Error bound explanation (answer given)
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I have been able to derive the interpolation polynomial $P_2(x)$ of degree two which interpolates $f(x) = sin x$, given the points $(0,0), left(frac{pi}{2}, 1right), (pi, 0).$
Solution:
$$P_2(x) = frac{4}{pi ^2}x(pi - x)$$
**Here is the question below I am having trouble with. I have the solution below I am just looking for some help with understanding it. **
Calculate $P_2left(frac{pi}{4}right)$, an approximation for $fleft( frac{pi}{4}right) = sin left( frac{pi}{4} right)$ and determine the error bound for the interpolation error at $x = frac{pi}{4}.$
Solution:
For the error bound we know that
$$E(f) = frac{(x-0)(x-frac{pi}{2})(x-pi)}{3!}f'''(z)$$
Using $f''' =-cos (x)$ and this the bound $|cos(z)|leq 1$ we obtain at $x = frac{pi}{4}$"
$$|E(f)|leq frac{pi ^3}{128}$$
How did they get:
$$|E(f)|leq frac{pi ^3}{128}$$
where is the $128$ coming from?
calculus linear-algebra numerical-methods
asked Nov 12 at 17:08
user123
46319
|
show 5 more comments
up vote
1
down vote
favorite
I have been able to derive the interpolation polynomial $P_2(x)$ of degree two which interpolates $f(x) = sin x$, given the points $(0,0), left(frac{pi}{2}, 1right), (pi, 0).$
Solution:
$$P_2(x) = frac{4}{pi ^2}x(pi - x)$$
**Here is the question below I am having trouble with. I have the solution below I am just looking for some help with understanding it. **
Calculate $P_2left(frac{pi}{4}right)$, an approximation for $fleft( frac{pi}{4}right) = sin left( frac{pi}{4} right)$ and determine the error bound for the interpolation error at $x = frac{pi}{4}.$
Solution:
For the error bound we know that
$$E(f) = frac{(x-0)(x-frac{pi}{2})(x-pi)}{3!}f'''(z)$$
Using $f''' =-cos (x)$ and this the bound $|cos(z)|leq 1$ we obtain at $x = frac{pi}{4}$"
$$|E(f)|leq frac{pi ^3}{128}$$
How did they get:
$$|E(f)|leq frac{pi ^3}{128}$$
where is the $128$ coming from?
calculus linear-algebra numerical-methods
asked Nov 12 at 17:08
user123
46319
Direct computation by plug $x =pi/4$ into $E(f)$… seems like that.
– xbh
Nov 12 at 17:13
@xbh why don't you calculate $f'''(z)$ in that calculation?
– user123
Nov 12 at 17:15
You can do that, but all that asked is a general bound for $E(f)$, so you could just use $1$ to bound the derivative $f'''$. If you like, you can explicitly calculate $f'''(pi/4)$ to get a more accurate bound. The answer is not unique, I think.
– xbh
Nov 12 at 17:18
1
Yeah, that is what I meant.
– xbh
Nov 12 at 17:26
1
You are welcome. Glad to help.
– xbh
Nov 12 at 17:27
|
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been able to derive the interpolation polynomial $P_2(x)$ of degree two which interpolates $f(x) = sin x$, given the points $(0,0), left(frac{pi}{2}, 1right), (pi, 0).$
Solution:
$$P_2(x) = frac{4}{pi ^2}x(pi - x)$$
**Here is the question below I am having trouble with. I have the solution below I am just looking for some help with understanding it. **
Calculate $P_2left(frac{pi}{4}right)$, an approximation for $fleft( frac{pi}{4}right) = sin left( frac{pi}{4} right)$ and determine the error bound for the interpolation error at $x = frac{pi}{4}.$
Solution:
For the error bound we know that
$$E(f) = frac{(x-0)(x-frac{pi}{2})(x-pi)}{3!}f'''(z)$$
Using $f''' =-cos (x)$ and this the bound $|cos(z)|leq 1$ we obtain at $x = frac{pi}{4}$"
$$|E(f)|leq frac{pi ^3}{128}$$
How did they get:
$$|E(f)|leq frac{pi ^3}{128}$$
where is the $128$ coming from?
