8\cdot 7\cdot 6)}{2^5} = 99225$$

That makes sense, but why don't you have the denominator: $$5!2^5$$?

Why doesn't a 5! appear in your computation?

 

[AKG]

It does, but it gets canceled out, i.e. 10!/5! = 10 x 9 x 8 x 7 x 6

 

[DivGradCurl]

Oh... sure. Thanks a lot!