# How many triangles in picture?

Can you tell me how many triangles are in this picture? I've counted 96, not sure that I got right answer.

-
Perhaps you could try counting them by using the symmetry of the shape (if you haven't already). –  Alice Nov 21 '13 at 13:50
Clearly there are 19 unique points. Number of triangles that can be formed from 19 points taking 3 at a time=$19\choose3$. Also note that some of the points lie on a straight line. When we took $19\choose3$ we included the ones that lie on a straight line. Perhaps finding the number of lines and subtracting that from $19\choose3$ will yield the answer. –  GTX OC Nov 21 '13 at 14:05
@GTXOC I see 19 unique points... –  apnorton Nov 21 '13 at 14:08
Alright I missed the ones in the middle of the base. –  GTX OC Nov 21 '13 at 14:09
@All: Not all triangles are valid even if the points could form one, consider lines between "star-corners". –  AlexR Nov 21 '13 at 14:11

Let's see! We'll break it up into the following cases (where an apex means one of the six points of the star):

A: Triangles that contain three apexes
There are 2 of these.

B: Triangles that contain a pair of opposite apexes
For each such pair, there are 2 of these, so 6 in total.

C: Triangles that contain a pair of non-opposite apexes
For each non-adjacent apex pair, there are three of these, so 18 in total.

D: Triangles that contain exactly one apex
For each apex, I count 10 such triangles, so 60 in total.

E: Triangles that contain no apex
For each edge of the internal hexagon, there are 3 triangles, so 18 in total.

I make that 2 + 6 + 18 + 60 + 18 = 104.

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I verified and this seems to be correct. –  AlexR Nov 21 '13 at 14:16
Look OK to me too, though categories $D$ and $E$ could be somewhat refined. –  Marc van Leeuwen Nov 21 '13 at 14:24
@Marc: I agree with you about category D. –  TonyK Nov 21 '13 at 14:29
A small reality check: If you dismiss the two big equilateral triangles, all other triangles belong to orbits of $6$ under rotations by $2\pi/6$, so $T-2$ must be divisible by $6$, which $T=104$ is but $T=96$ (the OP's original count) is not. –  Barry Cipra Nov 21 '13 at 14:54
For Category $D$, you can take the apex at the very top and count triangles that are and aren't symmetric across the "$y$"-axis. There are $2$ that are and $4$ pair that aren't. –  Barry Cipra Nov 21 '13 at 15:13

I only get 98. I am missing the 3rd triangle under section C. I only see 2 triangles that contain a pair of non-opposite apexes.

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# How to solve the following congruence's equation system?

Through many operations in an exercise I've reached this point

$$17≡4a+19b(mod$$ $$27)$$

$$8≡11a(mod$$ $$27)$$

I want to find both a and b to have the answer I need, I've been trying to use the Chinese rest theorem but found it doesn't work in my case, I'm starting converting the second one into

$$11a+27a≡1(mod$$ $$27)$$

Despite this I'm not sure if is correct and I still don't know how to clear this $$a$$ in the second equation in order to replace it in the first.

Any help will be really appreciated

There are a number of ways to go about this. We want to be able to "divide" by $$11$$, so we need $$8 + 27x = 11y$$. Solving this linear Diophantine equation (an elementary problem, feel free to Google) gives $$x = 5$$ and $$y = 13$$ as possible solutions. Then, \begin{align} 8 &\equiv 11a \mod{27}\\ 8 + 27\cdot 5 &\equiv 11a \mod{27}\\ 13(11) &\equiv 11a \mod{27}\\ 13 &\equiv a \mod{27} \end{align} Then $$a = 13 + 27k$$ for some $$k \in \mathbb{Z}$$. You can subtract $$4$$ times the last congruence above from $$17 \equiv 4a + 19b \mod{27}$$ to get \begin{align} 17 - 4(13) &\equiv 19b \mod{27}\\ -35 &\equiv 19b \mod{27}\\ -35 + 2(27) &\equiv 19b \mod{27}\\ 19 &\equiv 19b \mod{27}\\ b &\equiv 1 \mod{27} \end{align}

• But you should say how you computed $k$ such that $\,11\mid 8+27k,\,$ not simply pull it out of hat like magic! Else how is the OP supposed to figure out how to use the method for similar problems? – Bill Dubuque Oct 22 '18 at 19:11
• @BillDubuque I alluded to it in my explanation... Solving a linear Diophantine equation is usually a skill that's learned in sections immediately preceding problems like this. For the record though, the Euclidean algorithm with back substitution is the most systematic way of doing it (again, Google is your friend). – AlkaKadri Oct 22 '18 at 19:59
• Back-substitution is clumsy and error-prone. Better to use this version, or Gauss's algorithm below $$\!\bmod 11\!:\ \, 27x\!+\!8\equiv 0\iff x\equiv \dfrac{-8}{27}\equiv \dfrac{3}{-6}\equiv \dfrac{-1}{2}\equiv \dfrac{10}2\equiv 5\$$ In any case, it's not a good idea to say "Google it" in an answer. Google can locate all sorts of nonsense (low-quality, errorneous, crankish, etc). – Bill Dubuque Oct 22 '18 at 20:22
• @BillDubuque ah I see.. Yes you have a point there, I’ll certainly refrain from doing that in future answers. Thanks Bill! – AlkaKadri Oct 22 '18 at 20:25

Just so long as you avoid multiplying or dividing by a multiple of $$3$$ you can use ordinary arithmetic - so you can multiply the first by $$11$$ and the second by $$4$$ to isolate a term in $$b$$ using standard elimination. You avoid $$3$$ because the base $$27$$ has the prime factor $$3$$.

You can solve $$ax+by=1$$ for coprime $$a$$ and $$b$$ using the division algorithm to find their highest common factor (which is $$1$$ because they are coprime). Other methods are available, and sometimes work more quickly if you spot them. The division algorithm provides a proof that there is always a solution, and a systematic way of finding it.

• +1 Eliminating $a$ by cross multiplying yields $-2\equiv\!\!\overbrace{ 5}^{\large 44a}\!\!-7b\,\Rightarrow\, 7b\equiv 7\,\Rightarrow\, b\equiv 1\,$ (use least magnitude reps!), which is probably easier than computing inverses. – Bill Dubuque Oct 22 '18 at 19:01

Guide:

\begin{align} 27 &= 11(2)+5 \\ 11 &= 2(5)+1 \end{align}

\begin{align} 1 &= 11-2(5)\\ &= 11 - 2(27-2(11)) \\ &= 5(11)-2(27) \end{align}

Hence $$11^{-1} \equiv 5 \pmod{27}$$

Now you should be able to solve for $$a$$, substitute that to the first equation and solve for $$b$$ using similar procedure.

\begin{align*} 11a & \equiv 8 \pmod{27}\\ 55a & \equiv 40 \pmod{27}\\ a & \equiv 13 \pmod{27}. \end{align*} Now plug this in the first congruence to get $$b$$.

• But you should say how you computed the inverse of $11,\,$ not simply pull it out of a hat like magic! – Bill Dubuque Oct 22 '18 at 19:05
• @BillDubuque My intention was not to pull a magic for OP. But when OP says (in his question) that he/she tried using the Chinese Remainder Theorem, then I would assume familiarity of such a step (multiplying by an inverse) on his/her part. One can always seek more explanation if needed. – Anurag A Oct 23 '18 at 16:21
• But the question makes it clear the OP is having difficulty inverting $\,11\bmod 27.\,$ Giving the answer with no hint how it was computed is probably not going to help the OP get past their obstacles. – Bill Dubuque Oct 23 '18 at 16:36
• It is not that I disagree with your suggestion of adding more but it is a judgement call. When someone says CRT etc. and yet misses some basic step, then one can argue either way about his/her understanding of the basic material. Perhaps OP overlooked some simple step or perhaps OP has difficulty with the concept of inverse. I grapple with this on a daily basis :-) – Anurag A Oct 23 '18 at 17:00


# Difference between “ proof by reductio ad absurdum” and “proof by contradiction”?

I always thought that both “proof by reductio ad absurdum” and “proof by contradiction” mean the same, but now my professor asked this question on my homework and I don't know.

I believe that in both cases you assume the negation of the conclusion and develop a contradiction through the premises. This will imply the conclusion. Today I have a meeting with the assistant professor so I can clarify this, but I really would like to know what you guys think, or if possible it would be great if you point me into some good references.

UPDATE:

I just came from my extra help and the assistant professor explains the difference this way:

Reductio ad absurdum: $$\vDash [\neg p\to(q\wedge\neg q)]\to p$$

Proof by contradiction:

$$\vDash [\neg (p\to q) \to (r\wedge \neg r)]\to (p\to q)$$

And the examples of application were these:

Using proof by contradiction: $\sqrt2$ is irrational.( First suppose it is rational and derive a contradiction).

Using proof by reductio ad absurdum: If $f$ is differentiable on $(a,b)$ then $f$ is continuous on $(a,b)$. ( First we suppose that $f$ is differentiable on $(a,b)$ but not continuous on $(a,b)$ and derive a contradiction).

• – R. Suwalski Aug 28 '17 at 15:49
• I always thought of it as: reductio ad absurdum is proving $\lnot \phi$ by assuming $\phi$ and proving $\bot$ or $\psi \wedge \lnot \psi$. Proof by contradiction is proving $\phi$ by assuming $\lnot \phi$ and proving $\bot$ or $\psi \wedge \lnot \psi$. So, for example, in intuitionistic systems, reductio ad absurdum is still valid, but proof by contradiction wouldn't be as all it proves is $\lnot \lnot \phi$. (Not sure if this is officially valid, though.) – Daniel Schepler Aug 28 '17 at 15:55
• Usually (but the distinction is not so "stable") the proof by contradiction is of the form: "if from $A$ a contradiction follows, then $\lnot A$ can be inferred". In the indirect proof the assumption is $\lnot A$ and the conclusion inferred (through the contradiction) is $A$. – Mauro ALLEGRANZA Aug 28 '17 at 15:58
• Possible duplicate of Difference between proof of negation and proof by contradiction – Clement C. Aug 28 '17 at 16:08
• Ok. So "reductio ad absurdum" is synonymous with "indirect proof", and "proof of negation"? – Novato Aug 28 '17 at 16:17

## 1 Answer

Regarding the rule of indirect proof:

"if from assumption $\lnot A$ a contradiction follows, we can infer $A$",

we can see:

Sometimes the nomenclature RAA is used; it stands for reductio ad absurdum, the mediæval Latin name of the principle. [...] A genuine indirect proof in propositional logic ends with a positive conclusion.

The principle is equivalent to Double Negation elimination.

If we agree with this approach, proof by contradiciton is more general, because it applies also to inferences with negative conclusion, licensed by the principle of Negation Introduction:

"if from assumption $A$ a contradiction follows, we can infer $\lnot A$".


# Difference between lim as x→∞ and lim as |x|→∞

1. Sep 8, 2012

### Cristopher

I came across something I'd never seen before, the use of |x| instead of just ‘x’ in limits.
What is the difference between $\displaystyle\lim_{|x|\to\infty}x\sin\frac{1}{x}$ and $\displaystyle\lim_{x\to\infty}x\sin\frac{1}{x}$ ?
Is there any difference when evaluating them? Is that notation used only with infinity?

