Mixing
Geometrically Defining Discrete Integral
$begingroup$
We define the forward difference as an operator on real (or complex) functions as $D[f] = f(x+1) - f(x)$
It follows then that there is a forward anti-difference that can be defined, which we'll denote as
$${Large{mathfrak{D}}}_{a}^{b}[f] $$
Now the primary property that we care about in defining this antidifference is that $$ Dleft [{Large{mathfrak{D}}}_{a}^{x}[f] right] = f$$
(And vice versa). It turns out that when $b-a$ is an integer we can then easily define:
$${Large{mathfrak{D}}}_{a}^{b}[f] = f(a+1) + f(a+2) + ... f(b) = sum_{n=a+1}^{b} f(n) $$
And this definition meets our earlier critera.
My Question:
How do we define the forward anti-difference geometrically (as some property of the curve) for $a,b$ whose difference is NOT integral. In the case where $b-a$ is an integer, our definition effectively amounts to forming boxes of width $1$ and heights $f(a+1), f(a+2) ... f(b)$ under our function and computing the "signed" area of this collection of boxes. I want a similar "create shapes defined using the curve and compute their area" definition for $a,b$ whose difference is not integral.
Additional Insight:
Now if you are lucky, for a particular choice of function $g$ it might be that for any choice of $a,b$ where $b-a$ is integral you have some closed form $omega_g(a,b)$ such that
$$ {Large{mathfrak{D}}}_{a}^{b}[f] = omega_g(a,b)$$
And in this case, if the closed form has a natural definition for arguments $a,b$ where $b-a$ is not integral, then you can use it as a definition of the antidifference for arbitrary bounds. An example of this is to consider the function $f: mathbb{R} rightarrow mathbb{R}: f(x) = x$. Then for all $a,b$ where $b-a$ is an integer it is the case that:
$$ {Large{mathfrak{D}}}_{a}^{b}[x] = frac{1}{2}b(b+1) - frac{1}{2}a(a+1) $$
The closed form on the left hand side allows us to easily define the function for any pair of $a,b$ even when the difference isn't integral.
Now this works in practice it is unsatisfactory in theory because I don't have a natural geometric definition for $a,b$ when $b-a$ is not an integer (how do you describe that earlier formula in terms of boxes of fixed width or some other geometric construction, that generalizes to other functions? It's not obvious at all)
finite-differences
asked Jan 28 at 3:41
frogeyedpeasfrogeyedpeas
7,63172054
$endgroup$
add a comment |
$begingroup$
We define the forward difference as an operator on real (or complex) functions as $D[f] = f(x+1) - f(x)$
It follows then that there is a forward anti-difference that can be defined, which we'll denote as
$${Large{mathfrak{D}}}_{a}^{b}[f] $$
Now the primary property that we care about in defining this antidifference is that $$ Dleft [{Large{mathfrak{D}}}_{a}^{x}[f] right] = f$$
(And vice versa). It turns out that when $b-a$ is an integer we can then easily define:
$${Large{mathfrak{D}}}_{a}^{b}[f] = f(a+1) + f(a+2) + ... f(b) = sum_{n=a+1}^{b} f(n) $$
And this definition meets our earlier critera.
My Question:
How do we define the forward anti-difference geometrically (as some property of the curve) for $a,b$ whose difference is NOT integral. In the case where $b-a$ is an integer, our definition effectively amounts to forming boxes of width $1$ and heights $f(a+1), f(a+2) ... f(b)$ under our function and computing the "signed" area of this collection of boxes. I want a similar "create shapes defined using the curve and compute their area" definition for $a,b$ whose difference is not integral.
Additional Insight:
Now if you are lucky, for a particular choice of function $g$ it might be that for any choice of $a,b$ where $b-a$ is integral you have some closed form $omega_g(a,b)$ such that
$$ {Large{mathfrak{D}}}_{a}^{b}[f] = omega_g(a,b)$$
And in this case, if the closed form has a natural definition for arguments $a,b$ where $b-a$ is not integral, then you can use it as a definition of the antidifference for arbitrary bounds. An example of this is to consider the function $f: mathbb{R} rightarrow mathbb{R}: f(x) = x$. Then for all $a,b$ where $b-a$ is an integer it is the case that:
$$ {Large{mathfrak{D}}}_{a}^{b}[x] = frac{1}{2}b(b+1) - frac{1}{2}a(a+1) $$
The closed form on the left hand side allows us to easily define the function for any pair of $a,b$ even when the difference isn't integral.
