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# Limits: What Does “Approach” Mean?

#### (A new question of the week)

We’ve looked at the concept of limit of a function from several perspectives, including why they are needed, and what the definition means. Here we have a more fundamental question, which applies to both functions and sequences: What do we mean when we say a value approaches some number? More broadly, we’ll see how mathematicians have to rein in language by refining their definitions, and replacing words with symbols. This will be a long one, as we seek a helpful answer.

## Approach = “closer and closer” = “as close as we want”?

Anuj sent this question in mid-July:

Why approaching a value means that we can get arbitrarily close to that value?

Given two sets:

x = {3, 2.5, 2.04, 2.03, 2.02, 2.001, 2.0001, …}

y = {4, 6.25, 5.76, 4.25, 4.025, 4.001, 4.00001, .…}

As terms of the set x get closer and closer to 2, terms of the set y seem to be getting closer and closer to 4. But terms of set x also seem to be getting closer and closer to “values < 4 like 3.999, 3.8999, etc. Then why do we say the 4 is the value the terms of set y seems to be getting closer and closer to rather then some value <4?

And also, why do we need to check for arbitrary number of terms to prove the existence of a value the terms of a set are getting closer and closer to, rather than just looking at a bounded (finite) number of terms of set? Why can’t we prove the existence of a value the sequence is getting closer and closer just by checking finite number of terms of a sequence rather than looking at arbitrary number of terms of a sequence?

So, my question is that why approaching a value means that we can get arbitrarily close to that value, i.e., why getting closer and closer to a value means that we can get as close as we want to that value?

In book A Mathematical Bridge An Intuitive Journey in Higher Mathematics by Stephen Hewson

Aristotle: I see: all the terms in the sequence beyond a certain point are closer to ½ than any number I can specify. If I were to say that the limit were anything other than exactly one half, then eventually all of the terms beyond some point in your sequence would get closer to one half than they would to my other supposed limit. I thus concede the point: the sequence must tend to the limit of ½.

Here he is saying about the fact that a sequence gets closer and closer to many values until a certain point but diverges after this, it can’t be considered a limit value, but it gets closer and closer to ½ after any arbitrary point: that’s why it is the limit value. Can you explain to me why we must get closer and closer after any arbitrary point to consider it its limit value?

I also want to know what does approaching means and how epsilon and delta explain this idea of approaching a value?

At one level, the initial question could be just about language: “Approach” can mean merely “moving in the direction of”, so that his sequence (inaccurately called a “set’) of y “approaches” not only 4, but any number less than 4 (as one might “approach” a throne without going all the way up to it); but in this context we take it to mean “coming very close to” one specific number. Mathematics chooses a particular meaning of a word and makes it precise; in particular, we don’t really use “approach” as a technical term in itself, but use it only to informally explain what the formal definition (with epsilons and deltas) means.

Do, why do we want “approach” to mean this, and how does the formal definition accomplish it?

### Distance, limits, and approaching

Doctor Fenton answered, starting at a high level, likely because Anuj had mentioned “real analysis” in his title:

Hi Anuj,

Those are very good questions, and they lead to some very deep concepts.

As for what it means to “approach” something, we have the idea of distance on the number line (and more generally in sets such as the Euclidean plane or space). A distance (called a metric in math) is a function which assigns a unique non-negative number between each pair of elements of the set, with two requirements: (1) the positivity that I mentioned, i.e. if d is our metric, d(x,y) ≥ 0 for all x and y in the set; if x and y are not equal, then d(x,y) > 0, so if d(x,y) = 0,then x = y; and for any three elements x, y, and z,

d(x,y) ≤ (dx,z) + d(z,y) (the triangle inequality).

For the real numbers, the usual metric or distance between two numbers a and b is d(a,b) = |a-b|, the absolute value of the difference between a and b.

This is a formal definition of what a “distance” means; “approaching” says something about distance. But what, exactly?

You are talking about the mathematical idea of limits, and there are two different concepts of limits in math: limits of sequences; and limits of functions. These are closely related, but very different in some ways. Mathematically, a sequence is a function from the positive integers to the real numbers, while functions can have other domains. For simplicity, let’s stick to functions defined on a subset of the real numbers whose values are also real numbers.

Since positive integers are a subset of the real numbers, this restriction includes sequences; we will not go beyond real numbers here, though one could.

Now, to say that the numbers in a sequence, {an}, are “approaching” a number A means that the distance between an and A, |an-A|, gets close to 0, as the index n gets larger. (We also have to use the order property of real numbers, which allows us to say that for any two different numbers a and b, one number is less than the other: either a < b, or b < a.)

For sequences, we want to say that an approaches A if |an-A| approaches 0 as n approaches ∞, but “n approaches ∞” is a different kind of “approaching”. n approaches ∞ doesn’t mean that n keeps getting closer to another number, it just means that n keeps getting larger and larger, so that there there is no real number which is never exceeded.

Infinity is tricky, so “approaching infinity” needs its own definition!

### Different ways of approaching

We still have some clarifying to do! What does “gets close to” mean? It’s not as simple as you might think:

For the idea of a sequence of numbers an approaching a finite value A, the method of approaching can be quite complicated. an could steadily get closer to A, as n increases, such as the sequence an = 1-(1/(n+1)): a1 = 1/2, a2 = 2/3, a3 = 3/4, a4 = 4/5, and so on. Each number is steadily closer to A = 1: |an+1-A| < |an-A|. But we can also create sequences which for one index n get closer to A, but for a larger index may get further away, such as the sequence 2/3, 1/2, 4/5, 3/4, 6/7, 5/6, 8/9, 7/8, 10/11, and so on. The subsequence of odd elements of the sequence, 2/3, 4/5, 6/7, 8/9, … increase steadily towards 1, and so do the even elements 1/2, 3/4, 5/6, 7/8,…, but the whole sequence does not approach 1 steadily.

Here is what the first sequence lookw like:

Here each term is closer than the one before.

Here is the second sequence:

This alternates between two increasing sequences, though it could be expressed with a single formula, namely $$\frac{n-(-1)^n}{n-(-1)^n+1}.$$ The terms go back and forth, moving alternately closer and further, yet overall they approach 1. We’ll need a way to define “approach” that fits that!

We could also have a sequence that approaches 1 from both sides:

Here the distance from 1 decreases (or stays the same) at each step, though the direction alternates.

But all three of these approach 1 as n approaches infinity. So our understanding of “approaching” has to work for all of them.

### Asteroids, lightning, and epsilon: Standards of “closeness”

Then, there is the idea of “closeness”, when we say that an is getting close to A as n increases. “Close” is a vague concept, and its meaning depends upon the context. We describe asteroids that pass closer to the earth than the moon as being a “close approach”. The moon is about 400,000 km from earth, so an asteroid passing within 300,000 km of earth is “close”. But if lightning is striking 100 km from your house, you probably don’t consider that “close”, but if it is less than 1 km, you are probably concerned. To have meaning in a given context, we need a standard of “closeness”. In mathematics, that (arbitrary) standard of closeness is denoted ε (epsilon). Now we can say with precision whether an is close to A or not: an is close to A if |an-A| < ε, otherwise it is not close.

We need to be getting close, not just closer. Since “close” is a relative concept, it has no universal meaning; rather, whatever one chooses to call “close enough” is considered close for the moment. That is what epsilon does.

To get to the main point, saying that lim(n→∞) an = A is saying that if n is large enough, then an will be close to A. To make sense of this, we need to choose a standard of closeness to A, ε, and a standard of largeness for n, N, so that the statement “if n is large enough, then an will be close to A” can be given a precise meaning. Given ε, there is some N such that if “n is large enough“, i.e. n > N, then an will be “close” to A, |an-A| < ε.

For a function limit, lim(x→a) f(x)=L, we want to say that if x is close to a, then f(x) will be close to L. That means that for any given standard of closeness to L, ε, there will be a standard of closeness to a, δ, so that if |x-a| < δ, then |f(x)-L| < ε.

Does this help?

The sequence must not only get closer; it must get as close as you want – that is, there is no limit to how close it gets. (No pun intended. Well, maybe.)

This is a brief explanation of the epsilon-delta formulation of limits, which is explored in more detail at Epsilons, Deltas, and Limits — Oh, My!

## Getting closer

Anuj wasn’t satisfied yet, and restated his question:

Sorry but, I do not understand your answer. I meant to ask as stated above about what is approaching, is it just something getting closer to something until a certain point and how epsilon and delta formalize this notion of “approaching” something arbitrarily as limit and also is limit just a special case of approaching, where limit is just something toward which approach arbitrarily close.

Doctor Fenton tried again:

The idea of approaching in math is the same as in daily life. You get closer to something. If you are driving from one town to another, you will approach the destination, the distance between you and the destination is decreasing.

Note that this wording suggests always decreasing, which is not necessarily true in talking about limits. Rather, we mean that eventually the distance gets closer, and stays there.

When this is used in math, we are usually talking about an input/output relationship. There is some input which produces an output. You control the input, so you make the input approach something. For a sequence, you choose the index n, so you choose how far out in the sequence to go. For a function, you want the input to approach a number a, so again, you choose how to approach a. The value of x is what you choose, so you can choose how to approach a. If you choose to make x go away from a, then x won’t be approaching a. You can think of a point sliding along the graph of the function, getting closer and closer to the desired value a. You control the value of x, so you can make it approach a or not approach a.

The claim that

lim(x→a) f(x) = L

claims that the “closeness” of x to a causes the closeness of f(x) to the limiting value of L.

So we can move the input (x or n) toward the goal (a or ∞), and then look at how the output ($$f(x)$$ or $$a_n$$) approaches the expected limit, which we don’t have direct control over.

