text:
We implemented \texttt{FedRobust} in the Tensorflow platform \citep{abadi2016tensorflow} and numerically evaluated the algorithm's robustness performance against affine distribution shifts and adversarial perturbations. We considered the standard MNIST \citep{lecun1998mnist} and CIFAR-10 \citep{krizhevsky2009learning} datasets and used three standard neural network architectures in the literature: AlexNet \citep{krizhevsky2012imagenet}, Inception-Net \citep{szegedy2015going}, and a mini-ResNet \citep{he2016deep}.

\subsection{CIFAR-10 data: Experimental setup}
In the experiments, we simulated a federated learning scenario with $n=10$ nodes where each node observes $m=5000$ training samples. We also divided the extra $10,000$ samples in each dataset to two validation and test sets containing $5000$ samples each. For CIFAR-10 samples, we applied the sandard normalization and scaled and linearly mapped the pixel intensity values to interval $[-1,1]$. We applied batch normalization \cite{ioffe2015batch} in order to stabilize training and used the ADAM optimizer \citep{kingma2014adam} with stepsize value $10^{-4}$ and default beta parameters $\beta_1=0.9$ and $\beta_2=0.99$ to optimize the neural net's parameters for $T=100$ epochs ($10000$ iterations).

We did cross validation to choose $\lambda\in \{0.1,0.5,1,5,10,50\} $ and chose the $\lambda$-value resulting in the closest additive penalty $\frac{1}{n}\sum_{i=1}^n[ \Vert\Lambda^{i^*}-I\Vert_2^2 + \Vert\delta^{i^*}\Vert_2^2]$ to $10$ percent of the average sample norm, i.e. $\frac{0.1}{m}\sum_{i=1}^m\Vert\mathbf{x}^{\operatorname{val}}_i \Vert^2_2$, over the $m=5000$ validation samples. To perform GDA optimization, we applied two ascent steps per descent step with stepsize $\frac{1}{2\lambda}$. In order to simulate an affine distribution shift, we manipulated each $\tilde{\mathbf{x}}^i_{j}$ in the original training dataset via an affine transformation chosen randomly at each node:
\begin{equation}
    \mathbf{x}^i_j = (I_{d}+\tilde{\Lambda}^i)\tilde{\mathbf{x}}^i_{j}+\tilde{\delta}^i. \nonumber
\end{equation}
Here, each $\tilde{\Lambda}^i$ is a random matrix with i.i.d. Gaussian entries according to $\mathcal{N}(0,\frac{\sigma^2}{d})$, and $\tilde{\delta}^i$ is a random Gaussian vector according to $\mathcal{N}(0,{\sigma^2}I_{d})$ where we set $\sigma=0.01$. In  test time, we did not apply any random affine transformation to test samples and instead considered the following three scenarios: (1) no perturbation, (2) adversarial affine distribution shift obtained by optimizing the inner maximization in \eqref{eq:ERM} using projected gradient descent, (3) adversarial perturbations designed by the projected gradient descent algorithm. We used $100$ projected gradient steps with stepsize $0.1$.

We considered three baselines in the experiments: (1) \texttt{FedAvg} where the server node averages the updated parameters of the local nodes after every gradient step, (2) Distributed FGM training where the nodes perform fast adversarial training \citep{goodfellow2014explaining} by optimizing an $\ell_2$-norm bounded perturbation $\delta^{i}_{j}$ using one gradient step followed by projection onto the ball $\{\delta^{i}_{j}:\, \Vert \delta^{i}_{j} \Vert_2\le \epsilon_{\operatorname{fgm}} \}$, and (3) Distributed PGD training where each node preforms PGD adversarial training \citep{madry2017towards} similar to distributed FGM but uses $10$ projected gradient steps, each followed by projection onto $\{\delta^{i}_{j}:\, \Vert \delta^{i}_{j} \Vert_2\le \epsilon_{\operatorname{pgd}} \}$. We used the value $\epsilon_{\operatorname{fgm}} = \epsilon_{\operatorname{pgd}} = 0.05 \, \mathbb{E}[\Vert\mathbf{x}_i\Vert_2]$ in the experiments. We observed training instability after achieving perfect training accuracy for the baseline \texttt{FedAvg} algorithm, and hence performed early stopping to avoid the instability in the \texttt{FedAvg} experiments. We did not encounter the instability issue in \texttt{FedRobust} experiments.