Now this works in practice it is unsatisfactory in theory because I don't have a natural geometric definition for $a,b$ when $b-a$ is not an integer (how do you describe that earlier formula in terms of boxes of fixed width or some other geometric construction, that generalizes to other functions? It's not obvious at all)
finite-differences
asked Jan 28 at 3:41
frogeyedpeasfrogeyedpeas
7,63172054
$endgroup$
add a comment |
$begingroup$
We define the forward difference as an operator on real (or complex) functions as $D[f] = f(x+1) - f(x)$
It follows then that there is a forward anti-difference that can be defined, which we'll denote as
$${Large{mathfrak{D}}}_{a}^{b}[f] $$
Now the primary property that we care about in defining this antidifference is that $$ Dleft [{Large{mathfrak{D}}}_{a}^{x}[f] right] = f$$
(And vice versa). It turns out that when $b-a$ is an integer we can then easily define:
$${Large{mathfrak{D}}}_{a}^{b}[f] = f(a+1) + f(a+2) + ... f(b) = sum_{n=a+1}^{b} f(n) $$
And this definition meets our earlier critera.
My Question:
How do we define the forward anti-difference geometrically (as some property of the curve) for $a,b$ whose difference is NOT integral. In the case where $b-a$ is an integer, our definition effectively amounts to forming boxes of width $1$ and heights $f(a+1), f(a+2) ... f(b)$ under our function and computing the "signed" area of this collection of boxes. I want a similar "create shapes defined using the curve and compute their area" definition for $a,b$ whose difference is not integral.
Additional Insight:
Now if you are lucky, for a particular choice of function $g$ it might be that for any choice of $a,b$ where $b-a$ is integral you have some closed form $omega_g(a,b)$ such that
$$ {Large{mathfrak{D}}}_{a}^{b}[f] = omega_g(a,b)$$
And in this case, if the closed form has a natural definition for arguments $a,b$ where $b-a$ is not integral, then you can use it as a definition of the antidifference for arbitrary bounds. An example of this is to consider the function $f: mathbb{R} rightarrow mathbb{R}: f(x) = x$. Then for all $a,b$ where $b-a$ is an integer it is the case that:
$$ {Large{mathfrak{D}}}_{a}^{b}[x] = frac{1}{2}b(b+1) - frac{1}{2}a(a+1) $$
The closed form on the left hand side allows us to easily define the function for any pair of $a,b$ even when the difference isn't integral.
Now this works in practice it is unsatisfactory in theory because I don't have a natural geometric definition for $a,b$ when $b-a$ is not an integer (how do you describe that earlier formula in terms of boxes of fixed width or some other geometric construction, that generalizes to other functions? It's not obvious at all)
finite-differences
asked Jan 28 at 3:41
frogeyedpeasfrogeyedpeas
7,63172054
$endgroup$
We define the forward difference as an operator on real (or complex) functions as $D[f] = f(x+1) - f(x)$
It follows then that there is a forward anti-difference that can be defined, which we'll denote as
$${Large{mathfrak{D}}}_{a}^{b}[f] $$
Now the primary property that we care about in defining this antidifference is that $$ Dleft [{Large{mathfrak{D}}}_{a}^{x}[f] right] = f$$
(And vice versa). It turns out that when $b-a$ is an integer we can then easily define:
$${Large{mathfrak{D}}}_{a}^{b}[f] = f(a+1) + f(a+2) + ... f(b) = sum_{n=a+1}^{b} f(n) $$
And this definition meets our earlier critera.
My Question:
How do we define the forward anti-difference geometrically (as some property of the curve) for $a,b$ whose difference is NOT integral. In the case where $b-a$ is an integer, our definition effectively amounts to forming boxes of width $1$ and heights $f(a+1), f(a+2) ... f(b)$ under our function and computing the "signed" area of this collection of boxes. I want a similar "create shapes defined using the curve and compute their area" definition for $a,b$ whose difference is not integral.
Additional Insight:
Now if you are lucky, for a particular choice of function $g$ it might be that for any choice of $a,b$ where $b-a$ is integral you have some closed form $omega_g(a,b)$ such that
$$ {Large{mathfrak{D}}}_{a}^{b}[f] = omega_g(a,b)$$
And in this case, if the closed form has a natural definition for arguments $a,b$ where $b-a$ is not integral, then you can use it as a definition of the antidifference for arbitrary bounds. An example of this is to consider the function $f: mathbb{R} rightarrow mathbb{R}: f(x) = x$. Then for all $a,b$ where $b-a$ is an integer it is the case that:
$$ {Large{mathfrak{D}}}_{a}^{b}[x] = frac{1}{2}b(b+1) - frac{1}{2}a(a+1) $$
The closed form on the left hand side allows us to easily define the function for any pair of $a,b$ even when the difference isn't integral.
