plotting – plot an inverse function

I have a phrase that in it x is a function of y but I need to plot y as a function of x. I tried InverseFunction but because my phrase is not single valued in some points mathematica can not give me plot. After that I tried Parametric plot but these code has not any answer. Infact I wanna to inverse x and y axes. what should I do?

0.008 y + 
 125. (2. - 
    0.0000271476 (659.714/(1 - 0.169673 y) + (
       2 (2.30388*10^-6 + 0.169673 y)^2)/(1 - 0.169673 y)^2) - (
    0.00271476 (2.30388*10^-6 + 0.169673 y))/(1 - 0.169673 y))^2 y=x 






Plot({0.008 y + 
 125. (2. - 
    0.0000271476 (659.714/(1 - 0.169673 y) + (
       2 (2.30388*10^-6 + 0.169673 y)^2)/(1 - 0.169673 y)^2) - (
    0.00271476 (2.30388*10^-6 + 0.169673 y))/(1 - 0.169673 y))^2 y}, {y, 5.6, 6}) 

Derivative of Inverse of sum of matrices

Given is the function $f : mathbb{R}^p to mathbb{R}$ with

$$
f(x) = q(x)^{top} G^{-1} q(x)
$$

where $G = A + x_1 B_1 + ldots + x_p B_p$.

  • The matrices $A, B_1, ldots, B_p in mathbb{R}^{n times n}$ are all symmetric positive definit.

  • $q: mathbb{R}^n to mathbb{R}^n$ and the Jacobian $nabla q$ is known.

Is it possible to derive a closed form for $nabla f$? For me, the hard part is
$G^{-1}$. Any hints or suggestions are really appreciated!

real analysis – Definition of the flow of an ODE and its inverse

At lesson, the teacher considers a flow $Phi$ given by the solutions of the ode system for $tin(0, T)$ and $xinmathbb R^d$,
$$
begin{cases}
y'(s)=b(y(s), s),&sleq T\
y(t)=x
end{cases},qquad(star)
$$

that is $Phi(x, t, s)=y(s)$ solving $(star)$. He said that we will be mostly concerned with $Phi(cdot, 0, cdot)$. The field $b$ is assumed to be Lipschitz continuous in both variables and bounded.

Then, he intoduces the inverse $Psi$ of the above flow as follows: $Psi(x, 0, s)=y(s)$ satisfying
$$
begin{cases}
y'(s)=-b(y(s), t-s),&s<tleq T\
y(0)=x
end{cases},
$$

and he said that $Psi$ is such that
$$
Phi(Psi(x, 0, s), 0, s)=x,quad Psi(Phi(x, 0, s), 0, s)=x.qquad (starstar)
$$

I do not understand $(starstar)$. Can someone help me? Maybe is the definition of the inverse wrong?

Thank you

probability or statistics – How to find the inverse of CDF of geometric function?

The cdf of geometric distribution is given by $ F(x)=1-(1-p)^x$.
I want to calculate the inverse of it, for example, $F^{-1}(U)$

I am doing the following

f(x_) := 1 - (1 - p)^x;  InverseFunction(f(u))

But I do not get anything.

my ultimate goal is to generate sample of a random variable that has geometric distribution.

ds.dynamical systems – Showing that the inverse of a function is approximately equivalent to $frac{1}{n^{1/alpha}}$

I’m currently working with someone on my PhD, and last week they asked me to check that a certain approximation holds as an exercise. Unfortunately, I couldn’t figure out how to do it, and we’ve since moved on as we don’t have time to go back and look at this, but it’s still really bugging me.

The set up:

We have a piecewise map $f_alpha: (0,1) to (0,1)$ defined as
begin{align}
f_alpha(x) = begin{cases}
x(1 + (2x)^alpha), &x in (0, 1/2) \
2x – 1, &x in (1/2, 1)
end{cases}
end{align}

with $0 < alpha < 1$.

Picture of the map

Say we have a starting point $x_0$. We denote by $x_1$ the point in $(0,1/2)$ that is mapped by the left part of the map to $x_0$. We then define a sequence {$x_i$} such that $x_{n+1}$ is the point in $(0,1/2)$ that maps to $x_n$.

What I was asked to show was
begin{align}
frac{1}{2}|x_n| sim frac{1}{n^{1/alpha}}
end{align}

The problem is I don’t know how to go about showing this. I was hoping I could write something of the form $|x_n – frac{1}{n^{1/alpha}}|$ and show that it converges to zero. However, I can’t explicitly express the inverse of the LHS of the function, so I can’t explicitly define $x_n$ given $x_{n-1}$. Would showing that it diverges in the forward direction work? Alternatively, am I approaching this completely wrong?

