Right Riemann sum Error bound proof
$begingroup$
How do we prove the right Riemann sum error bound?
In wikipedia (https://en.wikipedia.org/wiki/Riemann_sum#Right_Riemann_sum) they have mentioned the following bound, but no proof.
$$left | int_a^bf(x)dx - A_{mbox{right}}right | leq frac{M_1(b-a)^2}{2n}$$
riemann-integration
edited Jan 15 at 9:57
pedroth
765
asked Jan 15 at 8:04
jnxdjnxd
63
$endgroup$
add a comment |
$begingroup$
How do we prove the right Riemann sum error bound?
In wikipedia (https://en.wikipedia.org/wiki/Riemann_sum#Right_Riemann_sum) they have mentioned the following bound, but no proof.
$$left | int_a^bf(x)dx - A_{mbox{right}}right | leq frac{M_1(b-a)^2}{2n}$$
riemann-integration
edited Jan 15 at 9:57
pedroth
765
asked Jan 15 at 8:04
jnxdjnxd
63
$endgroup$
add a comment |
$begingroup$
How do we prove the right Riemann sum error bound?
In wikipedia (https://en.wikipedia.org/wiki/Riemann_sum#Right_Riemann_sum) they have mentioned the following bound, but no proof.
$$left | int_a^bf(x)dx - A_{mbox{right}}right | leq frac{M_1(b-a)^2}{2n}$$
riemann-integration
edited Jan 15 at 9:57
pedroth
765
asked Jan 15 at 8:04
jnxdjnxd
63
$endgroup$
How do we prove the right Riemann sum error bound?
In wikipedia (https://en.wikipedia.org/wiki/Riemann_sum#Right_Riemann_sum) they have mentioned the following bound, but no proof.
$$left | int_a^bf(x)dx - A_{mbox{right}}right | leq frac{M_1(b-a)^2}{2n}$$
riemann-integration
riemann-integration
edited Jan 15 at 9:57
pedroth
765
asked Jan 15 at 8:04
jnxdjnxd
63
edited Jan 15 at 9:57
pedroth
765
asked Jan 15 at 8:04
jnxdjnxd
63
edited Jan 15 at 9:57
pedroth
765
edited Jan 15 at 9:57
pedroth
765
edited Jan 15 at 9:57
pedroth
765
765
asked Jan 15 at 8:04
jnxdjnxd
63
asked Jan 15 at 8:04
jnxdjnxd
63
asked Jan 15 at 8:04
jnxdjnxd
63
63
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This could be an exercise when learning the definite integrals. Here is a demo.
For convenience, suppose $f$ is differentiable on $[a,b]$. Let $x_j = a + j varDelta x$, where $varDelta x = (b-a)/n$, for $j =1,2,dots, n$. Then
$$
A _{text{right}} = sum_1^n f(x_j) varDelta x.
$$
Thus
begin{align*}
newcommandAbs[1]{leftvert #1 right vert}
&quad Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} \
&= Abs {sum_1^n int_{x_{j-1}}^{x_j} (f(x_j) - f(x) ),mathrm dx
} \
&= Abs { sum_1^n int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx
}, tag{Lagrange's MVT}
end{align*}
where $eta_j in (x_{j-1}, x_j)$ for each $j$.
Now use $-M_1 leqslant f' leqslant M_1$, and notice that $(x_j - x) geqslant 0$, we have
$$
-M_1 (x_j - x) leqslant f'(eta_j)(x_j -x) leqslant M_1 (x_j -x),
$$
hence
$$
-M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx leqslant int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx = int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx leqslant M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx,
$$
i.e.
$$
Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant M_1 int_{x_{j-1}}^{x_j} (x_j -x)mathrm dx = M_1 frac {(b-a)^2}{2n^2}.
$$
Therefore we have
$$
Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
$$
answered Jan 15 at 10:00
xbhxbh
6,1251522
$endgroup$
add a comment |
$begingroup$
This result is stated for a monotone function and uniformly spaced points. Assume first that $f$ is non-decreasing ( the proof for $f$ non-increasing is similar).
