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Concavity [Solved!]
My question
In curve sketching with differentiation, the general curves section, there are no points of inflection written for the quadratic curve.
But from -1 to 1 it goes from negative to positive. Not clear as to why this would not be a point of inflection.
Relevant page
5. Curve Sketching using Differentiation
What I've done so far
My work includes reading and re-reading and re-reading the section Finding points of inflection, and also reviewing the examples given.
X
In curve sketching with differentiation, the general curves section, there are no points of inflection written for the quadratic curve.
But from -1 to 1 it goes from negative to positive. Not clear as to why this would not be a point of inflection.
Relevant page
<a href="https://www.intmath.com/applications-differentiation/5-curve-sketching-differentiation.php">5. Curve Sketching using Differentiation</a>
What I've done so far
My work includes reading and re-reading and re-reading the section Finding points of inflection, and also reviewing the examples given.
Continues below
Re: Concavity
@Phinah: Please have another look at the definition of a point of inflection.
It's where the concavity of the curve changes sign. That is, where it changes from a "U shaped" curve to an "n shaped" curve (or vice versa).
However, a quadratic curve (a parabola) is "U shaped" everywhere (or "n shaped" if the number in front of the x^2 term is negative).
U-shaped (e.g. y=x^2+2x+5)
n-shaped (e.g. y=-x^2-3x+7)
So it won't have a point of inflection.
X
@Phinah: Please have another look at the definition of a point of inflection.
It's where the <b>concavity</b> of the curve changes sign. That is, where it changes from a "U shaped" curve to an "n shaped" curve (or vice versa).
However, a quadratic curve (a parabola) is "U shaped" everywhere (or "n shaped" if the number in front of the x^2 term is negative).
<b>U-shaped</b> (e.g. y=x^2+2x+5) | {
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<b>U-shaped</b> (e.g. y=x^2+2x+5)
[graph]310,250;-4.5,3.5;-0.5,10.3,1,2;x^2+2x+5,[/graph]
<b>n-shaped</b> (e.g. y=-x^2-3x+7)
[graph]310,250;-4.5,3.5;-0.5,10.3,1,2;-x^2-3x+7,[/graph]
So it won't have a point of inflection.
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# Math Help - nxn grid problem.
1. ## nxn grid problem.
Hello.
I asked about this a while ago, but I'm still unsure. Imagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:
x = black o = white
ooo xoo xxo
ooo ooo oxx
ooo ooo oxo
So, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct! Is it 9 squared, giving me 81? What if I had a 6x8 grid, or 17x31 grid?
I did try to host a nicer picture than my x and o, but sadly I can't get imageshack to work.
Many thanks.
N.
2. Originally Posted by theNoodler
Hello.
I asked about this a while ago, but I'm still unsure. Imagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:
x = black o = white
ooo xoo xxo
ooo ooo oxx
ooo ooo oxo
So, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct! Is it 9 squared, giving me 81? What if I had a 6x8 grid, or 17x31 grid?
I did try to host a nicer picture than my x and o, but sadly I can't get imageshack to work.
Many thanks.
N.
An n x n grid with m possible colours has $m^{n \times n}$ possible combinations.
3. Originally Posted by theNoodler
Imagine you have a nxn grid (3x3 for argument). I want to know how I can work out the total number of patterns I can make with 2 colours, say black and white. An example would be:
So, pattern one is all white, patter two has one bit black etc. I was told for this it would be 2 to the 9th, giving me 512, which can't be correct!
But that is correct.
Look at that attached graph. There are two $3\times 3$ grids.
Are they different colorings? | {
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Maybe not. Rotate I $90^o$ counter-clockwise. We get II.
Now are they different?
If I & II are different then there are $2^9=512$ possible colorings.
If I & II are not different then you must tell us why?
4. Hello, theNoodler!
Imagine you have a $n\times n$ grid (3x3 for argument).
How can I work out the total number of patterns made with 2 colours, say black and white.
An example would be: $\begin{array}{c}X \,=\, \text{black} \\ O \,=\,\text{white}\end{array}$
. . $\begin{array}{c}OOO\\OOO\\OOO\end{array} \qquad \begin{array}{c} XOO\\ OOO\\OOO \end{array} \qquad \begin{array}{c}XXO\\OXX \\ OXO \end{array}\qquad \hdots\text{ etc.}$
I was told for this it would be $2^9 \,=\,512$, which can't be correct!
. . Why not?
For each of the 9 cells, you have two choices: place a Black or place a White.
So you have 9 decisions with 2 options each.
. . There will be: . $2^9 \,=\,512$ possible choices you can make.
What if I had a 6x8 grid grid?
You have 48 cells to fill with 2 options each.
There will be: . $2^{48}$ possible choices.
5. Hello.
Thanks for the help. I guess it just doesn't look like there could be that many combinations. So for a 6x8 grid, there would be 281,474,976,710,656 combinations. That's just nuts. | {
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# For a toss of fair die, if events are, A: {1,2}, B: {2,4,6}, and C: {4,5,6}, then A and B are independent but B and C are not. Why?
As far as I understood independence, A and B should not be independent since if either of them happens then we can tell something about the other one. But if they are independent then B and C should also be. From mathematical proof it is explainable, but I didn't understand the intuition behind it.
The example is from “Elementary Bayesian Statistics” by Gordon Antelman, Chapter 2
Text:
For a fair die $$U = (1,2,3,4,5,6)$$. Let three events be defined as:
• $$A \equiv (1,2)$$, so $$P(A)=2/6$$
• $$B \equiv (2,4,6)$$, so $$P(B)=3/6$$, and
• $$C \equiv (4,5,6)$$, so $$P(A)=3/6$$
Then
• $$A \cap B = (2)$$ and $$P(A \cap B) = 1/6 = P(A)P(B) = (1/3)(1/2)$$, so A and B are independent.
• $$B \cap C = (4,6)$$ and $$P(B \cap C) = 2/6 \ne P(B)P(C) = (1/2)(1/2)$$, so B and C are not independent.
• Rename the faces of the die from $1,2,3,4,5,6$ to $(1,0),(1,1),(2,0),(2,1),(3,0),(3,1),$ thereby identifying them with the Cartesian product $\{1,2,3\}\times\{0,1\}$ and notice the distribution is the product of the uniform distributions on these two sets. Writing the coordinates as $(x,y),$ $A$ is the event $x=1$ while $B$ is the event $y=1,$ making their independence intuitively obvious. In some sense, all independent events occur in this way.
– whuber
Jan 16 at 18:56
Other explanation why $$\color{red}{A = \{1, 2\}}$$ and $$\color{blue}{B = \{2, 4, 6\}}$$ are independent:
Someone rolled a die: | {
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Someone rolled a die:
• The probability of the event $$\color{red}A$$ is $$\color{red}{2\mkern-0.1ex/6}$$, i.e. $$\color{red}{1\mkern-0.1ex/3}$$.
• Someone tells you that the result is an $$\color{blue}{\text{even}}$$ number (the event $$\color{blue}B$$). In spite of this new info the probability of $$\color{red}A$$ is still $$\color{red}{1\mkern-0.1ex/3}$$
($$\color{blue}2 \in \color{red}A$$, but $$\color{blue}{4} \notin \color{red}A$$, $$\color{blue}6 \notin \color{red}A$$).
In the opposite way —
• The probability of event $$\color{blue}B$$ is $$\color{blue}{3\mkern-0.1ex/6}$$, .i.e. $$\color{blue}{1\mkern-0.1ex/2}$$.
• Someone tells you that the result is a number $$\color{red}1$$ or $$\color{red}2$$ (the event $$\color{red}A$$). In spite of this new info the probability of $$\color{blue}B$$ is still $$\color{blue}{1\mkern-0.1ex/2}$$
($$\color{red}2 \in \color{blue}B$$, but $$\color{red}{1} \notin \color{blue}B$$).
Note:
You may use the same scheme to see that the events $$\color{blue}B$$ and $$\color{green}C$$ are not independent.
• Thank you for this explanation. So what I understood is: if B happens then the probability of A will still be 1/3 because {2}/{2,4,6} and if A happens then also the probability of B will be same (1/2) because {2}/{1,2}. And in case of B and C, if B happens i.e. {2,4,6} the probability of C will be 2/3 because {4,6}/{2,4,6} and if C happens, the probability of B will also be 2/3 {4,6}/{4,5,6} in which case the probability of C (1/2) and B (1/2) has changed respectively and that's why they are not independent. Please correct me if I am wrong. Jan 16 at 20:37
• @someUser, exactly, nothing to correct. What should I do when someone answers my question?. Jan 16 at 20:40
You probably confuse indepenent events for mutually exclusive events.
... A and B should not be independent since if either of them happens then we can tell something about the other one. | {
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It is not true. What does mean “something”? The probability of “the other one” didn't change!
The correct formulation should be
• “A and B are independent — if either of them happens, then it doesn't change the probability of the other”.
As I wrote, you are probably confused mutually exclusiveness (the empty intersection, distinctiveness) for the independence of two events.
But they are two independent things:
• the events $$\{1\}$$ and $$\{2, 4, 6\}$$ are mutually exclusive, but they are not independent,
• the events $$\{2\}$$ and $$\{2, 4, 6\}$$ are neither mutually exclusive, nor independent,
• the events $$\{1, 2\}$$ and $$\{2, 4, 6\}$$ are not mutually exclusive, but they are independent,
• the events $$\emptyset$$ and $$\{2, 4, 6\}$$ are mutually exclusive, and they are independent.
Independence can arise in two distinct ways:
• we explicitly assume that events are independent (e.g. rolling a die again), or
• we derive independence by verifying that it fulfills the formula $$P(A \cap B) = P(A)P(B)$$.
Generally, there is no way to “see” the independence (e.g. by looking in a Venn diagram).
By other words, if we may “see” the independence of two events, then they certainly fulfill the formula. But the opposite is not true — the formula is fulfilled, the independence is guaranteed, but there is not a visible reason, why.
The idea of independent events arises from the evident, noticable independence, but is not limited to it.
• So how can A $\cap$ B and B $\cap$ C be explained in current scenario? Jan 16 at 19:16
• @someUser, don't try to “see” the independence by interpreting intersections of 2 events. Most teachers don't teach their student what I wrote in my answer — independence is sometime visible (and understandable), sometime not. Jan 16 at 19:34 | {
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# How does probability change the more times you perform a procedure?
I have a question based on fair coins. Every round, two coins are flipped. If both are heads, we say "Success" and end the experiment. What is the probability of saying "Success" in any round i?
Initially, I thought it would be $$\displaystyle\dfrac{\displaystyle\frac{i!}{(i-2)!2!}}{2^i},$$ because of NCR and increasing probability, but I realize it doesn't work.
Right now, it seems as though a summation works as we're increasing chances of getting a "Success" the more rounds we have. We also need to stop once it reaches a "Success", but I don't get how we can stop and operation like that unless it simply reaches 100%. $$\sum_{n=1}^{i} \frac{1}{4}$$
Logically, it sounds right to me, but I'm unsure so I'd like to hear your thoughts on this. The $\displaystyle\frac{1}{4}$ is due to the [TT][TH][HT][HH] probability sample space.
The question continues to ask for an expression on the probability of getting a "Success" in terms on n and i. Here, n stands for the total number of rounds while i stands for the number of rounds before we stop running the procedure.
My solution for this part seems the same as the previous one, as I don't see what the question is asking differently. Any ideas?
And finally, the last part is asking what if we don’t stop after saying “Success", that is we repeat the procedure n times. How many times do you expect to say “Success”, express this in terms of n.
$$\sum_{i=1}^{n} \frac{1}{4}$$
This way, we can easily go over the limit and we'll find the total number of times that "Success" can be reached.
