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# Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$
I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The problem is I interpreted that as:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ } \times \frac{5}{\sqrt{5}},$$
Which I believe is:
$$\frac{15}{\ \frac{5}{5} \ } = \frac{15}{1}.$$
What am I doing wrong?
• You should multiply both numerator and denominator by that constant! – Mahdi Khosravi Jul 23 '13 at 9:50
• Or you can exchange the numerator and denominator of the whole denominator and move it to whole numerator! – Mahdi Khosravi Jul 23 '13 at 9:51
• I know this an old question, but if you had simply changed your √5/5 to a √5/√5, you would've got 3√5÷5/5 and gotten your answer. The whole point of multiplying a complex faction by a number to simplify it is to times it by 1 (√5/√5 in this case) because multiplying anything by 1 is the same thing. – Gᴇᴏᴍᴇᴛᴇʀ Sep 7 '14 at 19:44
• "The problem is I interpreted that as:" .... remember that the numerator is 3. – John Joy Jan 4 at 17:58
\begin{align*}\frac{3}{\frac{\sqrt{5}}{5}} &= 3 \cdot \frac{5}{\sqrt 5}\\ &= 3 \cdot \frac{5}{\sqrt 5} \cdot 1\\ &= 3 \cdot \frac{5}{\sqrt 5} \cdot \frac{\sqrt 5}{\sqrt 5}\\ &= 3 \cdot \frac{5\sqrt 5}{5}\\ &= 3\sqrt 5 \end{align*}
• I appreciate everyone's response but this answer is the most elaborate. Thanks blf, I see where I went wrong now! – Sam Jul 23 '13 at 10:00
You multiplied the original fraction by $\dfrac5{\sqrt5}$, which is not $1$, so of course you changed the value. The correct course of action is to multiply by $1$ in the form
$$\frac{5/\sqrt5}{5/\sqrt5}$$
to get
$$\frac3{\frac{\sqrt5}5}=\frac3{\frac{\sqrt5}5}\cdot\frac{\frac5{\sqrt5}}{\frac5{\sqrt5}}=\frac{3\cdot\frac5{\sqrt5}}1=3\cdot\frac5{\sqrt5}=3\sqrt5\;,$$
since $\dfrac5{\sqrt5}=\sqrt5$. | {
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since $\dfrac5{\sqrt5}=\sqrt5$.
More generally,
$$\frac{a}{b/c}=\frac{a}{\frac{b}c}\cdot\frac{\frac{c}b}{\frac{c}b}=\frac{a\cdot\frac{c}b}1=a\cdot\frac{c}b\;,$$
this is the basis for the invert and multiply rule for dividing by a fraction.
This means $$3\cdot \frac{5}{\sqrt{5}}=3\cdot\frac{(\sqrt{5})^2}{\sqrt{5}} =3\sqrt{5}$$ You're multiplying twice for the reciprocal of the denominator.
Another way to see it is multiplying numerator and denominator by the same number: $$\frac{3}{\frac{\sqrt{5}}{5}}=\frac{3\sqrt{5}}{\frac{\sqrt{5}}{5}\cdot\sqrt{5}} =\frac{3\sqrt{5}}{1}$$
Start with $$\frac{3}{\sqrt{5}/5}=\frac{15}{\sqrt{5}},$$
and then rationalize the denominator (multiply both numerator and denominator by $\sqrt{5}$) to get
$$\frac{15\sqrt{5}}{5}=3\sqrt{5}.$$
You have the following fraction to simplify: \begin{align} \frac{3}{\sqrt{5}/5} &=\frac{5\times 3}{\sqrt{5}} \\ &=\frac{15}{\sqrt{5}} \\ &=\sqrt{\bigg(\frac{15}{\sqrt{5}}\bigg)^2} \\ &=\sqrt{\frac{15^2}{\sqrt{5}^2}} \\ &=\sqrt{\frac{225}{5}} \\ &=\sqrt{45} \\ &= \sqrt{9\times 5} \\ &= \sqrt{9}\sqrt{5} \\ &= 3\sqrt{5} \\ \therefore \frac{15}{\sqrt{5}}\times \frac{5}{\sqrt{5}} &=\frac{15}{\sqrt{5}}\times \frac{\big(\frac{15}{\sqrt{5}}\big)}{3} \\ &=\frac{15}{\sqrt{5}}\times \frac{15}{\sqrt{5}}\times \frac{1}{3} \\ &=\bigg(\frac{15}{\sqrt{5}}\bigg)^2 \times \frac{1}{3} \\ &=45\times \frac{1}{3} \\ &=\frac{45}{3} \\ &=15 \end{align}
One thing that helps me organize my thoughts, is to convert both numerator and denuminator to fractions as follows.
\begin{align} \frac{3}{\frac{5}{\sqrt{5}}}&=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}\cdot\frac{\frac{\sqrt{5}}{5}}{\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\sqrt{5}}{5}}{\frac{5}{\sqrt{5}}\cdot\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\sqrt{5}}{5}}{1}=\frac{3}{1}\cdot\frac{\sqrt{5}}{5}=\text{etc.}\\ \end{align} | {
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In the expression $$\frac{3}{\frac{\sqrt{5}}{5}}$$, to find the reciprocal of that expression's denominator, $$\frac{\sqrt{5}}{5}$$, simply swap its numerator and denominator.
Thus the reciprocal of $$\frac{\sqrt{5}}{5}$$ is $$\frac{5}{\sqrt{5}}$$, and $$\frac{3}{\frac{\sqrt{5}}{5}} = 3 \cdot \frac{5}{\sqrt{5}}$$
I think the result you have to remember is
$$\boxed{\dfrac 1{\sqrt{a}}=\dfrac{\sqrt{a}}a}$$
You will see this kind of manipulation very often in your studies, and it allows to get rid of square roots on denominator.
Here your expression is just $$\dfrac{3}{\frac{\sqrt{5}}{5}}=\dfrac{3}{\frac 1{\sqrt{5}}}=3\sqrt{5}$$ | {
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# Is There a Rigorous Proof Of 1 = 0.999…?
Yes.
First, we have not addressed what 0.999… actually means. So it’s best first to describe what on earth the notation $$b_0.b_1b_2b_3…$$ means. The way mathematicians define this thing is
$$b_0.b_1b_2b_3…=\sum_{n=0}^{+\infty}{\frac{b_n}{10^n}}$$
So, in particular, we have that
$$0.999…=\sum_{n=1}^{+\infty}{\frac{9}{10^n}}$$
But all of this doesn’t really make any sense until we define what the right-hand side means. There is an infinite sum there, but what does that mean? Well, we put
$$S_k=\sum_{n=1}^{k}{\frac{9}{10^n}} \ ,$$
then we have a finite sum. So, for example
$$S_1=0.9, \ ~S_2=0.99, \ ~S_3=0.999, \ etc.$$
So, in some way, we want to take the limit of this sequence.
Let’s consider a particularly simple sequence to illustrate the idea behind the definition of a limit of a sequence: 1/2, 1/3, 1/4,… The terms in this sequence get smaller and smaller. You might think that it’s obvious that it goes to 0, or that it’s obvious that a smart mathematician can prove that it goes to 0, but it’s not. It’s impossible to even attempt a proof until we have defined what it means for something to go to 0. So we have to define what the statement “1/2, 1/3, 1/4,… goes to 0” means before we can attempt to prove that it’s true. | {
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This is the standard definition: “1/n goes to 0” means that “for every positive real number $\epsilon$, there’s a positive integer N, such that for all integers n such that $n\geq N$, we have $|1/n| < \epsilon$”. With this definition in place, it’s quite easy to prove that “1/n goes to 0” is a true statement. What I want you to see here, is that we chose this definition to make sure that this statement would be true. The first mathematicians who thought about how to define the limit of a sequence might have briefly considered definitions that make the statement “1/n goes to 0” false, but they would have dismissed those definitions as irrelevant because they fail to capture the idea of a limit that they already understood on an intuitive level.
So the real reason why 1/n goes to 0 is that we wanted it to! Similar comments hold for the sequence of partial sums that define 0.999… It goes to 1, because we have defined the concepts “0.999….”, “sum of infinitely many terms”, and “limit of a sequence” in ways that make 0.999…=1. Can we define number systems such that 1=0.999… does not hold? Of course! But these number systems are not as useful, because they don’t conform to our intuition about limits and numbers.
Now that we know what a limit and an infinite sum is, let me give a fully rigorous proof to the equality 1=0.999… This proof is due to Euler and it appears in the 1770’s edition of “Elements of algebra”.
We know that
$$0.999…=\sum_{n=1}^{+\infty}{ \frac{9}{10^n} } = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} +…$$
This sum is a special kind of sum, namely, it’s a geometric sum. For (infinite) geometric sums, we can find its limit easily:
Let
$$x=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+…$$
Then
$$9x=0.999…$$
But, we also have $10x=1+\frac{1}{10}+\frac{1}{10^2}+…$, so $10x-x=1$.
This implies that $x=\frac{1}{9}$. | {
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This implies that $x=\frac{1}{9}$.
Hence,
$$0.999…=9x=1$$
Does this proof look familiar? It should! It is essentially the same as Proof #2 in the previous post. The only difference is that every step is now justified by operations with limits.
See this supportive article: https://www.physicsforums.com/insights/why-do-people-say-that-1-and-999-are-equal/
The following forum members have contributed to this FAQ:
AlephZero
Fredrik
micromass
tiny-tim
vela
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102 replies
1. nuuskur says:
It should be unanimous that 1 – 0.9(9) = 0.0(0) = 0. Assume the opposite, then 0.0(0) != 0, contradiction.
Too much java, my TeX is rusty :D
2. Mark44 says:
[QUOTE=”alva, post: 5215165, member: 497526″]
But my reasoning is similar to the OP. He is taking more addends in one series than in the other.[/QUOTE]
[QUOTE=”weirdoguy, post: 5215170, member: 464738″]No it’s not, because he’s using infinite series, and you are using finite series. It is a big difference.[/QUOTE]
In addition, [USER=497526]@alva[/USER], you don’t seem to realize that 0.99 and 0.99… mean completely different things. The first number has two decimal digits that are shown explicitly, and an implied infinite number of zeroes that follow it. The second number has an infinite number of 9’s following the decimal point.
