## convex analysis – How to prove the projection theorem?

For any point $$y$$, the projection onto a nonempty and closed convex set X is defined as
$$Pi_X(y)=argmin_{xin X}frac{1}{2}|x-y|_2^2.$$
I am trying to prove for any point $$xin X$$,
$$langle y – Pi_X(y) , x – Pi_X(y) rangle le 0$$
First, let $$f(x)=frac{1}{2}|x-y|_2^2$$, $$nabla f(x)=x-y$$. $$f(x)$$ is a convex function because $$nabla^2f(x)=1ge0$$. So,
$$f(x)ge f(Pi_X(y))+nabla f(Pi_X(y))^T(x-Pi_X(y))=f(Pi_X(y))+(Pi_X(y)-y)^T(x-Pi_X(y))\ Rightarrow (Pi_X(y)-y)^T(x-Pi_X(y)) = langle Pi_X(y)-y, x-Pi_X(y) rangle le f(x) – f(Pi_X(y))\ Rightarrow langle y – Pi_X(y) , x – Pi_X(y) rangle ge f(Pi_X(y)) – f(x)$$
And because $$Pi_X(y)$$ is the projection of $$y$$ onto $$X$$, $$f(Pi_X(y)) le f(x)$$, $$f(Pi_X(y)) – f(x)le0$$. At this point, I can’t continue my proof. Is there something wrong with my process?

Posted on

## calculus and analysis – Plot of the derivative of a function

I have a function `functionSL` as a function of `t` (`t<0`) where I want to find the extremum of the function and also find at which `t` it occurs. I took the derivative of `functionSL` with `t` which I wrote it as the function `functionSLD`.

``````d = 3;
torootL(a_?NumericQ, t_?NumericQ, zl_?NumericQ, zh_?NumericQ) := a - NIntegrate((zl y^d)/Sqrt((1 - (zl/zh)^(d + 1) y^(d + 1)) (1 + t^2 (1 - (zl/zh)^(d + 1))^-1 - y^(2 d))), {y, 0, 1}, PrecisionGoal -> 6, Method -> "GlobalAdaptive")
zs(a_?NumericQ, t_?NumericQ, zh_?NumericQ) := zl /. FindRoot(torootL(a, t, zl, zh), {zl, 0.5, 0, 1})
intSL(a_?NumericQ, t_?NumericQ, zh_?NumericQ) := NIntegrate(With({b = zs(a, t, zh)/zh}, (((-1)/(d - 1)) (zs(a, t, zh)^(2 d) (1 + t^2 (1 - (zs(a, t, zh)/zh)^(d + 1))^-1))^-1 zs(a, t, zh)^(2 d)) x^d ((1 - (b x)^(d + 1))/(1 - (zs(a, t, zh)^(2 d) (1 + t^2 (1 - (zs(a, t, zh)/zh)^(d + 1))^-1))^-1 (zs(a, t, zh) x)^(2 d)))^(1/2) - ((b^(d + 1) (d + 1))/(2 (d - 1))) x ((1 - (zs(a, t, zh)^(2 d) (1 + t^2 (1 - (zs(a, t, zh)/zh)^(d + 1))^-1))^-1 (zs(a, t, zh) x)^(2 d))/(1 - (b x)^(d + 1)))^(1/2) + (b^(d + 1)x)/((1 - (b x)^(d + 1)) (1 - (zs(a, t, zh)^(2 d) (1 + t^2 (1 - (zs(a, t, zh)/zh)^(d + 1))^-1))^-1 (zs(a, t, zh) x)^(2 d)))^(1/2)), {x, 0, 1}, MinRecursion -> 20, MaxRecursion -> 20, AccuracyGoal -> 12, PrecisionGoal -> 10, Method -> {"GlobalAdaptive", "SingularityHandler" -> Automatic})
functionSL(a_?NumericQ, t_?NumericQ, zh_?NumericQ) := ((-((1 - (zs(a, t, zh)^(2 d) (1 + t^2 (1 - (zs(a, t, zh)/zh)^(d + 1))^-1))^-1 zs(a, t, zh)^(2 d)) (1 - (zs(a, t, zh)/zh)^(d + 1)))^(1/2)/(d - 1)) + intSL(a, t, zh) + 1)/(4 zs(a, t, zh)^(d - 1))
functionSLD(t_) := Evaluate(Derivative(0, 1, 0)(functionSLL)(0.01, t, 1))
``````

