Understanding Taylor Series and Error Bounds in Calculus

[zcd]

I'm doing some review over summer before starting college, and one of the practice exams has a question pertaining to the remainder of a taylor series

Homework Statement

Show that $$\left|\cos{(1+x)}-\{\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\}\right|<\frac{1}{15000}$$ for $$|x|<0.2$$

Homework Equations

The two equations I somewhat remember are
$$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{(n+1)}$$
and
$$|R_n(x)|\leq{}M_n\frac{r^{n+1}}{(n+1)!}$$
where M (iirc) is the largest value of the nth derivative in the area of convergence?

The Attempt at a Solution

Here's where I'm stuck; my class in high school never fully went over error of taylor approximation because it wasn't part of the curriculum, so we were given the equations without any actual application. To make matters worse, I last took AP calculus BC over a year ago, so this is still very rusty. Can anyone point me in the right direction?

 

[Tedjn]

I also don't recall the formulas either, although those look right. You can review the details at various places online, such as http://en.wikipedia.org/wiki/Taylor's_theorem#Estimates_of_the_remainder. You may also find relevant, if you are not already using it, the formula for cos(a+b).

 

[LCKurtz]

I'm doing some review over summer before starting college, and one of the practice exams has a question pertaining to the remainder of a taylor series

Homework Statement

Show that $$\left|\cos{(1+x)}-\{\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\}\right|<\frac{1}{15000}$$ for $$|x|<0.2$$

Homework Equations

The two equations I somewhat remember are
$$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{(n+1)}$$
and
$$|R_n(x)|\leq{}M_n\frac{r^{n+1}}{(n+1)!}$$
where M (iirc) is the largest value of the nth derivative in the area of convergence?

The Attempt at a Solution

Here's where I'm stuck; my class in high school never fully went over error of taylor approximation because it wasn't part of the curriculum, so we were given the equations without any actual application. To make matters worse, I last took AP calculus BC over a year ago, so this is still very rusty. Can anyone point me in the right direction?

If you expand cos(1+x) in a Taylor series about x = 0 up through the term containing x³ you will have:

cos(1+x) = (terms from constant through x³) + R₃

where R₃ is the error term if you stop with x³. So you will have

|cos(1+x) - (terms from constant through x³)| = |R₃|, where

$$R_3(x)=\frac{f^{(4)}(c)}{(4)!}(x-0)^{(4)}$$

If you write out the series you will see it is exactly what is given. All you have to do is estimate the remainder. It tells you that |x| ≤ .2 and you will find it easy to overestimate f⁽⁴⁾(x). Just be brave and work it out. Don't use any decimals, stick with fractions.

 

[zcd]

Still getting my feet wet again so correct me if I'm wrong :P

$$\left|\cos{(1+x)}-\{\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\}\right|=\left|\cos{(1)}((1-\frac{x^2}{2}+R_3(x))-(1-\frac{x^2}{2}))-\sin{(1)}((x-\frac{x^3}{3!}+R_3(x))-(x-\frac{x^3}{3!}))\right|$$
$$=\left|\cos{(1)}R_3(x)-\sin{(1)}R_3(x)\right|$$

since $$|R_n(x)|\leq{}M_n\frac{r^{n+1}}{(n+1)!}$$ and M_n=1 for sin and cos, $$|R_3(x)|\leq{}\frac{(\frac{1}{5})^4}{4!}=\frac{1}{15000}$$

plugging it all back in,
$$\left|\cos{(1)}\frac{1}{15000}-\sin{(1)}\frac{1}{15000}\right|<\frac{1}{15000}$$

is that right?

 

[LCKurtz]

Still getting my feet wet again so correct me if I'm wrong :P

$$\left|\cos{(1+x)}-\{\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\}\right|=\left|\cos{(1)}((1-\frac{x^2}{2}+R_3(x))-(1-\frac{x^2}{2}))-\sin{(1)}((x-\frac{x^3}{3!}+R_3(x))-(x-\frac{x^3}{3!}))\right|$$
$$=\left|\cos{(1)}R_3(x)-\sin{(1)}R_3(x)\right|$$

since $$|R_n(x)|\leq{}M_n\frac{r^{n+1}}{(n+1)!}$$ and M_n=1 for sin and cos, $$|R_3(x)|\leq{}\frac{(\frac{1}{5})^4}{4!}=\frac{1}{15000}$$

plugging it all back in,
$$\left|\cos{(1)}\frac{1}{15000}-\sin{(1)}\frac{1}{15000}\right|<\frac{1}{15000}$$

is that right?

The only thing right is your estimate for R₃. The rest is just jumbled confusion.

Just write the Taylor series for cos(1+x) using the terms up through x³ like I suggested. You know, calculate f(0), f'(0), f''(0) etc. and write the series. This doesn't involve the remainder. Rearrange it to get the expression you are given. Write it as suggested in my previous post.

 

[zcd]

I think I get it

$$\cos{(1+x)}=\cos{(1)}-x\sin{(1)}-\frac{x^2}{2}\cos{(1)}+\frac{x^3}{3!}\sin{(1)}+...$$
$$\cos{(1+x)}=\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})+R_3(x)$$
$$\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]=R_3(x)$$
$$\left|\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]\right|=\left|R_3(x)\right|$$
$$\left|\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]\right|\leq\frac{1}{15000}$$

how does the =< turn into <?

 

[LCKurtz]

I think I get it

$$\cos{(1+x)}=\cos{(1)}-x\sin{(1)}-\frac{x^2}{2}\cos{(1)}+\frac{x^3}{3!}\sin{(1)}+...$$
$$\cos{(1+x)}=\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})+R_3(x)$$
$$\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]=R_3(x)$$
$$\left|\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]\right|=\left|R_3(x)\right|$$
$$\left|\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]\right|\leq\frac{1}{15000}$$

how does the =< turn into <?

Good job. Look at the limits on x. Are you given strict inequality in your estimate?

 

[zcd]

Ahh I see. Thanks for all the help :)