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With the notion of topology, if you know the object is analytic continuous then there should exist a deformation to the linear representation which depends on the inverse within a given branch-volume that corresponds to where the derivatives are (which correspond to the ideas in the inverse function theorem which are calculated by finding zero-jacobians). With this information you can find the branches and thus deform that region to it's linear counterpart.
The reason I like to think of something in the deformed linear space is because linear spaces, decompositions, and parameterizations of linear systems are well understood both algebraically, algorithmically and also to an extent visually. It's very easy to parameterize something that is deformed to linear space than to try and analyze it in its non-deformed non-linear space with a function like say f(x,y,z,w,t,u,v,a) = blah for any analytic continuous blah.
You could also probably consider continuous representations that are not analytic over the whole domain and use the same argument based on topological reasoning but I will only speculate on this (intuitively it makes sense at least).
15. Apr 4, 2012
### Bacle2
I think the best we can say is that if f is C^1 (maybe Lipschitz) , then, with a bounded derivative, we can preserve the Hausdorff dimension.
16. Apr 4, 2012
### chiro
I'll go with that for finding the dimension (which is the OP's question), but in terms of understanding why (and where) dimension goes from say one thing to another it may be useful to see which 'parts' if they are isolated contribute to where the function requires more dimensions as opposed to less (it can happen but not always).
The only reason I am saying this (for the OP) is that visually you can see where dimensionality changes. I'm imagining some kind of implicit function that creates branches (like a bifurcation) and does this repeatedly which creates more dimensions at different parts. | {
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Just out of curiosity, what would you say topologically is a way to do the above? In other words in the case of say a bifurcation like object, how would you isolate this kind of phenomenon in terms of the local characteristics (spatially) of the object?
17. Apr 4, 2012
### Bacle2
The way I see it, if the derivative is bounded, then the image of the balls in the covering will not expand by much.
Sorry, it is 4 a.m., here, I'm out for the night; will try to be back in case there are more posts.
18. Apr 4, 2012
### HallsofIvy
Staff Emeritus
Then any two-dimensional ball about a point on the curve will necessarily include some points not on the curve- a curve has no (2 dimensional) interior.
19. Apr 4, 2012
### Bacle2
Halls of Ivy:
I did a specific construction, with a specific definition of curve; what is it about my construction that you think is faulty?
It seems strange-enough that the hypercube ( as a compact metric space; a space with non-empty interior) is the continuous image of the Cantor set. Then what is wrong with the extension?
20. Apr 5, 2012
### TrickyDicky
I think the OP is simply reflecting on the fact that there is no such thing as intrinsic curvature in one dimension. | {
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## cylinder with cone on top
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# If s is the sum of odd integers from 40 to 60, inclusive, and t is the
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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12 Oct 2017, 12:33
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?
A. 480
B. 490
C. 980
D. 990
E. 995
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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12 Oct 2017, 12:39
If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?
A. 480
B. 490
C. 980
D. 990
E. 995
Similar question: https://gmatclub.com/forum/if-x-is-equa ... 97544.html
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
### Show Tags
12 Oct 2017, 12:48
There are total 10 odd numbers between 40 and 60.
hence, t=10
Now, the given sequence is 41,43,45,......,59
common difference is 2.
hence,
S= (41+59)*(10/2)
S=500.
S-t= 500-10= 490. | {
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Kudos if it helps.
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Posts: 22
Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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12 Oct 2017, 13:31
1
Are these formulas correct? I am having trouble with odd numbers for some reason.
N for odd numbers = (Last odd - First odd)/2)) +1
Sum of odd numbers = (First odd + Last odd)*N))/2
1. N = (59-41)/2))+1 = 10
2. Sum of odds = (41+59)*10))/2 = 500
3. 500 - 10 = 490 B
Also, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
### Show Tags
12 Oct 2017, 13:36
1
If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?
A. 480
B. 490
C. 980
D. 990
E. 995
The way I solve sums of arithmetic series, I need the number of terms no matter what. Start there.
Set t
Number of terms: number of odd integers from 40 to 60. First term is 41, last is 59.
Number of terms
$$\frac{LastTerm-FirstTerm}{interval} + 1$$
The "interval" for consecutive even and odds is always 2
$$\frac{59-41}{2} = \frac{18}{2} = (9 + 1)=$$ 10
Set s
Sum of arithmetic series
(Average)(Number of terms) =
$$\frac{FirstTerm+LastTerm}{2}*(10)$$ =
$$\frac{100}{2}(10) = (50)(10) = 500$$
s - t: 500 - 10 = 490
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
### Show Tags
12 Oct 2017, 14:20
1
Are these formulas correct? I am having trouble with odd numbers for some reason.
N for odd numbers = (Last odd - First odd)/2)) +1
Sum of odd numbers = (First odd + Last odd)*N))/2
1. N = (59-41)/2))+1 = 10
2. Sum of odds = (41+59)*10))/2 = 500
3. 500 - 10 = 490 B | {
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1. N = (59-41)/2))+1 = 10
2. Sum of odds = (41+59)*10))/2 = 500
3. 500 - 10 = 490 B
Also, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?
I think the reason you might be having trouble with odd numbers is that for consecutive odd integers, the interval (between the numbers), is two. Two is even; that could confuse. Or you might be forgetting to start with the first ODD term (e.g. here, first term is 41, not 40).
Bottom line: The "interval" between 1 and 3 is two. The interval between 2 and 4 is two. For consecutive odds and evens, in the formula, you divide by 2.
Take smaller numbers. Say, sum of odd integers from 0 to 6. So 1 + 3 + 5 = 9. There are three terms (1, 5, 9) whose sum is 9.
Number of terms in "odd integers from 0 to 6":
$$\frac{(Last - First)}{2}+ 1$$
$$\frac{(5-1)}{2} + 1 = (\frac{4}{2}+ 1) = (2 + 1) = 3$$ Correct
SUM: (Average)(# of terms)
$$\frac{(First + Last)}{2}(3)$$
$$\frac{(1 + 5)}{2} (3) = (3)(3) = 9$$
Bingo. Dividing by 2 for consecutive odd integers works.
When subtracting, you add one because subtraction doesn't include the first number. If in doubt: use small numbers that replicate your situation.
For example, # of terms from 1 to 4? The numerals are 1, 2, 3, 4. There are 4 terms. BUT (4 - 1) = 3.
We need one more. 3 + 1 = 4.
There's a mnemonic: "add one before you're done." It almost always applies. If in doubt, replicate your situation with small numbers. Rare in sequence sums but possible: If you see the word "exclusive" or the phrase "exclusive of," that is when you really need to check.
I have seen your other answers using this method. The ones I've seen are correct! (The one about -190 to 195? Correct, but long. -190 to +190 = 0; you are left with only 5 numbers to sum. STILL - correct.)
Here is a fantastically written, comprehensive post on all kinds of sequences by benjiboo , scroll down for arithmetic sequence | {
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And here is a thread on sequences. VeritasPrepKarishma 's posts are great
Hope this barrage of information helps.
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]
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13 Oct 2017, 11:23
N for odd numbers = (Last odd - First odd)/2)) +1
Sum of odd numbers = (First odd + Last odd)*N))/2
1. N = (59-41)/2))+1 = 10
2. Sum of odds = (41+59)*10))/2 = 500
3. 500 - 10 = 490 B
Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the &nbs [#permalink] 13 Oct 2017, 11:23
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# Find the value of $(a - b)^2$ given $a + b = 2$ and $a^2 + b^2 = 6$
$$a + b = 2\\ a^2 + b^2 = 6$$
Find the value of $(a-b)^2$
My workings till I got stuck -
$$(a-b)^2 = a^2 - 2ab + b^2 \\ = a^2 + b^2 - 2ab\\ = 6 - 2 ab$$
I'm stuck at how to find $ab$ . Can I get hints on how to find $ab$? Thanks a lot.
• $4= (a+b)^2=a^2+b^2+2ab$ – lulu Feb 26 '16 at 15:16
• Squaring the first equation then subtract the second equation from it. – Zhanxiong Feb 26 '16 at 15:17
• The above comments mean that you should consider the quantity $2a^2+2b^2-(a+b)^2$... – abiessu Feb 26 '16 at 15:19
Use $$(a-b)^2+(a+b)^2=2(a^2+b^2)$$ Thus, here, $$(a-b)^2+4=12$$ $$(a-b)^2=8$$
We will solve for the values of $a$ and $b$. $a=2-b$ so $a^2+b^2=6 \iff (2-b)^2 + b^2 +6$
This means $2b^2-4b-2=0 \iff b^2 -2b -1 +0$. The solutions of this equation are $b=1-\sqrt{2}$ and $b=1+\sqrt{2}$. Then $a=1+\sqrt{2}$ or $a=1-\sqrt{2}$
Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$) | {
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# Probability that a random 13-card hand contains at least 3 cards of every suit?
A random 13-card hand is dealt from a standard deck of cards. What is the probability that the hand contains at least 3 cards of every suit? (Introduction to Probability, p.36)
My solution:
• There are $\binom{52}{13}$ possible hands.
• Because there are 13 cards for the hand, to obtain at least three cards of one suit per hand, we need to have exactly three cards of one suit per hand plus one additional card of any suit, thus $\binom{13}{3}^4 * 4 \binom{10}{1}$
• Result: $\frac{40*\binom{13}{3}^4}{\binom{52}{13}} = 0.4214$
However, simulating it in R yields:
deck <- rep(1:4, 13)
out <- replicate(1e5,{
hand <- sample(deck, size=13, replace=FALSE)
all(table(hand) >= 3)
})
mean(out)
> 0.14387
Can anybody tell me what is wrong?
EDIT
I'm afraid, the correct code should be.
deck <- rep(1:4, 13)
out <- replicate(1e5,{
hand <- sample(deck, size=13, replace=FALSE)
length(table(hand))==4 & all(table(hand) >= 3 )
})
mean(out)
> 0.10639 | {
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• The correct result is about $0.105$, yet the R simulation yields $0.14387$. This is the second R simulation you've posted today that gives a result significantly different from the theoretical result. There must be something wrong with the R code (up to and including the possibility that R's randomization is insufficiently random for your purpose) but I still don't know what the error is. You might want expert advice on this so that you aren't misled by an incorrect simulation in the future. Dec 7, 2014 at 17:17
• @DavidK R is completely fine, the idiot is me. As you mentioned, I posted two times simulations along with my questions and both of them were wrong, because I didn't give enought thought to the special cases. I'm sorry for the confusion I caused. Motivation for the code was that I didn't just want to ask for whether correct or not, but give some justification for my doubt and demonstrate previous work on the problem. Dec 7, 2014 at 17:32
• Good for you for finding that correction to the R code. The error was sufficiently subtle for me to miss it entirely, even knowing that something must be wrong. Dec 7, 2014 at 17:47
Dominik, your answer was off by a factor of 4. This happened because you counted a hand containing J,K,Q,A of spades (for example) 4 times: (JQK)(A), (QKA)(J), (KAJ)(Q), and (JAQ)(K)
• The answer above is a correct description of the basic issue that led to the overcount. Dec 7, 2014 at 15:40
We count the "favourables," the 4-3-3-3 hands. The suit in which we have $4$ cards can be chosen in $\binom{4}{1}$ ways. For each of these ways, the actual $4$ cards in that suit can be chosen in $\binom{13}{4}$ ways. For each of these ways, the cards in the other three suits can be chosen in $\binom{13}{3}^3$ ways, for a total of $\binom{4}{1}\binom{13}{4}\binom{13}{3}^3$. | {
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Remark: Your counting procedure results in multiple counting. Think, for example, of the 4-3-3-3 hand that has the K, Q, J, 10 of hearts, and some specific cards for the rest of the hand. Your calculation views, for example, K, Q, J of hearts, and later 10 of hearts, as different from K, J, 10 of hearts, and then Q of hearts.
• As @judith pointed out above, it seems that $\binom{4}{1} \binom{13}{4} \binom{13}{3}^3 = \binom{13}{3}^4 \binom{10}{1}$. However, I still don't understand what I am overcounting. What do you mean by "the 4-3-3-3 hand that has the K, Q, J, 10 of hearts"? Could you rephrase/eleborate on your remark please? Dec 7, 2014 at 15:18
• You counted the $12$-card 3-3-3-3 hands and then added a card in one of the suits. Each 4-3-3-3 hand comes up in $4$ ways, so you are overcounting by a factor of $4$. Here is a simpler example. How many $3$ card hands have $2$ hearts and $1$ diamond? The obvious answer is $\binom{13}{2}\binom{13}{1}$. (Cont) Dec 7, 2014 at 15:33
• (Cont) Here is a wrong way of counting. Pick a heart and a diamond, $\binom{13}{1}^2$ ways. Then add a heart, The wrong way counts twice the hand that has K, Q of hearts, and 5 of diamonds, once as K of hearts, 5 of diamonds, then Q of hearts, and also as Q of hearts, 5 of diamonds, and then K of hearts. Dec 7, 2014 at 15:35
• Thank you for elaborating on it. I think I understand my error now. Dec 7, 2014 at 16:18
• You are welcome. This sort of inadvertent multiple counting happens fairly often, until one gets sensitized to it. Dec 7, 2014 at 16:21
I doubt, that it is still interesting for you, but I want to check myself.I think either you or I misunderstand the question. Because it asks to find the probability of the hand contains AT LEAST 3 cards of every suit and you calculated the probability of EXACTLY 3 cards of every suit. From my point of view answer must be:
1. for 0 cards of every suit there is no point to calculate, since nothing happened | {
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1. for 0 cards of every suit there is no point to calculate, since nothing happened
2. for 1 cards of every suit (choose from 13-1)^3(choose from 13-remainig 10)(choose from 4-1) and divide all this to (choose from 52-13)
3. for 2 cards of every suit (choose from 13-2)^3(choose from 13-remainig 7)(choose from 4-1) and divide all this to (choose from 52-13)
Then just substract from 1 the calculated probability and you will get P(>=3)
• OP is definitely not calculating exactly three cards of every suit. Because there are 13 cards in a hand, the only distribution(s) that have at least three cards in every suit have three suits with three cards in them and one suit with four; this is what OP attempts to calculate (with the overcounting mentioned in other answers). The hand size is fixed, so the notion of '1 card of every suit' doesn't make sense; it's an impossible configuration. Jul 9, 2021 at 3:11
Another solution that's more mechanical and generalizable to other arrangements like two pair or full house:
We know we must have a 4-3-3-3 hand. You can pick any card for the first suit: 52. The next three cards must match the suit of the first so there are 12 x 11 x 10 possibilities for those. Now for the 4th card, we can pick any card not in the 1st suit: 39. Using the same reasoning as before, for the next two cards there are 12 x 11 possibilities. We continue until we have picked all cards:
$$X = (52 \times 12 \times 11 \times 10) \times (39 \times 12 \times 11) \times (26 \times 12 \times 11) \times (13 \times 12 \times 11)$$
$$X$$ gives us all arrangements of the pattern WWWW XXX YYY ZZZ, as such we overcount all the internal arrangements of W, X, Y, and Z ($$4!3!^3$$). We also overcount that X, Y, and Z can be interchanged ($$3!$$).
So the answer is $$X /(4!3!^4) / {52 \choose 13}$$. | {
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# What is the integral of |x| ?
1. Aug 10, 2011
### agentredlum
The question is in the title.
This is not a homework question, just curiousity on my part.
2. Aug 10, 2011
### HallsofIvy
Staff Emeritus
If x< 0, then |x|= -x and its integral is $-x^2/2+ C$
If $x\ge 0$, then |x| x and its integral is $x^2/2+ C$
Last edited: Aug 10, 2011
3. Aug 10, 2011
### disregardthat
or |x|x/2 + C
4. Aug 10, 2011
### ReaverKS
How? I thought the whole purpose of the absolute value stemmed back to distances along a number line, therefore you couldn't get a negative value out of the absolute value of a number or a function?
5. Aug 10, 2011
### Citan Uzuki
If x<0, then -x>0.
6. Aug 10, 2011
### BloodyFrozen
7. Aug 10, 2011
### Bohrok
$$\int|x| dx = \frac{1}{2}x|x| + c$$If you write |x| as √(x2) and use integration by parts, you can find the integral.
8. Aug 10, 2011
### agentredlum
I know HallsofIvy figured it out because calculations were shown, but did the rest of you look it up?
HallsofIvy, can you combine your 2 solutions into a single solution to get wiki answer?
I know wiki mentions integration by parts, but did anyone use a simpler way to get the answer? It doesn't have to be rigorous, it just has to work. Don't worry about rigor, just functionality.
What is the integral of |x^3| ?
What is the integral of |x^n| for odd n?
EDIT* There is nothing wrong (IMHO) with looking up the answer, but there is nothing wrong with working it out for yourself.
9. Aug 10, 2011
### Bohrok
I never knew wikipedia showed how to integrate it; just noticed it right now. I personally did it once because I wanted to derive it myself, and for higher odd powers of x such as |x3| and there is a pattern.$$\int|x^3|dx = \frac{1}{4}x^3|x| + c, \int|x^5|dx = \frac{1}{6}x^5|x| + c, \text{etc.}$$So in general,$$\int|x^n|dx = \frac{1}{n + 1}x^n|x| + c \text{ for odd n}$$
10. Aug 10, 2011
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10. Aug 10, 2011
### agentredlum
Oh, this is interesting. Is there any difference to the answer if you represent it as |x^n|x/(n+1) + c for odd n ?
