## inverse – How to get x value where area under curve is some number?

My goal is to get x value where area under the curve (from the x ~ infinity) is about 0.05. But the code which I have tried shows errors. How to correct it?

``````Solve(!(
*SubsuperscriptBox(((Integral)), (x), ((Infinity)))(
*FractionBox((
*SuperscriptBox((E), ((-x)/2))
*SuperscriptBox((x), (3/2))), (3
*SqrtBox((2 (Pi))))) (DifferentialD)x)) == 0.05, x)
``````

## plotting – Plot a functional relation involving a derivative and inverse

I have a function `f(x)` obtained by solving certain ODE. Thus, it is given as an interpolation function. I need to plot $$frac{dx}{df}$$ as a function of $$f$$.

Below I will give a simple analytical example just to explain what I mean. Let $$f(x)=arctan(x).$$ Then we have
$$frac{df}{dx}=frac{1}{1+x^2},quad text{or} quad frac{dx}{df}=1+x^2.$$

Now we express $$x$$ in terms of $$f$$, i.e., $$x=tan(f),$$ and substitute in the equation above:
$$frac{dx}{df}=1+x^2=1+tan(f)^2.$$

Thus, given $$f(x)=arctan(x)$$, I would like to get a plot of $$1+tan(f)^2.$$

One naive way to do it is to parametrize $$f$$ and $$frac{dx}{df}$$ in terms of $$x$$ and use `ParametricPlot`

``````ParametricPlot({f(x), 1/f'(x)}, {x, -10, 10}, AspectRatio -> 1)
``````

However, for my numerically defined function this does not work very well. Additionally, I would like to get the dependence in a functional form, the best would be again an interpolation function. How can I achieve this, maybe it is possible to formulate the problem as ODE and use `NDSolve`?

## co.combinatorics – What is the inverse Laplace transform of \$frac{(1/s)_{n}}{s} \$?

Introduction

So far, I have found (p. 5) the following generating functions of the unsigned Stirling numbers of the first kind:

$$begin{equation} tag{1} label{1} sum_{l=1}^{n} |S_{1}(n,l)|z^{l} = (z)_{n} = prod_{k=0}^{n-1} (z+k) =: g(z) , end{equation}$$ and $$begin{equation} tag{2} label{2} sum_{n=l}^{infty} frac{|S_{1}(n,l)|}{n!}z^{n} = (-1)^{l} frac{ln^{l}(1-z)}{l!} .end{equation}$$

Instead of summing the latter expression over $$n$$, I’m curious whether there is a simpler or different expression of the latter generating sum when summed over the other indices:

$$begin{equation} tag{3} label{3} f(z) := sum_{k=1}^{n} frac{|S_{1}(n,k)|}{k!}z^{k} . end{equation}$$ (One could also take the sum from $$k=1$$ to $$k=infty$$, as $$|S_{1}(n,k)| = 0$$ when $$k>n$$.)

Work so far

Approach 1

We see that $$(3)$$ is the egf version of the ogf in $$(1)$$. So one of the ways I’ve tried to find $$f(cdot)$$ is by converting the first equation to the third one by applying the inverse Laplace transform to $$g(1/s)/s$$.

From $$(1)$$, we see that $$g(1/s)/s = frac{(1/s)_{n}}{s}$$. The tricky part of find the inverse Laplace transform lies in the numerator. Therefore, I tried to express it in terms of other functions of which the inverse Laplace transform might be known.

For instance, note that the Chu-Vandermonde identity states: $$begin{equation} tag{4} label{4} _2F_1left(-n,b;c;1right) = frac{(c-b)_{n}}{(c)_{n}} . end{equation}$$

Now, set $$c=1$$ and $$b = 1 – 1/s$$. Then:

$$begin{equation} tag{5} label{5} _2F_1left(-n,1-1/s;1;1right) = frac{(1/s)_{n}}{n!} =: h(s) . end{equation}$$

If the inverse Laplace transform of $$h(cdot)$$ would be known, then I would only have to multiply by $$n!$$ and convolve it with $${ mathcal{L}^{-1} (1/s) } (t) = u(t)$$, where $$u(t)$$ is the unit step function.

However, I did not find an expression of the inverse Laplace transform of the hypergeometric function in $$(5)$$ in the “Tables of Laplace Transforms” by Oberhettinger and Badii (1973).