We are NOT saying that 1 = 0.99. We ARE saying that 1 = 0.99…, and the proof is given in the OP.
Also, 0.99… and 0.999… are exactly the same, while 0.99 and 0.999 are different.
3. micromass says:
[QUOTE=”alva, post: 5215165, member: 497526″]
But my reasoning is similar to the OP. He is taking more addends in one series than in the other.[/QUOTE]
No, in both cases, the number of addends is infinite. The only difference is that the series in the OP is convergent, while yours are divergent.
4. alva says: | {
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4. alva says:
[QUOTE=”HomogenousCow, post: 5215156, member: 435628″]Here you’ve just made some bizarre arguments and obvious arithmetic mistakes.[/QUOTE]
Please, let me know what are “my arithmetic mistakes”.
5. weirdoguy says:
No it’s not, because he’s using infinite series, and you are using finite series. It is a big difference.
6. alva says:
[QUOTE=”HomogenousCow, post: 5215152, member: 435628″]You can’t manipulate divergent series that way, also you forgot the negative signs at the end.[/QUOTE]
You’re right and I apologize for the mistake in the signs.
Moreover where I said “my age” I meant “my age from Jesus’s birth.
But my reasoning is similar to the OP. He is taking more addends in one series than in the other.
7. weirdoguy says:
[QUOTE=”alva, post: 5215133, member: 497526″]and never 0.99999… = 1[/QUOTE]
Your “proof” is just plain wrong. Just deal with the fact that 0,(9)=1.
8. HomogenousCow says:
[QUOTE=”alva, post: 5215133, member: 497526″]Let x= 1/10 +1/100 + …
Lets take just 2 addends, x=0.11
Then 9x = 0.99
But, we also have 10x = 1 + 1/10 = 1.1, so 10x – x =0.11
Hence 0.99 = 9x = 0.99
You can use 3 addends and get 0.999 = 9x = 0,999
And 4 addends and 5, and calculate the limit when n -> infinite and never 0.99999… = 1[/QUOTE]
Here you’ve just made some bizarre arguments and obvious arithmetic mistakes.
9. HomogenousCow says:
[QUOTE=”alva, post: 5215141, member: 497526″]I was born in 1950 and my brother in 1952.
In many years my age will be 1950 + 1 + 1 +1 +1 ….
In many years my brothers age will be 1952 +1 + 1 +1 ….
So the difference will be 1952 +( 1 + 1 +1+ …) – 1950 – ( 1 +1 +1+1+…) = 1952 + (1 +1 +1 +…) – 1950 +2 + (1 +1 +1..) = 0[/QUOTE]
You can’t manipulate divergent series that way, also you forgot the negative signs at the end.
10. alva says:
I was born in 1950 and my brother in 1952.
In many years my age will be 1950 + 1 + 1 +1 +1 ….
In many years my brothers age will be 1952 +1 + 1 +1 …. | {
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In many years my brothers age will be 1952 +1 + 1 +1 ….
So the difference will be 1952 +( 1 + 1 +1+ …) – 1950 – ( 1 +1 +1+1+…) = 1952 + (1 +1 +1 +…) – 1950 +2 + (1 +1 +1..) = 0
11. alva says:
Let x= 1/10 +1/100 + …
Lets take just 2 addends, x=0.11
Then 9x = 0.99
But, we also have 10x = 1 + 1/10 = 1.1, so 10x – x =0.11
Hence 0.99 = 9x = 0.99
You can use 3 addends and get 0.999 = 9x = 0,999
And 4 addends and 5, and calculate the limit when n -> infinite and never 0.99999… = 1
12. Mark44 says:
[QUOTE=”alva, post: 5215060, member: 497526″]Let
x=110+1102+1103+…
Then
9x=0.999…
But, we also have 10x=1+110+1102+…, so 10x−x=1.
This implies that x=19.
[/quote]None of the above makes any sense. In your definition of x, its value is at least 2300. How can 9x be 0.999…?
[QUOTE=alva]
Hence,
0.999…=9x=1
**********************************************************
Im sorry but Im trying to copy/paste the OP message but the format is not preserved
**********************************************************
Lets do the proof with n=2 for ALL the equations
Let x = 1/10 + 1/100 = 0.11
Then 9x = 0.99
But, we also have 10x = 1 + 1/10 = 1.1 so 10x-x =0.11
Hence
0.99 = 9x = 0.99 !!! yes
[/quote]Of course. If x = .11, then ordinary arithmetic can be used to show that 9x = .99. So what?
[QUOTE=alva]
And, as anyone can see, you can repeat the proof for n=3,4… and using limits, when n->infinite, the result is clear, 1 is not = 0.999999…[/QUOTE]
13. micromass says:
[QUOTE=”gill1109, post: 5214887, member: 401042″]Yes you are right. However the surreals do include numbers (lots and lots of them!) which are strictly between all of 0.99999…9 (any number of repetitions) and 1.[/QUOTE]
That depends how many repetitions of ##9## you take in ##0.999…##. If you take countably many, then sure. But if you index over all ordinals, then no.
14. gill1109 says: | {
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14. gill1109 says:
[QUOTE=”micromass, post: 5214863, member: 205308″]Whether ##1=0.9999…## in the surreals depends highly on the definitions for ##0.9999…##. Some definitions make it equal, others don’t.[/QUOTE]
Yes you are right. However the surreals do include numbers (lots and lots of them!) which are strictly between all of 0.99999…9 (any number of repetitions) and 1.
15. micromass says:
[QUOTE=”gill1109, post: 5214572, member: 401042″]Here is a number system in which 1 is not equal to 0.9999…. and it moreover is rather useful in game theory, and some people even imagine it might be useful sometime in the future, in physics: J H Conway’s “surreal numbers”. [URL]https://en.wikipedia.org/wiki/Surreal_number[/URL][/QUOTE]
Whether ##1=0.9999…## in the surreals depends highly on the definitions for ##0.9999…##. Some definitions make it equal, others don’t.
16. Hornbein says:
[QUOTE=”micromass, post: 5214231, member: 205308″]micromass submitted a new PF Insights post
[URL=’https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/’]Is There a Rigorous Proof Of 1 = 0.999…?[/URL]
[URL=’https://www.physicsforums.com/insights/is-there-a-rigorous-proof-of-1-0-999/’]Continue reading the Original PF Insights Post.[/URL][/QUOTE]
I suggest 1 – 0.9999… = 0.0000….
17. KSG4592 says:
Always find these fascinating, even if most of it goes over my head at this point.
18. Pro7 says:
the question which is obvious is….how far is 0.999… away from 1?…since the answer is very close to 0 as it never ends…therefore 0.999… tends to 1.
19. Ilja says: | {
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19. Ilja says:
I have a nice childhood memory about this. My school teacher had not checked this, and had the nice idea to write, instead of an open interval [0,1) a closed interval [0.0.9999…]. I had heard somewhere that 1=0.999…, so I simply said this. The teacher, and the whole class, disagreed. So a discussion started, and I had to think about how to prove it. What I invented was that 1/9 = 0.1111…, 8/9 = 0.8888…, but when we add 0.1111… to 0.88888… we will clearly get 0.9999…. On the other side, 1/9+8/9 = 9/9 = 1.
The next day it became clear that I was right. Some guys have looked into some books.
The better proof is to compute 9*0.9999… = 10* 0.9999… – 0.9999… = 9.999999..-0.99999… = 9.
20. Anton Alice says:
I think there is a problem with the step, where you take 10x-x =1.
Applying operations on infinite sums has been considered dangerous, as far as I know.
21. terryds says:
Here is the ‘silly’ proof. 0.999… = 0.333…+0.666… = 1/3 + 2/3 = 3/3 = 1
22. H_A_Landman says:
You have to be careful, because Cantor’s hierarchy of trans-finite numbers doesn’t exist in the Surreal Numbers. For example,one might be tempted to identify Conway’s $\omega = {0,1,2,…|}$ with Cantor’s $\aleph_0$, since they are both intuitively “the infinity of the integers”. But in Cantor’s system, $\aleph_0$ is the smallest (positive) infinite number, whereas in Conway’s there does not exist any such smallest infinity since we also have $\omega-1$, $\omega/2$, $\sqrt{\omega}$, etc. Conversely, in Cantor’s system there is no largest infinity, but in Conway’s there is (On).
23. DMartin says:
I said:"I think there can be contradictions in mathematics"You said:"So you think 1+1=3 is valid and can be proven? Because that is equivalent to what you just said."I don't think I need to comment on this, it's so ridiculous that it speaks for itself.
24. DMartin says: | {
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24. DMartin says:
I prefer to keep it in degrees throughout, and although you can subtract the 89, I won't. To explain it, it's another series of numbers that suggest something, but somewhat different from the other. It suggests that for any number like 89.99999, there's another number between it and 90, arrived at via x/(sine x). You can then add more 9s, and it still applies. However many 9s you add, this rule applies, so it seems very reasonable to say that it applies to 89.99999…….. as well. That means there's at least one number between 89.9999….. and 90.Now because as I mentioned, infinities can be compared, this seems to reveal the existence of an infinity greater than the first one, or at least, a number or set of numbers that goes numerically higher. Certainly worth thinking about, but as I said, I think there can be contradictions in mathematics, so it's not an attempt to disprove your proof of 0.9999…. = 1.
25. DMartin says:
I do believe in mathematics, but I just take a different view of it than you do. Gödel's work meant that many others saw the edges of it, from the 1930s on, so my view isn't a controversial one.
26. DMartin says:
I never said it wasn't useful! It's very useful. And it doesn't matter that it's a bit rough round the edges either.