Thanks.

2. Sep 8, 2012

### Vorde

It seems like it's implying that the limit as x goes to infinity is equal to the limit as x goes to negative infinity. But someone who has actually seen this notation before might know better, I'm just guessing.

3. Sep 8, 2012

### alberto7

I haven't seen limits as |x|→∞, but I have seen limits as x→±∞ or limits as x→∞ with this meaning. In the latter case they distinguished limits as x→+∞ and as x→-∞. I think they pictured ∞ as a single point outside the line, making it a circle, its Alexandroff compactification, considering limits as x→+∞ and as x→-∞ as the lateral limits towards ∞.

About the limits as |x|→∞, I think it could be generalized in the following way. We may say that f(x)→a as g(x)→b if, for any neighborhood U of a, there exists a neighborhood V of b such that f(g-1(V\{b}))⊆U. Also f(x)→a as g(x)→∞ if, for any neighborhood U of a, there exists M>0 such that f(g-1((M,∞)))⊆U, and similar definitions.

Last edited: Sep 8, 2012
4. Sep 8, 2012

### HallsofIvy

Staff Emeritus
I see two different ways of interpreting "limit as |x| goes to infinity". First would be that "limit as x goes to infinity" and"limit as x goes to -infinity" must be the same. The other would be that x represents a point in the plane or a complex number and x goes away from the origin in any direction.

5. Sep 8, 2012

### ObsessiveMathsFreak

The first notation only really makes sense if you consider 'x' to be a complex variable. In this case, taking the limit at infinity means asking whether the function approaches a fixed value no matter which direction you approach infinity from.

The second notion on the other hand can be used if x is real and you are simply going to "positive" infinity.

At infinity, functions of complex variables generally either have a finite limit, or else a pole. They could also have an essential singularity (O~o) as well. And if the function is multi-valued, it will almost always have a (irregular) branch point there.

6. Sep 9, 2012

### Cristopher

Thank you all.

Yes, I also thought that it says that the limit as x→+∞ and as x→-∞ are the same. Indeed, I did look at the graph of the funcion x sin(1/x) before asking, and these limits are both equal to 1. I also noticed there's symmetry in the graph, I thought it may have something to do with that, too. I posted the question in the hope of getting more info on the scope of this notation.

Just for reference I saw the notation here:
http://en.wikipedia.org/wiki/L%27Hopital%27s_rule
In the section ‘Other ways of evaluating limits’


# Factorial Trailing Zeroes

## # Solution

Brute force solution is to calculate $n!$ and keep dividing by and count number of $10$'s.

### # Number of 5's

Factor a number: $60 = 2 \times 2 \times 3 \times 5$. We can see $60$ only has 1 trailing zero b/c the number of pair of $2$ and $5$ is 1.

For a factorial, the number of 2's is definitely more than that of 5's.

Thus, the number of 5's is the number of trailing zeros.

Complexity:

• time: $O(\log n)$
• space: $O(1)$
def trailingZeroes(self, n: int) -> int:
cnt = 0
while n>0:
cnt += n//5
n //= 5
# this also works
# n //= 5
# cnt += n
return cnt

Last Updated: 9/21/2020, 4:44:16 PM


# The probability that in a game of bridge each of the four players is dealt one ace

The question is to show that the probability that each of the four players in a game of bridge receives one ace is $$\frac{24 \cdot 48! \cdot13^4}{52!}$$ My explanation so far is that there are $4!$ ways to arrange the 4 aces, $48!$ ways to arrange the other cards, and since each arrangement is equally likely we divide by $52!$. I believe the $13^4$ represents the number of arrangements to distribute 4 aces among 13 cards, but I don't see why we must multiply by this value as well?

Imagine $4$ groups of $13$ spots for the $4$ hands
$\small\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square$

An ace can be placed in each group in any of $13$ places, hence $13^4$
The aces' suits can be distributed between groups in $4!$ ways, hence $24$
The remaining cards can be placed in 48! ways,
and 52! is the unrestricted ways of placing the cards
thus (putting the terms in the order you have), $Pr = \dfrac{24\cdot48!\cdot13^4}{52!}$

SIMPLER WAY:

There is a much simpler way to get the same result.

We need an ace in each group of 13, how the rest of the cards go doesn't matter !

The first ace has to be in some group, each of the other aces have to fall in a different group, so the $2nd$ ace has $39$ permissible spots out of $51,$ and so on

thus $Pr = \dfrac{39}{51}\cdot\dfrac{26}{50}\cdot\dfrac{13}{49}$

There are two approaches to card-game questions like these. In the first approach, we temporarily assume that order of cards within the hand matters. In doing so, we treat our sample space as all ways of arranging the fifty-two cards in a row.

Approaching like this, we have the following steps for multiplication principle:

• Pick who gets which ace: $4!$ ways
• Arrange the remaining $48$ cards in a row and give them to the players (12 to north, the next twelve to east, the next twelve to south, etc...): $48!$ ways
• Pick where in the hand the ace goes for each player: $13^4$ ways

Note that the final step was necessary in order to have what we count cover all possible ways that the 52 cards be dealt such that every player receives at least one ace. If we hadn't multiplied by $13^4$, it would have been as though we only considered the ace being the first card in each players' hand. Since we are considering order important, the hand $A\spadesuit 2\spadesuit 3\spadesuit\dots$ is a different outcome than $2\spadesuit A\spadesuit 3\spadesuit\dots$

Since there are $52!$ number of equiprobable ways to deal the cards (where order matters) the probability is as given:

$$\frac{4!48!13^4}{52!}$$

My preference is instead to work in the situation that order doesn't matter.

Here, we break apart as multiplication principle:

• Choose which player gets which ace: $4!$ ways
• Choose twelve additional cards for each player: $\binom{48}{12,12,12,12}=\frac{48!}{12!12!12!12!}$ number of ways

There are a total of $\binom{52}{13,13,13,13}=\frac{52!}{13!13!13!13!}$ number of deals (where order within each hand doesn't matter) for a probability then of:

$$\frac{4!\binom{48}{12,12,12,12}}{\binom{52}{13,13,13,13}}$$

which after a short amount of manipulation you see is precisely the same answer as before.

There are $4!$ ways to distribute the aces, then $\frac{48!}{12!^4}$ ways to distribute the other $48$ cards, $12$ to a player.

Without worrying about the aces, there are $\frac{52!}{13!^4}$ ways to distribute $52$ cards, $13$ to a player.

Therefore, the probability is $$\frac{4!\frac{48!}{12!^4}}{\frac{52!}{13!^4}}=\frac{13^4}{\binom{52}{4}}\doteq0.1054982$$

• Being a bridge player, I am very interested by this answer. My question is : does this result means that if I have one ace and my partner has one ace, the probability that one of the opponents has the two other aces is 90% ? Or, is such a statement totally wrong ? – Claude Leibovici Jan 8 '16 at 4:54
• If you have one ace and your partner has one ace, then the other two players have two aces. It should be close to $50\%$ that one has both, but not quite. – robjohn Jan 8 '16 at 8:11
• To count the number of ways that one of them does not have both aces, note that there are $2!$ ways to distribute the aces, then $\frac{24!}{12!^2}$ ways to distribute the other $24$ cards, $12$ to a player. Without worrying about the aces, there are $\frac{26!}{13!^2}$ ways to distribute $26$ cards, $13$ to a player. Therefore, the probability is $$\frac{2!\frac{24!}{12!^2}}{\frac{26!}{13!^2}} =\frac{13^2}{\binom{26}{2}} =0.52$$ So the probability that one of them has both aces is $0.48$. – robjohn Jan 8 '16 at 8:11
• Thanks for tha answer ! I am real bad with probabilities and I totally misunterpreted the result of $1-0.105$. Shame on me !! Cheers – Claude Leibovici Jan 8 '16 at 8:28

Since the order doesn't matter, let's make hands one by one.

1. For the first, there are 4 Aces and I need to choose 1 to give to this person, $\binom{4}{1}$. Then I need to choose 12 cards from the 48 non-Aces to complete this hand, $\binom{48}{12}$. Thus the number of ways to make this hand is $$\binom{4}{1}\binom{48}{12}.$$

2. I already gave the first person an Ace, so I have three left and I need to choose one to give to this person. I also used up 12 non Aces on the last person, so there are 36 non Aces left and I need to choose 12. The number of ways to make this hand is thus, $$\binom{3}{1}\binom{36}{12}.$$

3. and 4. follow the same logic and so I have $$\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}$$ ways to make hands 3 and 4.

Finally, all the possible ways to make 4 13-card hands are $$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13} = \binom{52}{13,13,13,13} = \frac{52!}{13!\,13!\,13!\,13!}.$$

Lastly, putting it all together, the probability of each person getting an Ace is thus \begin{align*}\frac{\binom{4}{1}\binom{48}{12}\binom{3}{1}\binom{36}{12}\binom{2}{1}\binom{24}{12}\binom{1}{1}\binom{12}{12}}{\binom{52}{13,13,13,13}} &= \frac{4!}{1!3!}\cdot\frac{48!}{12!36!}\cdot\frac{3!}{1!2!}\cdot\frac{36!}{12!24!}\cdot\frac{2!}{1!1!}\\ &\qquad\times\frac{24!}{12!12!}\frac{1!}{1!0!}\cdot\frac{12!}{12!0!}\cdot\frac{13!\,13!\,13!\,13!}{52!}\\ &=\frac{24 \cdot 48! \cdot13^4}{52!}. \end{align*}

After the cards have been shuffled and cut, there are $\binom{52}4$ equally likely possibilities for the set of four positions in the deck occupied by the aces. Among those $\binom{52}4$ sets, there are $13^4$ which result in each player getting an ace; namely, make one of the $13$ cards to be dealt to South an ace, and the same fo West, North, and East. So the probability is $$\frac{13^4}{\binom{52}4}=\frac{4!\cdot48!\cdot13^4}{52!}=\frac{2197}{20825}\approx.1055$$

Suppose you are arranging the cards in rows of 13. Each row must have an ace; if we place the aces as the first cards, there are 4! ways to arrange them. Then, there are 48! ways to arrange the remaining 48 cards. However, this restricts the aces to being the first card; they can, in fact, be in any of 13 positions - for each of the 4 players. So, multiply by 13^4 to account for the 4 aces being in any of the 13 positions.


# Urn I contains $2$ white and $4$ red balls, whereas urn II contains $1$ white and $1$ red ball.

Urn I contains $2$ white and $4$ red balls, whereas urn II contains $1$ white and $1$ red ball. A ball is randomly chosen from urn I and put into urn II, and a ball is then randomly selected from urn II. What is

• the probability that the ball selected from urn II is white?
• the conditional probability that the transferred ball was white given that a white ball is selected from urn II?

I got the answer $a = \frac{4}{9}$ but I need help with $b$.
If I make $P(T) = \text{transfered ball}$ is white $P(T) = \frac13$ and P(W) = selected ball from urn 2 is white.

