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# An Introduction to Autoencoders and Variational Autoencoders

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## What is an Autoencoder?

An autoencoder is a neural network trained to compress its input and recreate the original input from the compressed data. This procedure is useful in applications such as dimensionality reduction or file compression where we want to store a version of our data that is more memory efficient or reconstruct a version that is less noisy than the original data.

Autoencoders consist of two main components to achieve the goal of reconstructing data:

• an encoder network that learns to create a compressed or encoded version of the input data.
• a decoder network that is trained to reconstruct the original data from the compressed version.

Mathematically, we can describe the encoder as a function f that takes an input x to create an encoding h.

h = f(x)

The decoder represents another function, d that takes h as an input to produce a reconstruction r.

r = d(h)

Essentially, the autoencoder is trained to learn the following mapping:

r = d(f(x))

As you can see, the autoencoder does not learn to exactly reconstruct the original input x but r, which is an approximation to x.
An important feature of autoencoders is that they are unable to learn the identity function and reconstruct the original input perfectly. Instead, they are designed in a way that only allows for approximate recreations.
If we allowed an autoencoder to reconstruct the original image, it might default to just learning the identity function. Then the encoding will look exactly like the input making the autoencoder essentially useless.

The hidden layers have fewer nodes than the input and output layers to prevent the networks from learning the identity function. The number of nodes in the output layer equals the number of nodes in the input layer to allow for a reconstruction that is similar to the original.

Reducing the number of hidden nodes available to fewer than the input forces the autoencoder to prioritize the features it wants to preserve. These features need to be useful for reconstruction, which induces the network to focus on useful properties and ignore the noise.
Let’s say we wanted to build an autoencoder to encode and reconstruct an image of 20×20 = 400 pixels. Since we don’t use convolutions, we require 400 neurons in the input and output layers. If we equipped the hidden layers with 400 neurons as well, the autoencoder could simply pass through the single pixels without learning anything about the relationship between the pixels. Let’s cut the number of neurons in the hidden layers down to 200. Now, the network is forced to learn a useful representation of the 400 pixels it receives from the input layer that can be processed by 200 neurons and which allows for a reconstruction of the full 400 pixels that is as accurate as possible.
We call networks, which do not have the capacity to learn the full mapping, under-complete. As we discuss later in this article, there are other ways to force autoencoders to not learn the full mapping.

Compared to deterministic methods for data compression, autoencoders are learned. This means they rely on features that are specific to the data that the autoencoder has been trained on. An autoencoder will perform poorly with data it hasn’t been trained on.

## What is a Variational Autoencoder?

A variational autoencoder, in contrast to an autoencoder, assumes that the data has been generated according to some probability distribution. The variational autoencoder tries to model that distribution and find the underlying parameters. Learning a probability distribution allows variational autoencoders to not only reconstruct data but to generate new data on the basis of that distribution.

To build a variational autoencoder in practice, we still encode the data using an encoder network. But instead of creating an encoding straight from the output of the encoder, we use the output to parameterize a Gaussian distribution. To generate the encoding, we sample the data from that distribution. The decoder then learns to recreate the image from the encoding. Due to the stochasticity inherent in the sampling process and thus in the encoding, the reconstruction will be similar but different from the original. For example, if you feed an image of a dog to the autoencoder, the reconstruction should be a different dog.

Mathematically, instead of defining the encoder as a function mapping x to a lower-dimensional encoding h, we define it as a conditional probability distribution that produces h given x.

p_{encoder} (h |x)

Similarly, we define the decoder as producing r given the encoding h.

p_{decoder}(r|h)

In practice, you

• feed your input x (such as an image) to the encoder
• train your encoder to produce μ and σ, the parameters of a Gaussian distribution. A Gaussian distribution is parameterized by a mean (μ) and a standard deviation (σ)
• randomly sample a term ε from a standard Gaussian distribution (with mean 0 and standard deviation 1)
\epsilon \sim N(0, 1)
• create the latent encoding vector h by multiplying σ with the randomly sampled ε and adding it to the mean μ. This creates the stochasticity that enables the variational autoencoder to generate new data. Note that there are different versions of this formula.
h = \mu + \sigma \odot \epsilon
• Lastly, you pass h to the decoder which produces r in an attempt to recreate x.

## Autoencoder Loss Functions

Just as other neural networks, autoencoders use cost functions to measure the reconstruction loss and reduce it through backpropagation with gradient descent.

The goal of the autoencoder is to reduce the loss between the original input x and the reconstruction r that has been generated by the encoder f and the decoder d.

L(x, r) = L(x, d(f(x)))

Several loss functions are applicable to variational autoencoders.

### Mean Squared Error

The mean squared error usually works pretty well in vanilla autoencoders. It measures the squared distances between an observation in x and in r.

L(x_i,r_i) = || x_i -r_i||^2

The cost across the entire dataset is the sum of the squared distances between the observations in x and r.

J(x, r) = \frac{1}{n}\sum^n_{i=1} (x_i - r_i)^2

### Binary Cross Entropy

If your data is binary or at least in the range [0,1], you can also use binary cross-entropy in vanilla autoencoders.

J(x, r) = - \sum^n_{i=1} x_i log(r_i) + (1-x_i) log(1-r_i)

### Variational Autoencoder Loss and Kullback Leibler Divergence

When training variational autoencoders, we do not aim to reconstruct the original input. Instead, the goal is to generate a new output that is reasonably similar to the input but nevertheless different. Accordingly, we need a way to quantify the degree of similarity rather than the exact difference. We can achieve this by measuring the difference between the probability distributions underlying the data generating process of the input and the output.

As the name implies, the Kullback-Leibler divergence measures the divergence between 2 probability distributions, P and Q.

D_{KL}(P || Q)

In a discrete data setting, the Kullback-Leibler divergence is calculated as follows:

D_{KL}(P||Q) = \sum_{i=1}^n P(x_i) log(\frac{Q(x_i)}{P(x_i)})

P(x) represents the distribution of the encoder p_{encoder} (h |x) that projects our input variable x into latent space to create the encoding h. As discussed in the section above, when training a variational autoencoder, we let the encoder produce the parameters σ and μ of a normal distribution.

The standard normal distribution is our prior distribution p(h) if we don’t know anything about the input data. To quantify the loss, we measure the divergence of p_{encoder} (h |x) from a standard normal distribution using the Kullback-Leibler divergence.

D_{KL}(p_{encoder} (h |x) || p(h)) =  D_{KL}(N(\mu, \sigma) || N(0,1))

In practice, we plug the values generated by the encoder into a derivation of the Kullback Leibler divergence for Gaussian distributions.

D_{KL}(p_{encoder} (h |x) || p(h)) =
\frac{1}{2}\left[-\sum_i\left(\log\sigma_i^2 + 1\right) + \sum_i\sigma_i^2 + \sum_i\mu^2_i\right]

The derivation goes beyond the scope of this post. If you are interested in it, here is the derivation in a post on Stats Stackexchange.

To calculate the full variational autoencoder loss, we calculate the loss between the original input and the reconstruction. Then we add the Kullback Leibler divergence between the probability distributions underlying the data generating process of the input and the output.

L = \frac{1}{n}\sum^n_{i=1} (x_i - r_i)^2 +  \frac{1}{2}\left[-\sum_i\left(\log\sigma_i^2 + 1\right) + \sum_i\sigma_i^2 + \sum_i\mu^2_i\right]

## Autoencoder Regularization

Autoencoders with an equal or higher number of neurons compared to the input layer have too much capacity and are known as overcomplete. As previously stated, we usually design autoencoders to be under-complete. By giving the hidden layers fewer neurons and thus less capacity, we prevent the autoencoder from simply copying the input to the output without learning anything about the underlying structure in the data.
In addition to the number of neurons in the hidden layers, the capacity of an autoencoder also depends on the number of hidden layers or depth of the network. Designing the capacity of autoencoders and variational autoencoders to fit the complexity of the data distribution is one of the major challenges in training those networks.

Regularization techniques can help us prevent an autoencoder from simply copying its input to its output, even in the over-complete case. Thus, regularization allows us to add more capacity to models by increasing their depth and breadth while enabling them to learn useful mappings.

Note that variational autoencoders are trained to maximize a probability distribution given the input data rather than learning to reconstruct the data directly. Remember that the generation of the encodings involves a random term. Accordingly, a variational autoencoder cannot learn a deterministic identity function which is why regularization is not necessary.

### Sparse Autoencoder

Sparsity in the context of neural networks means that we deactivate neurons and thus prevent them from transmitting an impulse. The underlying idea is that if we enforce a sparsity constraint on the hidden layers in an autoencoder, it shouldn’t be able to learn the identity function even if the layer is large.

