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How to develop a numerical differentiation formula using Taylor expansion/series?
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I have a problem understanding a numerical differentiation problem, using Taylor series.
It's an exercise and we should develop a 2nd order approximation formula, which uses just $f(x), f(x-h)$ and $f(x+3h)$. As help, the professor gave us a tip in the exercice, that we should start with:
$$f'(x) = fracaf(x+3h) + bf(x) + cf(x-h) h $$
Attached here is the problem AND the solution (which I think is wrong, or he made something wrong, at least I hope), and a legend for my questions.
I found other results, and here are my 3 questions:
- For point 1 (marked 1 on the screenshot), I found $27/6$ as factor there, not $9/2$
- My 2. question (marked 2 in screenshot) concerns the last term in the result. Shouldn't we multiply that big term with $h^3$?
- 3rd question (marked 3 in screenshot): Let's say the professor is right, and there are no typos or calculating errors. If we are dividing by $h$, these $h$'s on the nominator will get eliminated by 1 power. So the term (middle term), that he says should be 1, is multiplied by $h$, and if we divide by $h$ there won't be any $h$ anymore there. For the term after if, which is multiplied by $h^2$, and we are dividing by $h$, it will be then just $h$, which makes our formula of 1st order. Or did I understand something wrong?
Thank you for any help! Exam is tomorrow, and I am really freaking out about this.
derivatives numerical-methods taylor-expansion numerical-optimization
edited Jul 19 at 17:30
Lorenzo B.
1,5402418
asked Jul 19 at 17:15
ZelelB
1256
 |Â
show 1 more comment
up vote
1
down vote
favorite
I have a problem understanding a numerical differentiation problem, using Taylor series.
It's an exercise and we should develop a 2nd order approximation formula, which uses just $f(x), f(x-h)$ and $f(x+3h)$. As help, the professor gave us a tip in the exercice, that we should start with:
$$f'(x) = fracaf(x+3h) + bf(x) + cf(x-h) h $$
Attached here is the problem AND the solution (which I think is wrong, or he made something wrong, at least I hope), and a legend for my questions.
I found other results, and here are my 3 questions:
- For point 1 (marked 1 on the screenshot), I found $27/6$ as factor there, not $9/2$
- My 2. question (marked 2 in screenshot) concerns the last term in the result. Shouldn't we multiply that big term with $h^3$?
- 3rd question (marked 3 in screenshot): Let's say the professor is right, and there are no typos or calculating errors. If we are dividing by $h$, these $h$'s on the nominator will get eliminated by 1 power. So the term (middle term), that he says should be 1, is multiplied by $h$, and if we divide by $h$ there won't be any $h$ anymore there. For the term after if, which is multiplied by $h^2$, and we are dividing by $h$, it will be then just $h$, which makes our formula of 1st order. Or did I understand something wrong?
Thank you for any help! Exam is tomorrow, and I am really freaking out about this.
derivatives numerical-methods taylor-expansion numerical-optimization
edited Jul 19 at 17:30
Lorenzo B.
1,5402418
asked Jul 19 at 17:15
ZelelB
1256
thanks for looking up at the question. I tried as stated above, that I found for point 1 for example, on that point 27/6, and for point 2, I multiply the term there with h^3. He doesn't multiply it with nothing. Where are the h^3?
– ZelelB
Jul 19 at 17:32
1. $9/2=27/6$ is correct. 2.You forgot the factor $1/h$
– gammatester
Jul 19 at 17:38
But there is an error in 1: it should read $f'''(x_0)$ not $f'''(0)$.
– gammatester
Jul 19 at 17:45
you're right for 1, thanks. But for 2, how he forgot 1/h? I am saying he should multiply there with h^3. If you look at the line before, from which he's derivating that line. He is summing the h's, so a term multiplies h, another h^2, and another term multiplies h^3. Or am I wrong. Still don't get it. Where are the h^3 of line before gone? And where is here getting "-9/2" MINUS not +9/2.