### The output inspector: Who gets to decide what is “close”?

An example I like to use is to suppose that you are sitting at a computer where you can type inputs, and the computer displays an output. Other people will give you a number, and you can choose to type the number into the computer or not. You control the input. Suppose that the computer will not allow you to input the exact value 3, but only numbers close to 3. An output inspector looks at the output to decide if it is “acceptable”. You are claiming that if the input is close to 3, then the output will be close to 9. There are two standards of closeness being used: the closeness of the given number x to the desired input 3, which will be denoted by δ, and which you will use to decide whether to type the number into the computer, and the standard of closeness the output inspector will use, denoted by ε, to decide whether its output is acceptable. Can you guarantee that an input you allow will always produce an acceptable output?

If you tell the output inspector what input standard you are using, and if the inspector wants you to fail, then she can always guarantee that you will input a value that will produce unacceptable output. The inspector just needs to find a value which meets the input standard whose output does not meet the output standard.

This idea of an antagonist trying to make you fail is useful in thinking about proofs. To show that you can always succeed (something is always true) you need to show that an enemy can’t force you to fail. Here, if the enemy knew your standard of closeness, then she could choose an output standard by which some inputs you would permit, would not work.

For example:

Suppose the output inspector knows that the computer produces the output by doubling the input and adding 3. That is, if x is the input, then the output will be 2x+3. If you are given the number x, then you will determine the “input error” |x-3| and compare it to δ. If |x-3| < δ, you type it in. The “output error” will be |(2x+3)-9| = |2x-6| = 2|x-3|. That is, the output error will always be twice the input error. If you tell the output inspector that δ=0.1, then she knows that any input x between 2.9 and 3.1 will be allowed to go into the computer. But if 2.9 < x< 3.1, then 8.8 < 2x+3 < 9.2. If the inspector chooses ε = 0.1, then if the input error is larger than 0.05, the output error will be larger than 0.1, so the output will not be acceptable. For example, if x = 3.06, |x-3| = |3.06-3| = 0.06 < 0.1, so x = 3.06 passes the input standard, but the output 2(3.06)+3 = 9.12, which is unacceptable because |9.12-9| = 0.12 > 0.1.

So the decision about what counts as close must be first made by the inspector (closeness of the output); we can’t let the inspector choose epsilon knowing our delta!

So this is why epsilon comes first in our definition of a limit:

On the other hand, if the output inspector tells you ε, then since the output error is twice the input error, if you choose your input standard δ so that 2δ ≤ ε, then any input x meeting |x-3| < δ will produce an output whose error is |(2x+3)-9| = 2|x-3| < 2δ ≤ ε .

That is, no matter what output standard the inspector chooses, you can ALWAYS find an input standard δ which will ALWAYS produce acceptable output.

This is the key to the epsilon-delta definition of a limit, though this is an unusually simple example.

You always have to consider relationships where you have control of the input, so the question is whether your ability to control the input will allow you to control the output. The function f(x) whose output is 1 when x is irrational and 0 when x is rational, at any point a, if ε < 1 (e.g. ε=0.1) , then no matter what restriction you place on x, |x-a| < δ, i.e. x is in the interval a-δ<x<a+δ, there will be points in (a-δ,a+δ) where f(x) = 0 and points where f(x)=1, so f cannot approach any fixed value L.

Does this help?

That is an example where the limit does not exist, because the output varies too wildly.

## From dynamic image to arbitrary choice

I think I understand but just to clarify: approaching means getting closer to something until a certain point and limit is just a special way of approaching where the value we approach arbitrarily close is the limit and I read that this is how limits were discovered, that is the limit is defined by the value we get arbitrarily close to because that is something that will help find derivative at a point. Is this the correct interpretation of limit as the value we approach until a certain point and that is arbitrarily close, is that it?

Approaching arbitrarily closely is indeed to key idea: “Approaching” here doesn’t just mean going in that direction, but getting as close as the “output inspector” chooses to require.

Doctor Fenton replied:

I don’t understand what you mean when you say “approaching means getting closer to something until a certain point“. What does “until a certain point” mean?

There are two approaches occurring: in the limit of a sequence an, the index n “approaches ∞” while the numbers an approach A. That is a “dynamic” description: we think of either the index moving along a list of the positive integers in the sequence case, while the an move along the real axis, or x moving along the x-axis and y moving along the y-axis in the function case. (n approaching ∞ doesn’t mean that n is approaching any specific number, it just means that n steadily gets larger without bound). That is, no matter how large a positive integer N we choose, the index n will get larger than N, n > N, and stay larger than (once n is larger than N, all indices larger than n will also be larger than N).

Here is our sequence $$b_n$$, seen as moving to the right (n “approaching” infinity) and, irregularly, up toward the value $$A=1$$:

Here, $$\epsilon=0.22$$, and we see that for $$n>4$$, $$|b_n-1|<\epsilon$$. But it isn’t moving constantly up toward 1; in fact, it moved inside the green region, then back outside before staying in.

The same sort of thing would be happening with a function, as x approaches a. It might wiggle, but $$f(x)$$ will always be coming within whatever bounds you choose.

It’s hard to describe this dynamic motion directly, and that is what the “arbitrarily” close is meant to capture. Instead of describing some type of moving, we use the arbitrariness of the standard of closeness instead. You can think of “approaching” as describing having a sequence of such standards, each more restrictive. For example, the first standard will be ε1, the second standard ε2, and so on, with each standard smaller than the previous one: ε1 > ε2 > ε3 > ε4 > … (and this sequence approaches 0). To say that an approaches a limit A means that for each standard εk, there will be an index Nk such that for all indices n such that n > Nk, an will be in the interval (A-εk,A+εk). That is, n = Nk may not be the first time an gets inside the interval (A-εk,A+εk): it could have “two steps forward, one step back” behavior, moving closer to A, but then backing away, but there will be an index Nk which will be the last time an is outside the interval. After that, all the an will be inside the interval. Then we reduce the size of ε to ε(k+1) and find Nk+1 and repeat that process. That is what is meant by “approaching (arbitrarily) close”. We don’t have to talk about something “moving”.

Note how the idea of “approaching” had to be refined in order to define it clearly, and in doing so it has lost the literal idea of motion.

This idea applies to relationships: for a sequence, the relationship between the size of the index n and the size of the sequence value an; for functions, the size of x and the size of y=f(x).

Anuj wrote back,

OK, I understand now.


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## Bulletin of the Belgian Mathematical Society - Simon Stevin

### On secant spaces to Enriques surfaces

Andreas Leopold Knutsen

#### Abstract

We relate the minimal gonality of smooth curves in a complete, ample, base point free linear system $|L|$ on an Enriques surface to the existence of certain secant spaces on the image of the surface mapped by the adjoint system. We also explicitly compute the minimal gonality in terms of invariants of the line bundle $L$. In particular, we obtain a precise criterion for the variation of the gonality of the curves.

#### Article information

Source
Bull. Belg. Math. Soc. Simon Stevin, Volume 16, Number 5 (2009), 907-931.

Dates
First available in Project Euclid: 9 December 2009

https://projecteuclid.org/euclid.bbms/1260369406

Digital Object Identifier
doi:10.36045/bbms/1260369406

Mathematical Reviews number (MathSciNet)
MR2574369

Zentralblatt MATH identifier
1204.14014

#### Citation

Knutsen, Andreas Leopold. On secant spaces to Enriques surfaces. Bull. Belg. Math. Soc. Simon Stevin 16 (2009), no. 5, 907--931. doi:10.36045/bbms/1260369406. https://projecteuclid.org/euclid.bbms/1260369406

format: plain


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# Math Help - Strange Integration Problem

1. ## Strange Integration Problem

Hi, I'm doing a maths assignment and for part of I have to integrate a function that this segment is from.

$1/10x^2$

What is the procedure for doing that? Neither my textbook of tutoring sites have had examples that seem to be applicable for this function.

2. Originally Posted by Schermy
Hi, I'm doing a maths assignment and for part of I have to integrate a function that this segment is from.

$1/10x^2$

What is the procedure for doing that? Neither my textbook of tutoring sites have had examples that seem to be applicable for this function.
please clarify, and use parentheses in the future.

do you mean $\frac 1{10}x^2$ or $\frac 1{10x^2}$ ?

the first case is trivial, just apply the power rule: $\int x^n~dx = \frac {x^{n + 1}}{n + 1} + C$ for $n \ne -1$

the second case is also trivial, if you rewrite it. write it as $\frac 1{10}x^{-2}$ and apply the power rule

3. I'm kind of confused by your question? Do you want to integrate ? If so the answer is is -1/(10x). Oops jhevon beat me to it.

4. It was the former case and thanx very much for the answer, especially so promptly.

5. Originally Posted by Schermy
It was the former case and thanx very much for the answer, especially so promptly.

6. $\frac 1{10}x^3$ ???

7. Originally Posted by Schermy
$\frac 1{10}x^3$ ???
nope

again, the rule is: $\int x^n~dx = \frac {x^{n + 1}}{{\color{red}n + 1}} + C$ for $n \ne -1$

8. Try again it should be be $\frac{x^3}{30} + C$ do you see what you did wrong? oops I was to slow again.

9. don't forget the + C guys

sorry for being so fast 11rdc11

10. $
\frac{1}{10} \frac {x^{3}}{{3}} + C
$

P.S. Still learning the maths markup

11. Originally Posted by Schermy
$
\frac {x^3}{3} + C
$
don't forget the 1/10 that was there before

12. Originally Posted by Jhevon
don't forget the + C guys

sorry for being so fast 11rdc11

Lol thank you for reminding me to include my constant


text:
## The Annals of Probability

### Note on the $k$-Dimensional Jensen Inequality

Martin Schaefer

#### Abstract

Let $f$ be a measurable convex function from $R^k$ to $R^1$ and let $X_1, \cdots, X_k$ be real-valued integrable random variables. The best approximation for $f(EX_1, \cdots, EX_k)$ one can get by Jensen's inequality is $f(EX_1, \cdots, EX_k) \leqq \inf Ef(\mathbf{Z})$ where the infimum is taken over all $k-\dim$. random vectors $\mathbf{Z} = (Z_1, \cdots, Z_k)'$ such that $Z_i$ has the same distribution as $X_i (1 \leqq i \leqq k)$. An application is given in the case where $f(y)$ is the span of the vector $y$ which leads to a new approximation for $f(A\mathbf{u})$ where $A$ is a stochastic $(k \times m)$-matrix and $\mathbf{u}$ is an arbitrary element of $R^m$.