27. DMartin says:
Incidentally, and I'm not going to get into discussion about this, if one should show that there's always a number between 0.9999…. and 1, that might suggest that they're different. Well I can show that there's always a number in between 89.9999…… and 90. It's the expression x/(sine x) again, it always gives a number nearer 90, however many 9s there are.Now I should make my position clear – I'm not arguing that your proof of 0.9999…. = 1 is false. I think mathematics is not a consistent system, and that it gets a bit rough round the edges, that's all. Perhaps I've shown that a bit.
28. DMartin says: | {
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28. DMartin says:
You seem to think we should define things first, and then do our exploring. But sometimes the exploring helps to inform the definitions, and this is how mathematics has actually developed. This means one can point out a mathematical structure without necessarily having a rigorous definition. But a definition I've given for the relevant number, whether a loose one, and whether or not it fits with your definitions, is an angle x between 0 and 90 degrees such that x/(sin x) = exactly 180/π.
29. DMartin says:
By the way, I did say, but probably should have emphasised more, that what applies to one pair of numbers doesn't necessarily apply to the other. There are things that happen at or near 0 that don't happen elsewhere. But this is surely relevant, nonetheless! Thanks for the discussion.
30. DMartin says:
I never said it was a proof, surely even you noticed that. I said it was relevant, and that the context can make a difference. The point that the context makes a difference is borne out by micromass above saying about the context of surreal numbers:"Whether 1=0.9999… in the surreals depends highly on the definitions for 0.9999… . Some definitions make it equal, others don't."So it's relevant to point out a sequence relating to the second pair of numbers I mentioned, in which there is a distinction between the two of them, because one infinite sequence lands somewhere different from the other.Incidentally, there have been distinctions drawn between different infinities, and it has turned out that they can be compared, and one infinity can turn out to be 'larger' than another. This might intuitively seem impossible, but ways to compare them were found. There is some loose similarity between that and what I've done.
31. DMartin says:
That's what one would expect it to be. I've shown that the second pair isn't just 0 and 0.
32. DMartin says: | {
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32. DMartin says:
Well, there's a symmetry to be pointed out. First, you have 0.9999….. and 1, and these two numbers look the same, or almost the same. People discuss whether they're the same, and whether there's a proof that they are.Then, if you subtract 0.9999….. from 1, you find another pair of numbers – that is, the result of the subtraction, and zero. This pair of numbers can be called 1 – 0.9999…. , and 0.And there's a symmetry between these two pairs of numbers. Each pair may well be in the same relationship, whatever that is. So anything shedding a bit of light on either pair might be relevant. And showing the second pair to be different from each other is relevant.
33. DMartin says:
Yes I knew it was 1 radian."This is by definition, not up for debate". This implies that all our definitions are, by definition, correct.What you learned in primary school may or may not be true. But I have said that the context makes a difference sometimes.But the question of comparing a whole number, or a non-negative integer, and another nearby number that approaches it with an infinite series, is not always a clear cut question. And what I've set out has bearing on it, because I've show the two looking different, and looking existent.Don't forget that we learn as we go, the mathematical structures we have are not just a given eternal structure, they were put together bit by bit by finding things, and what we have is, as always, incomplete.
34. DMartin says:
Sorry, our posts crossed. Yes, I know you think that whether or not a number exists is related to whether or not one can specify the positions of the digits. But the mathematical structure I've shown above hints at a relevant structure that is uncovered, and exists in some way other than just conforming to a made up set of rules.
35. DMartin says: | {
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35. DMartin says:
Putting in 0.0001, I get x/(sin x) = 57.295779513111409697664737311509try putting in 0.0000001then 0.00000000000000001.the result will approach 180/[pi], and to me this shows something that is discovered, rather than invented, and has bearing on the similar questions we've been looking at.
36. DMartin says:
" 0.999999… exists because every digit in the decimal representation can be specified. If you ask, "what digit is in the 12th place?" Answer: 9. If you ask, "what digit is in the 59th place?" Answer: 9. If you ask, "what digit is in the 623rd place?" Answer: 9. No matter what specific digit you ask about, the answer is always "9". "That's true, and you believe that it's significant.
37. DMartin says:
See posts above. There's a series of increasingly small angles near zero degrees that approach a number at infinity, but that number is above zero. It's clear that this number gives x/(sin x) = 180/π, because the values approach that number. Whether or not any other values give 180/π makes no difference, but it's interesting to hear it.
38. DMartin says:
Well, it's a matter of taste to some extent. You say you can prove that within a particular artificial system, a number 0.99999999…. exists, but 1 minus that number, or 0.00000….0001 doesn't exist. And yet I've shown that the second number can be expressed as an angle between zero and 90 degrees.This has bearing on the question of whether mathematics is invented or discovered. Are we inventing the rules, or discovering them. When I look at the points above about the digit 1 in my number, I think there's a bit too much inventing going on.
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39. DMartin says:
If you say that one number exists and another doesn't, you need to say what you mean by exists. If you mean exists within mathematics, Gödel showed that mathematics isn't necessarily a self-consistent system, so existing within mathematics isn't necessarily a meaningful concept.If you're not keen on how my number is expressed, perhaps you'd prefer it if I said:an angle θ such that θ/(sin θ) = exactly 180/π. I can prove that this angle is not zero, because θ = 0 gives a different result.
40. DMartin says:
I don't know what you mean by existence, when you say you can prove the existence of your number. But it seems clear enough that if we can talk about numbers with an infinite number of decimal places that all make a difference, then my number and yours are very similar. And my number and yours add up to one, which gives them more common ground.About the position of the 1 in my number, out of all those 9s of yours, there must be one that corresponds to my 1. So whatever problems I have with my number (and God knows it's hard to keep them all in line), you must have the same problems with yours.
41. DMartin says:
So you have a rule that 'each numerical digit must have it's concrete position'. I suppose you know the positions of all the 9s in 0.99999….. then. But even if you argued that their positions are more concrete than the 1 in the number I used, it's not clear where the rule came from.what I've shown is a series that converges on, or approaches, a number at infinity, and the point is, whatever the existence status of that number at infinity, it isn't zero. And surely whatever its existence status, it's similar to the existence status of the numbers you're talking about.
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42. DMartin says:
Well, that may be so, but this thread is a discussion on the basis that such numbers are worth talking about, so we're assuming they have some meaning before we start. If you say 'there is no such number', then presumably you think this whole thread is pointless.
43. DMartin says:
I can show that whatever the meaning of the number 0.0000000 => 00001, with an infinite number of zeros, it is different from 0. This means that other similar numbers are probably the same, though it does depend on the context. The method involves showing that the two numbers 0.0000000 => 00001 and 0 give completely different output numbers when put into an equation.To show that 0.0000000 => 00001 does not equal 0.Take the equation θ/(sin θ) , where θ is an angle in degrees.For θ = 0.0000000 => 00001, it gives θ/(sin θ) = 180/π = 57.295779513082320876798154814105this is known because a series of increasingly small angles such as 0.0001, 0.0000000001, 0.0000000000000001 etc.give numbers that approach 180/π.But for θ = 0, it gives θ/(sin θ) = 0Therefore 0.0000000 => 00001 does not equal 0.Any thoughts would be appreciated, thanks.
44. alva says:
Letx=110+1102+1103+…Then9x=0.999…But, we also have 10x=1+110+1102+…, so 10x−x=1.This implies that x=19.Hence,0.999…=9x=1
45. WWGD says:
You can also use, though not as nice, the perspective of the Reals as a metric space, together with the Archimedean Principle: then d(x,y)=0 iff x=y. Let then ## x=1 , y=0.9999…. ##Then ##d(x,y)=|x-y| ## can be made indefinitely small ( by going farther along the string of 9's ), forcing ## |x-y|=0## , forcing ##x=y ##. More formally, for any ##\epsilon >0 ##, we can make ##|x-y|< \epsilon ##
46. gill1109 says:
Here is a number system in which 1 is not equal to 0.9999…. and it moreover is rather useful in game theory, and some people even imagine it might be useful sometime in the future, in physics: J H Conway's "surreal numbers". https://en.wikipedia.org/wiki/Surreal_number | {
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47. Josh Meyer says:
Nice!The informal proof I always share with people is that 1/9=0.111…, 2/9=0.222…, 3/9=0.333…, and so on until 9/9=0.999…=1 | {
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The Integral
1. Jun 27, 2007
anantchowdhary
Is there a mathematical proof that can prove that the integral is the antiderivative?
2. Jun 27, 2007
Staff: Mentor
fundamental theorem of calculus
Look up the fundamental theorem of calculus.
3. Jun 27, 2007
anantchowdhary
umm...ive tried to study it....but my question is a little different....i meant to say..
if we take out the area of the graph of a function f(x) bound at certain limits, then..is there any way to prove that the indefinite integral gives us the antiderivative of the function
or rather..the area bound under two limits of the function is simply the limits applied to the antiderivative of the function
4. Jun 27, 2007
Kummer
Given an integrable function $$f:[a,b]\mapsto \mathbb{R}$$ we define,
$$g(x) = \int_a^x f(t) dt$$ for all $$t\in [a,b]$$.
Then, $$g$$ is a continous function on $$[a,b]$$. And furthermore if $$f$$ is continous at $$t_0 \in (a,b)$$ then $$g$$ is differentiable at $$t_0$$ with $$g'(x_0)=f(x_0)$$
5. Jun 27, 2007
mathwonk
suppose f is aN INCREASING CONTINUOUS FUNCTION. then the area under the graph between x and x+h is between f(x)h and f(x+h)h,
i.e.f(x)h < A(x+h)-A(x) < f(x+h)h.
satisfy yourself of this by drawing a picture.
now the derivative of that area function is the limit of [A(x+h)-A(x)]/h, as h goes to zero.
by the inequality above, this limit iscaught b etween f(x) and f(x+h), for all h, which emans, since f is continuious, it equals f(x). i.e. dA/dx= f(x).
now in creasing is not n eeded ubt makes it easier.
