Maclaurin formula: finding a delta for a given error

[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.

 

[vela]

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.

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).

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.

Hint: Use the fact that x ≥ sin x when x ≥ 0.

 

[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.

 

[vela]

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.

 

[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?

 

[Mark44]

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?

What does this say - x⁶ ?

 

[vela]

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?

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

 

[Mark44]

the sine is always less than one

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

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

 

[vela]

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.

 

[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

 

[vela]

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?

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.

 

[Telemachus]

Thanks vela, this is really helpful and very useful.