#### Article information

Source
Ann. Probab., Volume 4, Number 3 (1976), 502-504.

Dates
First available in Project Euclid: 19 April 2007

https://projecteuclid.org/euclid.aop/1176996102

Digital Object Identifier
doi:10.1214/aop/1176996102

Mathematical Reviews number (MathSciNet)
MR400339

Zentralblatt MATH identifier
0342.26019

JSTOR
Schaefer, Martin. Note on the $k$-Dimensional Jensen Inequality. Ann. Probab. 4 (1976), no. 3, 502--504. doi:10.1214/aop/1176996102. https://projecteuclid.org/euclid.aop/1176996102


text:
×
Back to all chapters

# Implicit Differentiation

Not all equations can be written like y = f(x), so taking the derivative can be tricky. Save the mess and do it directly with implicit differentiation.

# Implicit Differentiation - Radical Functions

If $$\sqrt[3]{x^2}+\sqrt[3]{y^2}=4,$$ what is $$\displaystyle \frac{dy}{dx}?$$

If $$x=y\sqrt{11+y}$$, what is $$\displaystyle \frac{dy}{dx}?$$

Find $$\displaystyle \frac{dy}{dx}$$ for $$7\sqrt{x}+6\sqrt{y}=13y^2$$ at the point $$(1,1).$$

Given $f(x)=\sqrt[3]{\frac{x}{x^2+124}},$ what is the value of $$f'(1)?$$

If $$\sqrt{3x}+\sqrt{y}=\sqrt{5}$$, what is $$\displaystyle \frac{dy}{dx}?$$

×


text:
Table 1.

Univariate analysis of variance (ANOVA) results testing the effect of sex on several morphological traits of chela size for Cherax dispar

VariableSex F1,66P
Wrist depth 7.80 0.007*
Maximum depth 7.96 0.006*
Chela length 6.45 0.013
Propodus depth 4.69 0.034
Dactylus depth 5.11 0.027
Propodus length 8.52 0.0048*
Dactylus length 7.73 0.007*
VariableSex F1,66P
Wrist depth 7.80 0.007*
Maximum depth 7.96 0.006*
Chela length 6.45 0.013
Propodus depth 4.69 0.034
Dactylus depth 5.11 0.027
Propodus length 8.52 0.0048*
Dactylus length 7.73 0.007*

Analyses were performed on the relative size of morphological traits to chela size that was calculated as the first component derived from a principal component analysis (PCA) on the seven morphological variables (see Materials and methods for details). *Significant difference among groups taken after Bonferroni adjustment across the seven tests

Close Modal


text:
Problem of the Week

Updated at Mar 8, 2021 3:19 PM

To get more practice in equation, we brought you this problem of the week:

How would you solve the equation $$(\frac{u}{5}-5)\times \frac{5}{3-u}=21$$?

Check out the solution below!

$(\frac{u}{5}-5)\times \frac{5}{3-u}=21$

 1 Expand.$\frac{u}{3-u}-\frac{25}{3-u}=21$2 Join the denominators.$\frac{u-25}{3-u}=21$3 Multiply both sides by $$3-u$$.$u-25=21(3-u)$4 Expand.$u-25=63-21u$5 Add $$25$$ to both sides.$u=63-21u+25$6 Simplify  $$63-21u+25$$  to  $$-21u+88$$.$u=-21u+88$7 Add $$21u$$ to both sides.$u+21u=88$8 Simplify  $$u+21u$$  to  $$22u$$.$22u=88$9 Divide both sides by $$22$$.$u=\frac{88}{22}$10 Simplify  $$\frac{88}{22}$$  to  $$4$$.$u=4$Doneu=4


text:
# Wald's Sequential Probability Ratio Test¶

## Sequential test: draw data sequentially¶

Sequence of data $X_1, X_2, \ldots$.

Two hypotheses, $H_0$ and $H_1$, each of which completely specifies the joint distribution of the data.

Assume that the joint distributions under $H_0$ and $H_1$ are absolutely continuous with respect to each other, relative to some underlying measure.

Let $f_{0m}$ be the likelihood of $H_0$ for data $(X_j)_{j=1}^m$ and let $f_{1m}$ be the likelihood of $H_1$ for data $(X_j)_{j=1}^m$.

The likelihood ratio of $H_1$ to $H_0$ is $f_{1m}/f_{0m}$; this is (loosely speaking) the probability of observing $X_1, \ldots, X_m$ if $H_1$ is true, divided by the probability of observing $X_1, \ldots, X_m$ if $H_0$ is true.

The probability of observing the data actually observed will tend to be higher for whichever hypothesis is in fact true, so this likelihood ratio will tend to be greater than $1$ if $H_1$ is true, and will tend to be less than $1$ if $H_0$ is true. The more observations we make, the more probable it is that the resulting likelihood ratio will be small if $H_0$ is true. Wald (1945) showed that if $H_0$ is true, then the probability is at most $\alpha$ that the likelihood ratio is ever greater than $1/\alpha$, no matter how many observations are made. More generally, we have:

Wald's SPRT

For any $\alpha \in (0, 1)$ and $\beta \in [0, 1)$, the following sequential algorithm tests the hypothesis $H_0$ at level no larger than $\alpha$ and with power at least $1-\beta$ against the alternative $H_1$.

Set $m=0$.

1. Increment $m$
2. If $\frac{f_{1m}}{f_{0m}} \ge \frac{1-\beta}{\alpha}$, stop and reject $H_0$.
3. If $\frac{f_{1m}}{f_{0m}} \le \frac{\beta}{1-\alpha}$, stop and do not reject $H_0$.
4. If $\frac{\beta}{1-\alpha} < \frac{f_{1m}}{f_{0m}} < \frac{1-\beta}{\alpha}$, go to step 1.

## SPRT miracle¶

Don't need to know the distribution of the likelihood ratio $\mbox{LR}=\frac{f_{1m}}{f_{0m}}$ under the null hypothesis to find the critical values for the test.

## Connection to Gambler's ruin¶

For iid data, the likelihood ratio is a product of terms. On a log scale, it's a sum. Each "trial" produces another term in the sum, positive or negative. But—unlike the classical Gambler's Ruin problem in which the game is fair—the terms are not equal in magnitude: the steps are not all the same size.

## Likelihood ratio for two values of $p$ in iid Bernoulli trials¶

Suppose $X_1, X_2, \ldots,$ are iid $\mbox{Bernoulli}(p)$ random variables and let $p_1 > p_0$.

Set $\mbox{LR} \leftarrow 1$ and $j \leftarrow 0$.

• Increment $j$
• If $X_j = 1$, $\mbox{LR} \leftarrow \mbox{LR} \times p_1/p_0$.
• If $X_j = 0$, $\mbox{LR} \leftarrow \mbox{LR} \times (1-p_1)/(1- p_0)$.

What's $\mbox{LR}$ at stage $m$? Let $T_m \equiv \sum_{j=1}^m X_j$. $$\frac{p_{1m}}{p_{0m}} \equiv \frac{p_1^{T_m}(1-p_1)^{m-T_m}}{p_0^{T_m}(1-p_0)^{m-T_m}}.$$ This is the ratio of binomial probability when $p = p_1$ to binomial probability when $p = p_0$ (the binomial coefficients in the numerator and denominator cancel).

Wald's SPRT for $p$ in iid Bernoulli trials

Conclude $p > p_0$ if $$\frac{p_{1m}}{p_{0m}} \ge \frac{1-\beta}{\alpha}.$$ Conclude $p \le p_0$ if $$\frac{p_{1m}}{p_{0m}} \le \frac{\beta}{1-\alpha}.$$ Otherwise, draw again.

The SPRT approximately minimizes the expected sample size when $p \le p_0$ or $p > p_1$. For values in $(p_1, p_0)$, it can have larger sample sizes than fixed-sample-size tests.

## Simulation of SPRT for Binomial¶

Let's watch the SPRT in action for iid Bernoulli trials.

In [1]:
# This is the first cell with code: set up the Python environment
%matplotlib inline
from __future__ import print_function, division
import matplotlib.pyplot as plt
import math
import numpy as np
import scipy as sp
import scipy.stats
from scipy.stats import binom
import pandas as pd
# For interactive widgets
from ipywidgets import interact, interactive, fixed
import ipywidgets as widgets
from IPython.display import clear_output, display, HTML

In [2]:
def plotBinomialSPRT(n, p, p0, p1, alpha, beta):
'''
Plots the progress of the SPRT for n iid Bernoulli trials with probabiity p
of success, for testing the hypothesis that p=p0 against the hypothesis p=p1
with significance level alpha and power 1-beta
'''
fig, ax = plt.subplots(nrows=2, ncols=1, sharex=True)
trials = sp.stats.binom.rvs(1, p, size=n+1)  # leave room to start at 1
terms = np.ones(n+1)
sfac = p1/p0
ffac = (1.0-p1)/(1.0-p0)
terms[trials == 1.0] = sfac
terms[trials == 0.0] = ffac
terms[0] = 1.0
logterms = np.log(terms)
#
ax[0].plot(range(n+1),np.cumprod(terms), color='b')
ax[0].axhline(y=(1-beta)/alpha, xmin=0, xmax=n, color='g', label=r'$(1-\beta)/\alpha$')
ax[0].axhline(y=beta/(1-alpha), xmin=0, xmax=n, color='r', label=r'$\beta/(1-\alpha)$')
ax[0].set_title('Simulation of Wald\'s SPRT')
ax[0].set_ylabel('LR')
ax[0].legend(loc='best')
#
ax[1].plot(range(n+1),np.cumsum(logterms), color='b', linestyle='--')
ax[1].axhline(y=math.log((1-beta)/alpha), xmin=0, xmax=n, color='g', label=r'$log((1-\beta)/\alpha)$')
ax[1].axhline(y=math.log(beta/(1-alpha)), xmin=0, xmax=n, color='r', label=r'$log(\beta/(1-\alpha))$')
ax[1].set_ylabel('log(LR)')
ax[1].set_xlabel('trials')
ax[1].legend(loc='best')
plt.show()

interact(plotBinomialSPRT, n=widgets.IntSlider(min=5, max=300, step=5, value=100),\
p=widgets.FloatSlider(min=0.001, max=1, step=0.01, value=.45),\
p0=widgets.FloatSlider(min=0.001, max=1, step=0.01, value=.5),\
p1=widgets.FloatSlider(min=0.001, max=1, step=0.01, value=.6),\
alpha=widgets.FloatSlider(min=0.001, max=1, step=0.01, value=.05),\
beta=widgets.FloatSlider(min=0.001, max=1, step=0.01, value=.05)
)

Out[2]:
<function __main__.plotBinomialSPRT>

## Behavior¶

For $p_0 < p_1$,

• when $p \ge p_1$, the likelihood ratio is likely to cross the upper (green) line eventually and unlikely to cross the lower (red) line.
• when $p \le p_0$, the likelihood ratio is likely to cross the lower (red) line eventually and unlikely to cross the upper (green) line.
• the SPRT approximately minimizes the expected number of trials before rejecting one or the other hypothesis, provided $p \notin (p_0, p_1)$.

