Question

use mclaurin formula

f(x)= tan x up to the term x^5

Answer 1

i never really knew the difference between taylor and mcclaurin ... one has to do with 0 and the other with a

Answer 2

\[P(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...\]

Answer 3

so essentially, you want the first 5 derivatives of tan(x0

Answer 4

if you wan't want to calculate that derivative ... there is a trick

Answer 5

teach me that trick

Answer 6

tan(x) = sin(x)/cos(x)
let
\[ \tan x = \sum_{k=0}^\infty a_kx^k \]
then sin(x) = ??

Answer 7

if you wan't want to calculate that *without* derivative ... there is a trick

Answer 8

from the
tanx = \[\sum_{k=o}^{\infty} akx ^{k}\]
a is what

Answer 9

\(a_k \) is the coefficient in the power series. we do not know it's value yet.
do you know maclaurin series for sin(x) and cos(x)

Answer 10

\[\sin x = \sum_{k=0}^{\infty} (-1)^k \frac{ x^{2k++!} }{ (2k+1)!} + 0 x^{2k+1} \]

\[\cos x= \sum_{k=0}^{\infty} (-1)^K \frac{ x^2k }{ (2k)! } + 0 ( x^{2n} ) \]

Answer 11

don't put that O notation if use use than infinity as your upper limit.

Answer 12

\[ \sin x = \sum_{k=1}^\infty a_k x^k \times \cos x \\
\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!} \times \sum_{k=0}^{\infty } a_k x^k \]

Answer 13

just for the practice ... lets see if i can do this the "other" way ;)
\[f:~tan\]\[f':~sec^2\]\[f'':~2~sec^3~sin\]\[f''':~6~sec^4~sin^2~+~2~sec^2\]\[f'''':~24~sec^5~sin^3~+~16~sec^3~sin\]\[f^{(5)}:~120~sec^6~sin^4~+~120~sec^4~sin^2~+~16~sec^2\]

Answer 14

yeah you can do that too.

Answer 15

Here's the trick
\[ x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O[x]^8 = \left( 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O[x]^7 \right )\times \left( a_0 + a_1x + a_2 x^2 + a_3x^3 +...\right) \]

up to five terms ... your job is to compare coefficients and determine their value.

this will be yoru tan x

Answer 16

the max series is at a=0 .... or simply evaluated at 0

f:0,2,4 all go to zero

f:1,3,5 got to 1, 2, 16

\[P(x)=x+\frac{2}{3!}x^3+\frac{16}{5!}x^5+...\]

i am definitely not smart enough to work thru experiments method ;)

Answer 17

you are hard worker.

Answer 18

\[
x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O[x]^8 = \\
\left( 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O[x]^7 \right )\times \left( a_0 + a_1x + a_2 x^2 + a_3x^3 +a_4x^4 +a_5x^5 + ...\right) \]
there is no constant term on the left ... so \( a_0 *1 = 0 \) so \( a_0 = 0 \)

Answer 19

on the left there is x ... from the right, you get \(a_1 x \) ... so a1 = 1

Answer 20

keep doing it ... by watchig a2 and a4 = 0 and a3 = 1/3
a5 ... you have to determine by the combination of powers on the left.

or just try hard way amistre mentioned.

Answer 21

thank u very much... @experimentX and @amistre64
it very help me

Answer 22

youre welcome :)