– ZelelB
Jul 19 at 17:53
It should read $Big(frac92 a f'''(x_0)-frac16cf'''(x_0)Big)h^3$
– gammatester
Jul 19 at 18:12
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a problem understanding a numerical differentiation problem, using Taylor series.
It's an exercise and we should develop a 2nd order approximation formula, which uses just $f(x), f(x-h)$ and $f(x+3h)$. As help, the professor gave us a tip in the exercice, that we should start with:
$$f'(x) = fracaf(x+3h) + bf(x) + cf(x-h) h $$
Attached here is the problem AND the solution (which I think is wrong, or he made something wrong, at least I hope), and a legend for my questions.
I found other results, and here are my 3 questions:
- For point 1 (marked 1 on the screenshot), I found $27/6$ as factor there, not $9/2$
- My 2. question (marked 2 in screenshot) concerns the last term in the result. Shouldn't we multiply that big term with $h^3$?
- 3rd question (marked 3 in screenshot): Let's say the professor is right, and there are no typos or calculating errors. If we are dividing by $h$, these $h$'s on the nominator will get eliminated by 1 power. So the term (middle term), that he says should be 1, is multiplied by $h$, and if we divide by $h$ there won't be any $h$ anymore there. For the term after if, which is multiplied by $h^2$, and we are dividing by $h$, it will be then just $h$, which makes our formula of 1st order. Or did I understand something wrong?
Thank you for any help! Exam is tomorrow, and I am really freaking out about this.
derivatives numerical-methods taylor-expansion numerical-optimization
edited Jul 19 at 17:30
Lorenzo B.
1,5402418
asked Jul 19 at 17:15
ZelelB
1256
I have a problem understanding a numerical differentiation problem, using Taylor series.
It's an exercise and we should develop a 2nd order approximation formula, which uses just $f(x), f(x-h)$ and $f(x+3h)$. As help, the professor gave us a tip in the exercice, that we should start with:
$$f'(x) = fracaf(x+3h) + bf(x) + cf(x-h) h $$
Attached here is the problem AND the solution (which I think is wrong, or he made something wrong, at least I hope), and a legend for my questions.
I found other results, and here are my 3 questions:
- For point 1 (marked 1 on the screenshot), I found $27/6$ as factor there, not $9/2$
- My 2. question (marked 2 in screenshot) concerns the last term in the result. Shouldn't we multiply that big term with $h^3$?
- 3rd question (marked 3 in screenshot): Let's say the professor is right, and there are no typos or calculating errors. If we are dividing by $h$, these $h$'s on the nominator will get eliminated by 1 power. So the term (middle term), that he says should be 1, is multiplied by $h$, and if we divide by $h$ there won't be any $h$ anymore there. For the term after if, which is multiplied by $h^2$, and we are dividing by $h$, it will be then just $h$, which makes our formula of 1st order. Or did I understand something wrong?
Thank you for any help! Exam is tomorrow, and I am really freaking out about this.
derivatives numerical-methods taylor-expansion numerical-optimization
edited Jul 19 at 17:30
Lorenzo B.
1,5402418
asked Jul 19 at 17:15
ZelelB
1256
edited Jul 19 at 17:30
Lorenzo B.
1,5402418
edited Jul 19 at 17:30
Lorenzo B.
1,5402418
edited Jul 19 at 17:30
Lorenzo B.
1,5402418
1,5402418
asked Jul 19 at 17:15
ZelelB
1256
asked Jul 19 at 17:15
ZelelB
1256
asked Jul 19 at 17:15
ZelelB
1256
1256
thanks for looking up at the question. I tried as stated above, that I found for point 1 for example, on that point 27/6, and for point 2, I multiply the term there with h^3. He doesn't multiply it with nothing. Where are the h^3?
– ZelelB
Jul 19 at 17:32
1. $9/2=27/6$ is correct. 2.You forgot the factor $1/h$
– gammatester
Jul 19 at 17:38
But there is an error in 1: it should read $f'''(x_0)$ not $f'''(0)$.
