Mixing
Help for this problem involving rieman integral and partitions
If $f: I--->mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$
(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$ and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$
(b) If $P_1$ is a partitions obtained from $P$ by adding $k$ points to $P$, show that $L(P_1,f) ≤ L(P,f) + 2k||f|| ||P||$ and also that $U(P_1,f) ≥ U(P,f) - 2k||f|| ||P||$
So what i have intuitively and some hint of my professor is that
First, the partition of Lemma 7.1.2 is
$P':=[{x_0,x_1,...,x_{k-1},z,x_k,...,x_n}$]
So what i have so far is, and part of this, is a hint of the professor we have
$|m_k|$, $|m'_k|$, $|m''_k|$ $≤$ $||f||$ hence
$0 ≤ L(P',f) - L(P,f) = (m'_k-m_k)(z-x_{k-1})+(m''_k-m_k)(x_k-z)$ $≤$ $2||f||(x_k-x_{k-1} ≤ 2||f|| ||P||$
Where
$m_k:=inf[f(x):x in [x_{k-1},x_k]]$
$m'_k:=inf[f(x):x in [x_{k-1},z]]$
$m''_k:=inf[f(x):x in [x_z,x_k]]$
And for part b i think i need to use induction but i think i need to solve a first to solve b and i... im lost sincerely ): can someone help me?
calculus real-analysis riemann-integration riemann-sum partitions-for-integration
asked Nov 20 '18 at 20:58
Daniel ML
303
add a comment |
If $f: I--->mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$
(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$ and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$
(b) If $P_1$ is a partitions obtained from $P$ by adding $k$ points to $P$, show that $L(P_1,f) ≤ L(P,f) + 2k||f|| ||P||$ and also that $U(P_1,f) ≥ U(P,f) - 2k||f|| ||P||$
So what i have intuitively and some hint of my professor is that
First, the partition of Lemma 7.1.2 is
$P':=[{x_0,x_1,...,x_{k-1},z,x_k,...,x_n}$]
So what i have so far is, and part of this, is a hint of the professor we have
$|m_k|$, $|m'_k|$, $|m''_k|$ $≤$ $||f||$ hence
$0 ≤ L(P',f) - L(P,f) = (m'_k-m_k)(z-x_{k-1})+(m''_k-m_k)(x_k-z)$ $≤$ $2||f||(x_k-x_{k-1} ≤ 2||f|| ||P||$
Where
$m_k:=inf[f(x):x in [x_{k-1},x_k]]$
$m'_k:=inf[f(x):x in [x_{k-1},z]]$
$m''_k:=inf[f(x):x in [x_z,x_k]]$
And for part b i think i need to use induction but i think i need to solve a first to solve b and i... im lost sincerely ): can someone help me?
calculus real-analysis riemann-integration riemann-sum partitions-for-integration
asked Nov 20 '18 at 20:58
Daniel ML
303
You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
– Acccumulation
Nov 20 '18 at 23:07
Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
– Daniel ML
Nov 21 '18 at 1:08
add a comment |
If $f: I--->mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$
(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$ and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$
(b) If $P_1$ is a partitions obtained from $P$ by adding $k$ points to $P$, show that $L(P_1,f) ≤ L(P,f) + 2k||f|| ||P||$ and also that $U(P_1,f) ≥ U(P,f) - 2k||f|| ||P||$
So what i have intuitively and some hint of my professor is that
First, the partition of Lemma 7.1.2 is
$P':=[{x_0,x_1,...,x_{k-1},z,x_k,...,x_n}$]
So what i have so far is, and part of this, is a hint of the professor we have
$|m_k|$, $|m'_k|$, $|m''_k|$ $≤$ $||f||$ hence
$0 ≤ L(P',f) - L(P,f) = (m'_k-m_k)(z-x_{k-1})+(m''_k-m_k)(x_k-z)$ $≤$ $2||f||(x_k-x_{k-1} ≤ 2||f|| ||P||$
Where
$m_k:=inf[f(x):x in [x_{k-1},x_k]]$
$m'_k:=inf[f(x):x in [x_{k-1},z]]$
$m''_k:=inf[f(x):x in [x_z,x_k]]$
And for part b i think i need to use induction but i think i need to solve a first to solve b and i... im lost sincerely ): can someone help me?
calculus real-analysis riemann-integration riemann-sum partitions-for-integration
asked Nov 20 '18 at 20:58
Daniel ML
303
If $f: I--->mathbb R$ is bounded, let $||f||:= Sup {|f(x)|: x in I}$, and if $P =(x_0,...,x_n)$ is a partition of $I:=[a,b]$, let $||P||:=Sup [x_1-x_0,...,x_n-x_{n-1}]$
(a) If $P'$ is the partition obtained from $P$ as in the proof of Lemma 7.1.2, show that $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$ and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$
(b) If $P_1$ is a partitions obtained from $P$ by adding $k$ points to $P$, show that $L(P_1,f) ≤ L(P,f) + 2k||f|| ||P||$ and also that $U(P_1,f) ≥ U(P,f) - 2k||f|| ||P||$
So what i have intuitively and some hint of my professor is that
First, the partition of Lemma 7.1.2 is
$P':=[{x_0,x_1,...,x_{k-1},z,x_k,...,x_n}$]
So what i have so far is, and part of this, is a hint of the professor we have
$|m_k|$, $|m'_k|$, $|m''_k|$ $≤$ $||f||$ hence
$0 ≤ L(P',f) - L(P,f) = (m'_k-m_k)(z-x_{k-1})+(m''_k-m_k)(x_k-z)$ $≤$ $2||f||(x_k-x_{k-1} ≤ 2||f|| ||P||$
Where
$m_k:=inf[f(x):x in [x_{k-1},x_k]]$
$m'_k:=inf[f(x):x in [x_{k-1},z]]$
$m''_k:=inf[f(x):x in [x_z,x_k]]$
And for part b i think i need to use induction but i think i need to solve a first to solve b and i... im lost sincerely ): can someone help me?
calculus real-analysis riemann-integration riemann-sum partitions-for-integration
calculus real-analysis riemann-integration riemann-sum partitions-for-integration
asked Nov 20 '18 at 20:58
Daniel ML
303
asked Nov 20 '18 at 20:58
Daniel ML
303
asked Nov 20 '18 at 20:58
Daniel ML
303
asked Nov 20 '18 at 20:58
Daniel ML
303
asked Nov 20 '18 at 20:58
Daniel ML
303
303
You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
– Acccumulation
Nov 20 '18 at 23:07
Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
– Daniel ML
Nov 21 '18 at 1:08
add a comment |
You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
– Acccumulation
Nov 20 '18 at 23:07
Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
– Daniel ML
Nov 21 '18 at 1:08
You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
– Acccumulation
Nov 20 '18 at 23:07
You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
– Acccumulation
Nov 20 '18 at 23:07
Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
– Daniel ML
Nov 21 '18 at 1:08
Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
– Daniel ML
Nov 21 '18 at 1:08
add a comment |
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You seem to have concluded $L(P,f) ≤ L(P',f) ≤ L(P,f) + 2||f|| ||P||$, and $U(P,f) ≥ U(P',f) ≥ U(P,f) - 2||f|| ||P||$ follows similar logic. So what are you confused with for (a)?
– Acccumulation
Nov 20 '18 at 23:07
Because this is only a partial solution to (a) and i dont know the dateails or why is that inequality true
– Daniel ML
Nov 21 '18 at 1:08