# Why, in Logic, Does “False” Imply Anything?

In a class on symbolic logic, students are taught the truth tables that define the “logical connectives” ∧ (and), ∨ (or), ¬ (not), and → (if  … then). Everything makes sense until they are told that if p is false, then $$p\rightarrow q$$ is true whether or not q is true. How can we say that “If pigs fly, then 2 is even” is a true statement? Or, for that matter, “If pigs fly, then there’s a kangaroo in my pocket”? This is especially troublesome when students, naturally wanting brevity, read $$p\rightarrow q$$ as “p implies q “. How can nonsense imply anything?

## A quick survey

Before I get into full answers, let’s look at some examples of the question. First, in 1994, in the infancy of Ask Dr. Math, we got this question:

8th Grade Logic

Why in a conditional statement if the "p" in the hypothesis is false, then the entire statement is true?  Why isn't it undecided?

In 1997, in answer to a more general question about logic, Doctor Mike included an answer to the question, knowing it is common:

A False Statement Implies Any Statement

... In case you still are wondering about why a False implies anything, try this explanation on for size.

Like Doctor Ken in the 1994 answer, Doctor Mike here focused on an example, pointing out that if the condition of such a statement is false, then whatever happens, you couldn’t be convicted of lying, because you made no promises about what would happen in any situation that actually happens. In 2004, Jay asked a follow-up question that was added to the same page:

Why are the logical statements "false implies true" and "false implies false" always considered "true"?

I've read the previous note, but could you please give a more "formal" explanation?

Doctor Schwa replied with an answer relating the idea to set theory:

More formally, I'd say "implies" means the same as "subset" in set theory.  That is, when you say, "if it rains, then the ground gets wet," you mean, "the set of times when it rains is a subset of the set of times when the ground gets wet."

So, since the empty set is a subset of any set, a false statement implies any statement.

Formally, this is a good way to think of it; but it may not satisfy everyone – particularly since it is not obvious why the empty set can be a subset at all. (That’s another question we get from time to time.)

In 2005, Doctor Achilles answered this one:

Logic and Conditional Sentences

I have a question about conditional statements.  I am having a hard time understanding why two false statements in a conditional makes it true.

I tried to use different statements to create a truth table but I get stuck on the same concept.  I tried a sentence like "If a polygon is a square, then the sides are equal."  If I assume a rectangle, then it seems to me that the statement is undefined.  Being true or false does not even apply.

This is much like the 1994 question, and is entirely reasonable. Be sure to read his answer; for the sake of space I will be focusing on two more recent questions whose answers include much of what others have said, while going beyond.

## A deeper look

The Logic behind Conditional Statements

I've read questions of the same title "Why is "false implies true/false" always "true"?" and I can understand the reason in the "subset way".

Could you explain the reason just in the logic way?  Because when we say "work the same as subsets", we must prove it does work the same...

When I trust the set theory, I must trust logic first.  So I want to understand the reason just on the way logic goes.

"False implies true/false" is true...why?

So Keven doesn’t like Doctor Schwa’s approach, because it doesn’t directly relate to logic. Having recently taught the subject and thought about this issue, I had several things to say.

### 1. “If … then” doesn’t mean “implies”

First, don't use the word "implies" to talk about a conditional statement; A->B should be read merely as "if A then B".  "Implies" suggests a cause-and-effect relationship, or at least a logical connection of some sort.  But the conditional statement is not meant to suggest that (even though many examples given in texts look that way).  The statement "if A then B" says nothing more than "if A is true, then B is true"--not "if A is true, then it CAUSES B to be true".  It means that whenever we find that A is true, then we can know that B is true (or else A->B would have been false).

The problem is that we naturally tend to see a conditional statement as something more, because everyday usage leaks over into our logic, and the word “implies” reinforces that tendency. Think of logical statements as merely observations about what things happen to be together, not about causality, or underlying reasons, or even necessary connections (“these things always go together”). Although mathematicians can use the word “implication” for this, it is misleading if you don’t pay close attention to the definition.

In fact, a logical statement is not even an assertion that we make for some reason, so that we have some stake in its truth. It is just a statement that may be true or false in any given situation. If A is true but B is not, then the statement “if A, then B” is false, because if it were true, then B would have to be true.

But what if we don’t have enough evidence to judge whether the statement is true?

### 2. It is a mathematical definition, not everyday reasoning

Second, this conditional statement is in a sense just something mathematicians define for their own purposes, not something that necessarily agrees with the natural-language use of the phrase.  And what we need in logic (at least in traditional two-valued logic, as opposed to a logic that might include an "undetermined" value) is for every statement to be either true or false.  In this context, we have to choose what A->B will mean in every case--in order to define a complete truth table.

Some of the earlier questioners felt that “If FALSE, then …” should just be “undetermined” or “I don’t know”; and they are correct, in real life. But symbolic logic requires a logical truth value of T or F for every statement, so we don’t have that option. We need some choice, which could be purely arbitrary (like rounding up on 5), or might have a specific reason based on how we plan to use it.

Here I gave a familiar answer:

Traditionally, mathematicians subscribe to a sort of "innocent until proven guilty" rule: we can't say something is false just because there is no evidence; instead, when there is no evidence of truth or falsity, we say it is true.  This is what lies behind the related facts about sets: we say that the null set is a subset of any set because there is no evidence that it is not--there is no element in the null set that is NOT an element of the other set!

But still, why not say “guilty until proven innocent”?

### 3. This is what works in describing logical arguments

Ultimately, I realized, the reason for the choice comes from our application, not from the real world. One application of symbolic logic is to validate arguments, and here, if the truth value of a conditional statement were not defined as it is, then a valid argument would fail the test:

Let's consider an argument like this:

I have a cold.
If I have a cold, then my nose is running.
------------------------------------------
Therefore, my nose is running.

This has the form

A
A->B
----
B

To show that this is a valid argument, we write it as a single statement:

((A) ^ (A->B)) -> B

That is, the whole argument is a big conditional.  If its truth table is ALWAYS true, no matter what the truth values of A and B are, then we consider the argument valid.

That is, an argument is considered valid if the equivalent statement is a tautology.

I made truth tables for this (valid) argument, using every possible definition of the conditional, and showed that the only definition for which it becomes a tautology when it should is the accepted one.

Now the truth table accurately reflects the validity of the argument. And that, I think, is why we make this definition: it works.

Why does it work?  Because we want to say an argument is valid when the conclusion follows from the premise: if A is really true, then B had better be true.  We DON'T CARE what happens if the premise is false; the argument is still valid because it doesn't tell you what happens then.  There's the meaning behind that "innocent until proven guilty" idea.

## An example

In 2008, I answered another question, which gave me a chance to fill out a couple areas:

Logic Statement False Implies True

I am well familiar with the linguistic arguments which clarifies this somehow confusing concept.  Is there a deeper philosophical argument that touches on the underlying logic of this concept {logic axiom}?

The most disturbing thing about this logic axiom is that it eliminates {defeats} the logical contingency of the conclusion on the premise, which somehow goes against the very essence of logic! By layman definition, logic is something that allows "naturally" the consequence to flow from a premise.  When the same conclusion happens no matter what the premise is, the connectedness of the logic in between the conclusion and its premise loses significance, just like the definition of a function is defeated when one certain value from the domain point {maps} into two or more different values from the range.

If the moon is made of cheese, then I will go to the movie next week can rather best describe a sarcastic {insane} mode of thinking than a flow of "natural" logic especially when it can be said equally that if the moon is NOT made of cheese, I still go to the movie!  The dilemma of this concept, though I use it myself to prove some propositions like the empty set is a subset of every set, is that it kills the "natural" connectedness inherent in logic.

Unfortunately, the very definition of logic itself is so intuitive and vague in the same way the set or sanity is defined, otherwise undefined!  Even though I trained myself to live with this concept and I use it in my formal proofs, I try to avoid using it as much as possible.  It is like employing proof by contradiction.  I would rather prove directly.

Abe was thinking of conditional statements in cause-and-effect terms, as identifying “natural consequences”. He thought of logic as philosophy, and his vague conception led to confusion. (Logic is considered a part of philosophy not because it is deep in itself, but because we need to think clearly when we get to the deep stuff.) I repeated many of the ideas in the previous answer (which had not yet been archived so I couldn’t refer him to it): The conditional statement is not about cause-and-effect; the decision about truth value in the disputed cases depends on context; and one important context is judging the validity of an argument, which works if we take the “innocent until proven guilty” approach.

He responded with a comment that overstated the conclusion:

I found it rather interesting that there is indeed some philosophical underlying mode of thinking built into the definition of "P implies Q," that is, if there is no evidence that something is false, then we must assume that it is true.  This we may cast as the "default rule of the truth."
...
Now comes the interesting case which P is false and Q is true, yet we must assume that the implication is True ONLY for lack of better knowledge or evidence pointing to the other direction.  This is to me a philosophy or a mode of thinking which I accept as a sound one though it has some serious implications beyond math.

Since, as I have said, math has its own truth which need not agree with the real world, it doesn’t tell us anything about deeper philosophical ideas: The definition of the conditional statement is not about truth in general, but just about what we want a conditional statement to mean. He also tried to restate my reasoning, which led me to clarify it with a fully stated example:

Let's take an example.  I have a piece of paper here that is coated with a chemical that changes color.  I claim that if the paper is wet, it is red.  That is,

WET -> RED

(Note that I didn't say "wet implies red"; the word "implies", as I said before, is not really appropriate for this connective, as you'll see in a moment.)

Now let's consider what you might see when I show you the paper, taking the four cases in your order.

1. It's wet, and it's red.  That agrees with my statement, so you say my statement is true.  (You do not have enough evidence to conclude that my statement is ALWAYS true; you've just seen one case.  Maybe tomorrow it will be cooler, and you'll find that the paper is only red if it's wet AND warm.  That's why you can't say it's true that wet implies red, only that it is true in this instance that "if it's wet, then it's red."  Do you see the difference?)

2. It's dry, and it's blue.  You don't know that it would be red if it were wet; there's no evidence one way or the other.  So, simply by convention, you say that my statement is true, meaning that the evidence is consistent with that conclusion.  But you can't say that you've proved that wetness IMPLIES redness; all you can say is that it might.

3. It's wet, and it's blue.  That disproves my statement; we have a case where it is wet but NOT red.  My statement is definitely false. (This case IS enough to disprove the stronger claim that wet implies red; you have a counterexample.)

4. It's dry, and it's red.  Hmmm ... maybe it's ALWAYS red, and my statement was technically true but misleading; or maybe it's red for some other reason than wetness.  Or maybe it actually turns blue when it gets wet, and I just lied.  Again, you really don't know!  The evidence at hand deals only with the case where it's dry, and my statement is about what would be true if it were wet.  So you have to say that it's true, because you haven't disproved it, just like in case 2.

So your cases 2 and 4 are both "true" for the same reason, not for different reasons.  The evidence in both cases is consistent with my statement, so we call it true.

He concluded with a concise statement:

In sum, P implies Q is nothing more than a claim or a proposition.  We may uphold the rest of the logic table for P implies Q since the logic equivalence (truth value) for the remaining three cases does NOT contradict our claim about P implies Q, although not useful statements in some cases.  Thanks again for the great example.

One might say that “truth”, in the sense we need here, just means “non-contradiction”.

