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# Least linear square fit of multiple measurements with functions sharing parameters

1. Apr 6, 2009

### Sebastian.de

Dear All,

I would like to least square fit a number of measurements using several nonlinear functions with shared parameters (similar to the advanced fitting in Origin) using Mathematica.
Therefor I would be interested in an Algorithm like Gauss-Newton that fits several data-set to several functions (with shared parameters) "in parallel".
For example, I would like to fit the set of functions f1=a*exp(c1/x) ... fn=a*exp(cm/x) to a number of giving data-sets y1(x1,...,xm) ... yn(x1,...,xm). As indicated all "a" in f1...fn should have the same value while c1...cn differ.
Unfortunately, I can not find a suited algorithm on the Internet and I have a hard time adopting the algorithms fitting one function to one set of data.
I would appreciate any help, links or references that cover the given topic.

Thank you very much for your help!

Best regards,
Sebastian

2. Apr 8, 2009

### Wallace

Re: Least linear square fit of multiple measurements with functions sharing parameter

I may have misunderstood your question, but the simplest thing I can think of is to consider the sum of all your functions $$\chi^2$$'s as a single number and minimise that over all the parameters in all of your functions. Then this reduces to a regular multi-dimensional minimisation.

This assumes that you are weighting all your functions equally in the fit, but you could add weights to the different $$\chi^2$$'s if that was relevant to your problem.

3. Apr 11, 2009

### Sebastian.de

Re: Least linear square fit of multiple measurements with functions sharing parameter

Thank you very much for your reply. You didn't misunderstand my question. The approach you suggested seems perfectly fine to me. However, when I tried it I got stuck when generating the Jacobi-matrix.
I hardly have any experience in dealing with matrix calculation, so I am not sure if I ran into a big problem or if I am just to stupid to see the obvious.
The algorithms I found dealing with the least linear square problem generally proceeded in a way that minimizes the sum(f(x,parameters)-y)^2 (where f(x) is the fit function and y are the data points) by making up the Jacobi-matrix of (f(x,parameters)-y) with respect to the parameters. The square is then introduced by multiplying the Jacobi-matrix with its transpose.
If I understand your suggestion correct, I would end up minimizing a problem which looks like this: sum over i (sum(fi(xi,parameters-yi)^2). Having multiple sums of squares now instead of just one, I have not managed to figure out anything similar to the Jacobi-matrix used in the simpler problem. As a result, I was not able to use any of the algorithms I found.
I would be very grateful if you could point me towards any literature that deals with this problem in some detail. I am mildly optimistic that I would be able to write a script solving the problem with the help of a book that illustrates a applicable algorithm on an example.

Best regards,
Sebastian

4. Apr 11, 2009

### Wallace

Re: Least linear square fit of multiple measurements with functions sharing parameter

There are multi-dimensional minimisation algorithms that do not require you to calculate the Jacobian. The one that springs to mind is the downhill simplex method, but there are probably others. I've used simplex a fair bit and it usually works well.http://www.gnu.org/software/gsl/manual/html_node/Multidimensional-Minimization.html" [Broken] is a good implementation of this algorithm in the Gnu Science Libraries

Last edited by a moderator: May 4, 2017
5. Apr 14, 2009

### gjanik

Re: Least linear square fit of multiple measurements with functions sharing parameter

I happen to be working on an almost identical problem, although in my case the time constants are common and the amplitudes are not. I took the data sets and time shifted and concatenated them so that they became one long data set. I then created an analytic fit function consisting of a sum of exponentials, each multiplied by two UnitStep functions which "turn it on" during the appropriate time period and turn it off for all other times. Each exponential is multiplied by zero for all times except when it's data is active. Finally, I used the FindFit function to fit the single data set to the single fit function. It would be nice if they had a more straightforward way to do this. A quick Google showed me that other math packages, like SigmaPlot, handle this sort of problem directly. I have attached sample code if you are interested.

#### Attached Files:

• ###### double_fit_1a.zip
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6. Apr 18, 2009

### Sebastian.de

Re: Least linear square fit of multiple measurements with functions sharing parameter

Thank you very much to both of you. I think I have all the information I need now to fit the data.
Thanks again!