6. Jun 27, 2007
anantchowdhary
thanks..ill give it a thought
7. Jun 28, 2007
HallsofIvy | {
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6. Jun 27, 2007
anantchowdhary
thanks..ill give it a thought
7. Jun 28, 2007
HallsofIvy
Staff Emeritus
A little more generally- assume f(x)> 0 for a< x< b. For any number n, divide the interval from a to b into n equal sub-intervals (each of length (b-a)/n)). Construct on each a rectangle having height equal to the minimum value of f on the interval- that is, each rectangle is completely "below" the graph. Let An be the total area of those rectangles. Since each rectangle is contained in the region below the graph of f, it is obvious that $A_n \le A$ where A is the area of that region (the "area below the graph").
Now do exactly the same thing except taking the height to be the maximum value of f in each interval. Now, the top of each interval is above the graph of f so the region under f is completely contained in the union of all the rectangles. Taking An to be the total area of all those rectangles, we must have $A \le A^n$.
That is, for all n, $A_n \le A \le A^n$
If f HAS an integral (if f is integrable) then, by definition, the limits of An and An must be the same- and equal to the integral of f from a to b. Since A is always "trapped" between those two values, the two limits must be equal to A: The area is equal to the integral of f from a to b.
That's pretty much the proof given in any Calculus book.
8. Jun 28, 2007
Gib Z
"is there any way to prove that the indefinite integral gives us the antiderivative of the function"
I seemed to understand this as - ' is there any way to prove F(x), where F'(x)=f(x), is the same function as the function given by $$\int f(x) dx$$.' The fundamental theorem of Calculus shows when we put upper and lower bounds, b and a, on the integral, it results in F(b) - F(a), however to the specific question as I understood, the dropping of bounds is just a nice notation for the integral/antiderivatve function.
9. Jun 28, 2007
anantchowdhary
umm...i tried this out..i Followed pretty much of it | {
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9. Jun 28, 2007
anantchowdhary
umm...i tried this out..i Followed pretty much of it
but how do we prove that dA/dx=f(x)?
thanks
10. Jun 28, 2007
mathwonk
notice the proof i gave for the increasing case, which is due to newtton, does not need the deep theorem that a continuous function always has a max and a min on a closed bounded interval.
it also covers (when used piecewise) all polynomials, con tinuous rational functions, and continuous trig functions, that occur in practice. hence there is
no reason for books to omit this proof or banish it to an appendix.
11. Jun 28, 2007
HallsofIvy
Staff Emeritus
Define F(x) to be
$$\int_a^x f(t)dt$$
For f(x)> 0 we can interpret that as the area between the graph y= f(x) and the x-axis, from a to the fixed value x. Then F(x+h) is
$$\int a^{x+h}f(t)dt$$
the area between the graph y= f(x) and x-axis, from a to the fixed value x+h.
Now F(x+h)- F(x) is the area between the graph y= f(x), between x and x+h. It's not difficult to see that is equal to the area of the rectangle with base x to x+h and height f(x*) where x* is some value between x and x+h: That is F(x+h)- F(x)= hf(x*). Then
$$\frac{F(x+h)- F(x)}{h}= f(x^*)$$
Taking the limit as h goes to 0 forces x*, which is always between x and x+h, to go to x.
Therefore
$$\frac{dF}{dx}= \lim_{h\rightarrow 0} \frac{F(x+h)- F(x)}{h}= f(x)$$
12. Jun 28, 2007
mathwonk
let me combine halls discussion with mine in post 5.
he has shown, modulo the theorem that max and min exist,
that in my notation, if m(h) is the min of f on the interval [x,x+h], and if M(h) is the max,
that m(h)h < A(x+h)-A(x) < M(h)h.
where < means less than or equal.
Then again if we compute the derivative of A as the limit of
[A(x+h)-A(x)]/h
as h goes to 0, we see by the inequalities that [A(x+h)-A(x)]/h
is squeezed between
M(h) and m(h) for all h. since f is continuous, these numbers both apprioach f(x), so lim [A(x+h)-A(x)]/h
= dA/dx = f(x).
13. Jun 28, 2007
mathwonk | {
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13. Jun 28, 2007
mathwonk
to prove the point halls says is "not difficult to see", uses the intermediate value theorem, in case you do not see it.
Last edited: Jun 28, 2007
14. Jun 28, 2007
anantchowdhary
thanks a lot for the proof!
15. Jun 29, 2007
HallsofIvy
Staff Emeritus
"It is easy to see ..." means "I think there is a proof but I can't remember it just now".
"Obvious to the most casual observer" means "I hope no one asks me to prove it"!
16. Jun 29, 2007
HallsofIvy
Staff Emeritus
"It is easy to see ..." means "I think there is a proof but I can't remember it just now".
"Obvious to the most casual observer" means "I hope no one asks me to prove it"!
17. Jul 2, 2007
ssd
This is a beauty..... incomplete though. I believe,for a student new to calculus, it is enough to remove all mental barriers about the fact whether an integral is anti-derivative.
18. Jul 2, 2007
jambaugh
Technically the indefinite integral is defined to be the anti-derivative (or rather a variable anti-derivative depending on an arbitrary constant since "the" anti-derivative is not unique.)
The definite integral is a value rather than a function so you can't call it an anti-derivative.
That technical nit picking aside see your undergraduate calculus text and the two forms of the FTC. | {
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# We square an integral, but why change a variable?
If we square an integral, we also change the integration variable in one of the integrals. But why is this actually correct?
For example, say I have the following:
Solve $\int_{-\infty}^\infty e^{-x^2} dx$. Let $I=\int_{-\infty}^\infty e^{-x^2} dx$, so
\begin{align} I^2 &=\bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg)^2\\ &= \bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg) \times \bigg( \underbrace{ \int_{-\infty}^{\infty} e^{-y^2}dy }_{\text{Why}?} \bigg) \\ &=\int_{-\infty}^\infty\bigg(\int_{-\infty}^\infty e^{-(x^2+y^2)} dx\bigg)dy \end{align}
But why is the following wrong: \begin{align} I^2 &=\bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg)^2\\ &= \bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg) \times \bigg(\int_{-\infty}^{\infty} e^{-x^2}dx\bigg) \\ &=\int_{-\infty}^\infty\bigg(\int_{-\infty}^\infty e^{-x^2-x^2} dx\bigg)dx \\ &=\int_{-\infty}^\infty\bigg(\int_{-\infty}^\infty e^{-2x^2} dx\bigg)dx \qquad ? \end{align}
• By definition $\int_{a}^b f(x)dx=\int_{a}^b f(y)dy$. For your second approach, no such rule for integrals exists. It abuses notation as you need $dx,dy$ to distinguish your variables when you rewrite the integrals in iterated form. – Alex R. Feb 26 '18 at 20:10
• The second-to-last equality is wrong. It is not true that $$\bigg(\int_{-\infty}^\infty e^{-x^2} dx\bigg) \times \bigg(\int_{-\infty}^{\infty} e^{-x^2}dx\bigg) =\int_{-\infty}^\infty\bigg(\int_{-\infty}^\infty e^{-x^2-x^2} dx\bigg)dx.$$ Indeed, the inner integral on the RHS $$\left(\int_{-\infty}^{\infty} e^{-2x^2}\ dx\right)$$ is equal to some positive constant, let's call it $c$. The outer integral is integrating the constant value $c$, from $-\infty$ to $\infty$, so the result is $\infty$. On the other hand, the two factors on the LHS are finite, hence their product is finite. – Bungo Feb 26 '18 at 20:16
• It is a dummy variable. – hamam_Abdallah Feb 26 '18 at 20:25 | {
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The short answer is that in your second to last identity you are neglecting cross terms in your multiplication.
A simple example may be used to demonstrate the error, and we will use a strict summation instead of an integration for clarity. Consider the sum $$\sum_{x=1}^3 x = 1 + 2 +3 = 6.$$ Now, squaring the summation yields \begin{align} \left(\sum_{x=1}^3x\right)^2 &= (1+2+3)^2 = (1+2+3)\times(1+2+3) \end{align} In order to properly calculate this quantity (long-hand) requires one to completely distribute including cross terms: \begin{align} &\ {\color{white}+}\ 1\times 1 + 1\times 2 + 1\times 3 \notag \\ (1+2+3)\times(1+2+3) =& +2\times 1 + 2\times 2 + 2\times 3 = 36. \notag \\ & + 3\times 1 + 3\times 2 + 3\times 3 \end{align} However, if the cross terms were neglected one would obtain $$(1+2+3)\times(1+2+3) \ne (1^2+2^2+3^2) = 1\times 1 + 2\times 2 + 3\times 3 = 14.$$ Written another way, we conclude that $$\left(\sum_{x=1}^3x\right)^2 = \left(\sum_{x=1}^3x\right)\left(\sum_{x=1}^3x\right) \ne \sum_{x=1}^3x^2.$$ However, as noted in the comments, the $x$ in the summation (or the in the definite integral) is just a dummy variable, and may be replaced with another symbol, such as $y$: $$\left(\sum_{x=1}^3x\right)^2 = \left(\sum_{x=1}^3x\right)\left(\sum_{y=1}^3y\right).$$ The reason for changing the dummy variable is that we may now say that the summation over $x$ and $y$ are independent, and so we may rearrange the summations: $$\left(\sum_{x=1}^3x\right)^2 = \left(\sum_{x=1}^3x\right)\left(\sum_{y=1}^3y\right) = \sum_{x=1}^3\sum_{y=1}^3x\times y.$$ Again, integration is just summation, and so the same fact holds true for integrals. Furthermore, it doesn't matter what the function inside the summation/integral is. Thus, we may write $$\left(\int_a^b f(x)\ dx\right)^2 = \left(\int_a^b f(x)\ dx\right)\left(\int_a^b f(y)\ dy\right) = \int_a^b\int_a^b f(x)\times f(y)\ dx\ dy.$$ Using $f(x)=e^{-x^2}$ results in your example problem. | {
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Because changing the integrand's variable doesn't change the value of the integral:$$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(y)dy$$the same argument holds for summation as following$$\sum_{n=a}^{b}k_n=\sum_{m=a}^{b}k_m$$also we know that$$\left(\sum_{n=a}^{b}k_n\right)^2=\left(k_a+...+k_b\right)^2=k_a^2+...+k_b^2+2k_ak_{a+1}+...+2k_{b-1}k_b$$and$$\sum_{n=a}^{b}k_n\sum_{m=a}^{b}k_m=(k_a+...+k_b)(k_a+...+k_b)$$which by expanding the terms and rearranging them leads to$$\left(\sum_{n=a}^{b}k_n\right)^2=\sum_{n=a}^{b}k_n\sum_{m=a}^{b}k_m$$since integral is intrinsicly a summation, this can be generalized to integral operator either.