\begin{figure}
\centering
\includegraphics[width=\linewidth]{Cifar_affine.pdf}
\caption{Trained networks' test accuracy under affine distribution shifts in the CIFAR-10 experiments. Top row: constraining $\Vert\delta \Vert_2\le 1$ and changing maximum allowed $\Vert \Lambda - I\Vert_F$. Bottom row: constraining $\Vert \Lambda - I\Vert_F\le 0.4$ and changing maximum allowed $\Vert\delta \Vert_2$.}
\label{Fig: Cifar-affine}
\end{figure}

\subsection{\texttt{FedRobust} vs. \texttt{FedAvg} and adversarial training: Affine distribution shifts}

We tested the performance of the neural net classifiers trained by \texttt{FedRobust}, \texttt{FedAvg}, distributed FGM, and distributed PGD under different levels of affine distribution shifts. Figure \ref{Fig: Cifar-affine} shows the accuracy performance over CIFAR-10 with AlexNet, Inception-Net, and ResNet architectures.

As demonstrated, \texttt{FedRobust} outperforms the baseline methods in most of the experiments. The improvement over \texttt{FedAvg} can be as large as $54\%$. Moreover, \texttt{FedRobust} improved over distributed FGM and PGD adversarial training, which suggests adversarial perturbations may not be able to capture the complexity of affine distribution shifts. \texttt{FedRobust} also results in $4 \times$ faster training compared to distributed PGD. These improvements motivate \texttt{FedRobust} as a robust and efficient federated learning method to protect against affine distribution shifts.

\begin{figure}
\centering
\includegraphics[width=\linewidth]{Cifar_pgd.pdf}
\caption{Trained networks' test accuracy under PGD perturbations in the CIFAR-10 experiments. $X$-axis shows the maximum allowed $\ell_2$-norm for PGD perturbations.}
\label{Fig: Cifar-pgd}
\end{figure}

\subsection{\texttt{FedRobust} vs. \texttt{FedAvg} and adversarial training: Adversarial perturbations}

Figure \ref{Fig: Cifar-pgd} summarizes our numerical results of \texttt{FedRobust} and other baselines over CIFAR-10 where the plots show the test accuracy under different levels of $\ell_2$-norm perturbations. While we motivated \texttt{FedRobust} as a federated learning scheme protecting against affine distribution shifts, we empirically observed its robust performance against adversarial perturbations as well. The achieved adversarial robustness in almost all cases matches the robustness offered by distributed FGM and PGD adversarial training. These numerical results indicate that affine distribution shifts can cover the distribution changes caused by norm-bounded adversarial perturbations.
In summary, our numerical experiments demonstrate the efficiency and robustness of \texttt{FedRobust} against PGD adversarial attacks.

\begin{figure}
\centering
\includegraphics[width=.69\linewidth]{mnist_affine.pdf}
\caption{Trained networks' test accuracy under affine distribution shifts in the MNIST experiments. Top row: constraining $\Vert\delta \Vert_2\le 1$ and changing maximum allowed $\Vert \Lambda - I\Vert_F$, bottom row: constraining $\Vert \Lambda - I\Vert_F\le 0.6$ and changing maximum allowed $\Vert\delta \Vert_2$.}
\label{Fig: mnist-affine}
\end{figure}

\subsection{Numerical results for MNIST data}
We repeated the CIFAR experiments in Figures \ref{Fig: Cifar-affine} and \ref{Fig: Cifar-pgd} for the MNIST dataset. Figure \ref{Fig: mnist-affine} shows the numerical results under affine distribution shifts. The figure's top row includes the plots for fixed maximum delta norm $\Vert \delta\Vert_2\le 1$ and different levels of maximum allowed $\Vert \Lambda - I\Vert_F$, while in the bottom row we fix the maximum allowed linear shift $\Vert \Lambda - I\Vert_F\le 0.6$ and evaluate the test accuracy under different levels of $\Vert\delta\Vert_2$. As shown in the plots, \texttt{FedRobust} results in the best performance in most of the evaluations, which indicates the superior performance of \texttt{FedRobust} against affine distribution shifts. Figure \ref{Fig: mnist-pgd} shows the test accuracy of the trained networks under different levels of adversarial PGD perturbations. The figure's experiments again shows that \texttt{FedRobust} can effectively shield against PGD adversarial attacks and achieve a comparable performance to PGD and FGM adversarial training.

\begin{figure}
\centering
\includegraphics[width=.69\linewidth]{mnist_pgd.pdf}
\caption{Trained networks' test accuracy under PGD perturbations in the MNIST experiments. $X$-axis shows the maximum allowed $\ell_2$-norm for PGD perturbations.}
\label{Fig: mnist-pgd}
\end{figure}