### 2 thoughts on “Why, in Logic, Does “False” Imply Anything?”

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# Probability of meeting at the corner of city streets betwe

Two people agree to meet each other at the corner of two city streets between 1pm and 2pm. However, neither will wait for the other for more than 30 minutes. If each person is equally likely to arrive at any time during the one hour period, determine the probability that they will in fact meet.

This question is very similar to Probability of two people meeting during a certain time. But I'm afraid I'll have to ask again, because it's not quite what my query is about.

I've tried letting X and Y be the two random variables, and they are independent of each other. I tried phrasing it in terms of $P(|X-Y|<30)$ but I don't know how to solve this, as i don't know the individual probability of either X and Y? Any advice would be greatly appreciated. Sorry in advance if I seem to be repeating the question (in link) and for any wrong title or tag labelling.

I have also thought of another method:

If A arrives before B, then the probability that he arrives during first half an hour is 0.5 . He'll then wait for 0.5 hours. If He arrives during the last half an hour hours, with proability 0.5 . He then waits for 0.5 hours on average. Thus his total wait time is 0.5*0.5*2=0.5. For A and B to meet, B must arrive when A is waiting. Thus the probability that B arrived when A was waiting is 0.5 . Similarly if B arrives before A, the probability that they meet is 0.5 . Thus, Total probability that they meet is 1. Where did I go wrong? As the answer suggested is 0.75.

• Aren't $X$ and $Y$ both uniformly distributed between $0$ and $60$? (And I assume it is intended that they arrive independently, though it's not stated explicitly in the problem description.) – Brian Tung Apr 3 '16 at 22:24
• @BrianTung, However, if using uniform distribution, if I do invnorm on GDC, it doesn't mean that I can solve it? (as i don't know the mean and the variance?) Pls advise. – CCC Apr 3 '16 at 22:35
• Maybe I'm misunderstanding the question. Why isn't this question like the one you link to? – Brian Tung Apr 3 '16 at 22:37
• The variables are not normally distributed, so you cannot use mean and variance. You can integrate the probability density function over the region where the two people meet; since that function is constant and the region is more or less as depicted in that other question, the methods are essentially identical. – Brian Tung Apr 3 '16 at 22:48
• Probability is 75% according the following simulation (in R). "n <- 10^5 # number of trials a <- runif(n, min = 0, max = 60) b <- runif(n, min = 0, max = 60) sum(abs(a-b) < 30)/n" – snoram Apr 3 '16 at 23:06

The R code below essentially repeats @snoram's nice simulation, but using only 50,000 points. Then I go on to plot the points in the square with vertices at $(0,0)$ and $(60,60)$, with the points corresponding to the condition that the two people meet plotted in light blue.

From there, it is obvious that the two excluded regions (black) each have $1/8$th of the area. Because the joint distribution is uniform on the square this means that the probability of the 'condition' is $3/4.$

Of course, you can draw the boundaries of the condition, and thus solve the problem, without simulation. (It is $not$ necessary to know the distribution of $|X - Y|$.) In integral notation, you need $\int\int_C 1/60^2\,dx\,dy,$ where $C$ is the region corresponding to the condition. You'd have to break $C$ up into two parts in order to set numerical limits on the integral signs. But I think the geometrical argument suffices.

 m = 50000
x = runif(m, 0, 60);  y = runif(m, 0, 60)
cond = (abs(x-y) < 30);  mean(cond)


Prove that a_n tends to 0

Alexmahone

Active member
Assume $\frac{a_{n+1}}{a_n}\to L$, where $L<1$ and $a_n>0$. Prove that

(a) $\{a_n\}$ is decreasing for $n\gg 1$;

I've done this part.

(b) $a_n\to 0$ (Give two proofs: an indirect one using (a), and a direct one.)

Indirect proof: Assume for the sake of argument that $a_n$ does not tend to 0. Since the sequence is decreasing and bounded below by 0, it must converge to a positive value (call it $l$).

$\lim\frac{a_{n+1}}{a_n}=\frac{l}{l}=1$, a contradiction. So, $a_n\to 0$.

How do I go about the direct proof (one that doesn't use "proof by contradiction")?

Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
How do I go about the direct proof (one that doesn't use "proof by contradiction")?
Hint

$|a_{n}-0|<\epsilon\Leftrightarrow |a_{n+1}\cdot (a_n/a_{n+1})|<\epsilon \Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\epsilon$ and $a_{n+1}/a_n$ is bounded.

Alexmahone

Active member
Hint

$|a_{n}-0|<\epsilon\Leftrightarrow |a_{n+1}\cdot (a_n/a_{n+1})|<\epsilon \Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\epsilon$ and $a_{n+1}/a_n$ is bounded.
Is this a proof by induction that $|a_n|<\epsilon$? If so, how do we prove the base case for $n=1$?

Last edited:

HallsofIvy

Well-known member
MHB Math Helper
I would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.

Alexmahone

Active member
I would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.
I used a slightly different approach:

$L<\frac{L+1}{2}$

$\frac{a_{n+1}}{a_n}<\frac{L+1}{2}$ for $n\gg 1$ (Using the "sequence location theorem")

So, $a_{n+1}< \left(\frac{L+1}{2}\right)a_n$ for $n\ge N$

$a_{N+k}<\left(\frac{L+1}{2}\right)^k a_N$ for $k\ge 1$ (Can be proved using induction over $k$.)

Since $\frac{L+1}{2}<1$, $a_{N+k}\to 0$ as $k\to\infty$.

So, $a_n\to 0$.

Last edited:


# Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$

Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$ so another eigen value is $-2i$ another eigen value must be $0$ right? as we know these kind of matrix has eigen values $0$ or purely imaginary.

• Yes. $\quad\mu^{*} = \mu\quad$ and $\quad\mu^{*} = -\mu\quad$ $\Longrightarrow\quad\mu = 0$. – Felix Marin Sep 13 '13 at 19:50

Yes, the characteristic polynomial of a real $3\times 3$ matrix will be a real cubic polynomial. If not identically zero, the real skew matrix will have one pair of conjugate imaginary eigenvalues and a zero eigenvalue, each of multiplicity one.

• Does this eliminate the possibility of eigenvalues $2i,2i,-2i$? – Rebecca J. Stones Sep 13 '13 at 16:48
• @Rebecca: Yes, it does (for a 3x3 real skew matrix). – hardmath Sep 13 '13 at 17:17
• In case someone has doubts, the characteristic polynomial being a real cubic "eliminates the possibility" of three purely imaginary nonzero roots. Since these must occur in conjugate pairs, divding out a monic quadratic with one such pair as roots gives as quotient a real degree 1 polynomial having the third purely imaginary nonzero root (contradiction). – hardmath Sep 13 '13 at 21:50

In general, a real skew-symmetric $3 \times 3$ matrix $K$ looks like

$K = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}, \tag{1}$

so that the characteristic polynomial is

$\det(\lambda - K) = \lambda^3 + (a^2 + b^2 + c^2) \lambda, \tag{2}$

which is easily seen to have roots

$\lambda = 0, \pm i\sqrt{a^2 + b^2 + c^2}, \tag{3}$

which follow the general pattern the OP suggested; apparently the matrix she/he was thinking of satisfies $\sqrt{a^2 + b^2 + c^2} = 2$; but $0$ will always be an eigenvalue for such $K$ in any case.

Hope this helps. Cheers.

Since a real matrix $A$ is skew if, and only if, $A=-A^T$, since $\text{spec} (A)=\text{spec}(A^T)=-\text{spec}(-A^T)$ and since the characteristic polynomial has degree $3$ and non-real roots come in pairs, it follows that $0\in\text{spec}(A)\cap \text{spec}(A^T)$.

We know that a skew symmetric matrix $A=(a_{ij})_{3 \times 3}$ satisfies $A^T=-A$. This means that $a_{ii}=-a_{ii}$ or equivalently $a_{ii}=0$ for all $i \in \{1,2,3\}$.

We know the trace of $A$ is the sum of its eigenvalues, which is $0$ since the main diagonal of $A$ comprises of zeroes.

Since a $3 \times 3$ matrix has $3$ (not necessarily distinct) eigenvalues (given by the roots of the characteristic polynomial [a degree $3$ polynomial with leading coefficient either $1$ or $(-1)^3$, depending on how it's defined]), there is precisely one other eigenvalue $\lambda$ and we must have $2i-2i+\lambda=0$, which implies $\lambda=0$.

Here's another way of looking at it, which I thought of after posting my previous answer.

The approach here seemed sufficiently different that I felt a separate answer was merited. Hope I'm not overdoing it. Anyway . . .

First, note that for any skew-symmetric, real matrix, the only possible real eigenvalue is zero. For, if $K$ is real skew-symmetric, so that $K^T = -K$, and if $v \ne 0$ is an eigenvector with real eigenvalue $\lambda$, so that

$Kv = \lambda v, \tag{1}$

we have

$\langle Kv, v \rangle = \lambda \langle v, v \rangle, \tag{2}$

but

$\langle Kv, v \rangle = \langle v, K^Tv \rangle = \langle v, -Kv \rangle = -\langle Kv, v \rangle, \tag{3}$

or

$\lambda \langle v, v \rangle = -\lambda \langle v, v \rangle, \tag{4}$

and since $\langle v, v \rangle \ne 0$ this shows that

$\lambda = - \lambda, \tag{5}$

whence

$\lambda = 0. \tag{6}$

Here I take $\langle \cdot, \cdot \rangle$ to be the standard inner product on the real vector space on which $K$ operates.

Now argue as follows: since $2i$ is an eigenvalue of $K$, and the characteristic polynomial of $K$ is real, $-2i$ is an eigenvalue as well. This means

$\lambda^2 + 4 \mid p_K(\lambda), \tag{7}$

where

$p_K(\lambda) = det(\lambda - K) \tag{8}$

is the characteristic polynomial of $K$. But $\deg p_K(\lambda) = 3$ in the present case, so the quotient polynomial $q(\lambda)$ is of degree $1$, i.e. we must have

$p_K(\lambda) = (\lambda^2 + 4)q(\lambda) \tag{9}$

with

$q(\lambda) = \lambda - a, \tag{10}$

$a$ real. But this of course implies $a$ is a real eigenvalue of $K$, whence we must have

$a = 0. \tag {11}$

Well, I hope this answer is more than mere mathematical logorrhea, and it sheds a little more light on the subject at hand.

Cheers!


Math Help - Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity

1. Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity

Hey all,

I been playing around with limits, when i solve for the following function:

$\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$
But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks

2. Originally Posted by Oiler
Hey all,

I been playing around with limits, when i solve for the following function:

$\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$
But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks
????

If you graph that, you obtain a line y=1

3. Originally Posted by dwsmith
????

If you graph that, you obtain a line y=1
dwsmith, sorry just made some changes to the formula.

4. Originally Posted by Oiler
dwsmith, sorry just made some changes to the formula.
It is -1 when you go to negative infinity.

5. Originally Posted by Oiler
also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks
Let x = 0

0 = 2 is that true?

6. ofcourse, how dim of me. Thanks dwsmith.