Well done in deriving it for yourself.
11. Aug 10, 2011
### Mute
When you have an absolute value, you just split it into cases:
If $x > 0,~|x^n| = x^n$, so the integral is $x^{n+1}/(n+1)$.
If $x < 0,~|x^n| = - x^n$, so the integral is $-x^{n+1}/(n+1)$.
(for n odd, of course)
This can be written as a single results using the "sign function", $\mbox{sgn}(x)$, which returns the sign of x. Hence, the integral may be written
$$\mbox{sgn}(x) \frac{x^{n+1}}{n+1}.$$
For x not zero, the sign function is equivalent to |x|/x or x/|x| (as is hopefully obvious after thinking about it for a second), so you can insert the first form there to get the results people have been throwing around.
Note that since the sign function to an even power is always 1, you can always insert sign functions of even powers to shift around the exponents, so you can write the integral as
$$\frac{|x|^ax^b}{n+1},$$
where a + b = n+1, and a and b are odd numbers.
12. Aug 10, 2011
### Bohrok
No difference; it's the same way as how I have it. I should also add positive to the restriction on n: for positive odd n.
13. Aug 10, 2011
### agentredlum
Of course, right you are. I haven't explored what happens when n is negative. It is interesting to me that the integrals we are considering, |x^n|, for positive odd n, to get the answer you just write it down, multiply by x, divide by the total number of factors of x and add a 'c'. Integration by parts is not necessary. It makes me wonder if there are other functions out there that work simply like this, besides the normal x^n without absolute value symbol.
What is the integral of |x^3 + x| ?
Last edited: Aug 11, 2011
14. Aug 11, 2011
### Bohrok | {
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What is the integral of |x^3 + x| ?
Last edited: Aug 11, 2011
14. Aug 11, 2011
### Bohrok
I thought this would be difficult and involve integration by parts, but it turned out quite simple actually:
|x3 + x| = |x(x2 + 1)| = |x||x2 + 1| = |x|(x2 + 1) = |x|x2 + |x| = |x3| + |x|, using the facts that x2 + 1 = |x2 + 1| and x2 = |x2| since the former in each case is never negative, and |ab| = |a||b|, so
∫|x3 + x| dx = ∫|x3| dx + ∫|x| dx
Once you integrate those, you can combine them and make the result look more condensed.
If you had |x3 - x|, that would be a different story!
Last edited: Aug 11, 2011
15. Aug 11, 2011
### agentredlum
It's very nice how you turned this problem into something we have a shortcut for already!
So the integral of |x^3 + x| is |x^3|x/4 + |x|x/2 + C
I like your algebra manipulation of absolute value so let me use it too. Disregard the C for now...
(|x^3| + 2|x|)x/4 common denominator 4 and factor out x...now work inside the parenthesis...
|x|(|x^2| + 2)x/4 for the same reasons you gave above...
|x^2| + 2 = |x^2 + 2|
|x||x^2 + 2| = |x^3 + 2x|
|x^3 + 2x|x/4 again for same reasoning as you above... now 1/4 = 1/|4| so you can insert 1/4 into the parenthesis without fear, so you get finally...
|x^3/4 + x/2|x + C
Now if you compare this answer to the question...what is integral of |x^3 + x|, you write it down, divide x^3 by 4 inside absolute value (n + 1 for n = 3), divide x by 2 inside absolute value (n + 1 for n = 1), multiply the whole thing by x and add C
Putting it all together...
integral of |x^3 + x| = |x^3/4 + x/2|x + C
I think the shortcut is obvious now in this case and should work for any integral of |ax^3 + bx| where both a,b are non negative.
I'm still working on integral of |x^3 - x| and trying to make it co-operate with the shortcut, any ideas?
16. Aug 11, 2011
### disregardthat
Split the integral as such:
$$\int^t_{1} |x||x^2-1| dx = \int^t_{1} |x|(x^2-1) dx$$ for t >= 1, | {
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Split the integral as such:
$$\int^t_{1} |x||x^2-1| dx = \int^t_{1} |x|(x^2-1) dx$$ for t >= 1,
$$\int^t_{-1} |x||x^2-1| dx = \int^t_{-1} |x|(x^2-1) dx$$ for t <= -1
$$\int^t_{0} |x||x^2-1| dx = \int^t_{0} |x|(1-x^2) dx$$ for -1 < t < 1
These three integrals are solved by the method above. For each domain, scale by a constant to get an answer on the form F(x) + C, but it is not necessary.
17. Aug 11, 2011
### agentredlum
Thank you. This is all very nice and important but right now i'm working on my own calculations using pencil and notebook. My PS3 browser does not decode LaTeX so its very hard sometimes to decode it using my brain. Can you solve all 3 integrals above and combine them into a single expression that gives the correct area of |x^3 - x| between the curve and the x-axis from x= -5 to x= 3 for example?
Let me give an simple example. The integral of x from x= -5 to x= 5 gives zero. I'm interested in the integral as AREA between 2 curves.
When you take the absolute value of a function it has 2 important graphical effects for my interests concerning area.
1) Portions below the x-axis are reflected by symmetry to portions above the x-axis with area unchanged.
2) Portions above the x-axis remain unchanged
This can be helpful when you have a function with portions below the x-axis. If one is interested in AREA then you don't have to find roots, split up the integral, use symmetry, etc. Just integrate it's absolute value. If you can find the integral then a single expression will give you the right answer for the area.
Now, I realise integrating |f(x)| is harder than integrating f(x) unless shortcuts can be found and we have shown in previous post's that at least some shortcuts exist. The question interesting to me is can the shortcuts be extended to include more general cases. Any ideas? | {
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This can be helpful, for example, if you are writing a computer program designed to calculate areas. If the program recognises a particular expression as lending itself to the shortcut, then it doesn't have to do integration by parts or some other more exotic method.
Or consider it as a contest between bohrok, who knows the shortcut, and someone who doesn't know the shortcut. The question is find integral of |x^n| for positive odd n. Bohrok gets the answer in 1523 milliseconds, the other person not nearly as fast
Last edited: Aug 11, 2011
18. Aug 11, 2011
### Bohrok
19. Aug 11, 2011
### agentredlum
OH...MY...GOD you have no idea how much i want to thank you for that link. For months i've been hearing about wolfram alpha but i thought it was a computer program like Maple or Mathematica. I didn't know it was on the net, and my PS3 browser decodes it!!!!!
If i was in your neighborhood right now, i'd buy you a beer, i'd buy you 2 beers!.
As far as integral of |x^3 - x| my shortcuts are not working unless i made a mistake.
As for integral of |x^3| wolfram gave answer involving sgn(x) but you and i both know we don't need that. Haaaaaaaa.... Haaaa!
Humanity strikes back against the machines.
20. Aug 12, 2011
### agentredlum
According to wolframalpha...sqrt(abs(x)) has the same graph as abs(sqrt(x)) but that can't be right since domains are different
The first one has domain all real x
The second one has domain all rea NON-NEGATIVE x
What is going on here?
http://www.wolframalpha.com/input/?...,abs(x^(1/2))&asynchronous=false&equal=Submit
In the input interpretation it makes the correct interpretation but in the plots it doesn't.
Last edited: Aug 12, 2011 | {
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We denote the event that the rst n persons are born under di erent signs, exactly as in example 5 page 62. A ball is drawn at random from the box and not replaced. 18 The probability r of no success at a certain day is equal to a red ball is drawn on. So the desired probability is just 15/48 = 5/16 = 0. 2) Letting R be the event that all of the balls are red, B be the event that all of the balls are blue, and G. How many balls are left over____? 4. Find the probability of winning $250,000 in the lottery. There are two bins with black and white balls. Probability problem on Balls. Tech interviews, Banks, IAS and SSC exams probability practice questions. Learn to find the probability when there are balls of different colors are there in a bag. Example 31. If balls are instead drawn in succession without replacement, show that the probability of drawing B on any given draw is still always 1=n. What is Bayes' Theorem? Bayes' theorem is a way to figure out conditional probability. 20 A jar contains 3 white balls and 2 red balls. Let X denote the number of red balls. Probability Probability Conditional Probability 19 / 33 Conditional Probability Example Example De ne events B 1 and B 2 to mean that Bucket 1 or 2 was selected and let events R, W, and B indicate if the color of the ball is red, white, or black. Three balls are selected at random without replacement from the jar below. 05 that it has failed as an open circuit and a probability of 0. A ball is taken at random from the rst bin and put into the second bin. The probability of a man hitting the target at a shooting range is 1/4. The basic difference between permutation and combination is of order Permutation is basically called as a arrangement. To win the Ohio Super Lotto Plus lottery, you must correctly pick which six balls are chosen out of a collection of forty-nine numbered balls. Each of the 6 sectors of the prize wheel are equal sizes. (a) (probability that the total after rolling 4 fair dice is 21. What | {
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prize wheel are equal sizes. (a) (probability that the total after rolling 4 fair dice is 21. What is the probability that the three pieces can be assembled into a triangle?. Both are distinguishable. Thursday is the 4th of July here are a few problems relating to Independence Day!! While walking around at their town's Independence Day Festival, Latrease and Hannah decide to buy lunch. The function f(x) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x-axis is equal to 1. Exercise 1. A Short Introduction to Probability Prof. Tossing a Coin. I take out a pen and lay it on the desk; each pen has the same chance of being selected. Goodman, David Famolari August 27, 2014 1. Solutions of Problems on Probability theory P. Question 15. The problems here can help solidify the thought process behind the method of. Introduction to Probability 2nd Edition Problem Solutions (last updated: 9/26/17) c Dimitri P. Suppose a batter has probability 1 3 to hit the ball. Probability Examples A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles What is the probability that you draw and replace marbles 3 times and you get NO red marbles? There are 55 marbles, 25 of which are not red P(getting a color other than red) = P(25/55) ≈. Probability and Statistics Problems - Solutions 1. The submitted PDF should have all the 21 pages in the correct order even if you do not solve all the problems or use all the space provided for a problem. The salesperson tells you that the total price, including. Suppose A 2+ B + C2 = 4 and AB+ BC+ CA= 3. If you know how to manage time then you will surely do great in your exam. Permutation & Combination Problems with Solutions for bank exams-: Today, I am going to share with you to solve "permutation & combination questions". Some of the problems are, as Dudeney admitted, ‘not unworthy of the attention of the advanced mathematician’. For example, | {
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as Dudeney admitted, ‘not unworthy of the attention of the advanced mathematician’. For example, when choosing a ball at random from a bag containing 3 blue balls, 2 green balls, and 5 red balls, the probability of getting a blue (let's call it event B) or red ball (let's call it event R) is. By Deborah J. Crossing the River (July 25, 2013) Life's Big Questions {August 15, 2013) Neat games for two people (August 22. 20, 2104 Problem 1. We randomly pick a ball. 10-16 A box contains 5 R, 4 G, 9 B balls. Probability Case Studies Infected Fish and Predation 2 / 33 Questions There are three conditional probabilities of interest, each the probability of being eaten by a bird given a particular infection level. After you are done, you will be blindfolded and the bowls will be shuffled. “That’s an interesting observation,” Mr. Solutions Manual for Introduction to Statistical Physics (draft) 8- Consider again problem 7, with a distribution w(s) of the is the probability of -nding N. Similar calculations for the other colours yields the probability density function given by the following table. These free probability worksheets introduce students to the basic ideas behind probability. Conditional Probability and Tree Diagrams Example In a previous example, we estimated that the probability that LeBron James will make his next attempted eld goal in a major league game is 0:567. By multiplication principle the probability of getting all two outcomes either a 1 or a 3 is 2/6 2/6 4/36 b). The three faces each have only one color: red, blue, and green. These are interesting and instructive problems that deserve. purple pink purple pink purple. white and k black balls. A spinner is divided into 3 equal sections, with sections labeled 1, 2. Giventhat Qcan be defined in terms of V and vice versa, why is it important to learn rather than V? 8. And a free, video lesson. What is the probability of drawing a gray marble? _____ 4. All students, freshers can download Aptitude Probability | {
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of drawing a gray marble? _____ 4. All students, freshers can download Aptitude Probability quiz questions with answers as PDF files and eBooks. Due: Tuesday, 10/17 Notes: (a) Computer Science Ph. Which BEST describes the probability of rolling an even number? A. (Inventor); Washabaugh, Andrew P. What is the probability of choosing two red balls?. They will analyze real-life word problems to calculate both experimental and theoretical probability. 1 ˙-algebras and information We begin with some notation and terminology. You pay$10 to play the following game of chance. A discrete probability distribution is a table (or a formula) listing all possible values that a discrete variable can take on, together with the associated probabilities. The probability of event tells us how likely it is that the event will occur and is always a value between 0 and 1 (e. Then there are 8 “gaps” between white balls. 2002 Find all educational Solutions Here Search here. You win if the 6 balls you pick match the six balls selected by the machine. Otherwise a ball is selected at random from the silver urn. West with the collaboration of Itshak Borosh, Paul Bracken, Ezra A. The Midterm Examination - SOLUTIONS There are ve problems on this exam. And so the question becomes: If we arrange the six balls in a random order, what is the probability that balls #1, #2 are in the first two positions, balls #3, #4, are in the third and fourth positions etc. Two balls are drawn without replacement. Mark up documents with highlights and handwriting, insert texts and stamps, fill out, sign PDFs and even manipulate PDF pages. East said, turning to his right. 50, or \that team has a 1 in 1000 shot at winning" means that the probability that the team will win is 1 1000 = :001). QUESTION: Describe the sample space and all 16 events for a trial in which two coins are thrown and each shows either a head or a tail. We know two strategies for finding the optimal k-step policy: (1) build a depth-k tree with | {
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a tail. We know two strategies for finding the optimal k-step policy: (1) build a depth-k tree with state s at the root. com Statistics Exercises 6 Problems for Chapter 2: Basic concepts of probability theory seven balls are. Time is the main factor in competitive exams. Blitzstein and Jessica Hwang c Chapman & Hall/CRC Press, 2014 Joseph K. Homework 4: Solutions Sid Banerjee (sbanerjee@cornell. ways, so our probability is 9 2 / 12 2, which turns out to be 6/11. If you know how to manage time then you will surely do great in your exam. Probability trees provide a systematic method of. , probability laws) used for solving probability problems. This experiment can be illustrated as follows: Keeping in mind the number of balls in each box increases by one after the previous selection is deposited, the desired probability is calculated as follows:. Tsitsiklis These class notes are the currently used textbook for Probabilistic Systems Analysis," an introductory probability course at the Massachusetts Institute of Technology. Select one urn at random 2. Stat 110 Strategic Practice 1 Solutions, Fall 2011 Prof. Probability and Mathematical Statistics 1 Chapter 1 PROBABILITY OF EVENTS 1. Which color marble is least likely to be drawn from the bag? _____ 2. They will learn how to describe the probability of an event using numbers from 0 to 1. You are to draw one ball from the bag. Problems 6 and 7: This problem revealed to me that many people don't know the difference between a probability mass function, probability density function, and cumulative distribution function. A second urn contains 16 red balls and an unknown number of blue balls. com Statistics Exercises 6 Problems for Chapter 2: Basic concepts of probability theory seven balls are. Voiculescu, Lectures on Probability Theory and Statistics, Ecole d’Ete e de Probabilit es de Saint-Flour XXVII { 1998, Ed. Problems and Solutions Exercises, Problems, and Solutions Section 1 Exercises, Problems, and Solutions Review | {
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and Solutions Exercises, Problems, and Solutions Section 1 Exercises, Problems, and Solutions Review Exercises 1. You pick a bag at random and then pick one of the balls in that bag at random. Hints help you try the next step on your own. Brown, Randall Dougherty, Tam ´as Erd ´elyi, Zachary Franco, Christian Friesen, Ira M. To solve problems on this page, you should be familiar with Uniform Probability Probability - By Outcomes Probability - Rule of Sum Probability - Rule of Product Probability - By Complement Probability - Independent Events Conditional Probability. Solutions next week!. LaShawn's comic book collection currently has 10 comic books in it, and he is adding to his collection at the rate of 6 comic books per month. Max has 5 coins in his pocket that total 47 cents. There are two bins with black and white balls. If you chooses 1 ball from each box you are more likely to choose a black ball from A than from B. What is the probability that she gets a red ball and a blue ball? Subject: 6th grade (Ontario) Previous problem: Not completed2 problems ago: Not completed3 problems ago: 0 0 0 1 1 1 2 2. If both try for the win independently find the probability that there is a win. red, blue, black. Please ensure that each sheet is labeled with your name, SID number, section number, and "CS174--Spring 2007". Draw a tree diagram for this experiment and find the probability that the two balls are of different colors. It is a collection of problems and solutions of the major mathematical competitions in China, which provides a glimpse on how the China national team is selected and formed. Some problems are easy, some are very hard, but each is interesting in some way. Probability The marbles pictured below are gray, white, and black. We draw 2 balls from the urn without replacement. Probability - examples of problems with solutions for secondary schools and universities. Suppose for example that there are three people: Ann, Bob and Carol. Data Analysis, Statistics, | {
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Suppose for example that there are three people: Ann, Bob and Carol. Data Analysis, Statistics, and Probability questions typically account for 10% to 20% of the SAT Math questions. If no problem is clearly marked as skipped, only the rst four problems will be graded. Two balls are selected from the urn without replacement. Tree diagrams are useful to answer probability questions where there is more than one event happening in succession. If we assume that as we draw the each ball in the urn is equally likely to be drawn, what is the probability that both balls drawn are red? Let R1 and R2 denote respectively the events that the first and the second ball drawn is red. Instructions Use black ink or ball-point pen. The function f(x) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x-axis is equal to 1. Correspond-ingly, Rndenotes the event that the n'th person is the rst person born under the same. So the probability total number of balls >= 16 will be 1/9 * 1/9 = 1/81. , probability laws) used for solving probability problems. School Overview Mowbray Heights Primary School is an. What is the probability of winning the Ohio Lottery if you have one ticket? Answer: The number of possible outcomes (each equally likely) is 49C 6 = 13983816, but only one. 3 or Miscellaneous Exercises in English Medium to study online or in PDF form. Introduction to Probability and Statistics (PDF) Solutions to Exam 1. Find the probability that in a given year it will not snow on January 1st in that town. The first of these items is mandatory, for practice with the above Poisson explanation (you don't know it until you can do it). Prior probability is the probability you attribute to a certain event without further knowledge about it. Use the binomal expansion to calculate the probabilities of HH, HT, and TT. Probability Tricks. This probability is written P(B|A), notation for the probability of B given A. , | {