TITLE: What is the idea behind the $^2$ in the mean squared error?
QUESTION [3 upvotes]: Mean Squared Error: $\operatorname{MSE}=\frac{1}{n}\sum_{i=1}^n(\hat{Y_i} - Y_i)^2.$ <-- what is the purpose of the '$^2$'  in here?
Mean Absolute Error: $\mathrm{MAE} = \frac{1}{n}\sum_{i=1}^n \left| f_i-y_i\right| =\frac{1}{n}\sum_{i=1}^n \left| e_i \right|$

REPLY [1 votes]: In addition to the reasons pointed out to you by K.Rmth and Eckhard in their comments, consider that since $x^2 > |x|$ for $|x| > 1$ while $x^2 < |x|$ for $|x| < 1$ and thus the mean-square error penalizes large errors more that the mean absolute error does, but is more forgiving of small errors. Some people think that this emphasis on penalizing large errors more than small errors is a good thing; others don't.


TITLE: How to debug math?
QUESTION [61 upvotes]: May seem strange as I'm good in programming, but I just started diving into math. ATM I'm learning combinatorics at Khan Academy, and here's an example of a question that I struggled with (that's not the question):

The answer to «ways of getting $1$ head in $4$ flips» is $4$ (HTTT, THTT, TTHT, TTTH). But let's put it the other way round: «ways of getting $3$ tails in $4$ flips» (which is, obviously, the same question). It would be the number of places to put it: $4×3×2 = 24$, next divided by 3 to remove duplicates (i.e. TTTH is counted $3$ times): $24/3 = 8$. But $8 ≠ 4$! We used right calculations, reasoning, and got absolutely different answer, how is that possible?

I thought: there must be generic ways of debugging math problems, i.e. not just directly solving, but rather finding the problem when everything supposed to work, but it doesn't. I tried searching, and to my surprise found nothing, like nobody ever asked the question. Well, here's mine:
What are common ways to debug math problems; both generic and specialized would be nice to know. If you know any relevant articles, I'd be glad to read it too.

REPLY [11 votes]: Regarding math at the graduate level and beyond, these programming debugging techniques can be quite effective:

Rubber duck debugging:  Find someone/something to whom to explain the bug and your solution.  This is also effective for helping undergraduates find bugs in their solutions to word problems.  Code review, i.e., presenting your argument to an at least minimally interested colleague, finds many, many bugs by the same mechanism.
Bisection: Did the claim hold half-way through a construction?  Either way, next check at the appropriate quarter point.
Unit Testing:  In math, this comprises having a pool of "friendly" and a pool of "bizarre" objects to test claims on.  Examples:  Does this claim hold for $[0,1]$?  How about for the Cantor set?  How about for the uncountable product of copies of $[0,1]$?  Does it work for Abelian groups?  solvable groups?  finitely generated groups?  Lie groups? et al.
Precondition checking:  This theorem holds if that measure is $\sigma$-finite.  Is that measure $\sigma$-finite?  (This catches many, many bugs.  Notorious example involving a definition: manifolds are Hausdorff, but there are published papers that forget to check this.)  In fact, all theorems are (software "design by contract") contracts.


TITLE: (Functional Analysis) A Remark About Parseval's Identity
QUESTION [1 upvotes]: Currently I'm self studying functional analysis, namely Hilbert spaces. In the text, the author gives the following theorem:

Theorem A (Parseval's identity): Let $\{e_i\}_{i\geq 1}$ be an orthonormal system. Then $\{e_i\}_{i\geq 1}$ is a basis in $H$ if and only if for all $x\in H$
$$
||x||^2=\sum_{i\geq 1}|\langle x,e_i \rangle|^2.\tag{1}
$$

If it's not obvious, $H$ is a Hilbert space. Now, I understand the proof of Theorem A; however, the author follows up with the following remark which I don't understand completely:

Remark 1: If $(1)$ is true for a dense subset of $x$'s, it implies that $\{e_i\}_{i\geq 1}$ is a complete system which by Theorem B implies that $\{e_i\}_{i\geq 1}$ is a basis.

I'm not going to fully write out Theorem B, but it's just the fact that a complete orthonormal system in a Hilbert space is a basis for it. What I don't understand about Remark 1 is the "if $(1)$ is true for a dense subset of $x$'s, it implies that $\{e_i\}_{i\geq 1}$ is a complete system". Any help here will be appreciated.