– gammatester
Jul 19 at 17:45
you're right for 1, thanks. But for 2, how he forgot 1/h? I am saying he should multiply there with h^3. If you look at the line before, from which he's derivating that line. He is summing the h's, so a term multiplies h, another h^2, and another term multiplies h^3. Or am I wrong. Still don't get it. Where are the h^3 of line before gone? And where is here getting "-9/2" MINUS not +9/2.
– ZelelB
Jul 19 at 17:53
It should read $Big(frac92 a f'''(x_0)-frac16cf'''(x_0)Big)h^3$
– gammatester
Jul 19 at 18:12
 |Â
show 1 more comment
thanks for looking up at the question. I tried as stated above, that I found for point 1 for example, on that point 27/6, and for point 2, I multiply the term there with h^3. He doesn't multiply it with nothing. Where are the h^3?
– ZelelB
Jul 19 at 17:32
1. $9/2=27/6$ is correct. 2.You forgot the factor $1/h$
– gammatester
Jul 19 at 17:38
But there is an error in 1: it should read $f'''(x_0)$ not $f'''(0)$.
– gammatester
Jul 19 at 17:45
you're right for 1, thanks. But for 2, how he forgot 1/h? I am saying he should multiply there with h^3. If you look at the line before, from which he's derivating that line. He is summing the h's, so a term multiplies h, another h^2, and another term multiplies h^3. Or am I wrong. Still don't get it. Where are the h^3 of line before gone? And where is here getting "-9/2" MINUS not +9/2.
– ZelelB
Jul 19 at 17:53
It should read $Big(frac92 a f'''(x_0)-frac16cf'''(x_0)Big)h^3$
– gammatester
Jul 19 at 18:12
thanks for looking up at the question. I tried as stated above, that I found for point 1 for example, on that point 27/6, and for point 2, I multiply the term there with h^3. He doesn't multiply it with nothing. Where are the h^3?
– ZelelB
Jul 19 at 17:32
thanks for looking up at the question. I tried as stated above, that I found for point 1 for example, on that point 27/6, and for point 2, I multiply the term there with h^3. He doesn't multiply it with nothing. Where are the h^3?
– ZelelB
Jul 19 at 17:32
1. $9/2=27/6$ is correct. 2.You forgot the factor $1/h$
– gammatester
Jul 19 at 17:38
1. $9/2=27/6$ is correct. 2.You forgot the factor $1/h$
– gammatester
Jul 19 at 17:38
But there is an error in 1: it should read $f'''(x_0)$ not $f'''(0)$.
– gammatester
Jul 19 at 17:45
But there is an error in 1: it should read $f'''(x_0)$ not $f'''(0)$.
– gammatester
Jul 19 at 17:45
you're right for 1, thanks. But for 2, how he forgot 1/h? I am saying he should multiply there with h^3. If you look at the line before, from which he's derivating that line. He is summing the h's, so a term multiplies h, another h^2, and another term multiplies h^3. Or am I wrong. Still don't get it. Where are the h^3 of line before gone? And where is here getting "-9/2" MINUS not +9/2.
– ZelelB
Jul 19 at 17:53
you're right for 1, thanks. But for 2, how he forgot 1/h? I am saying he should multiply there with h^3. If you look at the line before, from which he's derivating that line. He is summing the h's, so a term multiplies h, another h^2, and another term multiplies h^3. Or am I wrong. Still don't get it. Where are the h^3 of line before gone? And where is here getting "-9/2" MINUS not +9/2.
– ZelelB
Jul 19 at 17:53
It should read $Big(frac92 a f'''(x_0)-frac16cf'''(x_0)Big)h^3$
– gammatester
Jul 19 at 18:12
It should read $Big(frac92 a f'''(x_0)-frac16cf'''(x_0)Big)h^3$
– gammatester
Jul 19 at 18:12
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
1) $frac 276 = frac 92$
2) yes there should be an $h^3$ factor multiplying that term. However, that term ultimately gets dropped (actually becomes the error estimate). And is not relevant for finding optimal values of $a, b, c$
3) yes the $h's$ cancel. That is kind of the point.