$P(W) = \frac49$ (from part $a$) I am looking for $P(T|W)$ by Bayes's formula - I get $$P(T|W) = P(T\cap W)/P(W)$$
then I get $$P(W|T) \cdot \dfrac{P(T)}{P(W)}$$.
we know $P(T)$ and $P(W)$ and $P(W|T) = \frac29$.
so i get $\frac29\dfrac{\frac13}{\frac49} = \frac16$ but the answer key says it's $\frac12$?

• We need to find $\Pr(T\cap W)$, and divide by $\Pr(W)$. The probability of $T$ is $2/6$. The probability of $W$ given $T$ is $(2/3)$. So the probability of $T\cap W$ is $(2/6)(2/3)$, which is $2/9$. Now divide by $4/9$ and simplify. Oct 2 '15 at 20:16

In Bayes's formula formula: $P(T|W) = \frac{P(T)*P(W|T)}{P(W)}$

But $P(T)*P(W|T)=P(T\cap W)$ because these events occur at the same time, that is:

1) we have chosen white ball from urn1, probability of which is: $P(T)=\frac{2}{6}$...and now we have two white balls and one red in urn 2.

2) then we have chosen the white ball from urn 2, which means: $P(W|T)=2/3$.

So, $P(T)*P(W|T)=\frac{2}{6}*\frac{2}{3}=\frac{2}{9}=P(T\cap W)$ $$P(T|W) = P(T\cap W)/P(W)$$ We know that $P(W)=\frac{4}{9}$.

So, $P(T|W)=\frac{2}{9}/\frac{4}{9}=\frac{1}{2}$.

• ok, but from bayes's formula isn't P(T∩W)/P(W) equal to (P(W|T)⋅P(T)) /P (W)? so shouldn't the answers be the same? Oct 2 '15 at 20:49
• they are equal.
– Jane
Oct 2 '15 at 20:53
• look at the corrections above
– Jane
Oct 2 '15 at 21:02
• actually we can just use the Bayes's formula without including the probability of events intersection
– Jane
Oct 2 '15 at 21:03


Question

# If $$f\left( x \right)$$ and $$g\left( x \right)$$ are two functions with $$g\left( x \right) =x-\dfrac { 1 }{ x }$$ and $$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$, then $$f^{ ' }\left( x \right)$$ is equal to

A
3x2+3
B
x21x2
C
1+1x2
D
3x2+3x4

Solution

## The correct option is A $$3{ x }^{ 2 }+3$$$$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$Writing $${ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$ using $${ \left( a-b \right) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\left( a-b \right)$$, we have$${ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } -3x\cdot \dfrac { 1 }{ x } \left( x-\dfrac { 1 }{ x } \right)$$$$\Rightarrow { x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$We have,$$f\left( g\left( x \right) \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$As $$g\left( x \right) =x-\dfrac { 1 }{ x }$$, this yields$$f\left( x-\dfrac { 1 }{ x } \right) ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$On putting $$x-\dfrac { 1 }{ x } =t$$, we get$$f\left( t \right) ={ t }^{ 3 }+3t$$Thus, $$f\left( x \right) ={ x }^{ 3 }+3x$$and $$f^{ ' }\left( x \right) =3{ x }^{ 2 }+3$$Mathematics

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# Thread: Asymptotes and intercepts: help checking work

1. ## Asymptotes and intercepts: help checking work

I had to find the vertical and horizontal asymptotes as well as the x an y intercepts..could someone please help me by checking my work...there are only 8 problems...you do not have to check them all, but I wouldreally appreciate the help because I do not know if I am doing these correctly....thank you

2. Hello, aikenfan!

You did really good work . . . with a few slips here and there.

$12)\;\;f(x)\:=\:-\frac{x+2}{x+4}$
That minus-sign is in front of the fraction . . .

For the x-intercept, we have: . $-(x +2)\:=\:0 \quad\Rightarrow\quad x + 2 \:=\:0$

. . $x \:=\:-2\quad\Rightarrow\quad (-2,0)$

$18)\;\;f(x)\:=\:\frac{x^2-25}{x^2+5x}$
VA: . $x^2+5x \:=\:0 \quad\Rightarrow\quad x(x+5)\:=\:0\quad\Rightarrow\quad x \:=\:0,-5$

Vertical asymptotes: . $x = 0,\;x = -5$

$43)\;\;f(x) \:=\:\frac{2x^2+1}{x}$
x-intercept: . $2x^2+1\:=\:0\quad\Rightarrow\quad x^2 \:=\:-\frac{1}{2}\quad\Rightarrow\quad x \:=\:\sqrt{-\frac{1}{2}}$ . . . not a real number.

There is no x-intercept.

$44)\;\;g(x) \:=\:\frac{1-x^2}{x}$
x-intercepts: . $1-x^2\:=\:0\quad\Rightarrow\quad x^2 \:=\:1\quad\Rightarrow\quad x \:=\:\pm1\quad\Rightarrow\quad(1,0),\;(-1,0)$

$45)\;\;h(x) \:=\:\frac{x^2}{x-1}$
y-intercept: . $y \:=\:\frac{0^2}{0-1} \:=\:0\quad\Rightarrow\quad (0,\,0)$

3. Thank you very much for all of your help...I just have one more quick question, if you don't mind...for number 18, how do i figure out the horizontal asymptote? I think it would be 1, but I'm not sure.

4. Hello again, aikenfan!

For number 18, how do i figure out the horizontal asymptote?
Yes, the horizontal asymptote is: $y = 1$

We want: . $\lim_{x\to\infty}\frac{x^2-25}{x^2+5x}$

Divide top and bottom by $x^2$: . $\lim_{x\to\infty}\frac{\frac{x^2}{x^2} - \frac{25}{x^2}}{\frac{x^2}{x^2} + \frac{5x}{x^2}}$ . **

Then: . $\lim_{x\to\infty}\frac{1 - \frac{25}{x^2}}{1 + \frac{5}{x}} \;=\;\frac{1-0}{1+0} \;=\;1$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Rule: Divide top and bottom by the highest power of $x$
. . . . in the denominator.

5. Note that we should find $\lim_{x \to - \infty} \frac {x^2 - 25}{x^2 + 5}$ as well. But in this case, it would give the same value, so that's fine


# Show that if $\phi_{X}(t)=1$ in a neighborhood of $0$, then $X=0$ a.s.

Let $$\phi(t),t\in\mathbb{R}$$, be the characteristic function of a random variable $$X$$. Show that if $$\phi(t)=1$$ in a neighborhood of $$0$$, then $$X=0$$ a.s.

The problem comes with the following hint: Show that $$1-Re(\phi(2t))\le4(1-Re(\phi(t)))$$ for $$t\in \mathbb{R}$$. I am stumped by this one, I am not even sure where to begin or how to prove\use use the hint, any help here would be greatly appreciated.

Here is another method: start by noticing that since $$\cos \leq 1$$, $$\Re \phi(t) = \mathbb{E}[\cos(tX)] \leq 1, \quad \forall t \in \mathbb{R}. \tag{1}$$ Now let $$\def\eps{\varepsilon} (-\eps, \eps)$$ be an interval on which $$\phi = 1$$. Then, using the hint, we have for $$t \in (-\eps, \eps)$$ $$0 \leq 1- \Re \phi(2t) \leq 4(1-\Re \phi(t)) = 0.$$ The first inequality follows from $$(1)$$ and the last equality from the assumption. This proves that $$\Re \phi(2t) = 1$$ for every $$t \in (-\eps,\eps)$$. Reiterating this process, we see that $$\Re \phi(t) = 1, \quad \forall t \in \mathbb{R}.$$ But recall that $$|\phi(t)| \leq 1$$. This forces $$\phi(t) = 1$$ for every $$t \in \mathbb{R}$$. Since the characteristic function determines the distribution, it follows that $$X = 0$$ a.s.

• Beautiful, thank you! Jun 17, 2020 at 21:49

Let's assume that $$\varphi(t) = 1$$ for any $$t \in [0,\delta]$$. Then in particular $$\varphi(\delta)=1$$. We'll show that it is the case that $$\mathbb P(X \in \{\frac{2k\pi}{\delta} : k \in \mathbb Z \}) = 1$$ . Let $$\mu_X$$ be distribution of $$X$$.

Note that $$\varphi(\delta)=1$$ means: $$0 = 1 -\varphi(\delta) = 1 - \int_{\mathbb R} \cos(\delta x) d\mu_X(x) = \int_{\mathbb R} (1 - \cos(\delta x)) d\mu_X(x)$$ Since $$1-\cos(\delta x) \ge 0$$, we must have $$\cos(\delta x) = 1$$ , $$x - d\mu_X$$ almost surely, so that $$x = \frac{2k\pi}{\delta}$$ , $$d\mu_X$$ almost surely, which means $$\mu_X( \{\frac{2k\pi}{\delta} : k \in \mathbb Z \})=1$$.

Now note that we have only countable many points in set $$\{\frac{2k\pi}{\delta} : k \in \mathbb Z \}$$. For every $$k \in \mathbb Z \setminus \{0\}$$ we can find such $$t_k \in (0,\delta)$$ that $$\frac{2k\pi}{\delta}$$ is not equal to $$\frac{2 m \pi}{t_k}$$ for any $$m \in \mathbb Z$$ (because for every $$m \in \mathbb Z$$ there is at most one $$s \in (0,\delta)$$ such that $$\frac{2m \pi}{s} = \frac{2k\pi}{\delta}$$, but we have only countable many $$m \in \mathbb Z$$, but continuum-many $$s \in (0,\delta)$$, so there exists such $$t_k$$) which means that $$\mu_X(\frac{2k\pi}{\delta}) = 0$$ (for that given $$k \in \mathbb Z \setminus \{0\}$$, because $$\varphi(t_k)=1$$, so $$\mu_X( \{ \frac{2m\pi}{t_k} : m \in \mathbb Z \}) = 1$$, too). Since $$k \in \mathbb Z \setminus \{0\}$$ was arbitrary, and there are only countable many of them, we have $$\mu_X( \{ \frac{2k \pi}{\delta} : k \in \mathbb Z \setminus \{0\} \} ) = 0$$, so that $$\mu_X(\{0\}) = 1$$ what was to be proven.

EDIT: If you're interested, here's an approach with your hint. Let's prove it beforehand. $$1 - Re(\varphi(2t)) = \int_{\mathbb R} (1-\cos(2tx))d\mu_X(x) = 2\int_{\mathbb R} (1 - \cos^2(tx))d\mu_X(x)$$

It would be sufficient to show $$1-\cos^2(s) \le 2(1- \cos(s))$$ which is equivalent to $$0 \le \cos^2(s) - 2\cos(s) + 1 = (\cos(s)-1)^2$$, so true. Hence $$1- Re(\varphi(2t)) \le 4\int_{\mathbb R}(1 - \cos(tx))d\mu_X(x) = 4(1-Re \varphi(t))$$

Having lemma, it is pretty easy. Note that you have such $$\delta$$, that $$\varphi(t) = 1$$ for any $$[-\delta,\delta]$$. Now let's prove it is also the case for any $$t \in [-2\delta,2\delta]$$ using hint: Take $$s \in [-2\delta,2\delta]$$. We have $$1 - Re(\varphi(2s)) \le 4(1 - Re(\varphi(s)) = 0$$ since $$s \in [-\delta,\delta]$$. Moreover, $$|\varphi(s)| \le 1$$, so $$\varphi(s) = 1$$. Use it again, to prove the fact for any $$[-2^k\delta,2^k\delta]$$ getting $$\varphi(t)=1$$ for any $$t \in \mathbb R$$.