I'll add here that $\{e_i\}_{i\geq 1}$ being a complete system in a normed space $X$ means the linear span
$$
\left\{\sum_{i=1}^{n}\alpha_i e_i:\text{ for all }n\in\mathbb{N},\text{ for all scalars }\alpha_i\right\}
$$
of the vectors $\{e_i\}_{i\geq 1}$ is a dense set in $X$.

REPLY [2 votes]: For $x$ in the dense set check that $\|x-\sum \langle x, e_i \rangle e_i\|^{2}=\|x\|^{2}-\sum |\langle x, e_i \rangle|^{2}=0$. This shows that $x-\sum \langle x, e_i \rangle e_i $ belongs to the closed subspace spanned by $e_i$'s. This automatically implies that any vector in the closure of the  dense set is also in the closed subspace spanned by $e_i$'s.


TITLE: $1 - x$ is a unit iff $x$ is nilpotent.
QUESTION [1 upvotes]: Suppose we have a $Z$ graded ring $R$ and $x \in R$ a homogeneous element of nonzero degree. Let $u = 1 - x.$ I want to show that $u$ is a unit iff $x$ is nilpotent. One direction is trivial. Now suppose $u$ is a unit. I am having trouble showing that $x$ is nilpotent.
I know that $u$ being a unit implies it is homogeneous. However, since $x = 1 - u$ and 1 is homogeneous of degree $0$ this must mean that $u$ must be homogeneous of degree $0$ as well as $x$ is homogeneous. But this implies that $x$ is homogeneous of degree 0 which is a contradiction. I feel as though I am missing something. Any hints would be appreciated!

REPLY [1 votes]: Hint: The key to the other direction was showing that if $x$ is nilpotent, then $1+x+\dots+x^{n-1}$ is the inverse of $1-x$. In this direction, you show that every possible inverse must look like this.
Let the degree of $x$ be $d$. Let $y$ be the inverse of $1-x$, and suppose that the $n^\text{th}$ component of $y$ in the grading is $c_n$. Then the $n^\text{th}$ component of $(1-x)y$ is $c_n-[x]c_{n-d}$ (where $[x]$ is the $d^\text{th}$ component of $x$). Since $(1-x)y=1$, this implies
$$
c_n-[x]c_{n-d}=\begin{cases}
  1 & n=0\\ 0 & \text{otherwise}
\end{cases}
$$
In particular, if $m$ is the smallest index for which $c_m\neq 0$, then it must be true that $c_m=1$, and that $m=0$. You can also show that $c_i=0$ for all $m<i<m+d$. You then have
$$
c_{d}-c_0[x]=0\implies c_{d}=[x]
$$
and you can proceed by induction to show that $c_{kd}=[x]^k$. Since you cannot have infinitely many nonzero coordinates, some $[x]^k$ must equal zero.


TITLE: Approach for solving recurrence relations using induction
QUESTION [0 upvotes]: This recurrence relation is for the famous problem The Tower of Hanoi.
We are given that: $T_n=2T_{n-1}+1$.
The problem here is, using induction, show that
$T_n=2^n – 1, n \ge 1$
To make things simpler, I'll give the vague solution that has been given to us by our teachers.
Solution that I can't seem to understand:
Observe that $T_1=1$. If $n\ge2,\,\,2T_{n-1}+1=2(2^{n-1}-1)+1=2^n-1=T_n$, verifying the formula.
Given the solution, I tried applying the induction method on it, but I just can't seem to solve this problem. Could somebody please look at the solution above and help me understand the approach to solve such problems?

Similarly there are other problems that have been mentioned in our books, but I think if I get the approach for one, I'll be able to solve the others. Listing one of them here just so that it may give you'll any hints.
Power set problem:
Given $S_n=2S_{n-1},n\ge1$ with $s_0=1$, using the recurrence relation for $S_n$, show that $Sn=2^n$, $n\ge0$.

REPLY [2 votes]: The solution given basically is induction, but with  steps and some text skipped.
As user @Henry points out, the standard weak induction proofs (which is what you’ll be majorly/exclusively doing in intro to induction) go like this: For a statement $P(n)$ which is to be proven,

Prove that $P(1)$ holds, i.e. the statement is true when $n=1$ is plugged in. (Base Case)
Assume that $P(k)$ holds for some positive integer $k$, i.e. the statement is true when $n=k$ is plugged in. (Induction Hypothesis)
Based on (2), prove that $P(k+1)$ holds, i.e. the statement is true when $n=k+1$ is plugged in.