$frac 1h [(a+b+c)f(x) + (3a+c)f'(x)h + (frac 92 a + frac 12 c) f''(x)h^2+epsilon h^3] = f'(x)$
Which can only be true if
$(a+b+c) = 0$ and $(3a+c) = 1$
as for $frac 92 a + frac 12 c = 0$ This makes your error term decrease at quadratic speed rather than linear speed.
answered Jul 19 at 18:56
Doug M
39.2k31749
but how does it drop completely? if there were the h^3 (which he forgot to write on the solution), it will be just h^2 (after dividing by h), and not drop completely.. I dont get it :(
– ZelelB
Jul 19 at 20:03
I thought, it's the most important term! We divide the h^3 by h, so we he h^2, and that's our 2nd order formula, he's looking for
– ZelelB
Jul 19 at 20:04
add a comment |Â
up vote
1
down vote
More generally,
suppose you want
to find
$f'(x) approx dfracaf(x+uh) + bf(x+vh) + cf(x+wh) h
$.
$f(x+rh)
=f(x)+rhf'(x)+r^2h^2f''(x)/2+r^3h^3f'''(x)/6+O(h^4)
$
so
$beginarray\
af(x+uh) + bf(x+vh) + cf(x+wh)
&=(a+b+c)f(x)+(au+bv+cw)hf'(x)\
&quad +(au^2+bv^2+cw^2)h^2f''(x)/2+(au^3+bv^3+cw^3)h^3f'''(x)/6\
&quad+O(h^4)\
endarray
$
Therefore you want
$0=a+b+c,
1=au+bv+cw,
0=au^2+bv^2+cw^2
$
and the error is
$(au^3+bv^3+cw^3)h^2f'''(x)/6
+O(h^3)
$.
This leads to Vandermonde determinants.
Your case is
$u=3, v=0, w=-1$,
so
$0=a+b+c,
1=3a-c,
0=9a+c
$.
Adding the second and third,
$a=1/12$,
so
$c = -9a
=-3/4$.
From the first,
$b = -a-c
=-1/12+3/4
=2/3
$.
The error coefficient is
$au^3+bv^3+cw^3
=27/12-3/4
=15/12
=5/4
$
so the error is
$(5/4)h^2f'''(x)/6+O(h^3)
=(5/24)h^2f'''(x)+O(h^3)
$.
answered Jul 19 at 18:27
marty cohen
69.2k446122
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
1) $frac 276 = frac 92$
2) yes there should be an $h^3$ factor multiplying that term. However, that term ultimately gets dropped (actually becomes the error estimate). And is not relevant for finding optimal values of $a, b, c$
3) yes the $h's$ cancel. That is kind of the point.
$frac 1h [(a+b+c)f(x) + (3a+c)f'(x)h + (frac 92 a + frac 12 c) f''(x)h^2+epsilon h^3] = f'(x)$
Which can only be true if
$(a+b+c) = 0$ and $(3a+c) = 1$
as for $frac 92 a + frac 12 c = 0$ This makes your error term decrease at quadratic speed rather than linear speed.
answered Jul 19 at 18:56
Doug M
39.2k31749
but how does it drop completely? if there were the h^3 (which he forgot to write on the solution), it will be just h^2 (after dividing by h), and not drop completely.. I dont get it :(
– ZelelB
Jul 19 at 20:03
I thought, it's the most important term! We divide the h^3 by h, so we he h^2, and that's our 2nd order formula, he's looking for
– ZelelB
Jul 19 at 20:04
add a comment |Â
up vote
2
down vote
1) $frac 276 = frac 92$
2) yes there should be an $h^3$ factor multiplying that term. However, that term ultimately gets dropped (actually becomes the error estimate). And is not relevant for finding optimal values of $a, b, c$
3) yes the $h's$ cancel. That is kind of the point.