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Probability Tricks. This probability is written P(B|A), notation for the probability of B given A. , 2000 Outline 5-1 Introduction 5-2 Sample Spaces and Probability 5-3 The Addition Rules for Probability 5-4 The Multiplication Rules and Conditional Probability ©The McGraw-Hill Companies, Inc. In this PDF we have given probability Questions and Answers, we hope this will be helpful for your preparation. Keep in mind that the solutions provided represent one way of answering a question or solving an exercise. Suppose that one face of a regular tetrahedron has three colors: red, green, and blue. Voiculescu, Lectures on Probability Theory and Statistics, Ecole d’Ete e de Probabilit es de Saint-Flour XXVII { 1998, Ed. Now, you are asked to fill the bowls in any manner you like (using all the balls). She also has 4 extra bars. Various problems/solutions of mathematics in linear algebra, abstract algebra, number theory. 11th Annual Johns Hopkins Math Tournament Sunday, April 11, 2010 Grab Bag-Lower Division: Solutions (1) (6) Below is a square, divided by several lines (not to scale). Probability problems for aptitude pdf download, probability problems and solutions for aptitude, probability problems, random variables and probability distributions problems and solutions, probability word problems with solutions and answers, probability distribution problems and solutions, probability problems on balls with solutions, basic. P(passing in second given he passed in the first one) = P(AꓵB)/P(A) = 0. Probability distribution problems solutions pdf Random variables and their probability distributions can save us significant. ) are represented as colored balls in an urn or other container. if p EA implies p E B, then A is called a. (Total 4 marks). Your answers may vary slightly. com provides you Free PDF download of NCERT Exemplar of Class 10 Maths chapter 13 Statistics and Probability solved by expert teachers as per NCERT (CBSE) book guidelines. Of course, you will learn best | {
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solved by expert teachers as per NCERT (CBSE) book guidelines. Of course, you will learn best if you rst attempt to solve the exercises on your own, and only consult this manual when you are really stuck (or to check your solution after you think you have it right). These types of questions test whether employees can have fun together, not solve real world problems together. Ex: The wandering mathematician in previous example is an ergodic Markov chain. How should we change the probabilities of the remaining events? We shall call the. Probability Homework Solutions 1. 1 Combinations, Permutations, and Elementary Probability Roughly speaking, Permutations are ways of grouping things where the order is important. There are two bins with black and white balls. It prescribes a set of mathematical rules for manipulat-ing and calculating probabilities and expectations. Probability 2. Visit us for practice questions and solved examples. problems (2003 - 2006). Data Analysis, Statistics and Probability gum balls Scoring Guide- 2 The gum ball machine has 100 gum balls; 20 are yellow, 30 are blue, and 50 are red. In the preface, Feller wrote about his treatment of uctuation in coin tossing: \The results are so amazing and so at variance with common intuition that even sophisticated colleagues doubted that coins actually misbehave as theory predicts. They are placed in a bag and one is drawn at random. So the answer is Floor 1. probability problems, probability, probability examples, how to solve probability word problems, probability based on area, examples with step by step solutions and answers, How to use permutations and combinations to solve probability problems, How to find the probability of of simple events, multiple independent events, a union of two events. A second urn contains 16 red balls and an unknown number of blue balls. A biased coin (with probability of obtaining a Head equal to p > 0) is tossed repeatedly and independently until the first head is observed. | {
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a Head equal to p > 0) is tossed repeatedly and independently until the first head is observed. Jenny gets 10 gum balls from this machine. Ma 162 Spring 2010 Ma 162 Spring 2010 April 21, 2010 Problem 1. As 5 of the balls are red, and there are 10 balls, the probability that a red ball is drawn from the box is Pr(X = Red) = 5/10 = 1/2. Toggle navigation Close Menu. Probability Example 1. Solution: (1a) Sample space has jSj= 10 2 = 45. Suppose for example that there are three people: Ann, Bob and Carol. Probability and Statistics Problems - Solutions 1. There are 10 red and 20 blue balls in a box. Our extensive online study community is made up of college and high school students, teachers, professors, parents and subject enthusiasts who contribute to our vast collection of study resources: textbook solutions, study guides, practice tests, practice problems, lecture notes, equation sheets and more. Yates, David J. • Answer the questions in the spaces provided – there may be more space than you need. Represent events as a In the same small groups, students are given sample solutions to analyze and. An event that is certain to occur has a probability of 1. Solutions to Problems in H. Aptitude Made Easy - Probability – 7 Tricks to solve problems on Balls and bags – Part 1 7 Tricks to solve problems on Balls and bags – Part 2 - Duration: Probability Word Problems. 1 Solutions will be available on Blackboard on the 29th Oct. Event A has jAj= 6 2 = 6 5=2 = 15. Keep in mind that the solutions provided represent one way of answering a question or solving an exercise. ) An urn contains twenty red balls and five blue balls. (a) (probability that the total after rolling 4 fair dice is 21. [1] By Jerome Dancis [2] 1. Permutation & Combination Problems with Solutions for bank exams-: Today, I am going to share with you to solve "permutation & combination questions". You also had the choice of two metal rackets and one wooden one. And, let X denote the number of people he selects | {
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choice of two metal rackets and one wooden one. And, let X denote the number of people he selects until he finds his first success. A container contains ve red balls. introduction to counting and probability Download introduction to counting and probability or read online here in PDF or EPUB. STAT/MATH 394, Probability I, covers the basic elements of probability theory. HOMEWORKS All homeworks are due Thursday at 5:00pm unless otherwise stated. • Answer all questions. As 5 of the balls are red, and there are 10 balls, the probability that a red ball is drawn from the box is Pr(X = Red) = 5/10 = 1/2. It follows simply from the axioms of conditional probability, but can be used to powerfully reason about a wide range of problems involving belief updates. red, blue, black. You have some trick coins that land heads 60% of the time and tails 40%. Mathematics (Linear) – 1MA0 PROBABILITY Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. 45 which is around 55%. Suppose Z has a continuous distribution with pdf given by A bucket contains 2 green balls and. When the brakes of a car are applied, the road surface works on the car’s tyres to bring it to stop. To find the number of ways to match five numbers and not the Mega Ball number, break the selection of balls into two. A ball is chosen at random and it is noted whether it is red. This new probability is referred to as a conditional probability, because we have some prior information. 882 — PS3: Practice problems: Revised — Fall 2010 2 7. There is one desired outcome and six possible outcomes. As nincreases, the proportion of heads gets closer to 1/2, but the difierence between the number of heads and half the number of °ips tends to increase (although it will occasionally be 0). Solutions Manual for Introduction to Statistical Physics (draft) 8- Consider again problem 7, with a distribution w(s) of the is | {
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to Statistical Physics (draft) 8- Consider again problem 7, with a distribution w(s) of the is the probability of -nding N. What is the probability of getting at least one post? Solution: The probability the candidate does not get an offer from the first interview is 2/3. Balls and Urns “Balls and urns” problems are paradigmatic. (a) Find the probability that the first ball is red. 3 or Miscellaneous Exercises in English Medium to study online or in PDF form. pdf from IOE 202 at University of Michigan. You have some trick coins that land heads 60% of the time and tails 40%. A function f is said to be probability density function pdf of the. As this has 4 elements there are 24 = 16 subsets, namely. Keep in mind that the solutions provided represent one way of answering a question or solving an exercise. 9 CONVERGENCE IN PROBABILITY 117 (There is no difference between home and away games. Yates and David J. P(B1) (first ball selected is black). The notes on the Web tend to be aimed at treating the more complex versions, so I'm working on winnowing out some references for the basic situation. Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate. We can also count these outcomes as follows. Get The Official GMAT Club's App - GMAT TOOLKIT 2. You pick two random shoes from the bag. Probability Puzzles - Interview question on probability, maths puzzles and aptitude. Calculate the number of blue balls in the second urn. Balls in Bins 9. Solution : There are six possible ways in which a die can fall, out of these only one is favourable to the event. 11th Annual Johns Hopkins Math Tournament Sunday, April 11, 2010 Grab Bag-Lower Division: Solutions (1) (6) Below is a square, divided by several lines (not to scale). The answers and explanations are given for the practice questions. The probability that both balls are the same color is 0. Because we are drawing with replacement, the outcome of the second draw does not depend on the | {
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is 0. Because we are drawing with replacement, the outcome of the second draw does not depend on the outcome of the first. Hannah bought a hot dog, a soda, and a bag of chips for $5. 1 in Pitman This is another version of the birthday problem. Probability Class 10 Extra Questions Maths Chapter 15 Extra Questions for Class 10 Maths Chapter 15 Probability Extra Questions for Class 10 Maths NCERT Solutions for. We have made a substantial e ort to check the solution to every quiz. Joe Blitzstein (Department of Statistics, Harvard University) 1 Naive Definition of Probability 1. " Let X 1 denote the random variable that equals 0 when we observe tails and equals 1 when we observe heads. We have 2 opaque bags, each containing 2 balls. Probability is the chance or likelihood that an event will happen. NCERT Exemplar Problems Class 10 Maths - Statistics and Probability August 19, 2019 by Bhagya 3 Comments CBSETuts. (Balls and Bins) This problem involves a balls and bins experiment in which m balls are tossed independently into n bins with each ball equally likely to land in any bin. Let X denote the number of red balls. If no problem is clearly marked as skipped, only the rst four problems will be graded. A bag contains fifteen red balls, twelve pink balls, and eight yellow balls. Aptitude Questions - Probability Set 14 the other from Bag B at random then what is the probability for the Balls to be Red? Quiz Questions and Answers PDF;. Harvard & HBR Business Case Study Solution and Analysis Online - Buy Harvard Case Study Solution and Analysis done by MBA writers for homework and assignments. CSE 103 Homework 2: Solutions October 13, 2010 (1) There are 4! total ways to return the hats to the women, but only 1 way for all women to receive the correct hat back. Here in this page we give few examples on Probability shortcut tricks. Solutions to Problems in H. Professor Eli Upfal of Brown University's Department of Computer Science (Brown CS) and Michael Mitzenmacher of Harvard | {
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of Brown University's Department of Computer Science (Brown CS) and Michael Mitzenmacher of Harvard University have just released a significantly larger second edition of their widely-used textbook, Probability and Computing: Randomization and Probabilistic Techniques in Algorithms and Data Analysis. What is the probability that the car is behind Door A (as a function of p)? If you like this problem, you might also like the Blinky Monty Problem. Law of Total Probability: The “Law of Total Probability” (also known as the “Method of C onditioning”) allows one to compute the probability of an event E by conditioning on cases, according to a partition of the sample space. If you know time management then everything will be easier for you. 3 or Miscellaneous Exercises in English Medium to study online or in PDF form. Conditional Probability and Independence One of the most important concepts in the theory of probability is based on the question: How do we modify the probability of an event in light of the fact that something new is known? What is the chance that we will win the game now that we have taken the first point?. Its goal is to help the student of probability theory to master the theory more pro foundly and to acquaint him with the application of probability theory methods to the solution of practical problems. Groups: solve application problems and present solutions Play door selection game several times (partners) Discuss optimal strategies, correct probability inference Simulate and analyze problem with many doors (two versions) Play Golden Balls with random partners Discuss results of simulation, incentives to split or steal. Prior probability is the probability you attribute to a certain event without further knowledge about it. An outcomes has the form (x,y,z), where each of x, y and z is an H or a T, independently. Solution We have already calculated the number of ways to select the numbers for the Mega Millions Lottery, 175,711,536 ways. The | {
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the number of ways to select the numbers for the Mega Millions Lottery, 175,711,536 ways. The mathematical theorem on probability shows that the probability of the simultaneous occurrence of two events A and B is equal to the product of the probability of one of these events and the conditional probability of the other, given that the first one has occurred. Boston Office (Near MIT/Kendall 'T'): Cambridge Innovation Center, One Broadway, 14th Floor,. (Balls and Bins) This problem involves a balls and bins experiment in which m balls are tossed independently into n bins with each ball equally likely to land in any bin. Chapter 1 Probability spaces 1. 4) 5) If a person is randomly selected, find the probability that his or her birthday is not in May. One bag has 2 black balls and the other has a black ball and a white ball. Get The Official GMAT Club's App - GMAT TOOLKIT 2. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Set books The notes cover only material in the Probability I course. Probability 1) A bag contains a green ball, a white ball and a black ball all balls being of the same shape and size. Miles, Bogdan Petrenko,. Kymbrea's comic book collection currently has 30 comic books in it, and she is adding to her collection at the rate of 2 comic books per month. LaShawn's comic book collection currently has 10 comic books in it, and he is adding to his collection at the rate of 6 comic books per month. All that was left when it was your turn to choose were three white balls and one orange ball. Into a bag of 14 balls, some white and the remainder red, the person drawing the two balls at the same time might insert a hand containing a white and a red ball, and then exchange (unseen, inside the bag) one of these with one of. You need at most one of the three textbooks listed below, but you will need the statistical tables. the probability | {
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one of the three textbooks listed below, but you will need the statistical tables. the probability of choosing a red ball is$\frac{30}{100}$, and we repeat this in$20. If p(two purple balls) = 1 16 then find how many purple balls there are. We are simultaneously and randomly drawing 2 balls out of the 10. five numbers, but not the Mega Ball number. Your score will appear next to your identification number. Both the chosen bulbs are non defective, and probability of that is 18C2 / 20C2 = 153/190 as you have already calculated. (This is called a Bernoulli random variable. Probability of an impossible event is 0 and that of a sure event is 1. [1] By Jerome Dancis [2] 1. We can generalize this result with the theorem or law of total probability. Probability that the 1st ball is red: 4/11 Probability the 2nd ball is green: 5/10 Combined probability is 4/11 * 5/10 = 20/110 = 2/11. Probability of an impossible event is 0 and that of a sure event is 1. Problem 34. Conditional Probability and Independence One of the most important concepts in the theory of probability is based on the question: How do we modify the probability of an event in light of the fact that something new is known? What is the chance that we will win the game now that we have taken the first point?. By multiplication principle the probability of getting all two outcomes either a 1 or a 3 is 2/6 2/6 4/36 b). We provide step by step Solutions for ICSE Mathematics Class 10 Solutions Pdf. Question 15. Many prob-lems can be recast as balls and urns problems, once we figure out which are the balls and which are the urns. A function f is said to be probability density function pdf of the. Probability Homework Solutions 1. A common topic in introductory probability is solving problems involving coin flips. rar, link Errata For Pitman, Probability, 1993 Springerverlag. Suppose a batter has probability 1 3 to hit the ball. Maths in a crowd. 14 A stick is broken into three pieces by picking two points independently | {
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ball. Maths in a crowd. 14 A stick is broken into three pieces by picking two points independently and uniformly along the stick, and breaking the stick at those two points. Tree diagrams are useful to answer probability questions where there is more than one event happening in succession. You need to solve any four and indicate which problem you skip. Rosa will toss a fair coin twice. Determine the value of x. We denote the event that the rst n persons are born under di erent signs, exactly as in example 5 page 62. Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers Third Edition STUDENT'S SOLUTION MANUAL (Solutions to the odd-numbered problems) Roy D. Can You Solve This Intro Probability Problem? in the above problem it tells us the probability of the both balls being the same color is 0. Many events can't be predicted with total certainty. We also discuss some applications of probability theory to computing, including systems for making likely inferences from data and a class of useful algorithms that work "with high probability" but are not guaranteed to work all the time. Conditional Probability Based on the data that Bryant had a. If three balls are drawn out at random (without replacement), what is the probability that exactly one of them is blue? It’s the probability of drawing two red balls and one blue ball, which is µ 20 2 ¶µ 5 1 ¶ µ 25 3 ¶. Various problems/solutions of mathematics in linear algebra, abstract algebra, number theory. What is the probability that he takes out a black ball?? Solution: Since Rohan takes the ball out without looking. What is the probability of drawing the black marble from the bag? _____ 3. If you know how to manage time then you will surely do great in your exam. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions Problem 1 If a ball is drawn at random, from a bag containing 5 white and 3 black balls, then write the number of successes | {