Standard activation functions such as the logistic sigmoid constrain the impulse to a value between 0 and 1. Generally speaking, if the output of a neuron is close to 0, it won’t transmit an impulse, while if it’s close to 1, it will. In a sparse layer, we want many of the neurons to have outputs close to 0 to prevent them from firing.

One way to enforce sparsity on the hidden layer is to penalize the average activations in the hidden layer through the cost function. Using the L1 norm, we can add a regularization term Ω that is a function of the encoding layer h and has the effect of shrinking many of the parameters in h to 0. With the hyperparameter λ, we can control the regularizing effect.

J(x, r) = \frac{1}{n}\sum^n_{i=1} (x_i - r_i)^2 + \lambda \Omega(h)

If you don’t know why L1 regularization can set parameters equal to 0, check out my post on regularization in machine learning.

An approach that is described in detail in this Stanford lecture enforces a penalty using the average activations in the hidden layer. The idea is that in a hidden layer you want to make sparse, the average activation \hat p across all neurons should be relatively small because most neurons produce no impulses. So you can set a relatively small target value p and use a measure of the difference as a penalty on the cost function. If many neurons in the hidden layer are active, the average activation will be relatively large, and thus the penalty on the cost is larger as well.

### Denoising Autoencoder

Denoising autoencoders work by corrupting the input source of the data. The autoencoder is then trained to reconstruct the original data from the corrupted input. For example, if you are training an autoencoder to reconstruct pictures from photos that have been blurred.

If x is our original input data, and \hat x is the corrupted version of x, the denoising autoencoder is trained to reduce the following loss:

L(x, r) = L(x, d(f(\hat x)))

## Why Make Autoencoder Deep?

The ability of neural networks to learn complex mappings can generally be improved by either increasing the number of neurons in the layers, or by increasing the number of layers. Since autoencoders consist of feedforward networks, the same reasoning applies to them.
In general, increasing the number of layers and thus making the network deeper as opposed to simply adding more neurons, can dramatically reduce the computational cost and training data required for learning certain functions. If you are interested in why that is the case, check out this paper.
It has also been determined experimentally that deep autoencoders are better at compressing data than shallow ones.


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Re: Calc - COS(), can this be fixed?

It is readily apparent that the COS() function is inaccurate (check COS(PI()/2).

I do not know if this is a system call, or an internal math library. Either way it can be fairly unhelpful for precision calculations. Also, SIN() on larger angles is not accurately handled (check: SIN(PI()) should return 0, it does not).

A work-around is to instead: 1) if angle > pi/2 then reduce it to QI and deal with sign issues separately 2) if COS(angle) is desired, call SIN(PI()/2 - angle) instead

Can the trig function accuracy be improved? Or, failing that, is there any hope of incorporating this into the program itself so that the function error is more transparent to the end-user? Any factor of pi/2 should be correct regardless of other floating point errors. (Many other functions you may care to perform rely upon factors of PI/2 cancelling out variables in those locations).

edit retag close merge delete

Sort by » oldest newest most voted

No, this is normal, and likely will not be changed. Any person doing floating-point maths using computers should be prepared to such stuff.

See here, and here, and here for some information on this.

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I have read this over before... oddly it means a great deal more now. I must have learned something along the way. ;) Thanks for the redirects! The others were also quite edifying!!

( 2017-08-25 17:53:52 +0100 )edit

The nonzero answer is cooct. The number pi is irrational, that means that it cannot be represented in any number system. Any software will therefore calculate the cosine or sine of an approximation of pi or pi/2. Still, many calculators do give result 0, maybe because they use less accuracy or have a better formula. Anyway, bugs should not be reported here.

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As for nonzero being correct, please refer back to sine of any angle = length of side opposite from the angle/length of the hypotenuse, cosine any angle = length of the side adjacent to the angle/length of the hypotenuse... These are the most basic definitions of these functions before any numbers get thrown in the mix. Good luck convincing anyone that zero divided by any number is anything other than zero or that any number divided by itself, is any other result than 1. The value of the "any number" whether symbolically implied or explicitly stated is irrelevant unless "any number" actually is 0, exactly.

You don't need 12 (or more) different power series expansions to accomplish trig functions. If you understand what the code names represent (sin(), cos()... et al) you know that any can be determined from the most accurate calculation of any two (one the inverse function of the other) power series expansion for angular values 0-pi/2 and a bit of mathematical shuffling to adjust for trig identity properties, identity reflection and quadrant locality of the input angle (that has been reduced to it's equivalent value below 2pi). This may be quite edifying if you really want a better explanation and more information.

Which led me to pointing out the SIN(PI()) problem, since the accuracy of the used expansion seemed adequate for display limitations at least up to sin(PI()/2), but not necessarily beyond that... Why? Because it means the value of PI() in this issue is not the relevant concern. You can easily see that PI() is accurate as a double. This means the accuracy of the trig functions only has single precision. You can correct the issue one of three ways. 1) Rewrite all the power series code to have double precision. 2) Limit calculations back into a range where errors due to foreshortened accuracy at least behave properly at non-value related boundaries. 3) Do both. Regardless of actual decimal values involved: sin(pi/2) and cos(0) will always = 1; sin(0) and cos(pi/2) will always = 0; sin(pi/4) and cos(pi/4) will always = SQRT(2) / 2. If you determine the floating point values of the definitions, and then compare results of calculations against them you can always limit return results to the definition accuracy vs floating point accuracy and thereby not carry errors through in those places. For trigonometry the 0 and 1 are needful, the point of symmetry quite helpful, others would be merely convenient but not required (yes there are more). (They used to do all this with tables and 16" or so of slide rule, if you can believe that. Can you say "a nightmare of work" for very limited accuracy? Oh yeah, in the snow, uphill both ways. -.O )

Finally, it isn't a bug report per se, as it isn't technically a programmatic bug, it's a request for more accurate handling of trig functions. Perhaps, this also ...

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I agree with most of your points. As to improving the code, you need to talk to the developers, not to the users who hang out here. Your chances to be noticed by the developers here are minimal.

( 2017-08-25 12:20:45 +0100 )edit

Well, joy for the understanding, heavy sigh for trying to contact the devs directly. Thanks much =D

( 2017-08-25 12:35:53 +0100 )edit

Heh, I stated something wrong anyway. I'll just share the formulas and result and you can choose to make sense of it ... or not. LOL

Formula: from SIN(PI()/2 - 0.000000037) to SIN(PI()/2 + 0.000000037) all return result of 1.00000000000000000000

Is the internal value of pi restricted to single in the functions or... what I said above? or something else I don't know about? Probably some combination that will make me feel foolish. Cheers!

( 2017-08-25 13:04:35 +0100 )edit

Wrapping the functions with ROUND(), rounding to the maximum precision ROUND(SIN();15) seems to get good results.

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Indeed :) Thank you for that. Sadly, it has come to me after some experimentation in VB that the range of accuracy I am dabbling in exceeds 15 digits by some as yet unknown amount, I have to write some code to be able to test my formulas in that medium. I believe it's not more than 3 to 5 digits, but playing around right in that range with the limitations there explains many things. The last article Mike offered explained a few things about the hardware I was not aware of also... so, ah well.

( 2017-08-25 22:59:07 +0100 )edit

There is no way to fix it, because LibreOffice does not make symbolic calculations. If you need such, you should use a CAS (computer algebra system).

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Stats

Asked: 2017-08-24 09:38:10 +0100

Seen: 195 times

Last updated: Aug 25 '17

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# Analysis (implicit and inverse func thrms)

1. Feb 16, 2008

### steviet

1. The problem statement, all variables and given/known data

The equations:

uz - 2e^(vz) = 0
u - x^2 - y^2 = 0
v^2 - xy*log(v) - 1 = 0

define z (implicitly) as a function of (u,v) and (u,v) as a function of (x,y),
thus z as a function f (x,y)

Describe the role of the inverse and implicit function theorems in the
above statement and compute

$$\partial$$z/$$\partial$$x(0,e).

(Note that when x=0 and y=e, u=e^2, v=1 and z=2)

2. Relevant equations

Implicit and inverse function theorems

3. The attempt at a solution

I'm finally starting to get a grasp on the 2 theorems and their
respective proofs (I hope), but as far as explaining their role in this
concrete example I'm a little lost.
If anyone can put me on the right path, it would be greatly appreciated.
(This is for an undergrad analysis course)

Last edited: Feb 16, 2008
2. Feb 16, 2008

### HallsofIvy

Staff Emeritus
Perhaps it would be a good idea to check on exactly what the "implicit function theorem" and ""inverse function theorem" say!