$frac 1h [(a+b+c)f(x) + (3a+c)f'(x)h + (frac 92 a + frac 12 c) f''(x)h^2+epsilon h^3] = f'(x)$
Which can only be true if
$(a+b+c) = 0$ and $(3a+c) = 1$
as for $frac 92 a + frac 12 c = 0$ This makes your error term decrease at quadratic speed rather than linear speed.
answered Jul 19 at 18:56
Doug M
39.2k31749
but how does it drop completely? if there were the h^3 (which he forgot to write on the solution), it will be just h^2 (after dividing by h), and not drop completely.. I dont get it :(
– ZelelB
Jul 19 at 20:03
I thought, it's the most important term! We divide the h^3 by h, so we he h^2, and that's our 2nd order formula, he's looking for
– ZelelB
Jul 19 at 20:04
add a comment |Â
up vote
2
down vote
up vote
2
down vote
1) $frac 276 = frac 92$
2) yes there should be an $h^3$ factor multiplying that term. However, that term ultimately gets dropped (actually becomes the error estimate). And is not relevant for finding optimal values of $a, b, c$
3) yes the $h's$ cancel. That is kind of the point.
$frac 1h [(a+b+c)f(x) + (3a+c)f'(x)h + (frac 92 a + frac 12 c) f''(x)h^2+epsilon h^3] = f'(x)$
Which can only be true if
$(a+b+c) = 0$ and $(3a+c) = 1$
as for $frac 92 a + frac 12 c = 0$ This makes your error term decrease at quadratic speed rather than linear speed.
answered Jul 19 at 18:56
Doug M
39.2k31749
1) $frac 276 = frac 92$
2) yes there should be an $h^3$ factor multiplying that term. However, that term ultimately gets dropped (actually becomes the error estimate). And is not relevant for finding optimal values of $a, b, c$
3) yes the $h's$ cancel. That is kind of the point.
$frac 1h [(a+b+c)f(x) + (3a+c)f'(x)h + (frac 92 a + frac 12 c) f''(x)h^2+epsilon h^3] = f'(x)$
Which can only be true if
$(a+b+c) = 0$ and $(3a+c) = 1$
as for $frac 92 a + frac 12 c = 0$ This makes your error term decrease at quadratic speed rather than linear speed.
answered Jul 19 at 18:56
Doug M
39.2k31749
answered Jul 19 at 18:56
Doug M
39.2k31749
answered Jul 19 at 18:56
Doug M
39.2k31749
answered Jul 19 at 18:56
Doug M
39.2k31749
39.2k31749
but how does it drop completely? if there were the h^3 (which he forgot to write on the solution), it will be just h^2 (after dividing by h), and not drop completely.. I dont get it :(
– ZelelB
Jul 19 at 20:03
I thought, it's the most important term! We divide the h^3 by h, so we he h^2, and that's our 2nd order formula, he's looking for
– ZelelB
Jul 19 at 20:04
add a comment |Â
but how does it drop completely? if there were the h^3 (which he forgot to write on the solution), it will be just h^2 (after dividing by h), and not drop completely.. I dont get it :(
– ZelelB
Jul 19 at 20:03
I thought, it's the most important term! We divide the h^3 by h, so we he h^2, and that's our 2nd order formula, he's looking for
– ZelelB
Jul 19 at 20:04
but how does it drop completely? if there were the h^3 (which he forgot to write on the solution), it will be just h^2 (after dividing by h), and not drop completely.. I dont get it :(
– ZelelB
Jul 19 at 20:03
but how does it drop completely? if there were the h^3 (which he forgot to write on the solution), it will be just h^2 (after dividing by h), and not drop completely.. I dont get it :(
– ZelelB
Jul 19 at 20:03
I thought, it's the most important term! We divide the h^3 by h, so we he h^2, and that's our 2nd order formula, he's looking for
– ZelelB
Jul 19 at 20:04
I thought, it's the most important term! We divide the h^3 by h, so we he h^2, and that's our 2nd order formula, he's looking for
– ZelelB
Jul 19 at 20:04
add a comment |Â
up vote
1
down vote
More generally,
suppose you want
to find
$f'(x) approx dfracaf(x+uh) + bf(x+vh) + cf(x+wh) h
$.