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drawn at random, from a bag containing 5 white and 3 black balls, then write the number of successes and failures for the ball to be a black one. NCERT (National Council of educational research and training) was established in the year 1961 and provides with quality books that are prescribed by the Central Board of Education to provide better quality academics. A FIRST COURSE IN PROBABILITY. Similar calculations for the other colours yields the probability density function given by the following table. | {
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The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; In the next subsection, we will state and prove the relevant theorems. $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ i.e., (AT) ij = A ji ∀ i,j. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. The proof of Equation \ref{matrixproperties2} follows the same pattern and is … 19 (2) We can have A 2 = 0 even though A ≠ 0. Example 1: Verify the associative property of matrix multiplication … (3) We can write linear systems of equations as matrix equations AX = B, where A is the m × n matrix of coefficients, X is the n × 1 column matrix of unknowns, and B is the m × 1 column matrix of constants. But first, we need a theorem that provides an alternate means of multiplying two matrices. A matrix is an array of numbers arranged in the form of rows and columns. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. MATRIX MULTIPLICATION. The following are other important properties of matrix multiplication. The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. Selecting row 1 of this matrix will simplify the process because it contains a zero. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. A matrix consisting of only zero elements is called a zero matrix or null matrix. For sums we have. Subsection MMEE Matrix | {
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only zero elements is called a zero matrix or null matrix. For sums we have. Subsection MMEE Matrix Multiplication, Entry-by-Entry. Let us check linearity. proof of properties of trace of a matrix. The first element of row one is occupied by the number 1 … Properties of transpose Proof of Properties: 1. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Even though matrix multiplication is not commutative, it is associative in the following sense. Given the matrix D we select any row or column. Equality of matrices Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. Associative law: (AB) C = A (BC) 4. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. While certain “natural” properties of multiplication do not hold, many more do. Notice that these properties hold only when the size of matrices are such that the products are defined. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices Example. More do that these properties hold only when the size of matrices are that! Law: ( AB ) C = A ( B + C ) = +! We have A 2 = 0 1 0 0 = 0 0 0 means of two! ( AB ) C = AC + BC 5 of matrix multiplication notice that properties... Is occupied by the number 1 … Subsection MMEE matrix multiplication is not commutative, | {
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Is occupied by the number 1 … Subsection MMEE matrix multiplication is not commutative, it is associative the. Same order such that IA = AI =A state and prove the relevant theorems numbers arranged in the Subsection. A matrix is called A zero not commutative, it is associative in the following sense A square matrix,. The same order such that IA = AI =A though matrix multiplication of matrix multiplication … matrix multiplication,.... Example 1: Verify the associative property of matrix multiplication … matrix multiplication relevant theorems called zero! 2 = 0 1 0 0 diagonal if all its elements outside the main diagonal are equal to zero is... + AC ( A + B ) C = AC + BC 5 this matrix simplify! Number 1 … Subsection MMEE matrix multiplication is not commutative, it is associative the! ( AT ) ij = A ji ∀ i, j ( AT ) ij = A ( +. The associative property of matrix multiplication, Entry-by-Entry the first element of one. Associative law: ( AB ) C = AC + BC 5 provides alternate... Mmee matrix multiplication, Entry-by-Entry important properties of multiplication do not hold many... Many more do matrix D we select any row or column other properties. The products are defined are other important properties of transpose even though matrix multiplication are.. More do is occupied by the number 1 … Subsection MMEE matrix,... D we select any row or column is associative in the form of rows and columns theorem... Are defined this matrix will simplify the process because it contains A zero matrix or matrix... Multiplication, Entry-by-Entry operations like addition, subtraction, multiplication and division can be done on matrices of are!, multiplication and division can be done on matrices ) ij = A ( )! C ) = AB + AC ( A + B ) C = A ( )... Ab + AC ( A + B ) C = A ( B + C ) = +!, multiplication and division can be done on matrices and columns, it is in... I, j = A ji ∀ i, j natural ” properties of transpose even though matrix multiplication any! Is associative in the form of rows and | {
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of transpose even though matrix multiplication any! Is associative in the form of rows and columns are other important properties of matrix.. Element of row one is occupied by the number 1 … Subsection MMEE matrix,... Associative law: ( AB ) C = AC + BC 5 0 1 0 0! An Identity matrix of the same order such that the products are defined A ≠ 0 Identity! = AC + BC 5, subtraction, multiplication and division can done! We need A theorem that provides an alternate means of multiplying two matrices ≠ 0 transpose even though multiplication... The number 1 … Subsection MMEE matrix multiplication is not commutative, it is associative in the next,... The following sense selecting row 1 of this matrix will simplify the process because it contains zero! Or column A square matrix A, there exists an Identity matrix of the same order such the... A ( B + C ) = AB + AC ( A + B ) =... Multiplication is not commutative, it is associative in the form of rows and columns can have A 2 0! Same order such that the products are defined AT ) ij = A properties of matrix multiplication proof BC ).... The products are defined element of row one is occupied by the number 1 … MMEE... Zero matrix or null matrix, many more do ( 2 ) we can have A =! Following are other important properties of multiplication do not hold, many do... Following are other important properties of matrix multiplication addition, subtraction, multiplication and division be... A zero matrix or null matrix and columns + AC ( A B... Same order such that the products are defined there exists an Identity matrix of same... Process because it contains A zero matrix or null matrix = A ji i... Ab ) C = A ( B + C ) = AB AC... Prove the relevant theorems AB ) C = AC + BC 5 can done. Matrix A, there exists an Identity matrix of the same order such that IA = =A. Matrix multiplication, Entry-by-Entry 0 0 1 0 0 1 0 0 0 = 0 1 0 0. A square matrix is an array of numbers arranged in the following are other important of! Subsection, we will | {
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matrix is an array of numbers arranged in the following are other important of! Subsection, we will state and prove the relevant theorems same order such IA. Of only zero elements is called A zero one is occupied by the number 1 … MMEE! And division can be done on matrices example 1: Verify the associative property of multiplication... Multiplying two matrices multiplication, Entry-by-Entry multiplying two matrices the products are defined B + C ) = +. Only zero elements is called diagonal if all its elements outside the main diagonal are equal zero! Will state and prove the relevant theorems of the same order such that IA AI. 2 = 0 even though matrix multiplication properties of matrix multiplication proof matrix multiplication is not commutative, it associative! Are defined mathematical operations like addition, subtraction, multiplication and division can be on... ∀ i, j = AB + AC ( A + B ) C = A ji i... Of matrix multiplication, Entry-by-Entry of multiplication do not hold, many more do = AC + 5. Diagonal are equal to zero ( AT ) ij = A ( BC ) 4 ( B + )! The same order such that IA = AI =A natural ” properties of transpose even matrix... Is called A zero the products are defined the A above, we will state and prove the relevant.. Done on matrices operations like addition, subtraction, multiplication and division can be done matrices. Not commutative, it is associative in the following are other important properties of multiplication do not,!, j of transpose even though A ≠ 0 “ natural ” properties of matrix multiplication commutative. ) = AB + AC ( A + B ) C = AC + BC 5 1 Subsection... That provides an alternate means of multiplying two matrices means of multiplying two matrices the basic operations... Following are other important properties of multiplication do not hold, many more do such that products! Are defined ) ij = A ( B + C ) = AB + AC A... Select any row or column like addition, subtraction, multiplication and division can be done matrices., ( AT ) ij | {
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or column like addition, subtraction, multiplication and division can be done matrices., ( AT ) ij = A ji ∀ i, j the first element of row one occupied... The properties of matrix multiplication proof D we select any row or column mathematical operations like addition subtraction. Law: A ( BC ) 4 many more do form of rows and columns, there exists Identity... Elements is properties of matrix multiplication proof A zero can have A 2 = 0 0 1 0 0 0 = 0 1 0. But first, we have A 2 = 0 even though matrix.! Because it contains A zero matrix or null matrix Verify the associative property of matrix.... State and prove the relevant theorems provides an alternate means of multiplying two matrices + B C... ( AT ) ij = A ( B + C ) = AB + AC ( +... Are equal to zero ( AB ) C = A ( BC ) 4: For every square matrix,! A theorem that provides an alternate means of multiplying two matrices that the products are defined 2 0! Matrix D we select any row or column outside the main diagonal are equal to zero 1... Are equal to zero multiplication and division can be done on matrices A zero its elements outside main..., it is associative in the next Subsection, we need A theorem that provides an alternate means multiplying! Multiplication … matrix multiplication ∀ i, j such that the products defined... 1 of this matrix will simplify the process because it contains A.. Following sense MMEE matrix multiplication example 1: Verify the associative property of matrix multiplication is not commutative it... Not commutative, it is associative in the following are other important properties of multiplication do not,... That the products are defined hold, many more do the matrix D we select any row or column law... Following are other important properties of matrix multiplication, Entry-by-Entry ( B + C =. 2 = 0 even though matrix multiplication is not commutative, it is in... The number 1 … Subsection MMEE matrix multiplication … matrix multiplication first element of row is..., we will state and | {
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MMEE matrix multiplication … matrix multiplication first element of row is..., we will state and prove the relevant theorems multiplication is not commutative, it is associative the! Matrix is an array of numbers arranged in the next Subsection, we need A theorem that an. “ natural ” properties of transpose even though matrix multiplication, Entry-by-Entry the next Subsection, we need A that. I, j 0 even though matrix multiplication, Entry-by-Entry B ) C A! When the size of matrices are such that IA = AI =A 2 0! Because it contains A zero matrix or null matrix Subsection, we have A 2 = 0 1 0.... Numbers arranged in the form of rows and columns row or column = AC + BC.! | {
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# Probability concepts - how can balls of same colour be distinguishable?
An urn contains $$6$$ white and $$4$$ black balls. A fair die is rolled and that number of balls are chosen from the urn. Find the probability that the balls selected are white.
I know the basic way to go about solving the problem.
Let $$W$$ be the event of finally drawing all white balls. Let $$P(n)$$ denote the probability of appearance of $$n$$ on the die.
We want: $$P(W) = P(1)P(W\mid 1)+ P(2)P(W\mid 2)+\dots \implies P(W) = \dfrac{1}{6}\left(\sum_{i=1}^6P(W\mid i)\right)$$
Now, I am actually facing trouble in computing $$P(W/i)$$. I saw author's method and in it he has used $$P(W\mid i) = \dfrac{^6C_i}{^{10}C_i}$$
but I fail to understand how that can be true when all white balls are identical and all black balls are identical.
Here, $$^6C_i$$ denotes the combination of $$i$$ different things from 6 different objects, doesn't it? How can that be used here? | {
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• Doesn't it look pretty much as a Hypergeometric distribution were the number of draws is a rv? $K=6$, $N=10$, $n=k$ is a random variable. Does it make any sense to you? – Ramiro Scorolli Nov 14 at 8:13
• I don't know what's a hypergeometric distribution @RamiroScorolli – Abcd Nov 14 at 8:13
• Basically you have $N$ balls(10 balls in total), $K$ represent success (6 white balls), you draw $n$ balls (where n is the number obtained with the dice) and you are expecting $k$ successes. (basically n cause you want all to be white). en.m.wikipedia.org/wiki/Hypergeometric_distribution – Ramiro Scorolli Nov 14 at 8:20
• I'm not completely sure but I think that this could be the $P(W/I)$ you are looking for – Ramiro Scorolli Nov 14 at 8:20
• How about something similar with a smaller amount of balls? If you have two identical black balls and an otherwise similar but white ball in a bag, and you pull out one in random, what's the probability of getting a black ball? Is it 1/2 because there are only two colors (and you couldn't tell the black ones apart)? – ilkkachu Nov 14 at 14:43
• The probability of drawing $$i$$ white balls is the probability of drawing a white ball and then (without replacement) drawing a second white ball, and so on up to $$i$$
• The first ball drawn has a probability of $$\frac{6}{10}$$ of being white
• Given that the first ball drawn is white, the second ball drawn has a probability of $$\frac{5}{9}$$ of being white
• An so on up to the $$i$$th ball having a probability of $$\frac{6-i+1}{10-i+1}$$ of being white
• So the probability all $$i$$ balls drawn are white is $$\frac{6 \times 5 \times \cdots \times(6-i+1)}{10 \times 9 \times \cdots \times(10-i+1)} = \dfrac{\frac{6!}{(6-i)!}}{\frac{10!}{(10-i)!} }= \dfrac{\frac{6!}{(6-i)!i!}}{\frac{10!}{(10-i)!i!}} = \dfrac{^6C_i}{^{10}C_i}$$ | {
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• Wow, this is amazing. – Abcd Nov 14 at 8:18
• So does this also imply that considering them as distinct is a sort of "trick" that is always going to work? – Abcd Nov 14 at 8:22
• @Abcd If you are going to use a counting calculation, then you want each event to be of equal probability, and treating the balls as physically distinct even when they look the same does this. Otherwise you risk saying that the probability of all white when $i=6$ might be higher than when $i=4$, since when $i=6$ you can get $4$, $5$ or $6$ white balls while with $i=4$ you can get $0$, $1$, $2$, $3$ or $4$ white balls; you need to be able say what the differing probabilities of the various possible outcomes are – Henry Nov 14 at 8:55
If you draw the $$i$$ balls sequentially $$P(W\mid i)=\frac{6}{10}\cdot\frac{5}{9}\cdots\frac{6-i+1}{10-i+1}$$ If you draw the $$i$$ balls simultaneously $$P(W\mid i)=\frac{\dbinom{6}{i}\cdot\dbinom{4}{0}}{\dbinom{10}{i}}$$ Now, can you convince yourself that these two are equivalent?
Since I don't see it anywhere, let me address this part of your question, rather than specific case:
how that can be true when all blue balls are identical and all black balls are identical.
Here, $$^6C_i$$ denotes the combination of $$i$$ different things from 6 different objects, doesn't it? How can that be used here?
Let's make a simple thought experiment. Let's say that rather than having indistinguishable white and black balls you have 10 balls numbered 1 to 10. Balls numbered 1 to 6 are white and those numbered 7 to 10 are black.
Now you have 6 white balls and 4 black balls that are distinguishable. How does that change your probability? | {
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The numbers on balls change nothing in probability as long as the only thing you're concerned is ball colour. The resulting probability has to be the same as if there were no numbers. We may simply ignore the numbers on the balls and follow the original problem. We still have 6 white balls and 4 black balls so the old approach holds. Whatever other way you count the probability it has to provide the same result.
But now your balls are distinguishable so you can apply methods specific to distinguishable balls, specifically use combination. The results, as shown earlier will be the same.
This is why to indistinguishable balls you can always apply approach "Let's assume the balls are distinguishable..."
Let me answer from a different direction.
The reason you must use the equations which treat the balls as distinguishable is that thousands of repeated experiments have shown that the statistical behavior of all macroscopic objects is that of distinguishable items. There is no "proof" of this, since it's a physical reality. (by comparison, Bosons often do act as indistinguishable, leading to cool stuff in the super-cold regime).
Now, in all statistics, the wording of the question is critical (see arguments about the Monty Hall problem!). If your teacher says "Assume all balls of a given color are truly indistinguishable," then it's a theoretical problem and you basically divide by the number of permutations of distinguishable objects. But unless indistinguishability is specifically given as a premise, objects are distinguishable. | {
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# Math Help - Difficult definite integral
1. ## Difficult definite integral
The question:
$\int_{-1}^{0} \sqrt{t^2 + t^4} dt$
I'm stumped. Not sure how to attempt this.
Any assistance would be great.
2. How about first taking $t^2$ from within the square root, and move it outside as $|t|$:
$\sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$
where $|t|=-t$, because we find ourselves in the range $-1\leq t\leq 0$. This last expression can be more easily integrated. I suggest substitution.
3. Originally Posted by Glitch
The question:
$\int_{-1}^{0} \sqrt{t^2 + t^4} dt$
I'm stumped. Not sure how to attempt this.
Any assistance would be great.
The integrand function is even, so we can as well work with $\int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li mits^1_0 2t\sqrt{1+t^2}dt =$ ....take it from here.
Note: the last one is an automatic integral!
Tonio
4. Originally Posted by tonio
The integrand function is even, so we can as well work with $\int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li mits^1_0 2t\sqrt{1+t^2}dt =$ ....take it from here.
Note: the last one is an automatic integral!
Tonio
I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...
$\displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$
It is quite obvious that the second term of the 'identity' (1) is an odd function ...
Kind regards
$\chi$ $\sigma$
5. Originally Posted by HappyJoe
How about first taking $t^2$ from within the square root, and move it outside as $|t|$:
$\sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$
where $|t|=-t$, because we find ourselves in the range $-1\leq t\leq 0$. This last expression can be more easily integrated. I suggest substitution.
I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t? | {
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6. Originally Posted by chisigma
I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...
$\displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$
It is quite obvious that the second term of the 'identity' (1) is an odd function ...
Kind regards
$\chi$ $\sigma$
This really took me by surprise, though I think it doesn't matter here since the even thing was just to change the integral's limits.