7. $\displaystyle\lim_{x\to +\infty}\frac{e^x-1}{e^x+1}=\frac{\infty}{\infty}$

Applying L'Hopitals Rule
$\displaystyle\Rightarrow\lim_{x\to +\infty}\frac{e^x}{e^x}=1$

8. Originally Posted by dwsmith
$\displaystyle\lim_{x\to +\infty}\frac{e^x-1}{e^x+1}=\frac{\infty}{\infty}$

Applying L'Hopitals Rule
$\displaystyle\Rightarrow\lim_{x\to +\infty}\frac{e^x}{e^x}=1$
L'Hôpital's Rule is overkill for this. XD

$\lim\limits_{x\to\infty}\dfrac{e^x-1}{e^x+1} = \lim\limits_{x\to \infty}\dfrac{1-e^{-x}}{1+e^{-x}} = 1$ since $e^{-x}\rightarrow0$ as $x\rightarrow\infty$.

9. Originally Posted by Chris L T521
L'Hôpital's Rule is overkill for this. XD

$\lim\limits_{x\to\infty}\dfrac{e^x-1}{e^x+1} = \lim\limits_{x\to \infty}\dfrac{1-e^{-x}}{1+e^{-x}} = 1$ since $e^{-x}\rightarrow0$ as $x\rightarrow\infty$.
Sometimes you just have to kill it.

10. $\displaystyle \lim_{x \to \infty}\frac{e^x - 1}{e^x + 1} = \lim_{x \to \infty}1 - \frac{2}{e^x + 1}$

$\displaystyle = 1 - 0$

$\displaystyle = 1$.

11. Originally Posted by dwsmith
Sometimes you just have to kill it.
But you never have to "over-kill". When you are showing someone how to do something, the simplest way is always best.

12. Originally Posted by Oiler
Hey all,

I been playing around with limits, when i solve for the following function:

$\displaystyle\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$

But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations?

also how does $\{e^{x}-1\}=\{1+e^{x}\}$ ?.

Thanks
For this fraction, the denominator is always 2 greater than the numerator.

However, the "ratio" gets closer and closer to 1, as we increase x above 0.

$x\rightarrow\infty$ means x increases without bound above 0.

Hence, as x "approaches infinity", you need infinitely many decimal places
to express the value of the fraction as a value "other than 1".

That's part of the concept of limits.

viz-a-viz

$\displaystyle\frac{e^1-1}{e^1+1}=0.46211715726$

$\displaystyle\frac{e^2-1}{e^2+1}=0.76159415596$

$\displaystyle\frac{e^{10}-1}{e^{10}+1}=0.99990920426$

....onward.

If $x<0$ and decreases without bound below 0.

$x=-y$

$x\rightarrow\ -\infty\Rightarrow\ y\rightarrow\infty$

$\displaystyle\lim_{x \to -\infty}\frac{e^x-1}{e^x+1}=\lim_{y \to \infty}\frac{e^{-y}-1}{e^{-y}+1}=\lim_{y \to \infty}\frac{\frac{1}{e^y}-1}{\frac{1}{e^y}+1}$

$\displaystyle\lim_{y \to \infty}\left[\frac{\frac{1}{e^y}}{\frac{1}{e^y}}\right]\;\frac{1-e^y}{1+e^y}\right]=-\lim_{y \to \infty}\frac{e^y-1}{e^y+1}=-1$

When you have the concept, you can apply the fast methods as you please later.

13. Originally Posted by HallsofIvy
But you never have to "over-kill". When you are showing someone how to do something, the simplest way is always best.
To me, that was mighty simply. The constant disappear and the exponentials are their same derivatives.


# Analytical Reasoning Question II

I have yet another analytical question that got me

A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?

Please kindly explain your final answer so those of us who are still trying to learn can grasp the logic easily. Thanks in advance!

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The number of possible combinations is $5! = 120$. Each digit will be in each position an equal number of times, namely $5!/5 = 24$ times.

So the sum of the digits in the one's place is

$24(1+3+5+7+9) = 24*25 = 600$

This will be the sum of the digits in each place (one's, ten's, etc.). So the total sum is:

$600*11,111 = 6,666,600$, regards, iyengar

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but let me ask,is my answer right ??,as i am not a person with formal training, – Iyengar Sep 7 '11 at 15:32
Why do you need this line? $24(1+3+5+7+9) = 24*25 = 600$. Why sum the numbers? I don't know the answer yet(trying not to look in the back of the book for the answers until I am sure I have this right).:) – user10695 Sep 7 '11 at 15:34
My question is, why add the individual allowable digits, rather than just adding the total occurrence for each? I was thinking more something like this $24(5) = 24*5 = 120$ – user10695 Sep 7 '11 at 15:35
For each decimal place, 24 of these are of the same digit. Thus for a decimal place the total value is 24 (1 + 3 + 5 + 7 + 9) = 600 To get the total value of all these numbers multiply the first result by 11111. – Iyengar Sep 7 '11 at 15:37

As a slight alternative to iyengar's answer:

There are $120$ such numbers and their average is $55555$ (note that for any of the form, there is also $111110$ minus that number) so the answer is $120\times 55555=6666600$.

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where did you get 55555 from? The question specifies that there are no repeats allowed. – user10695 Sep 7 '11 at 16:15
If you have for example $17935$ then you also have $111110-17935=93175$, so the average term is half $111110$, namely $55555$. – Henry Sep 7 '11 at 16:23
I am sorry, I still don't get the 111110 and the 55555. Why those? – user10695 Sep 7 '11 at 19:30

This is partly a slight variant of Henry’s answer and partly a further explanation in answer to one of your questions.

If $d$ is one of the allowable digits $1,3,5,7,9$, let $d' = 10 - d$; note that $d'$ is also allowable. Now let $abcde$ be any allowable number; then $a'b'c'd'e'$ is also allowable, and it must be a different number from $abcde$. (It would be the same number only if all of the digits were $5$, but that’s not allowed.) In this way you can pair up all of the allowable numbers. Since there are $5! = 120$ permutations of the $5$ allowable digits, there are $120$ allowable numbers, and hence there are $60$ of these pairs.

It shouldn’t be too hard to see that $abcde + a'b'c'd'e' = 111110$ no matter which allowable number $abcde$ you start with. That is, the numbers in each pair sum to $111110$. Since there are $60$ pairs, the grand total is $60 \cdot 111110 = 6,666,600$.

Henry did essentially the same thing, except that instead of looking at the sum of $abcde$ and $a'b'c'd'e'$, he looked at their average, $111110/2 = 55555$. Since all pairs have the same average, the entire collection of allowable numbers must also have that average. And if the $120$ allowable numbers have an average of $55555$, their total must be $120 \cdot 55555 = 6,666,600$.

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I need help seeing this [If $d$ is one of the allowable digits $1,3,5,7,9$, let $d' = 10 - d$; note that $d'$ is also allowable. ] please. Why is $10-d$ allowable? – user10695 Sep 7 '11 at 22:04
Also, how is $abcde + a'b'c'd'e' = 111110$. I don't get that please. Is this binary? – user10695 Sep 7 '11 at 22:06
@user10695: 1st question: Just do the arithmetic: if $d=1$, $10-d=9$; if $d=3$, $10-d=7$; and so on. Subtracting $1,3,5,7$, or $9$ from $10$ leaves $9,7,5,3$, or $1$, respectively. 2nd question: Not binary, just ordinary addition. $e+e'=10$, so you write down $0$ and carry $1$. Then $d+d'=10$, and the carry makes $11$, so you write down $1$ and carry $1$. The same thing happens in the remaining three columns, so you end up with a total of $111110$. – Brian M. Scott Sep 7 '11 at 22:15
Thanks! What I thought $abcde + a'b'c'd'e' = 111110$ meant was $(a*b*c*d*e) + (a'*b'*c'*d'*e')$ . Now I see what you mean. Thanks for that. – user10695 Sep 7 '11 at 22:33

why all this ?

it is so simple There are 5 digits and as we all know it will 5! numbers (1*2*3*4*5)=120 on a digit place every number will repeat 24 times that is 120/5 on ones place the sum will be 24(1+3+5+7+9)=25*24=600 If u need further explanation why 24(1+3+5+7+9+) see this: at every digit place 1 will repeat 24 times 24*1=24 3 will repeat 24 times 24*3=72 5 will repeat 24 times 24*5=120 7 will repeat 24 times 24*7=168 9 will repeat 24 times 24*9=216 total give 600

every digit place will sum to 600 unit place 600 tenth place 600 hundreds place 600 Thousands place 600 ten thousands place 600 -------------------------------- 6666600

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Use MathJax please. – SchrodingersCat Oct 28 '15 at 16:30


Homework Help: SOlving y-intercept of a sin function

1. Sep 8, 2010

t_n_p

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

y = -2sin2(x+[pi/6])+1 for x [-pi, pi]

3. The attempt at a solution

set y=0,

1/2 = sin2(x+[pi/6])
pi/6 = 2(x+[pi/6]), where pi/6 is the base angle

Now because of the 2 infront of the (x+[pi/6]), I consider twice the domain, i.e. [-2pi, 2pi].
pi/6 is positive, and sin is positive in 1 & 2 quadrants.

therefore,
2(x+[pi/6]) = pi/6, 5pi/6, -7pi/6, -11pi/6.

divide by 2:
(x+[pi/6]) = pi/12, 5pi/12, -7pi/12, -11pi/12.

subtract [pi/6] from both sides:
x = -pi/12, 3pi/12, -9pi/12, -13pi/12.

This is my problem. Is -13pi/12 an issue since it is outside the original domain, or is it still ok since it is inside the modified domain [-2pi,pi]?

the answer has x values of -pi/12, 3pi/12, -9pi/12, 11pi/12, so my only issue is the last value. What has gone wrong!?

2. Sep 8, 2010

CompuChip

-13pi/12 is a solution to the equation y = 0, but you are only asked to give the solutions between -pi and pi. However, if you add (or subtract) any multiple of 2pi to -13pi/12, you get an equivalent solution. In this case, you could find -13pi/12 + 2pi = 11pi/12, which is in the requested interval.

3. Sep 8, 2010

t_n_p

ok, so I ask this, is there anyway to get 11pi/12 using the method I described, or can it only be obtained by adding 2pi?

so hypothetically, if I wanted x-intercepts for all x,
I would do:
-pi/12 +/- n*2pi
3pi/12 +/- n*2pi
-9pi/12 +/- n*2pi
-13pi/12 +/- n*2pi

where n is any integer?

I'm also wondering about the validity of adding pi (rather than 2pi). The period in our case is 2pi/2 = pi, so why doesn't adding/subtracing pi from all our values yield another number of solutions?

4. Sep 8, 2010

CompuChip

The important thing in this type of problem, is to do everything in the right order.

Step 1: finding the base angle and writing down two solution sets
At the beginning of your post, you get to the above equation,
$$\sin 2(x + \pi/6) = \frac12 = \sin \pi/6$$
This equation has two "branches" of solutions, namely
$$2(x + \pi/6) = \pi / 6 + n \cdot 2\pi$$
and
$$2(x + \pi/6) = \pi - \pi / 6 + n \cdot 2\pi$$
for any integer n (= 0, 1, 2, 3, ..., -1, 2, -3, ...).
[If you draw the graph of the sine function or you look at its geometric meaning in a unit circle, you will see how the symmetry leads to the second equation].

You have to add the n*2pi here, because you want the sines of both sides to be equal, and if you add any multiple of 2 pi to any of them, this will be the case.