• Oops I see we arrived at the same conclusion! :) Your first method is more powerful however as it shows that if $\phi(t) = 1$ for one $t$ then the distribution is lattice. Jun 17, 2020 at 21:45
• This is an incredible answer, thank you very much! Jun 17, 2020 at 21:48
• Yes Michh, nice answer +1. This fact about $\varphi(t)=1$ is useful when you want to characterise all types of characteristic functions. It can be shown that either $|\varphi(t)| <1$ for every $t \in \mathbb R \setminus \{0\}$ or $|\varphi(t)|=1$ for every $t \in \mathbb R$ and then you have $X = a$ almost surely for some $a \in \mathbb R$ or finally $|\varphi(t)|=1$ for some $t \in \mathbb R_+$, but $|\varphi(s)|<1$ for $s \in (0,t)$. In the latter case it is exactly the distribution on $\{ \frac{2k\pi}{t} : k \in \mathbb Z \}$. Jun 17, 2020 at 21:49
• @Spider Bite, I should thank you, too, because I wasn't aware of such lemma. Nice one to have in mind ^^ Jun 17, 2020 at 21:55
• @Wave because since $\phi(\delta)=1$, then in particular it is a real number, hence imaginary part must vanish. May 9 at 14:31


Question

# Check which of the following are solutions of an equation $$x + 2y = 4$$? (i) (0, 2)    (ii) (2, 0)   (iii) (4, 0)   (iv)  ($$\sqrt 2, -3\sqrt 2$$)   (v) (1, 1)    (vi) (-2, 3)

Solution

## Given equation is$$x+2y=4$$(i) Put the value x=0 and y=2 we get$$0+2\times 2=4$$$$\Rightarrow4=4$$Then both sides are equal then (0,2) is the solution of equation x+2y=4(ii) Put the value x=2 and y=0 we get$$2+2\times 0=4$$$$\Rightarrow 2\neq 4$$ Then both sides are not  equal then (0,2) is not the  solution of equation x+2y=4(iii) Put the value x=4 and y=0 we get$$4+2\times 0=4$$$$\Rightarrow4=4$$Then both sides are equal then (4,0) is the solution of equation x+2y=4(iv)$$x+2y=4$$ Put $$x=\sqrt{2},y=-3\sqrt{2}$$ we get$$\sqrt{2}+2(-3\sqrt{2})=4$$$$\Rightarrow \sqrt{2}-3\sqrt{2}=4$$$$\Rightarrow -2\sqrt{2}\neq 4$$ Then both sides are not  equal then $$(\sqrt{2},-3\sqrt{2})$$ is not the  solution of equation x+2y=4(v) Put the value x=1 and y=1 we get$$1+2\times 1=4$$$$\Rightarrow 3\neq 4$$ Then both sides are not  equal then (1,1) is not the  solution of equation x+2y=4(vi) Put the value x=-2 and y=3 we get$$-2+2\times 3=4$$$$\Rightarrow 4= 4$$ Then both sides are   equal then (-1,3) is  the  solution of equation x+2y=4Mathematics

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# I A question about rolling motion of a wheel on a frictionless surface

1. Dec 4, 2017

### NoahCygnus

Let's imagine a situation where we have a wheel of mass $M$ and radius $R$ on a frictionless surface, and we apply a force $\vec{F}$ as shown in the diagram. The force will produce both linear acceleration of centre of mass $a$ and angular acceleration $\alpha$. The wheel starts to both translate and rotate, and we remove the external forcr. I know the condition for rolling is $v_{com} = R\omega$. The question is, most textbooks include a surface with friction , and I can't tell if without friction the above condition for rolling will be met or the wheel will slip. How can I know ?

Last edited: Dec 4, 2017
2. Dec 4, 2017

### BvU

As drawn, the wheel will slip: angular acceleration will exceed linear acceleration. If $F$ is applied below the c.o.m, the reverse. Somewhere there is a height where even without friction the rolling condition is met. What height ? A very nice exercise for someone lke you !

3. Dec 4, 2017

### kuruman

The problem without friction is equivalent to a puck on a frictionless table pulled by a string wrapped around its circumference. The tension will exert a torque about the CM that will start the wheel spinning around its center and it will also accelerate the CM. Use Newton's Second law for rotation and translation to analyze the motion and answer your question.

4. Dec 4, 2017

### NoahCygnus

If the wheel is a cylinder, and I apply the force above the centre of mass at $r$, causing a rotation about an axis passing through the cm, perpendicular the circular surfaces of the cylinder, then
$F = Ma_{cm}$
$a_{cm} = F/M$

$\tau = rF = I_{cm}\alpha$
$\alpha= \tau/I_{cm}$

The tangential acceleration of a point on the circumference will be, $a_{t} = R\alpha \Longrightarrow a_{t} = R(\tau/I_{cm}) \Longrightarrow a_{t} = R(rF/I_{cm})$

If the wheel is to roll then, $v_{cm} = v_{t} \Longrightarrow d(v_{cm})/dt = d(v_{t})/dt \Longrightarrow a_{cm} = a_{t}$
$F/M = R(rF/I_{cm}) \Longrightarrow F/M = R(rF/1/2 MR^2) \Longrightarrow 2r = R \Longrightarrow r = R/2$

So I have to apply the force at a height $R/2$, right?

And if I apply the force below the centre of mass, the wheel will slip, right?

5. Dec 4, 2017

### BvU

above the center of mass, yes. You can eliminate ambiguity by calling that a height ${3\over 2}R$
Not wrong, but incomplete. How about a height in the range $R$ to ${3\over 2}R$ ?

Well done !

(the same exercise for a sphere is worked out here under 'sweet spot' )

6. Dec 5, 2017

### NoahCygnus

I did the math for the range $R$ to $(3/2) R$ and found that the tangential acceleration of a point on circumference will always be greater than the linear acceleration of the centre of mass. As a result the body will skid. Can you please explain to me how friction prevents the skidding if I apply a force in that range?
Also I checked the article , it seems quite useful. I will read it once I am free.

7. Dec 5, 2017

### BvU

You mean the tangential acceleration required for the no-slipping condition, right ?
If height = $3/2\, R\$ gives no skidding, anything below means skidding will occur to increase angular velocity, and anything above means skidding will reduce it until rolling catches up.

8. Dec 6, 2017

### A.T.

Static friction is an additional force in your equation. Assume pure rolling and then solve for the friction force required to achieve it.

9. Dec 6, 2017

### NoahCygnus

Yes that's what I meant.
I understand. Let's talk about the direction of frictional force. I infer that if I apply a force below $(3/2)R$ , such that skidding occurs, the direction of frictional force will be opposite to the translation motion of centre of mass, as to reduce the speed of centre of mass and to provide a negative torque to the wheel, so to increase its angular speed (Case 1 in the diagram). And if I apply the force above centre of mass, slipping will occur, so friction will act in forward direction as to speed up the centre of mass, and provide a positive torque, so to lower the angular speed until the rolling condition is met (Case 2 in the diagram). Am I correct? Also I read somewhere once the rolling condition is met, frictional force vanishes, how is that possible? I am having a hard time understanding that.

10. Dec 6, 2017

### BvU

The trajectory of a point on the circumference of the rolling object is a cycloid: at the ground it moves down and up again, so not horizontally.

11. Dec 7, 2017

### A.T.

Yes, and the above also applies when no skidding occurs, due to sufficient static friction. The static friction has the direction that you describe above for kinetic friction, while its magnitude is whatever is required to ensure the rolling condition for linear and angular accelerations.

Pure rolling means no sliding, thus no kinetic friction, but you still can have static friction.

12. Dec 8, 2017

### NoahCygnus

Let's talk about the energy. If the object smoothly rolls down an incline in a conservative field, such that no sliding occurs, the mechanical energy will be conserved even if there is static friction because static friction is a non-dissipative force. Am I correct?
If that's the case how come in reality, a smoothly rolling wheel loses kinetic energy?

13. Dec 8, 2017


I think the proof would involve showing f⁻¹. Since f is surjective, there exists a 2A such that f(a) = b. Let f : A !B be bijective. So x 2 is not injective and therefore also not bijective and hence it won't have an inverse.. A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached. The range of a function is all actual output values. In order to determine if $f^{-1}$ is continuous, we must look first at the domain of $f$. Let f: A → B. We will de ne a function f 1: B !A as follows. 1.Inverse of a function 2.Finding the Inverse of a Function or Showing One Does not Exist, Ex 2 3.Finding The Inverse Of A Function References LearnNext - Inverse of a Bijective Function … Bijective. https://goo.gl/JQ8NysProving a Piecewise Function is Bijective and finding the Inverse This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. I've got so far: Bijective = 1-1 and onto. Then f has an inverse. Proof. Let f : A !B be bijective. In mathematical terms, let f: P → Q is a function; then, f will be bijective if every element ‘q’ in the co-domain Q, has exactly one element ‘p’ in the domain P, such that f (p) =q. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. Accelerated Geometry NOTES 5.1 Injective, Surjective, & Bijective Functions Functions A function relates each element of a set with exactly one element of another set. The function f: ℝ2-> ℝ2 is defined by f(x,y)=(2x+3y,x+2y). Let’s define $f \colon X \to Y$ to be a continuous, bijective function such that $X,Y \in \mathbb R$. Now we much check that f 1 is the inverse … The Attempt at a Solution To start: Since f is invertible/bijective f⁻¹ is … Yes. Click here if solved 43 Theorem 1. An example of a function that is not injective is f(x) = x 2 if we take as domain all real numbers. 1. The codomain of a function is all possible output values. it's pretty obvious that in the case that the domain of a function is FINITE, f-1 is a "mirror image" of f (in fact, we only need to check if f is injective OR surjective). the definition only tells us a bijective function has an inverse function. If we fill in -2 and 2 both give the same output, namely 4. Please Subscribe here, thank you!!! it doesn't explicitly say this inverse is also bijective (although it turns out that it is). Let f 1(b) = a. Since f is injective, this a is unique, so f 1 is well-de ned. A bijective group homomorphism $\phi:G \to H$ is called isomorphism. The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism. A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. is bijective, by showing f⁻¹ is onto, and one to one, since f is bijective it is invertible. It means that each and every element “b” in the codomain B, there is exactly one element “a” in the domain A so that f(a) = b. Let b 2B. Bijective Function Examples. Show that f is bijective and find its inverse. 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# Is there a continuous onto function $f:\Bbb{D} \rightarrow [-1,1]$?