Now this works in practice it is unsatisfactory in theory because I don't have a natural geometric definition for $a,b$ when $b-a$ is not an integer (how do you describe that earlier formula in terms of boxes of fixed width or some other geometric construction, that generalizes to other functions? It's not obvious at all)
finite-differences
finite-differences
asked Jan 28 at 3:41
frogeyedpeasfrogeyedpeas
7,63172054
asked Jan 28 at 3:41
frogeyedpeasfrogeyedpeas
7,63172054
asked Jan 28 at 3:41
frogeyedpeasfrogeyedpeas
7,63172054
asked Jan 28 at 3:41
frogeyedpeasfrogeyedpeas
7,63172054
asked Jan 28 at 3:41
frogeyedpeasfrogeyedpeas
7,63172054
7,63172054
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $f(x)$ is defined over $[0, 0.5] subset mathbb{R}$. Then there are no points for which $D[f](x)$ is defined. Thus it is impossible to define ${Large{mathfrak{D}}}_{0}^{x}[f]$ in such way that it will satisfy the "primary property". Mixing discrete and continuous things together doesn't quite work here.
But you can define the same thing for some essentially discrete operator.
For example, if $$hat{D}[f] = f(lfloor x+1 rfloor) - f(lfloor x rfloor)$$
then any function $F(x)$ which satisfies
$$
F(lfloor x+1 rfloor)-F(lfloor x rfloor)=f(lfloor x rfloor)
$$
works as an antiderivative. Then you can define
$$
{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F(b)-F(a)
$$
and obtain (almost) the desired property
$$hat{D}left [{Large{hat{mathfrak{D}}}}_{a}^{x}[f] right] = f(lfloor x rfloor).$$
Here $lfloor x rfloor$ denotes the integral part of $x$.
This may seem not so interesting, but it may lead to various geometrical meanings. For instance, consider two different antiderivatives
$$
begin{aligned}
&F_1(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor),\
&F_2(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor)+(x-lfloor x rfloor)f(lfloor x rfloor).
end{aligned}
$$
Then I leave it to you to find out which area corresponds to
$$
{}_{1}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_1(b)-F_1(a), \
{}_{2}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_2(b)-F_2(a).
$$
The second one may be of interest. Note that $F_2(x)$ is continuous.
answered Jan 28 at 7:22
ZeeklessZeekless
577111
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Suppose that $f(x)$ is defined over $[0, 0.5] subset mathbb{R}$. Then there are no points for which $D[f](x)$ is defined. Thus it is impossible to define ${Large{mathfrak{D}}}_{0}^{x}[f]$ in such way that it will satisfy the "primary property". Mixing discrete and continuous things together doesn't quite work here.
But you can define the same thing for some essentially discrete operator.
For example, if $$hat{D}[f] = f(lfloor x+1 rfloor) - f(lfloor x rfloor)$$
then any function $F(x)$ which satisfies
$$
F(lfloor x+1 rfloor)-F(lfloor x rfloor)=f(lfloor x rfloor)
$$
works as an antiderivative. Then you can define
$$
{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F(b)-F(a)
$$
and obtain (almost) the desired property
$$hat{D}left [{Large{hat{mathfrak{D}}}}_{a}^{x}[f] right] = f(lfloor x rfloor).$$
Here $lfloor x rfloor$ denotes the integral part of $x$.
This may seem not so interesting, but it may lead to various geometrical meanings. For instance, consider two different antiderivatives
$$
begin{aligned}
&F_1(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor),\
&F_2(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor)+(x-lfloor x rfloor)f(lfloor x rfloor).
end{aligned}
$$
Then I leave it to you to find out which area corresponds to
$$
{}_{1}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_1(b)-F_1(a), \
{}_{2}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_2(b)-F_2(a).
$$
The second one may be of interest. Note that $F_2(x)$ is continuous.
answered Jan 28 at 7:22
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Suppose that $f(x)$ is defined over $[0, 0.5] subset mathbb{R}$. Then there are no points for which $D[f](x)$ is defined. Thus it is impossible to define ${Large{mathfrak{D}}}_{0}^{x}[f]$ in such way that it will satisfy the "primary property". Mixing discrete and continuous things together doesn't quite work here.
But you can define the same thing for some essentially discrete operator.