Best regards,
Sebastian


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# Ngô Quốc Anh

## August 29, 2010

### Achieving regularity results via bootstrap argument, 4

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 2:36

Let us consider the following equation

$\displaystyle u(x) = \int_{{\mathbb{R}^n}} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} , \quad x \in {\mathbb{R}^n}$

for $n\geqslant 1$ and $0<\alpha. In this entry, by using boothstrap argument, we show that

Theorem. If positive function $u \in L_{loc}^\frac{2n}{n-\alpha}(\mathbb R^n)$ solves the equation, then $u \in C^\infty(\mathbb R^n)$.

In the process of proving the result, we need the following result

Proposition. Let $V \in L^\frac{n}{\alpha}(B_3)$ be a non-negative function and set

$\displaystyle \delta(V)=\|V\|_{L^\frac{n}{\alpha}(B_3)}$.

For $\nu >r>\frac{n}{n-\alpha}$, there exist positive constants $\overline \delta<1$ and $C \geqslant 1$ depending only on $n, \alpha, r$ and $\nu$ such that for any $0 \leqslant V \in L^\frac{n}{\alpha}(B_3)$ with $\delta(V) \leqslant \overline \delta$, $h \in L^\nu(B_2)$ and $0 \leqslant u \in L^r(B_3)$ satisfying

$\displaystyle u(x) \leqslant \int_{{B_3}} {\frac{{V(y)u(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} + h(x), \quad x \in {B_2}$

we have

$\displaystyle {\left\| u \right\|_{{L^\nu }({B_{1/2}})}} \leqslant C\left( {{{\left\| u \right\|}_{{L^r}({B_3})}} + {{\left\| h \right\|}_{{L^\nu }({B_2})}}} \right)$.

## August 22, 2010

### Liouville’s theorem and related problems

Filed under: Giải tích 7 (MA4247), PDEs — Tags: — Ngô Quốc Anh @ 6:35

The following theorem is well-known

Theorem (Liouville). Let $\Omega$ be a simply connected domain in $\mathbb R^2$. Then all real solutions of

$\displaystyle \Delta u +2Ke^u=0$

in $\Omega$ where $K$ a constant, are of the form

$\displaystyle u=\log\frac{|f'|^2}{\left(1+\frac{K}{4}|f|^2\right)^2}$

where $f$ is a locally univalent meromorphic function in $\Omega$.

In geometry, our PDE

$\displaystyle \Delta u +2Ke^u=0$

says that under the case $\Omega=\mathbb R^2$, it holds

$e^u|dz|^2=f^*g_K$

where $g_K$ denotes the standard metric on $\mathbb S^2$ with constant curvature $K$. Thus we have

Corollary. All solutions of the PDE in $\mathbb R^2$ with $K>0$ and

$\displaystyle \int_{\mathbb R^2} e^u<\infty$

are of the form

$\displaystyle u(x)=\log\frac{16\lambda^2}{\left(4+\lambda^2K|x-x_0|^2\right)^2},\quad \lambda>0, \quad x_0 \in \mathbb R^2$.

## August 19, 2010

### L^infinity-boundedness for a single solution of -Delta u = Vexp(u)

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 7:00

The aim of this entry is to derive the $L^\infty$-boundedness for a single solution of the following PDE

$\displaystyle -\Delta u = V(x) e^u$

over a domain $\Omega$. This elegant result had been done by Brezis and Merle around 1991 published in Comm. Partial Differential Equations [here].

There are two possible cases.

The case of bounded domain. Let us assume $u$ a solution of the following PDE

$\displaystyle\begin{cases}- \Delta u = V(x){e^u}& \text{ in }\Omega , \hfill \\ u = 0&\text{ on }\partial \Omega ,\end{cases}$

where $\Omega \subset \mathbb R^2$ is a bounded domain and $V$ is a given function on $\Omega$.

Theorem. If $V \in L^p$ and $e^u \in L^{p'}$ for some $1 then $u \in L^\infty$.

Proof. It first follows from the Brezis-Meler inequality that

$e^{ku} \in L^1, \quad \forall k>0$

which by the Holder inequality gives

$e^{u} \in L^r, \quad \forall r<\infty$.

Therefore, if $p<\infty$

$Ve^u \in L^{p-\delta}, \quad \forall \delta>0$

while if $p=\infty$

$Ve^u \in L^r, \quad \forall r<\infty$.

Thus, a standard $L^p$-estimate argument from the elliptic theory implies that $u$ is bounded.