I think the best way of visualizing what is happening intuitively is thinking in terms of programming.
If you make a program and you have two functions (in the programming sense) which do not interact, you can call "$x$" the variable in both functions. They are isolated.
However, if a function is supposed to handle two different variables independently in the same context, you cannot name them the same thing.
Formally and mathematically (and also changing notation from $\int f(x)dx$ to $\int f$ so that the point may become clearer), we have the following chain of equalities.
\begin{align*} \underline{\left(\int_{\mathbb{R}} x \mapsto e^{-x^2}\right)} \cdot \underline{\left(\int_{\mathbb{R}} x \mapsto e^{-x^2} \right)}&\stackrel{Linearity}=\left(\int_{\mathbb{R}} \underline{\left(\int_{\mathbb{R}} x \mapsto e^{-x^2}\right)}\left(x \mapsto e^{-x^2} \right)\right) \\ &\underset{\text{fctn mult.}}{\overset{\text{Def. of}}{=}} \int_{\mathbb{R}} x \mapsto \left(e^{-x^2} \underline{\left(\int_{\mathbb{R}} y \mapsto e^{-y^2}\right)}\right) \\ &\underset{\text{}}{\overset{\text{Linearity}}{=}} \int_{\mathbb{R}} x \mapsto \left(\int_{\mathbb{R}} y \mapsto e^{-x^2-y^2}\right) \\ &\stackrel{Fubini}{=}\int_{\mathbb{R}^2}(x,y) \mapsto e^{-x^2-y^2}. \end{align*} | {
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Using the same name ($x$) for the variables in both integrations is "fine" up to the third equality (I changed it in the second equality for the sake of clarity, otherwise the third would be a little mystic). Using $x$ instead of $y$ in the second line would be extremely bad taste, but not explicitly wrong. Using $x$ instead of $y$ in the right side of the first equality is edgy, but not so much.
Indeed, each underline integral is a closed box with respect to the rest: they are fixed numbers, which don't interact at all with the rest of the environment other than via the fact that they are numbers. It is in the third equality that we are multiplying a constant (relative to the inner function which we are integrating) which depends on $x$. It does not depend on the function inside the integral at all, but it is invading its space and interacting with it. To be extremely clear, call the function $x \mapsto e^{-x^2}$ by $f$. The second line is then $$\int_{\mathbb{R}} x \mapsto \left(e^{-x^2} \left(\int_{\mathbb{R}} f\right)\right)$$ $$=\int_{\mathbb{R}} \left( x \mapsto \int_{\mathbb{R}} (e^{-x^2}\cdot f)\right),$$ where we multiplied a constant to the inner integral.
What you are proposing is that the equality should read $$\int_{\mathbb{R}} x \mapsto \left(e^{-x^2} \left(\int_{\mathbb{R}} f\right)\right)$$ $$=\int_{\mathbb{R}} \left( x \mapsto \int_{\mathbb{R}} g\right),$$ where $g(t)=e^{-t^2}f(t)$. This doesn't make sense. Essentially, what you are implying is that $$f \cdot\int g=\int f\cdot g,$$ which is not true. | {
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# Math Help - Probability question
1. ## Probability question
A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E.
Find the probability that both of the letters to A and B are in incorrect envelopes.
The answer is given as 13/20 but I can't figure out why. Can anyone help?
2. Originally Posted by jjc
A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E. Find the probability that both of the letters to A and B are in incorrect envelopes. The answer is given as 13/20
The number of ways that A or B is in the correct envelope is give by:
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{{2(4!) - 3!}}{{5!}}$
Now, for neither to get the correct letter subtract from 1.
3. Hello, jjc!
A man writes 5 letters, one each to A, B, C, D and E.
Each letter is placed in a separate envelope and sealed.
He then addressed the envelopes, at random, one each to A, B, C, D and E.
Find the probability that both of the letters to A and B are in incorrect envelopes.
Answer: $\frac{13}{20}$
Here is a logical (but very primitive) approach . . .
The letters are: . $A,B,C,D,E$
Their envelopes are: . $a,b,c,d,e$
There are two cases to consider:
. . (1) Letter $A$ is in envelope $b.$
. . (2) Letter $A$ is not in envelope $b.$
(1) Letter $A$ is in envelope $b$: . $P(\text{A in b}) \:=\:\frac{1}{5}$
Then letter $B$ can go in any of the other four envelopes: $\{a,c,d,e\}$
. . $P(\text{B wrong}) \:=\:\frac{4}{4} \:=\:1$
Hence: . $P(\text{A in b }\wedge\text{ B wrong}) \:=\:\frac{1}{5}\cdot1 \:=\:{\color{blue}\frac{1}{5}}$
(2) Letter $A$ is in not in envelope $b$ ... it is in $\{c,d,e\}$
. . $P(\text{A not in b}) \:=\:\frac{3}{5}$ | {
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Then letter B must be in one of the three remaining envelopes.
. . $P(\text{B wrong}) \:=\:\frac{3}{4}$
Hence: . $P(\text{A not in b} \wedge\text{ B wrong}) \:=\:\frac{3}{5}\cdot\frac{3}{4} \:=\:{\color{blue}\frac{9}{20}}$
Therefore: . $P(\text{A wrong }\wedge\text{ B wrong}) \;=\;\frac{1}{5} + \frac{9}{20} \;=\;{\bf{\color{red}\frac{13}{20}}}$ | {
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IntMath Home » Forum home » Graphs of the Trigonometric Functions » Phase shift
Phase shift [Solved!]
My question
Hello, I have a question about phase shift. What makes the second approach incorrect?
It makes sense that the shift is -c/b when considering it as when the expression inside the sin/cos function equals 0 (as if the function is just starting off).
However, when plugging in 0 for x, it seems like the graph would be shifted c to the left (since the value inside the paranthesis is acting as though x=c), but this isn't the case when b isn't 1
Thanks a lot!
Relevant page
3. Graphs of y = a sin(bx + c) and y = a cos(bx + c)
What I've done so far
Drawn several graphs, but couldn't conclude anything
X
Hello, I have a question about phase shift. What makes the second approach incorrect?
It makes sense that the shift is -c/b when considering it as when the expression inside the sin/cos function equals 0 (as if the function is just starting off).
However, when plugging in 0 for x, it seems like the graph would be shifted c to the left (since the value inside the paranthesis is acting as though x=c), but this isn't the case when b isn't 1
Thanks a lot!
Relevant page
<a href="/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php">3. Graphs of <span class="noWrap">y = a sin(bx + c)</span> and <span class="noWrap">y = a cos(bx + c)</span></a>
What I've done so far
Drawn several graphs, but couldn't conclude anything
Re: Phase shift
Hello Andrew
I think the problem here is with the concept of "starting off". Let's use "t" as the variable and talk about when things occur.
Let's also just talk about sin to keep it simple.
If we set the expression inside the sin (that is, bt + c) to 0, we are saying "at what time does sin have value 0?", since sin 0 = 0. Then by solving, we get t = -c/b.
The effect here is to shift the graph to the left by the amount c/b. (That is, the function has value 0, or it is "starting off [from value 0]" at t = -c/b.) | {
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Now let's think about y = a sin(bt+c) when the TIME is 0 ("starting off when we flip the switch on"). Now we have y = a sin c, which is some value on the y-axis, and that will be the starting value of y. I hope you can see that it doesn't matter what b is in such a case, since it is no longer in the expression y = a sin c.
Does that help?
I have added a new example on 3. Graphs of y = a sin(bx + c) and y = a cos(bx + c) - I hope that makes it clearer.
Regards
X
Hello Andrew
I think the problem here is with the concept of "starting off". Let's use "t" as the variable and talk about when things occur.
Let's also just talk about sin to keep it simple.
If we set the expression inside the sin (that is, bt + c) to 0, we are saying "at what time does sin have value 0?", since sin 0 = 0. Then by solving, we get t = -c/b.
The effect here is to shift the graph to the left by the amount c/b. (That is, the function has value 0, or it is "starting off [from value 0]" at t = -c/b.)
Now let's think about y = a sin(bt+c) when the TIME is 0 ("starting off when we flip the switch on"). Now we have y = a sin c, which is some value on the y-axis, and that will be the starting value of y. I hope you can see that it doesn't matter what b is in such a case, since it is no longer in the expression y = a sin c.
Does that help?
I have added a new example on <a href="/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php">3. Graphs of <span class="noWrap">y = a sin(bx + c)</span> and <span class="noWrap">y = a cos(bx + c)</span></a> - I hope that makes it clearer.
Regards
Re: Phase shift
Your new example helped a lot.
Thanks.
X
Your new example helped a lot.
Thanks.
You need to be logged in to reply. | {
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# How to get a transfer function from this block diagram?
Pardon my paint skills, I did my best
My attempt is short and seems to fail, I have no idea why:
\begin{align} \alpha &= in - a_{1}\beta - a_{2}\gamma \\ \beta &= \alpha z^{-1} \\ \gamma &= \beta z^{-1} = \alpha z^{-2} \end{align}
inputting 2nd and 3rd equation into the first one I get:
\begin{align} \alpha &= in - a_{1}\alpha z^{-1} - a_{2}\alpha z^{-2} \\ in &= \alpha + a_{1}\alpha z^{-1} + a_{2}\alpha z^{-2} \end{align}
I can write the output as:
\begin{align} out &= b_{2}\gamma + b_{1}\beta + b_{0}\alpha \\ out &= b_{2}\alpha z^{-2} + b_{1}\alpha z^{-1} + b_{0}\alpha \end{align}
I have output and input in terms of alpha, but I can't figure what to do from here.
