TITLE: Is there a Gröbner basis analogue that exists for vector spaces?
QUESTION [1 upvotes]: Suppose I have a coordinate system $t_1,\ldots t_N$ with a lexicographical ordering. Let LT denote choosing the lowest term of a polynomial with respect to this ordering. e.g. LT$(t_1 + t_2)=t_2$.
Given a finite dimensional vector space V, define LT(V):=span$_\mathbb{C}\{LT(w)|w\in V\}$.
I want to find a basis $\{v_1,\ldots , v_n\}$ for V such that LT(V)=span$_\mathbb{C}\{LT(v_1),LT(v_2),\ldots ,LT(v_n)\}$.
I have never worked with Gröbner bases and know very little about them, but this question looks similar to them I think, only here I'm talking about subspaces instead of ideals.
Is there a well-known method for finding such a basis? I've been searching, but haven't found anything yet, so I thought I'd ask on here.
For example, by an algorithm I know $\{1,t_1+t_3,t_2,t_1t_2,t_1t_2(t_1+t_3),t_2(t_1+t_3),t_2(t_2t_3),t_1t_2(t_2t_3)\}$ forms a basis for the space of sections $H^0(X,L)$ of of particular variety $X$ and a line bundle $L$ over it. And for the vector spaces I'm working with I already know $dim(LT(H^0(X,L))=dim(H^0(X,L))$, so I know $\{1,t_3,t_2,t_1t_2,t_1t_2t_3,t_2t_3,t_2(t_2t_3),t_1t_2(t_2t_3)\}$ forms a basis for $LT(H^0(X,L)$.
But in general, if the vector space I begin with is n-dimensional, it may not be the case that the basis I choose has n distinct lowest terms, so I'm looking for a way to find a basis with n distinct lowest terms.

REPLY [6 votes]: Darij is right.  Computing Groebner basis for a linear ideal is the same as computing the row echelon form (this can be seen from the Buchberger algorithm).  And I think he meant "nonlinear" instead of "noncommutative".


TITLE: When are vector bundles on toric varieties also toric varieties?
QUESTION [23 upvotes]: Let $X$ be a toric variety, and $\pi:E\to X$ a vector bundle, say of rank $2$. You can think of $X=\mathbb P^1$.
When is the total space of $E$, or of $P(E)$, a toric variety? What do I need in order to lift the torus action on $X$ to a torus action on $E$, or on $P(E)$, so to get an open orbit? I ask this because I was reading this thesis, and on page $2$ I found the following statement, which I do not understand:

If a vector bundle over a toric variety splits as a sum of line bundles, then its projectivization admits a toric variety structure".

I want to understand at least the case of $\mathbb P^1$, where every vector bundle splits. But I have no idea why the splitting is so relevant in general.

REPLY [2 votes]: For the case $X = \mathbb{P}^1$, it perhaps desirable to view $X$ as the weighted projective space $X = \mathbb{P}[1,1]$. Then if $E$ is the bundle $\mathcal{O}(a_1) \oplus \cdots \oplus \mathcal{O}(a_r)$, the total space of $E$ may be identified with the weighted projective space $\mathbb{P}[1,1,a_1, \ldots, a_r]$. The fan for this total space can be described as having generators $v_0, v_1, w_1, \ldots, w_r$ with the relation $v_0 + v_1 + \sum_i a_i w_i = 0$.
For the case of more general $X$, I'd imagine you can do something similar: whereby the fan of the total space can be obtained from the fan of the base by adding a number of generator(s) determined by the rank of the bundle and subjecting these generator(s) to new relation(s) which are modifications of the relations from the fan of the base.


TITLE: LQR Controller for a nonlinear system - how to split the SS model as A and B matrices?
QUESTION [4 upvotes]: Say we have a nonlinear system that is represented by the following state space model:
$$\dot{x}_1=x_2$$
$$\dot{x}_2=f_1(x)+b_1(x)u$$
$$\dot{x}_3=x_4$$
$$\dot{x}_4=f_2(x)+b_2(x)u$$
The inputs to the LQR controller are ($A$, $B$, $Q$, $R$) but the nonlinear system does not have an $A$ matrix nor a $B$ matrix.
I remember reading something as lie matrix method to split the above state space model into two matrices as $A$ and $B$, but I cannot seem to find this source.
Can someone suggest a way to apply the LQR controller to the given nonlinear system?