For $p_1 < p_0$, the directions are reversed.

## It works for $\beta = 0$ too: $P$-values¶

The inequalities hold when $\beta = 0$ also. Then the likelihood ratio has chance at most $\alpha$ of ever being greater than $1/\alpha$, if in fact $p > p_0$.

Hence, $1/\mbox{LR}$ is the $P$-value of the hypothesis $p > p_0$. This can be used to construct one-sided confidence bounds for $p$ and other parameters. The next chapter does exactly that, to find a lower confidence bound for the mean of a nonnegative population.

## Dependent observations¶

As an illustration of SPRT in for obervations that are not iid, consider the following problem.

There is a population of $B$ items. Item $b$ has "weight" $N_b$ and "value" $a_b \in \{0, 1\}$. Let $N = \sum_{b=1}^B N_b$. (You might think of items as manufacturing lots, where $a_b = 1$ means the lot is acceptable and $a_b=0$ means the lot is defective.)

We want to test the hypothesis $H_0$ that $\frac{1}{N}\sum_b N_b a_b = 1/2$ against the alternative hypothesis $H_1$ that $\frac{1}{N}\sum_b N_b a_b = \gamma$, for some fixed $\gamma > 1/2$.

We will draw items sequentially, without replacement, such that the chance that item $b$ is selected in draw $i$, assuming it has not been selected already, is $N_b/\sum_{j \notin {\mathcal B_{i-1}}} N_j$, where ${\mathcal B_{i-1}}$ is the sample up to and including the $i-1$st draw, and ${\mathcal B_0} \equiv \emptyset$. That is, the chance of selecting an item is in proportion to its weight among the items that have not yet been selected.

Let $\mathbb B_i$ denote the item selected at random in such a manner in the $i$th draw.

The chance that the first draw ${\mathbb B_1}$ gives an item with value 1, i.e., $\Pr \{a_{\mathbb B_1} = 1\}$, is $\frac{1}{N}\sum_b N_b a_b$. Under $H_0$, this chance is $p_{01} = 1/2$; under $H_1$, this chance is $p_{11} = \gamma$.

Given the values of $\{a_{\mathbb B_k}\}_{k=1}^i$, the conditional probability that the $i$th draw gives an item with value 1 is

$$\Pr \{a_{\mathbb B_i} = 1 | {\mathcal B_{i-1}} \} = \frac{ \sum_{b \notin {\mathcal B_{i-1}}} N_b a_b}{\sum_{b \notin {\mathcal B_{i-1}}} N_b}.$$

Under $H_0$, this chance is

$$p_{0i} = \frac{N/2 - \sum_{b \in {\mathcal B_{i-1}}} N_b a_b}{N - \sum_{b \in {\mathcal B_{i-1}}} N_b}.$$

Under $H_1$, this chance is

$$p_{1i} = \frac{N \gamma - \sum_{b \in {\mathcal B_{i-1}}} N_b a_b}{N - \sum_{b \in {\mathcal B_{i-1}}} N_b}.$$

Let $X_k$ be the indicator of the event that the $k$th draw gives an item with value $1$, i.e., the indicator of the event $a_{\mathbb B_k} = 1$. The likelihood ratio for a given sequence $\{X_k\}_{k=1}^i$ is

$$\mbox{LR} = \frac{\prod_{k=1}^i p_{1k}^{X_k}(1-p_{1k})^{1-X_k}} {\prod_{k=1}^i p_{0k}^{X_k}(1-p_{0k})^{1-X_k}}.$$

This can be simplified: $p_{0k}$ and $p_{1k}$ have the same denominator, $N - \sum_{b \in {\mathcal B_{i-1}}} N_b$, and their numerators share a term. Define $N(k) \equiv \sum_{j = 1}^{k-1}N_b$ and $A(k) \equiv \sum_{j = 1}^{k-1}N_b a_b$. Then

$$\mbox{LR} = \prod_{k=1}^i \left ( \frac{N\gamma - A(k)}{N/2 - A(k)} \right )^{X_k} \left ( \frac{N-N\gamma - (N(k) - A(k))}{N-N/2 - (N(k)-A(k))} \right )^{1-X_k}$$$$= \prod_{k=1}^i \left ( \frac{N\gamma - A(k)}{N/2 - A(k)} \right )^{X_k} \left ( \frac{N(1-\gamma) - N(k) + A(k)}{N/2 - N(k) + A(k)} \right )^{1-X_k},$$

where the products are defined to be infinite if the denominator vanishes anywhere.

If $H_0$ is true, the chance that $\mbox{LR}$ is ever greater than $1/\alpha$ is at most $\alpha$.

In [ ]:


text:
# Math Help - limits

1. ## limits

can somebody please explain this problem to me:

lim [ (x(e^(-2x+1))) / ((x^2)+x)]
x-->0

i get 0/0 = undefined
but my teacher tells me that it is not correct.

thank you

2. Nevermind.

3. Originally Posted by rahul14
lim [ (x(e^(-2x+1))) / ((x^2)+x)]
x-->0
$\lim _{x \to 0} \frac{xe^{-2x+1}}{x^2+x}=\lim _{x \to 0}\frac{e^{-2x+1}}{x+1}\to e$

4. Originally Posted by rahul14
can somebody please explain this problem to me:

lim [ (x(e^(-2x+1))) / ((x^2)+x)]
x-->0

i get 0/0 = undefined
but my teacher tells me that it is not correct.

thank you
do you mean
$\lim_{x \to 0} \frac {x.e^{-2x+1}}{x^2+x}$

if yes , then
$\lim_{x \to 0} \frac {x.e^{-2x+1}}{x^2+x}=\lim_{x \to 0} \frac {x.e^{-2x+1}}{x(x+1)}$

$= \lim_{x \to 0} \frac {e^{-2x+1}}{x+1}= \frac {e^{0+1}}{0+1} =e$

5. Originally Posted by Defunkt
Yes, $\frac{0}{0}$ is correct. Have you learned l'Hôpital's rule?
No, 0/0 is NOT correct for the limit! It's not even a number. And I surely would not recommend "L'Hopital's rule" for something as simple as this.

$\frac{xe^{-2x+1}}{x^2+ x}= \frac{xe^{-2x+1}}{x(x+1)}$
which is equal to $\frac{e^{-2x+1}}{x+1}$ as long as x is not 0. They have the same limit so just put x= 0 in $\frac{e^{-2x+1}}{x+1}$


text:
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## Description

 Title: MICROWAVE OBSERVATION OF THE O2-CONTAINING COMPLEX, O2-HCl Author(s): Grubbs II, G. S. Contributor(s): Dawes, Richard; Persinger, Thomas D.; Moon, Nicole; Marshall, Frank E Subject(s): Fundamental interest Abstract: begin{wrapfigure}{r}{0pt}_x000d_ includegraphics[scale=0.4]{o2hcl.eps}_x000d_ end{wrapfigure} _x000d_ _x000d_ In the realm of small-molecule van der Waals interactions, there exists much experimental and theoretical data for most fundamental atmospheric components. For complexes containing O$_2$, however, there is actually very little experimental data. This is most likely due to the spin complications brought about by the $^{3}Sigma$ state of the molecule. In this talk, the authors will detail the first known measurement of the complex O$_2$-HCl along with experimental and theoretical analyses of the complex. Previously measured O$_2$-HFfootnote{S. Wu, G. Sedo, E. M. Grumstrup, and K. R. Leopold, textit{J. Chem. Phys.}, textbf{127} (2007) 204315.} analysis have been used as a guide and this talk will outline similarities and differences in the two species. _x000d_ Issue Date: 6/20/2017 Publisher: International Symposium on Molecular Spectroscopy Citation Info: APS Genre: CONFERENCE PAPER/PRESENTATION Type: Text Language: English URI: http://hdl.handle.net/2142/96973 DOI: 10.15278/isms.2017.TE04 Date Available in IDEALS: 2017-07-272018-01-29


text:
# Plus One Chemistry Solution Chapter3 Classification of Elements and Periodicity in Properties

A chemistry course is a demanding and important course to any level of science students and it is very useful to ensure good results by taking quality materials. Chapter 1 Some Basic Concepts of Chemistry. To make this chapter easy to study and learn, we are providing a complete set of solutions. By going through these solutions, students will be able to solve problems speedily and confidently.