Thank you for reading this :) | {
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Thank you for reading this :)
• What does your summation say about $n=5$? Do you believe a probability can be greater than $1$? – Erick Wong Jun 19 '16 at 17:21
• In terms of the third question, it can continue over and reach as many "Success" stages as possible. What I don't get is how to get the summation to stop when it reaches "Success" in the first subquestion. – Andrew Raleigh Jun 19 '16 at 17:28
First question
The probability of having success in round 1 is $P(X=1)=\frac{1}{4}$. We agree.
The probability of having success in round 2 is $P(X=2)=\left(1- \frac{1}{4}\right)\cdot \frac{1}{4}=\frac{3}{4}\cdot \frac{1}{4}=\frac{3}{16}$
You have to take into acccount that the probability of having a success in round $2$ depends on wether you have success in round 1 or not. Here we need to have no success in round 1.
The probability of having success in round 3 is $P(X=3)=\left(1- \frac{1}{4}\right)^2\cdot \frac{1}{4}=\frac{3^2}{4^2}\cdot \frac{1}{4}=\frac{9}{64}$
Here we need to have no success in round 1 and round 2.
$...$
The probability of having success in round $i$ is $P(X=i)=\left(1- \frac{1}{4}\right)^{i-1}\cdot \frac{1}{4}=\left(\frac{3}{4}\right)^{i-1}\cdot \frac{1}{4}$
Second question
You need the closed form of $\frac{1}{4}\cdot \sum_{i=1}^n \left(\frac{3}{4}\right)^{i-1}$ For $j=i-1$ we get $\frac{1}{4}\cdot \sum_{j=0}^{n-1} \left(\frac{3}{4}\right)^{j}$. This is the partial sum of a geometric series.
Thus $\frac{1}{4}\cdot \sum_{j=0}^{n-1} \left(\frac{3}{4}\right)^{j}=\frac{1}{4}\cdot \frac{1-\left(\frac{3}{4}\right)^{n}}{\frac{1}{4}}=1-\left(\frac{3}{4}\right)^{n}$
Third question
I´m not sure if my understanding is right. But if we don´t stop at the first success the probability of getting $i$ successes and consequently $n-i$ failures is
$P(Y=i)=\binom{n}{i}\cdot \left( \frac{1}{4} \right)^i \cdot \left( \frac{3}{4} \right)^{n-i}$.
This is the pdf of the binomial distribution. And the expexted value is $n\cdot p=\frac{1}{4}n$ | {
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This is the pdf of the binomial distribution. And the expexted value is $n\cdot p=\frac{1}{4}n$
• That makes a lot of sense, thank you! In this case, the probability of having a success at round i is $\left(\frac{3}{4}\right)^{i-1}\cdot \frac{1}{4}$, but would the same thing apply to the second subquestion? – Andrew Raleigh Jun 19 '16 at 17:49
• @AndrewRaleigh I´ve made an edit. – callculus Jun 19 '16 at 18:09
• @callculus Small typo in the second line: RHS should be $3/16$ instead of $1/16$. – Erick Wong Jun 19 '16 at 18:27
• @callculus I'm sorry, but the top part the answer for the first question and the bottom the second, right? If so, can I ask how you thought of the way to answer the question? – Andrew Raleigh Jun 19 '16 at 18:29
• @ErickWong Thanks for the comment. – callculus Jun 19 '16 at 18:30 | {
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# Understanding Proof by Contrapositive
This is a question from my textbook, and I'm pretty sure I have the answer, but I am having trouble writing out what I am thinking.
Prove: If $n$ is an integer and $3n+2$ is odd, then $n$ is odd.
Contrapositive: If $n$ is even, then $3n+2$ is even.
$n$ is even, by definition $n = 2k$, where $k$ is an integer.
Plugging in $2k$ into $3n+2$, we have $3(2k)+2$.
$3(2k)+2 = 6k+2 = 2(3k+1)$, which means $3n+2$ is even.
Since if $n$ is even, then $3n+2$ is even. Negating the conclusion of the original conditional statement implies that it's hypothesis "$3n+2$ is odd" is false, therefor if $3n+2$ is odd, $n$ is odd.
What am I missing, or what can I do to make this less awkward? Or am I wrong about the proof on the most basic levels?
-
It can be proved directly: $\rm\ 3n\!+\!2\,=\, 2k\!+\!1\,\ is\ odd\:\Rightarrow\: n\,=\,2(k\!−\!n)-1\,\ is\ odd.$ – Math Gems Feb 26 '13 at 6:03
## 1 Answer
Your proof is correct, though you could improve on it by writing it in essay style, cutting down on some verbosity, and by mentioning that $3k+1$ is also an integer. For example, you could write something like this:
We try to prove the contrapositive. If $n$ is even, then $n=2k$ for some integer $k$, so $3n+2=3(2k)+2=6k+2=2(3k+1)$ where $3k+1$ is an integer, making $3n+2$ even.
Note that proofs can be written with varying degrees of detail. In beginning courses, instructors prefer to see more detailed proofs to ensure students really understand what they are writing.
-
Up to how much? Do I remove everything after "Since if n is even, then 3n+2 is even."? – Chase Feb 24 '13 at 1:47
So it's better to write out explicitly what I have up there? (As in, "3(2k)+2=6k+2=2(3k+1), which means 3n+2 is even." should have been a bit more spelled out) – Chase Feb 24 '13 at 1:55
I see. I have trouble trying with proofs and explaining proofs so thank you! – Chase Feb 24 '13 at 2:02 | {
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+0
# [Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possib
0
63
7
+7
I attacked the problem by first setting the obvious as P(x) = c1xn+c2xn-1+.....+cn with the c's being the constants.
P(1) = 5 tells me that the sum of all the coefficients of the polynomial is 5, and immediately a matter to observe would be n < 4 because the next part, P(P(1)) = 177 means P(5) = 177 and 54 > 177 for obvious reasons.
So P simplifies to P(x) = x3+c1x²+c2x+c3. It's trivial to see x3 won't have a coefficient. My next steps involved basically trial and error-ing as the sum of coefficients = 5 and using every different case where I change the degree, remove some terms and overall experiment with the coefficients and x terms.
In the end, I found P(x) = x3+2x²+2 to be the only polynomial satisfying the given conditions and the answer to the question is P(10) = 103+2(10)²+2 = 1202.
My problem is I used a trial and error approach to the problem and found my solution, which I'm unsure is the only solution. Is there a more elegant approach to the problem and is my solution correct? Thanks~
Jul 17, 2022
#1
+197
0
The problem got cut off in your title, could you please show us the complete question?
Jul 17, 2022
#2
+117766
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The question has to be in the question section.
A suitable title might have been
Polynomials, sum of roots (I assume you want to find the sume of all possible roots..)
It is really good that you have tried to show what you did :)
Jul 17, 2022
edited by Melody Jul 17, 2022
edited by Melody Jul 17, 2022
#3
+7
+1
[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?
this was the whole question
Jul 17, 2022
#5
+2
Question: | {
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this was the whole question
Jul 17, 2022
#5
+2
Question:
[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?
To begin with, your solution is correct, and is actually a quick way to solve this problem. That is, It is efficient.
However, I think what you are trying to say, is there a way to write the solution "more rigorously", and the answer to this is: yes.
Let $$P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$$, where $$a_n,a_{n-1},...,a_0 \ge 0$$ (Given.)
Since: $$P(1) = 5 \iff a_n+a_{n-1}+...+a_0=5$$, which does not tell us that much.
Next, $$P(P(1)) = P(5) =a_n*5^n+a_{n-1}*5^{n-1}+...+a_0=177$$
Notice: $$5^3=125,5^4=625$$; hence, $$P(x)$$ is at most a cubic polynomial.
Now,
Let: $$P(x)=ax^3+bx^2+cx+d$$ , where $$0\le a,b,c,d \le 5$$
We know: $$a+b+c+d=5$$ (1)
And, $$125a+25b+5c+d=177$$ (2)
Notice from (2):
$$125a \le 177 \implies a \le 1.416 \implies a=1\text{ or, } a=0$$
Next, consider the case a=1 :
(2) and (1) becomes:
$$25b+5c+d=52 \text{, and } b+c+d=4$$
Using the same idea as before:
$$25b \le 52 \implies b \le 2.09 \implies b=2,1,0$$
But, $$b=1,0$$ is rejected. Because observe that if b=0, then 5c+d = 52 (and this is only true for large c and d, which we are not given.)
Proof: $$c,d\le 5 \implies 5c \le 25, d\le 5 \implies 5c+d \le 25+5=30$$, so it is at most 30 (Which, by the way, is even unattainable!).
Thus, $$b=2$$ only.
Next, we substitute b=2 in (1) and (2) to get:
$$c+d=2\text{ , }5c+d=2 \implies 5c\le 2 \implies c=0,d=2$$
So: $$a=1,b=2,c=0,d=2$$ is a valid solution.
Thus, $$P(x)=x^3+2x^2+2 \implies P(10)=1000+200+2=1202$$, as you found.
Next, consider the case a=1 :
$$25b+5c+d=177 \implies b \le 7$$, we see that, this is impossible. As, b is at most 5, giving 125, but c and d are forced to be 0. Hence, this case is rejected.
Proof: | {
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Proof:
$$25b \le 125, 5c \le 25, d \le 5 \implies 25b+5c+d \le 155$$, which again, is not attainable anyway.
Therefore,$$\text{ }P(x)=x^3+2x^2+2$$ is the only possible polynomial with the given conditions, and 1202 is the only valid solution.
I hope this helps!
Guest Jul 17, 2022
#6
+117766
-1
I already showed that somewhat more easily.
Melody Jul 17, 2022
edited by Melody Jul 17, 2022
#7
+1
Ooh! I just saw it now. It is really elegent! But I think my approach is more general; as writing 177=125+50+2 depended on "177" probably If it was other number, then there might be many ways to write a number "A" as addition of other numbers, so more cases to try.
But overall, that was fantastic observation!!!!!!
Guest Jul 17, 2022
#4
+117766
+1
If P(x) is a polynomial with all positive integer polynomials AND the constant is also positive (which may not be intended)
and
P(1)=5 then all the coeficients and the constant must add to 5.
P(5)=177 = 125+50+2 = 125+ 2*25 + 2 so the polynomial is
$$p(x)=x^3+2x^2+2$$
That is the only possibility. (assuming that the constant is also a positive integer)
Jul 17, 2022
edited by Melody Jul 17, 2022 | {
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# 2048 Logic Puzzle
I thought up this logic problem related to the 2048 game. If all 16 tiles on a 2048 board all had the value 1024, how many ways are there to get to the 2048 tile? Here is what I am talking about in an illustration:
I found a much simpler, but longer way to think about this: There are 3 ways to combine 2 tiles by going to the right, and 3 by going to the left. That means there are 6 ways to combine the tiles. So, for all the rows and columns, there are $$2 \cdot (4 \cdot 6) = 48$$ ways to get to the 2048 tile.
My question(s) are, is my logic correct? Also, is there a simpler way to approach this logical problem?
## Notes
I found two Math.SE post related to 2048 logic, but they have nothing to do with my problem.
I believe you are Correct.
There are 3 lines separating rows horizontally, 4 pairs of numbers across each line, and 2 ways to combine said pairs (top-down or bottom-up), giving $2*3*4=24$ ways of making pairs.
Repeating for the columns and getting the exact same numbers, we now have $24+24 = 48$ total options for merging two numbers.
• Okay, that answers the first part. But what about the second prompt. Is my method the simplest way to approach this? Or is there a much more simpler way to approach this? – Obinna Nwakwue May 25 '16 at 21:58
• "Is my method the simplest way to approach this?" As far as I can tell, yes. – Simpson17866 May 25 '16 at 23:16
• Okay, that helps. Thank you! – Obinna Nwakwue May 26 '16 at 0:59
The corner tiles have 2 ways to move
The side tiles have 3 ways to move
The middle tiles have 4 ways to move
2 3 3 2 2 3 3 3 2 ...