REPLY [5 votes]: LQR stands for linear quadratic regulator, where: linear refers to the linear dynamics of the system (which can both be invariant or variant in time); quadratic refers to the cost function which is an integral of a quadratic form, which the LQR minimizes; regulator refers to the goal of the control input to bring the system to zero.
Since your system is not linear you can't directly use LQR for that system. You would either have to resort to linearizing your model around an equilibrium point, or use another technique like some model predictive control (MPC). Linearization will in general only stabilize the system locally around the equilibrium point, while nonlinear MPC might be none convex and therefore potentially really hard to solve.


TITLE: Probability of a desired song being played at max the 4th song.
QUESTION [1 upvotes]: I have an MP3 player and I loaded my Mp3 player with 3 songs from Michael jackson, 5 songs from Cardi B, 5 songs from Snoop Dog and 7 songs from Korn and I play them in shuffe mode (every song is played once in random
order).
What is the probability that the  first song by Michael Jackson is played at the lastest as the fourth song?
So I think I should use this formula $$ P[A]= \sum_{i=4}^n P[A|B_i]P[B_i] = (3/20)+(3/19)*(17/18)+(3/18)(17/18)(16/17).... $$
And so on. Is this the right track?
Would love it if someone could provide some hints.

REPLY [3 votes]: The probability that the first song by Michael Jackson (in the shuffle ordering) is one of the first four selections is $1$ minus the probability that there aren't any of them in the first four selections; that is,
\begin{align}
P(\text{the first} & \text{ four selections include an MJ song}) \\
    & = 1-P(\text{the first four selections include no MJ songs}) \\
    & = 1-P(\text{the three MJ songs are all in the last $16$ selections}) \\
    & = 1-\frac{\text{the number of ways to choose three of the last $16$ selections}}{\text{the number of ways to choose three of all $20$ selections}}
\end{align}
Does that help?

Your approach can also work, but I don't understand where you came up with the numbers you use.  It seems to me that it should be
\begin{align}
P(\text{the first} & \text{ four selections include an MJ song}) \\
    & = P(\text{the first selection is an MJ song}) \\
    & + P(\text{the first selection isn't an MJ song, but the second is}) \\
    & + P(\text{the first two selections aren't MJ songs, but the third is}) \\
    & + P(\text{the first three selections aren't MJ songs, but the fourth is}) \\
    & = \frac{3}{20} + \frac{17}{20} \times \frac{3}{19}
      + \frac{17}{20} \times \frac{16}{19} \times \frac{3}{18}
      + \frac{17}{20} \times \frac{16}{19} \times \frac{15}{18} \times \frac{3}{17}
\end{align}
in which case you should end up with the same answer yielded by the first analysis.

Unless I completely misunderstand the question.


TITLE: When the intersection of projective varieties is finite
QUESTION [1 upvotes]: What I want to solve is

Let $C$ be a projective curve, $P\in C$. Then there is a binational morphism $f:C \longrightarrow C'$, $C'$ a projective plane curve, such that $f^{-1}(f(P))=\{P\}$.

Here, a projective curve means that $C\subseteq \mathbb{P}^{n+1}$ is projective variety whose dimension is 1, and projective "plane" curve is just a curve in $\mathbb{P}^2$. The outline of proof is following :
We can assume : Let $T,X_1,\cdots,X_n,Z$ be coordinates for $\mathbb{P}^{n+1}$; Then $C\cap V(T)$ is finite ; $C\cap V(T,Z)=\emptyset$; $P=[0:0:\cdots :0:1]$ and $k(C)$ (function field on C) is algebraic over $k(u)$, where $u=\bar{T}/\bar{Z}\in k(C)$.
But I can't understand this assumption, because basically I think I'm not sure when the intersection of two varieties is finite set.(In $\mathbb{P}^2$, I know this set is finite if the homogeneous polynomials have no common component, but I don't know the case for larger dimension) Can anyone explain why we can assume such things?

REPLY [0 votes]: $C\cap V(T)$ is a closed subset of $C$, which means it's either all of $C$ or a finite collection of points. This is developed in Hartshorne chapter I exercise 4.8, for instance. So if $C$ isn't contained in $V(T)$, then this intersection is just a finite collection of points; if $C\subset V(T)$, just go to the $\Bbb P^{n-1}$ which is $V(T)$, pick new coordinates, and try again.
Your larger question about when the intersection of two projective varieties is finite is more difficult. The best available thing here is that if $X,Y$ are closed subvarieties of $\Bbb P^n$ of dimension $r,s$ respectively, then every irreducible component of $X\cap Y$ has dimension at least $r+s-n$, and if $r+s-n\geq 0$ then this intersection is nonempty. This is only an existence result - you can certainly have a nonfinite intersection if $r+s-n <0$: think about two lines in $\Bbb P^3$, for instance. They can be the same line, intersect in a point, or be skew lines and have no intersection. But the intersection can only be finite if $r+s\leq n$.


