Question

Find the first four nonzero terms of the Taylor series about 0 for the function f(x)=(sqrt(1+x))*cos(3x)

Answer 1

cos(3x)=1-((9/2)*x^2)+((81/4)*x^4)-...

Answer 2

@Easyaspi314 can you help with this one by chance?

Answer 3

expand sqrt(1+x) and multiply them

Answer 4

use \((1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + ....\)

n = 1/2

Answer 5

i got the first 2 terms correct but the 3rd and 4th arent correct

the first 2 terms are 1+(1/2)x

Answer 6

shouldn't 2 be 2!?

Answer 7

\(\large f(x) = \sqrt{1+x} \times cos(3x)\)

\(\large f(x) =\left( 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + ... \right) \times \left( 1 - \frac{9}{2}x^2 + .... \right) \)

Answer 8

yes 2! = 2

Answer 9

multiply them above, and combine like terms

Answer 10

you just need first 4 terms, so upto x^3 oly u need...

Answer 11

(-1/8)*(9/2)=(9/16)x^2 for the third term and that isnt right

Answer 12

\(\large f(x) = \sqrt{1+x} \times cos(3x)\)

\(\large f(x) =\left( 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + ... \right) \times \left( 1 - \frac{9}{2}x^2 + .... \right) \)

\(\large f(x) = 1 + \frac{1}{2}x + \left(- \frac{1}{8}x^2 - \frac{9}{2}x^2\right)+ \left( \frac{1}{16}x^3 - \frac{9}{4}x^3 \right)+ ... \)

Answer 13

\(\large f(x) = \sqrt{1+x} \times cos(3x)\)

\(\large f(x) =\left( 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + ... \right) \times \left( 1 - \frac{9}{2}x^2 + .... \right) \)

\(\large f(x) = 1 + \frac{1}{2}x + \left(- \frac{1}{8}x^2 - \frac{9}{2}x^2\right)+ \left( \frac{1}{16}x^3 - \frac{9}{4}x^3 \right)+ ... \)

\(\large f(x) = 1 + \frac{1}{2}x + - \frac{37}{8}x^2 - \frac{35}{16}x^3 + ... \)

Answer 14

oooooooh!!!! omgsh ur a genius! thank you so much!

Answer 15

thats one way to expand using known series;
another way is u cud have directly computed the derivatives for f(x)... and plug them in the maclaurin series

Answer 16

np :)