Step 2: solving for x
Now you can go and solve the two equations for x. The first one leads to
$$x + \pi/6 = \pi / 12 + n \cdot \pi$$
(note that nothing strange happens here, it is just basic algebra: if you divide everything by 2, then n*2pi changes into n*pi automatically. As you remarked, this corresponds to the period) and
$$x = \pi / 12 - \pi / 6 + n \cdot \pi = - \pi/12 + n \cdot \pi$$
For the second branch you will get something similar, which I will leave up to you to work out (be careful with all the minus signs though)

Step 3: Find the specific solutions requested
You now have two equations which describe infinitely many solutions, although basically there are just two and all the others differ from them by an integer number of 2pi steps.
In this case, you are specifically asked to list only the solutions between -pi and pi. So you can go and plug in some numbers for n into
$$x = - \pi/12 + n \cdot \pi$$
For n = 0 you get - pi / 12 which is in the interval, so you can write that down. For n = 1 you get - pi / 12 + pi = 11 pi / 12, that is also OK. For n = 2 you go above pi, and for n = -1 you are below -pi, so this is all the solutions you get.

Do the same for the other "branch" of solutions, and you will find two more.

5. Sep 8, 2010

t_n_p

ok, that makes sense, but is this also feasible....

Follow what I did in my original post, where I found 4 solutions:

x = -pi/12, 3pi/12, -9pi/12, -13pi/12.

Why can't I just add multiples of the period to each of these values?

For x = -pi/12
x=-pi/12 + n*pi
test:
for n=-1, x=-13pi/12 (outside domain, invalid)
for n=0, x = -pi/12 (inside domain, valid)
for n=1, x= 11pi/12 (inside domain, valid)
for n=2, x= 23pi/12 (outside domain, invalid)

For x = 3pi/12
x=3pi/12 + n*pi
test:
for n=-2, x= -21pi/12 (outside domain, invalid)
for n=-1, x=-9pi/12 (inside domain, valid)
for n=0, x=3pi/12 (inside domain, valid)
for n=1, x= 15pi/12 (outside domain, invalid)

For x = -9pi/12
x=-9pi/12 + n*pi
test:
for n=-1, x=-21pi/12 (outside domain, invalid)
for n=0, x=-9pi/12 (inside domain, valid)
for n=1, x= 3pi/12 (inside domain, valid)
for n=2, x=15pi/12 (outside domain, invalid)

For x = -13pi/12
x=-13pi/12 + n*pi
test:
for n=-1, x=-25pi/12 (outside domain, invalid)
for n=0, x=-13pi/12 (outside domain, valid)
for n=1, x= -1pi/12 (inside domain, valid)
for n=2, x=11pi/12 (inside domain, valid)
for n=3, x=23pi/12 (outside domain, invalid)

then compiling all solutions that are valid I'm left with x=-9pi/12, 3pi/12, -pi/12, 11pi/12
as expected.

A little more time consuming but still works. My only question with the above method is, if the period is not pi, for example it is 2pi, will the same method work? I.e. if the period is pi, will adding n*2pi still provide the answers I am after?

6. Sep 8, 2010

CompuChip

Yes, that also works. But where do you get these four solutions from in the first place? Are they guesses or what?

7. Sep 8, 2010

t_n_p

I explained in the original post.

Now because of the 2 infront of the (x+[pi/6]), I consider twice the domain, i.e. [-2pi, 2pi]. I know the base angle is pi/6 and I know sin is positive in the 1st and 2nd quadrants. There will be 2 solutions per a revolution, then since there are 2 revolutions (domain is now -2pi to 2pi), there will be a total of 4 solutions.

That's the way I was taught. Since there is a factor of 2 out the front, I double the original domain. If there is a factor of 3 out the front, I triple the original domain so that it becomes -3pi to 3pi yielding a total of 6 solutions and so and so forth

8. Sep 8, 2010

CompuChip

Ah, I see now. So to be entirely clear:

* If you already find four solutions this way, then you don't need to work everything out like in post #5. All you have to do is shift solutions outside the requested range by an integer number of periods (in this case, pi).

* Whenever you want to do the n * (2)pi thing: if you already divided everything by 2, you use multiples of pi. You only shift by 2pi, if you haven't divided by the 2 (or 3, or whatever number) in front yet.

9. Sep 8, 2010

t_n_p

I definately prefer to do it the 1st star method.
Basically then, when I get a solution outside the bounds, I add or subtract 1 PERIOD (being careful to note what the period is, as it will change from problem to problem) to bring it back inside the bounds. E.g. if period is 500pi, then add/subtract 500pi from any values outside of the required bounds to bring it back inside.

This particular case had me stumped, since 3 of the solutions appear within the bounds, and only 1 is invalid. Obviously this is due to the value of the phase shift relative to the values obtained from the unit circle.

All makes sense now. Thanks


Finding the probability of a selecting at least 1 of an element.

If there are 18 red and 2 blue marbles what is the probability of selecting 10 marbles where there is either 1 or 2 blue marbles in selected set.

Also, it seems intuitive that the probability should be twice that of selecting a single blue marble in a set of 10 from 19 red and 1 blue, but I'm not sure.

Another approachment is through hypergeometric distribution.

• The probability that there is exactly one blue ball in the selected set is: $$P(A_1)=\dfrac{\color{blue}{\dbinom 2 1}\cdot \color{red}{\dbinom {18} {10-1}}}{\dbinom {20}{ 10} }$$

• The probability that there are exactly 2 blue balls in the selected set is:

$$P(A_2)=\dfrac{\color{blue}{\dbinom 2 2}\cdot \color{red}{\dbinom {18} {10-2}}}{\dbinom {20}{ 10} }$$

So, the probability you are looking for is $P(A_1)+P(A_2)\approx 0.763$

In this scenario, with no restrictions, there are 3 possible cases:

$\bullet$ When selecting 10, there were no blue marbles ($A_{none}$)

$\bullet$ When selecting 10, there was one blue marble ($A_1$)

$\bullet$ When selecting 10, there were two blue marbles ($A_2$)

As there are no other possibilities, it stands to reason then that these three events form a partition of our overall sample space. That is: $X = A_{none}\cup A_1 \cup A_2$ and each $A$ is pairwise disjoint from the others.

It follows then, that $1 = P(X) = P(A_{none}) + P(A_1) + P(A_2)$ by rule of sums.

The event you are interested in is that either one or two blue marbles were selected, namely $A_1\cup A_2$.

By the above, $P(A_1\cup A_2) = 1 - P(A_{none})$

To calculate $P(A_{none})$, for each of the 10 marbles selected, you will have selected red. So, multiply the probabilities for each step that you in fact picked red.

So: $P(A_{none}) = \dfrac{18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9}{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11} = \dfrac{10\cdot 9}{20\cdot 19}$

And finally $P(A_1\cup A_2) = 1 - P(A_{none}) = 1 - \frac{10\cdot9}{20\cdot 19}\approx 0.763$

In comparison, if there were 19 red and 1 blue, it winds up being $1 - \frac{10}{20} = 0.5$ The answer to the original question is in fact a bit more than 50% more likely.

• The answer you gave is .864 but I believe it is supposed to be .763. I tried to change it but it said the change need to be at least 6 characters.
– qw3n
Dec 4, 2014 at 14:30
• absolutely correct. My mistake, a little arithmetic mistake. I had originally calculated 9/(2*19) and forgot to subtract it from 1. I did the subtraction from 1 in my head after and somehow messed up the tenths digit. Thanks for catching that. Dec 4, 2014 at 14:39

Given $18$ red marbles and $2$ blue marbles:

• The number of ways to choose $10$ marbles is $\binom{20}{10}=184756$

• The number of ways to choose $10$ red marbles is $\binom{18}{10}=43758$

• Hence the probability to choose only red marbles is $\frac{43758}{184756}=\frac{9}{38}$

• Hence the probability to choose not only red marbles is $1-\frac{9}{38}\approx76.31\%$

Given $19$ red marbles and $1$ blue marble:

• The number of ways to choose $10$ marbles is $\binom{20}{10}=184756$

• The number of ways to choose $10$ red marbles is $\binom{19}{10}=92378$

• Hence the probability to choose only red marbles is $\frac{92378}{184756}=\frac{1}{2}$

• Hence the probability to choose not only red marbles is $1-\frac{1}{2}=50\%$

As you can see, the former is not twice the latter...


Total number of subsets of size atmost $k$

I was working on a problem that involved taking subsets of a multiset. I want to count the total number of distinct subsets of size at most $$k$$.

Example 1:

consider the multiset $$S = \{ 3, 3, 5, 7 \}$$ and $$k=3$$ then the answer should be $$12$$.

That is $$\{ \}, \{ 3 \}, \{ 3 \}, \{ 5 \}, \{ 7 \}, \{ 3, 5 \}, \{ 3, 7 \}, \{ 3, 5 \}, \{ 3, 7 \}, \{ 5, 7 \}, \{ 3, 5, 7\}, \{ 3, 5, 7 \} = 12$$ subsets.

Example 2:

$$S = \{ 2, 3, 5 \}$$ and $$k=2$$ then the answer should be $$7.$$

$$\{ \}, \{ 2 \}, \{ 3 \}, \{ 5 \}, \{ 2, 3 \}, \{ 2, 5 \}, \{ 3, 5 \} = 7$$ subsets

By using the formula from this answer I can count the total number of distinct subsets. How can I extend that formula for my constraints? Any idea?

• For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Thomas Shelby Sep 11 '19 at 17:04
• sorry for that.please edit my question if i make any mistake. – Midhun Manohar Sep 11 '19 at 17:07
• I do not follow your first example. Why do include $\{3\}$ twice? Why do you include $\{3,5\}$ twice? Why also did you not include $\{3,3\},\{3,3,5\}$ or $\{3,3,7\}$? If you intend for each $3$ to be considered distinct... then why do you use identical characters for them? It would have made more sense to never consider multisets in the first place and instead talk about the subsets of size at most $3$ from the set $\{3,\color{red}{X},5,7\}$ instead where we replaced the second three with $X$... at which point you have $\sum\limits_{i=0}^k\binom{n}{k}$ such subsets. – JMoravitz Sep 11 '19 at 18:33
• elements of the subset should be distinct – Midhun Manohar Sep 12 '19 at 5:04

Let me restate the problem, and see if you agree that it is the same thing. We have a finite collection of elements of types $$1,2,\dots,n$$ with $$a_j$$ elements of type $$j$$ for $$j=1,2,\dots,n$$, and we wish to know how many ways we can select at most $$k$$ objects, subject to the condition that no more than one element of any type is selected.

In your first example, we have $$2$$ elements of type "$$3$$" and one element of each of types "$$5$$" and "$$7$$".

Exactly $$k$$ elements may be selected in $$\sum a_{j_1}a_{j_2}\cdots a_{j_k}$$ ways, where the sum is over all $$k$$-tuples $$(j_1,j_2,...j_k)$$ with $$1\leq j_1

If you want at most $$k$$ objects, just add up the values for each nonnegative integer $$\leq k$$.

In you first example, with $$n=3$$, we have $$a_1=2,a_2=1,a_3=1$$. When $$k=0$$ we have an empty product, so the value is $$1$$. When $$k=1$$, we get $$2+1+1=4.$$ When $$k=2$$, we get $$2\cdot1+2\cdot1+1\cdot1=5$$. When $$k=3$$, we get $$2\cdot\cdot1=2$$. Altogether, we have $$1+4+5+2=12.$$

I gave a more concrete answer to a similar question a few days ago. Look at Find number of ways to select subset with distinct objects of at most K size..

