Right Riemann sum Error bound proof












1












$begingroup$


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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 9:57









      pedroth

      765




      765










      asked Jan 15 at 8:04









      jnxdjnxd

      63




      63






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          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
          $$
          Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
          } leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
          } leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
          $$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            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



            $$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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











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            2 Answers
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            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            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
            $$
            Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
            } leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
            } leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              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
              $$
              Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
              } leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
              } leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                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
                $$
                Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
                } leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
                } leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
                $$






                share|cite|improve this answer









                $endgroup$



                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
                $$
                Abs {-int_a^b f(x) ,mathrm dx + A_{text{right}}
                } leqslant sum_1^n Abs { int_{x_{j-1}}^{x_j} (f(x_j) - f(x))mathrm dx
                } leqslant n cdot M_1 frac {(b-a)^2}{2n^2} = frac {M_1(b-a)^2}{2n}.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 10:00









                xbhxbh

                6,1251522




                6,1251522























                    1












                    $begingroup$

                    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



                    $$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      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
















                    1












                    $begingroup$

                    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



                    $$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      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














                    1












                    1








                    1





                    $begingroup$

                    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



                    $$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$






                    share|cite|improve this answer











                    $endgroup$



                    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



                    $$sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - int_{a}^{b} f(x) , dx = sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - sum_{j=1}^nint_{x_{j-1}}^{x_j} f(x) , dx \ leqslant sum_{j=1}^n M frac{(b-a)^2}{2n^2} = M frac{(b-a)^2}{2n} $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 22 at 0:37

























                    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


















                    • $begingroup$
                      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


















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