This works due to the theorem called the First Principle of Finite Induction, whose proof using the Well-Ordering Principle I link here. It basically states that:

Let S be a set of positive integers with the following properties: (i) The integer $1$ belongs to S,(ii) Whenever the integer $k$ is in S, the next integer $k+1$ must also be in S. Then S is the set of all positive integers.

So a proof using induction would basically read like:
Let S be the set of all positive integers $n$ for which $P(n): T_n =2^n-1$ is true.
We know that $1\in S$, because $T_1=2^1-1$.
ASSUME that $k\in S$ for some positive integer $k$, i.e. $T_k=2^k-1$.
 Then, by the given recurrence, $T_{k+1}=2T_k+1$ we have $T_{k+1}=2^{k+1}-1$.
 This means that $k+1\in S$.
 By the First Principle of Finite Induction, $S$ is the set of all positive integers. But since S is the set of all integers for which $P(n)$ is true, it must mean that $P(n): T_n=2^n-1$ must be true for all positive integers $n$. Hence proved.
We generally omit all references to the set S in proofs by induction, and just go : P(1), check. Assume P(k) is true. P(k+1), check. Hence proved.
 I hope you can now write a proof using the three-step algorithm presented above.


TITLE: Proving two integrals are equal to each other
QUESTION [5 upvotes]: Question: How do you show that$$\int\limits_{0}^{\infty}\frac {e^{-ax}}{\sqrt{b+x}}\,\mathrm dx=\int\limits_{\sqrt{ab}}^{\infty}\frac 2{\sqrt{a}}e^{ab-t^2}\,\mathrm dt$$

This problem emerged when I was trying to prove an identity by working backwards, and I'm not sure how to tackle it. I need to find a way to get the limits of the right-hand side to zero to infinity but am unsure how to do that.
Should I make the substitution$$u^2=ab$$If so, how do I get rid of the $\sqrt{a}$ term in the denominator?
Note: Here, $a,b>0$ and are real numbers!

REPLY [6 votes]: Let $$t^2=ab+ax$$ Then $2t\,dt=a\,dx$. The integral becomes $$\int\limits_{\sqrt{ab}}^{\infty}\frac{e^{ab-t^2}}{\frac{t}{\sqrt{a}}}\frac{2t}{a}\,dt=\int\limits_{\sqrt{ab}}^{\infty}\frac{2}{\sqrt{a}}e^{ab-t^2}\,dt$$


TITLE: What direction does the angular velocity of an object lie on if linear velocity is in XY axis?
QUESTION [0 upvotes]: So,I was watching the solution of a question of rotational motion. The object's rotational motion was in XY plane. And the direction of angular velocity of the object was given to be along Z-axis in the solution. Now, I cannot understand how the angular velocity can possibly be along that direction. I expected it to be either along clockwise or anticlockwise direction, but on the XY plane either way.
So, please tell me where I am going wrong here. Also,I am a total beginner at this subject, so please frame your answers bearing  thing in mind.

REPLY [0 votes]: It is a convention. Any instantaneous rotation of a particle defines a plane, in the example X-Y. And each plane has one and only one normal direction associated.
If we define the angular velocity as a vector normal to the plane of rotation, the vector modulus encodes its magnitude, the vector direction encodes the plane of rotation, and where the arrow points encodes if the rotation is CW or CCW.


TITLE: Proofs of Karpelevich's results about eigenvalues of nonnegative matrices.
QUESTION [4 upvotes]: Are there any books or papers written in English that contain proofs of the results obtained in the following paper?

F.I. Karpelevich, On the characteristic roots of matrices with nonnegative elements, Izv. Akad. Nauk SSSR Ker. Mat., 1951, v.15, issue 4, 361-383.

The statement without proof of Karpelevich's main result can be found in theorem 5.1 of the paper On $p$th roots of stochastic matrices by Higham and Lin. According to the authors, more details of Karpelevich's results can be found in Minc's book Nonnegative Matrices, but again, without proofs.

REPLY [4 votes]: Update: I don't find any translation, but I have found a paper that restates Karpelevich's main theorem in a simpler form:
Ito, H., A new statement about the theorem determining the region of eigenvalues of stochastic matrices. Linear Algebra and its Applications, 246: 241–246 (1997).

Abstract
Let $M_n$ denote the set of points in the complex plane that are eigenvalues of $n$-dimensional stochastic matrices. The set $M_n$ is completely determined by the Karpelevich theorem, the statement of which, however, is lengthy and intricate. The paper shortens the presentation of the theorem.

