Question

Find the taylor series of tan(x) with a = 0;

Answer 1

is this in english?

Answer 2

Mathematics.

Answer 3

isnt it just taylor sin over cos?

Answer 4

cos=1
-sin=0
-cos=-1
sin=0
cos=1
-sin=0
-cos=-1
sin=0
cos=1

cos starts at 1 and are the evens

\[cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}...\]

and sin(x) is the derivative of cos(x)

\[cos(x)'=-(-\frac{2x}{2*1!}+\frac{4x^3}{4*3!}-\frac{6x^5}{6*5!}+\frac{8x^7}{8*7!}...\]
\[sin(x)=\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...\]

Answer 5

the other way is to try and determine a pattern for tan(x) from its derivatives

Answer 6

tan
sec^2
2 sec^2 tan
4 sec^2 tan^2 + 2sec^4

its gets messy I believe

Answer 7

the good news tho might be that tan0=0 so all the tan terms cancels out and your left with C sec^2(0) = 1