I took some sample values of `functionSLD` for some `t`,

``````In(44):= functionSLD(0)

Out(44)= -3.58024*10^-12

In(48):= functionSLD(-10)

Out(48)= 0.15527

In(90):= functionSLD(-15)

Out(90)= 0.0477369

In(91):= functionSLD(-16)

Out(91)= 0.041289

In(93):= functionSLD(-16.5)

Out(93)= 0.039934

In(59):= functionSLD(-17) // Quiet

Out(59)= 0.0424448

In(60):= functionSLLP(-17.5)

Power::infy: Infinite expression 1/0. encountered.

NIntegrate::izero: Integral and error estimates are 0 on all integration subregions. Try increasing the value of the MinRecursion option. If value of integral may be 0, specify a finite value for the AccuracyGoal option.

NIntegrate::izero: Integral and error estimates are 0 on all integration subregions. Try increasing the value of the MinRecursion option. If value of integral may be 0, specify a finite value for the AccuracyGoal option.

Power::infy: Infinite expression 1/0. encountered.

NIntegrate::izero: Integral and error estimates are 0 on all integration subregions. Try increasing the value of the MinRecursion option. If value of integral may be 0, specify a finite value for the AccuracyGoal option.

General::stop: Further output of NIntegrate::izero will be suppressed during this calculation.

Power::infy: Infinite expression 1/0. encountered.

General::stop: Further output of Power::infy will be suppressed during this calculation.

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

General::stop: Further output of FindRoot::lstol will be suppressed during this calculation.

FindRoot::jsing: Encountered a singular Jacobian at the point {zl} = {0.825567}. Try perturbing the initial point(s).

FindRoot::jsing: Encountered a singular Jacobian at the point {zl} = {0.825567}. Try perturbing the initial point(s).

FindRoot::jsing: Encountered a singular Jacobian at the point {zl} = {0.825567}. Try perturbing the initial point(s).

General::stop: Further output of FindRoot::jsing will be suppressed during this calculation.

In(61):= functionSLD(-17.5) // Quiet

Out(61)= \$Aborted
`````` I expect `functionSL` to have an extremum for at least two values (looking only at `t<0`) so that `functionSLD` has at least two roots, I think I got one root at `t=0` which is clear in the plot of $$frac{dS}{dt}$$ and it really confirms my expectations, the other is located somewhere else (it seems clear in the plot).

You can see that at `t=0`, `functionsSLD = -3.58024*10^-12` which is essentially zero, as `t` goes to lower values `functionSLD` rises and then goes down again and it looks like it is going to be essentially zero again but as you can see at `t=-17.5` I aborted the calculation (in the sample values of `functionSLD`) because it just takes so long and it seems like there is a problem.

In the end, what I want to see is a plot of `functionSLD(t)` vs. `t` ($$frac{dS}{dt}$$).

I would also like somebody to check my `NIntegrate Rules` if there is something wrong with it, or can it be improved. I added a singularity handler because an error occurred which I pasted in the above sample values code.

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## calculus and analysis – 1D Transient Heat Conduction Problem

I could really use some help with this problem regarding heat conduction through a steel bar

Solve the one-dimensional transient heat conduction problem as follows.
You insert a steel bar that had been sitting at room temperature at 72ºF into a
bed of hot coals at 800 ºF to a depth of 6 inches. The bar is 4 ft long and has a
diameter of 0.5 in. You keep the bar in the hot coals while you are holding on to
the end. How long before the temperature of the point where you are holding the
bar (5 inches from the end of the bar) reaches 130 ºF when you can no longer
hold on to the bar because it is too hot? Plot the temperature profile along the
length of the rod at that time. The bar is made of mild steel with the properties as follow:
`Thermal conductivity (k) in Btu/hr ft ºF = 26.0`
`Density (rho) in lbm/in3 = 0.284`
`Specific Heat (C) in Btu/lbm ºF = 0.122`

Any help with how to do this in Mathematica is appreciated, thanks!