Is there a continuous onto function $f:\Bbb{D} \rightarrow [-1,1]$ ,

where $\Bbb{D}$ is a closed unit disk in $\Bbb{R}^2$ ?

I think the answer is no, otherwise $f(\Bbb{D})=[-1,1]$, so removing one point in the domain is still connected but the image is not. Am I right? Any hint?

• Try $f(x,y) = x.$ – zhw. Jul 27 '18 at 5:40
• Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $\mathbb{D}$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail. – Suzet Jul 27 '18 at 5:42

The problem is, that because $f$ is not injective, we cannot assert that since $f(\mathbb D) = [-1,1]$, we have for all $x \in [-1,1]$ that $f(\mathbb D \backslash \{y\}) = [-1,1] \backslash \{x\}$ for some $y$. Indeed, what may happen is that many points of $\mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.

As you have seen, the first projection serves as an onto map.

If we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D \setminus \{x\}) = [-1,1] \setminus \{0\}$, and this is contradiction because $D \setminus \{x\}$ is connected regardless of what $x$ is , but the image is disconnected.

In fact, if $f$ is injective, then since $\mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.

Note that $\mathbb D$ is a subset of $\mathbb R^2$ and $[-1,1]$ is a subset of $\mathbb R$. Because $2 \neq 1$, we are inclined to believe that a homeomorphism is not possible.

This result, called the invariance of domain, holds in more generality : if a set in $\mathbb R^m$ is homeomorphic to some other set in $\mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.

Sure.

If you map each point to its radius $(x,y)\mapsto\sqrt{x^2+y^2}$, you have a continuous function $\mathbb{D}\to[0,1]$. A slight modification to $(x,y)\mapsto2\sqrt{x^2+y^2}-1$ should be what you want.

• Thank you sir! this one helps too! – user444830 Jul 27 '18 at 5:56


# How to find an $n$-digit number with $m$ number of factors?

I found on the Internet a question which I did not understand:

There is a $6$-digit number with $28$ distinct factors. What is the number?

I searched and found that

Number of factors of a number of the form ${p_1}^{m_1} \times {p_2}^{m_2} \times {p_3}^{m_3} \cdots$ is $(m_1 +1) \times (m_2 +1 ) \times (m_3 +1 ) \cdots$.

But I did not understand how one can answer this question.

Also, if there is some way, I would be glad to know

The general formula of $n$-digit number with $m$ number of factors?

Note by Vinayak Srivastava
3 months, 3 weeks ago

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Hey Vinayak, here is an explanation of the formula for the number of factors: Let $n = p_1^{m_1} \cdot p_2^{m_2} \cdots$. Every factor of $n$ is a product of the prime numbers risen to some power: $f = p_1^{n_1} \cdot p_2^{n_2} \cdots$ with $0 \leq n_1 \leq m_1, 0 \leq n_2 \leq m_2 ...$. For example let $n = 12 = 2^2 \cdot 3^1$. All the factors of 12 are:

$2^0 \cdot 3^0 = 1$

$2^1 \cdot 3^0 = 2$

$2^2 \cdot 3^0 = 4$

$2^0 \cdot 3^1 = 3$

$2^1 \cdot 3^1 = 6$

$2^2 \cdot 3^1 = 12$

As you can see, for each power of the prime numer $p_n$ there are $m_n + 1$ choices $0, 1, ... m_n$. Therefore, the number of factors is $(m_1 + 1) \cdot (m_2+1) \cdots$. I hope this helps!

- 3 months, 3 weeks ago

Thanks @Finnley Paolella ! Now can you please tell me how to find the general formula of $n$-digit number with $m$ number of factors?

- 3 months, 3 weeks ago

I don't know if there is a general formula, but my strategy would be the following: try to factor $m$ to find the power of prime numbers. For example: $28 = 4 \times 7$. A number of the form $n = p_1 ^3 \times p_2 ^ 6$ will always have 28 factors. And then we can try:

$2^3 \times 3^6 = 5832$ is too small

$2^3 \times 5^6 = 125000$

And that is a solution to the original question!

However, the solution is not unique. Here is another solution:

$2^6 \times 11^1 \times 443^1 = 311872$

Number of factors: $(6 + 1) \times (1+ 1) \times (1 + 1) = 28$

- 3 months, 3 weeks ago

Interestingly, although there are 3,013 six-digit numbers with 28 factors, there are some numbers of factors for which there is only one six-digit number with that many factors. These are, I believe, 7, 13, 19, 38, each of which has only one six-digit number with that many factors. (edit: there are probably a lot more solo numbers of factors >50 but I aggregated everything with 50 or more factors to save on processor time.)

- 3 months, 3 weeks ago

How did you calculate this?

- 3 months, 3 weeks ago

Slightly educated brute force. I wrote a bit of simple code that worked out how many (distinct) factors a number had, looped it from 100,000 to 999,999, let it churn for a couple of hours and graphed the results.

- 3 months, 3 weeks ago

WOW!!

- 3 months, 3 weeks ago

Hey Stef, that is amazing! Great visualisation :)

- 3 months, 3 weeks ago

Thanks.

- 3 months, 3 weeks ago

Really awesome stuff! Interestingly you can also see that prime number of factors such as 7 and 11 are very less, fantastic!

- 3 months, 3 weeks ago

Yeah. I'm a big fan of visualising data. You can see all the primes (two factors), all the squares (odd number of distinct factors), and a bit of the long tail of large numbers of factors (it was long tail before I cut it off).

- 3 months, 3 weeks ago

Stef, can you please share the process, I mean the code if you don't mind? It seems you used matplotlib to visualise after processing your data from code. Thanks!

- 3 months, 3 weeks ago

Happy to. It's not very optimised (after all, I only wanted to run it once) so please excuse the code. Also it's written in Lua which is quite like C, Java, JavaScript and Python. A few things to know, though:

1. Arrays (actually tables, but you can use them as arrays) are initialised with {} and referenced with []. Also they start at an index of 1. So, if a={10,11,12}, a[1]=10.
2. Instead of bracketing or indenting code, Lua uses the word end to end blocks of code so if ... then ... end, for ... do ... end etc. are common constructions.
3. You can put multiple statements on a line if you separate them with semi-colons.
4. Variables are weakly typed, automatically coerced and don't need declaring: every variable just holds the value of nil unless you give it a value.
5. Anything else just ask... :)
6. (edit) output was in comma separated format that I could paste straight into Apple Numbers as it takes CSV from the clipboard.
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 function factors(n) c=0 for f=1,n do if (n//f)==(n/f) then c=c+1 end if c==50 then break end end return c end -- set parameters min=100000 max=999999 step=1000 -- used to track progress ns=min+step -- used to track progress -- initialise array a={} for f = 1,50 do a[f] = 0 end -- calculate factors for f=min,max do if f==ns then print(f); ns = ns + step; end -- display progress n=factors(f) a[n] = a[n] + 1 end -- display output s="" for f=1,50 do s = s..f..","..a[f].."\n" end print(s)

- 3 months, 3 weeks ago

Lua??? Retro gaming?

- 3 months, 3 weeks ago

lol. Yeah. Lua. Despite having spent more time with other languages than I have with Lua, I can get an idea out of my head and running faster in Lua than any other language.

Maybe it's the way my brain works, maybe it's the way that Lua works but, if I don't need to share or integrate with anything else, Lua has become my go-to.

(edit) You know that moment when you've spent an hour debugging a bit of code only to realise that the language didn't work the way you expected it to...? I get that with C, JavaScript, Python and Swift. I don't get it with Lua.

- 3 months, 3 weeks ago

I like Lua. But my favourite language is c++. And I think Lua=C languages + Pascal :)

- 3 months, 3 weeks ago

I’ve not thought about Pascal in years. The whole do...end format is very Pascal-like. There’s one bit of syntax that I really miss from Pascal, and that’s using a:=b for assignment. No more accidentally assigning values in if statements. 😁

- 3 months, 3 weeks ago

Thanks Stef, i got the logic. Will see if i can run with my own code in python!

- 3 months, 3 weeks ago

Cool. Let me know how it goes?

- 3 months, 3 weeks ago

Also... check the min and max values. I was messing about with the code before I pasted it and didn't set them back to what they should have been (corrected in my code now).

- 3 months, 3 weeks ago

Sure, Thanks! Will paste it here if it does work :)

- 3 months, 3 weeks ago

Can I learn Lua :) (I don't know anything about programming)?

- 3 months, 3 weeks ago

Lua and Python are both good languages to learn how to program. Python is much more commonly used, especially on brilliant.org. Brilliant.org even has a Programming with Python course. If you want to learn Lua after that, the fundamentals will carry over.

So, despite my liking for Lua, I'd recommend learning Python. 🙃

- 3 months, 3 weeks ago

OK, although I purchased the Premium for math, I will do it if I have time from school!

- 3 months, 3 weeks ago

Can you help me? I wrote a C++ program, but the numbers are too big. I heard in python there are no limits :) I'm not familiar in python programming. I can solve it with hours of programming(vectors, save to txt, read from txt etc.), but I think you can help me.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 #include #include using namespace std; long long int power(int w, int b); int check_power(int a, int b, int c, int d); long long int solution(int n, int theta); int main() { for(int n=1;n<=50;n++){ int j=2*n; long long int a=solution(j,0), b=solution(j-1,1); if((a >possible; vector temp; temp.resize(4); for(int a=1;a<=j;a++) for(int b=a;b>0;b--) for(int c=b;c>0;c--){ int d=(j)/(a*b*c); if(a*b*c*d!=j||d>c||check_power(a-1,b-1,c-1,d-1)!=theta) continue; temp[0]=a; temp[1]=b; temp[2]=c; temp[3]=d; possible.push_back(temp); } if(possible.size()==0) return -1; for(int x=0;x<4;x++) minimum[x]=possible[0][x]; for(int x=1;x0) k*=power(w,r); } if(l>k) continue; for(int w=0;w<4;w++) minimum[w]=possible[x][w]; } long long int output=1; for(int x=0;x<4;x++) output*=power(x,minimum[x]-1); return output; }

- 2 months, 2 weeks ago

The program isn't complete. As you can see this works with only 2,3,5 and 7. I need to add other primes. This is the problem.

- 2 months, 2 weeks ago

I heard in python there are no limits :)

And the garbage collector is made of gold. :) In truth, I hear a lot of good things about Python but never really liked it much myself. Much of what I do these days most of what I do is in Lua, which has 64-bit numbers. If I need an integer bigger than that then I either look to see if I can simplify the problem or pull out my clunky-and-in-need-of-a-rewrite library that stores integers as strings and has functions that can add, subtract etc.

There are libraries that handle big numbers. You should be able to find one (or many) for C++. Or, if you go down the route of writing your own solution, that's quite rewarding, too.

(I just had a glance at the problem and don't see where big numbers come into it. Did you link the right problem? Or have I missed something?) (edit: looked at your output. OK... those numbers are bigger than I would have expected).