• it seems working.can you elaborate more – Midhun Manohar Sep 11 '19 at 17:30
• @Midhunmanohar What is it that you don't understand? – saulspatz Sep 11 '19 at 17:32
• i have never seen use of multiplier symbol ∏ like this.can you please elaborate the formula more – Midhun Manohar Sep 12 '19 at 6:06
• @Midhunmanohar That was my fault. I meant sum, not product. No wonder you didn't understand. I don't know where my mind was. Anyway, it's the sum over all such products. I've edited my answer with a link to another answer that may be easier to follow. – saulspatz Sep 12 '19 at 10:47


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# Verify proof of $f(x)=e^x$ if $f(x+y)=f(x)f(y)$ and $f'(x)$ exists for all $x$

This is exercise 6.26.8 from Tom Apostol's Calculus I, I'd like to ask someone to verify my proof. I'd be also interested in alternative proofs:

If $f(x+y)=f(x)f(y)$ for all $x$ and $y$ and if $f(x)=1+xg(x)$, where $g(x) \to 1$ as $x \to 0$, prove that (a) $f'(x)$ exists for every $x$, and (b) $f(x)=e^x$.

(a) $$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0}f(x)\frac{1 + hg(h) - 1}{h} = \lim_{h \to 0}f(x)g(h) = f(x)$$

(b) $$\left(\frac{f(x)}{e^x}\right)' = \frac{f'(x)e^x - f(x)e^x}{e^{2x}} = \frac{f(x) - f(x)}{e^x} = 0 \implies f(x) = ke^x \; \text, \; k\in \mathbb R$$ $$k = ke^0 = f(0) = \lim_{x \to 0}1 + xg(x) = 1 \implies f(x) = e^x$$

• Yes, it's fine. – egreg Dec 15 '15 at 0:42

Yes, it's fine. You have less computations if you set $F(x)=f(x)e^{-x}$, so $$F'(x)=f'(x)e^{-x}-f(x)e^{-x}=0$$ and $F$ is constant.
An alternative proof could be by observing that $f(x)=0$ for some $x$ implies $f$ constant $0$, which contradicts the assumptions. So we know $f(x)\ne0$ for all $x$ and differentiability (part a) implies $f$ is continuous, so everywhere positive. Then $$F(x)=\log f(x)$$ is well defined and $$F'(x)=\frac{f'(x)}{f(x)}=1$$


# Arranging cats and dogs - what is wrong with my approach

We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?

Since there are 5 spaces the cats can be in with the dogs fixed, there are $${5 \choose 3} * 4! * 3! = 1440$$ ways and this is the correct answer.

I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $$4 * 3$$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)

The other two dogs are free to go to any of the 4 spaces, with $$4^2$$ possibilities.

The cats can now be arranged in $$3!$$ ways.

So, our final answer should be $$3! * 4^2 * 4 * 3 = 1152$$

Where have I gone wrong?

• Is your problem arising as a consequence of a rainfall ? Mar 30, 2019 at 16:27

The second computation is missing a symmetry. Say your initial pattern is $$\underline {\quad}C_1\underline {\quad}C_2\underline {\quad}C_3\underline {\quad}$$

You then populate the spaces immediately to the right of $$C_1$$, and $$C_2$$. As:

$$\underline {\quad}C_1D_1\underline {\quad}C_2D_2\underline {\quad}C_3\underline {\quad}$$

So far so good. You still have $$D_3,D_4$$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?

Taking the two possible orders into account, we see that you are missing $$4\times 3!\times 4\times 3=288$$ cases. Adding them back gives you the desired result.

Phrased differently: once you have placed $$D_3$$ there are now five available spaces for $$D_4$$ (since $$D_4$$ might go either to the left or to the right of $$D_3$$). thus you should have had $$3!\times 4\times 5\times 4\times 3=1440$$

You can separate the two cases for the two last dogs: single dogs and double dogs.

Single dogs: $$P(4,2)=\frac{4!}{2!}=12.$$ Double dogs: $$P(2,2)\cdot C(4,1)=2\cdot 4=8.$$ Hence, there are $$12+8=20$$ (not $$4^2=16$$) ways to distribute the last two dogs.

The final answer is: $$3!\cdot 20\cdot 4\cdot 3=1440.$$


# Contour Plot not taking previously defined expressions?

I am new to Mathematica so my question might sound a bit silly, but I hope you help me.

While learning ContourPlot, I have learned that when I use the previously defined expressions as inputs onto it, it doesn't seem to work, and only putting them in manually seems to work.

For example, defining

f1 = x^2/9 + y^2/4 == 1
f2 = x^2 - 1 == y

and then evaluating

ContourPlot[{f1, f2}, {x, -3, 3}, {y, -3, 8}]

results in empty plot, while putting the expressions manually inside such as

ContourPlot[{x^2/9 + y^2/4 == 1, x^2 - 1 == y}, {x, -3, 3}, {y, -3, 8}]

results in what I want...

What's more is, this also seems to occur when using NSolve... Can anyone tell me what's going on?

• I think it has something to do with the HoldAll attribute of ContourPlot... It works with ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]. – MelaGo Oct 15 '19 at 5:44
• Thank you! Could you specify what you mean by that? – Danny Han Oct 15 '19 at 7:57
• @DannyHan As @MelaGo noted, ContourPlot has attribute HoldAll. Before evaluating anything (in particular, before inserting the definitions of f1 and f2), ContourPlot will look at the first argument and decide what type of plot you want - whether you're giving it a single curve or a list and whether you are giving it a function or an equation. (continued...) – Lukas Lang Oct 15 '19 at 8:25
• (...continued) Here, it sees {f1, f2} and decides to plot two functions. When it starts evaluating f1 and f2 for given x and y, it doesn't get a number, but True and False, so it doesn't plot anything (non-numerical results are simply ignored by all/most plotting functions). Evaluate forces evaluation to occur before ContourPlot, so ContourPlot sees a list of equations, which it can then correctly plot. – Lukas Lang Oct 15 '19 at 8:25
• @LukasLang Great explanation! Worth posting as an answer. – Alexey Popkov Oct 15 '19 at 10:24

As suggested, I'm expanding my comment into an answer

The problem here is the HoldAll attribute of ContourPlot. Like Plot and similar functions the process goes something like this:

• Look at the first argument, and decide what form it has:
• If it's a list, the user wants to plot multiple functions
• If it's an equation with ==, the user wants to plot the solution to that equation
• Start the evaluation at different points (this part is done recursively on many points)
• Set the values of the variables (second and third argument) to the correct values (similar to Block, as noted in the details section of ContourPlot)
• Evaluate the first argument and use the result

Now we see what the problem in your case is:

ContourPlot[{f1, f2}, {x, -3, 3}, {y, -3, 8}]

When ContourPlot decides on the type of plot you want, it decides on "plot contours for two functions", since all it sees at that point is {f1, f2}. Now, values like e.g. x=0,y=0 are assigned and f1,f2 are evaluated. The problem is that this results in e.g.

          {f1, f2}
(* --> *) {x^2/9 + y^2/4 == 1,x^2 - 1 == y}
(* --> *) {0^2/9 + 0^2/4 == 1,0^2 - 1 == 0}
(* --> *) {0 == 1,- 1 == 0}
(* --> *) {False, False}

And like most plotting functions, non-numeric results (like the False above, are simply discarded).

It is now also clear why Evaluate fixes the issue: It forces the first argument of ContourPlot to be evaluated before ContourPlot has a chance to look at it. So now the evaluation sequence goes like this:

          ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]
(* --> *) ContourPlot[{x^2/9 + y^2/4 == 1,x^2 - 1 == y}, {x, -3, 3}, {y, -3, 8}]

At this point ContourPlot examines the first argument and decides on "plot the solutions of two equations", which is what we want.

It should be noted that the only thing that ContourPlot needs to see are the lists (when multiple things are to be plotted) and the equations - everything else can be evaluated later. This means the following will also work:

g1 = x^2/9 + y^2/4
g2 = x^2 - 1

ContourPlot[{g1 == 1, g2 == y}, {x, -3, 3}, {y, -3, 8}]

## TL;DR;

To summarise, force evaluation of the first argument of ContourPlot using Evaluate to ensure that the right type of plot is chosen:

ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]

• OMG thank you so much! Now I understand :) – Danny Han Oct 15 '19 at 12:29


# Section12.5The Multivariable Chain Rule¶ permalink

The Chain Rule, as learned in Section 2.5, states that $\ds \frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)g'(x)\text{.}$ If $t=g(x)\text{,}$ we can express the Chain Rule as \begin{equation*} \frac{df}{dx} = \frac{df}{dt}\frac{dt}{dx}. \end{equation*}

In this section we extend the Chain Rule to functions of more than one variable.

It is good to understand what the situation of $z=f(x,y)\text{,}$ $x=g(t)$ and $y=h(t)$ describes. We know that $z=f(x,y)$ describes a surface; we also recognize that $x=g(t)$ and $y=h(t)$ are parametric equations for a curve in the $x$-$y$ plane. Combining these together, we are describing a curve that lies on the surface described by $f\text{.}$ The parametric equations for this curve are $x=g(t)\text{,}$ $y=h(t)$ and $z=f\big(g(t),h(t)\big)\text{.}$

Consider Figure 12.5.2 in which a surface is drawn, along with a dashed curve in the $x$-$y$ plane. Restricting $f$ to just the points on this circle gives the curve shown on the surface. The derivative $\frac{df}{dt}$ gives the instantaneous rate of change of $f$ with respect to $t\text{.}$ If we consider an object traveling along this path, $\frac{df}{dt}$ gives the rate at which the object rises/falls.

We now practice applying the Multivariable Chain Rule.

##### Example12.5.3Using the Multivariable Chain Rule

Let $z=x^2y+x\text{,}$ where $x=\sin(t)$ and $y=e^{5t}\text{.}$ Find $\ds \frac{dz}{dt}$ using the Chain Rule.

Solution

The previous example can make us wonder: if we substituted for $x$ and $y$ at the end to show that $\frac{dz}{dt}$ is really just a function of $t\text{,}$ why not substitute before differentiating, showing clearly that $z$ is a function of $t\text{?}$

That is, $z = x^2y+x = (\sin(t) )^2e^{5t}+\sin(t) .$ Applying the Chain and Product Rules, we have \begin{equation*} \frac{dz}{dt} = 2\sin(t) \cos(t) \, e^{5t}+ 5\sin^2(t) \,e^{5t}+\cos(t) , \end{equation*} which matches the result from the example.

This may now make one wonder “What's the point? If we could already find the derivative, why learn another way of finding it?” In some cases, applying this rule makes deriving simpler, but this is hardly the power of the Chain Rule. Rather, in the case where $z=f(x,y)\text{,}$ $x=g(t)$ and $y=h(t)\text{,}$ the Chain Rule is extremely powerful when we do not know what $f\text{,}$ $g$ and/or $h$ are. It may be hard to believe, but often in “the real world” we know rate–of–change information (i.e., information about derivatives) without explicitly knowing the underlying functions. The Chain Rule allows us to combine several rates of change to find another rate of change. The Chain Rule also has theoretic use, giving us insight into the behavior of certain constructions (as we'll see in the next section).

We demonstrate this in the next example.

