Hello, I know how to rederive & prove Taylor's Formula as being a simple application(adsbygoogle = window.adsbygoogle || []).push({});

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

Edit: Thinking about it I can use linear approximation:

f(x) = f(x₀) + f'(x₀)(x - x₀)

f(x) = sin(0) + cos(0)(x - 0)

f(x) = x

sin(0.1) = 0.100

and then use the error function thing but it seems to come out of nowhere, i.e. I made it up &

while it's good to have that backup I'm trying to learn how to get this thing right :p

Another Edit: I think I got it

sin(0.1) = sin(0) + cos(0)•(0.1) - sin(0)•(0.1)² /2 - cos(0)•(0.1)³/6 = 0 + 0.1 - 0 - (0.001)/6

Now, 0.001/6 would be around the third decimal place & that's what the question asked for.

I think that's right, please let me know as it seems correct :D

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# Homework Help: Taylor's Formula Question?

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