- 2 months, 2 weeks ago

A factor $f$ of a number $n = p_{1}^{m_{1}} \cdot p_{2}^{m_{2}} \cdot p_{3}^{m_{3}} \ldots$ must be a number with prime exponents which are less than or equal to the exponents of the number. To make things simple, here is an example: $60 = 2^2 \cdot 3^1 \cdot 5^1$. Without Brute-forcing, we can think logically here. If a factor $f$ can be represented here, say 15 which is equal to $3^1 \cdot 5^1$, the factor's prime exponents will be less than or equal to the prime exponents of $n$. In this case, they are equal

Taking this a step further, for each prime exponent say $m_{1}$, there will be $m_{1} + 1$ options to choose for a power. For the number 60 and specifically prime power of 2, we have $2^{m_{1}}, m_{1} \in \{0, 1, 2\}$ so there are 3 options.

Similarly each of the $m_{i^{th}}$ exponent will have $m_{i} + 1$ options.

The total options will thus equal the product of all $= (m_{1} + 1) \cdot (m_{2} + 1) \cdot (m_{2} + 1) \ldots$

- 3 months, 3 weeks ago

Thanks @Mahdi Raza! Also, as @Finnley Paolella stated, the answer to the question can be $125000$, can there be a general formula of that also?

- 3 months, 3 weeks ago

I don't think there is a closed-from general formula. It also depends on the base we choose, a number in base 2 will have more digits than a number in base 10. As I said in my comment, the general strategie will be the following:

1. factor $m$

2. try out different possibilites

- 3 months, 3 weeks ago

So if we are given 1 more condition, can there be a unique solution?

- 3 months, 3 weeks ago

The thing about those number theories problems is that it is hard to tell how many solutions there are, because prime numbers are more or less randomly distributed along the number line. So it can easily be that there is just one solution, especially if the number m can only be factored in a few different ways. For example, if we were to find a 5 digit number with 17 factors, there will be a unique solution! You can find it using the strategy I provided :)

- 3 months, 3 weeks ago

Is it $65536$?

- 3 months, 3 weeks ago

Yes ;)

- 3 months, 3 weeks ago

That was easy.. $2^{16}$

- 3 months, 3 weeks ago

So, the question which I gave is of no use?

- 3 months, 3 weeks ago

The problem hasn't got a unique solution. A more interesting problem would be: How many solutions are there? But I think this is a very time expensive problem (at least if you don't want to program it).

- 3 months, 3 weeks ago

I don't know programming :(

- 3 months, 3 weeks ago

I think it's possible that I don't understand the question. (edit: Narrator: "He didn't understand the question.")

The lowest composite number I can generate with $2$ distinct factors is $2 \times 3 = 6 ( = 3! )$

The lowest composite number I can generate with $3$ distinct factors is $2 \times 3 \times 4 = 24 ( = 4! )$

...

The lowest composite number I can generate with $n$ distinct factors is $(n+1)!$

...

The lowest composite number I can generate with $28$ distinct factors is $29! = 8841761993739701954543616000000$. This has a lot more than 6 digits. What is it that I don't understand here?

(Additional, just for laughs, if I insist that the $28$ factors are prime factors, I get: $2566376117594999414479597815340071648394470$.)

- 3 months, 3 weeks ago

@Stef Smith, Sir, I did not understand what you mean. Please elaborate!

- 3 months, 3 weeks ago

A misunderstading of how factors work. Some days my brain doesn't work so well. :)

- 3 months, 3 weeks ago

Oh no problem then :)

- 3 months, 3 weeks ago

Hey Stef, The number 24 has more than 3 factors. It has 8 factors: 1, 2, 3, 4, 6, 8, 12, 24. Therefore, applying the factorial function does not work because 29! factorial has way more factors than 29.

I hope this helps :)

- 3 months, 3 weeks ago

Ah... I understand. Thanks.

- 3 months, 3 weeks ago

So we're looking for something like 100032 which has the 28 factors (1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192, 521, 1042, 1563, 2084, 3126, 4168, 6252, 8336, 12504, 16672, 25008, 33344, 50016, 100032).

- 3 months, 3 weeks ago

I now think(thanks to @Finnley Paolella) that there are too many such numbers!

- 3 months, 3 weeks ago

125000?

- 3 months, 3 weeks ago

Yes, it is one of the answers I know of now, but as Finnley Paolella stated, it is not unique.

- 3 months, 3 weeks ago

- 3 months, 3 weeks ago

Great Question, But yes has way too many solutions.

- 3 months, 3 weeks ago

@Vinayak Srivastava,if product of factors are given,then a unique solution can be determined.

- 2 months, 3 weeks ago

- 2 months, 3 weeks ago

- 2 months, 3 weeks ago

- 2 months, 2 weeks ago

Read a little bit, but got bored, I don't know why!

- 2 months, 2 weeks ago

I started to read and found a programming task so I connected them and started to prove that to the first 50 numbers :)

- 2 months, 2 weeks ago

Great!

- 2 months, 2 weeks ago

Ohh! I forgot to post the results :)

- 2 months, 2 weeks ago

The same happens with me all time.😅

- 2 months, 2 weeks ago

Task: Find the least number n that can we represented as a product n = a ∙ b in k (1 ≤ k ≤ 50) ways. Products a ∙ b and b ∙ a are the same, all numbers are positive integers.

This isn't the same problem. For example 4 has 3 divisors(1,2,4), but we can write this in only two ways : 1x4 and 2x2.

The result is:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 1 1 2 4 3 12 4 24 5 36 6 60 7 192 8 120 9 180 10 240 11 576 12 360 13 1296 14 900 15 720 16 840 17 9216 18 1260 19 786432 20 1680 21 2880 22 15360 23 3600 24 2520 25 6480 26 61440 27 6300 28 6720 29 2359296 30 5040 31 3221225472 32 7560 33 46080 34 983040 35 25920 36 10080 37 206158430208 38 32400 39 184320 40 15120 41 44100 42 20160 43 5308416 44 107520 45 25200 46 2985984 47 9663676416 48 30240 49 233280 50 45360

- 2 months, 2 weeks ago


# What is the indefinite integral of $\int \left(-\frac{1}{5x}\right) dx$?

Why does $$\int \left(-\frac{1}{5x}\right) dx = -\frac{1}{5}\ln(x)+c$$ and not $$-\frac{1}{5}\ln(5x)+c$$ instead?

When I differentiate both solutions I get the same answer, $$-\frac{1}{5x}$$.

Can anyone give me an explanation as to why? Thank you.

• $\int -\frac{1}{5x} dx=-\frac{1}{5}\int\frac{1}{x}dx$ Jul 18, 2020 at 14:48
• The expressions are equivalent since $\ln(x)$ and $\ln(5x)=\ln(5)+\ln(x)$ differ by a constant. Jul 18, 2020 at 14:53
• Also note that you are implicitly assuming that the domain is the positive real numbers. You need a different antiderivative for the negative real numbers. Jul 18, 2020 at 14:55

## 5 Answers

Both are correct: on one side, $$\int -\frac{1}{5x}dx = -\frac15 \int \frac{1}{x}dx = -\frac15\ln(x) + c,$$ because of the linearity of integrals $$\int af(x) dx = a\int f(x) dx$$ for every constant number $$a$$ (in particular $$a = -\frac15$$ in this case).

On the other side, you might also change variable, and set $$u = 5x$$, if you wish, but in that case $$du = 5dx$$ and therefore $$\begin{split} \int - \frac{1}{5x} dx &= - \int \frac{1}{u} \frac{du}{5} \\ &= -\frac{1}{5} \ln(u) + c’ \\ &= -\frac15\ln(5x) + c’ = -\frac15 \ln(x) - \frac15\ln(5) + c’ \end{split}$$ Since $$\ln(5x) = \ln(5) + \ln(x)$$ it’s just a metter of choosing a different constant $$c$$ or $$c’$$.

They clearly both give the same function if you differentiate, for they differ for a constant value $$\ln(5)$$ whose derivative is zero.

• Thank you, this really helped. Jul 18, 2020 at 15:37
• We need to include the absolute value sign in $\ln|x|$ since we don't know if $x>0$. Jul 19, 2020 at 3:31
• @Axion004 in which case you need also to use two constants $c$ for $x>0$ and $x<0$ as they could be different. I have assumed $x>0$, albeit implicitly, in order to show why there are two possible solutions which look different but they are the same. Jul 19, 2020 at 9:15

Well, notice that:

$$\int-\frac{1}{\text{n}x}\space\text{d}x=-\frac{1}{\text{n}}\int\frac{1}{x}\space\text{d}x=\text{C}-\frac{\ln\left|x\right|}{\text{n}}\tag1$$

• Yeah i get that but is the other way still right? I mean, I got a different answer. Jul 18, 2020 at 14:51

Both answers are equivalent.

Through the logarithm product rule,

$$-\frac15\ln(5x) + C = -\frac15(\ln(x) + \ln(5)) + C$$

Then simplifying:

$$\text{LHS} = -\frac15\ln(x) + \left( -\frac15\ln(5) + C \right)$$

Since $$C$$ is a constant, $$-\dfrac15\ln(5) + C$$ is a different constant.

Then let $$-\dfrac15\ln(5) + C = C_1$$. Then

$$\text{LHS} = -\frac15\ln(x) + C_1$$

Which is the first expression in your question.

The reason why they both have the same derivative is because they differ by a constant.

Your answer is probably not considered the best answer because it is less “simplified” than the other.

$$\bullet$$ See this $$\int -\frac{dt}{5t} = -\frac{1}{5} \int \frac{dt}{t} = -\frac{1}{5} \ln\lvert t \rvert + C$$

And also this:

$$\bullet$$ Let $$u = 5t \implies du = 5 dt$$, and the substitution makes sense as the function is bijective on the whole of $$\mathbb{R}$$.

therefore, \begin{align*} -\frac{1}{5} \int \frac{du}{u} = -\frac{1}{5} \ln\lvert u \rvert + C = -\frac{1}{5} \ln\lvert 5t \rvert + C \end{align*} Now is it fine @mikejacob ?

$$\int\frac1x\,dx=\ln\lvert x\rvert+c=\ln\lvert x\rvert+\ln a+c^\prime=\ln\lvert ax\rvert+c^\prime$$

, where $$c,c^\prime\in\mathbb R,a\in\mathbb R^+$$.


5 Letter Arrangements of the word 'Statistics'

How many different 5-letter 'words' can be formed from the word 'statistics'?

I really am pretty stumped. I understand how to calculate more simpler questions in which each letter of the word is different using the Permutation formula n!/(n-r)! I can also deal with questions where there are repeating letters, but the 'new' words that are being created are the same length as the original word. But for this type of question where there are repeating letters and the 'new words' are shorter than the original, I don't know where to start. I don't need the precise answer - I just want to know how to get there. In fact, I know the answer (1390) but I cannot come up with a solution. I have tried using the permutation formula but it doesn't seem appropriate for this question.

• What is your progress on this problem? What are your ideas? – Sasha Patotski Oct 3 '13 at 23:05

I will give a basic counting solution. If you want an answer using generating functions, look at the answer to this question.

If all five letters are different (which happens when you use one of each), you get $5!=120$ words.