Thanks in advance.

inverse – InverseSeries slower than it should be

I am running into problems with InverseSeries taking far longer than it should. Given (for example)

$$F = q + a_2 q^2 + a_3 q^3 + ldots + a_{99} q^{99} + O(q^{100}),$$

Computing the inverse series is an easy inductive procedure; you start with $h =q + O(q^2)$; then once you have computed $h = q + b_2 q^2 + ldots b_{n-1} q^{n-1} + O(q^n)$ you write

$$F(q + ldots + b_n q^n) = q + b_n q^n + (text{some integer})q^n + O(q^{n+1})$$

and then you immediately solve for $b_n$. The most expensive part of this calculation is substituting a degree $d < 100$ polynomial into another degree $d < 100$ polynomial, which is annoying, but is something that mathematica can do quickly. But somehow whatever the code for InverseSeries does it freezes on examples that I can do faster “by hand”, where “by hand” means repeating the procedure above. What is going wrong, and how can it be fixed?

Here is an example which I can do manually in a few minutes but which Mathematica freezes on: (apologies I just cut and paste and I’m not sure of the correct formatting)

F=q + 5*q^2 + 8*q^3 - 91*q^4 - 937*q^5 - 4338*q^6 - 4296*q^7 + 100209*q^8 + 893854*q^9 + 
 3729551*q^10 + 988009*q^11 - 106732034*q^12 - 842141735*q^13 - 3136372148*q^14 + 
 1939998825*q^15 + 110833571873*q^16 + 783647004546*q^17 + 2565779096246*q^18 - 
 4488443604709*q^19 - 112739763063361*q^20 - 720080159676431*q^21 - 
 2023854259650355*q^22 + 6663444254248652*q^23 + 112683480468581322*q^24 + 
 653025143345851832*q^25 + 1515512580624420315*q^26 - 8476135390176280093*q^27 - 
 110899981154977421620*q^28 - 583930392628937740722*q^29 - 1044565210454102472039*q^30 + 
 9941896857454493006615*q^31 + 107623001789755733812129*q^32 + 
 514102735149021795473972*q^33 + 613792697659453617456234*q^34 - 
 11079610033717882616194102*q^35 - 103081343563886911581711082*q^36 - 
 444704193352528841884178970*q^37 - 225025293356332289039299015*q^38 + 
 11910949077592669155117835160*q^39 + 97495940945417413732877822907*q^40 + 
 376751398687168231753588890297*q^41 - 120771711888529263471915272617*q^42 - 
 12459713853379824931899133505129*q^43 - 91077401581362833228626720544419*q^44 - 
 311117312980830971962734331463980*q^45 + 423409533789796043944291041702618*q^46 + 
 12751209819173128030866296596556654*q^47 + 84023998931079059215982401415796850*q^48 + 
 248534954969155732476737675655795980*q^49 - 683386056103211869925369718093390968*q^50 - 
 12811678811355566322183515871298439783*q^51 - 76520094447189982535636680956140515685*
  q^52 - 189602835141152041220949710014366298265*q^53 + 
 901796156112889026951225880453991632299*q^54 + 
 12667783405502608595565452275522258404767*q^55 + 
 68734962279410595586298612084172931752196*q^56 + 
 134791811237603860485461449038540755090558*q^57 - 
 1080243554175113259472227932037669341843782*q^58 - 
 12346146389210270584499323364043494595441096*q^59 - 
 60821986510436232899477409897383013159169223*q^60 - 
 84453089234141305383131947547028579364218564*q^61 + 
 1220755361722927881921447396291832803501792011*q^62 + 
 11872945850923577492146096629138104329817610927*q^63 + 
 52918198782659044265844329535961836715539885129*q^64 + 
 38827109879617899252090203102633725191994493265*q^65 - 
 1325700320345933216980561394495920749772802020542*q^66 - 
 11273565472834002281261047695733288006882573779028*q^67 - 
 45144122775480096558532668590782893304307260671478*q^68 + 
 1946921721169086796443972909260898503226690770989*q^69 + 
 1397711549208204159945800378857903222775143338416122*q^70 + 
 10572298815650787143264952603576466566718787250535755*q^71 + 
 37603891277190420847613819836483962094154567835724926*q^72 - 
 37821086921685840599054422479757772091067018047483657*q^73 - 
 1439614451083277821545879752083980298083221704233081816*q^74 - 
 9792105697020965631846509911819444901839996690316011367*q^75 - 
 30385601492212559222034933849420980155613860977063995271*q^76 + 
 68828587955746985319973902536669959795661768467143495036*q^77 + 
 1454360270238898050230763204551740251603555283304218044880*q^78 + 
 8954418190489156428783367158778712189784710458271181026622*q^79 + 
 23561874662345057738637256592014339088905587514752725686907*q^80 - 
 95073548237398956870134671383103419296070440672983254195983*q^81 - 
 1444965649525281617723355435984923952001247508939637157703773*q^82 - 
 8078993303722001150427771574102727524401649171648024656431126*q^83 - 
 17190586988179334113503555950063459542575783017420671817190713*q^84 + 
 116720939563902411942410912309281786658001019096608596078968870*q^85 + 
 1414458400220043763667264074378644793595838587048619397064524313*q^86 + 
 7183809027731495420914353937307702698028195834348018120893712457*q^87 + 
 11315740144000203718321130213960736762158950882145625092638703481*q^88 - 
 133986776062069373176140968514177694241044451729485410879555592273*q^89 - 
 1365829577076882428189328294851928846429957728379702882979892304112*q^90 - 
 6285000176579497085055667164010443212571157903316261506956637435875*q^91 - 
 5968441318395573163766260626029750170233952666896285829048609490166*q^92 + 
 147128693911614522979104972693172044799519264544249762929510274217366*q^93 + 
 1301991842961016224683563531219134176695631970901573864797088538009930*q^94 + 
 5396830252406463811525778852570512932759858746657388951447519106376923*q^95 + 
 1167964618642564767028414545268490923294196688052413127194026692300170*q^96 - 
 156437015725997346523888239756083509066002272937792785486425464123464702*q^97 - 
 1225744012537375381434374076978042918332650567495871839872895404990084290*q^98 - 
 4531695465977987868416821546494004400866073156662768616479635202191331514*q^99 + 
 3077132211500782837488560708493387436849869673271024678631245962616281272*q^100 + O(q)^101;
InverseSeries(F) 