In particular you might want to determine where, if ever, the Jacobian is zero.


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Sunday
March 26, 2017

Posted by on .

A tank in the shape of an inverted right circular cone has height $10$ meters and radius $8$ meters. It is filled with $6$ meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is $\delta = 1470 kg/m^3$

• Calculus I (please clarify) - ,

What do the $signs mean before and after the numbers? What does$ \ mean in front of "delta"?

Does <$6$ meters> mean 6 cubic meters or a liquid height of 6 meters?


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## Thinking Mathematically (6th Edition)

$-15$
Change the operation to addition, and then change the sign of the second term to obtain the answer: $-7-8 = -7 + (-8) = -15$


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# How do you solve the differential equation given f'(s)=6s-8s^3, f(2)=3?

Nov 11, 2016

$f \left(s\right) = 3 {s}^{2} - 2 {s}^{4} + 23$

#### Explanation:

Let $y = f \left(s\right) \implies \frac{\mathrm{dy}}{\mathrm{ds}} = 6 s - 8 {s}^{3}$

This is a first order separable DE, so we can separate the variables as follows:

$\int \mathrm{dy} = \int 6 s - 8 {s}^{3} \mathrm{ds}$

Integrating gives:

$y = 3 {s}^{2} - 2 {s}^{4} + C$

We know $f \left(2\right) = 3 \implies 3 = \left(3\right) \left(4\right) - \left(2\right) \left(16\right) + C$
$\therefore 3 = 12 - 32 + C \implies C = 23$

Hence, the solution is:

$y = 3 {s}^{2} - 2 {s}^{4} + 23$


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# Tag Info

5

As K.G. and nightcracker note, the reason we don't recommend this method of password storage is that it becomes insecure if the secret $k$ is compromised. Given that the whole point of password hashing is to protect the passwords in the event that your server is compromised, it's generally not safe to assume that the compromise won't include the secret key ...

3

Quite a difficult question. What you seem to need is a one-way permutation $P$. Indeed, suppose you have it of width $d$, then consider the function $$F(K,S,R_S) = E_{K_2}(P(E_{K_1}(S,R_S))),$$ where $E$ is any good 64-bit block cipher (say, Simon) and $K_1,K_2$ are derived from $K$. This function $F$ should fulfill (2) because of the encryptions of both ...

2

What the specification is saying is that prior to processing, the message is padded to a full block length, with the empty message padded to a single block. The spec on page 4 describes the input into the algorithm as: Define $||a||_n = max\{1, \lceil|a|/n\rceil \}$, where the empty string counts as one block Let $m = ||M||_n$ Partition $M$ into $M[1] ... ... 1 The usual way to encode long random bitstrings, so that they can be easily memorized and/or entered by humans, is to break them into blocks of (typically) 10 to 12 bits and map each block to an entry in a fixed dictionary of common words. This approach is commonly used for secure passphrase generation, e.g. by Diceware, S/KEY and PGP. Assuming an 11-bit ... 1$Poly1305_{k,r}(N,M)$is a Carter-Wegman nonce-based MAC, whose security crucially depends on the uniqueness of nonce$N$for every message$M$. It is defined as $$Poly1305_{k,r}(N,M) = f(M,r) + AES_k(N),$$ where$f(M,r)$is a polynomial of$r$with coefficients derived from the binary representation of$M$, and$AES_k(N)$is the encryption of nonce$N$... 1$Poly1305_{{r,s}}(m)$is a one-time authenticator - it can be used to authenticate only a single message with any given key$(r,s)\$ without violating the security guarantees (the violation is immediate - only two authenticated messages with the same key are required to create a forgery according to the nacl docs). There are two 128 bit key values to this ...

Only top voted, non community-wiki answers of a minimum length are eligible


text: Modular arithmetic - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Modular arithmetic (/showthread.php?tid=435) Modular arithmetic - Stereotomy - 04/02/2010 I'll just preface this by saying I'm just a physics undergrad, so this might be a bit beyond my understanding, and I may well be missing something obvious or making a stupid mistake, but while playing around I noticed that it seems to be true that $^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$ $\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$ Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around? RE: Modular arithmetic - bo198214 - 04/02/2010 (04/02/2010, 07:16 PM)Stereotomy Wrote: $^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$ $\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$ Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around? I dont think it is true. For example: $7^7 \text{mod} 5 = 3\neq 2 = 7 \text{mod} 5$ RE: Modular arithmetic - Stereotomy - 04/03/2010 (04/02/2010, 10:57 PM)bo198214 Wrote: (04/02/2010, 07:16 PM)Stereotomy Wrote: $^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$ $\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$ Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around? I dont think it is true. For example: $7^7 \text{mod} 5 = 3\neq 2 = 7 \text{mod} 5$ Ah, good point, though $7^{7^{7}} \text{mod} 5 = 3$ Is true as well. In fact, thinking about it, the numbers I tried out with this all had b>a. Perhaps that's an additional condition that either b > a or m, n > 1? Just quickly tried this for a few low examples, a = 8, 9, 10, 11, and it seems to hold. RE: Modular arithmetic - bo198214 - 04/03/2010 (04/03/2010, 12:18 AM)Stereotomy Wrote: $7^{7^{7}} \text{mod} 5 = 3$ Is true as well. In fact, thinking about it, the numbers I tried out with this all had b>a. Perhaps that's an additional condition that either b > a or m, n > 1? There is an article which proves that ${^n a}$ (which is $a{\uparrow}^2 n$ in Knuth's arrow notation) finally will be constant for $n\to\infty$ mod any $M$, see this thread.


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Interview with Prof. George Lusztig

Interview Editorial Consultant: Tai-Ping Liu
Interviewers: Shun-Jen Cheng (SJC), Wei-Qiang Wang (WQW), David Vogan (DV)
Interviewee: George Lusztig (GL)
Date: January 6th, 2016
Venue: Institute of Mathematics, Academia Sinica

Prof. George Lusztig was born in Romania in 1946. He graduated from the University of Bucharest in 1968 and received his PhD in mathematics from Princeton in 1971. Currently, he is an Abdun Nur Professor at the Massachusetts Institute of Technology. He is a world leading expert in representation theory. He was elected as a fellow of the Royal Society in 1983, to the National Academy of Sciences in 1992, and was awarded the Shaw Prize in 2014.

SJC: First of all, George, thank you for allowing us to do this interview with you. Well, I have a couple of questions here that the editor-in-chief of MathMedia asked me to ask you. So let me do it one by one. The first question is when did you know you were interested in mathematics? At what age? What time?

GL: 14, maybe. I was interested in mathematics before but at 14 I decided to be really a mathematician.

SJC: Oh, really. So before you went to university, you already knew you wanted to be a mathematician.

GL: Yes.

SJC: I see. I was wondering because in Chinese culture, maybe you know that, the parents usually are…

GL: No, but this was because in 8th grade there was an Olympiad, this kind of national Olympiad which I had no idea about. I didn’t expect to do well, but I did. I decided to really spend lots of time on it.

WQW: When you were in junior high school? 14?

GL: That was the first year of high school.

SJC: Yes, I see. That’s right, first year of high school. The other question I have is about when you went to the University of Bucharest. How does its mathematics department compare to the mathematics department in the West? I mean, how is the program in the university?

GL: Well, that’s hard to compare. It’s very different. First, this undergraduate school was 5 years. But the last two years you have to specialize in something, which was like a master’s degree program maybe.

SJC: But I mean when you entered the university, do you only have to take mathematics courses, or do you have to take others?

GL: Yes, you have to go to mathematics department. You have to pass some entrance examination. You have to decide exactly which university and which subject you like.

SJC: So you don’t have to take other classes outside mathematics like in the U.S., where you have to take….

GL: No, you have to take Marxism, for example.

SJC: Marxism?

GL: Yes, every year, some kind of Marxism. Maybe in two years of the five.

SJC: What did you learn? I mean these courses, what do they teach…?

WQW: We have a similar course when we were in China. We learned the history of Chinese Communist Party.

SJC: Oh, really.

GL: Yes, there is something called dialectical materialism (辯證唯物主義). You know that word? That was some Marxist philosophy. And there’re some political economy or something. Some of these subjects, but also Marxism. Also, all the time we had to take some course on how to teach.

SJC: How to teach?

GL: Yes, because this department was basically preparing school teachers, so we had a class on how to teach. Otherwise, in the first year we had physics, and we had also mechanics.