If only two of the letters are equal (two 'S', two 'T', or two 'I', and $\binom{4}{3}$ ways to choose the remaining three), you have $\frac{5!}{2!}=60$ words for each of {'S', 'T', 'I'}. In total $720$ words.

If you have "two pairs" (there are three ways to get the pairs, and three ways to get the last letter in each case), there are $\frac{5!}{2!2!}=30$ words. In total for all these cases, there are $270$ words.

If you have "full house" (there are two ways to get three equal, and two ways for each of those to get the last pair), there are $\frac{5!}{3!2!}=10$ words. In total for these cases, there are $40$ words.

If you have three equal and two different (the triple can be had in two ways, and the remaining two can be had in $\binom{4}{2}=6$ ways for each triple), there are $\frac{5!}{3!}=20$ words. For all these cases, there are $240$ words.

Summing all of these cases gives $1390$ words.

• The answer given to the question in the textbook was 1390. Your answer makes sense to me but it seems that some multiple is missing from the calculations... I can't see where though. – frantastic Oct 4 '13 at 0:41
• @frantastic, I'm not in a position to review the argument carefully right now, but never discount the possibility that the textbook answer is wrong. Often, you will be able to find a list of errata for your textbook somewhere online (often on the writer's or publisher's website). – dfeuer Oct 4 '13 at 1:35
• @frantastic: I hade made an error in the "pair" case, forgetting to count the number of ways to choose the last three. In the last case, I had gotten 120 instead of 240 through bad mental calculation. – Mårten W Oct 4 '13 at 8:29


# Finding the probability by combination?

Question as follows A bag contains 50 tickets numbered 1,2,3,....,50 of which five are drawn at random and arranged in ascending order of the number appearing on the tickets ( x1 < x2 < x3 < x4 < x5 ). Find the probability that x3 = 30.

Total number of elementary events =50C5 Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 =29C2 ways and remaining two in 20C2 ways.

Therfore,favourable number of events = 29C2×20C2

Hence,required probability = (29C2×20C2)/50C5 =551 / 15134

Answer given is 551/15134 but in the solution they have taken the case where x2 will be less than x1. Is my assumption correct? if it is not then what could be the correct answer.

• You need to provide the full method, not just the numeric bottom line, otherwise your question is practically unreadable. P.S., I did not down-vote you, because I see good intentions and effort made in the question. Nevertheless, you need to clarify things here if you're hoping to get an answer. Jan 4, 2017 at 10:37
• The two numbers less than 30 you refer to are chosen, and "arranged in ascending order.." Jan 4, 2017 at 10:37
• the number of ways you can select to satisfy the condition is (any 2 from 1 - 29) x (any 1 from 30) x (any 2 from 31 - 50) = 29C2 x 1C1 x 20C2 ......... for the condition to be true, 30 had to be selected, and two that were smaller and two that were larger, the actual ordering leads to that condition, but is not part of the calculation
– Cato
Jan 4, 2017 at 10:41
• You are taking the question as wrong. Suppose you have option to pick 2 numbers out of 29. You picked 23, 12. In C(29,2) ways. Then you have to arrange them in ascending order. That is 12, 23. Only 1 way. So we have 1 * C(29,2). Similarly for last 2 numbers. Jan 4, 2017 at 10:47
• the order they are drawn doesn't matter, there will always be a lowest number, then a 2nd ranked number, then a 3rd ranked number. Can you see that if 30 was drawn, then there has to be exactly 2 drawn from 1-29? If exactly 1 was drawn from 1-29, then the condition would not be met. 1,2,30,31,32 is a set of tickets that meets the condition, I'm only counting that once
– Cato
Jan 4, 2017 at 10:49

Required probability = $\frac{\binom{29}{2} \times \binom{20}{2}}{\binom{50}{5}}$
Because $x_3$=30 is fixed. You pick two numbers from first 29 and arrange them in ascending order. Similarly 2 numbers from last 20 and arrange them in ascending.


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# set operations proofs

1 - 6 directly correspond is also in By deﬁnition of intersection, x ∈ A and x ∈ B. We can use the set identities to prove other facts about sets. &= \{x\mid x \in \overline{A}\wedge x \in \overline{B} )\} \\ Here are some basic subset proofs about set operations. Theorem: For any sets, $$|A\cap B|\le|A|$$ and $$|A\cap B|\le|B|$$. Note here the correspondence of Hence . $\bigcup_{i=1}^{n} S_i\,.$ This can also proven using set properties as follows. Since , . This proof might give a hint why the equivalences and set identities tables are so similiar. B . and between Back to Schedule Consider an arbitrary element x. Theorem: For any sets, $$\overline{A\cup B}= \overline{A} \cap \overline{B}$$. It's the table of logical equivalences with some search-and-replace done to it. 13. . Hence . A, and -------     De Morgan's Laws &= A\cap \overline{B}\cap A\cap \overline{C} \\ by the distribution Theorem: For any sets, $$A-B = A\cap\overline{B}$$. \begin{align*} Could have also given a less formal proof. ------- Associative Laws Since (from ), For example: Those identities should convince you that order of unions and intersections don't matter (in the same way as addition, multiplication, conjunction, and disjunction: they're all commutative operations). Theorem For any sets A and B, B ⊆ A∪ B. Proof… B ) B &= A\cap \overline{B}\cap \overline{C} \\ The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: ) by the definition of ( B - A ) . $$A\cup{U}= {U}\\A\cap\emptyset= \emptyset$$, $$(A\cup B)\cup C = A\cup(B\cup C)\\(A\cap B)\cap C = A\cap(B\cap C)$$, $$A\cup(B\cap C) =(A\cup B)\cap(A\cup C)\\A\cap(B\cup C) = (A\cap B)\cup(A\cap B)$$, $$\overline{A\cap B}=\overline{A} \cup \overline{B}\\\overline{A\cup B}= \overline{A} \cap \overline{B}$$, $$A\cup(A\cap B) = A \\ A\cap(A\cup B) = A$$, $$A\cup\overline{A} = {U}\\A\cap\overline{A} = \emptyset$$, Written $$A\cup B$$ and defined B . Then if , then since . A x\in S \wedge x\in\overline{S} \\ For example, (b) can be proven as follows: ------- Idempotent Laws by the definition of . Properties of Set Operation Subjects to be Learned . The properties 1 6 , and 11 A B Let x be an arbitrary element in the universe. Basic properties of set operations are discussed here. Furthermore a similar correspondence exists between \[\{1,2,3,4\}\cap\{3,4,5,6\} = \{3,4\}\,., For example, \begin{align*} Then by the definition of the operators, &= (A-B)\cap (A-C)\,.\quad{}∎ A. Proof: Let x ∈ A∩B. , These can also be proven using 8, 14, and 15. ( B - A ) Then . &= \{x\mid x\in A \wedge x\in \overline{B}\} \\ Let us prove some of these properties. This can also be proven in the similar manner to 9 above. \end{align*}. A • Applying this to S we get: • x (x S x S) which is trivially True • End of proof Note on equivalence: • Two sets are equal if each is a subset of the other set. but also for others. 8. ( cf. ) and implications If we need to do union/intersection of a lot of things, there is a notation like summation that is used occasionally. \mathbf{R} = \mathbf{Q} \cup \overline{\mathbf{Q}}\,.\], Written $$A\cap B$$ and defined ( cf. ) x 11. Hence. (See example 10 for an example of that too.). Here is an example. by the definition of . Be careful with the other operations. from the equivalences of propositional logic. Return to the course notes front page. Theorem: For any sets, $$|A\cup B|\ge|A|$$ and $$|A\cup B|\ge|B|$$. -------     Identity Laws 4. Next -- Recursive Definition x\in S \wedge x\notin{S}\,. As an example, we can prove one of De Morgan's laws (the book proves the other). 9. 7. Thus we see that these sets contain the same elements.∎, More Formal Proof: By definition of the set operations, the commutativity of Thus $$A-B\not\subseteq A$$. x \overline{\mathbf{Q}} = \mathbf{R}-\mathbf{Q} \,.\]. . A A \end{align*}\]. This section contains many results concerning the properties of the set operations. Proof for 9: Let x be an arbitrary element in the universe. &= \{x\mid x\in (A \cap \overline{B})\} \\ x\in S\cap\overline{S}\\ Also since , . Then there must be an element $$x$$ with $$x\in(A-B)$$, but $$x\notin A$$. Also .


# Antiderivatve to find a function such the the slope is true for all points $(x,y)$

"The graph of a certain function $f$ has the slope $4x^3-5$ at each point $(x,y)$ and the line $x+y=0$ is the tangent line to the graph. Find the function $f$."

I took the antiderivative to get $f(x)=x^4-5x+C$ but I'm not really sure how to get the initial conditions. I have that $x+y=0$ or $y=-x$ so the slope is $-1$. I can then plug it into $-1=4x^3-5$ and solve for $x$ to get $x=1$.

I can then plug that x value into $y=-x$ to get $y=-1$. Since the tangent line and the graph must share the same common point then $(1,-1)$ must be on the graph of $f(x)$. So I can solve for the initial condition which means that $-1=1-5+C$ or $3=C$ so $f(x)=x^4-5x+3$ . Is the method correct?

Your method, and your result, are correct. You can verify this solving the system of the line and the quartic: $$\begin {cases} y=-x\\ y=x^4-4x+3 \end{cases}$$ this gives the equation $$-x=x^4-5x+3 \iff x^4-4x+3=0 \iff (x-1)^2(x^2+2x+3)=0$$
That has a double root at $x=1$, so the line is tangent to the curve at this point.


Consider the problem of adding two $$n$$-bit binary integers, stored in two $$n$$ element arrays $$A$$ and $$B$$. The sum of the two integers should be stored in binary form in an $$(n + 1)$$ element array $$C$$. State the problem formally and write pseudocode for adding the two integers.

The problem can be formally stated as…

Input: Two $$n$$ bit binary integers stored in two $$n$$ element array of binary digits (either 0 or 1) $$A = \langle a_1, a_2, ... , a_n \rangle$$ and $$B = \langle b_1, b_2, ... , b_n \rangle$$.

Output: A $$(n + 1)$$ bit binary integer stored in $$(n + 1)$$ element array of binary digits (either 0 or 1) $$C = \langle c_1, c_2, ... , c_{n+1} \rangle$$ such that $$C = A + B$$.

We also assume the binary digits are stored with least significant bit first, i.e. from right to left, first bit in index $$1$$, second bit in index $$2$$, and so on. Why we are doing this is discussed after the pseudocode.

$$\textsc {Add-Binary }(A, B)$$\begin{aligned}1& \quad n=\textsc {Max}(A.length,\,B.length) \\2& \quad \text {let }C[n+1]\text { be }\text { new }\text { array } \\3& \quad carry=0 \\4& \quad \textbf {for }i=1\textbf { to }n \\5& \quad \qquad C[i]=\textsc {}(A[i]+B[i]+carry)\mod2 \\6& \quad \qquad carry=\lfloor \textsc {}(A[i]+B[i]+carry)/2\rfloor \\7& \quad C[n+1]=carry \\8& \quad \textbf {return }C \\\end{aligned}

#### Left to Right or Right to Left

An earlier version of the solution presented here assumed the least significant bit was stored in index $$n$$ instead of index $$1$$. Which made the solution not only wrong (it did not handle all possible cases properly), it also caused a great deal of confusion in the comments section.