Approach 2

Another approach I tried is to note that $$begin{equation} tag{6} label{6} g(1/s) = (1/s)_{n} = frac{Gamma(1/s + n)}{Gamma(1/s)} . end{equation}$$

Unfortunately, the inverse laplace transform of $$begin{equation} tag{7} label{7} q(s) := frac{Gamma(s+a)}{Gamma(s+b)} end{equation}$$ is only given by Oberhettinger and Badii (p. 308) in the case when $$Re(b-a) >0$$, which is not the case here.

Approach 3

Finally, I tried rewriting $$g(1/s)/s$$ as follows:

begin{align} g(1/s)/s &= frac{1}{s^{2}} cdot frac{1+s}{s} cdot frac{1+2s}{s} dots frac{1+ns}{s} \ &= frac{ prod_{k=1}^{n} (1+ks) }{s^{n+2}} \ &= frac{n! prod_{k=1}^{n}Big{(}s+frac{1}{k} Big{)} }{s^{n+2}}. end{align}

Observe that $${ mathcal{L}^{-1} s^{-(n+2)} }(t) = ( (n+1)! )^{-1} t^{n+1}$$, so we are left with finding the inverse Laplace transform of the numerator (and convolve afterwards). However, I have not been able to do so thusfar.

Questions

1. Is the inverse Laplace transform of $$frac{(1/s)_{n}}{s}$$ known, or can it be calculated somehow?
2. Are there already any other expressions known for $$f(cdot)$$ that can be found by other means than the ones I laid out so far?

N.B. this is a more elaborate version of a question I asked earlier on MSE.

## inverse trigonometric functions in traditional form

Is there a way to have ‘arcsin’ rather than ‘sin^-1’ as arcsin in TraditionalForm? I couldn’t find any way to set options for customizing the way in which expressions are turned into traditional form, and feel that ‘arcsin’ is widely more used among mathematicians than sin^-1.

## Is the inverse of MST cut property true? Why?

If we partition the nodes of a graph into sets A and B, there is an edge e of weight larger than any other edge crossing the cut between A and B, e would never be in the minimum spanning tree?

## Limit of two convergent sequences of complex numbers equals the limit of the arithmetical sum of their inverse productory

Down below i leave my quesiton. It’s pretty much self explaining.

Let $$(z_{n})_{n=0}^{+infty}$$ and $$(w_{n})_{n}^{+infty}$$ convergent sequences of complex numbers. Prove that

$$begin{equation*} lim_{n rightarrow +infty}frac{z_{0}w_{n}+z_{1}w_{n-1}+dots+z_{n}w_{0}}{n+1} =(lim_{k rightarrow +infty}z_{k})(lim_{m rightarrow +infty}w_{m}) end{equation*}$$

What i have got so far:

I have recognized that
begin{align*} frac{z_{0}w_{n}+z_{1}w_{n-1}+dots+z_{n}w_{0}}{n+1} = frac{1}{n+1}prod_{i=0}^{n}(z_{i}w_{n-i}) end{align*}
I am kind of stuck on here right now. I have tried to split the limit (using the product propertie) to try and use the following result:

begin{align*} lim_{nrightarrow +infty}frac{z_{0}+z_{1}+dots+z_{n-1}}{n} = lim_{k}z_{k}, end{align*}
for any $$(z_{n})_{n=0}^{+infty}$$ convergent sequence of complex numbers.

If i could prove that $$prod_{i=0}^{n}(z_{i}w_{n-i})$$ can be adapted to be written as $$v_{0}+v_{1}+dots+v_{n}$$ and that $$lim_{n}v_{n} = (lim_{k rightarrow +infty}z_{k})(lim_{m rightarrow +infty}w_{m})$$ the exercise would become a lot more easier.
I wasn’t able to do that, if anyone could help me with this one i would be really thankfull. Have a nice one guys.

## reference request – Hadamard’s global inverse function theorem for an open subset of \$mathbb{R}^n\$?

Hadamard’s global inverse function theorem states that: Let $$F : mathbb{R}^N to mathbb{R}^N$$ be a $$C^2$$ mapping. Suppose that $$F(0) = 0$$ and that the Jacobian determinant of $$F$$ is nonzero at each point. Further suppose furthermore that $$F$$ is proper (inverse image of compact set is compact). Then $$F$$ is one-to-one and onto.