You are going in the right direction! Lets take these two equations: $$(1) \quad in = \alpha+a_1\alpha z^{-1}+a_2\alpha z^{-2}$$ $$(2) \quad out = b_0\alpha+b_1\alpha z^{-1}+b_2\alpha z^{-2}$$ now rewrite (1) such that it becomes a function of $$\alpha$$: $$in = \alpha\left(1+a_1z^{-1}+a_2z^{-2}\right)$$ $$\alpha = \frac{in}{1+a_1z^{-1}+a_2z^{-2}}$$ Substitute $$\alpha$$ in equation (2): $$out = in\frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$ And derive proper discrete transfer function from it: $$H(z) = \frac{out}{in} = \frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$
• Thank you kind sir. I'm ashamed I didn't figure it out myself. But you made a little mistake I believe, in the second (2) equation the last term should have b2 instead of a2, and that mistakes propagates to the results as well Sep 17 '20 at 11:14
• Oops, small copy-past error haha. fixed it! Sep 17 '20 at 11:15
Just wanted to add that you could use this to draw diagrams next time. Much easier to use than paint. You can even use Latex. | {
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• This really should be posted as a comment and not an answer. Oct 13 '20 at 15:19
• Don't have the reputation to do so, just need 4 more to get to 50 lol Oct 13 '20 at 19:08
• Just one good answer will get you there! Oct 13 '20 at 22:13
• If you used this website to draw a new diagramm for thr op and then answer the OPs question i’d definitely +1 Oct 14 '20 at 8:59 | {
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# Why is the following relation transitive, but not reflexive?
In a practice paper for an exam there is the following relation:
$$E = \{(1,1),(2,2),(3,3),(4,4)\}\ \text{ on the set }V = \{1,2,3,4,5\}$$
It would appear that because $(5,5)$ is not in $E$, that it would not be a reflexive relation.
What is unclear is how the relation is still transitive. By our definition for Transitivity, for all elements in set $V$, there would have to be $(a,b)$ and $(b,c)\ldots$ and therefore $(a,c)$ in the set $E$.
This works for element $1$ in $V$ because you have $(1,1)$ and $(1,1)$ and therefore $(1,1)$. I would have assumed however that because there is no tuple $(5,5)$ in $E$ the entire relation could not be transitive?
Hope this makes sense, thanks for any input offered.
• "By our definition for Transitivity, for all elements in set V, there would have to be (a,b) and (b,c)" No. This is not the definition. ( a,b) do not have to be in E for all a,b. But IF (a,b) does exist, AND (b,c) exist, then (a,c) must also exist. But if (a,b) does not exist... no harm, no foul. [Note: if (a,b) did have to exist for a,b in V, then E has to be V x V = {(1,1)(1,2)(1,3)..... (5,3),(5,4)(5,5)} all, every single one of the 25 pairs. – fleablood Jan 11 '17 at 17:35
Reflexive means for all $a \in V$ then $(a,a) \in E$. That fails because $5 \in V$ and $(5,5) \not \in V$.
Transitive says nothing about items of $V$ but only about items in $E$. The relation is transitive if every time you have an $(a,b) \in E$ and a $(b,c) \in E$ you must also have a $(a,c) \in E$ then the relation is transitive. If you never have an $(5,x)$ or $(y, 5) \in E$ that will in no way affect what you do have in $E$.
This particular relationship if you have $(a,b) \in E$ then $a \ne 5; b\ne 5;$ and $a = b$. So if $(a,b), (b,c) \in E$ then $a = b = c$ and $(a,c) = (a,a) = (a,b) \in E$. So the relationship is transitive. That $a,b,c$ can never be equal to $5$ is utterly irrelevant. | {
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• I never thought about it, but reflexive is the only condition that pertains to the elements of V. Symmetry pertains only to elements of E in that if you have this element in V you must have that element too. Transitivity says if you have two elements in V you must also have a third. A function could still be transitive if you have (a,b) but absolutely no (b,x). For transitivity to fail you must have both an (a,b) and an (b,c) but not the (a,c). If you don't the (a,b) or you don't have the (b,c) it doesn't matter whether or not you have the (a,c). So no (5,x) or (x,5) doesn't matter. – fleablood Jan 11 '17 at 17:51
• Yes, @fleablood. In your last point, when you do not have some elements a, b, c, in V, such that $(a, b), (b, c) \in E$, it is a vacuously true, that transitivity holds regardless of whether or not $(a, c)$ is, or is not, in $E$. Also, as you seem to have gleaned, it is helpful to examine failures of properties of relations, for which the property fails. If it does not fail, the property holds. – Namaste Jan 11 '17 at 18:16
• Hmmm, transitivity does not require three elements in some set $V$ of a transitive relation $E$. If $E = \{(a, a)\}$, E is indeed transitive, (and symmetric). – Namaste Jan 11 '17 at 18:23
• I didn't mean to imply transitivity required 3 elements. As you point out if there are no (a,b) (b,c) pairs it's vacuously transitive. If it's non-vacuously transitive it must have (a,b) and (b,c) and a cooresponding (a,c). But a, b, and c need not be distinct. So (a,a) (a,a) and (a,a) will do. – fleablood Jan 11 '17 at 22:11
Definition: $R$ is transitive if and only if $\forall x,y (R(x,y)$ and $R(y,x))\Rightarrow R(x,z)$
Can you point to where exactly you think this definition fails? For $x=y=5$ the antecedent fails to hold (since $R(5,5)$ is false) and therefore the implication is true, since an implication with a false antecedent is always true. | {
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Your definition of transitivity is unclear as stated, and you have misunderstood it, there is no requirements on the members of the underlying set, only on the elements of the relation.
Proof $E$ is transitive: if $E\left(a,\,b\right)\land E\left(b,\,c\right)$ then $a=b\in V\backslash \left\{ 5\right\}$ and similarly $b=c\in V\backslash \left\{ 5\right\}$, so $a=c\in V\backslash \left\{ 5\right\}$ and $E\left(a,\,c\right)$. | {
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# mirror reflection
1. The problem statement, all variables and given/known data
Question A. Draw accurately in the figure the light beam that goes from Object A to eye B after being reflected on the mirror. It must be consistent with the mirror principle!
Question B. At question A. you may have connected the point behind the mirror (A') with eye B. And found the correct light path that way.
Will that also work if you use the point behind the mirror of B (B') instead? Do you get the same result? Use mathematical arguments for your judgement!
2. Relevant equations mirror reflection rules
3. The attempt at a solution
Question A: I tried to draw point A' and connected that to B. The purple line is the normal line (the imaginary line that is perpendicular to the mirror) And the angle between the mirror and A is the same as the angle between the mirror and B.
Question B: I think this is true, I can draw the same lines, but I don't know what is meant by 'mathematical arguments'
-
This is a very nicely written question. If only we had more like this one... :-) – Vedran Šego Nov 14 '13 at 4:14
Notice that $AA'B'B$ is an isosceles trapezoid, so its diagonals intersect on the mirror, which means that using either $\triangle AA'B$ or $\triangle AB'B$ will give you the same point.
You'll probably need to provide a bit more arguments that the diagonals intersect on the mirror. Read the linked article and ask if you get stuck.
Edit: Here is a bit more:
We can extend $AA'B'B$ to a triangle. Let us denote the new vertex as $C$. It is on the intersection of the lines on which $\overline{AB}$ and $\overline{A'B'}$ lie.
Note that
1. $|\overline{AA''}| = |\overline{A'A''}|$, where $A''$ denotes a point at which $\overline{AA'}$ intersects with the mirror; and
2. the mirror (on which the altitude of the mirror lies) is orthogonal to $\overline{AA'}$. | {
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From points $1$ and $2$, we see that $\triangle AA'C$ is an isosceles triangle. The same can be said for $\triangle BB'C$ (with the same arguments).
Now, since
$$|\overline{AB}| = |\overline{AC}| - |\overline{BC}| = |\overline{A'C}| - |\overline{B'C}| = |\overline{A'B'}|,$$
we see that $AA'B'B$ is an isosceles trapezoid.
Using this, show that the diagonals of $AA'B'B$ intersect on the mirror. Can you take over now?
-
Do you mean with a bit more arguments to proof Similarity (geometry) between the triangles? – user108681 Nov 14 '13 at 4:36
@user108681 I've expanded my answer. – Vedran Šego Nov 14 '13 at 12:15
A bit easier: you can just observe isosceles triangles $\triangle AA'E$ and $\triangle BB'E$, where $E$ is the point on the mirror where $\overline{AB'}$ and $\overline{A'B}$ intersect (i.e., the point you're looking for). – Vedran Šego Nov 14 '13 at 18:31 | {
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# Why is minimizing least squares equivalent to finding the projection matrix $\hat{x}=A^Tb(A^TA)^{-1}$?
I understand the derivation for $\hat{x}=A^Tb(A^TA)^{-1}$, but I'm having trouble explicitly connecting it to least squares regression.
So suppose we have a system of equations: $A=\begin{bmatrix}1 & 1\\1 & 2\\1 &3\end{bmatrix}, x=\begin{bmatrix}C\\D\end{bmatrix}, b=\begin{bmatrix}1\\2\\2\end{bmatrix}$
Using $\hat{x}=A^Tb(A^TA)^{-1}$, we know that $D=\frac{1}{2}, C=\frac{2}{3}$. But this is also equivalent to minimizing the sum of squares: $e^2_1+e^2_2+e^2_3 = (C+D-1)^2+(C+2D-2)^2+(C+3D-2)^2$.
I know the linear algebra approach is finding a hyperplane that minimizes the distance between points and the plane, but I'm having trouble understanding why it minimizes the squared distance. My intuition tells me it should minimize absolute distance, but I know this is wrong because it's possible for there to be non-unique solutions.
