If:
∞
sin(x) = ∑ (-1)^n • (x^(2n + 1))/(2n + 1)!
n = 0
∞
cos(x) = ∑ (-1)^n • (x^(2n))/(2n)!
n = 0
Then would the Taylor Series of tan(x) be the ratio of the Taylor Series of sine and cosine respectively??
#Taylor Series Inquiry
desert basalt
subtle bronzeBOT
shadow wedge
,w taylor expansion tan(x)
tawny plankBOT
shadow wedge
If:
∞
sin(x) = ∑ (-1)^n • (x^(2n + 1))/(2n + 1)!
n...
This appears to be mostly true, but be careful;
Desmos seems to bug out when graphing at the asymptotes
Wait
desert basalt
Yeah I’m trying to graph them right now
shadow wedge
Are you saying $\tan(x) = \sum_{n=0}^{\infty}\frac{\frac{\left(-1\right)^{n}x^{2n+1}}{\left(2n+1\right)!}}{\frac{\left(-1\right)^{n}x^{2n}}{\left(2n\right)!}}$
tawny plankBOT
King Leo
shadow wedge
If so, you must remember that every other term of sin(x) is 0, and so some terms are technically not accounted for. same for cos(x). so they dont share the same n
Idk if im wording that properly
desert basalt
I meant something like this
shadow wedge
I meant something like this
Oh then yes
desert basalt
**King Leo**
Idk if it’s the same as this
shadow wedge
Desmos seems to bug out when graphing at the asymptotes
_ _
desert basalt
Ok I was asking because chat gpt said no, Desmos looked weird at the asymptotes (sad it doesn’t support infinite sums)
Ok thank you @shadow wedge
.close
subtle bronzeBOT
Solved