With a uniform partition $x_j = a + jfrac{b-a}{n}$, we can apply the mean value theorem to obtain
$$f(x_j) = f(x) + f'(xi_j(x))(x_j - x)$$
where $xi_j(x)$ is between $x$ and $x_j$.
Integrating and using $f' geqslant 0$, we get the local error bound
$$f(x_j)(x_j - x_{j-1}) - int_{x_{j-1}}^{x_j} f(x) , dx = int_{x_{j-1}}^{x_j} f'(xi_j(x))(x_j - x) , dx leqslant M frac{(x_j - x_{j-1})^2}{2} = M frac{(b-a)^2}{2n^2}$$
and the global error bound is
$$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$
answered Jan 15 at 8:48
RRLRRL
50.8k42573
$endgroup$
$begingroup$
Thanks for the explanation. One doubt though, in the taylor linear approximation of f(xj) whats the domain of x?
$endgroup$
– jnxd
Jan 21 at 23:24
$begingroup$
I might be less confusing if I refer to the justification as the mean value theorem -- which is exact for any $x, x_j in [a,b]$ assuming $f$ is differentiable in $[a,b]$. I'm sure you know that $f(x) - f(y) = f'(c)(x-y)$ where $c$ is between $x$ and $y$.
$endgroup$
– RRL
Jan 22 at 0:36
$begingroup$
Thanks again. Seeing it as MVT helped me understand it better.
$endgroup$
– jnxd
Jan 22 at 21:49
add a comment |
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2 Answers
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2 Answers
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active
oldest
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active
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votes
$begingroup$
This could be an exercise when learning the definite integrals. Here is a demo.
For convenience, suppose $f$ is differentiable on $[a,b]$. Let $x_j = a + j varDelta x$, where $varDelta x = (b-a)/n$, for $j =1,2,dots, n$. Then
$$
A _{text{right}} = sum_1^n f(x_j) varDelta x.
$$
Thus
begin{align*}
newcommandAbs[1]{leftvert #1 right vert}
&quad Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} \
&= Abs {sum_1^n int_{x_{j-1}}^{x_j} (f(x_j) - f(x) ),mathrm dx
} \
&= Abs { sum_1^n int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx
}, tag{Lagrange's MVT}
end{align*}
where $eta_j in (x_{j-1}, x_j)$ for each $j$.
Now use $-M_1 leqslant f' leqslant M_1$, and notice that $(x_j - x) geqslant 0$, we have
$$
-M_1 (x_j - x) leqslant f'(eta_j)(x_j -x) leqslant M_1 (x_j -x),
$$
hence
$$
-M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx leqslant int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx = int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx leqslant M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx,
$$
i.e.
$$
Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant M_1 int_{x_{j-1}}^{x_j} (x_j -x)mathrm dx = M_1 frac {(b-a)^2}{2n^2}.
$$
Therefore we have
$$
Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
$$
answered Jan 15 at 10:00
xbhxbh
6,1251522
$endgroup$
add a comment |
$begingroup$
This could be an exercise when learning the definite integrals. Here is a demo.
For convenience, suppose $f$ is differentiable on $[a,b]$. Let $x_j = a + j varDelta x$, where $varDelta x = (b-a)/n$, for $j =1,2,dots, n$. Then
$$
A _{text{right}} = sum_1^n f(x_j) varDelta x.
$$
Thus
begin{align*}
newcommandAbs[1]{leftvert #1 right vert}
&quad Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} \
&= Abs {sum_1^n int_{x_{j-1}}^{x_j} (f(x_j) - f(x) ),mathrm dx
} \
&= Abs { sum_1^n int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx
}, tag{Lagrange's MVT}
end{align*}
where $eta_j in (x_{j-1}, x_j)$ for each $j$.