SJC: So this was the department of mathematics and mechanics, I see. So it’s like the Russian style. I see. That answers the second question here: If the educational system was influenced by the one in the Soviet Union, but obviously the answer is yes in this university.

GL: But one impact was that lots of Russian mathematics books were available for very low price, so I could easily buy. And they translated everything from the West. Bourbaki, all the volumes were translated. I could easily afford them, more easily than now.

SJC: Now, none of us can afford, especially Elsevier journals.

GL: Yes.

SJC: But I suppose you have to take Russian in university or in high school?

GL: We started in the 4th grade until 11th grade studying Russian and also the first two years in the university.

SJC: That’s quite a bit.

WQW: Russian is the official second language?

GL: No, we have to take two foreign languages. But Russian was compulsory and the second one was a choice of English, French or German.

SJC: Oh, I see. So which language did you choose?

GL: No, that was in theory, but it was the question of what are available. So not every school has all the choices. I took French actually.

SJC: So, George, during your undergraduate days in the University of Bucharest, were there any professors who have influenced you most?

GL: Yes, there is one called Teleman. There’re quite a few of them but they are all in the same family. They are all related to each other. So this professor had 4 brothers, all mathematicians and all descendants of mathematicians, and one of them was my professor.

SJC: I see. One of the questions is about the reason why you moved to the West. And how did you move to the West? So you went directly to the U.S. after you left Romania?

GL: No, in fact I went to the West once and then I returned to Romania. In 1968 I went to Italy to some summer school, then to England, which was not really allowed but I asked permission afterwards. Then I went to the embassy asking for the retroactive permission. Then I went back, and half a year later I went to Bonn. Then I went to the US. In fact, I had an invitation to Princeton because of Atiyah, whom I met in Oxford. That was the first time.

SJC: Yeah, that’s the next question. What was it like to study with Michael Atiyah?

GL: But of course I was not officially his student. Of course he is a very great mathematician and also extremely nice person. He helped me a lot. And we actually talked, talked probably every day, I think. It’s quite….

SJC: I thought probably not everybody knows that Atiyah at that time was in Princeton.

GL: The first time I met him it was in Oxford in 68 and he was moving to Princeton. As he’s moving, he could recommend some young people to the (Institute of) Advanced Study. So he recommended me to work with him. So 69 he was already in Princeton. But I didn’t go to Princeton directly, he asked me to go to England.

SJC: No, I thought you told me that you went to Italy first.

GL: No, that’s another trip, the first trip abroad. First Italy and then England and back to Romania. Then I received an invitation from Princeton which I got already when I was in Oxford. When I returned I applied for permission to go to Princeton, and it was refused.

SJC: It was refused! So what did you do?

GL: But I also got an invitation to Bonn, Arbeitstagung, you know Arbeitstagung?

SJC: Yes.

GL: That one was approved because that was short, that was 1 week. And that one was approved sort of at the local level. Long trip has to be approved in Bucharest. So simultaneously I got one that was approved and the other was not approved. So I went to Bonn and from there, not directly, because I didn’t have a U.S. visa.

SJC: But you have an invitation to Princeton?

GL: Yes, so I actually went to Canada first. There was a conference there and I applied for U.S. visa in Canada and stayed two months, then went to Princeton.

WQW: Then you spent one year in Princeton before going to Warwick?

GL: Two years.

DV: He had to get a PhD.

WQW: Yes, to have the right to teach. But your thesis was not quite on Lie theory, yet, right?

GL: No.

WQW: How did you slowly or quickly move into Lie theory, or finite groups of Lie type?

GL: Of course Atiyah he knew some Lie theory, and he knew a lot of classical group representation very very well. But he had low opinion of the exceptional groups; they are not important and can be ignored. So all I knew was in this index theory. Michael Atiyah is a topologist actually but then because of a lecture by Quillen, Quillen was also at Princeton at that time, he solved some problems in topology, using a method of representation of finite groups, in fact, using something called Brauer lifting of modular representation of finite group to a virtual representation, with complex coefficient, something like that was done by Green. Actually a very deep theorem, no other topologist was aware of such a thing. So he was quite amazing. So he used that and solved something in topology, then I was quite interested in that. Then I asked Atiyah whether this Brauer lifting was known more explicitly. Because he started with the natural representation of $GL_n(F_q)$ over a finite field, and in some way attached some complex representation, virtual representation of it, and I asked him if the components of the virtual representation are known? Atiyah told me that the characters are known, but the actual representations are not known. And the characters are known from Green’s work.

WQW: But in type A?

GL: That’s just standard representation of $GL_n(F_q)$. One of these components is a discrete series representation. And that was not known, only its character. I got interested. So I learned what I needed basically. Then I was more and more interested in representation theory.

WQW: At that time, was Green’s theory on characters for $GL_n(F_q)$ available already?

GL: Yes, that was (known since) 55. And the interesting part was that at the end of my stay, I got an offer to go to Warwick which was the place where Green was located. So Green was actually at Warwick. So I thought that it would be a good place as I wanted to solve this problem. But I was wrong because he was actually not in good health. He was almost never coming to the department. Did I answer your question?

WQW: Yes, but then you still continued working on that in Warwick?

GL: Oh yes, but Carter was there, so I talked to him and he explained various algebraic groups and we wrote some joint papers. Eventually, I could solve the problem.

WQW: At that stage, you were still just working on type A?

GL: No, so my first thing was I found some way to construct this discrete series of $GL_n(F_q)$, by some explicit model using Dirichlet sums. But then I immediately started to ask the same question for classical groups because they also have a natural representation. You cannot take Brauer lifting, so that I also found somewhat less explicitly. But by working on this, I started to understand some of the representation theory by looking at the Brauer lifting.

DV: So those particular representations of the classical groups are… I mean in terms of the technology at the time, they were inaccessible. But on the other hand, in terms of what you did over the next 10 years, these were extremely easy somehow. If you were thinking about how complicated the problems turned out to be, these were misleadingly simple. Is that…

GL: No, basically what was happening for classical groups in this Brauer lifting, so for symplectic, there are 2n components. You have 2n dimensional representation in modular setting and you lift this 2n dimensional representation. And one of them is a discrete series corresponding to Coxeter torus actually. But the other one is also difficult. They are not discrete series but you have to decompose using discrete series. In $GL_n$, all the other one are simply induced. They are not decomposed at all. In other classical groups, it’s non-trivial to decompose. I have to understand this Hecke algebra with unequal parameters which I have to use… So that was more complicated than $GL_n$. But this method would only construct one series but for classical groups there are several discrete series, several non-isomorphic tori. So in fact I did quite a lot of work and eventually I did something which I never published, I just wrote some very short summary about it, in fact it was quite a nice theory. It was never published because it became obsolete in some sense. But actually all this helped me quite a lot to gain a lot of experience and helped me to do the other things.

DV: So, it sounds like you probably did things like almost writing down character tables for some of these classical groups, and anyway computing very explicitly the characters of a lot of representations and understanding all the conjugacy classes. So do you enjoy that kind of… you know, everyone has to do some computations like that for mathematics, but some people like it more than other people. Do you enjoy that kind of work? You’ve done a lot of it.

GL: Yes, certainly I like computations.

WQW: Yes, those computations are something I’ve never really read, but I learned your braid group action computation. After 20 years, it’s still brutal to me. I taught this a couple of times, I always choose the easier types, but even those were pretty complicated. $G_2$ I’ve never tried, even today.

DV: So you talked about having to take a class and teaching in Bucharest. Did you learn from…?

GL: No.

DV: But you mentioned on the other hand, Teleman. How do you learn to write mathematics well, to do mathematics well? Whom have you followed, tried to be like?

GL: I did write some papers while I was a student there. But I don’t think I learned from Teleman at all. Maybe from Atiyah certainly. I tried to follow his example and even more Deligne I would say, certainly the way he writes.

WQW: Atiyah’s papers I read a few. I don’t remember your styles were very similar, let me put it that way.

GL: No.

DV: By now you’ve read and heard the mathematics of hundreds of people, you know, read papers and seen people lecture. Are there individuals that you think we should all be more like, best examples?

GL: Milnor is No. 1. I was always told he’s the greatest. His books were really absolutely the best. And his lectures were the best. He was my model for this one. I know him a lot, I was doing topology in the Institute and was close to him.

DV: So did you ever think about deliberately moving back towards topology? After you started working on finite groups, reductive groups and so on, it seems that you were solving things so fast but also opening up new problems so fast, from 1970 to 1980s. It’s hard to see why you would go back to do something else, because there’s still more.