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## calculus and analysis – Fourier Series of ODE

I am having trouble finding the Fourier series of a 2nd order ODE. Should I be using the piecewise function as well to set up the range for t?

Solve 𝑦′′ + 𝜔^2𝑦 = 𝑟(𝑡), where 𝑟(𝑡) = |𝑡|, -𝜋 < 𝑡 < 𝜋 using Fourier series

So far I have set up the ode and set equal to r(t)
`r(t)=y''(t)+omega^2*y(t)`
`Plot(r(t),{t,-Pi,Pi})`
Any help with the mathematica code would be greatly appreciated. How can I find An, Bn with the function being an ODE

## algebra precalculus – How to find the distribution of cases in the AstraZeneca/Oxford Phase 3 study interim analysis?

I’m trying to figure out whether it’s possible to find how the 131 cases from an interim analysis of the AstraZeneca/Oxford Phase 3 study are distributed among the two arms of two dosing regimens.

Vaccine Efficacy is abbreviated as VE.

Facts from the AstraZeneca November 23rd Press Release, as I understand them:

• There are two dosing regimens. Let’s call them Regimen A and Regimen B.
• Regimen A (2741 participants) has a VE of 90%
• Regimen B (8895 participants) has a VE of 62%
• The combined VE across the two regimens is 70%.
• There were a total of 131 cases in the two regimens.

This is how the AstraZeneca Phase III Study Protocol defines Vaccine Efficacy (VE):

“VE is calculated as RRR = 100*(1-relative risk), which is the incidence of infection in the vaccine group relative to the incidence of infection in the control group expressed as a percentage.”

Side Note: I think they actually made a mistake with the word “relative risk” inside the formula. I think “relative risk” must be “risk ratio”. I looked up the notion of “relative risk” and it”s far more complicated than their own description of the formula. The word “risk ratio” I got from Lesson 3: Measures of Risk – Principles of Epidemiology in Public Health Practice

Given he definition of VE and the total number of cases, I think it must follow that there were 30 cases in the vaccine group, and 101 cases in the placebo group, because this is the only proportion that results in a vaccine efficacy of 70%.

`1-(30/101) = 0.702 = 0.70`

Now the question is, how are these cases (30 in the vaccine group, 101 in the placebo group) spread along the two regimens?

I tried to write down the formulas, as best as I could. I used the following for the variable names:

• `P`: The amount of cases in the placebo group P.
• `V`: The amount of cases in the vaccine group V.
• Three suffixes: `t` for Total, `a` for Regimen A, `b` for Regimen B.

`Pt = 101 = Pa + Pb`
`Vt = 30 = Va + Vb`
`(1-Va/Pa) = 0.90` —> `Va/Pa = 0.1`
`(1-Vb/Pb) = 0.62` —> `Vb/Pb = 0.38`

Is this making sense?
Can this be solved?

If so, is there one solution, or multiple solutions?

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## calculus and analysis – Simplify kronecker delta in the derivative of a summation

new to Mathematica and tackling an economics maximization problem. When I take the derivative of the utility function s.t. the budget constraint, it shows it with a changed index and kronecker deltas in it.

I understand that the derivative changes depending on whether the indices are equal.
How do I simplify that Kronecker delta in order to have the derivative taken at t=k(1)?

$$L=sum _{t=0}^{infty } beta ^t left{left(frac{text{Cs}(t)^{1-text{sigma}}}{1-text{sigma}}-frac{NN(t)^{text{phi}+1}}{text{phi}+1}right)-text{budget} lambda (t)right};$$

budget = P(t)*Cs(t) + Q(t)*B(t) – B(t – 1) – W(t)*NN(t) – T(t);

Take the derivative with respect to B(t)

D(L(Cs, l, B, P, Q, W, l, T), B(t))

And the result has this shape:

$$left{sum _{K(1)=0}^{infty } -beta ^{K(1)} lambda (K(1)) (Q(K(1)) delta _{t,K(1)}-delta _{t,K(1)-1})right}$$

Posted on

## real analysis – Is this set connected in \$mathbb{R}^3\$

The following question was part of my analysis quiz any I was unable to solve 1 option . So, I am posting it here.