TITLE: Determining the time dependent wave function for the first two stationary states in infinite square well
QUESTION [0 upvotes]: I am working on problem 2.5 in Griffiths Quantum Mechanics, and I am trying to work through part b. This question provides that a particle is in the infinite square well and has its initial wave function as an even mixture of the first two stationary states: $$\psi(x,0) = A[\psi_1(x) + \psi_2(x)]$$
part a asks for normalization, which gives us $$A = \frac{1}{\sqrt{2}}$$ then part b asks for $\psi(x,t)$.
So, the confusion I am having is with the constants $c_n$. In the section prior to this, on the infinite square well, my understanding is that we can find the solution of the time-independent Schrodinger equation, tack on the relevant time dependence, and write any solution as a linear combination of the product of the solution to the time-independent solution and the time dependence term.
I tried to find $c_n$ by using that $$c_n = \frac{\sqrt{2}}{\sqrt{a}}\int _0^a\sin\left(\frac{n\pi}{a}x\right)\psi (x,0)dx$$ and to note that since these are the first and second stationary states, we know that they take the form $$\frac{\sqrt{2}}{\sqrt{a}}\sin\left(\frac{n\pi}{a}x\right)dx$$ Then, we compute the above integral to find $c_n$. Then, I planned to write the general solution appealing to the formula $$\psi(x,t) = \sum^{\infty}_{n=1}c_n\frac{\sqrt{2}}{\sqrt{a}}\int _0^a\sin\left(\frac{n\pi}{a}x\right)e^{-i\left(\frac{n^2\pi^2\hbar}{2ma^2}t\right)}dx$$However, the solution just takes the states, tacks on the time dependence, and that's it. My question is why we don't need to solve for the $c_n$ constants here?
To clarify, the solution presents the following: $$\frac{1}{\sqrt{a}}e^{-i\omega t}\left[\sin{\frac{\pi x}{a}} + \sin{\frac{2\pi x}{a}}e^{-3i\omega t}\right]$$
Where do the $c_n$ constants go/why don't we need them?

REPLY [1 votes]: Your approach is correct but you're missing the final step.
What happens when you calculate $c_n$, i.e. when you evaluate the integral? In this case $\psi(x,0)$ is completely known and it can be calculated. Which values of $c_n$ are non-zero?
Is there a way you can read off $c_n$ just by looking at the form of $\psi(x,0)$?
Also: don't forget your $dx$'s, it will bite you in the butt later

Edit: I will put my response in my answer otherwise it will get too long. You are indeed trying to calculate
$$c_n=\int\mathrm d x\,\psi_n^*(x)\psi(x,0)$$
where $$\psi_n(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$$
The following formula allows you to evaluate this integral:
\begin{equation}\int_0^a\mathrm d x\,\frac{2}{a}\sin\left(\frac{m\pi x}{a}\right)\sin\left(\frac{n\pi x}{a}\right)=\cases{1&$m=n$\\0&$m\neq n$}\label{lol}\tag{1}\end{equation}
Now, the function you are integrating is of the form $\psi(x,0)=\alpha\, \psi_1(x)+\beta\,\psi_2(x)$ where $\alpha^2+\beta^2=1$ due to normalization. So you can plug it and work it out.
But there is some nice linear algebra going that you shouldn't miss. Because of equation (1) we can treat the space of wavefunctions as a vector space. We can identify the basis functions as basis vectors: $$\psi_i(x)\leftrightarrow \vec e_i$$, we can decompose a general vector in terms of basis vectors (provided the basis vectors form a basis):
$$\vec v=\sum_{i=1}^Nv_i\vec e_i$$
and we can find each of these coefficients by taking the dot (inner) product
$$v_i=\langle \vec e_i,\vec v\rangle=\vec e_i\cdot\vec v$$
For the wavefunctions this dot product is replaced by an integral but it turns out a equation like (1) is enough to use a lot of the functionality that you know from vectors (linear algebra).
If we now are given a vector of the form $\vec v=\alpha\, \vec e_1+\beta\,\vec e_2$ my question is then what are $v_1,v_2$?