These solutions are prepared by expert Chemistry teachers from basic level chemistry. After completing your study for Plus One Chemistry Chapter 1 Some Basic Concepts of Chemistry students can have a look at these solutions. In order to download these chemistry textbook solutions in pdf format follow the link given below

 Board SCERT, Kerala Text Book NCERT Based Class Plus One Subject Chemistry Textbook Solution Chapter Chapter 1 Chapter Name Some Basic Concepts of Chemistry Category Plus One Kerala

## Kerala Syllabus plus one Chemistry Textbook Solution Chapter  3 Classification of Elements and Periodicity in Properties

### Chapter 1  Some Basic Concepts of Chemistry Textbook Solution

Question 1.
What is the basic theme of organisation in the periodic table?
Solution.
By 1865, number of identified elements was 63 and with such a large number of elements it was very difficult to study individually the chemistry of all these elements and their innumerable compounds individually. To ease out this problem, scientists searched for a systematic way to organize their knowledge by classifying the elements. Here comes the basis of periodic table. Various elements have now been divided into different groups on the basis of similarities in chemical properties. This has made the study simpler as now the properties of elements are studied in groups rather than individually.

Question 2.
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to it?
Solution.
Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing atomic weights in such a way that the elements with similar properties occupied the same vertical column or group. He realized that some of the elements did not fit in with his scheme of classification if the order of atomic weight was strictly followed. So, he did not stick to his criteria, he ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together.

Question 3.
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
Solution.
According to Mendeleev’s periodic law, the physical and chemical properties of the elements are periodic functions of their atomic masses, but according to modern periodic law, the physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 4.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Solution.
In the sixth period, the orbitals to be filled are 6s, 4f, 5d and 6p. The complete filling of these orbitals require 2 + 14 + 10 + 6 = 32 electrons, hence, the sixth period of the periodic table should have 32 elements.

Question 5.
In terms of period and group where would you locate the element with Z= 114?
Solution.
The outermost electronic configuration of element (114Z) is
[Rn] 5f14 6d107s27p2. It has n = 7, so period → 7 It belongs to p-block so,
group number = 10 + 4 = 14.

Question 6.
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Solution.
17Cl → It belongs to the third period. So, outermost shell is n = 3. Its configuration is [Ne] 3s23p5.
Therefore, its atomic number = 17.

Question 7.
Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
Solution.
(i) Lawrencium  →  103Lr
(ii) Seaborgium  →  106Sg

Question 8.
Why do elements in the same group have similar physical and chemical properties?
Solution.
In a group, the chemical properties of the elements remain nearly the same due to same valence shell configuration.

Question 9.
What does atomic radius and ionic radius really mean to you?
Solution.
Atomic radius: It is the distance between the centre of the nucleus and outermost shell where electrons are present.
Ionic radius : It is the distance between the nucleus and outermost shell of an ion.

Question 10.
How do atomic radius vary in a period and in a group? How do you explain the variation?
Solution.
Variation of atomic radii in a period :
As we move from left to right across a period, there is regular decrease in atomic radii of representative elements. This can be explained on the basis of effective nuclear charge which increases gradually in a period, i.c, electron cloud is attracted more strongly towards nucleus as the effective nuclear charge becomes more and more along the period. The increased force of attraction brings contraction in size.

Variation of atomic radii in a group : Atomic radii in a group increase as the atomic number increases. The increase in size is due to extra energy shell which outweighs the effect of increased nuclear charge.

Question 11.
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F
(ii) Ar
(iii) Mg2+
(iv) Rb+
Solution.
Isoelectronic species are those which have same number of electrons.
(i) F  has 10 electrons. Therefore, the species N3+, O2-, Ne, Na+, Mg2+, etc., are isoelectronic with F .
(ii) Ar has 18 electrons. Therefore, the species P3-, S2-, Cl, K+, Ca2+, etc., are isoelectronic with Ar
(iii) Mg2+ has 10 electrons. Therefore, the species N3-, O2-, Ne, Na+, etc., are isoelectronic with Mg2+.
(iv) Rb+ has 36 electrons. Therefore, the species Br, Kr, Sr2+, etc., are isoelectronic with Rb+

Question 12.
Consider the following species :
N3-, O2-, F, Na+, Mg2+ and Al3+
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Solution.
(a) All these are isoelectronic species as they are having same number of electrons i.c., 10.
(b) As Z/e decreases, size increases so, order should be,
N3- > O2- > F  > Na+ > Mg2+ > Al3+.

Question 13.
Explain why cations are smaller and anions larger in radii than their parent atoms?
Solution.
A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 14.
What is the significance of the terms – ‘isolated gaseous atom’and’ground state’while defining the ionization enthalpy and electron gain enthalpy? [Hint: Requirements for comparison purposes.]
Solution.
In the definition of ionization enthalpy and electron gain enthalpy, isolated gaseous atom is required for comparison purposes. Ionization energy is the minimum amount of energy required to remove most loosely bound electron from an isolated atom in the gaseous state of an element so as to convert it into gaseous monovalent positive ion. Electron gain enthalpy is the energy change accompanying the process of adding an electron to a gaseous isolated atom to convert it into a negative ion, i.e., a monovalent anion.

Both the above mentioned processes are carried out on an isolated gaseous atom, which in turn is obtained from either the excitation of a ground state atom (in case the element is monoatomic) or atomisation of polyatomic elements.

The force with which an electron is attracted by the nucleus is appreciably affected by presence of other atoms in the neighbourhood. Since in the gaseous state the atoms are widely separated, therefore these interatomic forces are minimum.
The term ground state means that the atom must be present in the most stable state.

Question 15.
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1. [Hint: Apply the idea of mole concept to derive the answer]
Solution.
Energy of the electron in the ground state of H-atom, E1 = -2.18 x 10-18 J
Ionisation energy = E – En
Ionisation enthalpy per mole of atomic hydrogen = (E – E1)NA
= [0 – (- 2.18 x 10-18)] x 6.023 x 1023
= 2.18 x 6.023 x 105 J/mol = 13.13 x 105 j/mol
= 1.313 x 106 J/mol

Question 16.
Among the second period elements the actual ionization enthalpies are in the order Li <B <Be< C< 0 < N< F < Ne. Explain why?
(i) Be has higher ∆iH than B
(ii) O has lower ∆iH than N and F?
Solution.
(i) An s-electron is attracted to the nucleus more than a p-electron. In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron. The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2 p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium. Therefore, it is easier to remove the 2p-electron from boron as compared to the removal of a 2s-electron from beryllium. Thus, boron has a smaller first ionization enthalpy than beryllium.
(ii) O has lower ionisation energy than N because N (1s2 2s2 ${ 2p }_{ x }^{ 1 }\quad { 2p }_{ y }^{ 1 }\quad { 2p }_{ z }^{ 1 }$) has extra stable electronic configuration whereas O (1s2 2s2 ${ 2p }_{ x }^{ 1 }\quad { 2p }_{ y }^{ 1 }\quad { 2p }_{ z }^{ 1 }$) does not. O has lower ionisation energy than F because O has larger size than F.

Question 17.
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Solution.
The electronic configurations of Na and Mg are :
Na : 1s2 2s2 2p6 3s1
Na+ : 1s2 2s2 2p6
Mg : 1s2 2s2 2p6 3s2
Mg+ : 1s2 2s2 2p6 3s1
Mg2+ : 1s2 2s2 2p6

The 1st ionization enthalpy of Na is lesser than that of Mg because Mg has an extra stable configuration and smaller size, so, a larger amount of energy would be required to remove an electron from the 3s orbital, which has a pair of electrons.

The 2nd ionization enthalpy of Na is more than that of Mg because Na+ has an extra stable configuration (complete octet), whereas Mg+ does not have an extra stable configuration.

Question 18.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Solution.
We have to consider two factors :
(i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other. The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of shielding or screening of the valence electron from the nucleus by the intervening core electrons. As we descend the group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels. The increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron requires less energy down a group.

Question 19.
The first ionization enthalpy values (in kJ mol-1) of group 13 elements are :
B         Al      Ga    In       Tl
801    577    579   558   589
How would you explain this deviation from the general trend?
Solution.
(i) Al has lower ionization enthalpy than B due to larger size.
(ii) Ga has slightly higher ionization enthalpy than Al due to ineffective shielding by 3d electrons.
(iii) In has lower ionization enthalpy than Ga due to larger size.
(iv) Tl has higher ionization enthalpy than In due to ineffective shielding by 4f electrons.

Question 20.
Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F
(ii) F or Cl
Solution.
(i) O or F : F has more negative electron gain enthalpy than O due to smaller size, higher nuclear charge and greater possibility of attaining the nearest stable noble gas configuration by gaining one electron.
(ii) F or Cl : Cl has more negative electron gain enthalpy because in F the incoming electron is added to the smaller n = 2 quantum level and suffers significant repulsion from the other electrons present in this level. In Cl, the added electron goes to n = 3 quantum level and occupies a larger region of space and electron-electron repulsion experienced is far less.

Question 2.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Solution.
The second electron gain enthalpy of O is +ve. This is because energy has to be supplied to convert O(g) to O2-(g) in order to overcome the repulsive forces.

Question 22.
What is the basic difference between the terms electron gain enthalpy and electronegativity?
Solution.

 Electron gain enthalpy Electronegativity 1. It provides a measure of the ease with which an atom adds an electron to form an anion. It is a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself. 2. It has an absolute value. It is not a measurable quantity. 3. Its periodicity is not regular in a period or in a group. The periodicity is regular in a period but not so regular in a group. 4. Its units are electron volts per atom or kilojoules per mole or kilocalories per mole. It has no units but is merely a number.

Question 23.
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Solution.
The statement is not correct because electronegativity of an element varies with the state of hybridisation and oxidation state of the element.

Question 24.
Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron.
Solution.
The distance between the nucleus and the outermost shell of an ion is known as ionic radius.
(a) The gain of an electron leads to the formation of an anion. The radius of the anion is larger than the atomic radius of its parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge which acts on more electrons so, each electron is held less tightly and thereby the electron cloud expands.

(b) The removal of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same and since it is now acting on lesser number of electrons and pulls them closer, the ion is smaller.

Question 25.
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Solution.
Two isotopes of the same element have the same first ionization enthalpies because of same effective nuclear charges.
e.g., $_{ 17 }^{ 35 }{ Cl }$ and $_{ 17 }^{ 37 }{ Cl }$ have same ionization enthalpies.