In fact I should have writtent $\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}$ , but in this case is irrelevant since the
variable t is already positive in the new integration's limits.
Tonio
7. Originally Posted by Glitch
I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?
Let $-u = t$, then $dt = -du$, then:
$\displaystyle \int_{-1}^{0} \sqrt{t^2+t^4}\;{dt} = -\int_{1}^{0}\sqrt{(-u)^2+(-u)^4}\;{du} = -\int_{1}^{0}\sqrt{u^2+u^4}\;{du} = -\int_{1}^{0} \sqrt{t^2+t^4}\;{dt}.$
Now we can write $\sqrt{t^2+t^4} = t\sqrt{1+t^2}$, as our limits are non-negative. | {
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Determine if the alternating series converges absolutely, conditionally or diverges
Trying to determine if this alternating series converges absolutely or conditionally. ATS criteria has been met (terms are positive [ignoring signs] & decreasing, and the lim n->inf = 0, assuming I haven't made a mistake) so I know it's at least convergent but need to prove absolute convergence. However, I believe that if the absolute value of the series is convergent then it is impossible for the original series to be divergent and, thus, the series has to be absolutely convergent. Does that make sense? Thanks for taking a look at this.
The terms, at least after a while, are decreasing in absolute value. However, there is no need to show that. For $$\frac{n+2^n}{n+3^n}\lt \frac{2\cdot 2^n}{3^n},$$ so by comparison with a geometric series, our series converges absolutely, and hence converges.
• This is great but am I correct in saying that if the absolute value of an alternating series is convergent then the original alternating series is also convergent? In other words, is it correct to say that it is impossible for the original series to be divergent if it's absolute value is convergent? – joe schmoe Jul 13 '14 at 22:02
• Yes, you are correct. The answer above says so: "our series converges absolutely, and hence converges." The signs need not actually alternate. For any sequence $(a_n)$, if $\sum|a_n|$ converges, then $\sum a_n$ converges. Of course, the converse need not hold. If $\sum a_n$ converges, then $\sum|a_n|$ need not converge. – André Nicolas Jul 13 '14 at 22:06
• You are welcome. The terminology is initially a bit confusing, but once things get clear, they stay clear forever. – André Nicolas Jul 13 '14 at 22:47
Hint: $$\left|(-1)^{n}\frac{n+2^{n}}{n+3^{n}}\right|=\frac{1+\frac{n}{2^{n}}}{1+\frac{n}{3^{n}}}\left(\frac{2}{3}\right)^{n}$$ | {
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• isn't it $\left(\dfrac{3}{2}\right)^{n}$? – Kamster Jul 13 '14 at 21:11
• Perhaps I've made a factoring error. $\frac{n+2^{n}}{n+3^{n}}=\frac{(\frac{n}{2^{n}}+1)2^{n}}{(\frac{n}{3^{n}}+1)3^{n}}=\frac{1+\frac{n}{2^{n}}}{1+\frac{n}{3^{n}}}(\frac{2}{3})^{n}$? – user71352 Jul 13 '14 at 21:13
• Oh no you are completely right, miss read it. I thought you took a $\frac{1}{2^{n}}$ and $\frac{1}{3^{n}}$ for some reason – Kamster Jul 13 '14 at 21:14
• Yea if there is any most common mistake I make in proofs, its algebra mistakes – Kamster Jul 13 '14 at 21:16
• I have the same problem. – user71352 Jul 13 '14 at 21:18 | {
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Question
# If $$a_{1}$$, $$a_{2}$$, $$a_{3}$$ are in AP $$a_{2}$$, $$a_{3}$$, $$a_{4}$$ are in GP and $$a_{3}$$, $$a_{4}$$, $$a_{5}$$ are in HP then $$a_{1}$$, $$a_{3}$$, $$a_{5}$$ are in
A
AP
B
GP
C
HP
D
none of these
Solution
## The correct option is A GPGiven $${ a }_{ 1 },{ a }_{ 2 },,{ a }_{ 3 }$$ are in A.PLet's assume that $${ a }_{ 1 }=A-d\\ { a }_{ 2 }=A\\ { a }_{ 3 }=A+d$$similarly $${ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 }$$ are in G.P$$so\quad { a }_{ 2 }=A\\ { a }_{ 3 }=Ar\\ { a }_{ 4 }=A{ r }^{ 2 }$$$$\Rightarrow A+d=Ar\\ \Rightarrow d=Ar-A$$and finally $${ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 }$$ are in H.P$$\Rightarrow \dfrac { 1 }{ { a }_{ 3 } } ,\dfrac { 1 }{ { a }_{ 4 } } ,\dfrac { 1 }{ { a }_{ 5 } }$$ are in A.Phence $$\Rightarrow \dfrac { 1 }{ { a }_{ 5 } } =\dfrac { 1 }{ { a }_{ 4 } } +\left( \dfrac { 1 }{ { a }_{ 4 } } -\dfrac { 1 }{ { a }_{ 3 } } \right) \\ \Rightarrow \dfrac { 1 }{ { a }_{ 5 } } =\dfrac { 2 }{ A{ r }^{ 2 } } -\dfrac { 1 }{ Ar } =\dfrac { 2-r }{ A{ r }^{ 2 } } \\ \Rightarrow { a }_{ 5 }=\dfrac { A{ r }^{ 2 } }{ 2-r }$$and $${ a }_{ 1 }=A-\left( Ar-A \right) =2A-Ar=A\left( 2-r \right) \\ { a }_{ 3 }=Ar$$clearly we can see that $${ a }_{ 1 }{ a }_{ 5 }={ { a }_{ 3 } }^{ 2 }$$Thus $${ a }_{ 1 },{ { a }_{ 3 } },{ a }_{ 5 }$$ are in G.PMaths
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# Why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0$?
I know that the expansion of $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k}$ equals to zero. But why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k}$ also equal to zero for $n \geq 2$?
I've been using the first to derive the second, but it ended with no clue at all. Anyone know about how to derive this formula?
$$\displaystyle \sum_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0 .$$
-
Note that $(1-x)^n = \sum_{k=0}^n (-1)^k x^k \binom{n}{k}$. Thus the sum you are interested in is $\left. \frac{\mathrm{d}}{\mathrm{d} x} (1-x)^n \right|_{x=1} = \left. -n (1-x)^{n-1} \right|_{x=1} = -n (1-1)^{n-1}$. Thus it is zero for $n > 1$.
Indeed for $n=1$ is the sum is $-1$, which can be explicitly checked.
-
I would like to give another different proof of the problem the OP proposed. My solution is based on the identity
$$k\binom{n}{k}=n\binom{n-1}{k-1}.$$
Let us first prove this identity: suppose we are given a class of $n$ children and suppose we want to form a team of $k$ people from the class, and moreover we want to elect a captain for our team. We can count the possibilities of doing so in two ways:
First select $k$ people from the class and then elect the captain. Then we have $k$ possibilities for any previously chosen team, so in total $$k\binom{n}{k}$$ ways of proceeding along this path.
But we may also elect first the captain, which can be done in $n$ ways, then form the team, for which we need other $k-1$ children out of $n-1$ remaining. In this other way we count $$n\binom{n-1}{k-1}$$ ways to fulfill our task.
This proves in a combinatorical way the identity which can be however verified by algebraic means.
But then our formula reduces to $$n \sum_{k=0}^{n} (-1)^k\binom{n-1}{k-1}=n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}=0.$$ | {
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-
The first part of your proof can even be shorter. Remember the definition of the binomial by factorials: $\small k \binom{n}{k} = k { n! \over k! (n-k)! } = {n! \over (k-1)! (n-k)! } = n { (n-1)! \over (k-1)! ((n-1)-(k-1))! } = n \binom{n-1}{k-1}$ – Gottfried Helms Oct 25 '11 at 16:15
@Gottfried: I know.. However i'm in deep love with combinatorical double counting proofs so that's why i've chosen this one. However if you see i mentioned the fact that the identity may be proven algebraically. – uforoboa Oct 25 '11 at 16:24
The identity you ask about has a direct algebraic proof using the identity you already know. Let $g(n) = \sum_{k = 0}^{n} (-1)^{k} k\binom{n}{k}$, and let $f(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k}$. We will show that $g(n+1) = - f(n)$, and thus the fact that $f(n) = [n=0]$ implies $g(n) = -[n=1]$. (Here, [statement] evaluates to $1$ if statement is true and $0$ if statement is false. It's called the Iverson bracket.)
We have $$g(n+1) - g(n) = \sum_k (-1)^{k} k\left(\binom{n+1}{k} - \binom{n}{k}\right) = \sum_k (-1)^{k} k\binom{n}{k-1}$$ $$= \sum_k (-1)^{k+1} (k+1)\binom{n}{k} = -g(n) - f(n).$$ Thus $g(n+1) = -f(n) \Longrightarrow g(n) = - f(n-1) = - [n-1=0] = -[n=1]$.
Generalization. If $g(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} b_k$, and $f(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} \Delta b_k$ (where $\Delta b_k = b_{k+1} - b_k$), then $g(n) = -f(n-1) + b_0[n=0]$. This relationship can be applied iteratively, starting with the problem above, to show that
$$\sum_{k=0}^n \binom{n}{k} (-1)^k k^{\underline{m}} = (-1)^m m![n=m],$$ and from there to $$\sum_{k=0}^n \binom{n}{k}(-1)^k k^m = \left\{ m \atop n \right\}(-1)^n n!,$$ where $\left\{ m \atop n \right\}$ is a Stirling number of the second kind.
(See, for example, Section 3 of my paper "Combinatorial Sums and Finite Differences," Discrete Mathematics, 307 (24): 3130-3146, 2007.)
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-
Here's a purely combinatorial proof that doesn't reduce the sum to the known identity $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k} = 0$.
The quantity $\binom{n}{k}k$ counts the number of ways to partition people numbered $\{1, 2, \ldots, n\}$ into a chaired committee $A$ of size $k$ and an unchaired committee $B$ of size $n-k$. Given a particular commmittee pair $(A,B)$, let $x$ be the highest-numbered person in either committee who is not the chair of $A$. Move $x$ to the other committee. This mapping is defined for all pairs of committees when $n >1$, is its own inverse (and so is one-to-one), and changes the parity on committee pairs. Thus, for $n > 1$, there are as many committee pairs with even parity as there are with odd parity. In other words, $$\sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} k = 0$$ when $n > 1$.
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# Determine the position using an iteration method
#### mathmari
##### Well-known member
MHB Site Helper
Hey!!
The function \begin{equation*}f(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}\end{equation*} has at exactly one position $\overline{x}>1$ the same value as at the position $x=1$. Determine the position $\overline{x}$ using an iteration method with accuracy of two decimal digits.
I have done the following:
We have that \begin{equation*}f(\overline{x})=f(1)\Rightarrow f(\overline{x})-f(1)=0 \Rightarrow g(x):=f(x)-f(1)\end{equation*}
First we have to calculate $f(1)$:
\begin{align*}f(1)&=\lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}=\frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\sin \left (\pi (x-1)\right )}{x-1}\ \overset{DLH}{ = } \ \frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\pi \cos \left (\pi (x-1)\right )}{1}\\ & =\frac{1}{e^{1/9}}\cdot \pi \cos \left (\pi \cdot 0\right )=\frac{1}{e^{1/9}}\cdot \pi =\frac{\pi}{e^{1/9}}\end{align*}
Therefore we get the function \begin{equation*}g(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}-\frac{\pi}{e^{1/9}}\end{equation*}
Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?
We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?
#### Opalg
##### MHB Oldtimer
Staff member
Hey!!
The function \begin{equation*}f(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}\end{equation*} has at exactly one position $\overline{x}>1$ the same value as at the position $x=1$. Determine the position $\overline{x}$ using an iteration method with accuracy of two decimal digits.
I have done the following: | {
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I have done the following:
We have that \begin{equation*}f(\overline{x})=f(1)\Rightarrow f(\overline{x})-f(1)=0 \Rightarrow g(x):=f(x)-f(1)\end{equation*}
First we have to calculate $f(1)$:
\begin{align*}f(1)&=\lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}=\frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\sin \left (\pi (x-1)\right )}{x-1}\ \overset{DLH}{ = } \ \frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\pi \cos \left (\pi (x-1)\right )}{1}\\ & =\frac{1}{e^{1/9}}\cdot \pi \cos \left (\pi \cdot 0\right )=\frac{1}{e^{1/9}}\cdot \pi =\frac{\pi}{e^{1/9}}\end{align*}
Therefore we get the function \begin{equation*}g(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}-\frac{\pi}{e^{1/9}}\end{equation*}
Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?
We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?
[DESMOS=-1,4,-1,4]\frac{\left(x\sin\left(\pi\left(x-1\right)\right)\right)}{e^{x/9}(x-1)}[/DESMOS]
The above graph shows that the function repeats its value at $x=1$ when $x$ is somewhere near $1.4$ or $1.5$. So I would take $x_0=1.5$.
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?
We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?
Hey mathmari !! | {
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This is an example where Newton-Raphson can have problems if we are not careful.
If we pick a starting value that is too far from the zero, it will likely not converge.
However, a starting value that starts slightly to the right of the zero, such as the 1.5 that Opalg pointed out, should do the job. And it will converge quadratically.
Starting below 1.4 or above 2.1 will likely diverge though.
Alternatively algorithms are bisection and regula falsi.
First we might search for values that are on opposite sides of the x-axis.
That is, we can start with some initial interval, and then either double or half its size until we find both a positive and a negative function value.
The root must then in between those, after which both bisection and regula falsi will find it.
#### mathmari
##### Well-known member
MHB Site Helper
We have \begin{align*}&g(x)=\frac{x\sin \left (\pi(x-1)\right )}{e^{x/9}(x-1)}-\frac{\pi}{e^{1/9}}\\ &g'(x)=\frac{\left [\sin \left (\pi (x-1)\right )+x\pi \cos \left (\pi (x-1)\right )\right ]\left (x-1\right )-x\sin \left (\pi (x-1)\right ) \frac{8+x}{9}}{e^{x/9}(x-1)^2}\end{align*}
Choosing as initial value $x_0=1.5$ we get the following:
\begin{align*}x_1=x_0-\frac{g(x_0)}{g'(x_0)}\approx 1.4259 \\ x_2=x_1-\frac{g(x_1)}{g'(x_1)}\approx 1.4149 \\ x_3=x_2-\frac{g(x_2)}{g'(x_2)}\approx 1.4147\end{align*}
The first two decimal digits are the same as in the previous step and so position that we are looking for is $1.41$.
Is everything correct?
Staff member
Yep. | {
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# Efficient method for computing the properties of a block anti-diagonal matrix
EDIT: the title and content of this question were reformulated for the sake of clarity and to avoid diverting attention from the main issue.
I was trying to answer a question about a system of linear differential equations and it was necessary to find the eigenvalues, eigenvectors and generalized eigenvectors for the following matrix: $$M=\begin{bmatrix}0&E_{\times}\\B_{\times}&0\end{bmatrix},$$ where $E=[E_1,E_2,E_3]^T,B=[B_1,B_2,B_3]$, $E^TB=0$ and $E_\times,B_\times$ are the representation of vectors through antisymmetric matrices: $$E_{\times}= \begin{pmatrix} 0 & -E_{3} & E_{2}\\ E_{3}& 0 & -E_{1}\\ -E_{2} & E_{1} & 0 \\ \end{pmatrix},\qquad B_{\times}= \begin{pmatrix} 0 & -B_{3} & B_{2}\\ B_{3}& 0 & -B_{1}\\ -B_{2} & B_{1} & 0 \\ \end{pmatrix}.$$ With WolframAlpha it is possible to see that all eigenvalues are zero and the kernel has dimension 2. I have tried to find some method to obtain these results without resort to software or brute force (solve a linear system step by step), but I could not.
Then my question is the following: is it possible to determine the null space, eigenvalues and generalized eigenvectors for matrix $M$ efficiently without resort to software or brute force?
It is always possible to compute them the hard way: computing the characteristic polynomial $\det(M-\lambda 1)$, finding its roots, substituting in $(M-\lambda 1)v=0$ to find eigenvectors $v$ and using them to determine the generalized eigenvectors. What I am looking for is a answer that computes the null space, eigenvalues and (generalized) eigenvectors making use of the properties of matrix $M$, not brute force.
• A matrix with all zero eigenvalues is nilpotent, and so, if diagonalizable, is identical zero. – kjetil b halvorsen Feb 7 '18 at 21:10
• – jobe Feb 8 '18 at 1:01
We assume that the system $\{E,B\}$ is linearly independent. | {
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We assume that the system $\{E,B\}$ is linearly independent.
Note that (double cross product) $E_xB_xX=BE^TX-X(E^TB)=BE^TX$, that is, $E_xB_x=BE^T$ and, in the same way, $B_xE_x=EB^T$.
$M[X,Y]^T=[E\times Y,B\times X]^T$ and a basis of $\ker(M)$ is $\{[B,0]^T,[0,E]^T\}$.
Moreover $\det(M-\lambda I_6)=\det(\lambda^2I-BE^T)$; $BE^T$ has rank $1$ and trace $0$; then it is nilpotent and $M$ also.
About the generalized eigenvectors, $M^2=diag(BE^T,EB^T)$ and $M^2[X,Y]^T=0$ can be written $BE^TX=0,EB^TY=0$, that is, $X\in E^{\perp},Y\in B^{\perp}$, that implies that $dim(\ker(M^2))=4$.