text:
{\bf Problem.} A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other.  How many distinct license plates are possible?
{\bf Level.} Level 5
{\bf Type.} Counting & Probability
{\bf Solution.} Since the two letters have to be next to each other, think of them as forming a two-letter word $w$. So each license plate consists of 4 digits and $w$. For each digit there are 10 choices. There are $26\cdot 26$ choices for the letters of $w$, and  there are 5 choices for the position of $w$. So the total number of distinct license plates is $5\cdot10^4\cdot26^2 = \boxed{33,\!800,\!000}$.

set_name: MATH


TITLE: Find the average of $P$ defined on the Fat Cantor set
QUESTION [1 upvotes]: Suppose we have $P:A\to[0,1]$, where $A$ is the fat cantor set denoted as $C$.
We produce $C$ by removing $1/4$ of $[0,1]$ around mid-point $1/2$
$$C_{1}=[0,3/8]\cup [5/8,1]$$
$$C_{1,1}=[0,3/8] \ \ \ \ \ \ C_{1,2}=[5/8,1]$$
We repeat, removing $1/16$ of $[0,1]$ around each midpoint of remaining intervals $C_{1,1}$ and $C_{1,2}$.
$$C_{2}=[0,5/32]\cup[7/32,12/32]\cup[5/8,25/32]\cup[27/32,1]$$
$$C_{2,1}=[0,5/32] \ \ \ \ \ \ \ C_{2,2}=[7/32,12/32] \ \ \ \ C_{2,3}=[5/8,25/32] \ \ \ C_{2,4}=[27/32,1]$$
Repeat the process until $\lim\limits_{n\to\infty}C_{n}=C$, where $\lim\limits_{n\to\infty} 1/4^n$ of $[0,1]$ around each midpoint of remaining intervals are removed.
I want to find the average, in terms of integration, such that it is equivelant to
$$\lim_{n\to\infty}\sum_{i=1}^{2^{n}} P(t_{n,i})(1/2^n)$$
Where $\bigcup\limits_{i=1}^{2^{n}}C_{n,i}=C_n$ and $t_{n,i}\in C_{n,i}$

REPLY [3 votes]: If $P$ is continuous, the average should be $\dfrac{1}{m(C)} \int_C P(x)\; dx$ where $m$ is Lebesgue measure.
EDIT:
It looks to me like if $C(i)$ is the $i$'th iteration starting at $C(0)=[0,1]$, $a_i = \int_{C(i)} x^2\; dx$ has
generating function
$$ g(x) = \sum_{i=0}^\infty a_i x^i = {\frac {35\,{x}^{3}-1066\,{x}^{2}+7456\,x-8192}{ 3 \left( x-1
 \right)  \left( x-2 \right)  \left( x-16 \right)  \left( x-8 \right)
 \left( x-32 \right) }}$$
and $\lim_{i \to \infty} a_i$ is $-$ the residue of this at $x=1$, namely $19/105$.  Dividing by the measure of $C$, namely $1/2$, I get the average for $P(x)=x^2$ to be $38/105$.


TITLE: Singular Lebesgue-Stieltjes measure
QUESTION [0 upvotes]: I read that if $F:\mathbb{R}\to\mathbb{R}$ is a non-decreasing singular function, i.e. a non-decreasing continuous function such that $F′(x)=0$ almost everywhere, then the Lebesgue-Stieltjes measure $\mu_F$, defined as the Lebesgue extension of the interval measure $m(a,b)=m[a,b)=m(a,b]=m[a,b]=F(b)−F(a)$, is such that $\mu_F(A)=0$ for the set $A:=\{x\in\mathbb{R}:F′(x)=0\}$ (which is complement of a set $B$ of Lebesgue measure $\mu(B)=0$) where its derivative is 0.
By definition of $\mu_F$ I understand that
$$\mu_F(A)=\inf\sum_k F(b_k)−F(a_k)$$
where $[a_k,b_k]$ are intervals containing $A$, but I do not know how to see that this infimum is 0. If we could find such intervals such that $F′$ is continuous on $[a_k,b_k]$ the desired property would follow from $F(b_k)−F(a_k)=\int_a^b F′(t)dt=0$ where the integral is Riemann's. By using the Lebesgue integral instead, I can only see that, since $F$ is monotonic, $F(b_k)−F(a_k)\geq\int_{[a_k,b_k]}F′d\mu=0$ which does not help me to show what I want... Does anybody know how to prove that such $μ_F$ is zero where $F′$ is zero? I thank you so much!