$f(x+rh)
=f(x)+rhf'(x)+r^2h^2f''(x)/2+r^3h^3f'''(x)/6+O(h^4)
$
so
$beginarray\
af(x+uh) + bf(x+vh) + cf(x+wh)
&=(a+b+c)f(x)+(au+bv+cw)hf'(x)\
&quad +(au^2+bv^2+cw^2)h^2f''(x)/2+(au^3+bv^3+cw^3)h^3f'''(x)/6\
&quad+O(h^4)\
endarray
$
Therefore you want
$0=a+b+c,
1=au+bv+cw,
0=au^2+bv^2+cw^2
$
and the error is
$(au^3+bv^3+cw^3)h^2f'''(x)/6
+O(h^3)
$.
This leads to Vandermonde determinants.
Your case is
$u=3, v=0, w=-1$,
so
$0=a+b+c,
1=3a-c,
0=9a+c
$.
Adding the second and third,
$a=1/12$,
so
$c = -9a
=-3/4$.
From the first,
$b = -a-c
=-1/12+3/4
=2/3
$.
The error coefficient is
$au^3+bv^3+cw^3
=27/12-3/4
=15/12
=5/4
$
so the error is
$(5/4)h^2f'''(x)/6+O(h^3)
=(5/24)h^2f'''(x)+O(h^3)
$.
answered Jul 19 at 18:27
marty cohen
69.2k446122
add a comment |Â
up vote
1
down vote
More generally,
suppose you want
to find
$f'(x) approx dfracaf(x+uh) + bf(x+vh) + cf(x+wh) h
$.
$f(x+rh)
=f(x)+rhf'(x)+r^2h^2f''(x)/2+r^3h^3f'''(x)/6+O(h^4)
$
so
$beginarray\
af(x+uh) + bf(x+vh) + cf(x+wh)
&=(a+b+c)f(x)+(au+bv+cw)hf'(x)\
&quad +(au^2+bv^2+cw^2)h^2f''(x)/2+(au^3+bv^3+cw^3)h^3f'''(x)/6\
&quad+O(h^4)\
endarray
$
Therefore you want
$0=a+b+c,
1=au+bv+cw,
0=au^2+bv^2+cw^2
$
and the error is
$(au^3+bv^3+cw^3)h^2f'''(x)/6
+O(h^3)
$.
This leads to Vandermonde determinants.
Your case is
$u=3, v=0, w=-1$,
so
$0=a+b+c,
1=3a-c,
0=9a+c
$.
Adding the second and third,
$a=1/12$,
so
$c = -9a
=-3/4$.
From the first,
$b = -a-c
=-1/12+3/4
=2/3
$.
The error coefficient is
$au^3+bv^3+cw^3
=27/12-3/4
=15/12
=5/4
$
so the error is
$(5/4)h^2f'''(x)/6+O(h^3)
=(5/24)h^2f'''(x)+O(h^3)
$.
answered Jul 19 at 18:27
marty cohen
69.2k446122
add a comment |Â
up vote
1
down vote
up vote
1
down vote
More generally,
suppose you want
to find
$f'(x) approx dfracaf(x+uh) + bf(x+vh) + cf(x+wh) h
$.