Now use $-M_1 leqslant f' leqslant M_1$, and notice that $(x_j - x) geqslant 0$, we have
$$
-M_1 (x_j - x) leqslant f'(eta_j)(x_j -x) leqslant M_1 (x_j -x),
$$
hence
$$
-M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx leqslant int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx = int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx leqslant M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx,
$$
i.e.
$$
Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant M_1 int_{x_{j-1}}^{x_j} (x_j -x)mathrm dx = M_1 frac {(b-a)^2}{2n^2}.
$$
Therefore we have
$$
Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
$$
answered Jan 15 at 10:00
xbhxbh
6,1251522
$endgroup$
add a comment |
$begingroup$
This could be an exercise when learning the definite integrals. Here is a demo.
For convenience, suppose $f$ is differentiable on $[a,b]$. Let $x_j = a + j varDelta x$, where $varDelta x = (b-a)/n$, for $j =1,2,dots, n$. Then
$$
A _{text{right}} = sum_1^n f(x_j) varDelta x.
$$
Thus
begin{align*}
newcommandAbs[1]{leftvert #1 right vert}
&quad Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} \
&= Abs {sum_1^n int_{x_{j-1}}^{x_j} (f(x_j) - f(x) ),mathrm dx
} \
&= Abs { sum_1^n int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx
}, tag{Lagrange's MVT}
end{align*}
where $eta_j in (x_{j-1}, x_j)$ for each $j$.
Now use $-M_1 leqslant f' leqslant M_1$, and notice that $(x_j - x) geqslant 0$, we have
$$
-M_1 (x_j - x) leqslant f'(eta_j)(x_j -x) leqslant M_1 (x_j -x),
$$
hence
$$
-M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx leqslant int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx = int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx leqslant M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx,
$$
i.e.
$$
Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant M_1 int_{x_{j-1}}^{x_j} (x_j -x)mathrm dx = M_1 frac {(b-a)^2}{2n^2}.
$$
Therefore we have
$$
Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
$$
answered Jan 15 at 10:00
xbhxbh
6,1251522
$endgroup$
This could be an exercise when learning the definite integrals. Here is a demo.
For convenience, suppose $f$ is differentiable on $[a,b]$. Let $x_j = a + j varDelta x$, where $varDelta x = (b-a)/n$, for $j =1,2,dots, n$. Then
$$
A _{text{right}} = sum_1^n f(x_j) varDelta x.
$$
Thus
begin{align*}
newcommandAbs[1]{leftvert #1 right vert}
&quad Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} \
&= Abs {sum_1^n int_{x_{j-1}}^{x_j} (f(x_j) - f(x) ),mathrm dx
} \
&= Abs { sum_1^n int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx
}, tag{Lagrange's MVT}
end{align*}
where $eta_j in (x_{j-1}, x_j)$ for each $j$.
Now use $-M_1 leqslant f' leqslant M_1$, and notice that $(x_j - x) geqslant 0$, we have
$$
-M_1 (x_j - x) leqslant f'(eta_j)(x_j -x) leqslant M_1 (x_j -x),
$$
hence
$$
-M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx leqslant int_{x_{j-1}}^{x_j} f'(eta_j) (x_j - x)mathrm dx = int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx leqslant M_1 int_{x_{j-1}}^{x_j} (x_j - x) mathrm dx,
$$
i.e.
$$
Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant M_1 int_{x_{j-1}}^{x_j} (x_j -x)mathrm dx = M_1 frac {(b-a)^2}{2n^2}.
$$
Therefore we have
$$
Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
} leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
} leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
$$
answered Jan 15 at 10:00
xbhxbh
6,1251522
answered Jan 15 at 10:00
xbhxbh
6,1251522
answered Jan 15 at 10:00
xbhxbh
6,1251522
answered Jan 15 at 10:00
xbhxbh
6,1251522
6,1251522
add a comment |
add a comment |
$begingroup$
This result is stated for a monotone function and uniformly spaced points. Assume first that $f$ is non-decreasing ( the proof for $f$ non-increasing is similar).