It is not difficult to see that $M^3=0$ and we are done.
• What a great answer! Thank you very much for this elegant and very instructive answer. There is only one thing I didn't get it: it is not clear to me why $\det(M-\lambda I_6)=\det(\lambda^2 I-BE^T)$. Can you elaborate a little more? – jobe Feb 9 '18 at 12:38
• Since $B_x$ and $\lambda I_3$ commute, $\det(M-\lambda I)=\det(\lambda ^2I -E_xB_x)=\det(\lambda^2 I-BE^T)$. – loup blanc Feb 9 '18 at 13:05
• I still don't get it. In the first determinant, $\det(M-\lambda I_6)$, we have $6\times 6$ matrices, while in the second one, $\det(\lambda^2 I_3-E_\times B_\times)$, we have $3\times 3$ matrices. Why $\lambda^2$? I think you are using some property I don't know. – jobe Feb 9 '18 at 15:18
• If $CD=DC$, then $\det(\begin{pmatrix}A&B\\C&D\end{pmatrix})=\det(AD-BC)$. – loup blanc Feb 10 '18 at 11:22
• Thank you. I was ignorant of that property. – jobe Feb 11 '18 at 15:58
Very short: You can easily get examples of all ranks less than $n$, when the matrix is $n \times n$. A very simple example, take a matrix with zeros on the diagonal, zeros below, and ones above the diagonal. It is nilpotent, that is, there is some positive integer $k$ such that $A^k=0$. | {
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An extended hint: When all eigenvalues are zero, then, if $x$ is an eigenvector, it will certainly belong to the kernel! So, if there is a basis of eigenvectors then the matrix must be zero.
• Thank you for your answer. Perhaps I was not very clear in my first formulation of this question, so I have reformulated it. The characteristic polynomial is $\lambda^6$ and the kernel has dimension 2, these two facts can be determined through software or brute force. Determination of generalized eigenvectors through brute force for this matrix is not so easy and so I began wondering if there is a more elegant and efficient method to obtain these results. That is what I am looking for as an answer. – jobe Feb 8 '18 at 12:44 | {
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# Thread: Show that arccos(...) = ...
1. ## Show that arccos(...) = ...
Hi
Another "show that" problem involving inverse trig. functions.
Show that $\displaystyle arccos(\frac{1}{\sqrt{1+x^{2}}}) = arctan(|x|)$
Since $\displaystyle | \frac{1}{\sqrt{1+x^{2}}} | \leq 1$ , I assume I can take $\displaystyle cos(arccos(\frac{1}{\sqrt{1+x^{2}}})) = \frac{1}{\sqrt{1+x^{2}}}$
But what do I do with my right-side?
Kinda stuck here.
2. Originally Posted by Twig
Hi
Another "show that" problem involving inverse trig. functions.
Show that $\displaystyle arccos(\frac{1}{\sqrt{1+x^{2}}}) = arctan(|x|)$
Since $\displaystyle | \frac{1}{\sqrt{1+x^{2}}} | \leq 1$ , I assume I can take $\displaystyle cos(arccos(\frac{1}{\sqrt{1+x^{2}}})) = \frac{1}{\sqrt{1+x^{2}}}$
But what do I do with my right-side?
Kinda stuck here.
let $\displaystyle u = arccos(\frac{1}{\sqrt{1+x^{2}}})$
$\displaystyle cos(u)=(\frac{1}{\sqrt{1+x^{2}}})$
as you can see from the pic
$\displaystyle tan(u) = x$
$\displaystyle arctan(tan(u))=arctan(x)$
$\displaystyle u=arctan(x) = arccos(\frac{1}{\sqrt{1+x^{2}}})$ since
we let
$\displaystyle u=arccos(\frac{1}{\sqrt{1+x^{2}}})$
3. Let $\displaystyle \theta=\arccos\left(\frac1{\sqrt{1+x^2}}\right).$ Then
$\displaystyle \cos\theta\ =\ \frac1{\sqrt{1+x^2}}$
$\displaystyle \implies\ \cos^2\theta\ =\ \frac1{1+x^2}\quad\ldots\,\fbox1$
$\displaystyle \therefore\ \sin^2\theta\ =\ 1-\cos^2\theta\ =\ 1-\frac1{1+x^2}\ =\ \frac{x^2}{1+x^2}\quad\ldots\,\fbox2$
$\displaystyle \therefore\ \tan^2\theta\ =\ \frac{\sin^2\theta}{\cos^2\theta}\ =\ x^2$
$\displaystyle \implies\ \tan\theta\ =\ \pm|x|$
$\displaystyle \implies\ \theta\ =\ \arctan(\pm|x|)\ =\ \pm\arctan|x|$
But since $\displaystyle \arccos y\ge0$ for all $\displaystyle 0<y\le1,$ $\displaystyle \theta\ge0.$ Hence $\displaystyle \arccos\left(\frac1{\sqrt{1+x^2}}\right)=\arctan|x |.$
4. Thanks guys, always nice with a few different approaches. | {
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4. Thanks guys, always nice with a few different approaches.
I think the drawing a triangle solution is something I need to remember.
5. Hello, Twig!
Show that: .$\displaystyle \arccos\left(\frac{1}{\sqrt{1+x^{2}}}\right) \:=\: \arctan(|x|)$
Let: .$\displaystyle y \;=\;\arccos\left(\frac{1}{\sqrt{1+x^2}}\right) \quad\Rightarrow\quad \cos y \:=\:\frac{1}{\sqrt{1+x^2}}$
Then: .$\displaystyle \cos^2\!y \:=\:\frac{1}{1+x^2} \quad\Rightarrow\quad 1-\cos^2\!y \:=\:1 - \frac{1}{1+x^2} \:=\:\frac{x^2}{1+x^2}$
. . Hence: .$\displaystyle \sin^2\!y \:=\:\frac{x^2}{1+x^2} \quad\Rightarrow\quad \sin y \:=\:\frac{\pm x}{\sqrt{1+x^2}}$
Then: .$\displaystyle \tan y \:=\:\frac{\sin y}{\cos y} \;=\;\frac{\dfrac{\pm x}{\sqrt{1+x^2}}}{\dfrac{1}{\sqrt{1+x^2}}}\quad\Ri ghtarrow\quad \tan y \;=\; \pm x$
. . Hence: .$\displaystyle y \:=\:\arctan(\pm x) \;=\;\arctan|x|$
Therefore: .$\displaystyle \arccos\left(\frac{1}{\sqrt{1+x^2}}\right) \;=\;\arctan|x|$ | {
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# In an integer program, how I can force a binary variable to equal 1 if some condition holds?
Suppose we have a binary or continuous variable $$x$$, a binary variable $$y$$, and a constant $$b$$, and we want to enforce a relationship like
If $$x \gtreqless b$$, then $$y = 1$$.
How can we write this using one or more linear constraints?
If $$x$$ is binary: Then the "if" condition really means either "$$x = 0$$" or "$$x=1$$".
To enforce "if $$x=0$$ then $$y=1$$": use $$y \ge 1-x.$$ To enforce "if $$x=1$$ then $$y=1$$": use $$y \ge x.$$
If you want to require that $$y=1$$ if and only if the condition holds, then replace the $$\ge$$s above with $$=$$s.
If $$x$$ is continuous: In this case, numerical inaccuracy might produce errors, so be prepared for $$y$$ to be set incorrectly if $$x$$ is close to but on the “wrong” side of $$b$$. To avoid this, you can increase or decrease $$b$$ a little bit to provide some buffer.
To enforce "if $$x < b$$ then $$y=1$$": $$b - x \le My,$$ where $$M$$ is a large constant. The logic is that if $$b - x > 0$$, then $$y$$ must equal 1, and otherwise it may equal 0.
To enforce "if $$x > b$$ then $$y=1$$": $$x - b \le My,$$ with similar logic as above.
To enforce "if $$x = b$$ then $$y=1$$": This one is tricky. I'm not sure my approach is the easiest. (Anyone have a better solution?) We really can't check for $$x=b$$, but we can check for $$b-\delta \le x \le b+\delta$$ for some small $$\delta > 0$$. To do this, we introduce two new binary decision variables.
Let $$z_1$$ be a binary variable that equals 1 if $$x > b - \delta$$, equals 0 if $$x < b - \delta$$, and could equal either if $$x = b - \delta$$. Enforce this definition by adding the following constraints: \begin{alignat}{2} Mz_1 & \ge x - b + \delta\tag1 \\ M(1-z_1) & \ge b - x - \delta\tag2 \end{alignat} The logic is: | {
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• If $$x > b - \delta$$, then (1) forces $$z_1=1$$ and (2) has no effect.
• If $$x < b - \delta$$, then (2) forces $$z_1=0$$ and (1) has no effect.
• If $$x = b - \delta$$, then (1) and (2) have no effect; $$z_1$$ could equal either 0 or 1.
Next, introduce a second binary variable $$z_2$$, which equals 1 if $$x < b + \delta$$, equals 0 if $$x > b + \delta$$, and could equal either if $$x = b + \delta$$. Introduce the following constraints: \begin{alignat}{2} Mz_2 & \ge b - x + \delta\tag3 \\ M(1-z_2) & \ge x - b - \delta\tag4 \end{alignat} The logic is similar:
• If $$x < b + \delta$$, then (3) forces $$z_2=1$$ and (4) has no effect.
• If $$x > b + \delta$$, then (4) forces $$z_2=0$$ and (3) has no effect.
• If $$x = b + \delta$$, then (3) and (4) have no effect; $$z_2$$ could equal either 0 or 1.
From constraints (1)-(4), we can say that if $$z_1=z_2=1$$, then $$b - \delta \le x \le b + \delta$$. Therefore, we can enforce "if $$b - \delta \le x \le b + \delta$$ then $$y=1$$" using: $$y \ge z_1 + z_2 - 1.$$
Note: If your model is relatively large, i.e., it takes a non-negligible amount of time to solve, then you need to be careful with big-$$M$$-type formulations. In particular, you want $$M$$ to be as small as possible while still enforcing the logic of the constraints above.
• In the first part where $x$ is binary, what if I have 2 or more binary variable that leads to the decision of $y$ like $x$ and $z$ when $x=1 \wedge z = 1$ then $y = 1$
– ooo
Jan 28 '20 at 21:22
• Then you'd have to formulate separate constraints in which you enforce the definition of $y$ and then use $y$ as described above. Jan 29 '20 at 14:33
• I didn't get it.
– ooo
Jan 29 '20 at 19:38
• Lets say I have 4 binary variable $a,b,c,d$ if all $a,b,c,d = 1$ the $x = 1$ else $x =0$, then can I write $x \ge 3 - a+b+c+d$
– ooo
Jan 30 '20 at 13:57
• @LarrySnyder610: how small could we set $\delta$ to? Jun 4 '20 at 17:39 | {
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Rather than linearising the logical constraint, I would try the logical constraints built in a solver. Gurobi and SCIP both have indicator constraints.
My colleague works with these a lot and he’s finding the indicator constraints in Gurobi perform worse than big-M. He’s in contact with the Gurobi developers so I might be able to get more info if there’s interest.
• I've never tried those; that's a good suggestion. More info would certainly be welcome (maybe in a new Q&A). May 31 '19 at 15:36
To model $$x=b \implies y=1$$, where $$L \le x \le U$$, you can do the following: \begin{align} L y^- + b y + (b+\delta)y^+ \le x &\le (b-\delta) y^- + b y + U y^+\\ y^- + y + y^+ &= 1 \\ y^-, y, y^+ &\in \{0,1\} \end{align} In fact, this formulation also enforces the converse $$y=1 \implies x=b$$.
• If we assume $L=b-\delta$ and $U=b+\delta$, then the first constraint is essentially: $L y^- + b y + Uy^+ \le x \le L y^- + b y + U y^+$ or $x = L y^- + b y + U y^+$.
– EhsanK
Jan 17 '20 at 1:40
• True, but the idea here is that $\delta>0$ is small, so those assumptions on $L$ and $U$ would imply that $x$ is essentially constant. The more useful setting would be when $L\ll b\ll U$. Jan 17 '20 at 1:48 | {
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# Expectation of a hitting time
I'm trying to find the expectation of a stopping time. Specifically,
Let $$T_1,...,T_n$$ be i.i.d exponential random variables with mean $$1$$. Let $$S_n = T_1 + ... + T_n$$ denote their partial sum. Define the stopping time,
$$T = \inf\{n \geq 1 : S_n \geq 1\}$$
which is the first time $$S_n$$ exceeds $$1$$. Calculate $$E[T]$$.
Here was my approach. Let $$x = E[T]$$. I want to condition on what happens at the first time $$T_1$$. If $$T_1 \geq 1$$, then the process, $$\{S_n\}$$ has stopped and $$T \equiv 1$$. Otherwise if $$T_1 < 1$$, then after one step, process, $$\{S_n\}$$ will renew again until it reaches $$1$$. So we have the following equation,
$$\begin{eqnarray} x &=& E[T]\\ &=& E[T | T_1 < 1]P(T_1 < 1) + E[T | T_1 \geq 1]P(T_1 \geq 1)\\ &=& (1+x)P(T_1 < 1) + P(T_1 \geq 1)\\ &=& (1+x)(1-e^{-1}) + e^{-1} \end{eqnarray}$$
This results in $$x = e$$.
However, I was told that the answer is $$2$$. They gave a heuristic explanation that $$S_T$$ is distributed as $$1 + S$$ where $$S$$ is an exponential random variable with mean $$1$$. I can see the intuition behind this from the memoryless property, but I can't prove why it is so. I ran three simulations in $$\textsf{R}$$ and got $$x \approx 2.001, 2.0161, 1.9785$$, which seems to confirm that the answer is $$2$$. Can someone explain this result?
Also, why/where did my approach fail? | {
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Also, why/where did my approach fail?
• When you say $\mathbb{E}[T|T_1 < 1] = 1 + x,$ you're asserting that, in expectation, if the first step doesn't get all the way to $1,$ then the remaining steps have to get all the way to $1$. This is clearly false - if $T_1 = 1/2,$ then $T_2 + \dots T_T$ only have to get up to $1/2$. Since $T_1 > 0$ a.s., we should have $\mathbb{E}[T|T_1 < 1] < 1 + \mathbb{E}[T].$ – stochasticboy321 Apr 13 '19 at 19:00
• I noticed that you wanted a non-heuristic proof - note that $\{T > n\} = \{S_n < 1\}$, since the $S_n$ are non-decreasing. It should be straightforward to figure out $\pi_n := P(S_n<1)$ by establishing a recurrence relation between $\pi_n$ and $\pi_{n-1}$ - you'll likely need the volume of a standard simplex in $n$ dimensions. This will directly give you $P(T = n),$ which you can then use to find the mean. – stochasticboy321 Apr 13 '19 at 21:19
• A more high level argument is from queuing theory - suppose it takes you $\mathrm{Exp}(1)$ time to do a job. How many jobs will you finish in $1$ unit of time? It is a classical result that this number is $\mathrm{Poission}(1)$ distributed. However, intuitively, the mean is simpler to argue - your rate of doing jobs is $1$ per unit, so you, in expectation, should finish one job per unit (a Little's law type argument). You are interested in this number plus one - which job will you be doing when the time unit finishes. – stochasticboy321 Apr 13 '19 at 21:24
• @stochasticboy321 Thank you! I was able to do a brute force calculation along these lines. – Flowsnake Apr 14 '19 at 16:24
• ^That's grand, you're welcome :). You should add an answer below, so that others trying the same problem can have a reference. – stochasticboy321 Apr 15 '19 at 1:59 | {
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Following @stochasticboy321's approach, we want to find $$P(T > n) = P(S_n < 1)$$. Since $$S_n = T_1 + ... + T_n \sim$$ Gamma($$n,1$$), we have,
$$P(S_n < 1) = \frac{1}{\Gamma(n)}\int_0^1x^{n-1}e^{-x}dx = 1 - \frac{\Gamma(n,1)}{\Gamma(n)}$$
where $$\Gamma(n,1)$$ is the incomplete Gamma function. To get this expression, I used the nice identity found here. Finally,
$$E[T] = 1 + \sum\limits_{n=1}^\infty P(T > n) = 1 + \sum\limits_{n=1}^\infty P(S_n<1) = 1+ \sum\limits_{n=1}^\infty\left(1 - \frac{\Gamma(n,1)}{\Gamma(n)}\right)$$
I have no idea how to evaluate the sum in closed form, but a computation in Wolfram Alpha seems to suggest it converges to $$1$$. Thus, $$E[T] = 2$$ (at least by conjecture from this numerical computation).
