Set your calculator to "Radian mode"! sin(.1) is, in fact, 0.099...
Hello, I know how to rederive & prove Taylor's Formula as being a simple application
of Integration By Parts to ∫f'(x) dx = f(b) - f(a) leading to
f(b) = f(a) + f'(a)(b - a) + ½f''(a)(b - a)² + ... + Rⁿ
and that we can approximate f(x) by setting a = 0 & b = x
f(x) = f(0) + f'(0)x + ½f''(0)x² + ... + Rⁿ
and I understand that Rⁿ lies between the lower & upper bound, i.e.
∃ c ∈ (a,b) : f ⁿ (c)(b - a)ⁿ/n! = Rⁿ
& f ⁿ (c)(b - a)ⁿ/n! = Rⁿ
but I'm confused on how to apply all of this.
For example, f(x) = sinx
okay, I can create the taylor polynomial
sinx = x - x³/3! + x^5/5! - ... + Rⁿ
and the question is compute sin(0.1) to 3 decimals.
Well
a = 0,
x = 0 + 0.1 = 0.1
If I compute it to 3 decimals I get
sin(0.1) = (0.1) - (0.1)³/3! + (0.1)^5/5! = 0.099...
Would this be the 3 decimals? Like, this is not correct at all compared to my calculator's answer for sin(0.1).
the way the book has done it has me really confused.
When I add the error on I get no significant change,
I'm confused as to what the point of the formula is here.
I used my calculator for this btw (isn't this equation supposed to do away with calculators? ) ?
Now, my book has done something very different & I don't understand.
(Serge Lang: A First Course in Calculus - page 309)
Lang ignores calculating the Rⁿ term, and hasn't shown me how to get it yet, but
uses the sin/cosine natural upper bound of 1, so Rⁿ ≤ Mⁿ|xⁿ|/n!
where Mⁿ = 1 as an upper bound.
(As a side note, this comes from the proof that ∃ c ∈ (a,b) : f ⁿ (c)(b - a)ⁿ/n!
where he's chosen f ⁿ (v)(b - a)ⁿ/n! = Mⁿ = 1, i.e. v being the maximum,
is the logic correct or am I missing something? He's just squeezing in Rⁿ underneath)
With this he just calculates
Rⁿ ≤ Mⁿ|xⁿ|/n! -----------> Rⁿ ≤ 1|(0.1)³|/3!
Rⁿ ≤ (0.001)/6
Now, he says that "such accuracy would put us in the required range of accuracy.
Hence we can just use the first term of Taylor's formula."
sin(0.1) = 0.100 ± (0.001)/6
I have no idea how or why he just did all that, non whatsoever...
I don't understand how I could get an incorrect answer by using the x/3! & x/5!
because the added values are supposed to give a better approximation, not a
worse one