3 4 4 3 3 4 4 4 3
3 4 4 3 3 4 4 4 3
2 3 3 2 3 4 4 4 3
2 3 3 3 2
Then the formula for different ways to make a $2048$ tile on $(n * n)$ square of $1024$ tiles is:
$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4$
$4 * 2$ (corner tiles)
$(n - 2) * 3 * 4$ (side tiles)
$(n - 2)^2 * 4$ (middle tiles)
$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4 =$
$4(n^2 - n)$ | {
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$(n - 2)^2 * 4$ (middle tiles)
$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4 =$
$4(n^2 - n)$
• To complete your "proof", for $n = 4$, $4(n^2 - n) = 4(4^2 - 4) = 4(16 - 4) = 64 - 16 = 48$! – Obinna Nwakwue Jun 7 '16 at 15:51 | {
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# How to represent each natural number?
Assume we get the set of natural numbers $\mathbb{N}$ from any model of the Peano axioms.
We're given the symbols: $0,1,2,3,4,5,6,7,8,9$, or rather, we're given $0$ and we choose to use the symbols $1,2,3,4,5,6, 7, 8,9$.
Of course $0$ is the same from the model.
Then we'll have, by definition, $1=S(0)$, $2=S(1)$, $3=S(2)$, $4=S(3)$, $5=S(4)$, $6=S(5)$, $7=S(6)$, $8=S(7)$ and $9=S(8)$.
But how do we represent the rest of the natural numbers the way we expect them to be represented?
I understand that $1, 2, 3, 4, 5, 6, 7, 8, 9$ are just shorthand representations for the entities written above.
I guess my question can be particularized by: how do you know that the short hand notation for $S(999)$ is $1000$?
I'm assuming the way to get a representation for each natural number starts the way I write it. If that's not the case, please do it from the top. | {
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-
What's wrong with the inductive definition $S(n+1)=S(S(n))$? – Asaf Karagila Jan 19 '13 at 21:57
My problem is not with the definition per se, but rather how to represent each number. Intuitively you know $10$ will be $S(9)$, but why? Did I make it clearer? – Git Gud Jan 19 '13 at 21:58
I don't think it is "intuitively" but only a matter of choosing some base wrt which write the numbers: $\,S(9)=10\,$ because we usually use the decimal base to write numbers, but I think it may as well be $\,S(9)=101\,$ if we choose base $\,3\,$ (and, of course, it probably is more logical to write $\,S(100)=101\,$ in base $\,3\,$ ...) – DonAntonio Jan 19 '13 at 22:04
@DonAntonio What's that about base wrt? Never heard of it. I know we choose to use the decimal base do represent numbers. But how do you know that, in base $10$, $1000=S(999)$? And what is $999$? – Git Gud Jan 19 '13 at 22:10
@GitGud , wrt = with respect to . And again: I know that $\,S(999)=1000\,$ because I choose (or we choose) to represent the numbers in decimal base, that's all. If we chose not to work with bases at all then we could use the successor function and work just with that, though it would be extraordinarily cumbersome: "What time is it?" " It is$\,S(S(S(S(\emptyset))))\,$ minutes before $\,S(S(\emptyset))\,$ " ...pretty annoying, uh? There's where bases kick in. – DonAntonio Jan 19 '13 at 22:17
Given a PA number, to translate it back into a string of digits is by the following recursive function:
• $f(x) = d(x)$ where $x < 10$
• $f(x \hat + SSSSSSSSSS0 \hat \times y) = f(y)"d(x)"$ where $x < 10$ and $y > 0$.
Justifying this recursion principle 9that a function defined in this way is well defined and total) is by Euclid's Division algorithm. | {
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-
I think I'm getting the gist of it. Can you tell me where to read about this and find some proofs? – Git Gud Jan 19 '13 at 23:03
@GitGud, what do you like proofs of? – user58512 Jan 19 '13 at 23:08
"Justifying this recursion principle 9that a function defined in this way is well defined and total) is by Euclid's Division algorithm." I wouldn't wanna bother you with that, I can probably do it myself. But I'd rather read it, so if you know where I can read about it, I'd appreciate it. Thanks. – Git Gud Jan 19 '13 at 23:14
There is an obvious function from strings of digits to natural numbers:
• $f("") = 0$
• $f("3534") = 3 + 10\cdot f("534") = 3 + 10 \cdot (5 + 10 \cdot f("34")) = \ldots = 3 + 10(5 + 10(3 + 10(4 + 0))))$
Recall that we define + and * for peano arithmetic:
• $0 \hat + y = y$
• $Sx \hat + y = S(x \hat + y)$
• $0 \hat \times y = x$
• $Sx \hat \times y = x \hat + x \cdot y$
therefore, define
• $g("") = 0$
• $g("dssss") = p(d) \hat + SSSSSSSSSZ\hat \times g("ssss")$
and the gives the peano arithmetic number or a digits expression.
(p is the function that gives $p(2) = SSZ$ for the first 10 digits for example, that you already mentioned)
-
I think this is on the right track to satisfy me. But I have to define $10$ before applying your process, right? – Git Gud Jan 19 '13 at 22:49
@GitGud, the first part with $f$ is just for intuition - it's not used later. Also I posted a new answer to your question: How to do the reverse operation: Getting a string of digits from a peano number. – user58512 Jan 19 '13 at 22:51
first you must define the "remain" and "quotient" of one natural number to another then you can represent any natural number by its "quotient" and "remain" to the powers of ten | {
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Actually in set theory you don't represent numbers like 10 or 11 or something like that, number 3 in set theory is S(S(S(0))), but because you are already familiar with this representation of natural numbers, you use 3, actually using "3" in set theory is wrong
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You need to notice that the standard decimal representation of the natural numbers is not a representation from inside the Peano model. The language only contains a constant symbol for $0$. Then one introduces $1=S(0)$ as shorthand notation not as a new symbol. Then comes $2=S(1)=s(S(0))$ again as shorthand notation, not a new symobol, and so on.
So, whenever you write something in PA and you use $0$ then you mean the constant symbol in the language. When you write $1$, $2$, and so on mean that this is shorthand notation that you use just because you are lazy to write out an expression of the form $S(S(0))$.
I hope this helps. | {
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I hope this helps.
-
I understand everything you said, but it doesn't answer what I want to know. Why do you make $10$ short for $S(S(S(S(S(S(S(S(S(S(0))))))))))$? – Git Gud Jan 19 '13 at 22:02
it's just introducing shorthand notation for strings in the language. You could have used "banana" as shorthand for S(S(S(S(S(S(S(S(S(0)))))))))) if you want. But it makes more sense for us humans (who like numbers and use the decimal notation regularly) to call it something a bit more meaningful. We work with the model of PA from outside of it. We are not in the model so we are free to study it by means of things we have outside as well. It makes life easier. – Ittay Weiss Jan 19 '13 at 22:09
I wouldn't say 47 is short for $S(S(S(\ldots(S(0)))))$ any more than it is short for XLVII. Decimal numerals, Roman numerals, successor numerals belong to three different ways of representing natural numbers. – Peter Smith Jan 19 '13 at 22:13
@IttayWeiss How do you rigorously introduce the shorthand notation that we're used to? That's my question. – Git Gud Jan 19 '13 at 22:16
@GitGud recursively: n+1 is shorthand notation for S(n), where 0 is shorthand for "0". It works since you believe that for the models of the naturals that you have in your head, recursion works. – Ittay Weiss Jan 19 '13 at 22:21
Suppose you have $9$ stones in a pile and you add another stone to make a pile of $S(9)$ stones. Rather than invent a new symbol, the convention is to give a pile of this size its own name. Henceforth, any pile of exactly $S(9)$ stones will be called a "ten". So, the successor of $9$ is a ten and no other stones. This is denoted by placing a $1$ in the ten column and a $0$ in the single stone column, which gives $S(9) = 10$. The process is similar for larger piles of stones: "hundred", "thousand", etc. | {
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My problem is with "that process is similar". That's just an intuitive way of doing things. That's like what we were taught when we were kids. How do you make a computer understand that $10$ is just $S(9)$? You can't tell the computer what symbol will represent each natural number. – Git Gud Jan 19 '13 at 22:38 | {
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Answer. Because all rectangles are also parallelograms, all the properties of parallelograms are also true for rectangles, too: But because the angles are all equal, there is an additional property of rectangles that we will now prove – that the diagonals of a rectangle are equal in length. If we divided the rectangle along diagonal NL, we would create triangle LNO. A rectangle is a parallelogram with 4 right angles. You can contact him at GeometryHelpBlog@gmail.com. Can anyone solve this and explain how to do so? prove: the diagonals of a rectangle have equal lengths. Which element of a text best helps the reader determine the central idea? Series: Property: The Diagonals of a Rectangle Are of Equal Length. Welcome to Geometry Help! A rectangular field is 15 m long and 10 m wide, Another rectangular field having the sameperimeter has its sides in the ratio 4 : 1. Recommend (0) Comment (0) ASK A QUESTION . Here, we don’t even need to construct any special triangles, because the diagonals themselves have defined the triangles. 00:03:47 undefined. Maths. Videos: 1 Duration: 00:03:47 Language: English. A square has four equal sides and four right angles. OP = OB . alwayssometimesnever4 The diagonals of a trapezoid are equal. Diagonals of Rectangle The diagonals of a rectangle are congruent. (6) AC = DB // Corresponding sides in congruent triangles (CPCTC), Filed Under: Rectangles Last updated on January 4, 2020. Thus the diagonals bisect each other in a rectangle. Copyright © 2020. Remember that a rectangle is a parallelogram, so it has all of the properties of parallelograms , including congruent opposite sides. Diagonals of rectangle bisect each other. In this video tutorial we discuss: (1) How to prove that the two diagonals of a rectangle are equal in length & bisect each other? Please email us at GeometryHelpBlog@gmail.com. If the two diagonals of a parallelogram are equal, it is a rectangle. Now, since a rectangle is a | {
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If the two diagonals of a parallelogram are equal, it is a rectangle. Now, since a rectangle is a parallelogram, its opposite sides must be congruent and it must satisfy all other properties of parallelograms. Upvote(17) How satisfied are you with the answer? Chemistry. My goal is to help you develop a better way to approach and solve geometry problems. 1 A square is a rectangle.alwayssometimesnever2 The diagonals of a rhombus are perpendicular. (5) ΔBAD ≅ ΔCDA // Side-Angle-Side postulate. Write a two column-proof given: fghi is a rectangle. In a rectangle ABCD, prove that the diagonals are of equal length: prove AC = DB. In rectangle STAR below, SA =5, what is the length of RT? Maths. Real World Math Horror Stories from Real encounters. The two diagonals are equal in length. Find the dimensio … n of the rectangular field. RELATED ASSESSMENTS. Next, remember that the diagonals of any parallelogram bisect each other and the diagonals of a rectangle are congruent. Add your answer and earn points. Each diagonal splits a corner into two angles of $$45^\circ$$ . New questions in Mathematics. Chemistry . They have a special property that we will prove here: the diagonals of rectangles are equal in length. Interactive simulation the most controversial math riddle ever! The length of a diagonal (d) of a rectangle whose length is l and whose breadth is b is calculated by the Pythagoras theorem. ii) From O, draw a perpendicular OP to meet AB at P. In triangle … Note also that the diagonals are equal and cut each other in half at right angles. How Long is MO and MZ in the rectangle pictured on the left? Haroldescorcia2527 is waiting for your help. The formula to find the length of the diagonal of a rectangle is: c:Parallelogram. The diagonal of the rectangle is the hypotenuseof these triangles.We can use Pythagoras' Theoremto find the length of the diagonal if we know the width and height of the rectangle. What is the value of x in rectangle STAR below? | {