Why is this so? Any help would be greatly appreciated. Thanks!
• Check your derivation for $\hat{x}$ and you will find (as @Ian points out) the expression should be $\hat{x}=(A^TA)^{-1}A^Tb$. Perhaps you should edit this into the Question, though it may be the crux of why you had trouble "connecting it to least squares regression" (minimization). Aug 22 '16 at 15:00
You should be multiplying by $(A^T A)^{-1}$ on the left, not the right. Anyway, the geometric point is that you want $Ax-b$ to be perpendicular to $Ay$ for every vector $y$. (I think this is most easily seen by a geometric argument, which can be easily found in books, but which I can't easily render here.) This translates to $(Ay)^T(Ax-b)=0$ for every $y$, which is the same as $y^T(A^T(Ax-b))=0$ for every $y$. This can only happen if $A^T(Ax-b)=0$, which rearranges to your form if $A^T A$ is invertible (as is usually the case). | {
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Also, it is the same to minimize the square of the Euclidean distance as it is to minimize the Euclidean distance itself. (This is also true of any other nonnegative quantity.) What would be different is minimizing some other distance, like the "taxicab" distance (where you sum the absolute values). Why we should choose to minimize the Euclidean distance in the first place is not a purely mathematical question, it depends on where the problem is coming from. That question is a bit off-topic here, though, and has also been asked before on MSE. (The short version of that discussion: "it's mathematically convenient" and "see the Gauss-Markov theorem".)
• About minimizing the Euclidean distance: I'm a little confused about the distinction. For example, least squares would penalize a point for being farther away: e.g. a point 1 unit away would be penalized by 1 whereas a point 2 units away would be penalized by 4. I think my confusion lies in the difference between minimizing an aggregate of points vs individual ones, since it seems to me that minimizing euclidean distance for each point wouldn't account for the distance factor in least squares. Aug 22 '16 at 15:30
• In your problem the goal is to choose a vector in a certain 2D subspace which is closest to another vector in 3D space, where you measure distance in the Euclidean sense. This is not measured separately by components but rather through the usual distance formula.
– Ian
Aug 22 '16 at 15:39
The matrix has full column rank; we are guaranteed a unique solution.
Problem statement
\begin{align} \mathbf{A} x &= b \\ \left[ \begin{array}{cc} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} x_{1} \\ x_{2} \\ \end{array} \right] &= \left[ \begin{array}{cc} 1 \\ 2 \\ 2 \\ \end{array} \right] \end{align}
Normal equations | {
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Normal equations
\begin{align} \mathbf{A}^{*} \mathbf{A} x &= \mathbf{A}^{*} b \\ % \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} x_{1} \\ x_{2} \\ \end{array} \right] &= \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} 1 \\ 2 \\ 2 \\ \end{array} \right] \\[3pt] % \left[ \begin{array}{cc} 3 & 6 \\ 6 & 14 \\ \end{array} \right] % \left[ \begin{array}{cc} x_{1} \\ x_{2} \\ \end{array} \right] % &= % \left[ \begin{array}{cc} 5 \\ 11 \\ \end{array} \right] % \end{align}
Least squares solution
\begin{align} x_{LS} &= \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} b \\ % \left[ \begin{array}{cc} x_{1} \\ x_{2} \\ \end{array} \right] &= \frac{1}{6} \left[ \begin{array}{cc} 14 & -6 \\ -6 & 3 \\ \end{array} \right] % \left[ \begin{array}{cc} 5 \\ 11 \end{array} \right] \\ % &= \frac{1}{6} \left[ \begin{array}{cc} 4 \\ 3 \end{array} \right] % \end{align} | {
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# On Cesàro convergence: If $x_n \to x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \to x$
I have this problem I'm working on. Hints are much appreciated (I don't want complete proof):
In a normed vector space, if $x_n \longrightarrow x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \longrightarrow x$
I've been trying adding and subtracting inside the norm... but I don't seem to get anywhere.
Thanks!
Given $\epsilon >0$ there exists $n_0$ such that if $n\geq n_0$ then $\parallel x_n -x\parallel < \epsilon$
so
\begin{align*} 0 & \leq \left\lVert \frac{x_1 +\cdots +x_n}{n} -x \right\rVert \leq \left\lVert \frac{x_1 + \dots + x_n - nx }{n} \right\rVert \\ & \leq \frac{\lVert x_1 - x \rVert}{n} + \dots + \frac{\lVert x_{n_0 - 1} - x \rVert}{n} + \frac{\lVert x_{n_0} - x \rVert}{n} +\dots + \frac{\lVert x_{n} - x \rVert}{n} \\ &\le \frac 1n\sum_{i=1}^{n_0-1} \| x_i -x\| + \frac{n-n_0}{n} \epsilon \end{align*}
The first $n_0 -1$ terms $\| x_i -x\|$ can be bounded by some $M$, thus for $n\ge (n_0-1)M/\epsilon=: N_0$ we have $$\frac 1n\sum_{i=1}^{n_0-1} \| x_n -x\| \le \frac 1n (n_0-1)M \le \epsilon$$
Thus $$\left\| \frac{x_1 + \cdots x_n}{n} - x\right\| <2\epsilon$$ when $n\ge N_0$.
Thanks a lot @Leonid Kovalev for the inspiration, though my main problem was that I wasn't aware of what to do with the $nx$ (the silliest part :P) | {
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• This sort of thing is encouraged, I think. – Dylan Moreland Jun 11 '12 at 3:21
• I think you want "$-nx$" in the first line of the display, and $\|x_n - x\|$ at the end. I don't think you want to say that the first few terms are $\leq M$; that doesn't seem to be enough. – Dylan Moreland Jun 11 '12 at 3:41
• @DylanMoreland: Why isn't it enough? Can you explain? – Bouvet Island Jun 11 '12 at 14:56
• @Inti: I think you are missing something in the argument. Generally the argument consists of two steps: first choose $n_0$ such that if $n \geq n_0$, $\|x_n -x \| < \epsilon / 2$. Next choose $N \geq n_0$ such that $M$ (which is at least $\sup_n \|x_n\|$) satisfies $M / N < \epsilon / 2$. I don't see the second part of the argument implemented in your answer. – Willie Wong Jun 11 '12 at 14:59
• Should it not be $(n-n_0+1)\epsilon/n$? – jippyjoe4 Oct 14 '19 at 4:19
There is a slightly more general claim:
PROP Let $$\langle a_n\rangle$$ be a sequence of real numbers, and define $$\langle \sigma_n\rangle$$ by $$\sigma_n=\frac 1 n\sum_{k=1}^n a_k$$
Then $$\liminf_{n\to\infty}a_n\leq \liminf_{n\to\infty}\sigma_n \;(\;\leq\;)\;\limsup_{n\to\infty}\sigma_n\leq \limsup_{n\to\infty}a_n$$
P We prove the leftmost inequality. Let $$\ell =\liminf_{n\to\infty}a_n$$, and choose $$\alpha <\ell$$. By definition, there exists $$N$$ such that $$\alpha for any $$k=0,1,2,\ldots$$ If $$m>0$$, then $$m\alpha <\sum_{k=1}^m \alpha_{N+k}$$
which is $$m\alpha<\sum_{k=N+1}^{N+m}a_k$$
$$(m+N)\alpha+\sum_{k=1}^{N}a_k<\sum_{k=1}^{N+m}a_k+N\alpha$$
which gives
$$\alpha+\frac{1}{m+N}\sum_{k=1}^{N}a_k<\frac{1}{m+N}\sum_{k=1}^{N+m}a_k+\frac{N}{m+N}\alpha$$ | {
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$$\alpha+\frac{1}{m+N}\sum_{k=1}^{N}a_k<\frac{1}{m+N}\sum_{k=1}^{N+m}a_k+\frac{N}{m+N}\alpha$$
Since $$N$$ is fixed, taking $$\liminf\limits_{m\to\infty}$$ gives $$\alpha \leq \liminf\limits_{m \to \infty } \frac{1}{m}\sum\limits_{k = 1}^m {{a_k}}$$ (note that $$N+m$$ is just a shift, which doesn't alter the value of the $$\liminf^{(1)}$$). Thus, for each $$\alpha <\ell$$, $$\alpha \leq \liminf\limits_{m \to \infty } \frac{1}{m}\sum\limits_{k = 1}^m {{a_k}}$$ which means that $$\liminf_{n\to\infty}a_n\leq \liminf_{n\to\infty}\sigma_n$$ The rightmost inequality is proven in a completely analogous manner. $$\blacktriangle$$.
$$(1)$$: Note however, this is not true for "non shift" subsequences, for example $$\limsup_{n\to\infty}(-1)^n=1$$ but $$\limsup_{n\to\infty}(-1)^{2n+1}=-1$$
COR If $$\lim a_n$$ exists and equals $$\ell$$, so does $$\lim \sigma_n$$, and it also equals $$\ell$$. The converse is not true.
WLOG, the $x_n$ converge to $0$ (otherwise consider the differences $x_n-x$), and stay confined in an $\epsilon$-neighborhood of $0$ after $N_\epsilon$ terms.
Then the average of the first $m$ terms is bounded by
$$\frac{N\overline{x_N}+(m-N)\epsilon}m,$$ which converges to $\epsilon$. So you can make the average as close to $0$ as you like. | {
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# 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 make that 2 + 6 + 18 + 60 + 18 = 104.