$f(x+rh)
=f(x)+rhf'(x)+r^2h^2f''(x)/2+r^3h^3f'''(x)/6+O(h^4)
$
so
$beginarray\
af(x+uh) + bf(x+vh) + cf(x+wh)
&=(a+b+c)f(x)+(au+bv+cw)hf'(x)\
&quad +(au^2+bv^2+cw^2)h^2f''(x)/2+(au^3+bv^3+cw^3)h^3f'''(x)/6\
&quad+O(h^4)\
endarray
$
Therefore you want
$0=a+b+c,
1=au+bv+cw,
0=au^2+bv^2+cw^2
$
and the error is
$(au^3+bv^3+cw^3)h^2f'''(x)/6
+O(h^3)
$.
This leads to Vandermonde determinants.
Your case is
$u=3, v=0, w=-1$,
so
$0=a+b+c,
1=3a-c,
0=9a+c
$.
Adding the second and third,
$a=1/12$,
so
$c = -9a
=-3/4$.
From the first,
$b = -a-c
=-1/12+3/4
=2/3
$.
The error coefficient is
$au^3+bv^3+cw^3
=27/12-3/4
=15/12
=5/4
$
so the error is
$(5/4)h^2f'''(x)/6+O(h^3)
=(5/24)h^2f'''(x)+O(h^3)
$.
answered Jul 19 at 18:27
marty cohen
69.2k446122
More generally,
suppose you want
to find
$f'(x) approx dfracaf(x+uh) + bf(x+vh) + cf(x+wh) h
$.
$f(x+rh)
=f(x)+rhf'(x)+r^2h^2f''(x)/2+r^3h^3f'''(x)/6+O(h^4)
$
so
$beginarray\
af(x+uh) + bf(x+vh) + cf(x+wh)
&=(a+b+c)f(x)+(au+bv+cw)hf'(x)\
&quad +(au^2+bv^2+cw^2)h^2f''(x)/2+(au^3+bv^3+cw^3)h^3f'''(x)/6\
&quad+O(h^4)\
endarray
$
Therefore you want
$0=a+b+c,
1=au+bv+cw,
0=au^2+bv^2+cw^2
$
and the error is
$(au^3+bv^3+cw^3)h^2f'''(x)/6
+O(h^3)
$.
This leads to Vandermonde determinants.
Your case is
$u=3, v=0, w=-1$,
so
$0=a+b+c,
1=3a-c,
0=9a+c
$.
Adding the second and third,
$a=1/12$,
so
$c = -9a
=-3/4$.
From the first,
$b = -a-c
=-1/12+3/4
=2/3
$.
The error coefficient is
$au^3+bv^3+cw^3
=27/12-3/4
=15/12
=5/4
$
so the error is
$(5/4)h^2f'''(x)/6+O(h^3)
=(5/24)h^2f'''(x)+O(h^3)
$.
answered Jul 19 at 18:27
marty cohen
69.2k446122
answered Jul 19 at 18:27
marty cohen
69.2k446122
answered Jul 19 at 18:27
marty cohen
69.2k446122
answered Jul 19 at 18:27
marty cohen
69.2k446122
69.2k446122
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thanks for looking up at the question. I tried as stated above, that I found for point 1 for example, on that point 27/6, and for point 2, I multiply the term there with h^3. He doesn't multiply it with nothing. Where are the h^3?
– ZelelB
Jul 19 at 17:32
1. $9/2=27/6$ is correct. 2.You forgot the factor $1/h$
– gammatester
Jul 19 at 17:38
But there is an error in 1: it should read $f'''(x_0)$ not $f'''(0)$.
– gammatester
Jul 19 at 17:45
you're right for 1, thanks. But for 2, how he forgot 1/h? I am saying he should multiply there with h^3. If you look at the line before, from which he's derivating that line. He is summing the h's, so a term multiplies h, another h^2, and another term multiplies h^3. Or am I wrong. Still don't get it. Where are the h^3 of line before gone? And where is here getting "-9/2" MINUS not +9/2.
– ZelelB
Jul 19 at 17:53
It should read $Big(frac92 a f'''(x_0)-frac16cf'''(x_0)Big)h^3$
– gammatester
Jul 19 at 18:12