REPLY [1 votes]: If you had $\mu_F(A)>0$, then also $\mu_F(K)>0$ for a compact subset $K\subset A$ by regularity. For each $x\in K$, find an interval $I_x$ centered at $x$ with $\mu_F(I_x)<\epsilon |I_x|$, and cover $K$ by finitely many of these. By deleting intervals if necessary, we can also assume that no point of $\bigcup I_{x_j}$ lies in more than two intervals, and we can keep the diameter of this union bounded.
It now follows that
$$
\mu_F(K)\le\mu_F\left( \bigcup I_{x_j}\right) \le \epsilon \sum |I_x| \le 2\epsilon \left| \bigcup I_{x_j} \right| \le C\epsilon .
$$
Since we can do this for any $\epsilon>0$, the claim follows.


TITLE: Find $\int_0^1 f(x) \,dx \simeq A_0 f(x_0) + A_1 f(x_1)$ that is exact for $f(x) = a e^x + b \cos(\pi x /2)$?
QUESTION [0 upvotes]: I am reading through some exercises of the chapter I am reading about numerical integration. I stumbled upon the following question:

Find the formula $\int_0^1 f(x) \,dx \simeq A_0 f(x_0) + A_1 f(x_1)$ that is exact for functions of the form $f(x) = a e^x + b \cos(\pi x/2)$

I know that for any polynomial of degree $\leq n$ its integral can be exactly estimated (that is, is equal to) by the formula
$$
\int_a^b f(x) \,dx = \sum_{i=0}^{n} A_i f(x_i)
$$
where $A_i = \int_a^b \ell (x)\, dx$ (Lagrange fundamental polynomial) for $n+1$ distinct nodes $x_i \quad 0 \leq i \leq n$.
However, the $f$ in the exercise is clearly not a polynomial. So no idea how to find such a formula that is exact as with numerical integration for polynomials. Can you help?

REPLY [1 votes]: You can just set for instante $x_0=0$ and $x_1=1$ and compute $A_0,A_1$ such that $$
\int_0^1 e^x dx = A_0 e^0 + A_1 e^1, \quad \int_0^1 \cos \frac{\pi x}{2}\, dx = A_0 \cos \frac{\pi \cdot 0}{2} + A_1 \cos \frac{\pi \cdot 1}{2}.
$$
This way you obtain the quadrature
$$
Q(f)=\frac{2}{\pi} f(0) + \frac{e \pi -2-\pi }{e \pi } f(1).
$$
If you require the formula to be exact to an extended set of functions, you'll be able to remove the initial arbitrarity in the choice of $x_0,x_1$. For instance, if you take $x_0 \approx 0.221611$, $x_1 \approx 0.79518$, $A_0 \approx 0.513634$, $A_1 \approx 0.486366$, the formula will be exact for functions of the form
$$
f(x)= a e^x + b \cos \frac{\pi x}{2} + c \sin \frac{\pi x}{2} + d.
$$


TITLE: Can 2 different random variables have the same CDF?
QUESTION [6 upvotes]: I'm looking for proof that two different random variables can have the same Cumulative Distribution Function; in other words, I'd like to disprove that a CDF uniquely defines a random variable. (Probably a silly question, but I couldn't find anything usable so far).

REPLY [4 votes]: Try $X$ such that $P(X=0)=P(X=1)=\frac12$ and $Y=1-X$. Then $Y\ne X$ in the sense that $P(Y=X)=0$ but the distributions of $X$ and $Y$ coincide.
For continuous random variables, choose $X$ uniform on $(0,1)$ and $Y=1-X$, or $X$ standard normal and $Y=-X$.


TITLE: Let B be a bilinear form on $M_n(\Bbb{C})$, $B(X,Y)=n\text{tr}(XY)-\text{tr}(X)\text{tr}(Y)$. Find $\text{rad }V$.
QUESTION [0 upvotes]: There is another part of the problem which asks-
Let $V_1=\{X\in V|\text{tr}(X)=0\}$ (here $V=M_n(\Bbb{C})$). Prove that $V=V_1\oplus \text{rad }V$.
I can prove that $V_1$ is non-degenerate. And B is symmetric, hence reflexive. Thus $V=V_1\oplus V_1^{\bot}$.
$\text{rad }V=\text{rad }V_1^{\bot}$.
Now, my claim is $\text{rad }V=V_1^{\bot}$. To prove this I have $\text{rad }V_1^{\bot}=V_1^{\bot}$ or equivalently $B(X,Y)=0\ \forall X,Y\in V_1^{\bot}$
So, my target is show that for any $X,Y\in V_1^{\bot}$, we have $B(X,Y)=0$.
But I don't know how to prove this statement. Can anybody prove this clam? Thanks for assistance in advance.