##### Example12.5.4Applying the Multivarible Chain Rule

An object travels along a path on a surface. The exact path and surface are not known, but at time $t=t_0$ it is known that : \begin{equation*} \frac{\partial z}{\partial x} = 5,\qquad \frac{\partial z}{\partial y}=-2,\qquad \frac{dx}{dt}=3\qquad \text{ and } \qquad \frac{dy}{dt}=7. \end{equation*}

Find $\frac{dz}{dt}$ at time $t_0\text{.}$

Solution

We next apply the Chain Rule to solve a max/min problem.

##### Example12.5.5Applying the Multivariable Chain Rule

Consider the surface $z=x^2+y^2-xy\text{,}$ a paraboloid, on which a particle moves with $x$ and $y$ coordinates given by $x=\cos(t)$ and $y=\sin(t)\text{.}$ Find $\frac{dz}{dt}$ when $t=0\text{,}$ and find where the particle reaches its maximum/minimum $z$-values.

Solution

We can extend the Chain Rule to include the situation where $z$ is a function of more than one variable, and each of these variables is also a function of more than one variable. The basic case of this is where $z=f(x,y)\text{,}$ and $x$ and $y$ are functions of two variables, say $s$ and $t\text{.}$

##### Example12.5.8Using the Multivarible Chain Rule, Part II

Let $z=x^2y+x\text{,}$ $x=s^2+3t$ and $y=2s-t\text{.}$ Find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}\text{,}$ and evaluate each when $s=1$ and $t=2\text{.}$

Solution
##### Example12.5.9Using the Multivarible Chain Rule, Part II

Let $w = xy+z^2\text{,}$ where $x= t^2e^s\text{,}$ $y= t\cos(s)\text{,}$ and $z=s\sin(t)\text{.}$ Find $\frac{\partial w}{\partial t}$ when $s=0$ and $t=\pi\text{.}$

Solution

# Subsection12.5.1Implicit Differentiation

We studied finding $\frac{dy}{dx}$ when $y$ is given as an implicit function of $x$ in detail in Section 2.6. We find here that the Multivariable Chain Rule gives a simpler method of finding $\frac{dy}{dx}\text{.}$

For instance, consider the implicit function $x^2y-xy^3=3\text{.}$ We learned to use the following steps to find $\frac{dy}{dx}\text{:}$ \begin{align*} \frac{d}{dx}\Big(x^2y-xy^3\big) \amp = \frac{d}{dx}\Big(3\Big)\\ 2xy + x^2\frac{dy}{dx}-y^3-3xy^2\frac{dy}{dx} \amp = 0\\ \frac{dy}{dx} = -\frac{2xy-y^3}{x^2-3xy^2}. \end{align*}

Instead of using this method, consider $z=x^2y-xy^3\text{.}$ The implicit function above describes the level curve $z=3\text{.}$ Considering $x$ and $y$ as functions of $x\text{,}$ the Multivariable Chain Rule states that $$\frac{dz}{dx} = \frac{\partial z}{\partial x}\frac{dx}{dx}+\frac{\partial z}{\partial y}\frac{dy}{dx}. \label{eq_mchain1}\tag{12.5.1}$$

Since $z$ is constant (in our example, $z=3$), $\frac{dz}{dx} = 0\text{.}$ We also know $\frac{dx}{dx} = 1\text{.}$ Equation (12.5.1) becomes \begin{align*} 0 \amp = \frac{\partial z}{\partial x}(1) + \frac{\partial z}{\partial y}\frac{dy}{dx} \Rightarrow\\ \frac{dy}{dx} \amp = -\frac{\partial z}{\partial x}\Big/\frac{\partial z}{\partial y}\\ \amp = -\frac{\,f_x\,}{f_y}. \end{align*}

Note how our solution for $\frac{dy}{dx}$ in Equation <<Unresolved xref, reference "eq_mchain2"; check spelling or use "provisional" attribute>> is just the partial derivative of $z$ with respect to $x\text{,}$ divided by the partial derivative of $z$ with respect to $y\text{.}$

We state the above as a theorem.

We practice using Theorem 12.5.10 by applying it to a problem from Section 2.6.

##### Example12.5.11Implicit Differentiation

Given the implicitly defined function $\sin(x^2y^2)+y^3=x+y\text{,}$ find $y'\text{.}$ Note: this is the same problem as given in Example 2.6.7 of Section 2.6, where the solution took about a full page to find.

Solution

# Subsection12.5.2Exercises

Terms and Concepts

In the following exercises, functions $z=f(x,y)\text{,}$ $x=g(t)$ and $y=h(t)$ are given.

1. Use the Multivariable Chain Rule to compute $\lz{z}{t}\text{.}$

2. Evaluate $\lz{z}{t}$ at the indicated $t$-value.

In the following exercises, functions $z=f(x,y)\text{,}$ $x=g(t)$ and $y=h(t)$ are given. Find the values of $t$ where $\frac{dz}{dt}=0\text{.}$ Note: these are the same surfaces/curves as found in Exercises 12.5.2.712.5.2.12.

In the following exercises, functions $z=f(x,y)\text{,}$ $x=g(s,t)$ and $y=h(s,t)$ are given.

1. Use the Multivariable Chain Rule to compute $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}\text{.}$

2. Evaluate $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ at the indicated $s$ and $t$ values.

In the following exercises, find $\lz{y}{x}$ using Implicit Differentiation and Theorem 12.5.10.

In the following exercises, find $\lz{z}{t}\text{,}$ or $\plz{z}{s}$ and $\plz{z}{t}\text{,}$ using the supplied information.


# In a geometric progression, we know the partial sums $S_2 = 7$ and $S_6 = 91$. Find $S_4$.

In a geometric progression, $$S_2 = 7$$ and $$S_6 = 91$$. Evaluate $$S_4$$. Alternatives: 28, 32, 35, 49, 84.

Here's what I tried so far:

$$S_2 = \frac{a_1(1-r^2)}{1-r} \implies 1-r = \frac{a_1(1-r^2)}{7} \\ S_6 = \frac{a_1(1-r^6)}{1-r} \implies 1-r = \frac{a_1(1-r^6)}{91}$$

Then: $$\frac{1-r^2}{1} = \frac{1-r^6}{13} \\ r^6 - 13r^2 + 12 = 0$$

Now i can't solve this equation, perhaps there's an easier way…

• The formulas you're using are for the sum of the first ever-so-many terms of the progression... is that what the $S_n$ are, the partial sums of the progression? Or are they the terms in the progression themselves? – Eevee Trainer May 7 '19 at 4:39
• $S_n$ is the sum of the first n terms of the progression. – rodorgas May 7 '19 at 4:41
• Given $r\ne\pm1$, you could have simplified $(1-r^6)/(1-r^2)=1+r^2+r^4$ – J. W. Tanner May 7 '19 at 5:04
• You absolutely should explain such key pieces in the question body. Those are easy to miss in comments. Also, a series has infinitely many terms. Your sums don't. Read the tag descriptions before using them. – Jyrki Lahtonen May 7 '19 at 5:58
• What are $S_2$ and $S_6$, precisely ? – Yves Daoust May 7 '19 at 6:07

Let $$x=r^2$$ then we see $$x^3-13x+12=0$$ so $$x^3-x-12x+12=0$$ so $$x(x-1)(x+1)-12(x-1)=0$$ so $$(x-1)(x^2+x-12)=0$$ so $$(x-1)(x+4)(x-3)=0.$$

• Ooops, sorry, my bad. I will delete this comment. – Yves Daoust May 7 '19 at 6:29

Here's how to do it with only a quadratic equation (of sorts).

Denote the geometric sequence $$a_1, a_2=a_1r, a_3=a_1r^2, a_4=a_1r^3, a_5=a_1r^4, a_6=a_1r^5...$$

Then $$S_2=a_1(1+r), S_4=a_1(1+r+r^2+r^3)=a_1(1+r)(1+r^2)$$,

and $$S_6=a_1(1+r+r^2+r^3+r^4+r^5)=a_1(1+r)(1+r^2+r^4).$$

$$\dfrac {S_6}{S_2}=\dfrac{91}7=13=1+r^2+r^4$$ so $$(r^2)^2+(r^2)-12=(r^2-3)(r^2+4)=0$$

so $$r^2=3$$ so $$S_4=S_2(1+r^2)=7(1+3)=28.$$

$$\frac{S_6}{S_2}=\frac{r^6-1}{r^2-1}=r^4+r^2+1=13$$ and

$$\frac{S_4}{S_2}=\frac{r^4-1}{r^2-1}=r^2+1.$$

With $$s:=r^2+1$$, we have $$s(s-1)+1=13$$

giving the two solutions

$$S_4=4\cdot 7$$ and $$S_4=-3\cdot7.$$

• $-21$ was not one of the alternatives offered by OP; presumably the progression is in real numbers, not complex – J. W. Tanner May 7 '19 at 6:39
• @J.W.Tanner: then $-21$ is rejected. – Yves Daoust May 7 '19 at 6:40

$$\begin{cases} S_2=a_1+a_1r=a_1(1+r)=7\\ S_6=a_1+a_1r+\cdots+a_1r^5=91\end{cases}\\ S_6-S_2=a_1r^2(1+r+r^2+r^3)=a_1r^2\cdot \frac{1-r^4}{1-r}=84\\ \frac{S_6-S_2}{S_2}=\frac{a_1r^2(1-r^4)}{a_1(1+r)(1-r)}=r^2(1+r^2)=12 \Rightarrow r^2=3$$ Hence: \begin{align}S_4&=a_1(1+r+r^2+r^3)=\\ &=a_1(1+r+r^2(1+r))=\\ &=a_1(1+r)(1+r^2)=\\ &=S_2(1+r^2)=\\ &=7(1+3)=\\ &=28.\end{align}


# Find a_n

#### anemone

##### MHB POTW Director
Staff member
Let $a_0=2$, $a_1=3$, $a_2=6$, and for $n \ge 3$, $a_n=(n+4)a_{n-1}-4na_{n-2}+(4n-8)a_{n-3}$.

The first few terms are $2,\;\;3,\;\;6,\;\;14, \;\;40, \;\;152, \;\;784, \;\;5168,\;\; 40576, \;\;363392$.

Find with proof a formula for $a_n$ of the form $a_n=c_n+d_n$, where $c_n$ and $d_n$ are well-known sequences.

Last edited:

#### mente oscura

##### Well-known member
Let $a_0=2$, $a_1=3$, $a_2=6$, and for $n \ge 3$, $a_n=(n+4)a_{n-1}-4na_{n-2}+(4n-8)a_{n-3}$.

The first few terms are $2,\;\;3,\;\;6,\;\;14, \;\;40, \;\;152, \;\;784, \;\;5168,\;\; 40576, \;\;363392$.

Find with proof a formula for $a_n$ of the form $a_n=c_n+d_n$, where $c_n$ and $d_n$ are well-known sequences.
Hello.

I guessed the sequences, because I have not been able to prove it.I thought that it could be a factor involved, and, then, by differences I found with another string.

$$a_n=2^n+n!$$

Regards.