Let V be the span of (1,1,1) and (0,1,1) . Let $$u_1 =(0,0,1) , u_2 =(1,1,0)$$ and $$u_3 =(1,0,1)$$ . Then Is $$(mathbb{R}^3$$V )$$bigcup$$ { $$t u_1 + (1-t) u_3 : 0leq t leq 1$$} connected?

I think It will be connected because If I remove V from $$mathbb{R}^3$$ then also it is connected amd adding anything makes it connected . But answer is disconnected !

Why so?

## calculus and analysis – Remove complex part of a solution

I have a function `intSL(a,zl,zh)` where I want to compute its derivative with respect to `zl` given some values of `a` and `zh`. I also want to plot it wrt to `zl` with the range `0 < zl < 1`.

However, there are some `zl` where my function displays an error message and produces a complex value. When I compare results that are complex valued to the result that are not, it seems like their corresponding real parts has a trend and so I think I could just eliminate the imaginary part (I think???), but I do not know how to do it.

Here is my code,

``````d = 3;
torootL(a_?NumericQ, t_?NumericQ, zl_?NumericQ, zh_?NumericQ) := a - NIntegrate((zl y^d)/Sqrt((1 - (zl/zh)^(d + 1) y^(d + 1)) (1 + t^2 (1 - (zl/zh)^(d + 1))^-1 - y^(2 d))), {y, 0, 1})
tz(a_?NumericQ, zl_?NumericQ, zh_?NumericQ) := t /. FindRoot(torootL(a, t, zl, zh), {t, 1, -1000, 10000})
intSL(a_?NumericQ, zl_?NumericQ, zh_?NumericQ) := NIntegrate(With({b = zl/zh}, (((-1)/(d - 1)) (zl^(2 d) (1 + tz(a, zl, zh)^2 (1 - (zl/zh)^(d + 1))^(-1)))^(-1) zl^(2 d)) x^d ((1 - (b x)^(d + 1))/(1 - (zl^(2 d) (1 + tz(a, zl, zh)^2 (1 - (zl/zh)^(d + 1))^(-1)))^(-1) (zl x)^(2 d)))^(1/2) - ((b^(d + 1) (d + 1))/(2 (d - 1))) x ((1 - (zl^(2 d) (1 + tz(a, zl, zh)^2 (1 - (zl/zh)^(d + 1))^(-1)))^(-1) (zl x)^(2 d))/(1 - (b x)^(d + 1)))^(1/2) + (b^(d + 1)x)/((1 - (b x)^(d + 1)) (1 - (zl^(2 d) (1 + tz(a, zl, zh)^2 (1 - (zl/zh)^(d + 1))^(-1)))^(-1) (zl x)^(2 d)))^(1/2)), {x, 0, 0.98, 0.99, 1}, MaxRecursion -> 20, PrecisionGoal -> 6)
``````