## August 17, 2010

### Evaluate complex integral via the Fourier transform

Filed under: Giải tích 7 (MA4247) — Tags: — Ngô Quốc Anh @ 5:56

As suggested from this topic, we are interested in evaluating the following complex integral

$\displaystyle G(t)=\mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {{{\left( {\frac{{\sin x}} {x}} \right)}^2}{e^{itx}}dx}$.

The trick here is to use the Fourier transform. Thanks to ZY for teaching me this interesting technique.

In $\mathbb R$, the Fourier transform of function $f$, denoted by $\mathcal F[f]$, is defined to be

$\displaystyle \mathcal F[f](y) = \int_{ - \infty }^\infty {f(x){e^{ - 2\pi ixy}}dx}$.

If we apply the Fourier transform twice to a function, we get a spatially reversed version of the function. Precisely,

$\displaystyle\begin{gathered} \mathcal{F}\left[ {\mathcal{F}[f]} \right](z) = \int_{ - \infty }^\infty {\mathcal{F}[f](y){e^{ - 2\pi iyz}}dy} \hfill \\ \qquad\qquad= \int_{ - \infty }^\infty {\mathcal{F}[f](y){e^{2\pi iy( - z)}}dy} \hfill \\ \qquad\qquad= {\mathcal{F}^{ - 1}}\left[ {\mathcal{F}[f]} \right]( - z) \hfill \\ \qquad\qquad= f( - z) \hfill \\ \end{gathered}$

where $\mathcal F^{-1}$ denotes the inverse Fourier transform.

## August 16, 2010

### The Moser-Trudinger inequality for domains with holes

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 2:42

In this entry, we are interested in the following result

Theorem (Moser-Trudinger’s inequality for domains with holes). Let $\Omega$ be a bounded smooth domain in $\mathbb R^2$. Let $S_1$ and $S_2$ be two subsets of $\overline \Omega$ satisfying

${\rm dist}(S_1,S_2) \geqslant \delta_0>0$

and let $\gamma_0$ be a number satisfying $\gamma_0 \in \left(0,\frac{1}{2}\right)$. Then for any $\varepsilon>0$, there exists a constant $c=c(\varepsilon, \delta_0, \gamma_0)>0$ such that

$\displaystyle\int_\Omega {{e^u}} \leqslant C\exp \left[ {\frac{1}{{32\pi - \varepsilon }}\int_\Omega {{{\left| {\nabla u} \right|}^2}} + C} \right]$

holds for all $u \in H_0^1(\Omega)$ satisfying

$\displaystyle\frac{{\int_{{S_1}} {{e^u}} }}{{\int_\Omega {{e^u}} }} \geqslant {\gamma _0}, \quad \frac{{\int_{{S_2}} {{e^u}} }}{{\int_\Omega {{e^u}} }} \geqslant {\gamma _0}$.

## August 13, 2010

### An application of the (Moser-)Trudinger inequality to the mean field equations

Let $(M,g)$ be a compact Riemannian surface with the volume $|M|$. The simplest form of the mean field equation studied in the contexts of the prescribing Gaussian curvature, statistical mechanics of many vortex points in the perfect fluid and self-dual gauss theories is given by

$\displaystyle - {\Delta _g}u = \lambda \left( {\frac{{{e^u}}}{{\int_M {{e^u}d{v_g}} }} - \frac{1}{{|M|}}} \right), \quad \text{ on } M$

with

$\displaystyle\int_M {ud{v_g}} = 0$

where $\lambda$ is a real number.

The mean field equation has a variational structure, and $u$ is a solution if and only if it is a critical point of

$\displaystyle {J_\lambda }(v) = \frac{1}{2}\int_M {{{\left| {\nabla v} \right|}^2}d{v_g}} - \lambda \log \int_M {{e^v}d{v_g}}$

defined for $v \in H^1(M)$ with

$\displaystyle\int_M {vd{v_g}} = 0$.

It is worth noticing from this entry that so far our Moser-Trudinger’s inequality is just for $\mathbb S^2$

Theorem (Moser-Trudinger’s inequality for $\mathbb S^2$). There are constants $\eta>0$ and $c=c(g)>0$ such that for each $p \geqslant 2$

$\displaystyle \log \overline\int_{{\mathbb{S}^2}} {{e^{2u}}d{v_{{g_c}}}} \leqslant \left[ {\frac{1}{{4\pi }}\int_{{\mathbb{S}^2}} {{{\left| {\nabla u} \right|}^2}d{v_{{g_c}}}} + 2\overline u } \right] + {C_2}$

for all $u \in W^{1,2}(\mathbb S^2)$.