For example, if $$hat{D}[f] = f(lfloor x+1 rfloor) - f(lfloor x rfloor)$$
then any function $F(x)$ which satisfies
$$
F(lfloor x+1 rfloor)-F(lfloor x rfloor)=f(lfloor x rfloor)
$$
works as an antiderivative. Then you can define
$$
{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F(b)-F(a)
$$
and obtain (almost) the desired property
$$hat{D}left [{Large{hat{mathfrak{D}}}}_{a}^{x}[f] right] = f(lfloor x rfloor).$$
Here $lfloor x rfloor$ denotes the integral part of $x$.
This may seem not so interesting, but it may lead to various geometrical meanings. For instance, consider two different antiderivatives
$$
begin{aligned}
&F_1(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor),\
&F_2(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor)+(x-lfloor x rfloor)f(lfloor x rfloor).
end{aligned}
$$
Then I leave it to you to find out which area corresponds to
$$
{}_{1}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_1(b)-F_1(a), \
{}_{2}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_2(b)-F_2(a).
$$
The second one may be of interest. Note that $F_2(x)$ is continuous.
answered Jan 28 at 7:22
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Suppose that $f(x)$ is defined over $[0, 0.5] subset mathbb{R}$. Then there are no points for which $D[f](x)$ is defined. Thus it is impossible to define ${Large{mathfrak{D}}}_{0}^{x}[f]$ in such way that it will satisfy the "primary property". Mixing discrete and continuous things together doesn't quite work here.
But you can define the same thing for some essentially discrete operator.
For example, if $$hat{D}[f] = f(lfloor x+1 rfloor) - f(lfloor x rfloor)$$
then any function $F(x)$ which satisfies
$$
F(lfloor x+1 rfloor)-F(lfloor x rfloor)=f(lfloor x rfloor)
$$
works as an antiderivative. Then you can define
$$
{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F(b)-F(a)
$$
and obtain (almost) the desired property
$$hat{D}left [{Large{hat{mathfrak{D}}}}_{a}^{x}[f] right] = f(lfloor x rfloor).$$
Here $lfloor x rfloor$ denotes the integral part of $x$.
This may seem not so interesting, but it may lead to various geometrical meanings. For instance, consider two different antiderivatives
$$
begin{aligned}
&F_1(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor),\
&F_2(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor)+(x-lfloor x rfloor)f(lfloor x rfloor).
end{aligned}
$$
Then I leave it to you to find out which area corresponds to
$$
{}_{1}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_1(b)-F_1(a), \
{}_{2}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_2(b)-F_2(a).
$$
The second one may be of interest. Note that $F_2(x)$ is continuous.
answered Jan 28 at 7:22
ZeeklessZeekless
577111
$endgroup$
Suppose that $f(x)$ is defined over $[0, 0.5] subset mathbb{R}$. Then there are no points for which $D[f](x)$ is defined. Thus it is impossible to define ${Large{mathfrak{D}}}_{0}^{x}[f]$ in such way that it will satisfy the "primary property". Mixing discrete and continuous things together doesn't quite work here.
But you can define the same thing for some essentially discrete operator.
For example, if $$hat{D}[f] = f(lfloor x+1 rfloor) - f(lfloor x rfloor)$$
then any function $F(x)$ which satisfies
$$
F(lfloor x+1 rfloor)-F(lfloor x rfloor)=f(lfloor x rfloor)
$$
works as an antiderivative. Then you can define
$$
{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F(b)-F(a)
$$
and obtain (almost) the desired property
$$hat{D}left [{Large{hat{mathfrak{D}}}}_{a}^{x}[f] right] = f(lfloor x rfloor).$$
Here $lfloor x rfloor$ denotes the integral part of $x$.
This may seem not so interesting, but it may lead to various geometrical meanings. For instance, consider two different antiderivatives
$$
begin{aligned}
&F_1(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor),\
&F_2(x)=f(0)+f(1)+dots+f(lfloor x-1 rfloor)+(x-lfloor x rfloor)f(lfloor x rfloor).
end{aligned}
$$
Then I leave it to you to find out which area corresponds to
$$
{}_{1}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_1(b)-F_1(a), \
{}_{2}{Large{hat{mathfrak{D}}}}_{a}^{b}[f]=F_2(b)-F_2(a).
$$
The second one may be of interest. Note that $F_2(x)$ is continuous.
answered Jan 28 at 7:22
ZeeklessZeekless
577111
answered Jan 28 at 7:22
ZeeklessZeekless
577111
answered Jan 28 at 7:22
ZeeklessZeekless
577111
answered Jan 28 at 7:22
ZeeklessZeekless
577111
577111
add a comment |
add a comment |
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