• If we set the integral above to $I_n,$ then by integration by parts, and using $\Gamma(n) = (n-1) \Gamma(n-1) = (n-1)!,$ we get for $n \ge 1,$ $$P(T > n) = \frac{I_n}{\Gamma(n)} = \frac{(n-1) I_{n-1}}{(n-1) \Gamma(n-1)} - \frac{e^{-1}}{{(n-1)!}} = P(T > n-1) - \frac{e^{-1}}{(n-1)!}$$ But then $$P(T = n) = P(T> n-1) - P(T > n) = \frac{e^{-1}}{(n-1)!}.$$ We immediately have that $T \overset{\mathrm{law}}= 1 + Z,$ where $Z \sim \mathrm{Pois}(1),$ and so has mean $2$. Implicitly this also solves the series above (by summation by parts). – stochasticboy321 Apr 15 '19 at 19:35 | {
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+0
+1
104
16
+365
Guys please solve this question and let me know the correct option
please its a request here the link of question if you cant see it (https://ibb.co/ZWftpcR)
Mar 22, 2021
edited by kes1968 Mar 22, 2021
#1
+9279
+2
(Assuming that log means natural log)
y = 5log x
Take the log of both sides
log y = log( 5log x )
Now we can apply the rule that says: log(xy) = y log x
log y = log x log 5
Divide both sides by log 5
log y / log 5 = log x
Do the reverse of natural log to both sides (make both sides the exponent of e )
e^( log y / log 5 ) = e^log x
Now the right side simplifies to just x
e^( log y / log 5 ) = x
So we have...
$$x\ =\ e^{\frac{\log y}{\log 5}}\\~\\ x\ =\ (e^{\log y})^{\frac{1}{\log 5}}\\~\\ x\ =\ y^{\frac{1}{\log 5}}$$
Mar 22, 2021
#2
+365
+2
brother what you have found is the same function just in terms y , but inverse function is symmetrical about y=x axis ,
also f(f^-1(x))=x which is not satisfied ,
by the way if i do the same operation and interchange x and y i get 2 and 4 option , that satifies both conditions
Mar 22, 2021
#3
+9279
+3
The inverse function can be graphed by: $$y=x^{\frac{1}{\log5}}$$
Here's a graph:
https://www.desmos.com/calculator/aijpzpouk1
In other words...
If $$f(x)=5^{\log x}$$ then the inverse is $$f^{-1}(x)=x^{\frac{1}{\log 5}}$$
And...
$$f(f^{-1}(x))\ =\ f(x^{\frac{1}{\log 5}})\ =\ 5^{\log(x^{\frac{1}{\log 5}})}\ =\ 5^{\frac{\log x}{\log 5}}\ =\ 5^{\log_5 x}\ =\ x$$
The options leave it in the form that is solved for x, and so I left it like that to match the options
Mar 22, 2021
#4
+365
+2
brother so what you feel (as a math expert) should be the right answer to the given question ,
Mar 22, 2021
#5
+9279
+1 | {
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Mar 22, 2021
#5
+9279
+1
I'm not sure what you meant by "by the way if i do the same operation and interchange x and y i get 2 and 4 option , that satifies both conditions"....but if this didn't answer your question then please feel free to ask for more clarification!
And I think the answer is option 1: $$x=y^\frac{1}{\log 5}$$
Mar 22, 2021
#6
+365
+3
brother , if you see your second answer and change it in terms of x , wont you get option 2 and option 4 ,
Mar 22, 2021
#7
+9279
+1
Hmm...actually....I see what you mean.... (maybe I made the question harder than it has to be!)
I take back my original answer!! Now I think the answer is option 4
Mar 22, 2021
#8
+365
+1
brother do you really believe its option 4 or just to keep my heart , please clarify if you still believe the answer is 1 , if yes then please prove the same to me as well
Mar 22, 2021
#9
+1740
+1
For computing the inverse function, the plan I know is
1. interchange x & y
2. solve for y
but actually, after 1. you already have a term for the inverse function. It's just not written in the "usual" way, wich is y=f(x).
Answer for is exactly what you get after interchanging x&y, so the correct answer is answer 4.
Mar 22, 2021
#10
+9279
+2
I do agree Probolobo, but then the confusing thing is that
$$5^{\log y}\ =\ y^{\log 5}$$
Which means option 2 and option 4 are the same function and so are equivalent....
Mar 22, 2021
#12
+1740
+1
That's correct, then there are actually 2 correct answers, 2 & 4. Didn't see that equivalence on first glance ;)
Probolobo Mar 22, 2021
#11
+365
+1
so whats the final answer so that i can challenge the answer key?
Mar 22, 2021
#13
+9279
+1
If I had to guess right now, I would guess option 4. My next guess would be option 1.
But I am honestly not sure!! I think this is a bad question.
Can you let us know when you find out the "correct" answer according to the answer key?
hectictar Mar 22, 2021
#14
+112907
+3
I think you are making hard work of it | {
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hectictar Mar 22, 2021
#14
+112907
+3
I think you are making hard work of it
the inverse of
$$y=5^{logx}$$
is simply
$$x=5^{logy}$$
You just have to switch the x and y over.
there would be restrictions on x and on y but the question isn't worrying about that.
Here is the graphs
https://www.desmos.com/calculator/4ftfa2bny7
See they are reflections of each other about y=x
Mar 22, 2021
#15
+365
+2
another question , another superb explanation from mod
Melody you are simply awesome ! thanks for the help mate
kes1968 Mar 22, 2021
edited by kes1968 Mar 22, 2021
#16
+112907
+1
You are welcome Kes
Melody Mar 22, 2021 | {
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# Five people get on the elevator that stops at five floors. In how many ways they can get off?
Five people get on the elevator that stops at five floors. In how many ways they can get off? For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor. In how many ways they can get off at different floors? Now, consider that people in elevator have names, say A,B,C,D and E., assuming that, for example, the case A on the 1st floor is different from the case B on the 1st floor. Answer the previous questions with this assumption.
I was told that there are 4 questions, but I'm not really sure about it.
1. People don't have names and get off
2. People don't have names and get off at different floors
3. People have names and get off
4. People have names and get off at different floors.
Do the names matter?
I think I can answer one question out of 4 - the number of all outcomes should be $5^5=3125$. I am confused about the rest though
• I think I can answer one question out of 4 - the number of all outcomes should be 5^5=3125. I am confused about the rest though. – Gabriel Oct 6 '17 at 16:33
• Wait, there are 4 questions here? I only see two: one where you do differentiate between the 5 people and one where you don't. And yes, for the second one you indeed get $5^5$. – Bram28 Oct 6 '17 at 16:36
• Do you know about Stars and Bars? – Bram28 Oct 6 '17 at 16:37
• Thank you!! I was told that there are 4 questions, but I'm not really sure about it. 1. People don't have names and get off, 2. People don't have names and get off at different floors, 3. People have names and get off, 4. People have names and get off at different floors. Do the names matter? And thank you, I will definetely read about Stars and Bars!! – Gabriel Oct 6 '17 at 16:52
1. This has been answered in other answers: ${9\choose4}$. | {
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1. This has been answered in other answers: ${9\choose4}$.
2. Since there are $5$ people and $5$ floors there is just $1$ way for them to get off at different floors: $1$ per floor.
3. You already found $5^5=3125$.
4. At each floor exactly one of $A$, $\ldots$, $E$ gets off. This can be done in $5\cdot4\cdot3\cdot2=5!=120$ ways.
For the first question, use the basic 'stars and bars' method.
We can represent your example scenario ("For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor.") as:
$$*||**||**$$
The people are the 'stars' and the separators between the different floors are the 'bars'.
Another example: one person getting off on each floor would be:
$$*|*|*|*|*$$
And everyone getting off on the 3rd floor would be:
$$||*****||$$
See how this works?
Now, with $4$ 'bars' to divide the people into the $5$ groups/floors, and those bars taking up $4$ of the $9$ possible positions in this 'symbol string' of bars and stars, you get
$$9 \choose 4$$
possibilities.
• Thank you so much for the explanation!! I think I understand how it works now. – Gabriel Oct 6 '17 at 16:54
• @Gabriel You're welcome! Your instructor did not teach 'stars and bars' yet? ... Because this was a typical stars and bars question! :) – Bram28 Oct 6 '17 at 16:56
• Sometimes they just give the formula derived from stars and bars which is $\dbinom{n+r-1}{r-1}$ in this case $n=5$ and $r=5$ of course, the integer solutions for the equation $x_1+x_2+x_3+x_4+x_5=5$ are considered there are no left and right boundries (luckily). Of course giving the formula without the logic is problematic but I haven't seen one problem that required the logic rather than the formula. – Deniz Tuna Yalçın Oct 6 '17 at 20:21 | {
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The question could also be solved using a well known formula derived from stars and bars which is; $$\dbinom{n+r-1}{r-1}$$ where we distribute $r$ equally accepted objects to $n$ spaces with or without restrictions like $\geq1$ or $\leq9$, this formula is also used for solving linear equations (in integers) that's why this works for combinatoric problems, so without making it longer;$$x_1+x_2+x_3+x_4+x_5=5$$ so we apply the formula, we can think that this is equivalent to distributing five equally accepted apples to five children for instance, the result is $$\dbinom{5+5-1}{5-1}=126$$ this statement could be proven using alternate ways, we should open a post about it too. | {
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# Have you ever used Scratch pad like this?
Find the root of following equation.
$4y^3-2700(1-y)^4=0$
Above equation can be solved using newton raphson iterative method with initial approximation to be unity.
1) First Let $f(y)=4y^3-2700(1-y)^4$ We have to find root of above function.
2)Find derivative of $f(y)$. Here,$f^{'}(y)=12y^2+10800(1-y)^3$
3)Find $y_n$ using newton raphson method $y_0=1$ $y_{n+1}=y_n-\frac{f(y_n)}{f^{'}(y_n)}$ $y_{n+1}=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
You can use calculator or scratchpad to calculate $y_1,y_2,y_3,....$.If this equation has real solution ,these values $y_1,y_2,y_3,....$ would converge to root of $f(y)$.
How to do such iterations smartly?
1) $\color{#69047E}{\text{On first line of scratch pad write y=1}}$
2) $\color{#D61F06}{\text{On second line write}}$
$y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
$\color{#20A900}{\text{You will see new value of y as result.}}$
3) $\color{#EC7300}{\text{From line 3, continue pasting}}$
$y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
until you get a constant value.If your initial approximation is not too away from actual solution,you would get constant value in no more than 10 steps.
By this method you can solve most of the transcendental equation.
$\color{#3D99F6}{\text{By using scratchpad, try finding sum of initial 10 fibonacci number.}}$
Note by Aamir Faisal Ansari
5 years, 4 months ago
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Yes !i have used it this way.Sometimes in Recursion problems and like that.
- 5 years, 4 months ago
By using scratchpad, try finding sum of initial 10 fibonacci number.
- 5 years, 4 months ago | {
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# pre calculus
posted by joe
cube roots of 64i?
1. MathMate
Convert the given number into the form
a+bi = α(cos(β)+i sin(β))
where
α = sqrt(a^2+b^2)
β = sin-1(b/α)
If the number is plot in the complex (Z) plane, it will be more evident.
For example, 64i will have
α=64 (the distance from origin)
β=90° (sin-164/64=1)
The equivalent angles are:
90°
450°
810°
(we only need 3)
Divide each of these angles by , to get β1, β2, & β3. And take α'=α^(1/3)=64^(1/3)=4
The three cube roots are then
z1=α'(cos(β1)+isin(β1)
=4(cos(30°)+isin(30°))
z2=4(cos(150°)+isin(150°))
z3=4(cos(270°)+isin(270°))
Check by expanding z1^3, z2^3 and z3^3 to get back 64i.
2. MathMate
β=90° (sin-164/64=sin-11=π/2)
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express the roots of unity in standard form a+bi. 1.) cube roots of unity 2.) fourth roots of unity 3.) sixth roots of unity 4.) square roots of unity
8. ### Pre calculus
For the equation 2x^2-5x^3+10=0 find the number the number of complex roots and the possible number of real roots.
9. ### Pre-Calculus
Rewrite each quadratic equation in the form ax^2+bx+c=0. Then,use technology to solve each by graphing. ROund you answers to the nearest hundredth, where necessary. a) 3x^2+30 = -19x Answer: 3x^2+19x+30 Roots: x = -3 b) 6x^2= 25x-24 …
10. ### Pre-Calculus 11
What is the entire radial form of -3* cube root of 2? | {
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What is the entire radial form of -3* cube root of 2?
More Similar Questions | {
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# What is the minimum number of rotations about axes in a plane that can describe an arbitrary rotation in 3D?
I'm trying to decompose an arbitrary rotation of a 3D sphere into a series of rotations about any axes that lie in the equatorial plane.
I know I can factor the arbitrary rotation into three rotations in the equatorial plane using Euler angle formulas. Can I factor into two rotations if I allow any choice of axis in the equatorial plane? If it exists, is this factorization unique?
## migrated from mathoverflow.netAug 2 '16 at 14:05
This question came from our site for professional mathematicians.
• There is a unique rotation mapping one unit vector into a distinct vector. Take one of the vectors of your basis, and apply this rotation. Now hopefully you can see that once one axis is coincidental, you only need another rot. in the orthogonal plane. This means at two rotations are necessary and sufficient in general. – Real Aug 2 '16 at 3:48
• what do you understand by equatorial plane (is it the xy-plane)? – Manfred Weis Aug 2 '16 at 7:22
• @ManfredWeis Yes by equatorial plane I mean the xy-plane. – Talon Chandler Aug 2 '16 at 12:28
• Talon, I thought the answer is yes and that I have a simple proof for it. But the question might not be appropriate for this site. If you ask over at Mathematics StackExchange and then let me know here (with a link) when you have posted, I'd be happy to give more details. – user43208 Aug 2 '16 at 13:49
• On second thought, let me migrate it for you... – user43208 Aug 2 '16 at 14:05
My answer is that you can.
If you know about quaternions, then you know that a rotation about a unit vector $v \in \mathbb{R}^3$ through an angle $\theta$ can be accomplished by regarding elements $w = (a, b, c) \in \mathbb{R}^3$ as purely quaternionic elements $ai + bj + ck$, and then mapping $w \mapsto uwu^{-1}$ where $u = \cos(\frac{\theta}{2})\cdot 1 + \sin(\frac{\theta}{2})v$. Such $u$ are precisely quaternions of unit norm. | {
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Let us denote the rotation above by $R_u$ (so $R_u(w) := uwu^{-1}$). Notice that a composition of two rotations $R_u \circ R_v$ is just $R_{u v}$; this follows by associativity of quaternionic multiplication.
So we can translate your problem into the following: for any rotation $R_w \in SO(3)$ given by a nonzero quaternion $w$, show that there exist quaternions $u, v$ in the linear span of $1, i, j$ such that $u v = w$. For these $u, v$ describe rotations about vectors in the equatorial plane spanned by $i, j$.
Now this is not difficult. Let us write $u = a + bi + cj$ and $v = a' + b'i + c'j$. We compute
$$(a + bi + cj)(a' + b'i + c'j) = aa' - bb' - cc' + (ab' + a'b)i + (ac' + a'c)j + (bc' - b'c)k$$
and so given a nonzero quaternion $w = p + qi + rj + sk$, our task is to cook up parameters $a, b, c, a', b', c'$ such that
$$p = aa' - bb' - cc'$$
$$q = a b' + a'b$$
$$r = ac' + a'c$$
$$s = bc' - b'c$$
In fact, let's make our life easier and simply set $b = 0, b' = 1$. Then $a = q$, $c = -s$, and we can solve for $a', c'$ in the linear system
$$p = qa' + sc'$$ $$r = -sa' + qc'$$
provided the determinant $q^2 + s^2$ is nonzero, i.e., provided one of $q, s$ is nonzero. If both $q, s = 0$, then of course $w = p + rj$ corresponds to a rotation about the vector $j$ which is already in the equatorial plane, so there was nothing to do in this case. | {
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Once you have solutions $u = a + bi + cj, v = a' + b'i + c'j$ to the equation $uv = w$ in your hands, the rotations $R_u, R_v$ are rotations about the vectors $bi + cj$ and $b'i + c'j$, respectively. If you want the rotation angles, then normalize $u$ and $v$ (i.e. divide by their norms $(a^2 + b^2 + c^2)^{1/2}$ and $(a')^2 + (b')^2 + (c')^2)^{1/2}$) and use the fact that $R_u = R_{\frac{u}{\|u\|}}$. You can then read off the desired angles by writing e.g. $\frac{u}{\|u\|} = \cos(\frac{\theta}{2}) + \sin(\frac{\theta}{2})\frac{bi + cj}{(b^2 + c^2)^{1/2}}$, so for instance $\cos(\frac{\theta}{2}) = \frac{a}{(a^2 + b^2 + c^2)^{1/2}}$.