With a uniform partition $x_j = a + jfrac{b-a}{n}$, we can apply the mean value theorem to obtain
$$f(x_j) = f(x) + f'(xi_j(x))(x_j - x)$$
where $xi_j(x)$ is between $x$ and $x_j$.
Integrating and using $f' geqslant 0$, we get the local error bound
$$f(x_j)(x_j - x_{j-1}) - int_{x_{j-1}}^{x_j} f(x) , dx = int_{x_{j-1}}^{x_j} f'(xi_j(x))(x_j - x) , dx leqslant M frac{(x_j - x_{j-1})^2}{2} = M frac{(b-a)^2}{2n^2}$$
and the global error bound is
$$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$
answered Jan 15 at 8:48
RRLRRL
50.8k42573
$endgroup$
$begingroup$
Thanks for the explanation. One doubt though, in the taylor linear approximation of f(xj) whats the domain of x?
$endgroup$
– jnxd
Jan 21 at 23:24
$begingroup$
I might be less confusing if I refer to the justification as the mean value theorem -- which is exact for any $x, x_j in [a,b]$ assuming $f$ is differentiable in $[a,b]$. I'm sure you know that $f(x) - f(y) = f'(c)(x-y)$ where $c$ is between $x$ and $y$.
$endgroup$
– RRL
Jan 22 at 0:36
$begingroup$
Thanks again. Seeing it as MVT helped me understand it better.
$endgroup$
– jnxd
Jan 22 at 21:49
add a comment |
$begingroup$
This result is stated for a monotone function and uniformly spaced points. Assume first that $f$ is non-decreasing ( the proof for $f$ non-increasing is similar).
With a uniform partition $x_j = a + jfrac{b-a}{n}$, we can apply the mean value theorem to obtain
$$f(x_j) = f(x) + f'(xi_j(x))(x_j - x)$$
where $xi_j(x)$ is between $x$ and $x_j$.
Integrating and using $f' geqslant 0$, we get the local error bound
$$f(x_j)(x_j - x_{j-1}) - int_{x_{j-1}}^{x_j} f(x) , dx = int_{x_{j-1}}^{x_j} f'(xi_j(x))(x_j - x) , dx leqslant M frac{(x_j - x_{j-1})^2}{2} = M frac{(b-a)^2}{2n^2}$$
and the global error bound is
$$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$
answered Jan 15 at 8:48
RRLRRL
50.8k42573
$endgroup$
$begingroup$
Thanks for the explanation. One doubt though, in the taylor linear approximation of f(xj) whats the domain of x?
$endgroup$
– jnxd
Jan 21 at 23:24
$begingroup$
I might be less confusing if I refer to the justification as the mean value theorem -- which is exact for any $x, x_j in [a,b]$ assuming $f$ is differentiable in $[a,b]$. I'm sure you know that $f(x) - f(y) = f'(c)(x-y)$ where $c$ is between $x$ and $y$.
$endgroup$
– RRL
Jan 22 at 0:36
$begingroup$
Thanks again. Seeing it as MVT helped me understand it better.
$endgroup$
– jnxd
Jan 22 at 21:49
add a comment |
$begingroup$
This result is stated for a monotone function and uniformly spaced points. Assume first that $f$ is non-decreasing ( the proof for $f$ non-increasing is similar).
With a uniform partition $x_j = a + jfrac{b-a}{n}$, we can apply the mean value theorem to obtain
$$f(x_j) = f(x) + f'(xi_j(x))(x_j - x)$$
where $xi_j(x)$ is between $x$ and $x_j$.
Integrating and using $f' geqslant 0$, we get the local error bound
$$f(x_j)(x_j - x_{j-1}) - int_{x_{j-1}}^{x_j} f(x) , dx = int_{x_{j-1}}^{x_j} f'(xi_j(x))(x_j - x) , dx leqslant M frac{(x_j - x_{j-1})^2}{2} = M frac{(b-a)^2}{2n^2}$$
and the global error bound is
$$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$
answered Jan 15 at 8:48
RRLRRL
50.8k42573
$endgroup$
This result is stated for a monotone function and uniformly spaced points. Assume first that $f$ is non-decreasing ( the proof for $f$ non-increasing is similar).