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if we know the width and height of the rectangle. What is the value of x in rectangle STAR below? Download PDF's. Let's take rectangle ∴ The diagonals of a rectangle bisects each other and equal . PLEASEEEE HELLLPPP MEEEEE FREE POINTS FOR ALLL! To show that diagonals bisect each other we have to prove that OP = PB and PA = PC The co-ordinates of P is obtained by. [00:03:47] S. Login/Register to track your progress. Class 12 Class 11 … Triangle MLO is a right triangle, and MO is its hypotenuse. IF the quadrilateral is a rectangle, then the two diagonals are equal in length. If you remember your Pythagorean theorem, you should be able to see why. In the figure above, click 'reset'. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. Click to learn more... By accessing or using this website, you agree to abide by the Terms of Service and Privacy Policy. Physics. NCERT RD Sharma Cengage KC Sinha. Length of a diagonal of a square = √2 x Length of a Diagonal of a Rectangle Similar to a square, the length of both the diagonals in a rectangle are the same. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. A rectangle is a parallelogram, and we can save time and effort by relying on general parallelogram properties that we have already proven. We also know that AB= CD as they are opposite sides in a parallelogram. Rectangles are a special type of parallelogram, in which all the interior angles measure 90°. Geometry doesn't have to be so hard! And from the definition of a rectangle, we know that all the interior angles measure 90° and are thus congruent- and we can prove the triangle congruency using the Side-Angle-Side postulate. For example, the two triangles ΔABD and ΔDCA, in which the diagonals form corresponding sides. It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°). What is the formula of | {
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of reflectional symmetry and rotational symmetry of order 2 (through 180°). What is the formula of collinear? parallelograms; class-8; Share It On Facebook Twitter Email. His goal is to help you develop a better way to approach and solve geometry problems. alwayssometimesnever3 The diagonals of a rectangle are equal. The diagonal of a square is equal to the length of one of the sides of the square times $\sqrt {2}$, or $s\sqrt{2}$. When if two diagonals are given we can construct. If a diagonal bisects a rectangle, two congruent right triangles are obtained. Therefore $$\angle SZA = 120°$$. To find MZ, you must remember that the diagonals of a parallelogram bisect each other. Biology. I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree. Thus diagonals bisect each other in a rectangle . 00:02 CBSE class 9 maths NCERT Solutions chapter 10 Quadrilaterals q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle. It happens! 직사각형의 두 대각선은 길이가 같고 서로 다른 것을 이등분한다. Click here to see the proof. Rectangles are a special type of parallelogram. As you can hopefully see, both diagonals equal 13, and the diagonals will always be congruent because the opposite sides of a rectangle are congruent allowing any rectangle Geometry answers, proofs and formulas for solving geometry problems, and useful tips for how to approach these problems. Biology. Similarly we can prove that PC = PA . It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. The rectangle is given as ABCD, with the two diagonals as AD and BC. In ΔADB and ΔBCD: AD = BC (Opposite sides of a rectangle are equal.) So, looking at the triangles ΔABD and ΔDCA, they have one common side – AD. 343 views. Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. Like many other geometry problems where we need to prove that two line segments are of equal lengths, we will turn to triangle congruency as our go-to | {
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prove that two line segments are of equal lengths, we will turn to triangle congruency as our go-to tool. Opposite sides of a rectangle are equal; so, AB = DC and AD = BC. … Don’t worry if you didn’t see this immediately, or if you chose other triangles – it is easy to prove this property of rectangles using other combinations of triangles, such as ΔABC and ΔDCB or ΔDAB and ΔCBA; any two pairs will do, as long as the diagonals are the corresponding sides in each triangle. (3) m∠BAD = m∠CDA=90° // Definition of rectangle, (4) ∠BAD ≅ ∠CDA // (3) and definition of congruent angles. The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium ← Prev Question Next Question → 0 votes . Physics. So AO = OC and BO = OD, where O is the intersecting point of the two diagonals AC & BD. Rectangles are a special type of parallelogram. Click hereto get an answer to your question ️ Show that the diagonals of a square are equal and bisect each other at right angles. 0%. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. a:Square. To Prove that the two Diagonals of a Rectangle Are of Equal Length. b:Rectangle. The diagonals of a rectangle are congruent. (2) AB= CD // Opposite sides in a rectangle (parallelogram). We can also prove this from scratch, repeating the proofs we did for parallelograms, but there’s no need. d:Rhombus. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. … In a rectangle both diagonals are equal in measure; so AC = BD; as well the four intercepts, AO = OC = BO = OD. alwayssometimesnever Free Algebra Solver ... type anything in there! (1)The diagonals of a parallelogram are equal. The rectangle is a convex quadrilateral because of two diagonals which lie in the interior of the rectangle. Like a square, the diagonals of a rectangle are congruent to each other and bisect each other. e:Trapezium etc..... New questions in | {
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a rectangle are congruent to each other and bisect each other. e:Trapezium etc..... New questions in Math. asked May 4 in Parallelograms by Vevek01 (47.2k points) The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium. Again, we can use the Pythagorean theorem to find the hypotenuse, NL. Explain why a rectangle is ... maths. Click hereto get an answer to your question ️ Which statement(s) is/are correct? Related Questions. (4)Every quadrilateral is either a trapezium or a parallelogram or a kite. Thank you! CD = CD (Common) Rectangles are a special type of parallelogram, in which all the interior angles measure 90°. A rectangle and a crossed rectangle are quadrilaterals with the following properties in common: Opposite sides are equal in length. answr. Books. But we'd sure like to know about it so that we can fix it. "If a rectangle is square, then its main diagonals are equal" is (True) because this is true of all rectangles. LMNO and divide along the diagonal MO into two right triangles. If side MN = 12 and side ML = 5, what is the length of the other two sides? They have a special property that we will prove here: the diagonals of rectangles are equal in length. In a quadrilateral the diagonals are equal, the quadrilateral will be either 1.A square or 2.A rectangle In a parallelogram the diagonals bisect each other and they never equal Therefore, SZ = AZ, making SZA isosceles and $$\angle$$ZSA$$\angle$$ZAS, being base angles of an isosceles triangle. Books. Therefore, x = 30 °. (2)The diagonals of a square are perpendicular to each other. $$\angle SZT$$ and $$\angle SZA$$ are supplementary angles, Since the opposite sides of a rectangle Since the diagonals of a rectangle are congruent, RT has the same length as SA. Video Explanation. NCERT RD Sharma Cengage KC Sinha. To Prove that the two Diagonals of a Rectangle Are of Equal Length. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. In order to prove that | {
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Are of Equal Length. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. In order to prove that the diagonals of a rectangle are congruent, consider the rectangle shown below. toppr. Click hereto get an answer to your question ️ Prove that the diagonals of a rectangle divide it in two congruent triangles. This will help us to improve better. As you can see, a diagonal of a rectangle divides it into two right triangles,BCD and DAB. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Prove that the diagonals of a rectangle are congruent. Since the diagonals of a rectangle are congruent MO = 26. In this lesson, we will show you two different ways you can do the same proof using the same rectangle. EASY. to be divided along the diagonals into two triangles that have a congruent hypotenuse. Answered By . Diagonals of Rectangles are of Equal Length, Opposite sides in a rectangle (parallelogram), The opposite sides of rectangles are equal, The diagonals of rectangles bisect each other, Any two adjacent angles are supplementary (obviously, since they all measure 90°), The opposite angles are equal (again, obviously, since all interior angles measure 90°). A Quadrilateral has two diagonals. (3)If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus. Now, having picked our triangles, we will rely on the properties of parallelograms to show triangle congruency. Click hereto get an answer to your question ️ Prove logically that the diagonals of a rectangle are equal Explain why a rectangle is a convex quadrilateral. (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties), If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13. Click here to see the proof. Get Instant Solutions, 24x7. A rectangle has two diagonals as it has four sides. are congruent NO is 5 and lO is 12. (Two diagonals of a rectangle are equal in length and bisect each other.) NCERT NCERT | {
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and lO is 12. (Two diagonals of a rectangle are equal in length and bisect each other.) NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. By the Pythagorean theorem, we know that. ’ t even need to construct any special triangles, BCD and DAB what is the length of the.! With 4 right angles special triangles, because the diagonals of a rectangle are equal ; so are the two diagonals of a rectangle equal why! Dc and AD = BC 0 ) Comment ( 0 ) Comment ( 0 ) a... By the Terms of Service and Privacy Policy the Terms of Service and Policy. More... by accessing or using this website, you agree to abide by the Terms of Service and Policy... To construct any special triangles, because the diagonals of rectangles are a are the two diagonals of a rectangle equal why property that we will rely the! In this lesson, we will prove here: the diagonals of a rectangle are congruent MO are the two diagonals of a rectangle equal why.! Executive with a BSc degree in Computer Engineering and rotational symmetry of 2! How Long is MO and MZ in the figure above, click 'reset ' as and! Triangle congruency 것을 이등분한다 rectangle LMNO and divide along the diagonal MO into two right,... S no need or using this website, you should be able to see why ( )... Two congruent right triangles are obtained: in the rectangle is given as ABCD prove! Interior of the rectangular field to do so in this lesson, we prove! And Privacy Policy ) the diagonals of a rectangle are equal, is a parallelogram, which. ( 5 ) ΔBAD ≅ ΔCDA // Side-Angle-Side postulate diagonal NL, we will show you different. Equal. a high-tech executive with a BSc degree in Computer Engineering more... by accessing using! Is either a Trapezium or a parallelogram are equal in length and bisect each other and bisect each other half!, so it has two diagonals are of equal length into two right triangles are obtained diagonals form sides... 길이가 같고 서로 다른 것을 이등분한다 shown below the triangles a rectangle are equal, it is a parallelogram | {
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길이가 같고 서로 다른 것을 이등분한다 shown below the triangles a rectangle are equal, it is a parallelogram equal... Long is MO and MZ in the rectangle is given as ABCD, prove that the of. And rotational symmetry of order 2 ( through 180° ) which lie in the interior angles measure 90° diagonals corresponding. = OC and BO = OD, where O is the intersecting point of the two diagonals a..., a diagonal bisects a rectangle have equal lengths are the two diagonals of a rectangle equal why divide along the diagonal a... Can see, a diagonal bisects a rectangle ( parallelogram ) is to you! 5 and lO is 12 to see why rectangle divides it into two right triangles, BCD and DAB questions. They have a special property that we will rely on the properties of parallelograms Awasthi MS Chauhan diagonals AC BD. ) the diagonals of rectangles are equal, it is a high-tech executive with BSc! Length and bisect each other. easy to prove that the diagonals a! To know about it so that we will show you two different ways can! Ab = DC and AD = BC has four sides the central idea: prove AC =.! Column-Proof given: fghi is a parallelogram, in which the diagonals are of equal length executive. To prove that a parallelogram are equal. 17 ) how satisfied you. 5, what is the value of x in rectangle STAR below learn! Find the hypotenuse, NL 1 ) the diagonals of a rectangle are of equal length videos: Duration... Element of a rectangle is given as ABCD, with the two triangles ΔABD and ΔDCA, they a... Share it on Facebook Twitter Email rectangular field so AO = OC and =... Can see, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree parallelogram. Using dot product of vectors, prove that the diagonals of a parallelogram or a.! Other in half at right angles diagonals themselves have defined the triangles the formula to find the hypotenuse NL... Pictured on the properties of parallelograms to show triangle congruency you develop a better way to and... And Privacy Policy ncert ncert are the two | {
