# Thread: The Remainder Estimation Theorem Problem

1. ## The Remainder Estimation Theorem Problem

Use the Remainder Estimation Theorem to find an interval containing x=0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval.

f(x)=cosx p(x)=1 - x^2/2! + x^4/4!

2. Originally Posted by soleilion
Use the Remainder Estimation Theorem to find an interval containing x=0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval.

f(x)=cosx p(x)=1 - x^2/2! + x^4/4!
This is an Alternating series so the error in less than next term so in this case we have

$|\frac{x^6}{6!}|<.0001$

$|\frac{x^6}{720}|<.0001$

$\frac{x^6}{720} < .0001 \iff < x^6 < 0.0720 \iff x < .650$

3. Originally Posted by TheEmptySet
This is an Alternating series so the error in less than next term so in this case we have

$|\frac{x^6}{6!}|<.0001$

$|\frac{x^6}{720}|<.0001$

$\frac{x^6}{720} < .0001 \iff < x^6 < 0.0720 \iff x < .650$

I think you just use p(x), how about f(x), it does not matter?

By the way, I think the number should be less than 0.0005, not 0.0001

4. Originally Posted by soleilion
I think you just use p(x), how about f(x), it does not matter?

By the way, I think the number should be less than 0.0005, not 0.0001
Not when using this theorem.

Theorem:

if $S_n=\sum (-1)^nb_n$ is a sum of an alternating series
that satisfies

$0 \le b_{n+1} \le b_n$ and $\lim_{n \to \infty}b_n =0$

Then the remainder $|R_n| \le b_{n+1}$

The question said
to find an interval containing x=0
It is not unique.
we could use any decimal that doesn't change in the thousandths column

5. Originally Posted by TheEmptySet
Not when using this theorem.

Theorem:

if $S_n=\sum (-1)^nb_n$ is a sum of an alternating series
that satisfies

$0 \le b_{n+1} \le b_n$ and $\lim_{n \to \infty}b_n =0$

Then the remainder $|R_n| \le b_{n+1}$

The question said

It is not unique.
we could use any decimal that doesn't change in the thousandths column
so I have to prove f(x) is converges, then I use
then, I can get x
right ?

6. The Macclaurin Series for cosine converges for all real x

$\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$

So by the theorm above we use the forth term (n=3)

$b_3=-\frac{x^6}{6!}$ we set this less than the error

$|b_3| < .0001$ and solve for x.

I hope this clears it up.

Good luck.

7. Originally Posted by TheEmptySet
The Macclaurin Series for cosine converges for all real x

$\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$

So by the theorm above we use the forth term (n=3)

$b_3=-\frac{x^6}{6!}$ we set this less than the error

$|b_3| < .0001$ and solve for x.

I hope this clears it up.

Good luck.
yes!!!
I understood this
Thnak you

8. Originally Posted by TheEmptySet
The Macclaurin Series for cosine converges for all real x

$\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$

So by the theorm above we use the forth term (n=3)

$b_3=-\frac{x^6}{6!}$ we set this less than the error

$|b_3| < .0001$ and solve for x.

I hope this clears it up.

Good luck.

sorry, I still have a question
How do you know b3<=0.0001
why not b2 or b4?

9. Originally Posted by soleilion
sorry, I still have a question
How do you know b3<=0.0001
why not b2 or b4?

You were using the fist three terms of the series n=0,1,2

$P(x)=\underbrace{1}_{n=0} - \underbrace{\frac{x^2}{2}}_{n=1} + \underbrace{\frac{x^4}{24}}_{n=2}$

so the next term would be when n=3.

Good luck.

10. Originally Posted by TheEmptySet
You were using the fist three terms of the series n=0,1,2

$P(x)=\underbrace{1}_{n=0} - \underbrace{\frac{x^2}{2}}_{n=1} + \underbrace{\frac{x^4}{24}}_{n=2}$

so the next term would be when n=3.

Good luck.

I know this
I mean how do you know b3 less than 0.0001
because b3 includes x, you do not know what x is

11. Originally Posted by soleilion
I know this
I mean how do you know b3 less than 0.0001
because b3 includes x, you do not know what x is
We set the term $b_3$ less then the error (x and all)

and solve the inequality for x. This gives us the distance x can be from the center of the series and get the accuracy we want. The Series is centered at zero.

since $x<.65$ we get the set $(-.65,.65)$

lets do a quick check

p(.65)=.796187

cos(.65)=.79608

The error is within what we wan't yeah!!

12. Originally Posted by TheEmptySet
We set the term $b_3$ less then the error (x and all)

and solve the inequality for x. This gives us the distance x can be from the center of the series and get the accuracy we want. The Series is centered at zero.

since $x<.65$ we get the set $(-.65,.65)$

lets do a quick check

p(.65)=.796187

cos(.65)=.79608

The error is within what we wan't yeah!!
Sorry, I did not understand

13. Originally Posted by TheEmptySet
We set the term $b_3$ less then the error (x and all)

and solve the inequality for x. This gives us the distance x can be from the center of the series and get the accuracy we want. The Series is centered at zero.

since $x<.65$ we get the set $(-.65,.65)$

lets do a quick check

p(.65)=.796187

cos(.65)=.79608

The error is within what we wan't yeah!!
why did I miss type??????????????????????

I type it again

can you see this the #2 post

I mean we can set x^8/8! less than .0001

also we can set x^4/4! less than .0001

we can get two different answer

How can we know x^6/6! is the right one?

14. Originally Posted by soleilion
why did I miss type??????????????????????

I type it again

can you see this the #2 post

I mean we can set x^8/8! less than .0001

also we can set x^4/4! less than .0001

we can get two different answer

How can we know x^6/6! is the right one?
$\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=\underbrace{1-\frac{x^2}{2}+\frac{x^4}{4!}}_{p(x)}-\frac{x^6}{6!}+\frac{x^8}{8!}-...$

according to the AST remainder theorem we use the next term in the series

$-\frac{x^6}{6!}$ this is when n=3 in the sum we set that term less than the error.

15. Originally Posted by TheEmptySet
$\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=\underbrace{1-\frac{x^2}{2}+\frac{x^4}{4!}}_{p(x)}-\frac{x^6}{6!}+\frac{x^8}{8!}-...$

according to the AST remainder theorem we use the next term in the series

$-\frac{x^6}{6!}$ this is when n=3 in the sum we set that term less than the error.