## August 11, 2010

### Two ways to present SU(3) Toda systems

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 4:13

This entry is concerned with SU(3) Toda systems in non-abelian relativistic self-dual gauge theory. Usually, they are given by

$\displaystyle\begin{gathered} - {\Delta _g}{u_1} = 2{\lambda _1}\left( {\frac{{{e^{{u_1}}}}}{{\int_M {{e^{{u_1}}}d{v_g}} }} - \frac{1}{{|M|}}} \right) - {\lambda _2}\left( {\frac{{{e^{{u_2}}}}}{{\int_M {{e^{{u_2}}}d{v_g}} }} - \frac{1}{{|M|}}} \right), \hfill \\ - {\Delta _g}{u_2} = - {\lambda _1}\left( {\frac{{{e^{{u_1}}}}}{{\int_M {{e^{{u_1}}}d{v_g}} }} - \frac{1}{{|M|}}} \right) + 2{\lambda _2}\left( {\frac{{{e^{{u_2}}}}}{{\int_M {{e^{{u_2}}}d{v_g}} }} - \frac{1}{{|M|}}} \right), \hfill \\ \end{gathered}$

on $M$ with

$\displaystyle\int_M {{u_1}d{v_g}} = \int_M {{u_2}d{v_g}} = 0$

where $(M,g)$ denotes a compact Riemannian surface and $\lambda_1$, $\lambda_2$ are non-negative constants.

We now consider function $h$ satisfying

$\displaystyle - \Delta h =\frac{1}{|M|} \quad \text{ in } M$

The existence of such a $h$ will be considered later.

We now introduce

$\displaystyle\begin{gathered} {v_1} = {u_1} - \log \int_M {{e^{{u_1}}}d{v_g}} - (2{\lambda _1} - {\lambda _2})h, \hfill \\ {v_2} = {u_2} - \log \int_M {{e^{{u_2}}}d{v_g}} - ( - {\lambda _1} + 2{\lambda _2})h. \hfill \\ \end{gathered}$

## August 8, 2010

### The nodal set of eigenfunctions

Filed under: PDEs — Ngô Quốc Anh @ 17:21

An eigenfunction $\varphi$ is meant to be a solution of Dirichlet’s problem

$\begin{cases}-\Delta \varphi = \lambda\varphi,&\text{ in } \Omega,\\\varphi=0,&\text{ on } \partial\Omega,\end{cases}$

where

$\displaystyle\Delta = \sum\limits_{k = 1}^n {\frac{{{\partial ^2}}}{{\partial x_k^2}}}$

is the Laplacian, $\Omega$ is a bounded smooth domain in $\mathbb R^n$, and $\lambda$ is a constant (i.e. the corresponding eigenvalue).

It is well known that the first eigenfunction is positive in $\Omega$, and all higher eigenfunctions must change sign.

Definition. The nodal set of an eigenfunction $\varphi$ is defined to be

$\displaystyle N = \overline {\left\{ {x \in \Omega :\varphi (x) = 0} \right\}}$.

To gain a better understanding of nodal sets let us consider a few examples.

Example 1. Let us consider the case when $\Omega=(0,l)$ where the first four eigenfunctions had been shown.

Red points are nodal sets. In this case, we simply call nodal nodes. This comes from the fact that all eigenvalues $\lambda_k$ and eigenfunctions $\varphi_k$ are already known

$\displaystyle{\varphi _k}(x) = \sqrt {\frac{2}{l}} \sin \left( {\frac{{n\pi }}{l}x} \right), \quad k \in \mathbb{N}$.

## August 6, 2010

### An upper bound for solutions via the maximum principle

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 1:29

It is known that [here] the following PDE

$\displaystyle\begin{cases}\Delta u =e^u, & {\rm in } \, \mathbb R^2,\\\displaystyle\int_{\mathbb R^2}e^u<\infty.\end{cases}$

has no $C^2$ solution. However, this is no longer true if we replace the whole space by a ball of radius $R$, say $B_R(0)$. In this entry, we show that if $u \in C^2(\overline B_R)$ is a solution of

$\Delta u=e^u, \quad {\rm in }\; B_R$

then

$\displaystyle u(0) \leqslant \log 8- 2\log R$.