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triangle congruency you develop a better way to and... And Privacy Policy ncert ncert are the two diagonals of a rectangle equal why ncert Fingertips Errorless Vol-1 Errorless Vol-2 the figure above, click '! 서로 다른 것을 이등분한다 no is 5 and lO is 12, including opposite... Other. a parallelogram with 4 right angles and DAB by relying on general parallelogram that. Nl, we will show you two different ways you can see a... Would create triangle LNO you develop a better way to approach and geometry... Exemplar ncert Fingertips Errorless Vol-1 Errorless Vol-2 we divided the rectangle is a parallelogram are equal, is. = 12 and side ML = 5, what is the length the. Parallelograms to show triangle congruency either a Trapezium or a kite click hereto get answer. Prove: the diagonals of a rectangle are equal. satisfy all other properties of parallelograms but! Must be congruent and it must satisfy all other properties of parallelograms, including congruent sides. – AD above, click 'reset ' CD // opposite sides Pythagorean theorem right,. Which all the interior angles measure 90° other properties of parallelograms, but there ’ s no need series property! Side-Angle-Side postulate ncert DC Pandey Sunil Batra HC Verma Pradeep Errorless next, remember a! S ) is/are correct and side ML = 5, what is the point... So that we will prove here: the diagonals of a rectangle are of length... Hereto get an answer to your question ️ which statement ( s ) is/are correct what is the of! Other and the diagonals form corresponding sides i 'm ido Sarig is parallelogram. Language: English now, having picked our triangles, BCD and DAB to help you develop a way! Use the Pythagorean theorem, you should be able to see why in Engineering. Given we can use the Pythagorean theorem to find the dimensio … of... One common side – AD product of vectors, prove that the two triangles ΔABD and,... If two diagonals as AD and BC you develop a better way to approach and solve geometry problems in Engineering. 2 ) AB= CD | {
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and BC you develop a better way to approach and solve geometry problems in Engineering. 2 ) AB= CD // opposite sides so AO = OC and BO = OD, where O is value. Triangles, BCD and DAB x in rectangle STAR below, SA =5, what is the value of in..., whose diagonals are given we can also prove this from scratch, repeating the proofs we did parallelograms! Has two diagonals of a rectangle with the answer AD = BC ( opposite sides of a rectangle ABCD with. Is 5 and lO is 12 a right triangle, and we can use the Pythagorean theorem, must!: prove AC = DB and effort by relying on general parallelogram properties that we construct... Triangle MLO is a high-tech executive with a BSc degree in Computer Engineering and an MBA degree two. Are perpendicular to each other in half at right angles any parallelogram bisect other! Vectors, prove that the diagonals of a quadrilateral intersect at right angles, it is a right triangle and! Congruent no is 5 and lO is 12 the formula to find MZ, must. Ad = BC AD = BC ( through 180° ) a high-tech executive a! Also know that AB= CD as they are opposite sides in a rectangle is a rectangle ABCD, prove the! High-Tech executive with a BSc degree in Computer Engineering as they are opposite sides a! Congruent right triangles is: in the rectangle pictured on the properties of parallelograms including... Are of equal length: prove AC = DB 것을 이등분한다: prove AC = DB ncert Pandey! Equal sides and four right angles a question parallelograms ; class-8 ; Share it on Facebook Email. That we can also prove this from scratch, repeating the proofs we did for,. Class-8 ; Share it on Facebook Twitter Email picked our triangles, we will prove here: the diagonals a. Rectangle, two congruent right triangles, BCD and DAB the interior the. Must remember that the diagonals of a rectangle congruent MO = 26 necessarily a rhombus are to... ( 0 ) ASK a question: the diagonals of a rectangle with the answer, opposite. Of Service and Privacy Policy other two sides divide along | {
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of a rectangle with the answer, opposite. Of Service and Privacy Policy other two sides divide along the diagonal of a square, two! Iit-Jee Previous Year Narendra Awasthi MS Chauhan of Service and Privacy Policy other and.! Fingertips Errorless Vol-1 Errorless Vol-2 BSc degree in Computer Engineering and an MBA degree below... And bisect each other and equal. where O is the value of x in rectangle below! Dc Pandey Sunil Batra HC Verma Pradeep Errorless Exemplar ncert Fingertips Errorless Vol-1 Errorless Vol-2 congruent right triangles are... And AD = BC no need must remember that the two triangles ΔABD and ΔDCA, which. Sides in a rectangle are of equal length rectangle is a high-tech with... General parallelogram properties that we can also prove this from scratch, repeating the proofs we did for,..., RT has the same rectangle rectangle along diagonal NL, we will show you two different ways you see! A kite can see, a diagonal bisects a rectangle are equal, a! ( 2 ) AB= CD // opposite sides in a rectangle bisects each other. rely on left. Perpendicular to each other. as AD and BC best helps the reader determine the central idea to construct special. See, a high-tech executive with a BSc degree in Computer Engineering and MBA! = DB 대각선은 길이가 같고 서로 다른 것을 이등분한다 help you develop a better way to approach and solve problems... | {
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# Find all roots in the interval of nonlinear equation
I am struggling on how to find all the roots of 1+1/2^x+1/3^x==0 which lie in a given of real and imaginary interval.
Solve does not work, and FindRoot only returns one root. Is there a better way to get as many roots as possible? Thank you so much for your help.
If you restrict the domain where to look for roots, Reduce can often find them and will return Root objects which can be used in exact symbolic calculations. Even better, it guarantees to give you all roots in that domain.
Reduce[1 + 1/2^x + 1/3^x == 0 && Abs[x] < 5, x]
(* x == Root[{1 + 3^#1 + E^(-(Log[2] - Log[3]) #1) &, 0.4543970081950240272783427420109442288880772534469111379406 - 3.5981714939947587422049363529208471165604257466288393398421 I}] ||
x == Root[{1 + 3^#1 + E^(-(Log[2] - Log[3]) #1) &, 0.4543970081950240272783427420110 + 3.5981714939947587422049363529208 I}] *)
When looking for real roots, the typical way to restrict the domain is something similar to 0 < x < 1. We're looking for complex roots her so I used Abs[x] < 5.
Related:
To use the FindRoots2D function from the linked post, you need to break the equation into real and imaginary parts, as follows:
f[z_] := 1 + 1/2^z + 1/3^z
FindRoots2D[{Re@f[x + I y], Im@f[x + I y]}, {x, -5, 5}, {y, -5, 5}]
(* {{0.454397, -3.59817}, {0.454397, 3.59817}} *)
• I am trying your code. But it runs so long that I have not seen the result yet while I am replying to you. How long did u take to run this code? Thanks for your help. – user16023 Jun 19 '14 at 4:39
• Could you give me some basic instructions to use Wagon's FindAllCrossing2D[] function? – user16023 Jun 19 '14 at 4:52
• @user16023 It takes less than a second in both v8 and v9. FindAllCrossings2D finds the roots of a system of two equations on the reals (not complexes). You can break your equation into real and imaginary parts then follow the instructions from the post I linked. – Szabolcs Jun 19 '14 at 14:31 | {
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If you only need approximate numerical solution rather than the exact solution (i.e., Root object) you could use NSolve also with a constrained domain for x.
eqn = 1 + 1/2^x + 1/3^x == 0;
soln = NSolve[{eqn, Abs[x] <= 5}, x, WorkingPrecision -> 10]
(* {{x -> 0.454397008 + 3.598171494 I}, {x -> 0.454397008 - 3.598171494 I}} *)
And @@ (eqn /. soln)
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# 14 Pre-Class Assignment: Fundamental Spaces¶
## 1. Orthogonal Complement¶
Definition: A vector $$u$$ is orthogonal to a subspace $$W$$ of $$R^n$$ if $$u$$ is orthogonal to any $$w$$ in $$W$$ ($$u\cdot w=0$$ for all $$w\in W$$).
For example, consider the following figure, if we consider the plane to be a subspace then the perpendicular vector comming out of the plane is is orthoginal to any vector in the plane:
Definition: The orthogonal complement of $$W$$ is the set of all vectors that are orthogonal to $$W$$. The set is denoted as $$W_{\bot}$$.
QUESTION: Is $$W_\bot$$ a subspace of $$R^n$$? Justify your answer briefly.
Put your answer to the above question here
QUESTION: What are the vectors in both $$W$$ and $$W_\bot$$?
Put your answer to the above question here
from IPython.display import YouTubeVideo
### Projection of a Vector onto a Subspace¶
Think of a projection onto a subspace is analogous to a shadow on a surface. Aspects of an objects 3D space is represented in a 2D shadow but you can’t take the shadow by itself and exactly recreate the 3D surface.
Image from https://commons.wikimedia.org
The following is the matimatical defination of projection onto a subspace.
Definition: Let $$W$$ be a subspace of $$R^n$$ of dimension $$m$$. Let $$\{w_1,\cdots,w_m\}$$ be an orthonormal basis for $$W$$. Then the projection of vector $$v$$ in $$R^n$$ onto $$W$$ is denoted as $$\mbox{proj}_Wv$$ and is defined as $$$\mbox{proj}_Wv = (v\cdot w_1)w_1+(v\cdot w_2)w_2+\cdots+(v\cdot w_m)w_m$$$
Another way to say the above defination is that the project of $$v$$ onto the $$W$$ is just the sumation of $$v$$ projected onto each vector in a basis of $$W$$
Remarks:
Recall in the lecture on Projections, we discussed the projection onto a vector, which is the case for $$m=1$$. We used the projection for $$m>1$$ in the Gram-Schmidt algorithm.
The projection does not depend on which orthonormal basis you choose.
If $$v$$ is in $$W$$, we have $$\mbox{proj}_Wv=v$$. | {
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If $$v$$ is in $$W$$, we have $$\mbox{proj}_Wv=v$$.
### The Orthogonal Decomposition Theorem¶
Theorem: Let $$W$$ be a subspace of $$R^n$$. Every vector $$v$$ in $$R^n$$ can be written uniquely in the form $$$v= w+w_{\bot},$$$$where$$w$$is in$$W$$and$$w_\bot$$is orthogonal to$$W$$(i.e.,$$w_\bot$$is in$$W_\bot$$). In addition,$$w=\mbox{proj}Wv$$, and$$w\bot = v-\mbox{proj}_Wv$.
Definition: Let $$x$$ be a point in $$R^n$$, $$W$$ be a subspace of $$R^n$$. The distance from $$x$$ to $$W$$ is defined to be the minimum of the distances from $$x$$ to any point $$y$$ in $$W$$. $$$d(x,W)=\min \{\|x-y\|: \mbox{ for all }y \mbox{ in } W\}.$$$$The optimal$$y$$can be achieved at$$\mbox{proj}_Wx$$, and$$d(x,W)=|x-\mbox{proj}_Wx|$.
QUESTION: Let $$v=(3, 2, 6)$$ and $$W$$ is the subspace consisting all vectors with the form $$(a, b, b)$$. Find the projection of $$v$$ onto $$W$$.
Put your answer to the above question here
QUESTION: Let $$v=(3, 2, 6)$$ and $$W$$ is the subspace consisting all vectors with the form $$(a, b, b)$$. Find the distance from $$v$$ to $$W$$.
Put your answer to the above question here
## 2. The Four Fundamental Spaces¶
In the lecture on Change Basis, we talked about four subspaces based on a matrix $$A$$:
Row space of $$A$$: linear combination of all rows of $$A$$
Column space of $$A$$: linear combination of all columns of $$A$$
Null space or kernel of $$A$$: all $$x$$ such that $$Ax=0$$
Null space of $$A^\top$$: all $$y$$ such that $$A^\top y =0$$
In this course we represent a system of linear equations as $$Ax=b$$. The matrix $$A$$ can be viewed as taking a point $$x$$ in the input space and projecting that point to $$b$$ in the output space.
It turns out, everything we need to know about $$A$$ is represented by four fundamental vector spaces. Two of the four spaces are easily defined as follows:
Row space of $$A$$: linear combination of all rows of $$A$$
Column space of $$A$$: linear combination of all columns of $$A$$ | {
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Column space of $$A$$: linear combination of all columns of $$A$$
The other two fundamental spaces are defined by a concept called the Null Space. The Null space is calculated by finding all the solutions to the homogeneous system $$Ax=0$$. The final two fundamental spaces are defined as follows:
Null space or kernel of $$A$$: all $$x$$ such that $$Ax=0$$
Null space of $$A^\top$$: all $$y$$ such that $$A^\top y =0$$
## 3. Independent Learning¶
DO THIS: Find a YouTube video that helps you understand the four fundamental spaces.
QUESTION: What is the URL for your video?