Here is an example of the evaluation of the derivative,

``````intSLL(zl_) := Evaluate(D(intSL(0.01, zl, 1), zl))

In(58):= intSLL(0.3)

During evaluation of In(58):= FindRoot::jsing: Encountered a singular Jacobian at the point {t} = {-5.99038*10^-9}. Try perturbing the initial point(s).

During evaluation of In(58):= FindRoot::jsing: Encountered a singular Jacobian at the point {t} = {-5.99038*10^-9}. Try perturbing the initial point(s).

During evaluation of In(58):= FindRoot::jsing: Encountered a singular Jacobian at the point {t} = {-5.99038*10^-9}. Try perturbing the initial point(s).

During evaluation of In(58):= General::stop: Further output of FindRoot::jsing will be suppressed during this calculation.

During evaluation of In(58):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(58):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(58):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(58):= General::stop: Further output of FindRoot::lstol will be suppressed during this calculation.

During evaluation of In(58):= FindRoot::reged: The point {10000.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(58):= FindRoot::reged: The point {10000.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(58):= FindRoot::reged: The point {10000.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(58):= General::stop: Further output of FindRoot::reged will be suppressed during this calculation.

Out(58)= 0.0149288

In(57):= intSLL(0.4)

During evaluation of In(57):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(57):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(57):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(57):= General::stop: Further output of FindRoot::lstol will be suppressed during this calculation.

During evaluation of In(57):= FindRoot::jsing: Encountered a singular Jacobian at the point {t} = {-6.30218*10^-9}. Try perturbing the initial point(s).

During evaluation of In(57):= FindRoot::jsing: Encountered a singular Jacobian at the point {t} = {-6.30218*10^-9}. Try perturbing the initial point(s).

During evaluation of In(57):= FindRoot::jsing: Encountered a singular Jacobian at the point {t} = {-6.30218*10^-9}. Try perturbing the initial point(s).

During evaluation of In(57):= General::stop: Further output of FindRoot::jsing will be suppressed during this calculation.

During evaluation of In(57):= FindRoot::reged: The point {10000.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(57):= FindRoot::reged: The point {10000.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(57):= FindRoot::reged: The point {10000.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(57):= General::stop: Further output of FindRoot::reged will be suppressed during this calculation.

Out(57)= 0.00650273

In(55):= intSLL(0.5)

Out(55)= 0.00353467

In(56):= intSLL(0.6)

During evaluation of In(56):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(56):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(56):= FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

During evaluation of In(56):= General::stop: Further output of FindRoot::lstol will be suppressed during this calculation.

During evaluation of In(56):= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in y near {y} = {0.984824086012257}. NIntegrate obtained 0.0293841 -0.119444 I and 0.0017267736029825806` for the integral and error estimates.

During evaluation of In(56):= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in y near {y} = {0.984824086012257}. NIntegrate obtained 0.0293841 -0.119444 I and 0.0017267736029825806` for the integral and error estimates.

During evaluation of In(56):= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in y near {y} = {0.984824086012257}. NIntegrate obtained 0.0293841 -0.119444 I and 0.001726773577114529` for the integral and error estimates.

During evaluation of In(56):= General::stop: Further output of NIntegrate::ncvb will be suppressed during this calculation.

During evaluation of In(56):= FindRoot::reged: The point {1.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(56):= FindRoot::reged: The point {1.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(56):= FindRoot::reged: The point {1.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(56):= General::stop: Further output of FindRoot::reged will be suppressed during this calculation.

During evaluation of In(56):= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In(56):= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In(56):= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In(56):= General::stop: Further output of NIntegrate::slwcon will be suppressed during this calculation.

Out(56)= 0.00225041 + 8.62966*10^-11 I

In(59):= intSLL(0.7)

During evaluation of In(59):= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In(59):= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in y near {y} = {0.99762205831973036103399454788132061366923153400421142578125000000}. NIntegrate obtained 0.00330503 -0.0437457 I and 0.0008224862881109875` for the integral and error estimates.

During evaluation of In(59):= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In(59):= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in y near {y} = {0.99762205831973036103399454788132061366923153400421142578125000000}. NIntegrate obtained 0.00330503 -0.0437457 I and 0.0008224862881109875` for the integral and error estimates.

During evaluation of In(59):= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In(59):= General::stop: Further output of NIntegrate::slwcon will be suppressed during this calculation.

During evaluation of In(59):= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in y near {y} = {0.99762205831973036103399454788132061366923153400421142578125000000}. NIntegrate obtained 0.00330503 -0.0437457 I and 0.0008224862758539473` for the integral and error estimates.

During evaluation of In(59):= General::stop: Further output of NIntegrate::ncvb will be suppressed during this calculation.

During evaluation of In(59):= FindRoot::reged: The point {1.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(59):= FindRoot::reged: The point {1.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(59):= FindRoot::reged: The point {1.} is at the edge of the search region {-1000.,10000.} in coordinate 1 and the computed search direction points outside the region.

During evaluation of In(59):= General::stop: Further output of FindRoot::reged will be suppressed during this calculation.

Out(59)= 0.0016139 + 2.5821*10^-6 I
``````

To summarize the sample result, `intSLL(zl)` produces the values for the corresponding `zl`

``````0.0149288                         zl=0.3   (with error)
0.00650273                        zl=0.4   (with error)
0.00353467                        zl=0.5
0.00225041 + 8.62966*10^-11 I     zl=0.6   (with error)
0.0016139 + 2.5821*10^-6 I        zl=0.7   (with error)
``````

You can see the trend of the values of the real part.