Question 26.
What are the major differences between metals and non-metals?
Solution.

 Metals Non-metals 1. They are usually solids at room temperature. Mercury is an exception. They may be solids or gases at room temperature. 2. They have high melting and boiling points. They have low melting and boiling points. 3. They are good conductors of heat and electricity. They are not good conductors of heat and electricity. 4. They are malleable and ductile. They are neither malleable nor ductile. 5. They have 1 to 3 electrons in their valence shells. They have 4 to 7 electrons in their valence shells. 6. They have low ionisation energies. They have high ionisation energies. 7. They have low electron affinities. They have high electron affinities. 8. They are electropositive. They are electronegative. 9. They are brittle and non-lustrous. They are hard and lustrous. 10. They are sonorous. They are non sonorous.

Question 27.
Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
Solution.
(a) Fluorine
(b) Magnesium
(c) Oxygen
(d) Group 17 (Halogens) :

• F, Cl – Non metals and gases
• Br – Non metal and liquid
• I – Shows metallic lustre.

Question 28.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Solution.
The trend Li < Na < K < Rb < Cs is observed for chemical reactivity because upon descending the group the ionization energy of alkali metals decreases i.e., it is easy for them to lose an electron from their valence shell and attain the nearest stable noble gas configuration. The trend F > Cl > Br > I is observed for chemical reactivity in halogens because the standard reduction potential decreases as we descend the group. F being the most electronegative readily accepts an e~ to complete its octet and hence the trend.

Question 29.
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Solution.
General outer electronic configuration :

Question 30.
Assign the position of the element having outer electronic configuration
(i) ns2 np4 for n = 3
(ii) (n – 1)d2ns2 for n = 4, and
(iii) (n-2)f 7(n-1)d1ns2 for n = 6, in the periodic table.
Solution.

Question 31.
The first (∆iH1) and the second (∆iH1) ionization enthalpies (in kJ mol-1) and the (∆H1) electron gain enthalpy (in kJ mol-1) of a few elements are given below:

Which of the above elements is likely to be:
(a) the least reactive element
(b) the most reactive metal
(c) the most reactive non-metal
(d) the least reactive non-metal
(e) the metal which can form a stable binary halide of the formula MX2(X = halogen)
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X= halogen)?
Solution.
(a) V : The element V has highest first ionization enthalpy and positive electron gain enthalpy hence, it is least reactive.
(b) II : The element II has the least first ionization enthalpy hence, it is most reactive metal.
(c) III; The element III has very high negative electron gain enthalpy hence, it is most reactive non-metal.
(d) IV : Element IV has high negative electron gain enthalpy but ionization energy is not that high hence, it is least reactive non-metal.
(e) VI : The first and second ionization energies of element VI indicate that it can form a stable binary halide.
(f) I : The element 1 has very low value of first ionization energy but very high second ionization energy. Hence, it will form a stable covalent halide of the formula MX.

Question 32.
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Solution.
(a) Li2O
(b) Mg3N2
(c) AlI3
(d) SiO2
(e) PF5
(f) LuF3

Question 33.
In the modern periodic table, the period indicates the value of
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
Solution.
(c) Principal quantum number

Question 34.
Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (/) for the last subshell that received electrons in building up the electronic configuration.
Solution.
(b): The d-block has 10 columns, because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.

Question 35.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Solution.
(c): Nuclear mass

Question 36.
The size of isoelectronic species – F, Ne and Na+ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same.
Solution.
(a): Nuclear charge (Z).

Question 37.
Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Solution.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Question 38.
Considering the elements B, Al, Mg, and K, the correct order of their metallic character is
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Solution.
(d) K > Mg > Al > B

Question 39.
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is
(a) B>C>Si>N>F
(b) Si>C>B>N>F
(c) F>N>C>B>Si
(d) F > N > C > Si > B
Solution.
(c) F>N>C>B>Si

Question 40.
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) CI>F>0>N
(d) O > F > N > Cl
Solution.
(b) F > O > Cl > N

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# When does replacing one random variable with another not change the joint distribution

Suppose $X_1, \dots, X_n$ are random variables with some joint distribution.

I wonder if there is some concept or condition for that if replacing $X_i$ with some other random variable $Y$, then $X_1, \dots, X_{i-1}, Y, X_{i+1}, \dots, X_n$ still have the same joint distribution?

One condition I know is that when $X_1, \dots, X_n, Y$ are independent, and $X_i$ and $Y$ are identically distributed. I wonder if there are weaker conditions?

Thanks!

## 1 Answer

I think what you're looking for is that $X_1, \ldots, X_n, Y$ are exchangeable.

• Thanks! Yes, it is one sufficient condition. I did think of that, but forgot to mention it in my question. Note that exchangebility isn't weaker than "X1,…,Xn,Y are independent, and Xi and Y are identically distributed." Both conditions may be stronger than necessary. – Tim Oct 19 '12 at 0:36


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5: pH Measurement and Its Applications (Experiment ...

In general we can say that an acid-base indicator changes color at a pH determined by the value of $$K_{ai}$$ or pK ai for that particular indicator. The five indicators you will use in this experiment, their color transitions, and their respective values of $$\text{p}K_{ai}$$ are given in Table 1.

Alkalize For Health - Saliva pH Test - Measure your ...

If OH-ions outnumber the H + ions then the water is alkaline. The pH scale goes from 0 to 14 and is logarithmic, which means that each step is ten times the previous. In other words, a pH of 4.5 is 10 times more acid than 5.5, 100 times more acid than 6.5 and 1,000 times more acid than 7.5.

Experiment 13: pH and its Relationship to Acids and Bases ...

This is the result of the sum: pH + pOH = 14. Using this expression, one can determine the pOH if the pH is known. A solution at pH 5.5 would thus have a pOH of 8.5. Weak acids such as acetic acid (vinegar) and weak bases such as ammonia only ionize partially in water giving a much lower concentration of $$\ce{H+}$$ and $$\ce{OH^{-}}$$ ions.

Study of Physico-Chemical Characteristics of Wastewater in ...

Generally, the wastewater collected at the monitored sites is slightly alkaline. The pH varies between 6.8 and 8.3—average value 7.82—thus the pH values are within the accepted range for Danube River according to the Romanian law , which is between 6.5 and 9.0.


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# Pascal’s Wager and the Space of Possible Gods

, , Comments Off on Pascal’s Wager and the Space of Possible Gods

Space is big. Really big. You just won’t believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it’s a long way down the road to the chemist, but that’s just peanuts to space.
– Douglas Adams, The Hitchhiker’s Guide to the Galaxy

Julia Galef spoke with Amanda Askell about Pascal’s Wager on Rationally Speaking (podcast, blog). Pascal’s wager being that if you’re among the faithful you shall be infinitely rewarded, if that faith is the right model of the world. Whereas if you’re an atheist you get nowt, if you’re correct about how the world is. So the expected utility of being numbered among the faithful is argued to be higher than that of not being.

Askell makes a number of defences of Pascal’s Wager. I’m going to make two arguments in response. The first is that the appeal to expected utility theory is flawed; in principle because the space of possible gods is not defined and, in the general case, will not support Askell’s proposed defence. The second is that because reward and punishment from a jealous god is asymmetric the calculus could well point the other way. Afterwards I sketch out why the first problem seems pretty hard to avoid.

## Gods of the continuum

Askell’s first defence (about 3 mins 45 into the podcast) relates to a many gods argument. Criticism: the Wager is unconvincing as adherents of other faiths can make similar claims, so being presented with the wager does not help guide our action. Defence: perhaps, but if a god does exist then you have a higher chance of picking the right faith by picking a faith rather than none, and therefore of enjoying infinite utils. Hence you should just pick one. Furthermore if we have any information which makes one deity more likely than another (e.g. we had a good vibe about in a dream, a fortune cookie etc) then the symmetry is broken and we should pick the most likely god to believe in.

The mathematical issue is that this relies on the space of possible gods being finite. In the interest of brevity, this section contains a quick version of a counter argument; with further elaboration below.

Suppose that for every real number, $r$, in the interval $(0,1) \in \mathcal R$ the god $G_r$  holds the number $r$ to be the Real Number. $G_r$ has decreed that on the Day of Observation believers in $r$ shall be risen to spend eternity in heaven. Unbelievers and heretics shall be annihilated. How do we assess the expected utility?

If we are completely ignorant of the Real Number, then a uniform distribution on $(0,1)$ would be appropriate for our model. Let $u(s,r)$ be the utility we receive by believing in $s$ when $r$ is the Real Number. Now compute the expected utility of arbitrarily picking a god $G_s$ to believe in. Here we’re even assuming that there really is a god $G_r$ (though the result would obviously still hold with some mass on an atheist universe):

$$\mathbb E[u(s,r)] = \mathbb E[u(s,r)|r \neq s] P(r \neq s) + \mathbb E[u(s,r) | r=s]P(r=s) = 0 \cdot 1 + \infty \cdot 0 = 0$$

Where the first expectation was equivalent to the integral of a constant, $0$, over the domain of a random variable, $(0,1)$,  since the removal of a set of measure zero $\{r\}$ from the domain has no effect on the integral. While $P(r=s)=0$ since again $\{r\}$ was a set of measure zero. Let $a(r)$ represent the utility of atheism if $r$ is the Real Number, so that the utility of being a non-believer then is:

$$\mathbb E[a(r)] = 0$$

Hence, in a case of uncountably many possible gods, picking a god randomly to believe in has no higher utility than atheism. Ignoring all short run costs and benefits of either policy. It’s worth repeating that this was not a hard atheist argument. We not only supposed a non-zero probability density for every possible god, but even assumed one of the theologies was correct!

Next, Askell’s asymmetry argument. Suppose we have information that we are more likely to find the Real Number in one place than another. Then decision theory directs us to choose the wager for a god $G_r$ which we believe maximises our chances. This defence breaks on a modified example also. For suppose that instead of being uniformly ignorant of the Real Number, we instead believed it to lie in the vicinity of $\frac{1}{2}$. We could model this with a Beta($2,2$) distribution and exactly the same outcome would obtain. For the same reason as above regarding the measure of a particular point in a continuum. Indeed the result should hold up for any continuous density on $(0,1)$ that was non-zero on every element.