Note that this paper does not prove Karpelevich's theorem. It only proves that the author's restatement of the theorem is equivalent to Karpelevich's original formulation.


TITLE: If all balls at $x$ and $y$ are isometric is there an isometry sending $x$ to $y$?
QUESTION [20 upvotes]: Let $(X,d)$ be a metric space and $x,y \in X$. Assume that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric.
Is it true that there exists an isometry of $X$ sending $x$ to $y$?

REPLY [20 votes]: There is a 5-point example $\ X := \{x\ y\ a\ b\ c\},\ $ with (symmetric) metrics $\ d\ $ as follows:
$$d(x\ y) = d(a\ b) = 1$$
$$d(x\ a) = d(y\ b) = 2$$
$$d(x\ b) = d(y\ a) = 3$$
$$d(x\ c) = d(y\ c) = 6$$
$$d(a\ c) = 5\qquad\qquad d(b\ c) = 4$$
Oviously, $\ B_r(x)\ $ and $\ B_r(y)\ $ are isometric for every $\ r>0,\ $ while there is no isometry $\ f:X\rightarrow X\ $ for which $\ f(x)=y$.

REMARK 1  This example is ironic because while the respective balls are isometric, the isometry of the balls doesn't respect the centers. In this sense every bounded (and especially--finite) required example would be ironic.
REMARK 2  Number $\ 5\ $ is minimal.


TITLE: x≡1 (mod 8) x≡9 (mod 12) has solution x = x_0. How many solutions mod 24 are there to the system of congruences?
QUESTION [0 upvotes]: Say,
x≡1 (mod 8)
x≡9 (mod 12)
has solution x = x_0. How many solutions mod 24 are there to the
system of congruences?
I am worried this question is too easy to be true. That is why I am confused. Don't know where to start.

REPLY [1 votes]: If $\,\color{#c00}{a\equiv b}\pmod n\,$ then $\ \begin{eqnarray}x\equiv \color{#c00}a\!\!\!\pmod n\\ x\equiv b\!\!\!\pmod{\! m}\end{eqnarray}$ $\!\iff\!$ $\begin{eqnarray}x\equiv \color{#c00}b\!\!\!\pmod n\\ x\equiv b\!\!\!\pmod{\! m}\end{eqnarray}$ $\!\color{#0a0}\iff\!$ $x\equiv b\pmod{\!{\rm lcm}(n,m)}$
where, above, we have employed the basic lcm law: $\,\ n,m\mid x-b \color{#0a0}\iff {\rm lcm}(n,m)\mid x-b$
Remark $\ $ This is a special case of the uniquitous constant case optimization of CRT. You can find many examples of this and further discussion in various prior posts.


text:
\section{Reversal of Number Multiplied by 11}
Tags: Reversals, 11, Reversal of Number Multiplied by 11

\begin{theorem}
Let $n \in \N$ be a [[Definition:Natural Number|number]] for which, when written in [[Definition:Decimal Notation|decimal notation]], no two adjacent [[Definition:Digit|digits]] [[Definition:Integer Addition|total]] to more than $9$.
Let $n'$ denote the [[Definition:Reversal|reversal]] of $n$.
Then $n \times 11$ is the [[Definition:Reversal|reversal]] of $n' \times 11$.
\end{theorem}

\begin{proof}
{{ProofWanted|Trivial but tedious, and it does not interest me.}}
\end{proof}

config: wiki


TITLE: Why $\nu>-1$ for Bessel function orthogonality
QUESTION [2 upvotes]: I'm physics student and not very good at proof. My mathematical physics textbook states the orthogonality of Bessel functions,
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
basically same as the equation (53) in the linked page.
However, textbook says that there is a condition that $\nu>-1$ but doens't explain why. Wikipedia also states the same equation and condition but doesn't explain why.
I can guess that it's related to the convergence of the integral but I'm not sure how to show it.
Can somebody help me?

REPLY [1 votes]: Use the limiting form $J_{\nu}(az) \sim \left(\frac{az}{2}\right)^{\nu+1}/\Gamma(\nu+1).$ Then for $z$  near $0$ the integrand behaves like
$$z J_{\nu}(az)J_{\nu}(bz) \sim f(\nu,a,b) \left(\frac{z}{2}\right)^{2\nu+1}$$
where $f(\nu,a,b)$ is independent of $z$.
Therefore the integral converges if $2\nu + 1 > -1 $ or $\nu >-1.$