REPLY [1 votes]: $X$ is in rad $V$ if for all $Y$, we have $B(X,Y) = 0$.  That is: for all $Y$, we have
$$
n\operatorname{tr}(XY) - \operatorname{tr}(X)\operatorname{tr}(Y) = 0 \iff\\
\operatorname{tr}(n\,XY - \operatorname{tr}(X)Y) = 0 \iff\\
\operatorname{tr}((n\,X - \operatorname{tr}(X)I)Y) = 0.
$$
In other words: if $C(X,Y)$ denotes the bilinear form $C(X,Y) = \operatorname{tr}(XY)$, then $X$ is in rad $V$ if and only if $C(n\,X - \operatorname{tr}(X)I,Y) = 0$ for all $Y$.  Because $C$ is a non-degenerate bilinear form over $V$, this is equivalent to saying that $n\,X - \operatorname{tr}(X)I = 0$.  This occurs if and only if $X$ is a multiple of $I$ (the identity matrix).
So, we find that the rad of $V$ is the one-dimensional space $\{a\,I: a \in \Bbb C\}$.


TITLE: I don't understand dividing in congruence equations.
QUESTION [1 upvotes]: I need to find the smallest possible value of x when:
x = 2 mod3 &
x = 3 mod5
My first step was to put it in the form 1 + 3a = 2mod5. It then becomes 3a = 1 mod 5 (presumably because 1 mod 5 = 1?). However the next step should be a = 2 mod5 but I don't understand the step to get to 2 mod5.

REPLY [0 votes]: You want to solve $3a \equiv 1 \mod 5$ so ...
1) Intuition.
$1 \equiv 6 \mod 5 \equiv 2*3$ so $ a = 2$.
2) Just keep adding $5$ until you get something.
$3a \equiv 1 \mod 5$.  but $3\not \mid 1$.  So add $5$.
$3a \equiv 1+5 = 6 \mod 5$.  And $3|6$ so $a = \frac 63 = 2$.
3) you can do it that hard way:
....
$3a = 1\equiv 5 \implies 3a = 1 + 5k \implies$
$3a = 3k + 1 + 2k \implies 1+2k \equiv 0 \mod 3 \implies 2k \equiv -1 \mod 3$
$\implies 2k = 3m - 1 = 2m + m- 1 \implies$
$m-1 \equiv 0 \mod 2 \implies m \equiv 1 \mod 2$
So let $m = 1$.
Then $2k = 3m - 1 =3-1 = 2 \implies $k = 1$
So $3a = 1 + 5k = 1 + 5 = 6 \implies$
$a = 2$.
The above seems pointlessly tedious, but it is basically Euclid's Algorithm and if $\gcd (3,5) = 1$ willl always work.
For harder cases it is useful but for simple cases your intuition should just tell you.
Less obvious.
$74a \equiv 1 \mod 981$.
Intuition?  I have none.
Add $981$ so $74a \equiv 1 + 981 \equiv 982 \mod 981$ ... we could be here all night....
$74a = 1 + 981k = 1 + 13*74k + 19k$ so $19k + 1 \equiv 0 \mod 74$
$19k \equiv -1 \mod 74$ so $19k = -1 + 74m = -1 + 4*19m -2m$
$-1 - 2m \equiv 0 \mod 19$ so $2m \equiv -1 \mod 19$ so $2m = -1 + 19n$
so $2m = - 1 + n + 2*9n$
So $n-1 \equiv 0 \mod 2$
So $n \equiv 1 \mod 2$.
Let $n = 1$.  $2m = -1 + 1 + 2*9*1 = 18$ so $m = 9$.
So $19k = -1 + 74m= -1 + 74*9 = 1+666$ (ooh, scary) $ = 665 = 19*35$
So $k = 35$ and $74a = 1 + 981k = 1 + 981*35 = 34336=464*74$
So $a \equiv 464\mod 981$