## Jealousy on Olympus

4. What about the atheist-loving god? [IC] The basic idea: Suppose there’s a god that sends all non-believers to heaven and all believers to hell. Given the logic of Pascal’s wager, I ought not to believe in God. Answer: If it’s rational for you to think that disbelief in God (or cars, or hands) will maximize your chance of getting into heaven, then that’s what you ought to do under PW. What’s the evidence for the belief-shunning God? Possibly: ‘Divine hiddenness’ plus God making us capable of evidentialism. The evidence against? God making us capable of performing expected utility calculations, all the historical testimonial evidence for belief-loving Gods. I suspect the latter will outweigh the former. But if you’re making this objection you’re already on my side really: we’re now just quibbling about what God wants us to do.

The problem here stems from an assumption that God is not jealous. In addition to, “all the historical testimonial evidence for belief-loving Gods,” there is a good deal of historical testimonial evidence for heresy hating gods. For example let us modify our previous set of gods to become jealous in a first approximation of Dante’s Inferno. There, god really hates heretics (among many other groups). Virtuous atheists sit about in Limbo.

So let’s suppose that each god, $G_r$ declares that on the Day of Outcomes believers in the Real Number $r$ shall be infinitely rewarded ($+\infty$). Heretics (adherents to $s \neq r$) shall be consigned to watch I’m a Celebrity! Get Me Out of Here on loop eternally ($- \infty$). While atheists shall be left in Limbo ($0$).

What’s the expected utility of different belief policies for these jealous gods? If we assume all numbers are symmetric then a uniform distribution is appropriate. Well, for the atheist it goes unchanged. For randomly picking a god $V_r$?

$$\mathbb E[u(s,r)] = \mathbb E[u(s,r)|r \neq s] P(r \neq s) + \mathbb E[u(s,r) | r=s]P(r=s) = -\infty \cdot 1 + \infty \cdot 0 = -\infty$$

Because the probability of picking exactly the right god is zero, whereas there is a continuum of ways to end up watching reruns of I’m a Celebrity! until long after the stars go out and the universe turns cold. So, while this might be characterised as quibbling about what god wants us to do, it really matters if we’re taking the wager seriously.

## Discrete gods

The root cause of the first issue is that we don’t actually know what the space of all possible theologies looks like. So, for a start, we can’t actually evaluate the expectation as supposed by Askell. Really, I should just object that the space is not well defined, and then shoot down any attempt to define it satisfactorily However, if we suppose that we’re satisfied with the definition of the space we can look again at the mechanics of Askell’s many gods defence. We’re asked to choose one belief from among the union, $U$, of a (presumably) countable set of theologies, $G$, and a set containing atheism, $A$. It is assumed that a PMF exists over this union such if we restrict our choice of beliefs to the subset $G$, the redistribution of probability mass from $A$ to $G$ increases the probability mass of each god in $G$. The problem is that if any individual item in $G$ has no mass to begin with such a rescaling achieves nothing. So, this issue is a generalisation of the false dichotomy: we’re asked to choose an option from a finite set when the actual set of choices is actually much larger.

In case the example I chose with gods labelled by real numbers seems too artificial, we can consider a more ‘realistic’ scenario that gets us to the same place. Suppose a religion consists of a set of beliefs. What is the space of all possible beliefs? Well as a first stab, we could say that it is the union of all sentences in all languages. There exist plenty of formal languages with a countably infinite number of possible sentences. For example, the number of correct C++ programs is at least countably infinite. Therefore the union of all sentences in all languages is at least countably infinite.

Next, let’s say that a theology is a set of beliefs. Then the space of all possible theologies therefore has a cardinality like the powerset of all possible beliefs. Moreover, the powerset of a countably infinite set is uncountable. An uncountable set has no probability mass function that is nonzero for all of its elements. Hence no probability mass function exists for the set of all possible theologies (defined in this manner) which is non-zero over all of its elements. Therefore redistributing whatever mass we gave to atheism over possible theologies has no effect on their mass and Askell’s defence fails again as in the case of the real numbers.

## Conclusion

We have backed the proposed defences of Pascal’s wager based on expected utility theory into something of a corner. The wriggle room that I can see left is to either:

Assert that the space of all possible theologies is countable, requiring by extension an assertion that the space of all possible beliefs is finite. However this has a zero prior for me, since, for example, I already have a several beliefs about every real number $r$, e.g. that a larger real number $s$ exists.

Assert that some arbitrary subset of possible theologies deserves to be treated as atoms with positive mass (e.g. the historically observed religions). But this is just begging the question (what about the historically unobserved but possible religions).

Assert that gods will reward beliefs that are ‘close enough’ to being correct in some sense to be defined. But without a proper framework (e.g. some weird  filtration) it will not be possible to evaluate the expected utility of the decision. However, it seems that this would resolve as deciding which belief sets have a higher likelihood of reward. This would seem to look something like smaller belief sets having a higher likelihood than larger ones by virtue of being closer to a larger number of other belief sets. Which would lead us again to the same sample space, but with a non-uniform distribution over it, and still no mass function.

Neither is it obvious that the historical religions fit the ‘close enough’ model c.f. the Inferno. Indeed we also have to think about which belief sets would be ‘close enough’ to satisfy deities that we can have no knowledge of, e.g. all religions occurring in the (possibly infinite) universe outside our light cone. We should then reason towards a set of beliefs maximally likely to be rewarded. It’s not clear that such a process would end up anywhere different than secular reasoning about morality would lead us to; particularly if we take the Inferno‘s treatment of virtuous pagans as a model. I’d be interested to see someone try.

Also, everything I’ve said here is agnostic with respect to what ‘god’ actually means. E.g. the simulation hypothesis being correct, and the extra-galactic grad student rewarding all simulants with some particular belief set.

We’ve made it through to the end without griping about infinite expected utils too!

Posted in Thought things..

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# Not Everything is a Function in Haskell

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Logo + Screenshot: Shuang, Rimu. "Untitled." Mar 19, 2014. Under a GNU General Public License v3.0 (NOT a CC-BY 3.0 License).

### moving, django, hakyll

A common misconception in Haskell is that everything is a function (a nice example of this is seydar on Reddit asking “Would it be heresy to henceforth call ‘y’ a function with no arguments…” here. I’m still learning Haskell, and I found at the very beginning that this was a nice way of thinking about things in Haskell. After all it tied together in a very nice way. All of a sudden it made perfect sense to pass functions to other functions because it was the exact same thing as passing non-function arguments. The syntax of Haskell also started making a ton of sense. If f x y = z has type a -> b -> c and f x = z has type a -> c, then naturally f = z has type c. So then if we have something such as (+) 3 2 which has type Int -> Int -> Int and 3 just has type Int, then maybe it’s just a nullary function!

Aha, they’re all the same thing; they’re all functions! But they’re not. Conal Elliott has a nice article about why this isn’t the case and goes into the motivations that he thinks are behind such thinking.

I realized today, as I progress further along the road of learning Haskell, that the monadic behavior behind what we usually write using do notation in Haskell, provides nice justification for why not everything in Haskell is a function. In particular, if, for example 3 was a function, then in Haskell, we would represent it as () -> Int, to describe the fact that it took no arguments. So what’s wrong with that? Maybe it gunks up our type system a little to say that Int and () -> Int can both be used to describe the same thing (even though they themselves might be distinct from one another), but Haskell’s already confusing enough right?

Well let’s take a look at the following short function

counter :: Int -> IO ()
counter x = putStrLn (show x) >>= (\y -> counter (x + 1))

Equivalently we might write

counter :: Int -> IO ()
counter x = do
putStrLn (show x)
counter (x + 1)

This function takes a number and then counts up forever, printing the result of each computation to the screen, which would yield something like the following if started at 1.

$1$ 2
$3$ 4
\$ 5
...

Note that (\y -> counter (x + 1)) has type () -> IO (). If we were to state that () -> a and a can refer to the same things, then this would mean that in fact (\y -> counter (x + 1)) could also be said to have type IO (). Yet counter (x + 1) has type IO () and it is most emphatically not (\y -> counter (x + 1)). In fact you’ll get a type error, and for good reason, because from the perspective of monads, it makes no sense to bind a result of IO () to something of type IO ().

Hence although it might be nice mathematically for everything in Haskell to be a function (and in fact it seems there are mathematical frameworks in which this is possible), this is not the case in real life.


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# High School Chemistry/Chemical Quantities & the Mole

A mole (moL) is a standard SI unit for the amount of a substance. This "mole" unit is used to measure mass, volume, and number of particles. It represents very large numbers. The mole is represented by Avogrado's number, which is 6.02 x 1023. This was named after Italian scientist Amedeo Avogadro for his helpful clarifications on the differences between atoms and molecules.

Any mole of a substance contains Avogrado's number. For example, the number of eggs in one dozen, as we know it, is 12 eggs. Now, how many eggs are in one mole? How many planes are in one mole? How many books are in one mole? Surprisingly enough, they all have the same answer: 6.02 x 1023. If you were to be asked: How many eggs are in 2 moles? -- the answer is 1.2 x 1024 (6.02 x 1023 times 2).

## Mole Hill

Not much explanation is needed here, but, the base-line:

• GOING UP = Divide
• GOING DOWN = Multiply

To calculate the grams (or mass) of a compound:

• Co2O3

Get the atomic mass for both cobalt and oxygen. NOTE: BRINClHOF elements, when alone, live in pairs (they are diatomic), so when calculating oxygen gass (O2), you must multiply oxygen's atomic mass by two, which gives you 31.9988.

• 2 • 58.933195 = 117.86639
• 3 • 15.9994 = 47.9982

Add the atomic masses, and now you've got your mass in grams (remember sig-figs):

• 165.8646 grams

## Chemical Quantities

### Percent Composition

#### Empirical Formula[1] → %

• Determine the molar mass.