With a uniform partition $x_j = a + jfrac{b-a}{n}$, we can apply the mean value theorem to obtain
$$f(x_j) = f(x) + f'(xi_j(x))(x_j - x)$$
where $xi_j(x)$ is between $x$ and $x_j$.
Integrating and using $f' geqslant 0$, we get the local error bound
$$f(x_j)(x_j - x_{j-1}) - int_{x_{j-1}}^{x_j} f(x) , dx = int_{x_{j-1}}^{x_j} f'(xi_j(x))(x_j - x) , dx leqslant M frac{(x_j - x_{j-1})^2}{2} = M frac{(b-a)^2}{2n^2}$$
and the global error bound is
$$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$
answered Jan 15 at 8:48
RRLRRL
50.8k42573
edited Jan 22 at 0:37
answered Jan 15 at 8:48
RRLRRL
50.8k42573
answered Jan 15 at 8:48
RRLRRL
50.8k42573
answered Jan 15 at 8:48
RRLRRL
50.8k42573
50.8k42573
$begingroup$
Thanks for the explanation. One doubt though, in the taylor linear approximation of f(xj) whats the domain of x?
$endgroup$
– jnxd
Jan 21 at 23:24
$begingroup$
I might be less confusing if I refer to the justification as the mean value theorem -- which is exact for any $x, x_j in [a,b]$ assuming $f$ is differentiable in $[a,b]$. I'm sure you know that $f(x) - f(y) = f'(c)(x-y)$ where $c$ is between $x$ and $y$.
$endgroup$
– RRL
Jan 22 at 0:36
$begingroup$
Thanks again. Seeing it as MVT helped me understand it better.
$endgroup$
– jnxd
Jan 22 at 21:49
add a comment |
$begingroup$
Thanks for the explanation. One doubt though, in the taylor linear approximation of f(xj) whats the domain of x?
$endgroup$
– jnxd
Jan 21 at 23:24
$begingroup$
I might be less confusing if I refer to the justification as the mean value theorem -- which is exact for any $x, x_j in [a,b]$ assuming $f$ is differentiable in $[a,b]$. I'm sure you know that $f(x) - f(y) = f'(c)(x-y)$ where $c$ is between $x$ and $y$.
$endgroup$
– RRL
Jan 22 at 0:36
$begingroup$
Thanks again. Seeing it as MVT helped me understand it better.
$endgroup$
– jnxd
Jan 22 at 21:49
$begingroup$
Thanks for the explanation. One doubt though, in the taylor linear approximation of f(xj) whats the domain of x?
$endgroup$
– jnxd
Jan 21 at 23:24
$begingroup$
Thanks for the explanation. One doubt though, in the taylor linear approximation of f(xj) whats the domain of x?
$endgroup$
– jnxd
Jan 21 at 23:24
$begingroup$
I might be less confusing if I refer to the justification as the mean value theorem -- which is exact for any $x, x_j in [a,b]$ assuming $f$ is differentiable in $[a,b]$. I'm sure you know that $f(x) - f(y) = f'(c)(x-y)$ where $c$ is between $x$ and $y$.
$endgroup$
– RRL
Jan 22 at 0:36
$begingroup$
I might be less confusing if I refer to the justification as the mean value theorem -- which is exact for any $x, x_j in [a,b]$ assuming $f$ is differentiable in $[a,b]$. I'm sure you know that $f(x) - f(y) = f'(c)(x-y)$ where $c$ is between $x$ and $y$.
$endgroup$
– RRL
Jan 22 at 0:36
$begingroup$
Thanks again. Seeing it as MVT helped me understand it better.
$endgroup$
– jnxd
Jan 22 at 21:49
$begingroup$
Thanks again. Seeing it as MVT helped me understand it better.
$endgroup$
– jnxd
Jan 22 at 21:49
add a comment |
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