To this purpose, let us recall the following

The Maximum Principle. Let assume $U\subset \mathbb R^2$ be open and bounded. We consider an elliptic operator $L$ of the form

$\displaystyle Lu = - \sum\limits_{i,j = 1}^2 {{a^{ij}}\frac{{{\partial ^2}u}}{{\partial {x_i}\partial {x_j}}} + \sum\limits_{k = 1}^2 {{b^k}\frac{{\partial u}}{{\partial {x_k}}}} } + cu$

where coefficients are continuous and the standard uniform ellipticity condition holds.

## August 2, 2010

### The Brezis-Merle inequality

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 0:25

I am going to talk about uniform estimates and blow-up phenomena for solutions of

$\displaystyle -\Delta u=V(x)e^u$

in two dimensions done by Brezis and Merle around 1991 published in Comm. Partial Differential Equations [here]. As a first step, I am going to derive some inequality that we need later.

Assume $\Omega \subset \mathbb R^2$ is bounded domain and let $u$ be a solution of

$\displaystyle -\Delta u=f(x)$

together with Dirichlet boundary condition. Here function $f$ is assumed to be of class $L^1(\Omega)$.

Theorem (Brezis-Merle). For every $\delta \in (0,4\pi)$ we have

$\displaystyle\int_\Omega {\exp \left[ {\frac{{(4\pi - \delta )|u(x)|}}{{{{\left\| f \right\|}_1}}}} \right]dx} \leqslant \frac{{4{\pi ^2}}}{\delta }{\rm diam}{(\Omega )^2}$

where $\|\cdot\|_1$ denotes the $L^1$-norm and $u$ a solution to our PDE.

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# 9. The Forced Response - Second Order Linear DEs

As in first order circuits, the forced response has the form of the driving function. For a constant driving source, it results in a constant forced response.

For non-constant driving functions e.g. when

E = E_0 sin omega t,

the complete response of a circuit is the sum of a natural response and a forced response.

Note: Such solutions can also be obtained using the Laplace transformation method (which we meet later) when initial conditions are given.

## Constant Forced Response

### Example 1

In a RLC series circuit, R = 10\ Omega, C = 0.02\ "F", L = 1\ "H" and the voltage source is E = 100\ "V". Solve for the current i(t) in the circuit given that at time t = 0, the current in the circuit is zero and the charge in the capacitor is 0.1\ "C".

Using L(di)/(dt)+Ri+1/Cq=E from Section 8, we can write the DE in i and q as follows:

(di)/(dt)+10i+50q=100\ \ \ (1)

Differentiating gives a 2nd order DE in i:

(d^2i)/(dt^2)+10(di)/(dt)+50i=0

Auxiliary equation: m^2+10m+50=0

Solution is: m_1=-5+5j, \ m_2=-5-5j

So

i=e^(-5t)(A\ cos 5t+B\ sin 5t)

Now

[i]_(t=0)=A=0

(This means at t = 0, i = A = 0 in this case.)

So

i=e^(-5t)B\ sin 5t

We need to find the value of B.

Differentiating gives:

(di)/(dt)=e^(-5t)(5 B\ cos 5t)+ (B sin 5t)(-5e^(-5t))

=5Be^(-5t)(cos 5t-sin 5t)

At t = 0, (di)/(dt)=5B

Returning to equation (1): (di)/(dt)+10i+50q=100

Now, at time t = 0,

[(di)/(dt)]_(t=0)+10(0)+50(0.1)=100

So [(di)/(dt)]_(t=0)=95=5B

Therefore B = 19.

So we have our solution for the current i:

i=19e^(-5t)sin 5t

Here is the graph of the current at time t.

Graph of i(t)=19e^(-5t)sin 5t.

## Natural and Non-Constant Forced Response

### Example 2

Here is an example showing the effect of a forced response. We consider the "natural" case first (with constant EMF).

### 2a. Natural Response

In an RLC circuit we have L = 1\ "H", R = 10\ Omega and C = 0.0025\ "F" and at t = 0, the current is 0 and i'(0) = 0.1\ "A/s".

Solve for i.

In this case, R^2 = 10^2 = 100 and (4L)/C = (4xx1)/0.0025 = 1600, so R^2<(4L)/C.

We have an example of Case 3, Under-damped, from Section 8.

