# Homework Help: Maclaurin formula: finding a delta for a given error

1. Aug 13, 2010

### Telemachus

Approximate the function $$f(x)=\sin(x)$$ using the corresponding Maclaurin polynomial: $$P_5(x)$$, in a bound $$\epsilon(0,\delta)$$. Determine a value of $$\delta>0$$, so that the rest $$R_5(x)$$ verifies $$|R_5(x)|<0.0005$$ for all $$x\in{\epsilon(0,\delta)}$$

Well, the first thing that puzzles me a bit is that the order for $$P_5$$ and $$R_5$$ is the same. I assume that it is a mistake, and that the polynomial must be $$P_4$$, or the rest $$R_6$$. I have chosen to leave the degree of the polynomial as it was and turn up one grade to the rest.

So I have to find a delta for which any value within the environment $$(0,\delta)$$ is less than the error 0.0005

So I did:

$$R_6(x)=|\displaystyle\frac{\sin(\alpha)x^6}{6!}|<0.0005$$ $$0<\alpha<x$$

So I must find x:

$$R_6(x)=|\sin(\alpha)x^6|<0.36$$ $$0<\alpha<x$$

I must find x that satisfies the equation: $$\sin(x)x^6<0.36$$

I don't know how to go ahead.

Bye there.

Last edited: Aug 13, 2010
2. Aug 13, 2010

### vela

Staff Emeritus
I think you're just getting confused by the notation. The usual notation is

$$R_n(x) = f(x)-P_n(x)$$

The remainder term turns out to depend on xn+1 and f(n+1)(x).
Hint: Use the fact that x ≥ sin x when x ≥ 0.

3. Aug 13, 2010

### Telemachus

I don't get it. I know that 1≥sin(x) how should I use the fact you mentioned? I wouldn't noted it if you weren't.

Thanks vela.

4. Aug 13, 2010

### vela

Staff Emeritus
The fact that x ≥ sin x for x ≥ 0 is one of those things you pick up along the way, typically when you do a problem like this one.

You're trying to solve for x, right? And the difficulty is that you can't solve an equation of the form xn sin x = c. So you use the usual approach. Instead of solving the equation exactly, you look for an upper bound that you can solve that gives you a good estimate. In this case, you know that x6 sin x ≤ x7.

5. Aug 13, 2010

### Telemachus

Thanks man!

One doubt. I was thinking that as the sine is always less than one I could use x⁶ as an upper bound too. Right?

6. Aug 13, 2010

### Staff: Mentor

What does this say - x⁶ ?

7. Aug 13, 2010

### vela

Staff Emeritus
Yes, you can. So what is the difference between the two choices? Why would you choose one over the other?

8. Aug 13, 2010

### Staff: Mentor

sin(x) = 1 for x = pi/2 + 2k*pi, k an integer.

What is x⁶? I can't read the exponent.

9. Aug 13, 2010

### vela

Staff Emeritus
Telemachus's idea is to replace sin x by 1, so he'd end up with the inequality x^6 ≤ 0.36.

It's weird. His exponent is in a smaller font than mine.

10. Aug 13, 2010

### Telemachus

Sorry, I was studying :P

x⁶ is $$x^6$$

I think I would choose $$x^6$$ cause its a better approximation than $$x^7$$, so I got $$|x<\sqrt[6]{\frac{9}{25}}$$

Edit: Ok, I've seen that I was wrong, and that $$x^7$$ is a better approximation. But now I have more doubts, damn. For this: x ≥ sin x, or even using x^6 ≥ x^6 sin x, couldn't it happen that using that approximation I would get outside the boundary I was looking for? In this case I've noted that it works, but in general I would like to know what could happens. I mean, while using the substitution it could happens that the new value could be bigger than the error I was looking for, right? cause, I know that $$x^7\geq{x^6 sin x}$$ and so $$x^8\geq{x^6 sin x}$$ it is, and even for any n>6 $$x^n\geq{x^6}$$ right?

Mmm ok, I am wrong for values of x, such that $$x\leq{1}$$. I've tried with x^8 and it seems to work too, and its a better approximation for the boundary I am looking for.

http://www.wolframalpha.com/input/?i=x^6+sin(x)+=+9/25

From here I get x ~~ 0.8807530939284029...

And http://www.wolframalpha.com/input/?i=(9/25)^(1/8)

0.88011173679339339727

Last edited: Aug 13, 2010
11. Aug 13, 2010

### vela

Staff Emeritus
Nope. As long as your steps are reversible, which they usually are when you're doing these algebraic manipulations, you can start with the condition you found for x and deduce that the error will be within the given bound.

12. Aug 13, 2010

### Telemachus

Thanks vela, this is really helpful and very useful.