#### MarkFL

##### Administrator
Staff member
Here is my solution:

Let's rewrite the recurrence as:

$$\displaystyle a_{n}-na_{n-1}=4\left(a_{n-1}-(n-1)a_{n-2} \right)-4\left(a_{n-2}-(n-2)a_{n-3} \right)$$

Now, if we define:

$$\displaystyle b_{n}=a_{n}-na_{n-1}$$

We may write the original recursion as:

$$\displaystyle b_{n}=4b_{n-1}-4b_{n-2}$$

This is a linear homogenous recursion with the repeated characteristic root $r=2$. Hence the closed form for $b_n$ is:

$$\displaystyle b_{n}=(A+Bn)2^n$$

Using the given initial values, we find:

$$\displaystyle b_1=a_1-a_0=3-2=1=(A+B)2$$

$$\displaystyle b_2=a_2-2a_1=6-6=0=(A+2B)4$$

From this 2X2 linear system, we find:

$$\displaystyle A=1,\,B=-\frac{1}{2}$$

Hence:

$$\displaystyle b_{n}=2^n-n2^{n-1}=a_{n}-na_{n-1}$$

Now, we may arrange this as:

$$\displaystyle a_{n}-2^{n}=n\left(a_{n-1}-2^{n-1} \right)$$

This implies one solution is $$\displaystyle c_n=2^n$$

If we define:

$$\displaystyle d_n=a_{n}-2^{n}$$

we then have:

$$\displaystyle d_{n}=nd_{n-1}\implies d_n=Cn!$$

And so by superposition we have the general form:

$$\displaystyle a_n=2^n+Cn!$$

Using the initial value we obtain:

$$\displaystyle a_0=2=2^0+C0!=1+C\implies C=1$$

Hence:

$$\displaystyle a_n=2^n+n!$$

#### anemone

##### MHB POTW Director
Staff member
Hello.

I guessed the sequences, because I have not been able to prove it.I thought that it could be a factor involved, and, then, by differences I found with another string.

$$a_n=2^n+n!$$

Regards.
Well, even though you didn't provide any proof, your answer is correct and since you've been actively engaged with our site for quite some time and solving many challenge problems in the Challenge Questions and Puzzles sub-forum, I would give allowance to you and hence I would declare it here that you got full mark for that!

Thanks for participating, mente!

Here is my solution:

Let's rewrite the recurrence as:

$$\displaystyle a_{n}-na_{n-1}=4\left(a_{n-1}-(n-1)a_{n-2} \right)-4\left(a_{n-2}-(n-2)a_{n-3} \right)$$

Now, if we define:

$$\displaystyle b_{n}=a_{n}-na_{n-1}$$

We may write the original recursion as:

$$\displaystyle b_{n}=4b_{n-1}-4b_{n-2}$$

This is a linear homogenous recursion with the repeated characteristic root $r=2$. Hence the closed form for $b_n$ is:

$$\displaystyle b_{n}=(A+Bn)2^n$$

Using the given initial values, we find:

$$\displaystyle b_1=a_1-a_0=3-2=1=(A+B)2$$

$$\displaystyle b_2=a_2-2a_1=6-6=0=(A+2B)4$$

From this 2X2 linear system, we find:

$$\displaystyle A=1,\,B=-\frac{1}{2}$$

Hence:

$$\displaystyle b_{n}=2^n-n2^{n-1}=a_{n}-na_{n-1}$$

Now, we may arrange this as:

$$\displaystyle a_{n}-2^{n}=n\left(a_{n-1}-2^{n-1} \right)$$

This implies one solution is $$\displaystyle c_n=2^n$$

If we define:

$$\displaystyle d_n=a_{n}-2^{n}$$

we then have:

$$\displaystyle d_{n}=nd_{n-1}\implies d_n=Cn!$$

And so by superposition we have the general form:

$$\displaystyle a_n=2^n+Cn!$$

Using the initial value we obtain:

$$\displaystyle a_0=2=2^0+C0!=1+C\implies C=1$$

Hence:

$$\displaystyle a_n=2^n+n!$$
Well done, MarkFL! I just love to read your solution posts because they are always so nicely written and well explained! Bravo, my sweetest global moderator!


# A group of order $8$ has a subgroup of order $4$

Let $G$ be a group of order $8$. Prove that there is a subgroup of order $4$.

I know that if $G$ is cyclic then there is such a subgroup (if $G=\langle a\rangle$ then the order of $\langle a^2\rangle$ is $4$). But how do I prove this when $G$ is not cyclic? Also, I know that $G$ has an element of order $2$, because the order of $G$ is even. I suspect that assuming all elements of $G$ are of order $2$ somehow leads to a contradiction but am unable to show it. Is this correct or is there a different approach that I'm missing? thanks

• The group might have all elements of order $2$, but then it is abelian (standard exercise), and the result is easy. Jul 11 '13 at 13:01
• A more general question is answered here: math.stackexchange.com/questions/306343/…? Jul 11 '13 at 13:07
• @GerryMyerson Though that requires quite a bit more than this special case. Jul 11 '13 at 13:11
• @Tobias, yes, which is why I'm not suggesting closing this question as a duplicate. But if OP wants to see something a little more hefty, I've pointed to a place to look. Jul 11 '13 at 13:12

You already noted that if $G$ has an element of order $8$ or $4$ then we are done.

Thus we can assume all elements have order $2$ (except the identity element). Then $G$ is abelian (this is a standard exercise, and I am certain it has been asked on MSE several times).

Let $a$ and $b$ be distinct elements of order $2$. Now it is straightforward to check that $\{e,a,b,ab\}$ is a subgroup of order $4$ (where $e$ is the identity element of $G$).

If $G$ is abelian so according to the Fundamental theorem for finite abelian groups we have: $$G\cong~~~\mathbb Z_8,~~\mathbb Z_4\times\mathbb Z_2,~~\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$$ So lets assume that $G$ is not abelian. If there is an element in $G$ which is of order $8$ then we have a contradiction. If all non trivial elements of $G$ be of order $2$ so again we have $G=\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$ which is a new contradiction, so we at least know that there is an elemnt of order $4$. I think we can stop here cause you wanted to know that. For the next you can use the subgroup generated by $x$ to show that $G=\langle y,x\rangle$ in which $y\in G-\langle x\rangle$ and of course $$G\cong \text{Q}_8,~~\text{or}~~\text{D}_8$$

• @amWhy: Yes Amy. What can I say about the event for my small job here? My English is not good as I can defend my self. :( Some one did the wrong job and someone else should pay the price. I think to myself I have been suggesting everyone I know here to be honest with policy but you see that I am a clueless victim. Sorry for saying these upsetting words but I don't have a tribune for saying it. I have just you to hear my stroy. Sorry Amy. Jul 12 '13 at 1:53
• @BabakS.: Sorry that you are going through this! I cannot even understand what happened, how is it possible to lose 5k rep in a day? It just does not make sense and must be a bug of some kind! Hope it gets resolved and it is not likely not something you did my friend! as amWhy said, hang in there! Jul 12 '13 at 4:10
• @amWhy: I am warned in a kind way. I don't know what happened because as I read somewhere in Meta, we cannot trace who downvote or upvote others. They think, I have another account and that is why I have more than 5k. Yes, I have and I told rob about it but that account was lost. I have nothing from that account. Even, I wanted rob to merge it for me. I have just 2 questions there. It is ridiculous for me having another active account just making plus. Jul 12 '13 at 15:10
• @amWhy:Where will I reach by doing this? More than you or Brian?? Where? What benefit will be for me? ISI Papers? Noble Prize? Believe me, I want to devote all reputations. A real sale is on the way. Sorry my friend. Jul 12 '13 at 15:11
• @BabakS. I defended your honor: I told Alexander Gruber that you are the most humble, gracious, and honest person that I know: and I said that with all sincerity. I think we all know you too well: I told him that you have had students who participate, and that I've see you suggest to them (at least, e.g., Flashdesign, that they accept others' answers...that you tend to answer them only when no one else has. I suspect that you've had an admiring student or two who enthusiastically upvoted some of your posts too eagerly, not knowing the consequences. Jul 12 '13 at 15:16

This is a direct consequence of the following theorem.

Let $G$ be a group of order $p^{n}$ for $p$ a prime. Then for each $m$ with $0 \leq m \leq n$, $G$ has a subgroup of order $p^{m}$.

The proof is here.

• This is the result pointed to by Gerry Meyerson. The proof takes considerably more than is needed for this special case. Jul 11 '13 at 13:24

I would consider elements of different orders and take cases accordingly.

You already are happy with what to do if there is an element of order 8 (and hence the group is generated by this element and cyclic).

Next, if there is a generator of order four, there clearly can only be one other generator, and this must be of order two. If we have $a$ and $b$ of orders 2 and 4 respectively you can easily check check what all the elements are, and it should be obvious what the subgroup of order 4 is.

Lastly, if you have a generator of order two, you could have another generator of order 4, but we have already considered this. The only other alternative is to have two more generators of order 2, which must commute (else the group they generate would be too large) and again, if you consider the abelian group generated by $a$, $b$ and $c$, all of order two, it should be clear how you can generate a group of order four.

Note that this is not the most elegant solution, but it is (probably) the most concrete. And please ask if anything is not clear.

• "if there is an element of order four, there clearly can only be one other element" Huh? Surely, that's not what you meant to write. Jul 11 '13 at 13:13
• If there is an element of order $4$, you are clearly done. And the fact that the elements of order $2$ will commute follows only once you use that all the elements have order $2$. Jul 11 '13 at 13:13
• @GerryMyerson sorry, I meant to talk about generators, not elements. Thank you! Jul 11 '13 at 13:15
• A group of order $8$ cannot have a generator of order $4$. Jul 11 '13 at 13:17
• When you say G is generated by a, b and c you mean G = {e, a, b, c, ab, ac, b*c...}? Jul 11 '13 at 13:36


# How do I find the antiderivative of $5\sin(4x)$?

This is actually a part of a bigger problem, which involves using the Mean Value Theorem for Integrals. The question is to find, $f_{ave}$:

$f(x) = 5 \sin 4x$ for $x\in [−π, π].$

Using the theorem, I have gotten it down to:

$$\frac{1}{2\pi} \int_{-\pi}^\pi5\sin(4x)dx = 5\sin(4c)$$

I know that I have to find the antiderivative and then solve for c, but I don't know how to find the antiderivative. Any help will be appreciated.

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HINT: Do a substitution; set $u=4x$, and solve the integral. –  Arturo Magidin Feb 20 '12 at 3:31
Am I overall on the right track? –  user754950 Feb 20 '12 at 3:34
@user754950: Yes, what you are doing will lead to an answer, once you "solve" the integral. And the hint I gave you should help you solve that integral. –  Arturo Magidin Feb 20 '12 at 3:38
A solution method surprisingly not listed below is integration by parts. I know,it would initially have resulted in a more complicated integral then the other methods here-but it probably would result in the same solution more directly then the methods below. I suggest it just to make sure the list of proposed solutions is as complete as possible. –  Mathemagician1234 Feb 20 '12 at 7:40

1st you can pull the $5$ out in front of the integral sign to give you $\frac{5}{2\pi}$. Then the antiderivative of $\sin(4x)$ is $-\frac{1}{4}\cos(4x)$ because the antiderivative of $\sin(x)$ is $-\cos(x)$ [recall that the derivative of $\cos(x)$ is $-\sin(x)$] then you multiply that by the reciprocal of the constant associated with $x$ [meaning $1$ over $4$ or $\frac{1}{4}$]. Make sure you evaluate the problem from $-\pi$ to $\pi$.