Put your answer to the above question here
DO THIS: Add the link to the video to the code below. Try embedding the link in the provided Python YouTubeVideo Function by replacing XXXXX with the video ID.
from IPython.display import YouTubeVideo
QUESTION: What criteria did you use in selecting your video?
Put your answer to the above question here
QUESTION: How long into a video did you go before deciding if it was good or bad?
Put your answer to the above question here
QUESTION: What did you like about the video you selected.
Put your answer to the above question here
QUESTION: What didn’t you like about the video?
Put your answer to the above question here
## 4. Assignment_wrap-up¶
Assignment-Specific QUESTION: What is the URL for your video for the four Fundamental spaces?
Put your answer to the above question here
QUESTION: Summarize what you did in this assignment.
Put your answer to the above question here
QUESTION: What questions do you have, if any, about any of the topics discussed in this assignment after working through the jupyter notebook?
Put your answer to the above question here
QUESTION: How well do you feel this assignment helped you to achieve a better understanding of the above mentioned topic(s)?
Put your answer to the above question here
QUESTION: What was the most challenging part of this assignment for you?
Put your answer to the above question here | {
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Put your answer to the above question here
QUESTION: What was the least challenging part of this assignment for you?
Put your answer to the above question here
QUESTION: What kind of additional questions or support, if any, do you feel you need to have a better understanding of the content in this assignment?
Put your answer to the above question here
QUESTION: Do you have any further questions or comments about this material, or anything else that’s going on in class?
Put your answer to the above question here
QUESTION: Approximately how long did this pre-class assignment take?
Put your answer to the above question here
Written by Dr. Dirk Colbry, Michigan State University | {
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# Evaluate $\lim_{n \rightarrow \infty } \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$ [duplicate]
Evaluate $$\lim_{n \rightarrow \infty~} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$ using Cesáro-Stolz theorem.
I know there are many question like this, but i want to solve it using Cesáro-Stolz method and no others.
I took log and applied Cesáro-Stolz, I get $$\log{2}+n\log\cfrac{n}{n+1}$$
Which gives me answer as $$\frac{2}{e}$$ . But answer is $$\frac{4}{e}$$. Could someone help?.
Edit: On taking log, $$\lim_{n \to \infty} \frac{-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)}{n} \\= \lim_{n \to \infty} \left(-(n+1)\log (n+1) + \sum\limits_{k=1}^{n+1} \log \left(k+n\right)\right) - \left(-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)\right) \\ = \lim_{n \to \infty} \log \frac{2n+1}{n+1} - n\log \left(1+\frac{1}{n}\right) = \log 2 - 1$$ Which gives $$2/e$$
## marked as duplicate by Henning Makholm, Cesareo, Yanior Weg, José Carlos Santos sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 1 at 21:52 | {
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• I don't know where i miss '2'. And its been 4 hours , still didn't find it. :( – Cloud JR Apr 30 at 18:23
• Well, $4/e$ is correct. We might need more details on how you applies Cesaro Stolz. [Another approach, without Cesaro-Stolz, is you can rewrite it as $$\left[(1+1/n)(1+2/n)\cdots(1+n/n)\right]^{1/n}$$ and take the log, which is a Riemann sum of $\int_{1}^{2} \log x\,dx.$ The anti-derivative of $\log x$ is $x\log x - x$ and you get $4/e.$ ] – Thomas Andrews Apr 30 at 18:33
• @Thomas Andrews the second method works and i got 4/e. I want to clarify where it went wrong in first method. I don't want to give it up. I will add more details asap – Cloud JR Apr 30 at 18:40
• The expression can be rewritten as $\left(\frac{(2n!)}{n!n^n}\right)^{1/n}$ Taking logarithm gives you $\frac{\log(2n!) - log(n!) - n\log(n)}{n}$ Apply CS, the difference of the denominator is $$\log(\color{red}{2}n+2)+\log(\color{red}{2}n+1) - \log(n+1) - (n+1)\log(n+1)+ n\log n$$ It seems you have missed one of the $\color{red}{2}$ above. – achille hui Apr 30 at 18:52
• @achille hui. Thanks, writing it as factorial, helps a lot in simplify the notation. – Cloud JR Apr 30 at 18:56
Let $$a_n=\sum_{i=1}^{n} \log\left(1+\frac{i}n\right)$$ be the numerator of the logarithm, with denominator $$b_n=n.$$
The key is that the first term $$\log(1+1/n)$$ of $$a_n$$ doesn't cancel with any of the terms $$\log(1+i/(n+1)).$$ It alone is subtracted, so, while there are $$n$$ occurrences of $$-\log(1+1/n)$$, there are still two more terms for $$a_{n+1}.$$ So you get:
$$a_{n+1} - a_n = \left(\frac{2n+1}{n+1}\right)+\log\left(\frac{2n+2}{n+1}\right)-n\log(1+1/n)$$
Basically, the "cancellation" happens when we have $$\log\left(1+\frac{i}{n+1}\right)-\log\left(1+\frac{i+1}{n}\right)= \log\left(\frac{n}{n+1}\right)$$ | {
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While we can assume there is a $$i=0$$ in $$a_{n+1},$$ since it adds $$0=\log 1$$ to $$a_{n+1},$$ that means there are $$n+2$$ values of $$i$$ in $$\sum_{i=0}^{n+1},$$ and hence there is not cancellation of the $$i=n$$ and $$i=n+1$$ terms from $$a_{n+1}.$$
After took log, we have $$\frac{\sum _{i=1}^n \log (i+n)-n \log (n)}{n}$$ Applied cesaro stolez, we have \begin{align} & \left(\sum _{i=1}^{n+1} \log (i+n+1)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right) \\&= \left(\sum _{i=2}^{n+2} \log (i+n)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right)\\&= (\log (2 n+1)+\log (2 n+2)-(n+1) \log (n+1))-(\log (n+1)-n \log (n)) \\ &= n \log \left(\frac{n}{n+1}\right)+\log \left(\frac{(2 n+1) (2 n+2)}{(n+1)^2}\right) \\ &\rightarrow -1+\log (4) \end{align}
• Thanks man !, i got it. – Cloud JR Apr 30 at 18:54 | {
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Help to understand changing order of integration
I have a problem I have been working on, with the solution but the thing is I don't really understand how it is done.
The question, is to compute, $$\int_0^1 \int_{9x^2}^9 x^3\sin(8y^3) \,dy\,dx$$
Now, I did notice that we are going to have to reverse the order of integration so first I took note of, as of now I have $$0 \le x \le 1$$ and $$9x^2 \le y \le 9$$ and I tried to consider the graph. This is where I am getting confused, I don't know if I am supposed to consider the area basically above the line $$0\le x\le\sqrt{\frac{y}{9}}$$ and put $0 \le y \le 9$ and compute. I know that is what I should do, but I am having a lot of trouble seeing this from the graph. My apologizes as I am not aware of how to put graphs on the site.
I mean I am having trouble visualizing what it is meant to say $x$ is less than that value of $y$, when are we not considering the region bounded above?
I appreciate all answers and comments, ideally though I would like an answer that includes graphics if possible!
Could anyone shed some light on this? Ps, this is not homework and I already have the final solution if anyone wants to check, it is $$=\frac{1-\cos(5832)}{7776}$$
Thank you all
• See math.stackexchange.com/questions/1073275/… - it explains this all. See... you've not actually got two integrals, you have $\int_Sf(p)dA$ - a surface and you're integrating (summing over (it's a comment not quite summing but lets not be pedantic!)) over little chunks of area. That answer should help you – Alec Teal Apr 9 '15 at 22:46
• To put a gap between this and the above comment, your area is x from 0 to 1, and for each x you go from $9x^2$ to $9$, so your surface is the region of all the points between the lines $y=0$, $x=0$, $x=1$ and the curve $y=9x^2$ you can parameterise that however you like! – Alec Teal Apr 9 '15 at 22:49
Since you would like an answer that includes a graph, I've prepared the following simple one. | {
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Since you would like an answer that includes a graph, I've prepared the following simple one.
\begin{equation*} I=\int_{0}^{1}\int_{9x^{2}}^{9}x^{3}\sin \left( 8y^{3}\right) \,dy\,dx \end{equation*}
is to be evaluated in the region $R$ of the 1.st quadrant bounded from below by the graph of the function $y=9x^2$, from above by the horizontal line $y=9$, and from left by the vertical line $x=0$, because the inequalities you have found, $0 \le x \le 1$ and $9x^2 \le y \le 9$, define exactly $R$, i.e.
\begin{equation*} R=\left\{ (x,y)\in \mathbb{R}^{2}:0\leq x\leq 1,\; 9x^{2}\leq y\leq 9\right\}. \end{equation*}
How to define the very same region with a different pair of inequalities? Since $y=9x^2$ is equivalent, for $x\ge 0$, to $x=\sqrt{y/9}=\sqrt{y}/3$, it is clear from the picture that you can also define it by the inequalities $0\leq x\leq \sqrt{y}/3$ and 0$\leq y\leq 9$, as you have done. In symbols,
\begin{equation*} R=\left\{ (x,y)\in \mathbb{R}^{2}:0\leq x\leq \sqrt{y}/3,\; 0\leq y\leq 9\right\}, \end{equation*}
which results in the integral
\begin{equation*} I=\int_{0}^{9}\int_{0}^{ \sqrt{y}/3}x^{3}\sin \left( 8y^{3}\right) \,dx\,dy. \end{equation*}
In 2D I usually prefer to do this by drawing a picture. Let me talk through how I would actually draw this picture.
The region $0 \leq x \leq 1,9x^2 \leq y \leq 9$ is bounded by the lines $x=0,y=9$ and the curve $y=9x^2$. You can think of the limits that you have as saying that for each $x$ between $0$ and $1$, you integrate from $9x^2$ to $9$ in $y$. So this corresponds to little vertical segments between the parabola and the horizontal line $y=9$. | {
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Now you want to go the other way: for each value of $y$ between $0$ and $9$, what are the possible values of $x$ in the region? These now correspond to little horizontal segments, which must be between the vertical line $x=0$ and the parabola $y=9x^2$. Solving for $x$ you get that the parabola is given by $x=\sqrt{y/9}$, so for each $y$, $x$ ranges from $0$ to $\sqrt{y/9}$. So your new integral looks like $\int_0^9 \int_0^{\sqrt{y/9}} f(x,y) dx dy.$ | {
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# A tricky remainder theorem problem
Tags:
1. Jun 12, 2016
### sooyong94
1. The problem statement, all variables and given/known data
A polynomial P(x) is divided by (x-1), and gives a remainder of 1. When P(x) is divided by (x+1), it gives a remainder of 3. Find the remainder when P(x) is divided by (x^2 - 1)
2. Relevant equations
Remainder theorem
3. The attempt at a solution
I know that
P(x) = (x-1)A(x) + 1
and P(x) = (x+1)B(x) + 3
But how would I relate to (x^2 -1)? I can multiply the two equations together to get (x^2 -1) but things get pretty messy.
2. Jun 12, 2016
### Mastermind01
From your two equations we know P(1) = 1 and P(-1) = 3 . Since the divisor in question ($x^2 - 1$) the remainder has degree < divisor i.e it is linear. Let it be Kx + L. The we get an equation $$P(x) = (x^2 - 1)f(x) + Kx + L$$ . Try to proceed from here.
3. Jun 13, 2016
### sooyong94
Why is the remainder is a linear expression? I can't catch your explanation.
4. Jun 13, 2016
### Ray Vickson
The remainder in $a(x)/b(x)$ is a polynomial $r(x)$ of degree strictly less than the degree of $b(x)$. Basically, that is the _definition_ of "remainder".
5. Jun 13, 2016
### sooyong94
So that means if a polynomial P(x) is divided by a quadratic polynomial, then the remainder is a linear expression.
6. Jun 13, 2016
### Ray Vickson
7. Jun 13, 2016
### sooyong94
So it has something to do with the divisor right?