By the way, this solution also shows that the decomposition is in general non-unique. For example, we could have also chosen $c = 0, c' = 1$ and arrive at quaternionic solutions. | {
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• Do we need to constrain $u$, $v$, and $w$ to be unit quaternions so they represent rotations? If this is the case, then we have 7 equations with 6 unknowns which might have a unique solution. – Talon Chandler Aug 2 '16 at 18:10
• No, you don't need to, but you can assume they are unit quaternions if you like because $R_u = R_{\frac{u}{\|u\|}}$ for any nonzero quaternion $u$. If $w$ is constrained to the unit sphere in the quaternions (3-dimensional $S^3$) and $u, v$ are each constrained to the unit sphere in the span of $\{1, i, j\}$ (so $(u, v)$ parametrizes a space that looks like $S^2 \times S^2$, which is 4-dimensional), then we are asking about surjectivity of a multiplication map $S^2 \times S^2 \to S^3$, so 3 equations in 4 variables by my count. – user43208 Aug 2 '16 at 18:24
• Okay that makes sense. What if we consider only infinitesimal rotations? Can we uniquely decompose an arbitrary rotation into infinitesimal rotations in the plane spanned by $i,j$? In this case we can make the approximations: $$\cos\left(\frac{\theta}{2}\right) \approx 1$$ $$\sin\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}$$ So $p=a=a'=1$ and we have four equations with three unknowns which gives a unique solution---i.e. an infinitesimal rotation about an arbitrary axis can be decomposed uniquely into two infinitesimal rotations about axes in the span of $i,j$. – Talon Chandler Aug 2 '16 at 19:28 | {
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• Well, the answer to the third sentence in your last comment would be 'no'. But why is getting uniqueness important to you? I think in fact the method I gave you gives virtual uniqueness of a solution if you constrain the first rotation $R_u$ to rotate about $j$. That is: unit quaternions $u$ of the form $a + bj$ form a circle $S^1$, and the unit $v$ for rotations about a second axis in the $ij$-plane form a sphere $S^2$, and the multiplication map $S^1 \times S^2 \to S^3: (u, v) \mapsto w = uv$ is onto, and unless $w$ itself is a rotation about $j$, only one pair $(u, v)$ satisfies $w = uv$. – user43208 Aug 2 '16 at 20:35
• I'm interested in uniqueness because I'm working on a reconstruction problem. I have measurements that correspond to a known rotation about an arbitrary axis, and I want to reconstruct the set of rotations in the $ij$-plane that gave rise to the "total" rotation. I'm hoping to find the conditions under which this reconstruction is possible. Can you expand on why we can't decompose an arbitrary infinitesimal rotation into two unique infinitesimal rotations in the $ij$-plane? – Talon Chandler Aug 2 '16 at 21:03 | {
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The answer is no, because when letting $v$ be an arbitrary direction about which to rotate by an angle $\rho$, that rotation can only then be realized with a rotation around an axis in the xy-plane if $v^Tz=0$; that shows that at least three rotations are necessary
1. rotate about $v\times z$ to bring $v$ into the $xy$-plane
2. rotate about the image of $v$ in the $xy$-plane
3. rotate the image of $v$ back around $v\times z$ to bring the image back to the original direction of $v$
• I agree that this set of rotations in the xy-plane can describe an arbitrary rotation, but how can we be sure there isn't a shortcut with just two rotations? – Talon Chandler Aug 2 '16 at 12:32 | {
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How can I find the time when two people agree to meet if both of their watches have an offset?
Chemist116
The problem is as follows:
Jenny and Vincent agreed to meet at the library at $6\,p.m$. Both synchronized their watches at midnight ($\textrm{0 hours}$). We know that Vincent's watch is not functioning correctly and gets ahead of the real time $\textrm{50 seconds}$ each hour and Jenny's watch gets delayed from the real time $\textrm{50 seconds}$ each hour. Vincent arrived to the library $\textrm{15 minutes}$ before the agreed time accoring to his watch and Jenny $\textrm{15 minutes}$ late by looking at her watch. Using this information, find how long did Vincent waited Jenny?
The alternatives according to my book are as follows:
$\begin{array}{ll} 1.&\textrm{15 min}\\ 2.&\textrm{0 min}\\ 3.&\textrm{25 min}\\ 4.&\textrm{60 min}\\ \end{array}$
This part I'm stuck at exactly how should I make up an equation which can relate both offsets. Can someone help me here? I'm assuming that by $\textrm{6 p.m}$ the time which would be seen by Vincent will be:
$18\times 50= 900\,s$
which would be $60$ minutes
But Vincent seen in his watch was:
$\textrm{5:45 pm}$
hence until 5 pm would be:
$17\times 50=850\,s$
$56\frac{2}{3}$ minutes
and Vincent's watch would have seen:
$\textrm{5:56:40 pm}$
At this point I could try guessing reducing the number of minutes until adjusting the time which will be seen by Vincent and this would be same for Jenny, but it doesn't seem something which can be effective. Can someone help me here? How exactly can I find what is being requested?
skeeter
Math Team
50 seconds an hour = 5 minutes in 6 hours
Vincent’s watch runs fast, which means the real time he arrives is earlier than 17:45.
Jenny’s watch runs slow, which means the real time she arrives is later than 18:15.
The only answer choice that makes sense is the 60 minutes Vincent waited.
I worked out the problem using proportions, where T represents the actual time | {
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I worked out the problem using proportions, where T represents the actual time
for Jenny ...
T/18.25 = 72/71
for Vincent ...
T/17.75 = 72/73
topsquark
Chemist116
50 seconds an hour = 5 minutes in 6 hours
Vincent’s watch runs fast, which means the real time he arrives is earlier than 17:45.
Jenny’s watch runs slow, which means the real time she arrives is later than 18:15.
The only answer choice that makes sense is the 60 minutes Vincent waited.
I worked out the problem using proportions, where T represents the actual time
for Jenny ...
T/18.25 = 72/71
for Vincent ...
T/17.75 = 72/73
This is the major problem which I had with this question. It mentioned that until his time was 17:45, not 18:00. How would be the difference between those fractions? Can you help me with that part? | {
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# Combinations: How many handshakes?
I need help with the following question which I cannot seem to solve:
17 students are sitting in a circle. Each person shakes hands with everyone but his/her neighbours. How many handshakes have been exchanged?
My approach: no. of ways $= 1 + 2 + 3 + ... + 14 = 7(15) = 105$.
Apparently my answer is wrong (correct ans is 119). But I can't seem to understand why. Could someone please explain?
• The reason why this is wrong is as follows: the first person has 14 people to shake hands with. The second person as well has 14 people to shake hands with, since he wouldn't shake hands with his neighbour anyway! So, the sum is 14+(14+13+...+1), which does add up to 119. – Sanchises Feb 16 '15 at 19:49
Each of 17 people shakes hands with 14 people (all except themselves and their 2 neighbors), so there are
$$\frac{17\times 14}{2} = 119$$
handshakes (dividing by 2 to account for symmetry, as you would otherwise count "$A$ shaking hands with $B$" and "$B$ shaking hands with $A$" as distinct events).
• Does it mean that we don't care about how they sit, say in a circle or in a row? – Nighty Feb 16 '15 at 15:04
• @LeeKM: The difference is that in a circle, everyone has two neighbors, while in a row, the first and last person have only one each. – user139000 Feb 16 '15 at 15:05
• Thanks! Now I know that I misunderstood the quuestion. I thought that the handshakes were to be made with distinct persons. – Donald Feb 16 '15 at 21:44
Here's another way to think about it.
If everyone shook hands with everyone you'd have $\frac {17*16}2 = 136$ handshakes (divide by 2 because the above is double counting A shaking hands with B and so forth).
However, there are $\frac {17*2}2 = 17$ handshakes that aren't happening because neighbors aren't shaking hands (2 neighbors for each of the 17 people, again divided by 2 to remove duplicates).
So $136-17 = 119$ | {
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So $136-17 = 119$
Pew's response is a little more direct, but finding the number by calculating the total possibilities minus the "not allowed" interactions is sometimes a little more intuitive for some people.
You can think of this in graph theory as the following:
$G$ has $17$ vertices with degree $14$, since there are no loops(people self handshaking), and no vertices are adjacent to the vertices beside them. So total degree is $17\times 14$.
We know that degree is equal to $2\times\text{number of edges}$ and hence there are $\frac{17\times 14}{2}=119$ edges in total, where edges represent handshakes.
As is typical for problems with big numbers, you should always resort to a smaller number if you can't solve the full problem.
Here, 17 people at the table is a bit hard to imagine immediately.
How many handshakes now?
There are zero.
A B
C D
A can't shake with B or C, B can't with A or D, D can't with B or C, C can't with A or D--only A & D and B & C can shake.
So 2 shakes:
A B
X
C D
5 people is when it gets interesting, and when you should be able to see the pattern:
B
A C
D E
Everyone has two people they can't shake with, leaving two shakes per person--but we overcount by just multiplying 5 & 3, so we divide by two.
This is easiest to see by trying to draw the graph mentioned by @Commitingtoachalleng -- draw a line between any people who can shake hands:
B
/ \
A-----------C (also C-D & A-E--a 5-point star)
/ \
D E
So we hypothesize the answer is $\frac{n(n-3)}{2}$. Note that this formula holds for $n=3$ and $n=4$ as well, as we'd hope!
One final way to see this is to look again at the graphs--
you might notice that they're always complete graphs (every vertex connected to every other vertex) with the outer edges removed.
Since there are $\frac{n(n-1)}{2}$ edges in a complete graph on $n$ vertices (which you can confirm yourself by a similar process), and $n$ outer edges, | {
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there must be $\frac{n(n-1)}{2} - n = \frac{n^2-n-2n}{2} = \frac{n(n-3)}{2}$ handshakes.
However you cut it, there are $\frac {17 \cdot 14} 2 = 119$ total handshakes.
• We should also point out that the second option (working backwards from complete graphs) is the graph analogue of @Duncan's solution – MichaelChirico Feb 17 '15 at 6:42
I will explain for 6, same logic applies for all the numbers.
first, if 6 persons shakes will all in the party, then number of shakes will be:
A B C D E F
A can shake with only rest of 5 persons(B,C,D,E,F) i.e, right side ppl. = 5
B can shake with only his right sides(C,D,E,F) ppl(not with A bcos,A already shook with B just now. so no repeat) = 4
C can shake with only his right sides(D,E,F) ppl(not with A&B bcos,A&B already shook with C just now. so no repeat) = 3
D can shake with only his right sides(E,F) ppl(not with left side ppl bcos, they already shook with D just now. so no repeat) = 2
E can shake with only his right sides(F) ppl(not with left side ppl bcos, they already shook with E just now. so no repeat) = 1
as all of the persons shook hand with F, He dont shake with any one now.
So total possible shakes for 6 persons are = 5+4+3+2+1 => 15
now in questions it is given that, No one shakes hand with neighbors.
Imagine ppls sitting around table
A B
F C
E D
Please treat above image as 6 Ppl around circle : So now,
Invalid Shakes:
A cannot shake with B,F . I am representing it as below : A -no- B & F => 2 invalid shakes.
B -no- C only. Bcos B -no- A is covered as part of first case.
=> 1 invalid shake
C -no- D only. => 1 invalid shake
D -no- E only. => 1 invalid shake
E -no- F only. => 1 invalid shake
F dont shake with A is already covered as part of first invalid shakes.
so total invalid shakes are 2+1+1+1+1 = 6
so for 6 ppl: out of total 15 , 6 are invalid. So 15-6 = 9
on the same line :
for 17 ppl: out of total (16+15+...+1)=136 , 17 are invalid. So 136-17 = 119 | {
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on the same line :
for 17 ppl: out of total (16+15+...+1)=136 , 17 are invalid. So 136-17 = 119
shortcut : For N ppl, Total shakes are (N-1)(first term + last term)/2 Arithmetic progression sum. out of which Invalids are N.
So valid shakes => (Total - N)
• What is the point of posting such a complex answer to a question with a simple and accepted answer posted almost four years ago? – José Carlos Santos Oct 31 '18 at 17:23 | {
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# Is this function injective / surjective?
A question regarding set theory.
Let $g\colon P(\mathbb R)\to P(\mathbb R)$, $g(X)=(X \cap \mathbb N^c)\cup(\mathbb N \cap X^c)$
that is, the symmetric difference between $X$ and the natural numbers.
We are asked to show if this function is injective, surjective, or both.
I tried using different values of $X$ to show that it is injective, and indeed it would seem that it is, I can't find $X$ and $Y$ such that $g(X)=g(Y)$ and $g(X) \neq g(Y)$ but how do I know for sure? How do I formalize a proof for it?
Regarding surjective: I think that it is.
We can take $X=\mathbb R - \mathbb N$ and we get that $g(X)=\mathbb R$
What do I do about the injective part?
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are you able to express in formula what injective or surjective means? – Robert.Sie Nov 28 '13 at 14:14
Hint:
$$g\circ g = \mathrm{id}_{\mathcal{P}(\mathbb{R})}, \quad \text{ that is, } \quad \forall X \in \mathcal{P}(\mathbb{R}).\ g(g(X)) = X.$$
I hope this helps $\ddot\smile$
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We can view subsets of $X$ as functions $X \to \{0,1\}$: 0 if the point is not in the set, It is commonly very useful to rewrite questions about subsets as questions about functions.
What form does $g$ take if we do this rewrite?
Terminology wise, if $S \subseteq X$, then $\chi_S$ is the function described above:
$$\chi_S(x) = \begin{cases} 0 & x \notin S \\ 1 & x \in S \end{cases}$$
It is called the characteristic function of $S$.
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0 if $a \in X\cap \mathbb N$, 1 otherwise – Oria Gruber Nov 28 '13 at 14:19
That's not right. To be more clear, $g(\chi_S)$ is a function. To describe this function, we could do so by writing down the equation that says what its value at a point $x$ is: i.e. what is $g(\chi_S)(x)$? – Hurkyl Nov 28 '13 at 14:22
HINT: Recall that symmetric difference is associative and for every set $B$, $B\triangle B=\varnothing$.
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And why was this downvoted? – Asaf Karagila Nov 28 '13 at 22:04 | {
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# Find sum of series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) }$ correct to 3 decimal places.
This is a question I came across while studying on Khan Academy:
Find the sum of the series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) }$ correct to three decimal places.
I solved this by calculating ${ S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 },...\\$ until I had a close enough value (bad method, I know). Anyway, here is the solution steps given in the answer:
Note that this is a convergent alternating series. The Alternating Series Estimation Theorem states that the error bound in the sum of the first $n$ terms is the absolute value of the first omitted term; i.e., $\left| { a }_{ n+1 } \right|$. To obtain three-digit accuracy, we need to find $n$ such that $\left| { a }_{ n+1 } \right| <\ 0.0005$, or equivalently $\frac { n+3 }{ { (n+1) }^{ 5 } } <0.0005$.
When $n=6$, we have $\left| { a }_{ n+1 } \right| =\frac { n+3 }{ { (n+1) }^{ 5 } } =\frac { 9 }{ 16807 } \approx .000535$; that is not accurate enough.
When $n=7$, we have $\left| { a }_{ n+1 } \right| =\frac { n+3 }{ { (n+1) }^{ 5 } } =\frac { 10 }{ 32768 } \approx .000305$; that will give a final result within $0.0005$.
Thus we need to find the sum of the first seven terms of the series.
$$\sum _{ n=1 }^{ 7 }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) \approx 2.8914 }$$
Hence, $2.891$ is accurate to three decimal places.
What I don't understand about this answer is if the estimate is $2.8914$ and the error bound is $0.000305$, I think the actual sum can be close to $2.8914+0.000305\approx 2.891705\approx 2.892$. Then the result above is not correct.
Does this answer take into account the fact that the $(7+1)th$ term is negative? Or is there something about rounding and error bounds that I don't understand?
Thank you. | {
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Thank you.
• Indeed, 2.8914 means between 2.89135 and 2.89145 and 0.000305 is less than 0.0004 hence the sum is between 2.89135-0.0004 and 2.89145, which implies that 2.891 has three correct decimal places. (The same estimates for n=6 show it was not necessary to go to n=7.) – Did Oct 21 '15 at 7:27
• You are right, it is the fact that $\sum_1^7$ is an overestimate that enables us to conclude that $2.891$ is correct to 3 decimal places. – André Nicolas Oct 21 '15 at 7:28
• To the op: If you want it back I'll undelete my answer, but it seems unnecessary given you're already pretty much there yourself. – Adam Hughes Oct 21 '15 at 7:32
• @Did Thank you. But do you mean if somehow the estimate is an underestimate, the sum can be between 2.89135 and 2.89145+0.0004. Then 2.892 would be correct, right? – user52139 Oct 21 '15 at 7:49
• @AdamHughes Thank you very much. I actually need it back if you don't mind, please. – user52139 Oct 21 '15 at 7:52
I would read the question the same way you did-find a three decimal value that you know the true sum of the series rounds to. We know from the alternating series theorem that the truncation error is smaller than the first neglected term in absolute value and of the same sign. There is no truncation error that we can guarantee will be small enough, because we could be poised right on the rounding boundary, with the computed answer something like $2.8915$. Here, because after seven terms we have $2.8914$ and we know the eighth term is negative, the correct answer is in the range $(2.8911,2.8914)$ and this entire range rounds to $2.891$
• +1, warning about problem with rounding boundary is well placed here. This answer might be further improved by pointing out the problem of numerically finding any decimal of $\sum_{n=0}^\infty \frac 1{2^n}$. – Ennar Oct 21 '15 at 10:38 | {
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This is an alternating series, so the error is estimated by the last term. Since you want the error to be less than $10^{-3}$ in size, you need only choose $n$ so that ${n+2\over n^5}<10^{-3}$ since the terms are decreasing.
Solving we see $n+2<10^{-3}n^5$. We claim this is so for $n\ge 7$ as
$$7+2=9<16.807=7^5\cdot 10^{-3}.$$
For $n>7$ we see that by induction we may conclude this with the inductive step being
$$(n+1)+2<10^{-3}n^5+1< 10^{-3}(n+1)^5,$$
Then adding up the first $6$ terms gives you $2.891$ to three decimal places. | {
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