Once again, using L(di)/(dt)+Ri+1/Cq=E we can write:

(di)/(dt)+10i+400q=E\ \ \ (1)

where E is a constant.

Differentiating gives:

(d^2i)/(dt^2)+10(di)/(dt)+400i=0

We are given:

i(0)=0

i'(0)=0.1

The auxiliary equation this time is: m^2+10m+400=0

Solution is: m_1=-5+19.365j, \ m_2=-5-19.365j

So since it is the under-damped case, we have

i=e^(-5t)(A cos 19.365 t+ {:B sin 19.365t)

Since i(0)=0, we substitute and get:

0=A

So

i=e^(-5t)(B sin 19.365t)\ \ \ (2)

Differentiating (2) gives:

(di)/(dt)=e^(-5t)(19.365B cos 19.365t) + {: (B sin 19.365t)(-5e^(-5t))

=e^(-5t)(19.365B cos 19.365t - {:5B sin 19.365t)

Since i'(0)=0.1, we substitute and obtain:

0.1=(19.365B)

So

B=0.1/19.365 = 0.00516

So we conclude:

i=0.00516 e^(-5t)(sin 19.365t)

Here is the graph of the solution for the natural response case:

Graph of i(t)=0.00516 e^{-5.0t} sin 19.36\ t.

Notice the signal decays quickly.

### 2b. Forced Response

Let us now take the same RLC circuit we had in Example 2a, but now we have a non-constant EMF (electromotive force) of:

E = -0.08\ cos\ 2.5t

Using the following result (from Section 8)

L(di)/(dt)+Ri+1/Cq=E

with

L = 1, R = 10 and 1/C = 1/0.0025 = 400,

we differentiate throughout with respect to t to give the following 2nd order DE, with initial conditions shown:

(d^2i)/(dt^2)+10(di)/(dt)+400i=0.2\ sin\ 2.5t

i(0)=0

i'(0)=0.1

I'm using Scientfic Notebook (a computer algebra system) to obtain the solution. (It's pretty ugly to work this one out on paper.)

Here is the solution that Scientific Notebook gives us.

i(t) =  -2.244 xx 10^{-4}\sin 19. 37t\cos 21.87t   +\ 5.132 xx 10^{-5}\sin 19.37t\sin 21.87t +\ 2.815 xx 10^{-4}\sin 19.37t\cos 16.87t  -\ 8.345 xx 10^{-5}\sin 19.37t\sin 16.87t  -\ 8. 345 xx 10^{-5}\cos 19.37t\cos 16. 87t  -\ 2.815 xx 10^{-4}\cos 19. 37t\sin 16.87t  +\ 5.132 xx 10^{-5}\cos 19.37t\cos 21.87t  +\ 2.244 xx 10^{-4}\cos 19.37t\sin 21.87t  +\ 5.107 xx 10^{-3}e^{-5.0t} \sin 19.36t  +\ 3. 212 xx 10^{-5} e^{-5.0t}\cos 19.36t

I hope you can see why I choose to use a computer algebra system to solve this kind of problem!

Easy to understand math videos:
MathTutorDVD.com

Here is the graph for the Forced Response solution we found just now:

You can see after the initial spike, the current settles down into a sinusoidal pattern.

We will also use the Laplace Transform in a later section to solve this type of DE.

Search IntMath

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

### Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.


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## Cryptology ePrint Archive: Report 2019/509

New Slide Attacks on Almost Self-Similar Ciphers

Orr Dunkelman and Nathan Keller and Noam Lasry and Adi Shamir

Abstract: The slide attack is a powerful cryptanalytic tool which has the unusual property that it can break iterated block ciphers with a complexity that does not depend on their number of rounds. However, it requires complete self similarity in the sense that all the rounds must be identical. While this can be the case in Feistel structures, this rarely happens in SP networks since the last round must end with an additional post-whitening subkey. In addition, in many SP networks the final round has additional asymmetries -- for example, in AES the last round omits the MixColumns operation. Such asymmetry in the last round can make it difficult to utilize most of the advanced tools which were developed for slide attacks, such as deriving from one slid pair additional slid pairs by repeatedly re-encrypting their ciphertexts.