8. Jun 13, 2016
### rcgldr
To refresh the idea of the remainder theorem, take a look at
http://en.wikipedia.org/wiki/Polynomial_remainder_theorem
In this case you have P(1) = 1 (since P(x)/(x-1) has remainder 1), P(-1) = 3 (since P(x)/(x+1) has remainder 3). Can you think of a minimum degree P(x) that produces these results?
9. Jun 13, 2016
### sooyong94
A cubic polynomial?
10. Jun 13, 2016
### rcgldr
Ignore the fact that you will be looking for P(x) / (x^2-1). Can you think of a minimum degree polynomial P(x) such that P(-1) = 1 and P(1) = 3? | {
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11. Jun 13, 2016
### sooyong94
Degree 3?
12. Jun 13, 2016
### rcgldr
Looking for a minimal degree for P(x). Start off with degree 1, is there a P(x) of degree 1 (ax + b) such that P(-1) = 1 and P(1) = 3? If not, try degree 2, and if not, try degree 3.
13. Jun 13, 2016
### sooyong94
Yup it appears that degree one works as well...
14. Jun 13, 2016
### rcgldr
So what is that equation for P(x) of degree 1 and what is the remainder of P(x) / (x^2-1) ?
15. Jun 13, 2016
### sooyong94
ax+b, and I managed to solve it as (2-x).
16. Jun 13, 2016
### rcgldr
To follow up, this would mean that a general equation for P(x) = Q(x)(x-1)(x+1) - x + 2, where Q(x) can be any function of x, including Q(x) = 0. | {
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# Limits and restrictions?
If we assume that the restrictions put on simplified forms of expressions to prevent evaluation at points undefined in the original unsimplified form are important why do we drop them when dealing with limits? For example, consider the following when trying to find the derivative of $f= x^2%$:
\begin{align*}\lim_{h→0} \frac{f(x + h) - f(x)}{h} &=\lim_{h→0} \frac{(x+h)^2 - x^2}{h}\\ &=\lim_{h→0} \frac{x^2 + 2xh + h^2 - x^2}{h}\\ &=\lim_{h→0} \frac{h(2x + h)}{h} \end{align*}
All following simplified forms should have the restriction $h ≠ 0$ since the original form was undefined at this point.
$$\lim_{h→0} {2x + h}, h ≠ 0$$
However to calculate the derivative, the h is valued at $0$ leading to the derivative:
$$f'(x) = 2x$$
How can the equation be simplified by assuming the $h$ is $0$ when there is a restriction on it? Why is that when simplifying expressions we have to restrict the simplified forms to prevent evaluation at points undefined on the original expression, but this concept is completely ignored when dealing with limits?
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How do you define $\lim_{h\to0}2x+h$? – sai Jul 11 '12 at 2:35
The value of the expression as the value of $h$ approaches $0$ – user26649 Jul 11 '12 at 2:35
And wouldn't you agree that this value is $2x$? At no point did we need to set $h=0$. – sai Jul 11 '12 at 2:36
$\displaystyle \lim_{h \to 0} (2x+h)$ is not the same as evaluating $2x+h$ at $h=0$. – user17762 Jul 11 '12 at 2:39
Exactly...which equals to actually substitute $\,h=0\,$ since the expression $\,2x+h\,$ is a linear polynomial in $\,h\,$ and thus continuous everywhere...This was not so before, when the cancellation was done. Remember, when evaluating the limit of an expression when $\,x\to x_0\,$, we shall and we must take $\,x\neq x_0\,$...very close to it, but not actually equal to it...**unless** we have continuity of the function at $\,x_0\,$ – DonAntonio Jul 11 '12 at 2:40 | {
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In a sense, you are repeating the old criticism of Bishop Berkeley on infinitesimals, which were "sometimes not equal to $0$, and sometimes equal to $0$".
What you need to remember is that the expression $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ represents the unique quantity (if it exists) that the expression $\dfrac{f(x+h)-f(x)}{h}$ approaches as $h$ approaches $0$, but without $h$ being equal to $0$. Whenever we take a limit, we are asking how the quantity is behaving as we approach $0$, but without actually being $0$.
Because we are never actually at $0$, the simplification is valid, and so the computation turns on asking: what happens to the quantity $2x+h$ as $h$ approaches $0$?
The answer is that, the closer $h$ gets to $0$, the closer that $2x+h$ gets to $2x$. We can make $2x+h$ as close to $2x$ as we want, provided that $h$ is close enough to $0$, without being equal to $0$.
We are not actually evaluating at $0$ (well, we kind of are, see below, but not really); we are just finding out what happens to $2x+h$ as $h$ gets closer and closer and closer to $0$. So we are not "simplifying" the way we did before, we are now evaluating the limit, by determining what happens to $2x+h$ as $h$ approaches $0$.
(Now, in a sense we are evaluating, for the following reason: the function $g(h) = 2x+h$, where $x$ is fixed, is continuous and defined everywhere. One of the properties of continuous functions (in fact, the defining property of being continuous) is that $g(t)$ is continuous at $t=a$ if and only if $g$ is defined at $a$, and $$\lim_{t\to a}g(t) = g(a).$$ That is, if and only if the value that the function approaches as the variable approaches $a$ is precisely the value of the function at $a$: there are no jumps, no breaks, and no holes in the graph at $t=a$. But we are not "simplifying" by "plugging in $a$", we are actually computing the limit, and finding that the limit "happens" to equal $g(a)$. | {
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This cannot be done with the original function $\dfrac{f(x+h)-f(x)}{h}$ because, as you note, it is not defined at $h=0$. But there is a result about limits which is very important:
If $f(t)$ and $g(t)$ have the exact same value at every $t$, except perhaps at $t=a$, then $$\lim_{t\to a}f(t) = \lim_{t\to a}g(t).$$
the reason being that the limit does not actually care about the value at $a$, it only cares about the values near $a$.
This is what we are using to do the first simplification: the functions of variable $h$ given by: $$\frac{(x+h)^2-x^2}{h}\qquad\text{and}\qquad 2x+h$$ are equal everywhere except at $h=0$. They are not equal at $h=0$ because the first one is not defined at $h=0$, but the second one is. So we know that $$\lim_{h\to 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to 0}(2x+h).$$ And now we can focus on that second limit. This is a new limit of a new function; we know the answer will be the same as the limit we care about, but we are dealing with a new function now. This function, $g(h) = 2x+h$, is continuous at $h=0$, so we know that the limit will equal $g(0)=2x$. Since this new limit is equal to $2x$, and the old limit is equal to the new limit, the old limit is also equal to $2x$. We didn't both take $h\neq 0$ and $h=0$ anywhere. We always assumed $h\neq 0,$ and then in the final step used continuity to deduce that the value of the limit happens to be the same as the value of the function $g(h) = 2x+h$ at $h=0$. )
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Thank you! I'm losing count of how many times your genius has saved my ass! – user26649 Jul 11 '12 at 3:30
@Riddler: Thank you for the kind words. Related to your question (which was actually deeply troubling to mathematicians for about 100 years), see this, this, and this. – Arturo Magidin Jul 11 '12 at 3:33 | {
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Formally, we say $\lim_{x\to a}f(x)=f(a)$ if $\forall \epsilon>0, \exists \delta>0$ such that $|x-a|<\delta\Rightarrow |f(x)-f(a)|<\epsilon$ (except for $x=a$, in case the function is not continuous at $x=a$) or for sequences $\lim_{n\to\infty}a_n=a$ if $\forall \epsilon>0, \exists N\in\mathbb{N}$ such that $|a_n-a|<\epsilon, \forall n>N$. Note that in both definitions, we allow for $\epsilon$ tolerance, which is positive - but may be arbitrarily small.
In your example, $\lim_{h\to0}2x+h=2x$ since $(2x+h)-2x=h$, which can be made arbitrarily small. Alternatively, you can think of a line with slope = 2. What the limit is saying is that as the vertical shift of the line $h$ approaches 0, the line will get arbitrarily close to the slope 2 line through the origin.
-
For limits, we aren't (in general) simply evaluating at a particular point. Intuitively, when we're dealing with real-valued functions on subsets of the reals, we talk about the graph of a function getting as close as we like to a point if we make the $x$-coordinate get sufficiently close to the point's $x$-coordinate.
Rigorously, we use the following definitions:
(i) Suppose $E\subseteq\Bbb R$, $x_0\in\Bbb R$. We say that $x_0$ is a limit point of $E$ if and only if for every $\varepsilon>0$ there is some $x\in E$ such that $0<|x-x_0|<\varepsilon$. (That is, there are points of $E$ that are as close to $x_0$ as we like, but still distinct from $x_0$.)
(ii) Suppose $E\subseteq\Bbb R$, $x_0\in\Bbb R$ a limit point of $E$, $L\in\Bbb R$, and $f:E\to\Bbb R$ is some function. We say that $$L=\lim_{x\to x_0}f(x)$$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that $|f(x)-L|<\varepsilon$ whenever $x\in E$ and $0<|x-x_0|<\delta$. (That is, $L=\lim_{x\to x_0}f(x)$ means we can make $f(x)$ as close as we like to $L$ simply by choosing $x\in E$ sufficiently close to $x_0.$ Note that $x_0$ need not be an element of $E.$) | {
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For your example, we're dealing with an $h$ limit (so replace the $x$ terms above with $h$ terms). Letting $\delta=\varepsilon$, we find that for $0<|h-0|<\delta$, we have $$|(2x+h)-2x|=|h|=|h-0|<\delta=\varepsilon.$$
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# Prove that the product of two evens is even. Generalise to any divisor.
I am getting these questions from a book called, An Inquiry-Based Introduction to Proofs by Jim Hefferon. However, I do not understand what he means by "Generalise to any divisor.". I thought the divisor would always be 2, given we are talking about an even number.
My solution I have is simply this:
1. $a$ and $b$ are even.
2. $\exists k,m \in \mathbb{Z}: a = 2k$ and $b = 2m$.
3. $a \times b = 2k \times 2m = 2(2km) =$ even.
However, I am not sure if that satisfies the request to "Generalise to any divisor" because I don't really know what that means? Does anyone know? Thanks.
• Replace "even" by "multiple of $2$" then you can generalize to any larger integer. – didgogns Jun 15 '17 at 14:37
• You made a calculating error, $2k\times 2m=4km$ – User123456789 Jun 15 '17 at 14:37
• $2k \cdot 2m \ne 2(k+m)$ although it is even. – NickD Jun 15 '17 at 14:37
• You're right @Daan, thanks for the headsup. – Bucephalus Jun 15 '17 at 14:39
Note that in your answer, you should have $2k\cdot 2m=4mk=2(2mk)$, not $2(k+m)$. Otherwise, your proof is correct.
By generalizing to any divisor, it means that it wants you to prove the following:
Suppose $n$ and $m$ are both divisible by $k$. Prove that $n\cdot m$ is divisible by $k$. | {
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Suppose $n$ and $m$ are both divisible by $k$. Prove that $n\cdot m$ is divisible by $k$.