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alright, I did as you said and I got: -1/(4pi) = sin(4c). Is this correct so far? If so, how do I solve for c? does it involve sine inverse? –  user754950 Feb 20 '12 at 3:50
I am not really sure where you get the 5sin(4c) from. Can you explain that so I can better understand your goal? –  Jared Feb 20 '12 at 4:09
Take a look at the graph of your equation on a graphing calculator. Note that the region you are integrating cancels out, but if you change your limits so you are integrating from 0 to pi/4, you can multiply the integral by 8 to get a non-zero, non-negative value. –  Jared Feb 20 '12 at 4:23
I'm not sure how to put integral symbols in the text; they will be written out in words with curly brackets. Now that you have redefined your limits, your integral problem should read (8/2pi){integral from 0 to pi/4}5sin(4x)dx = (4/pi){integral from 0 to pi/4}5sin(4x)dx. Now since the 5 is a constant, move it outside the integral: (20/pi){integral from 0 to pi/4}sin(4x)dx. Use u-substitution so u=4x and du=4dx, which means (1/4)du=dx. Since you changed from 4x to u, also change your interval: from 4*0=0 to4*(pi/4)= pi. So now you have (1/4)(20/pi){integral from 0 to pi}sin(u)dx. –  Jared Feb 20 '12 at 4:31
Now use the antiderivative of sin(x) I mentioned earlier to integrate. You have F(u)=-(5/pi)cos(u){from 0 to pi}. Plug in your intervals so you get F(pi)-F(0)=(final answer). That is [-(5/pi)(-1)]-[-(5/pi)(1)]=[5/pi]-[-5/pi]=(5/pi)+(5/pi)=(10/pi). I hope this is the final answer you are looking for. Remember to add units if your teacher requires that. –  Jared Feb 20 '12 at 4:37

For what it's worth:

You can use the "guess and check" method as follows.

The derivative of $-\cos x$ is $\sin x$. So, perhaps the antiderivative of $5\sin(4x)$ is $$-5\cos(4x).$$

Does this work? Let's check:

The derivative of our guess has to be $5\sin(4x)$; but, $${d\over dx} \bigl(-5\cos (4x)\bigr)=5\sin(4x)\cdot 4= 5\cdot 4\sin(4x).$$ Hmm, it's not quite right, we do not want that "4" there on the right hand side, that arose from the chain rule. But, if we introduced a multiplicative factor of $1\over4$ in our guess for the antiderivative, things would work out: $${d\over dx} \bigl(-{5\over4}\cos (4x)\bigr)={5\over4} \sin(4x)\cdot 4= 5 \sin(4x).$$

So, indeed, an antiderivative of $5\sin(4x)$ is $-{5\over4}\cos(4x)$.

More generally:

If $F(x)$ is an antiderivative of $f(x)$, then

$\ \ \$1) $cF(x)$ is an antiderivative of $cf$ (since $(cf)'=c f'$)

$\ \ \$2) For $c\ne0$, ${1\over c}F(cx)$ is an antiderivative of $f(cx)$ (by the chain rule).

Of course, you can use the "substitution method" for integrals for your problem.

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I think "guess and check",although useful in some differential equations problems,leads to sloppy habits. But it does work in this case,so you can't argue with success......... –  Mathemagician1234 Feb 20 '12 at 7:42
I would view adjusting constants at the end as a quite efficient method. –  André Nicolas Feb 20 '12 at 21:53

From the basic theory of primitives you can check that

$$\int {f\left( {ax} \right)dx = \frac{1}{a}F\left( {ax} \right) + C}$$

So you can use this and put

$$5\int {\sin \left( {4x} \right)dx = - \frac{5}{4}\cos \left( {4x} \right) + C}$$

Alternatively $\sin x$ is odd, you will have that the integral over any symmertric interval around the origin will be zero, that is

$$\int\limits_{ - \pi }^\pi {\sin \left( {4x} \right)dx} = 0$$

So you problem ultimately is finding $c\in[-\pi,\pi]$ for

$$5\sin \left( {4c} \right) = 0$$

which has solutions. $(0,\pm\pi/4,\pm\pi/2,\pm 3\pi/4,\pm\pi)$

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You may want to add "Alternatively" before "Since $\sin x$ is odd..." since what follows there is really an alternate way of solving the problem that does not require finding an antiderivative for $5\sin(4x)$. –  Arturo Magidin Feb 20 '12 at 4:03
@ArturoMagidin Indeed. Thanks. –  Pedro Tamaroff Feb 20 '12 at 4:06


What can be understood by $A-B = B-A$ in set theory?

What can be understood by $\rm A-B = B - A$ in set theory?

What does this tell us about the characteristics of $A$ and $B$ and their relationship? I'm quite confused as I am new to set theory.

Remember that "$A-B$" means "The set of things in $A$ but not in $B$." So "$A-B=B-A$" means "The things in $A$ but not $B$ are exactly the things in $B$ but not $A$."

Now, for which $A$ and $B$ is this equation true? As always, when you're trying to understand new abstract concepts (in this case, set difference and Boolean operations in general) it's best to try some examples. Does the equation $A-B=B-A$ hold for $A=\{1, 2, 3\}=B$? What about $A=\{1,2,3\}, B=\{1, 2\}$? What about $A=\{1, 2, 3\},B=\{2, 3, 4\}$?

Based on these examples, you should be able to make a good guess at what the answer should be. Now, try to prove it! (As usual, this will look like "Assume $x\in A-B$. Then [stuff]. So $x\in B-A$. etc.")

($A-B=B-A$) means that the set of everything in $A$ which is not in $B$ equals the set of everything in $B$ which is not in $A$.

This is possible only when $\underline{(\phantom{A=B})}$ because:

For any element $x$ of $A-B$, we have $x\in A$ and $x\notin B$.

For any element $x$ of $B-A$, we have $x\in B$ and $x\notin A$.

But there is no element that can be in $A$ and not in $A$, and in $B$ and not in $B$.

Therefore $A-B$ is $\underline{\phantom{\quad\emptyset\quad}}$ , as is $B-A$.   Meaning...

You can work it out using a Venn diagram.

The above picture (sourced from Wikipedia) shows that $A \cup B$ can be divided into three regions: $A - B$ (the bit purely in orange), $A \cap B$ (the bit that's orange and blue), $B - A$ (the bit that's purely in blue).

What your equation is saying is that the bit purely in orange is equal to the bit purely in blue. Since the remaining bit in orange overlaps with the thing in blue, and the remaining bit in blue overlaps with the thing in orange, these two sets must be equal.

More formally, we see that $$A = (A - B) \cup (A \cap B)$$ $$B = (B - A) \cup (A \cap B)$$

Since $A - B = B - A$, we have that \begin{align} A &= (A - B) \cup (A \cap B)\\ &= (B - A) \cup (A \cap B)\\ &= B\end{align}

In fact, the equation $A - B = B - A$ is equivalent to the one $A = B$. Akiva explains below.

• And conversely… – Akiva Weinberger Aug 23 '17 at 13:31
• @AkivaWeinberger I don't understand your comment. What are you trying to say? – man and laptop Aug 23 '17 at 13:33
• You should mention the converse as well. $A-B=B-A$ implies $A=B$, and conversely, $A=B$ implies $A-B=B-A$ (because they'd both be $\emptyset$). – Akiva Weinberger Aug 23 '17 at 13:35
• Sure ${}{}{}{}$ – Akiva Weinberger Aug 23 '17 at 13:48

Two sets $X,Y$ are equal if and only if for all elements $z$ it holds that $z\in X\iff z\in Y$. If $z\in A-B$ then $z\in A$ but $z\notin B$ and if $z\in B-A$ then $z\in B$ but $z\notin A$. It seems impossible that an element could belongs to both $A-B$ and $B-A$, doesn't it?

$A - B$ can be rewritten as $A - (A \cap B)$.

$B - A$ can be rewritten as $B - (A \cap B)$.

Given: $A - (A \cap B) = B - (A \cap B)$.

which means: $A$ and $B$ should be equal and $A - B = B - A = \emptyset$

• This is an interesting exposition; why did someone downvote? – Ari Brodsky Aug 23 '17 at 8:52
• @AriBrodsky No idea :) I thought my answer was to the point and complete. :) – Sarthak Mittal Aug 23 '17 at 9:03
• Almost likely the downvotes come from people who disagree with spoonfeeding OP. This is likely homework or a learning exercise and by doing it for OP you do not allow them to learn. – Ander Biguri Aug 23 '17 at 9:48
• @AnderBiguri yeah I understand that! just tried to post a simplified answer. :) – Sarthak Mittal Aug 23 '17 at 11:52

OP says

I'm quite confused as I am new to set theory.

So wordy exposition really can't be harmful, especially since the OP asked "what can be understood", leaving the door open for a lengthy response.

If $a$ and $b$ are numbers, then $a - b = a$ is only possible if $b = 0$.

If $A$ and $B$ are sets, then $A - B = A$ if $B = \emptyset$, but that is not the end of the story. We know that $A - B \subseteq A$, and after some thought we come up with

Definition: An element $\hat a \in A$ is said to be safe from $B$ if $\hat a \in A-B$.

Exercise 1: $\hat a \in A$ is safe from $B \Leftrightarrow \hat a \notin B$.

Exercise 2: $A - B = A \Longleftrightarrow$ all elements in $A$ are safe from $B$.

Exercise 3: $A - B = A \Leftarrow \Rightarrow A \cap B = \emptyset$.

Exercise 4: $B - A = B \Leftarrow \Rightarrow A \cap B = \emptyset$.

Exercise 5: $A - B = A \Leftarrow \Rightarrow$ all elements in $B$ are safe from $A$.

So, we can summarize as follows,

Proposition 1: $A - B = A \Leftarrow \Rightarrow B - A = B \Leftarrow \Rightarrow A \text{ and } B\;$ are disjoint sets.

We see that there is certainly some symmetry going on here.

If $a$ and $b$ are numbers, then $a - b = b - a \Leftarrow \Rightarrow a = b$. Does this at least now carry over into set theory? What could go wrong with so much symmetry?

Proposition 2: Suppose for two sets $A$ and $B$,

$\tag 1 A - B = B - A$ Then $A = B$ and $A - B = B - A = \emptyset$.
Proof
Set $C = A - B = B - A$ and let $c \in C$. Then since $C = B - A$, $\,c \in B$. But we also have, $c \in A - B$, i.e. $c \in A$ is safe from $B$. By exercise 1, $c \notin B$, a contradiction.

So $C = \emptyset$. So, $A - B = \emptyset$. But this means no element in $A$ is safe from $B$, or, by exercise 1, every element in $A$ must be be in $B$, Same argument shows that every element in $B$ must be in $A$, so $A = B$. $\qquad \blacksquare$

All this symmetry here is really exciting, and before you know it you'll really appreciate the following material.

Definition: The symmetric difference ${\displaystyle A\,\triangle \,B}$ of two sets $A$ and $B$ is given by

$\tag 2 {\displaystyle A\,\triangle \,B=(A - B)\cup (B - A)}$

Exercise 6: Show that $A\,\triangle \,B=(A\cup B) - (A\cap B)$.

Exercise 7: Show that the following equalities are all equivalent statements,

$\qquad {\displaystyle A\,\triangle \,B = \emptyset}$
$\qquad A = B$
$\qquad A - B = B - A$

Note: The definition "$\hat a \in A$ is said to be safe from $B$" was made to aid intuition and you won't find it by googling. But see symmetric difference.