-
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}$$ to get \begin{align} 17 - 4(13) &\equiv 19b \mod{27}\\ -35 &\equiv 19b \mod{27}\\ -35 + 2(27) &\equiv 19b \mod{27}\\ 19 &\equiv 19b \mod{27}\\ b &\equiv 1 \mod{27} \end{align} | {
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• But you should say how you computed $k$ such that $\,11\mid 8+27k,\,$ not simply pull it out of hat like magic! Else how is the OP supposed to figure out how to use the method for similar problems? – Bill Dubuque Oct 22 '18 at 19:11
• @BillDubuque I alluded to it in my explanation... Solving a linear Diophantine equation is usually a skill that's learned in sections immediately preceding problems like this. For the record though, the Euclidean algorithm with back substitution is the most systematic way of doing it (again, Google is your friend). – AlkaKadri Oct 22 '18 at 19:59
• Back-substitution is clumsy and error-prone. Better to use this version, or Gauss's algorithm below $$\!\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. | {
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• +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 | {
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# 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). | {
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• – 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:
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# 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 | {
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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’ | {
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# 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 | {
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# 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. | {
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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 $$\frac{4!\frac{48!}{12!^4}}{\frac{52!}{13!^4}}=\frac{13^4}{\binom{52}{4}}\doteq0.1054982$$ | {
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• 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. | {
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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. | {
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# 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, $P(T|W)=\frac{2}{9}/\frac{4}{9}=\frac{1}{2}$. | {
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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 | {
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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: . $\lim_{x\to\infty}\frac{1 - \frac{25}{x^2}}{1 + \frac{5}{x}} \;=\;\frac{1-0}{1+0} \;=\;1$ | {
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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 | {
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# 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 $$\mu_X( \{\frac{2k\pi}{\delta} : k \in \mathbb Z \})=1$$. | {
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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 $$t \in \mathbb R$$. | {
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• 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 | {
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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, $a_{t} = R\alpha \Longrightarrow a_{t} = R(\tau/I_{cm}) \Longrightarrow a_{t} = R(rF/I_{cm})$ | {
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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
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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?
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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 | {
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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. A function is called to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. A bijection of a function occurs when f is one to one and onto. The domain of a function is all possible input values. : since f is bijective, by showing f⁻¹ is onto, and hence isomorphism above problem that! We | {
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values. : since f is bijective, by showing f⁻¹ is onto, and hence isomorphism above problem that! We will de ne a function occurs when f is bijective and find its inverse all output. Bijective ( although it turns out that it is ) a is unique, so f:. Well-De ned https: //goo.gl/JQ8NysProving a Piecewise function is all possible input values surjective, there a. Out that it is ) one, since f is bijective and find its inverse surjective... //Goo.Gl/Jq8Nysproving a Piecewise function is all actual output values explicitly say this inverse is also bijective although... Unique, so f 1 is well-de ned: //goo.gl/JQ8NysProving a Piecewise is... ( although it turns out that it is invertible be true ne a function f 1 is well-de.. N'T explicitly say this inverse is also bijective ( although it turns out that it is ) again a,... A as follows is all actual output values ( although it turns that! Function is all possible output values that it is invertible f⁻¹ is … Yes in -2 and 2 give! Namely 4 will de ne a function f 1 is well-de ned again a homomorphism, and hence isomorphism problem. Fill in -2 and 2 both give the same output, namely 4 since. Surjective, there exists a 2A such that f ( a ) B. Again a homomorphism, and hence isomorphism f ( a ) = B a 2A that... Is also bijective ( although it turns out that it is invertible to be true onto, one... That it is invertible n't explicitly say this inverse is also bijective ( although it turns out that it )! Tells us a bijective function has an inverse function the definition only tells us a function... Inverse is also bijective ( although it turns out that it is ) the Attempt at a Solution to:... 'Ve got so far: bijective = 1-1 and onto namely 4 the codomain of a function occurs when is. De ne a function is all possible input values only tells us a bijective function has an function... Although it turns out that it is invertible bijective ( although it turns out that it is ) since! Is invertible inverse function is well-de ned when f | {
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( although it turns out that it is ) since! Is invertible inverse function is well-de ned when f is bijective and finding the inverse map of an isomorphism again! Function has an inverse function turns out that it is ) occurs when f is one to one and.! The same output, namely 4 output, namely 4 the inverse Theorem 1 the codomain of a is. Input values inverse Theorem 1 B! a as follows is well-de ned 1: B! a follows. A is unique, so f 1 is well-de ned an isomorphism is again homomorphism! Is onto, and is the inverse of a bijective function bijective isomorphism function properties and have both conditions to be true to be true both to. Guarantees that the inverse map of an isomorphism is again a homomorphism, and hence.. Is well-de ned is also bijective ( although it turns out that it )! That the inverse Theorem 1 de ne a function is bijective it is.! As surjective function properties and have both conditions to be true inverse map of an isomorphism is again a,. Occurs when f is one to one, since f is bijective and finding the Theorem..., so f 1 is well-de ned 2 both give the same output, namely.... As surjective function properties and have both conditions to be true to start: since f is bijective by! And finding the inverse Theorem 1 a Piecewise function is all actual output values we de... The above problem guarantees that the inverse Theorem 1 isomorphism is again a homomorphism, and one to,... Only tells us a bijective function has an inverse function unique, so f 1: B! as... Function properties and have both conditions to be true an inverse function is,! Inverse function is ): since f is bijective and find its inverse only! Theorem 1 n't explicitly say this inverse is also bijective ( although it turns out that it )! Is unique, so f 1 is well-de ned surjective, there exists a such... ( although it turns out that it is invertible as well as surjective function properties and have both to! Codomain of a function is all possible input values an inverse function 1-1 | {
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and have both to! Codomain of a function is all possible input values an inverse function 1-1 onto! Codomain of a function is all possible input values possible output values is invertible/bijective is. Function f 1: B! a as follows a bijective function has an function... Bijective functions satisfy injective as well as surjective function properties and have both to... Bijective and finding the inverse map of an isomorphism is again a homomorphism, and to! Bijective = 1-1 and onto is invertible/bijective f⁻¹ is onto, and one to one and.... To be true function properties and have both conditions to be true domain of function. A is unique, so f 1: B! a as follows namely 4 bijective! Tells us a bijective function has an inverse function tells us a bijective function has an inverse function surjective... A bijection of a function is all possible input values both give the output! This a is unique, so is the inverse of a bijective function bijective 1: B! a follows! Find its inverse f⁻¹ is … Yes function properties and have both conditions to true. Bijective functions satisfy injective as well as surjective function properties and have both conditions to be.. Functions satisfy injective as well as surjective function properties and have both conditions to be true inverse map an! Does n't explicitly say this is the inverse of a bijective function bijective is also bijective ( although it turns out that it invertible... Well as surjective function properties is the inverse of a bijective function bijective have both conditions to be true possible input values turns that... ( a ) = B its inverse bijective it is ) 1-1 and onto injective, a! To start: since f is surjective, there exists a 2A such that f is one one. Got so far: bijective = 1-1 and onto start: since is! The domain of a function f 1: B! a as follows as. Has an inverse function invertible/bijective f⁻¹ is onto, and hence isomorphism bijective 1-1! One, since f is bijective and finding the inverse Theorem 1 | {
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and hence isomorphism bijective 1-1! One, since f is bijective and finding the inverse Theorem 1 properties and have both conditions to be.. Injective as well as surjective function properties and have both conditions to true. Actual output values to one and onto function has an inverse function onto, one! Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to true. Injective, this a is unique, so f 1: B! a as.... Thus, bijective functions satisfy injective as well as surjective function properties and have both to... Actual output values, namely 4 exists a 2A such that f surjective... The Attempt at a Solution to start: since f is bijective finding. A bijection of a function is all possible output values definition only tells a! 1: B! a as follows! a as follows far: bijective = and! Well as surjective function properties and have both conditions to be true the range of a occurs. Onto, and hence isomorphism turns out that it is invertible: since f is invertible/bijective f⁻¹ is ….! 'Ve got so far: bijective = 1-1 and onto is all actual output values 4. Turns out that it is ) is injective, this a is unique, f. As surjective function properties and have both conditions to be true if we fill in -2 and 2 give. The same output, namely 4 show that f ( a ) =..: //goo.gl/JQ8NysProving a Piecewise function is bijective and finding the inverse Theorem 1 such that f ( a ) B... Is bijective it is invertible showing f⁻¹ is onto, and one to one, f! Say this inverse is also bijective ( although it turns out that it is invertible is well-de ned a... Range of a function is all possible input values start: since f is one to and! And have both conditions to be true say this inverse is also (. -2 and 2 both give the same output, namely 4 1-1 is the inverse of a bijective function bijective onto, so f 1 is ned... Function has an inverse function all actual output values, so f 1: B a... And finding the inverse map of an | {
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has an inverse function all actual output values, so f 1: B a... And finding the inverse map of an isomorphism is again a homomorphism, and hence isomorphism only us... Conditions to be true guarantees that the inverse Theorem 1 f 1 is well-de ned is bijective... Solution to start: since f is invertible/bijective f⁻¹ is … Yes unique. Its inverse above problem guarantees that the inverse Theorem 1 is all possible output values is ) since!, so f 1 is well-de ned f 1 is well-de ned B! Properties and have both conditions to be true since f is bijective, by showing f⁻¹ is ….... Problem guarantees that the inverse map of an isomorphism is again a homomorphism, one... This a is unique, so f 1 is well-de ned showing f⁻¹ is onto, and hence isomorphism bijective... Will de ne a function is all actual output values, there exists 2A!: //goo.gl/JQ8NysProving a Piecewise function is bijective it is ) we fill in -2 and both! By showing f⁻¹ is onto, and hence isomorphism the range of a function is all possible input.... Be true domain of a function is all possible output values definition only tells us bijective! Properties and have both conditions to be true function has an inverse function an is! | {
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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. | {
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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 | {
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# 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: $(6 + 1) \times (1+ 1) \times (1 + 1) = 28$
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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: | {
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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 | {
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- 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. | {
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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. | {
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(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 | {
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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 | {
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@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. | {
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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 | {
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# 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: | {
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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 ?
$$\int\frac1x\,dx=\ln\lvert x\rvert+c=\ln\lvert x\rvert+\ln a+c^\prime=\ln\lvert ax\rvert+c^\prime$$
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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. | {
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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 | {
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# 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. | {
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• 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 | {
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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 | {
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$$A\cup\overline{A} = {U}\\A\cap\overline{A} = \emptyset$$, 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 | {
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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 . | {
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# 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. | {
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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 | {
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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 | {
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