GL: That’s still the case. But it’s also the case that, while I was in Princeton, I thought I would be a topologist, I knew I had some contact with Daniel Sullivan. Actually I had the impression that he’s solving all the main problems in topology. So, that was the incentive for me to move away from topology. In fact, I always, actually even now I think representation theory has more interesting problems than topology. I think topology is very useful as a tool.

WQW: They were more interested in classical groups.

GL: But not so interested in as a goal in itself.

DV: I don’t remember who talked about mathematics as some kind of garden, and it was very important to put fertilizer on the ground and so on.

GL: Hironaka.

DV: I think that’s right. But there were some parts of mathematics that were the flowers. So maybe the groups and representations are some of the flowers. There are whole lot of really new ideas in mathematics that you’ve been closely involved with creating. The impression I have is that most of them came when there was a problem that you need to solve and there was no way to solve it using mathematics that was around. And so you made something new that was designed to solve an old problem. Is that…do you…ever just find things that look like they will be interesting ideas or, I don’t know. Somehow it feels like the Kac-Moody Lie algebras were just kind of made because after the Serre relations were written down, you could say, “Well, we don’t need this to be a finite dimensional algebra and everything, all the formalism can still be done.” And sort of feels to me like there is not so much reason to go do this mathematics that you can do. But the new things you brought in to the subject they always seem to be designed to tell you about old problems, at least at first.

GL: Actually, I had the impression that I always have some goals to go to, but always happened that I got some luck that things happened. For example, at some point, I really wanted to understand how to classify representations of classical groups. After the paper with Deligne, I wanted to know; that was really an important problem, I thought. Because $GL_n$ was known, (for classical groups) how to classify that was not known at all. I didn’t come to up with any new methods, I already knew I had this technique from my paper with Deligne, but then I met these Korean mathematicians, Lee and Chang, anyone knows these two people?

DV: The Lie group guy?

GL: Yes, and there’s another, he had a collaborator, who did character of $G_2$ over finite fields. And I met them in Vancouver and Chang told me that he had a student who just computed Iwahori Hecke algebra of classical type with arbitrary unequal parameters, he had construction and degrees. And this guy actually left mathematics. So that came exactly the right time. So then I studied that and I could see that was somehow what I was missing almost. So I think it went like that several times. I just got something I was missing, just by some accidents, it happened that it was provided to me.

WQW: But at that time if there were not available, probably you would be just forced to work out it yourself.

GL: Possibly. It all came one after another, then I understood these classical groups, and then immediately I found some preprint that somebody found the generic degree for Hecke algebra of (exceptional) type E8.

WQW: You were not a good student of Atiyah….

GL: What? No, all I really wanted to classify…. That I got some idea of how to construct this non-abelian Fourier transform matrix, because they had this list of degrees of E8, you could see they all have some leading terms which have some coefficients, one over 120 and one over 24. So then, I saw some patterns in that. But that came exactly at the right time.

WQW: Those you all computed by hand, right?

GL: No, I didn’t do the computation, someone did that.

SJC: But I am saying that if that were not available, chances are that you probably would compute yourself, no?

DV: Well, you have friends you pushed around. I mean you need all these branching tables for Weyl group representation eventually. You didn’t compute those yourself. But you were responsible.

GL: Yes, and even before that, the first time I used computer was to compute the fake degrees, you know that? I did fake degrees for classical groups in this work. I want to know for exceptional groups. And for these, I found some guy in Warwick and told him the rule to compute, so he did this, that was the first time I used computer.

WQW: I think I know the name. Beynon? He’s not a mathematician.

GL: He’s not a mathematician. He was in computer science.

WQW: So for the exceptional type sometimes without computer, even you feel it is hopeless to compute.

GL: Yes, computer made a big difference. Because maybe in the 60s I think, it was considered that classical groups are easy and exceptionals are difficult somehow. Now it’s reversed I think, because of the computer.

DV: I want to ask something that seems completely different but maybe it’s not completely different. So this was the administrator in the math department of MIT, Joan Johnson. Of course, many people in the department travel a lot. So they come and talk to her about places they travel. She told me what she liked best was when you talked to her about places you travelled. Because she said that “He sees everything, and he remembers everything.” Do you know that? I mean obviously you know you see some mathematical things more than other people do, but are you aware of it?

GL: I am surprised!

DV: I think all these things you were saying about fortunate accidents that something was available, the world of what’s available in mathematics is enormous. The question of seeing the right thing that’s available that’s not so easy necessarily.

GL: But actually quantum groups, at some point I got interested in because Borel wrote me a letter. He said there is this work of Michio Jimbo he thought that I might be interested because he had something that looked like Hecke algebra. He knew I was interested in Hecke algebra. He said Jimbo actually had Hecke algebra and it was directly involved. So he pointed out this thing, then I studied and I gave a course.

SJC: When was it? Which year was it?

GL: 86.

WQW: The paper was published in 86. I think among us, we probably know a little bit more about your work on quantum group than the early work on finite group Lie type. Even though I browsed your earlier books once or twice, I am still not sure how much I really read.

GL: But I think that the finite groups of Lie type were more exciting in some sense, than the quantum groups because you didn’t know what was the result, there was a mystery.

WQW: Right, those questions in a way are 100-year-old, the finite groups of Lie type. In a way it is an old classical problem. Somehow, can you say that introducing geometric method there is one of the keys for finite groups of Lie type?

GL: No, but certainly without geometric method you cannot do anything.

DV: What’s amazing is as you said Green in the 1950s did make this complete list of the irreducible representations of $GL_n$. So people who work hard enough can… you can do anything somehow. But it was the geometric methods that you and Deligne brought that made some of those results clear, made it clear what had to be true, not just the results of some terrible calculations.

GL: Oh, Green, all his papers were really amazing. He was the only person who knew modular representation very well, so he could use method of Brauer. He was student of (Phillip) Hall. So he knew all about Hall polynomials which he used to define Green polynomials. So he knew all these things. Because before him Steinberg was working on this, he only managed to do $GL_4$, from that work he was then completely exhausted, that of $GL_4$.

WQW: I am sure that the symmetric polynomials, Hall-Littlewood, and all these things, how they get into play, probably, one really need some expertise for those things. But back to quantum groups, for quite a few years from 86 to 94, you were quite focused on quantum groups. It’s a major part of your work.

GL: Yes, almost 10 years. I was doing two theories, there are really two kinds of things. One is quantum groups at a root of unity, trying to understand how they can help to do modular representations, and also this canonical basis.

SJC: That’s the time when we were students there.

WQW: Yes. Those things back then, I didn’t understand.

GL: Pardon?

WQW: The canonical basis. I am sure I took lecture notes from you.

GL: But (now) you know that very well.

WQW: But only now I really need it. The last 5 or 10 years I have been trained very well by my students. Somehow they forced me, different students for different purposes re-taught me.

GL: Actually Borel was correct. He could see that knowing Hecke algebra will help me. I can transport some of the thing to quantum groups, that’s exactly correct.

DV: So you talked about working with Atiyah and Deligne. These people who could think of as in some ways teachers as well as collaborators. So how do you like working with your students? I mean, you know there’re two things you can do. You can write a beautiful book and send it to the publisher, or you can have a student. This is very different kinds of satisfaction. Do you have good experiences?

GL: Yes, I have wonderful experiences. I have some really good students which I was extremely happy to have them. One example is Spaltenstein. Xuhua He is another one. I have several similar students.… But not all of them, some of them were really good.

DV: So have there been students that really changed the way you do mathematics, changed how you think about some of the problems?

GL: Yes, Spaltenstein I’ve learned a lot from him, maybe more than from many other students. He was actually very early, maybe in the very beginning, so it was in England.

SJC: He was your first or second student?

GL: Immediately, I started with three students. Maybe he came one year later. Maybe, he was the second. The first one is de Concini.

SJC: de Concini, that’s right.

WQW: Looking at the publication list, it’s very easy to see that most of your papers are single-authored. Then I am talking about this working style and collaborators. Is that basically the way you choose or is that just the most natural way? How do you feel about it?

GL: Yes, I usually work by myself. There were some exceptions, but I think my natural way would be to write it by myself.

WQW: Even those long computations, you prefer to write it by yourself.

SJC: But George said he likes computations.

DV: I don’t know if collaborating with Deligne spoils you for anybody else.

WQW: I don’t know, probably more people tend to have more collaborators than just working alone all their long career?

DV: I have asked the people of math reviews because they could easily do statistics about whether the number of coauthors on papers is actually increasing. I have the impression, yes, that it is.

GL: Even 4… I see papers of 4 authors that are also becoming quite usual. At least I’ve seen in this conference.