probability distributions – Confusion about Exponential versus Inverse Exponential Distributed random variable

Note: I have asked this question on another forum as we well, but have not received a response.
https://math.stackexchange.com/questions/3878100/confusion-about-exponential-versus-inverse-exponential-distributed-random-variab

I have an expression with random variables $h sim exp(lambda)$ and $g sim exp(gamma)$, and have the expression of the form.
$$h = (frac{a}{b}) frac{1}{g}$$

The CDF of h is

$$ = E_g (frac{a}{bg})$$
where $E$ represent the expected value with respect to $g$.

Should I now consider $g$ as an exponentially distributed random variable or Inverse exponential distributed random variable.

In the case of inverse exponential, I know that the expectation does not exist, then how can I solve the problem.

PS> The actual equation is a bit complex, but can be easily written in the format shared above.

inverse inequality is symmetric matrices

My question follows from https://math.stackexchange.com/questions/3857976/inverse-inequality-of-symmetric-matrix. Suppose we assume that $A$ and $B$ are two positive definite matrices with positive entries and $Ageq B $ entry wise.
Can we say that $A^{-1}leq B^{-1}$ entry wise?
I tried with numeric examples in Matlab, but I am not getting any counter-example.
Any help would be really great.

calculus – If $f, g$ are continuous and $f(x) ge g(x)$ for all $x in [a, b]$, then $int_{a}^{b} f(x) dx ge int_{a}^{b} g(x) dx$. Is the inverse true?

The proof for the first statement is trivial using the definition of Riemann sums, since

begin{equation}
int_{a}^{b} f(x) dx = lim_{n to infty} sum_{i=1}^{n} f(x_i^*) Delta x ge lim_{x to infty}sum_{i=1}^{n}g(x_i^*)Delta x = int_{a}^{b} g(x) dx
end{equation}

for any sample point $x_i^*$.

But is the inverse true? Namely, if $int_{a}^{b} f(x) dx ge int_{a}^{b} g(x) dx$ for all $x in (a, b)$, then $f(x) ge g(x)$. I couldn’t think of any counterexamples where this is true, and trying to adapt the above ‘proof’ did not work well since we cannot conclude that $f(x) ge g(x)$ for every $x in (a,b)$ simply from the sums alone.

Is there a mistake in my thinking? Otherwise, a relevant counter-example (if it is wrong) or a hint for a proof would be greatly appreciated!

calculus and analysis – Series of inverse functions, unclear numerical constant

I was answering another question here and came up with this simple illustrative example that should have an analytic solution. Indeed it has, but I do not understand it. In particular, where 85 is coming from?

g[x_]:=BesselJ[0,x]
f[x_]:=Exp[x]
Series[f[InverseFunction[g][y]],{y,0,0}]
Out[1]= E^-BesselJZero[0,85]+O[y]^1