MgSO4 = (1 x 24.3050) + (1 x 32.065) + (4 x 15.9994) = 120.37 g/moL

• Divide the mass of each element (mass IN THE COMPOUND - not atomic mass) by the overall mass of the compound.

${\displaystyle {\tfrac {24.3050}{120.37}}}$ = 0.20192
${\displaystyle {\tfrac {32.065}{120.37}}}$ = 0.26635
${\displaystyle {\tfrac {63.9976}{120.37}}}$ = 0.53167

• Times the numbers by 100, and round to the hundreths place.

0.20192 x 100 = 20.19%
0.26635 x 100 = 26.64%
0.53167 x 100 = 53.17%

• Add all to 100 to see if these percentages are correct (also, use common sense).

20.19 + 26.64 + 53.17 = 100

• Finished! MgSO4 is made out of 20.19% magnesium, 26.64% sulfur, and 53.17% oxygen.

#### % → Empirical Formula

4 steps route
1. Change percents straight to grams (57.61% → 57.61 grams).
2. Divide the grams by the atomic mass of the element to make them into moles.
3. Divide by the smallest number to make the numbers whole.
4. If this does not work, then multiply the numbers by either 2 or 3. You must do this to all the other numbers too.
Example

Convert the percent composition to empirical formula: 69.55% O 30.45% N (O?N?)

• Convert the percents straight to grams.

69.55g
30.45g

• Divide the numbers by the atomic mass of the element.

69.55g/15.9994amu = 4.347moL
30.45g/14.0067amu = 2.174moL

• Divide the numbers by the lowest number, which is 2.174moL.

2.174/2.174 = 1
4.347/2.174 = 2

• Although "4.347/2.174 = 1.999", we can round it to 2 -- when numbers get really close to a whole number, like in this case, it is fine to round it.

O2N

## Empirical Formula & Molecular Formula

The empirical formula is the basic ratio (lowest whole number ratio) of atoms in a chemical compound. For example, NO2 is the empirical formula, as it is to the simplest terms (lowest whole number ratio). For ionic compounds, the formula unit is, most of the time, usually the empirical formula, while for molecular compounds, the formula does not always indicate the actual number of atoms in each molecule.

The molecular formula is the formula that is any whole number ratio of atoms in a chemical compound (example: N2O4).

Simply simplify.

### Empirical Formula → Molecular Formula

In order to complete this process, you must know the molar mass of the molecular formula (remember, the empirical formula is the lowest whole number ratio while the molecular formula is any whole number ratio).

1. Calculate the molar mass of the empirical formula.
2. Divide M.F./E.F (molar masses).
3. Multiply the empirical formula by the number you get from dividing the M.F. by the E.F., and your molecular formula is done.

EX1: What is the molecular formula of the following molecule given the empirical formula is CH2O and the molar mass of the molecular formula is 90g/moL?

Calculate the molar mass of CH2O.

• 1 x 12.0107 = 12.0107
• 2 x 1.00794 = 2.01588
• 1 x 15.9994 = 15.9994
• (with rounding): 30g/moL

Divide the M.F. by the E.F.

• 90/30 = 3

The M.F. is three times bigger than the E.F., so this is your final answer:

• C3H6O3

### Hydrates

A hydrate is an ionic compound (salt) with water molecules loosely attached to its crystal structure. The formula for a hydrate shows the ratio of water molecules to formula units.

If you are asked how to get the percentage of water from a hydrate, simply:

1. Calculate the molecular mass of the water, so (example) "10H2O" --> "18.012 x 10 = 180.12"
2. Divide the molecular mass of the water by the OVERALL molecular mass of the hydrate, so "180.12/(180.12 + [OVERALL molecular mass of hydrate])".
3. Multiply the number by a 100 to get your percent.

When you are approached with a question along the lines of: "A 5.0 sample of a hydrate of MgCl2 was heated and only 4.2g of anhydrous salt remained", you must subtract the overall mass of the hydrate by the grams of the anhydrous salt that has remained, so: 5.0 - 4.2, which will give you .8g. This .8g represents the water (H2O).

When you are asked to find the number of water molecules in a hydrate, such as "When 5.00grams of FeCl3 * xH2O are heated, 3.00 grams of the salt remained. Find the chemical formula.", you must:

• Subtract the grams of salt that has remained by the overall mass before the heating.

5.00 - 3.00 = 2.00g of H2O

• Divide 2.00grams of H2O by the molar mass of water.

2.00g/18.02g = 0.111moL

• Now, you must find the molar mass of the salt. So, calculate the molar mass of FeCl3 and divide 3.00g from 162.206g (molar mass of FeCl3) to get your mole value.

3.00g/162.206g = 0.0185moL

• Divide these two numbers by the smallest number, which is 0.0185moL.

0.0185/0.0185 = 1
0.111/0.0185 = 6

• You have gotten your answer (note that 0.111moL belongs to water). Your full formula is FeCl3 * 6H2O.

FeCl3 * 6H2O


text:
# Paths with Spaces, I Give Up

I’ve wanted to look into the Pedestal framework for creating web-based applications in Clojure. However, one of the requirements is Leiningen 2.2.0 or greater. And, as I’ve written before, version 2.2 will not install on my system because of spaces in the path of the user home directory. (”C:\Users\David Clark” on my system.)

My user profile name is “david”. That’s what I use to sign on with. The fact that my home directory uses “David Clark” is an unfortunate result of how the computer was set up at the factory when I custom ordered it. This has been a periodic pain in the ass ever since I got the system because folks who develop tools primarily on other operating systems, just can’t seem to deal with spaces in file paths. Even after all these years. It’s a kind of snobbery that I just find annoying and trite. Windows exists. Get over it.

Rather than try to fix every program that screws this up, I decided to change my user account and profile to just use “C:\Users\david” as my home directory. Not so easy.

To start with, I followed the directions here. That got me most of the way there. Then I checked all of my environment variables and made changes there as needed. Leiningen installation still failed. In a couple places, it was still trying to put stuff in the “C:\Users\David Clark” directory or sub-directories.

Most of these tools use the Java system property variable user.home to determine what the home directory is. I wrote the following simple program to check what Java was reporting.

/*
* Program to display a few user-related properties in the
* Java environment.
*/
package userproperties;

/**
*
* @author David Clark
*/
public class UserProperties {

/**
* @param args the command line arguments. Not used.
*/
public static void main(String[] args) {
String userHome = System.getProperty("user.home");
System.out.println("User Home: " + userHome);
String userDir = System.getProperty("user.dir");
System.out.println("User Dir: " + userDir);
}
}

Unfortunately, the program reported that my home directory was still “C:\Users\David Clark”. Also, in a Windows terminal, the command echo %USERNAME% still shows the user name as “David Clark”, not “david”.

After a little more research, I came across this page that showed how to change some registry variables for the Windows Explorer Shell. Making those changes seems to have taken care of most things. At least Leiningen installs and runs again.

Fixing the %USERNAME% environment variable required a little more work. Running lusrmgr.msc and changing the appropriate Name field under the Users group followed by a reboot seems to take care of that.

So, I think things are now the way I want them. Everything seems to work as expected, or I’ve built in a big problem that will bite me in the ass at some inopportune time in the future ;-)

One other thing of note is that the registry itself contains a warning not to use those keys but either SHGetFolderPath or SHGetKnownFolderPath function instead. Maybe Java will catch up some day.


text:
# Variance and Expected Value Algebra

Linearity of Expected Value: Suppose $X$ and $Y$ are random variables and $a$ and $b$ are scalars.  The following relationships hold:

$E[aX+b] = aE[X]+b$

$E[aX+bY] = aE[X] +bE[Y]$

Variance:

$Var(aX+bY) = a^2Var(X)+2abCov(X,Y)+b^2Var(Y)$

Suppose $X_i$ for $i=\left\{1\ldots n\right\}$ are $n$ independent identically distributed (iid) random variables.  Then $Cov(X_i,X_j) = 0$ for $i\ne j$ and

$\displaystyle Var\left({\sum_{i=1}^n X_i}\right) = \sum_{i=1}^n Var(X_i)$

Example:

$X$ is the stock price of AAPL at market close.  $Y$ is the sum of closing AAPL stock prices for 5 days.  Then

$\begin{array}{rl} Var(Y) &= \displaystyle \sum_{i=1}^5 Var(X_i) \\ &= 5Var(X) \end{array}$.

Contrast this with the variance of $Z = 5X$.  In other words, $Z$ is a random variable that takes a value of 5 times the price of AAPL at the close of any given day.  Then

$\begin{array}{rl} Var(Z) &= Var(5X) \\ &=5^2Var(x) \end{array}$

The distinction between $Y$ and $Z$ is subtle but very important.

Variance of a Sample Mean:

In situations where the sample mean $\bar{X}$ is a random variable over $n$ iid observations (i.e. the average price of AAPL over 5 days), the following formula applies:

$\begin{array}{rl} Var(\bar{X}) &= \displaystyle Var\left(\frac{1}{n} \displaystyle \sum_{i=1}^n X_i\right) \\ &= \displaystyle \frac{nVar(X)}{n^2} \\ &= \displaystyle \frac{Var(X)}{n} \end{array}$

Filed under Probability

### 2 responses to “Variance and Expected Value Algebra”

1. I understand that E(XY) = E(X)E(Y) only if X and Y are independent. Suppose you had 3 independant random variables X,Y, and Z and you wanted E[X(Y+Z)]. Does this obey a distributive law, ie does E[X(Y+Z)] = E(X)E(Y)+E(X)E(Z)?
If so, what if it you wanted:
E{(aX+b)[(cY+d)+(mZ+n)]} ,
could you just multiply through like in regular algebra?

• uclatommy

Yes, you can distribute inside the brackets to get E[XY + XZ]. Then by linearity, this is equal to E[XY] + E[XZ]. If X and Y are independent, then E[XY] = E[X]E[Y]. And similarly if X and Z are independent, E[XZ] = E[X]E[Z].