What I was wondering was can we replace the domain $$mathbb{R}^N$$ with an open subset $$U$$? in other words does the following version of Hadamard’s theorem hold?

Let $$U subset mathbb{R}^N$$ be a unit $$L^2$$ ball centered at $$0$$ and $$F : U to mathbb{R}^N$$ be a $$C^2$$ mapping. Suppose that $$F(0) = 0$$ and that the Jacobian determinant of $$F$$ is nonzero at each point of $$U$$. Further suppose furthermore that $$F$$ is proper (inverse image of compact set is compact). Then $$F$$ is one-to-one and surjective onto its image.

I have searched around but I could only find Hadamard’s theorem for $$mathbb{R}^N$$. If someone is familiar with this topic I appreciate any comments or references.

## Getting inverse of x^2 with positive x

My goal it to get inverse of x^2 for positive x.
But the result shows always minus -Sqrt[y].
How to correct

``````u[x_] := x ^2
\$Assumptions = x > 0;
Refine[InverseFunction[u][y], y [Element] Reals]
``````

I have tried this code also but is also shows -Sqrt[y].

``````Refine[x /. Solve[y == u[x], x][[1]], y [Element] Reals]
``````

## Asymptotic expansion around infinity for inverse cdf of normal distribution

I’m trying to get a asymptotic expansion as $$xrightarrowinfty$$ for a particular expression. I have

``````f(x_) := 1/x*InverseCDF(NormalDistribution(0, 1), 1 - Exp(-x^(1/4)));
``````

As $$xrightarrowinfty$$, we have $$f(x)rightarrow 0$$ but I am interested in how it behaves at $$infty$$. I tried using `AsymptoticSolve` and `Series` but neither seems to get me a nice expansion.

How can I proceed with this?

## artificial intelligence – Proving facts in inverse reinforcement learning

I was going through paper titled “Algorithms for Inverse Reinforcement Learning” by Andrew Ng and Russell.

It states following basics:

• MDP $$M$$ is a tuple $$(S,A,{P_{sa}},gamma,R)$$, where

• $$S$$ is a finite seto of $$N$$ states
• $$A={a_1,…,a_k}$$ is a set of $$k$$ actions
• $${P_{sa}(.)}$$ are the transition probabilities upon taking action $$a$$ in state $$s$$.
• $$R:Srightarrow mathbb{R}$$ is a reinforcement function (I guess its what it is also called as reward function) For simplicity in exposition, we have written rewards as $$R(s)$$ rather than $$R(s,a)$$; the extension is trivial.
• A policy is defined as any map $$pi : S rightarrow A$$

• Bellman Equation for Value function $$V^pi(s)=R(s)+gamma sum_{s’}P_{spi(s)}(s’)V^pi(s’)quadquad…(1)$$

• Bellman Equation for Q function $$Q^pi(s,a)=R(s)+gamma sum_{s’}P_{sa}(s’)V^pi(s’)quadquad…(2)$$

• Bellman Optimality: The policy $$pi$$ is optimal iff, for all $$sin S$$, $$pi(s)in text{argmax}_{ain A}Q^pi(s,a)quadquad…(3)$$

• All these can be represented as vectors indexed by state, for which we
adopt boldface notation $$pmb{P,R,V}$$.

• Inverse Reinforcement Learning is: given MDP $$M=(S,A,P_{sa},gamma,pi)$$, finding $$R$$ such that $$pi$$ is an optimal policy for $$M$$

• By renaming actions if necessary, we will assume
without loss of generality that $$pi(s) = a_1$$.