Here is why assuming least significant bit in index $$n$$ will make the problem unnecessarily complicated.

Consider the following two binary additions

The one on the left adds $$111_b$$ and $$1_b$$ to $$1000_b$$. In this case, $$n = 3$$ and we end up with final array $$C$$ of length $$n + 1 = 4$$. The array indices are shown in blue with the assumption of storing bits as we they appear visually from left to right, i.e. most significant bit in first index and least significant bit in last index, $$n$$.

In this particular case there is no complication, and we could have just designed our pseudocode to iterate from the opposite direction, from $$n$$ downto $$1$$, and stored the result of the addition in line 4 in $$C[i + 1]$$ and the final $$carry$$ in line 6 in $$C[1]$$.

However, note that for the addition on the right, $$110_b$$ and $$1_b$$ sums up to $$111_b$$, and we end up with final array $$C$$ of length $$n = 3$$. In this case, $$C[1]$$ is empty (highlighted with light red), and we are left with the additional task of shifting all the elements to the left to meet our initial assumption of having most significant bit at index $$1$$.

One can argue that having zero in the first index is not a deal breaker, but depending on the use case it might add up to redundant work. For example, let’s say we need to repeatedly do this addition. And every time we end up with a case like the right one. Then we would keep on adding redundant zeroes in the beginning of the resulting array.

#### Python Code

# Bitwise binary addition of two lists # containing binary digits with least significant bit first def AddBinary(A, B): carry = 0 n = max(len(A), len(B)) C = [0 for i in range(n + 1)] for i in range(n): # One of A and B has length less than n # We need to treat index out of bound for that array # This is not explicitly handled in pseudocode a = A[i] if i < len(A) else 0 b = B[i] if i < len(B) else 0 # Modulo for sum and integer division for carry C[i] = (a + b + carry) % 2 carry = (a + b + carry) // 2 C[n] = carry return C # Utility function for converting decimal to binary def DecimalToBinary(num): if num == 0: return [0] ret = [] while num > 0: ret.append(num & 1) num = num >> 1 return ret # Utility function for converting binary to decimal def BinaryToDecimal(lst): num = 0 for i in range(len(lst)): num += lst[i] << i return num # Test import random num_failed = 0 total_tests = 100 for i in range(total_tests): # Create two random integers a = random.randint(0, 10000) b = random.randint(0, 10000) a_ = DecimalToBinary(a) b_ = DecimalToBinary(b) sum = BinaryToDecimal(AddBinary(a_, b_)) # Check if the sum is correct if sum != (a + b): num_failed += 1 print(f"#{i:<2} {a} + {b} = {sum} [Expected {(a + b)}]") if num_failed > 0: print(f"\nFailed {num_failed}/{total_tests}") else: print(f"Passed {total_tests}/{total_tests} tests")


# Finding the values of $5a+b$ if $ax^2+bx+10$ does not have $2$ distinct roots

If $ax^2+bx+10=0$ does not have two distinct real roots where( $a$ and $b$ are real) then the possible values of $5a+b$ from the given 4 options(as obviously there can be infinite possible values..but I only need to find out which of the four given possible values is one of these ...in short it is a Multi choice-multi-correct question) are-( there can be more than one option correct)

1) -3

2) -2

3) -1

4) 0

my attempt Here are some conclusions I have drawn-

• it is given that it does not have two distinct real roots so either( both roots are imaginary and conjugates of each other)- $D(b^2-4ac)<0$ (that is as $c=10$ so $b^2-40a<0$...or another case is both roots become equal that is $b^2-40a=0$.

• also I tried some manipulations like subtracting $f(3)-f(2)$ to get $5a+b$ ....now I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..which is not allowed (as real roots should not be distinct)so I think option (D) or zero shouldnt be possible but the answers given are $-2,-1,0$ (Edit:apparently $0$ is not the answer as pointed out be Tim phan below..but I still need help in proving it is equal to $-1,-2$)....also I am unable to get other answers I have tried getting $5a+b$ by keeping $x=5$ to get $5(5a+b+2)$ but I don't know whether it will be equal to zero or less than or greater than zero... As I don't know whether $5$ is a root of $f(x)$ or not..neither it is given in question...

Any help??

• I think you need more constrains. Here is why. Like you pointed out, essentially, you are solving a system of 2 equations with 3 unknown. $b^2 - 4 a c < 0$ and $b + 5a = N$. – Paichu Nov 19 '15 at 3:28
• In the attempted solution, you refer to $b^-40c$. Did you omit saying that $c=10$? Are $a,b,c$ restricted in some way, say to integers? – André Nicolas Nov 19 '15 at 3:30
• If that is the case, then you can try going about it in this way. Assume $b + 5a = 0$, then $b = -5a$ and $a = -\frac{1}{5}b$. Also, we have that $b^2 < 4ac$. Combine these together you will get two equations: $b^2 < -\frac{4}{5}b c$ and $25a^2 < 4a c$. You then proceed with 4 cases $a,b>0$, $a,b<0$, $a>0,b<0$, $a<0,b>0$. If all of them lead to contradiction, then it is not possible. – Paichu Nov 19 '15 at 3:42
• Why didn't you write $ax^2+bx+10$, instead of "$c$". Tbh, it looks like you were attempting to make it confusing. – YoTengoUnLCD Nov 19 '15 at 3:50
• Let $a=1$ and $b=1.01, 1.02, 1.03, 1.04,\dots$. Don't have distinct real roots, indeed don't have real roots at all. – André Nicolas Nov 19 '15 at 4:21

If you write $5a + b = k$ and use this to get rid of the $a$ in the equation $b^2 - 40a \leq 0$, you get:

$$b^2 - 40\left(\frac{k - b}{5}\right) \leq 0$$

$$b^2 + 8b \leq 8k$$

Complete the square:

$$(b + 4)^2 \leq 8k + 16$$

$$(b + 4)^2 \leq 8(k + 2)$$

Since $b$ is arbitrary, the only restriction is that the left hand side is nonnegative. So you get:

$$0 \leq 8(k + 2)$$

$$-2 \leq k$$

Hence $k = 0, -1, -2$ are all possible.

The possible values (from those listed) are $0$, $-1$, and $-2$, obtained by $(a,b) = (0,0)$, $(1,-6)$, and $\left(\frac{2}{5},-4\right)$. (The first two are not unique, but the third one is; see below.)

Now, if $ax^2+bx+10$ does not have 2 distinct roots, then the discriminant is nonpositive, i.e. $b^2-40a\le 0$. Hence, $40a\ge b^2$, necessarily implying that $a\ge 0$. Furthermore, $$b^2\le 40a\implies |b|\le\sqrt{40a}\implies b\ge -\sqrt{40a}$$ and hence $5a+b \ge 5a-\sqrt{40a}$. It can be checked that $$\min\limits_{a\ge 0}{\left(5a-\sqrt{40a}\right)} = -2$$ achieved at $a = \frac{2}{5}$. Hence, if $ax^2+bx+10$ does not have 2 distinct roots, then $$5a+b\ge -2$$ with equality if and only if $a = \frac{2}{5}$ and $b = -\sqrt{40a} = -4$.

• Why is this reasoning of my wrong....I tried some manipulations like subtracting $f(3)-f(2)$ to get $5a+b$ ....now I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..which is not allowed (as real roots should not be distinct)so I think option (D) or zero shouldnt be possible – Freelancer Nov 19 '15 at 8:16
• @Freelancer why can't we have $f(3)-f(2) = 0$? For example, if $f$ is constant, then this is true, and there's nothing saying it can't (since if it were constant, it would take the constant value $10$, which is nonzero) – Joey Zou Nov 19 '15 at 8:17
• I think since it is given that the equation does not have two distinct real roots so $f(3)-f(2)$ or $5a+b$ cannot be equal to zero unless both $f(3),f(2)$ become equal to zero..(which can be only possible when both 3,2 are roots are of the equation...which is not allowed (as real roots should not be distinct here) – Freelancer Nov 19 '15 at 8:19
• @Freelancer who cares if $f$ doesn't have two distinct roots? It says very little about the difference between its values. For example, $f(x) = x^2+1$ has no real roots, yet $f(1) - f(-1) = 0$, $f(2)-f(-2) = 0$, $f(3) - f(-3)=0$, and so on... – Joey Zou Nov 19 '15 at 8:21

I found that if a = 1/5 and b = -1 then 5a+b = 0. Did i break any rules?

• You found one solution ! Are there others ? – Tom-Tom Nov 19 '15 at 8:35

Assume $b + 5a = 0$, then $b = -5a$ and $a = -\frac{1}{5}b$. Also, we have that $b^2 < 4ac$. Combine these together to obtain $b^2 < -\frac{4}{5}b c$ and $25a^2 < 4a c$.

(1) If $a>0, b>0$, then we have: $b<-\frac{4}{5}c$< which implies $c<0$. False.

(2) If $a<0, b<0$, then we have: $-25|a|>4c$ which implies $c<0$. False.

If $a>0, b<0$, you have the same thing as in (1) and if If $a<0, b>0$, you have the same thing as in (2). Thus it is not possible for $b + 5a = 0$. Proceed similarly.

Second part: let $D>0$, assume $5a + b =-D$, so $b = -(D+5a)$. Then for this equation to not have two distinct real roots: $b^2 \leq 40a$. replacing $a$ in to obtain: $(D+5a)^2 < 40a$ Notice that this implies $a > 0$. Simplify: $$D + 5a < \pm\sqrt{40 a}$$ Let $u = \sqrt{a}$, then we have $5u^2 \mp \sqrt{40}u + D < 0$ is the condition, which is if we can find such $u$ under some condition on $D$, then the same condition is also the condition for $ax^2 + bx + 10 = 0$ to not have two distinct real roots. Solve this to get $$u=\frac{1}{10}\left(\sqrt{40} - \sqrt{40 - 20 D}\right)$$ this has a solution if $D \leq 2$, which means if $D \leq 2$, then the $ax^2 + bx + 10 = 0$ does not have two distinct real roots. Thus the possible solutions are $5a + b = -1, -2$, or (3) and (2). Please check my algebra carefully.

• You can't do that yet, since you do not know if $a$ is positive or negative, which can affect the inequality sign. – Paichu Nov 19 '15 at 3:58
• I edit the second part there. Check the algebra carefully though. Not super logical, but it is one way of doing that. – Paichu Nov 19 '15 at 4:52
• You're right. I just fixed it. – Paichu Nov 19 '15 at 5:16
• Also how can we assume $5a+b=-D$ ?? – Freelancer Nov 19 '15 at 5:17
• $D\le 2$ actually means $D = -1$ or $D=-2$. Also, how does $a>0,b<0$ give the same thing as in (1)? – Joey Zou Nov 19 '15 at 7:50