SJC: Yes, that’s right, three or four. That’s quite common because communication becomes so easy.

WQW: That’s true. Also, to be efficient. A single person has single expertise. We don’t learn things that quickly even if we know what to learn. To me, those collaborators, they were amazing, bringing in wonderful, different techniques. Those things for myself would take much longer or forever to learn.

DV: But there’re various jokes. I don’t remember any of them properly. A camel is a horse designed by a committee. There’re problems with collaboration.

WQW: Okay, I am sure that applies to some other collaboration.

DV: Or just to any collaboration with 4 people.

SJC: It’s a committee.

GL: I’ve never been in any team of 4, maybe three was maximum.

DV: You told me once that when you were in school at Bucharest.

GL: Not in school, but in university.

DV: Yes, you were on the crew team that you rowed. So how did that go with studying, with doing mathematics?

GL: Not so much in Bucharest. That was in last two years, actually last two years of high school I did rowing. That was not a contradiction. I went twice a week or something. I was very bad at physical exercise. Somehow I got this. I don’t know how I started but it improved my health and so it was quite good.

DV: But then you stopped.

GL: Then I went to university, and continued in the first year. It’s not so convenient. Actually you had to go far away. I did it for one year but it was not so enjoyable. In my hometown, we have a river and it’s quite convenient, and I liked it.

DV: So have you found other kinds of activities? I mean do you find it necessary to do things other than mathematics in order to do mathematics?

GL: Yes, like yoga, for example, I think I couldn’t function without that.

SJC: But when doing yoga, do you think of mathematics?

GL: No.

SJC: But you found it easy to just shut it off? For example, I mean when you’re ready to go to sleep, you could shut it off?

GL: No, the problem is when I go to sleep, I can. But then I often wake up in the middle of the night, then always some problems come then I cannot sleep for 2 hours. But when I go to sleep, it’s no problem, because I was tired.

DV: So you said you like to read Milnor’s books. Are there non-mathematical books that you find very worthwhile?

GL: Yes.

DV: Love in the Time of Cholera?

GL: Yes, that’s good. I like that. You like it?

DV: Well, I just remember 20 years ago or something that you mentioned this.

GL: Yes, I mentioned this.

DV: See you have to be careful of what you said. We’re all writing things down.

GL: Yes, this particular writer I like, I read several of his books.

SJC: One Hundred Years of Solitude?

GL: Yes.

WQW: I think you (Cheng) probably introduced those books to me.

SJC: Fantastic books. Marquez.

DV: Do you ever read fiction and so on, other than in English now?

GL: Yes, actually I have now read some Balzac’s books, in French, sometimes I read in Romanian, sometimes I read in Italian, but that was less. But in Italian I read quite a few books.

SJC: You said you went to Italy only for a short period.

GL: One year.

DV: But the language is not so distant from Romanian?

GL: Yes, but you have to learn, it’s not…

DV: Yes, I understand it is not a dialect.

GL: But you can also interpolate to some extent.

DV: Did any of your old Russian math books come with you? Do you have them?

GL: No, they came with my sister. My sister came later she brought them, so I got all the books.

DV: Oh, wonderful.

GL: But I can’t read literature in Russian. I’m not at that level… But actually one of the books I read recently which I thought extremely good which I really recommend. Its writer is called Vasily Grossman, some Russian, and he has this book. The subject is the battle of Stalingrad(史達林格勒戰役).

SJC: The battle of Stalingrad. I see.

GL: And it’s just really a wonderful book. It has some similarity with War and Peace. It also talks about some battles in which the existence of the country involved was in question.

DV: So what about music? How much do you have opportunities to listen to music?

GL: My wife Gongqing is playing every day, practicing piano.

SJC: It’s great to have live music at home, I think. Do you read before you go to sleep? Does it sort of relax you?

GL: No, not before.

WQW: You yourself also wrote several books. How did you enjoy that experience, writing books?

GL: They are not really books. The first one was on Discrete Series of $GL_n$, it was not supposed to be a book, it was supposed to be a paper, and I sent it to Annals of Mathematics. They proposed it to be a book instead of a paper. So I did that.

DV: The characters of reductive groups over finite fields?

GL: Also, I considered that is a paper, not a book. Because if it were a book, I should try to make it easy to read, which I didn’t. So I wrote it such that I finished it as soon as possible. So I write everything and don’t care so much about the readers.

DV: I was going to ask how many books you’ve written. But I guess it’s an easy calculation that just you write papers and the lengths are normally distributed and any paper of length more than 200 pages is a book.

WQW: But his Introduction to Quantum Group must be a book.

GL: It’s a type of… But …

SJC: It’s an introduction…

DV: See with the paper version, there was no introduction to it.

WQW: But this one you did intend to write as a book from the beginning.

GL: Yes.

DV: And Heckle algebras with unequal parameters?

GL: That was requested because I gave this Borel lecture. They asked me to, but I gave some lectures in MIT so I already had some notes. I just had to expand them. So actually maybe (Introduction to) Quantum Group is probably the only one supposed to be some kind of book, not a paper.

DV: That’s interesting.

WQW: Yes, this Introduction to Quantum Group takes many practices to get into it. I mean it takes introduction to Introduction.

SJC: Yes, I was wondering about the title. I mean you intentionally…the title is called “Introduction to Quantum Groups.”

DV: This is the title in the tradition of André Weil’s Basic Number Theory. I assume that you believe in the importance of these excellent books for getting the next generation of mathematicians into these problems. You said this most exciting or this extremely exciting work was about the characters of finite Chevalley groups. So Roger Carter wrote an account of that, which I think has been helpful to a lot of people. Do you think about these things? Do you feel that, do you see it’s not your job to write the books like that? I mean the quantum groups’ book is a little bit like that.

GL: I know, but probably the representation theory of finite groups, I feel, I don’t understand it well enough. So there is something which still needs to be done, I will think about that and when they are done I would write some nice exposition, but I don’t feel ready for that. I think it’s not yet complete. But also I don’t think I am good at this.

DV: You just have to learn LaTeX.

WQW: I think George’s using TeX, right?

DV: AMS TeX, yeah.

GL: No, but for example I am terrible at proofreading. So I’ve made mistakes.

DV: In these four books we have talked about there are 7 misprints….

WQW: I think your formulas we can trust. That’s not only my impression. I have used various things in a very concrete way.

DV: Sometime around 1980, you let me photocopy some pages of calculations you‘ve done about nilpotent elements and so on, and exceptional groups. It was wonderful to have these calculations. They were perfect. The most terrible thing about them was that you wrote in this light-blue ink and photocopy machines didn’t like this at all. So it’s very hard to read the photocopies, but they were very reliable.

WQW: Nowadays do you still write most of your work in hand-written notes or do you type up very quickly and throw away hand written notes along the way?

GL: No, I type. I certainly don’t write the whole paper, maybe just do some computations, then if I have enough, I go type them.

WQW: When we learned LaTeX versus the AMSTeX, the advantage of labelling automatically persuades us that LaTeX is the way we want to choose. Somehow this cross reference never bothers you? Because some of us sometimes need to totally change the order, of course, the labels of the equations would be all wrong. Automatically LaTeX would take care of that. How does that work for you in AMSTeX?

GL: Because I have a different way of numbering. Because I have some sections that I number formulae just by A, B, C, D, it doesn’t matter if I change the number of the section, the (labels of) formulae don’t need to be changed. So section number is not included in the formula. It’s only a letter. Then I refer to section 1t then a or something.

WQW: The quantum group book is really marvelous. The more I understand the subject, the more I appreciate it. Also, I almost find no mistakes. That’s one book in the last ten years I got to know reasonably well.

SJC: I think both of us own two copies of it.

WQW: One copy at home, one copy in office.

GL: But I think that quantum group is relatively easy. I think representations of finite groups are much more exciting and interesting. Quantum group is relatively easy.

WQW: That’s why I can understand it a little bit.

SJC: You mentioned about Atiyah and Deligne. Are there any other mathematicians you admire? Besides those two, are there any other you admire?

GL: Gauss, for example.

SJC: Oh, that’s one. We admire him as well. Somebody in Lie theory?

DV: Have you read papers by Lie or Élie Cartan?

GL: Lie certainly not, but Élie Cartan I read something.

DV: And Hermann Weyl?

GL: I have his book on classical groups that I read most of it.

DV: We are talking about the styles of mathematics that maybe, I don’t know if it is Roger Howe who said about this Weyl book that it has a million dollars’ worth of mathematics all in pennies.

GL: So maybe Chevalley I might mention. In fact, when I was at Warwick University, they had a library, and this Chevalley seminar was available, which was hard to get and it was not officially published somehow.