Posted on

## real analysis – Subsets \$Y\$ of a partially ordered set \$L\$ need not have least upper bounds nor greatest lower bounds

I am currently studying the textbook Principles of Program Analysis by Flemming Nielson, Hanne R. Nielson, and Chris Hankin. Appendix A Partially Ordered Sets says the following:

Partially ordered set. A partial ordering is a relation $$sqsubseteq : L times L rightarrow { text{true}, text{false} }$$ that is reflexive (i.e. $$forall l : l sqsubseteq l$$), transitive (i.e. $$forall l_1, l_2, l_3 : l_1 sqsubseteq l_2 land l_2 sqsubseteq l_3 Rightarrow l_1 sqsubseteq l_3$$), and anti-symmetric (i.e. $$forall l_1, l_2 : l_1 sqsubseteq l_2 land l_2 sqsubseteq l_1 Rightarrow l_1 = l_2$$). A partially ordered set $$(L, sqsubseteq)$$ is a set $$L$$ equipped with a partial ordering $$sqsubseteq$$ (sometimes written $$sqsubseteq_L$$). We shall write $$l_2 sqsupseteq l_1$$ for $$l_1 sqsubseteq l_2$$ and $$l_1 sqsubset l_2$$ for $$l_1 sqsubseteq l_2 land l_1 not= l_2$$.

A subset $$Y$$ of $$L$$ has $$l in L$$ as an upper bound if $$forall l^prime in Y : l^prime sqsubseteq l$$ and as a lower bound if $$forall l^prime in Y : l^prime sqsupseteq l$$. A least upper bound $$l$$ of $$Y$$ is an upper bound of $$Y$$ that satisfies $$l sqsubseteq l_0$$ whenever $$l_0$$ is another upper bound of $$Y$$; similarly, a greatest lower bound $$l$$ of $$Y$$ is a lower bound of $$Y$$ that satisfies $$l_0 sqsubseteq l$$ whenever $$l_0$$ is another lower bound of $$Y$$. Note that subsets $$Y$$ of a partially ordered set $$L$$ need not have least upper bounds nor greatest lower bounds but when they exist they are unique (since $$sqsubseteq$$ is anti-symmetric) and they are denoted $$bigsqcup Y$$ and $$sqcap Y$$, respectively.

It is this part that I am unsure about:

Note that subsets $$Y$$ of a partially ordered set $$L$$ need not have least upper bounds nor greatest lower bounds but when they exist they are unique (since $$sqsubseteq$$ is anti-symmetric) and they are denoted $$bigsqcup Y$$ and $$sqcap Y$$, respectively.

This isn’t obvious to me. Can someone please explain this / make it clear ?

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## fa.functional analysis – The best bound of the integral of a nondecreasing real function in a closed interval

Let $$F:(0,1)to (0,1)$$ be a nondecreasing function. Given the definite integrals
$$begin{equation} int_a^1 F(x)~dx=I_1 ~text{and}~int_b^1 F(x)~dx=I_2, end{equation}$$
where $$a and $$I_2le I_1le I_2+(b-a)F(b)$$. Consider the integral
$$begin{equation} G(t)=int_t^1 F(x)~dx, ~tin(a,b). end{equation}$$

How to find the best upperbound and lowerbound (represented by $$I_1$$ and $$I_2$$) of $$G(t)$$?

“The best bound” means that for given values $$I_1$$ and $$I_2$$, we can find an admissible function $$F$$ such that the bound of $$G(t)$$, $$G(a)=I_1$$ and $$G(b)=I_2$$ are achievable.

Personally, I think solving this problem will be useful in estimate the integral of the distribution function.

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