calculus linear-algebra numerical-methods
asked Nov 12 at 17:08
user123
46319
I have been able to derive the interpolation polynomial $P_2(x)$ of degree two which interpolates $f(x) = sin x$, given the points $(0,0), left(frac{pi}{2}, 1right), (pi, 0).$
Solution:
$$P_2(x) = frac{4}{pi ^2}x(pi - x)$$
**Here is the question below I am having trouble with. I have the solution below I am just looking for some help with understanding it. **
Calculate $P_2left(frac{pi}{4}right)$, an approximation for $fleft( frac{pi}{4}right) = sin left( frac{pi}{4} right)$ and determine the error bound for the interpolation error at $x = frac{pi}{4}.$
Solution:
For the error bound we know that
$$E(f) = frac{(x-0)(x-frac{pi}{2})(x-pi)}{3!}f'''(z)$$
Using $f''' =-cos (x)$ and this the bound $|cos(z)|leq 1$ we obtain at $x = frac{pi}{4}$"
$$|E(f)|leq frac{pi ^3}{128}$$
How did they get:
$$|E(f)|leq frac{pi ^3}{128}$$
where is the $128$ coming from?
calculus linear-algebra numerical-methods
calculus linear-algebra numerical-methods
asked Nov 12 at 17:08
user123
46319
asked Nov 12 at 17:08
user123
46319
asked Nov 12 at 17:08
user123
46319
asked Nov 12 at 17:08
user123
46319
asked Nov 12 at 17:08
user123
46319
46319
Direct computation by plug $x =pi/4$ into $E(f)$… seems like that.
– xbh
Nov 12 at 17:13
@xbh why don't you calculate $f'''(z)$ in that calculation?
– user123
Nov 12 at 17:15
You can do that, but all that asked is a general bound for $E(f)$, so you could just use $1$ to bound the derivative $f'''$. If you like, you can explicitly calculate $f'''(pi/4)$ to get a more accurate bound. The answer is not unique, I think.
– xbh
Nov 12 at 17:18
1
Yeah, that is what I meant.
– xbh
Nov 12 at 17:26
1
You are welcome. Glad to help.
– xbh
Nov 12 at 17:27
|
show 5 more comments
Direct computation by plug $x =pi/4$ into $E(f)$… seems like that.
– xbh
Nov 12 at 17:13
@xbh why don't you calculate $f'''(z)$ in that calculation?
– user123
Nov 12 at 17:15
You can do that, but all that asked is a general bound for $E(f)$, so you could just use $1$ to bound the derivative $f'''$. If you like, you can explicitly calculate $f'''(pi/4)$ to get a more accurate bound. The answer is not unique, I think.
– xbh
Nov 12 at 17:18
1
Yeah, that is what I meant.
– xbh
Nov 12 at 17:26
1
You are welcome. Glad to help.
– xbh
Nov 12 at 17:27
Direct computation by plug $x =pi/4$ into $E(f)$… seems like that.
– xbh
Nov 12 at 17:13
Direct computation by plug $x =pi/4$ into $E(f)$… seems like that.
– xbh
Nov 12 at 17:13
@xbh why don't you calculate $f'''(z)$ in that calculation?
– user123
Nov 12 at 17:15
@xbh why don't you calculate $f'''(z)$ in that calculation?
– user123
Nov 12 at 17:15
You can do that, but all that asked is a general bound for $E(f)$, so you could just use $1$ to bound the derivative $f'''$. If you like, you can explicitly calculate $f'''(pi/4)$ to get a more accurate bound. The answer is not unique, I think.
– xbh
Nov 12 at 17:18
You can do that, but all that asked is a general bound for $E(f)$, so you could just use $1$ to bound the derivative $f'''$. If you like, you can explicitly calculate $f'''(pi/4)$ to get a more accurate bound. The answer is not unique, I think.
– xbh
Nov 12 at 17:18
1
1
Yeah, that is what I meant.
– xbh
Nov 12 at 17:26
Yeah, that is what I meant.
– xbh
Nov 12 at 17:26
1
1
You are welcome. Glad to help.
– xbh
Nov 12 at 17:27
You are welcome. Glad to help.
– xbh
Nov 12 at 17:27
|
show 5 more comments
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Direct computation by plug $x =pi/4$ into $E(f)$… seems like that.
– xbh
Nov 12 at 17:13
@xbh why don't you calculate $f'''(z)$ in that calculation?
– user123
Nov 12 at 17:15
You can do that, but all that asked is a general bound for $E(f)$, so you could just use $1$ to bound the derivative $f'''$. If you like, you can explicitly calculate $f'''(pi/4)$ to get a more accurate bound. The answer is not unique, I think.
– xbh
Nov 12 at 17:18
1
Yeah, that is what I meant.
– xbh
Nov 12 at 17:26
1
You are welcome. Glad to help.
– xbh
Nov 12 at 17:27