DV: Yes, there are a bunch of these French seminars you could find in the university libraries and a few places.

GL: So these I studied.

WQW: These are the lecture notes for those seminars?

GL: No, they are not lectures, they are exposé, by Chevalley, some by Borel, some by other people. But the seminar was led by Chevalley, they are about the classification of reductive groups.

WQW: Do you still have any plan to write another book?

GL: No, but if I understand the representations (of the finite groups of Lie type)

SJC: There would be an introduction to it.

DV: Do you know this paper of Macdonald about classifying…rewriting Green’s work about $GL_n(F_q)$ in terms of Weil groups and, well, representations of Weil groups.

GL: No, I don’t know.

DV: He proves that representations of $GL_n(F_q)$ are in one-to-one correspondence with equivalence classes of n-dimensional representations of p-adic Weil group. The p-adic field should have residue field $F_q$. And you just want Weil group representations that are trivial on the Weyl inertia and the equivalence relation he puts on these representations is equivalence when restricted to the inertia sub-group. So essentially, it’s just the representation of the inertia sub-group modulo Weyl inertia, and except that there has to exist the extension to the Weil group. He proves also those were in one-to-one correspondence with $GL_n(F_q)$ representation. Anyway, the reason I asked this is I just learned about this paper a week or 2 ago, and it illuminates from 1980s or something. The formulation makes complete sense for any finite Chevalley group.

GL: No, but it need the special classes which its formulation does not…. So this will not give a correct…

DV: See, that’s apparently what Bill Casselman always said that it can’t possibly be right, but I am not convinced. I mean there are some l-packet stuff that has to be…Maybe it is quite difficult or impossible to describe l-packets, but it seems to me that those ideas could provide a way to say some of the things that you did.

GL: I don’t think so, I think you have to pronounce this very special representation, without that you cannot hope to look at the classification.

DV: I don’t think so. We can talk about it. So what pleases you the most about doing mathematics? If you think about what you are going to do over the next five years, you could hope to have really wonderful students, you could hope to teach a really excellent class, you could hope to solve some serious problems and write great papers, all these things...It’s reasonable for you to hope to do. These are all things you’ve done. What do you aim at?

GL: Teaching, yeah, but I still get most of my fun from writing papers. I don’t think I will solve big problems, I just have some fun doing some smaller things.

WQW: But in general, you yourself are more of a problem solver, or a theorist? Of course, this is very hard to classify. But some people still have tendency to solve problems, outstanding problems. They set some goals.

GL: No, actually I like to find conjectures, finding a pattern of conjectures. Probably that’s more enjoyable than solving….

WQW: In particular, solving your own conjecture?

DV: Gelfand, I think, said that making precisely correct conjectures, not only is more fun, it’s more important. Anybody can write proofs somehow, but only if the conjecture is just exactly right and interesting.

GL: That’s what I like about conjectures.

WQW: Probably they shouldn’t be too easy to prove.

DV: Well, you know you can… sometimes if you formulate the conjecture properly, then it “is” easy to prove. That’s maybe one of the best kind is when you say something in a way that wasn’t said before, so that...

GL: No, but sometimes I find if you find a good conjecture which clearly looks like correct, then I am not even that much interested in. The proof is not so important, I think.

DV: Yes, absolutely. Maybe the Riemann Hypothesis is an extreme example. It’s clear that it’s worthwhile to do number theory, using it to see what that tells you, because it’s very likely to be true. It tells you an enormous amount. I think that was the most of the questions I have on my list.

SJC: Okay, I think that’s about it. So George, thank you very much again.

GL: Thank you.

• Shun-Jen Cheng is a faculty member at the Institute of Mathematics, Academia Sinica.
• Weiqiang Wang is a faculty member at the University of Virginia.
• David Vogan is a faculty member at the Massachusetts Institute of Technology.


text: Lottery Tickets. 1 decade ago. You have been selected to explore how percent is used in people's daily lives, and to give examples of how it is used in yours. Sometimes the most amazing abilities of the human brain are revealed exactly when things go wrong with it. Having information about the percent composition of a compound will allow you to determine the mole:mole ratio of elements present in a compound. A Teacher can teach a class of students and at the same time a Student can attend multiple classes. Because the normal distribution approximates many natural phenomena so well, it has developed into a standard of reference for many probability problems. What are practical uses for percent composition? You can see how much kool aid powder is in a given amount of kool aid drink, in percentage. For example, if a 10-ounce box of cereal costs $3 and a 20-ounce box of cereal costs$5, the 20 ounce box is the better value because each ounce of cereal is cheaper. Solving Ratio and Percent Problems Using Proportional Relationships - 7.RP.3 Students are able to extend their knowledge of ratios and proportions to applications in real life situations. This means, I get $10.50 off! For one thing, I think that it's important in the creation of jewellery, as metals such as gold is not pure gold. Example 2 : Find the mass percentage of 6 g sodium hydroxide dissolved in 50 g of water. Real life problems on percentage will help us to solve different types of problems related to the real-life situations. 1 Answer. This is a very common use of percentages in real life. Convert the amount of cake Matt consumed into a fraction i n simplest form. Basically any time you are doing one computation and then doing another computation, that is composition of functions. For a solution, mass percent equals the mass of an element in one mole of the compound divided by the molar mass of the compound, multiplied by 100%. Percent composition is a starting point which can be used to find empirical formulas which can lead to molecular formulas for compounds with covalent bonding. Lv 6. Once you get to the till to pay for the items, you will be charged sales tax, a fixed percentage of the price of the items. There are important real functions and compositions thereof governing the world we live in, as well. Its specific focus is to present a simple template that students can use to translate and solve percent word or story problems of … A) 4 and 1/2 B) 9/20 C) 4 and 5/10 D) 45/100. Association : - Student -Teacher. Learn about hydrates in everyday life. Favorite Answer. Real Life Examples If you want to put tires on cars and you have 8 cars with no wheels and 48 tires, then cars will be your limiting reagent as you need 4 tires per car. Both objects can be created and destroyed independently. .25 x 42 = 10.50. Let's look at some examples. 1 decade ago. Converting a Fraction to a Percentage in a Real-World Situation. For example, a farmer would want to know the percentage of nutrients in a fertilizer they are using. In a survey one in five people said they preferred a certain brand of fast food. Mass percent composition describes the relative quantities of elements in a chemical compound. These include ﬁnding tax, tip, percent increase and decrease, commission, and simple interest. Example Percent composition is important because it shows how much a certain substance is comprised of one component or other. 0 0. Mass Percentage Examples Example 1 : Ordinary bleach is 5.25% NaOCl by mass, which means each 100 g of bleach contains 5.25 g NaOCl. The Task You will explore some online resources for percent concepts and applications. Percent by mass = "mass of solute"/"mass of solution" × 100 % EXAMPLE What is the percent by mass of rubbing alcohol in a solution that contains 275 g of rubbing alcohol in 500 g of solution? 2. While looking at the prices of various groceries, you can easily illustrate ratios using two different boxes of cereal. This lesson will present real world examples that involve solving for a percent of change. We have moved all content for this concept to for better organization. Answer Save. Thus, it helps in chemical analysis of the given compound. When would you use percent composition in your daily life? Whether you are shopping, working, or interested in sports, you have used percentages. When working with percents less than 1 and greater than 100, we follow the same rules that we learned for percents between 1 to 100. Please update your bookmarks accordingly. To understand the procedures follow step-by-step explanation so that you can solve any other similar type of percentage problems. Before multiplying 25 by 42, I need to convert the percentage into a decimal by moving the decimal point over two spots to the left (.25). The percent composition of any compound expresses its composition in terms of all the elements present. Reactants may be impure, reactions may not go to completion, or given reactions may have to compete with several smaller side reactions. Now I am set to find out the discount total. PERCENT BY MASS (m/m) Percent by mass (m/m) is the mass of solute divided by the total mass of the solution, multiplied by 100 %. This helps us a lot to gauge the amount of time our phone Introduction Percent is a part of our daily lives. Relevance. The formula for Percentage Composition . Real Life Percentage Example. The percentage composition of a given element is expressed using the following formula, $$\%C_{E}=\frac{g^{E}}{g^{T}}\times 100$$ This lesson will present real world examples that involve solving for a percent of change. Actually, I’m going to read “live composition of real functions” out of and into this. However, ideal conditions don't exist in real life. Once you have this, you can find the ratio of ions in a compound. In spring, the farmer may want to use a fertilizer that has a high percentage of nitrogen, because it … 9 Real Life Examples Of Normal Distribution The normal distribution is widely used in understanding distributions of factors in the population. If we are given two functions, it is possible to create or generate a “new” function by composing one into the other. Mass percent composition is also known percent by weight. Then by taking f(g(x)), where x is the date, one gets a real life composite function. Have a look at the food labels on something like juice or milk and see where it says "Percentage of recommended daily intake" for a certain portion or serving size. Example 1: Write each percent as a decimal: %, 1.8%, 0.32%, 235%. Solution. Marco ate 0.45 of his birthday cake. In this lesson, we are solving real world problems related to conversion between percentages and decimals. It is abbreviated as w/w%. Example 1 Lizzy bought some fabric that was 1.75 meters long. For example, if a city has a population of one lakh, and the death rate in car accidents is 500. Any higher and we could We explain Percent of Change Real World Problems with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. The percent value can also be found by multiplying first instead of later, so in this example, the 50 would be multiplied by 100 to give 5,000, and this result would be divided by 1250 to give 4%. First one’s life insurance cost goes up depending on the age of the buyer, that in turn is dependent on the current date. This is a simple example where f(x) = the cost and x is the age, and g(x) = the age of the man dependent on the date: x. Percent Composition Time for you to try!!! For example, the … The next two sections describe percent composition as it relates to you. 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You will places tires on all of the cars and then when all of the cars have tires, if there are excess tires, then the cars are the limiting … Winning or losing a lottery is one of the most interesting examples … Write this figure as a percentage. -- Created using PowToon -- Free sign up at http://www.powtoon.com/youtube/ -- Create animated videos and animated presentations for free. For example, to find 50 apples as a percentage of 1250 apples, one first computes the ratio 50 / 1250 = 0.04, and then multiplies by 100 to obtain 4%. Chris K. Lv 4. Answer Save. There's percentage of it that is an alloy. So, the chance of being killed in a crash is 500/1 lakh is 0.05%. Leah went 7/8 at the plate in last night's baseball game. As a real life percentage example, here is a message I received and which invited me to buy books at an online reseller. $\begingroup$ There are so many real-world examples of this it's hard to pick one. It often shows it in percentage. We just finished our stoichiometry unit in my high school chem class and I need to know atleast two different practical uses of percent composition. $\endgroup$ – 6005 Mar 18 '14 at 15:45 Aggregation : - Teacher - Department. Uses of percentage in our daily life Example one Example 5 example 4 Smartphone battery: You must have noticed the icon on the top of your smartphone screen, the one that shows battery. 7. This unit is intended to scaffold students in the thinking process of solving word problems that involve percents. Take, for example, savants - people who have mental abilities that could only be characterized as superhuman (like having photographic memory, playing music perfectly after hearing it just once, or doing complex mathematical calculations in one's head) but otherwise severely disabled … Percent composition plays an important role in everyday life. The grocery store is a good source of ratios in real life. 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Chapter 7: Inertia