In this paper we overcome this "last round problem" by developing four new types of slide attacks. We demonstrate their power by applying them to many types of AES-like structures (with and without linear mixing in the last round, with known or secret S-boxes, with 1,2 and 3 periodicity in their subkeys, etc). In most of these cases, the time complexity of our attack is close to $2^{n/2}$, which is the smallest possible complexity for slide attacks. Our new slide attacks have several unique properties: The first attack uses slid sets in which each plaintext from the first set forms a slid pair with some plaintext from the second set, but without knowing the exact correspondence. The second attack makes it possible to create from several slid pairs an exponential number of new slid pairs which form a hypercube spanned by the given pairs. The third attack has the unusual property that it is always successful, and the fourth attack can use known messages instead of chosen messages, with only slightly higher time complexity.

Category / Keywords: secret-key cryptography / Slide attack, 1-KSA, 1K-AES, Slid sets, Hypercuber of slid pairs, suggestive structures, Substitution slide

Date: received 16 May 2019, last revised 4 Oct 2019

Contact author: orrd at cs haifa ac il

Available format(s): PDF | BibTeX Citation

Short URL: ia.cr/2019/509

[ Cryptology ePrint archive ]


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If its eccentricity is the maximum value of the function,
Question:

Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$ be a given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum

value of the function, $\phi(t)=\frac{5}{12}+t-t^{2}$, then $a^{2}+b^{2}$ is equal to:

1. (1) 145

2. (2) 116

3. (3) 126

4. (4) 135

Correct Option: , 3

Solution:

The given ellipse is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b)$

Length of latus rectum $=\frac{2 b^{2}}{a}$

$\Rightarrow \frac{2 b^{2}}{a}=10 \Rightarrow b^{2}=5 a$……(i)

Now $\phi(t)=\frac{5}{12}+t-t^{2}$

$\phi^{\prime}(t)=1-2 t=0 \Rightarrow t=\frac{1}{2}$

$\phi^{\prime \prime}(t)=-2<0 \Rightarrow$ maximum

$\Rightarrow \phi(t)_{\max }=\frac{5}{12}+\frac{1}{2}-\frac{1}{4}=\frac{8}{12}=\frac{2}{3}$

Since, $\phi(t)_{\max .}=$ eccentricity

$\Rightarrow e=\frac{2}{3}$

Now, $b^{2}=a^{2}\left(1-e^{2}\right)$

$5 a=a^{2}\left(1-\frac{4}{9}\right) \Rightarrow 5 a=\frac{5 a^{2}}{9} \Rightarrow a^{2}-9 a=0$

$\Rightarrow a=9 \Rightarrow a^{2}=81$ and $b^{2}=45$

$\therefore a^{2}+b^{2}=81+45=126$


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# Metropolis sampling (symmetric proposal distribution)

• Can Metropolis sampling be used in conjunction with Gibbs sampling? So for example, if I have three parameters of interest, but only two of them have full conditionals that are known distributions, can I sample using Gibbs for those two parameters and Metropolis using the other parameter for each iteration?
• If the parameter of interest only takes positive values, would it be wrong (in terms of producing an answer, not efficiency) to use a normal proposal distribution centered on the previous iteration's value of the parameter?

## 1 Answer

• Yes. In fact, you can consider a Gibbs sampler to be a special case Metropolis Hasting sampler.

• If the parameter of interest is strictly positive, using a normal proposal will still work (any negative proposals will be automatically rejected). However, there is a very good chance that using the normal distribution for the proposed log of the parameter will actually be more efficient, as the posterior of the log-parameter is often approximately normal.


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# Given a regression based model with many feature variables; what tools would you utilize to figure out which feature variables add the most variance?

Given a hypothetical dataset {S} with 100 X feature variables and 10 predicted Y variables.

X1 ... X100 Y1 .... Y10
1 .. 2 3 .. 4
4 .. 3 2 .. 1

Let's say I want to improve the accuracy of Y1. I am prepared to constraint/remove the input variables in order to increase the accuracy. How would I go about finding the culprits for making Y1 more variable than needed?

E.g. I find that X49 adds the biggest swing in variance with Y1 and after constraining it Y1 is fitted better.

How would I go about finding it's X49?

EDIT: I'm asking for approaches towards sensitivity analysis. Not deciding which variables need to be removed. Let's assume all 100 X variables are important but some need to be constrained (e.g. X49)

There might be a smarter method but I would simply try to fit a model without $$X_i$$ for every feature $$X_i$$ (and also a reference model with all the features). By contrast the model where $$X_{49}$$ is removed should obtain the lowest variance if $$X_{49}$$ is responsible for a lot of variance.