Paper then states following theorem, its proof and a related remark:

Theorem: Let a finite state space $$S$$, a set of actions $$A={a_1,…, a_k}$$, transition probability matrices $${pmb{P_a}}$$, and a discount factor $$gamma in (0, 1)$$ be given. Then the policy $$pi$$ given by $$pi(s) equiv a_1$$ is optimal iff, for all $$a = a_2, … , a_k$$, the reward $$pmb{R}$$ satisfies $$(pmb{P}_{a_1}-pmb{P}_a)(pmb{I}-gammapmb{P}_{a_1})^{-1}pmb{R}succcurlyeq 0 quadquad …(4)$$
Proof:
Equation (1) can be rewritten as
$$pmb{V}^pi=pmb{R}+gammapmb{P}_{a_1}pmb{V}^pi$$
$$thereforepmb{V}^pi=(pmb{I}-gammapmb{P}_{a_1})^{-1}pmb{R}quadquad …(5)$$
Putting equation $$(2)$$ into $$(3)$$, we see that $$pi$$ is optimal iff
$$pi(s)in text{arg}max_{ain A}sum_{s’}P_{sa}(s’)V^pi(s’) quad…forall sin S$$
$$iff sum_{s’}P_{sa_1}(s’)V^pi(s’)geqsum_{s’}P_{sa}(s’)V^pi(s’)quadquadquadforall sin S,ain A$$
$$iff pmb{P}_{a_1}pmb{V}^pisucccurlyeqpmb{P}_{a}pmb{V}^piquadquadquadforall ain Atext{\} a_1 quadquad …(6)$$
$$iffpmb{P}_{a_1} (pmb{I}-gammapmb{P}_{a_1})^{-1}pmb{R}succcurlyeqpmb{P}_{a} (pmb{I}-gammapmb{P}_{a_1})^{-1}pmb{R} quadquad text{…from (5)}$$
Hence proved.

Remark: Using a very similar argument, it is easy to show (essentially by replacing all inequalities in the proof above with strict inequalities) that the condition $$(pmb{P}_{a_1}-pmb{P}_a)(pmb{I}-gammapmb{P}_{a_1})^{-1}pmb{R}succ 0$$ is necessary and sufficient for $$piequiv a_1$$ to be the unique optimal policy.

I dont know if above text from paper is relevant for what I want to prove, still I stated above text as a background.

I want to prove following:

1. If we take $$R : S → mathbb{R}$$—there need not exist $$R$$ such that $$π^*$$ is the unique optimal policy for $$(S, A, T, R, γ)$$
2. If we take $$R : S × A → mathbb{R}$$. Show that there must exist $$R$$ such that $$π^*$$ is the unique optimal policy for $$(S, A, T, R, γ)$$.

I guess point 1 follows directly from above theorem as it says “$$pi(s)$$ is optimal iff …” and not “unique optimal iff”. Also, I feel it also follows from operator $$succcurlyeq$$ in equation $$(6)$$. In addition, I feel its quite intuitive: if we have same reward for any given state for every action, then different policies choosing different actions will yield same reward from that state hence resulting in same value function.

I dont feel point 2 is correct. I guess, this directly follows from the remark above which requires additional condition to hold for $$pi$$ to be “uinque optimal” and this condition wont hold if we simply define $$R : S × A → mathbb{R}$$ instead of $$R : S → mathbb{R}$$. Additionally, I feel, this condition will hold iff we had $$=$$ in equation $$(3)$$ instead of $$in$$ (as this will replace all $$succcurlyeq$$ with $$succ$$ in the proof). Also this also follow directly from point 1 itself. That is we can still have same reward for all actions from given state despite defining reward as $$R : S × A → mathbb{R}$$ instead of $$R : S → mathbb{R}$$, which is the case with point 1.

Am I correct with the analysis in last two paragraphs?

Update

After some more thinking, I felt I was doing it all wrong. Also I feel the text from the paper which I specified is of not much help in proving these two points. So let me restate new intuition for proofs for the two points:

1. For $$R: Srightarrow mathbb{R}$$, if some state $$S_1$$ and next state $$S_2$$ has two actions between them, $$S_1-a_1rightarrow S_2$$ and $$S_1-a_2rightarrow S_2$$, and if optimal policy $$π_1^*$$ chooses $$a_1$$, then $$π_2^*$$ choosing $$a_2$$ will also be optimal, thus making NONE “uniquely” optimal since both $$a_1$$ and $$a_2$$ will yield same reward as reward is associated with $$S_1$$ instead of with $$(S_1,a_x)$$.

2. For $$R: (S,A)rightarrowmathbb{R}$$, we can assign large reward say $$+∞$$ to all actions specified in given $$π^*$$ and $$-∞$$ to all other actions. This reward assignment will make $$π^*$$ a unique optimal policy.

Are above logics correct and enough to prove given points?