# 7.5 Inertia Intro: Parallel Axis Theorem

There are two great uses for the parallel axis theorem:

1. Finding the inertia of a complex object with multiple parts.

2. Rotating an object about an axis other than through the center of mass (y’)

To begin with, the parallel axis theorem is equal to the inertia about the center of mass (Icm) plus the distance between the axes of rotation squared times the mass.

$$I=I_{cm}+md^2$$

#### Example 1:

For a disk, the distance between axes y and y’ is d and the $I_{cm} = \frac{1}{2}mr^2$.

$$I=I_{cm}+md^2 = \frac{1}{2}mr^2+mr^2 = \frac{3}{2}mr^2$$

You will need the table of common geometric shapes in the previous section to find the Icm for each object.

• I is the moment of inertia of an object with respect to an axis from which the center of
mass of the object is a distance d.
• ICM is the moment of inertia of the object with respect to an axis that is parallel to the first axis and passes through the center of mass.
• m is the mass of the object.
• d is the distance between the two axes.

The parallel axis theorem relates the moment of inertia ICM of an object, with respect to an axis through the center of mass of the object, to the moment of inertia I of the same object, with respect to an axis that is parallel to the axis through the center of mass and is at a distance d from the axis through the center of mass.

A conceptual statement made by the parallel axis theorem is one that you probably could have arrived at by means of common sense, namely that the moment of inertia of an object with respect to an axis through the center of mass is smaller than the moment of inertia about any axis parallel to that one. As you know, the closer the mass is “packed” to the axis of rotation, the smaller the moment of inertia; and; for a given object, per definition of the center of mass, the mass is packed most closely to the axis of rotation when the axis of rotation passes through the center of mass. The PAT is visually shown below, as z represents the axis on which the objects COM rotates about, z’ is the axis it is now going to rotate about, and d is the distance between these two axis’.

Source: Calculus-Based Physics 1, Jeffery W. Schnick.   https://openlibrary.ecampusontario.ca/catalogue/item/?id=ce74a181-ccde-491c-848d-05489ed182e7

Steps for finding the MMOI of an object

1. Determine the shape of the object (or shapes, if it a composite object).
2. Determine which axis the object is rotating about.
3. Find the center of mass for each individual shape.
4. Find the ICM  (inertia about its center of mass) for each shape.
5. Determine the distance from the CM of the shapes to the axis of rotation.
6. Use the Parallel Axis Theorem to find the inertia for each shape.
7. Add up all these individuals inertias to find IT.

#### Example 2:

Find the moment of inertia of a uniform rod with ICM=0.05kgm2, L=0.08m, and mass=0.250kg, with respect to an axis that is perpendicular to the rod and passes through at 1/4 of the length of the rod.

We know the distance (d) to be L/4 = 0.08m / 4 = 0.02m away from the z axis. Here we present the solution to the problem:

I=ICM + md2

I=0.05 kgm2 + (0.250kg)(0.02m)2

I=0.0501 kgm2

#### Example 3

A dumbbell consists of two .2 meter diameter spheres, each with a mass of 40 kg, attached to the ends of a .6 meter long, 20 kg slender rod. Determine the mass moment of inertia of the dumbbell about the y axis shown in the diagram.

Organize the known and unknown data in a table to complete as you go:

Find the center of mass:

• For each sphere, the rcm is 0.3m + 1/2 radius = 0.3m + 1/2 (0.2m) = 0.4m
• For the bar, the rcm is at 0.

Find the inertia about the center of mass for each shape separately. Use the rectangle equation:

For the sphere:

$$I_{cm-sph}=\frac{2}{5}mr^2 =\frac{2}{5}*(40kg)*(0.1m)^2 \\ \qquad \quad =0.16kgm^2$$

For the rod,

$$I_{cm-rod}=\frac{1}{12}ml^2 = \frac{1}{12}*(20kg)*(0.6m)^2\\\qquad \quad = 0.6 kgm^2$$

Next the parallel axis theorum is needed to change the axis of rotation from the cm of the sphere to the system cm. The distance between axes of rotation is 0.3m + 1/2 radius = 0.3m + 1/2 (0.2m) = 0.4m

$$I_{o-sph}=I_{cm-sph}+md^2 = 0.16 kgm^2 + (40kg)*(0.4m)^2 \\\qquad \quad = 6.56kgm^2$$

Finally, add up the parts: the 2 spheres and the cm of the rod:

$$I_{total} = 2 * I_{o-sph}+I_{cm-rod}\\\qquad \quad = 2*(6.56 kgm^2) + (0.6 kgm^2)\\\qquad \underline{I_{total} = 13.72 kgm^2}$$

Key Takeaways

Basically: The parallel axis theorem helps you to find the inertia about a different axis of rotation than the cg, and it lets you combine multiple objects.

Application: Find the total inertia of shape with multiple objects.

Looking Ahead: This will be used throughout dynamics.


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# Confindence Intervals for Ordinal Variable

I count how many days within a month people have executed a certain activity. The maximum is 30 (i.e., the person executed the activity each day), the minimum is 0 (i.e., the person never executed the activity). I get the following distribution:

I want to calculate the 95% confidence intervals for three proportions, namely for low (0-9), medium (10-19), and high (20-30) activity. Total n = 683, the first group (low) contains 311 cases. From this question I understand that the binomial proportion confidence intervals apply and:

$$\hat p \pm 1.96 \sqrt{\frac{\hat p (1 - \hat p)}{n}}$$

However, my data seems to be skewed towards the edges, especially having excessive zeros. My question: Is the described approach still valid for calculating 95% CI in my scenario?

• Assuming this is a sample from a much larger population, you can simply put $\hat p =\frac{311}{683}$ onto your expression. The uncertainty is related to who you sampled rather the pattern of answers. Nov 25 '21 at 23:24