Use the MacLaurin series for f(x).
Compute the 10th derivative of
f(x)=arctan(x^2/4)
at x=0.
f^(10)(0)=

Question

Use the MacLaurin series for f(x).
 Compute the 10th derivative of
 f(x)=arctan(x^2/4)
 at x=0.
 f^(10)(0)=

amistre64 · Answer

well, the 0 will cancel out all the x parts; and your left with the constant

amistre64 · Answer

$$\frac{d^n}{dx^n}x^a=aPn\ x^{a-n}$$if i see it correctly

anonymous · Answer

arctan(x)=x-x^3/3+x^5/5-...+$$(-1)^{k}x ^{2k+1}/(2k+1)$$

amistre64 · Answer

$$y = tan^{-1}(x^2/4)$$
$$tan(y)=x^2/4$$
$$sec^2(y)=x/2$$
$$y'=x cos^2(y)/2$$
$$cos^2(y)=\frac{16}{x^2+16}$$
$$y' = \frac{8x}{x^2+16}$$

gotta dbl chk that to be sure tho

amistre64 · Answer

x^4 not x^2 :)

amistre64 · Answer

16+x^4 = 16($1+\frac{x^4}{16}$)



             8x/2^0-x^5/2^1+x^9/2^5 - x^13/2^9
            -------------
$1+\frac{x^4}{16}$ ) 8x
            (8x+x^5/2)
            -----------
                  -x^5/2
                 (-x^5/2-x^9/2^5)
                 -------------
                              x^9/32
                             (x^9/32+x^13/2^9)

$$16\left(8x+\sum_{inf}^{n=1}\frac{(-1)x^{4n+1}}{2^{4n-3}}\right) $$

maybe ....

amistre64 · Answer

i spose 8x/16 = x/2 tho so id have to adjust for that

anonymous · Answer

that's so complicated;(

amistre64 · Answer

lol, im just trying to brush up on somethings with this really

anonymous · Answer

i knew an example like this.
$$e ^{-x ^{2}}=\sum_{k=0}^{\infty} (-1)^{k}x ^{2k}/(k!)$$
=1-x^2+x^4/2!-x^6/3!+x^8/4!+...
$$f ^{7}(0)=0$$
$$f ^{8}(0)/8!=1/4!$$
$$f ^{8}(0)=8!/4!$$

anonymous · Answer

but when i applied to my question,it is not correct,not sure whether it is a calculation problem or...

anonymous · Answer

$$arctanx=\sum_{0}^{\infty}(-1)^{k}x ^{2k+1}/(2k+1)$$

amistre64 · Answer

im trying to get the power series that is the derivative of arctan(x^2/4) so that I can determine the power series of arctan(x^2/4) thru integration of the polynomial

amistre64 · Answer

$$\frac{d}{dx}tan(x^2/4)=\frac {8x}{16+x^4}\to \ \frac{x}{2+\frac{x^4}{2^3}}$$
$$\frac{x}{2+\frac{x^4}{2^3}}=\sum_{n=0}^{inf }\frac{(-1)^nx^{4n+1}}{4n+1}$$
$$\int \sum_{n=0}^{inf }\frac{(-1)^nx^{4n+1}}{4n+1} dx = \sum_{n=0}^{inf }\frac{(-1)^nx^{4n+2}}{(4n+1)(4n+2 )}=tan^{-1}(\frac{x^2}{4})$$

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum+%28%28-1%29%5En+x%5E%284n%2B2%29%2F%28%284n%2B2%29%284n%2B1%29%29%29+from+0+to+inf well, im close

anonymous · Answer

is it you are right?

amistre64 · Answer

well, i got the power series for the 1st derivative right :)

anonymous · Answer

i think the answer should be 40320.
x^2/2-x^6/30+x^10/90
f^(10)(0)/10!=1/90

amistre64 · Answer

forgot to type in the denominators 2^(4n+1)

amistre64 · Answer

im sure it could very well be; but i havent gotten that far yet :)

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum+%28%28-1%29%5En+x%5E%284n%2B1%29%2F2%5E%284n%2B1%29%29+from+0+to+inf

anonymous · Answer

but in this case,it will be equal to 0!!

amistre64 · Answer

$$f'(x)=g(x)=\sum_{n=0}^{inf }\frac{(-1)^nx^{4n+1}}{4n+1}$$
$$\frac{d}{dx}x^k=k x^{k-1}$$
$$\frac{d^2}{dx^2}x^k=k(k-1) x^{k-2}$$
$$\frac{d^3}{dx^3}x^k=k(k-1)(k-2) x^{k-3}$$
$$\frac{d^n}{dx^n}x^k=^kP_n\ x^{k-n}$$
$$\frac{d^n}{dx^n}\left( g(x)=\sum_{n=0}^{inf }\frac{(-1)^n}{4n+1}x^{4n+1} \right)$$

amistre64 · Answer

the 9th derivative of g(x) is the 10th derivative of f(x)  :)

amistre64 · Answer

all that typing and the site crashes on me :/

amistre64 · Answer

$$\frac{d^9}{dx^9}\ g(x)=c_0\ ^{1}P_{9}\ x^{1-9}-c_1\ ^{5}P_{9}\ x^{5-9}+c_2\ ^{9}P_{9}\ x^{9-9}...$$

$$\frac{d^9}{dx^9}\ g(x)= c_2\ ^{9}P_{9}+0...=9.8.7.6.5.4.3.2.1\frac{1}{2^9}$$

amistre64 · Answer

9.8.7.6.5.4.3.2.1
  222       22   2


 9.7.6.5.3.1
------------
    2.2.2

 9.7.3.5.3
---------
    2.2

81*35/4

maybe :)

amistre64 · Answer

woohoo!! http://www.wolframalpha.com/input/?i=10th+derivative+arctan%28x%5E2%2F4%29 2835/4 = 81*35/4

anonymous · Answer

Thank you so much!!

anonymous · Answer

:)