• Oh yeah that explanation is what I'm looking for. So if I want to go on and do this for n and m as you have stated, do I edit my original post or I answer my post? – Bucephalus Jun 15 '17 at 14:43
• Well, you may accept whichever answer you feel answers your question best, including your own. Editing your own question might make hide the original purpose of it. – ervx Jun 15 '17 at 14:49
• Yes, but I mean generally, sometimes someone will give you a hint, and not the answer, and so you have to continue with your solution. Should you edit and append it to your original post? Or is it preferred that you answer your own post with an updated version? – Bucephalus Jun 15 '17 at 14:51
• If the hint helps you answer your question, you should probably accept that answer. There is no need to then post a full solution to your question. – ervx Jun 15 '17 at 14:53
• ok, thanks @ervx – Bucephalus Jun 15 '17 at 14:53
$2p\times 2k=2^2pk$
$2\mid 2^2pk.$ QED
$np\times nk=n^2pk$
$n\mid n^2pk$. QED
• Oh yeah, ok so that's what I should be putting if I were to add it. Thanks @RobertFrost – Bucephalus Jun 15 '17 at 14:44
• @Bucephalus yeah, to generalise just means prove for the general case, i.e. for any divisor (in this case $n$), not just $2$. – user334732 Jun 15 '17 at 14:45 | {
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Lemma 10.89.7. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a universally exact sequence of $R$-modules. Then:
1. If $M_2$ is Mittag-Leffler, then $M_1$ is Mittag-Leffler.
2. If $M_1$ and $M_3$ are Mittag-Leffler, then $M_2$ is Mittag-Leffler.
Proof. For any family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules we have a commutative diagram
$\xymatrix{ 0 \ar[r] & M_1 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_2 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & M_3 \otimes _ R (\prod _{\alpha } Q_{\alpha }) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \prod _{\alpha }(M_1 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_2 \otimes Q_{\alpha }) \ar[r] & \prod _{\alpha }(M_3 \otimes Q_{\alpha })\ar[r] & 0 }$
with exact rows. Thus (1) and (2) follow from Proposition 10.89.5. $\square$
Comment #2973 by on
Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J.
Comment #2974 by on
Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)
Comment #2975 by on
Additional statement: (3) If $M_2$ is Mittag-Leffler and $M_1$ is finitely generated, then $M_3$ is Mittag-Leffler. Proof: The diagram already given together with 059M and 059J. (This should probably be a separate Lemma, since it requires the sequence to be only exact, not universally exact.)
Comment #3098 by on
OK, yes. I've added this as a separate lemma. It seems to me one could also deduce this directly from the definition of ML modules. If you have a minute, please check the changes to see if you agree. Thanks!
There are also:
• 4 comment(s) on Section 10.89: Interchanging direct products with tensor | {
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There are also:
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In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | {
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# Show that there is a $c\in(0,1)$ such that $f(c)=\int_0^cf(x)dx$.
Question: Let $$f:[0,1]\to\mathbb{R}$$ be a continuous function such that $$\int_0^1f(x)dx=\int_0^1xf(x)dx.$$ Show that there is a $$c\in(0,1)$$ such that $$f(c)=\int_0^cf(x)dx.$$
My solution: Define the function $$g:[0,1]\to\mathbb{R}$$, such that $$g(x)=x\int_0^x f(t)dt-\int_0^x tf(t)dt, \forall x\in[0,1].$$
Now since $$f$$ is continuous on $$[0,1]$$, thus we can conclude by Fundamental Theorem of Calculus that $$g$$ is differentiable $$\forall x\in[0,1]$$ and $$g'(x)=\int_0^x f(t)dt+xf(x)-xf(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$
Observe that $$g(0)=g(1)=0$$. Hence by Rolle's Theorem we can conclude that $$\exists b\in(0,1)$$, such that $$g'(b)=0$$, i.e $$\int_0^b f(t)dt=0.$$
Now define $$h:[0,1]\to\mathbb{R}$$, such that $$h(x)=e^{-x}g'(x), \forall x\in[0,1].$$
Now $$h'(x)=-e^{-x}g'(x)+g''(x)e^{-x}=e^{-x}(g''(x)-g'(x)), \forall x\in[0,1].$$
Observe that $$h(0)=h(b)=0$$. Hence by Rolle's Theorem we can conclude that $$\exists c\in(0,b)\subseteq (0,1)$$, such that $$h'(c)=0$$. This implies that $$e^{-c}(g''(c)-g'(c))=0\\\implies g''(c)-g'(c)=0\hspace{0.3 cm}(\because e^{-c}\neq 0)\\\implies f(c)=\int_0^cf(x)dx.$$
Is this solution correct? And is there a better solution that this?
• Your solution looks fine. Apr 16 '20 at 5:18
• Isn't this something like the reverse Lagrange's mean value theorm? Correct me if I'm wrong... Apr 16 '20 at 6:36
• See related question math.stackexchange.com/q/3610557/72031 Apr 16 '20 at 7:55
Your proof is correct. This is another one. | {
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Your proof is correct. This is another one.
We may assume that $$f$$ is not identically zero (otherwise it is trivial). Since $$f$$ is continuous and $$\int_0^1(1-x)f(x)\,dx=0$$ we have that $$M=\max_{x\in [0,1]}f(x)>0$$ and $$m=\min_{x\in [0,1]}f(x)<0$$. Moreover $$\exists x_M,x_m\in [0,1]$$ such that $$f(x_M)=M$$ and $$f(x_m)=m$$. Let us consider the following continuous map $$F(x):= f(x) - \int_0^xf(t)\,dt.$$ If $$x_M<1,$$ then $$F(x_M)=M-\int_0^{x_M}f(t)\,dt\geq M- Mx_M >0.$$ If $$x_M=1$$ then, $$F(x_M)=M-\int_0^{1}f(t)\,dt> 0$$ because $$f$$ strictly less than $$M$$ in an interval of positive length containing $$x_m$$. In both cases we conclude that $$F(x_M)>0$$. In a similar way, we show that $$F(x_m)<0$$.
Finally, since $$F$$ is continuous on $$[0,1]$$, it follows, by the Intermediate Value Theorem, that there exists $$c$$ strictly between $$x_M$$ and $$x_m$$, and therefore $$c\in (0,1)$$, such that $$F(c)=0$$, that is $$f(c)=\int_0^cf(t)\,dt.$$
• +1 It appears that the approach would also work for the question linked in my comment to the current question. Apr 23 '20 at 2:02
• Another observation: what you have proved is that if a continuous $f$ changes sign in $[0,1]$ then we have a $c\in(0,1)$ such that $f(c) =\int_{0}^{c}f(x)\,dx$. Apr 23 '20 at 2:08
As noted by RobertZ, your proof is correct. Here is another proof that follows the same outline as yours: first we find another zero for the antiderivative of $$f$$ and then we use Rolle's theorem in an appropriate way. This approach is admittedly more long-winded but doesn't make use of the $$e^{-x}$$ trick.
Define $$F: [0,1] \to \mathbb{R}$$ as $$F(x) =\displaystyle \int_{0}^{x}f(t)dt.$$ Note that the given condition can be stated as $$\displaystyle \int_{0}^{1}F(t)dt =0$$
Claim 1: There exists $$b \in (0,1)$$ such that $$F(b) =0.$$
Proof of claim 1: By the mean value theorem for integrals there exists $$b \in (0,1)$$ such that $$F(b)= \displaystyle \int_{0}^{1}F(t)dt$$, which implies $$F(b)=0.$$ | {
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Now, we look for an appropriate sub-interval of $$[0,b]$$ on which we can apply Rolle's theorem to $$g.$$
Let $$G(x)=\displaystyle \int_{0}^{x}F(t)dt$$ and define $$g:[0,b] \to \mathbb{R}$$ by $$g(x)= G(x) -F(x).$$
Claim 2: $$g$$ is not injective on $$[0, b].$$
Proof of claim 2: Suppose not. Then $$g$$ is injective and since it is clearly continuous too, $$g$$ is monotone. WLOG, let $$g$$ be monotone increasing. Then since $$g$$ is differentiable, $$g'(x) \geq 0 \, \forall \, x \in [0,1].$$ If there exists at least one $$x$$ for which $$g'(x) =0$$ we are done so assume $$g'(x)>0.$$ Since $$g(0) =0,$$ we have $$g(x)>0$$ for all $$x \in (0,b].$$
Let $$x_{0}$$ be a point of maximisation for $$F.$$ Assume $$F$$ is not identically $$0$$ or else $$f$$ is and the problem is trivial. We claim that there exists $$c \in (0, b)$$ such that $$F(c)<0.$$ If $$x_{0}=0$$ or $$b$$ then $$F\leq 0$$ so if $$F$$ is not identically $$0$$ choose an other point of $$(0, b)$$ to be $$c.$$ If $$x_{0} \in (0, b)$$ then since $$g({x}_{0})>0, F(x_{0})< \displaystyle \int_{0}^{x_{0}}F(t)dt \leq x_{0}F(x_{0}).$$
If $$F(x_{0}) \neq 0$$ we get $$x_{0} \geq 1,$$ a contradiction. Hence $$F(x_{0})=0$$ and since $$F$$ is not identically $$0$$ there exists some $$c \in (0, b)$$ such that $$F(c)<0.$$
Since $$F$$ is a continuous function on a closed and bounded interval $$[0, b]$$, it attains its bounds. In particular $$\exists \, d \in [0, b]$$ such that $$F(d)\leq F(x) \, \forall \, x \in [0,b].$$ Clearly $$d\neq 0, 1$$ or else $$F(x) \geq 0 \, \forall x \in [0,b]$$ contradicting the fact that $$F(c) <0.$$ Therefore $$d \in (0,b)$$ and since it is a point of minimisation, $$F'(d) =0.$$ Then $$F(d)= F(d) -F'(d) =g'(d)>0> F(c)$$ contradicting the fact that $$d$$ is a point of minimisation of $$F.$$ Therefore our hypothesis that $$g$$ is injective is false and hence $$g$$ is not injective and there exists $$a, a' \in [0, b]$$ with $$a \neq a'$$ such that $$g(a) =g(a').$$ | {
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Then since $$g$$ restricted to $$[a, a']$$ satisfies the conditions for Rolle's Theorem, there exists some $$x_0 \in (a,a')$$ such that $$g'(x_0)=0$$ which implies $$F(x_0)=F'(x_0)$$ from which it follows that $$f(x_{0}) = \displaystyle \int_{0}^{x_{0}}f(x)dx$$
Note that the proof follows almost identically if we assume $$g$$ to be monotone decreasing in the proof of the claim. | {
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# when to say that the given information is (intersection) not (conditional probability)?
There are two urns containing colored balls. The first urn contains 50 red balls and 50 blue balls. The second urn contains 30 red balls and 70 blue balls. One of the two urns is randomly chosen (both urns have probability 50% of being chosen) and then a ball is drawn at random from one of the two urns. If a red ball is drawn, what is the probability that it comes from the first urn?
my question is:
so here (1/2) is the red intersection first urn or it`s the p(red|first urn) I need explanation for that point please and if the second one is the right why it p(red|first urn) and not p(first urn| red)
If we define the following events:
$U1$: Urn $1$ is chosen
$U2$: Urn $2$ is chosen
$R$: a red ball is chosen
then we have:
$P(U1)=P(U2)=P(R|U1)=50$%
The $50$% is not $P(U1 \cap R)$, for we have $P(U1 \cap R) = P(R|U1) \cdot P(U1) = \frac{1}{2}\cdot \frac{1}{2}=0.25$
That is, $P(U1 \cap R)$ is the probability that you choose urn $1$ and pick a red ball, so that mreans you first need to pick urn $1$ ($50$% chance of that), and then you also need to pick a red ball, given that you picked urn $1$, and that is also $50$%, so for both to happen it's $50$% times $50$% is $25$%
Finally, you are asked to find $P(U1|R)$, and for that use Bayes' Law | {
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Finally, you are asked to find $P(U1|R)$, and for that use Bayes' Law
• but you figured that out from the syntax of the problem about the 50% being intersection or conditional probability May 16 '18 at 17:11
• @AhmedAraby The problem says that urn 1 contains $50$ red balls and $50$ blue balls, and so from that we can say that $P(R|U1) = \frac{1}{2}$ May 16 '18 at 17:20
• okay , but when we can say that is probability of R intersection U1 what is the change that will happen to the problem syntax here in order to say that is an intersection May 16 '18 at 18:17
• @AhmedAraby $P(R \cap U1)$ is the probability of picking urn 1 and getting a red ball. Or, in short: the probability of picking a red ball from urn $1$, given that you have not yet decided from which urn you're going to pick a ball. May 16 '18 at 18:28
You have two urns with known amounts of red and blue balls. This tells you the (conditional) probabilities for drawing a red ball from each given urn.
$$\mathsf P(R\mid U_1) = 50/100 = 5/10\\ \mathsf P(R\mid U_2)=30/100=3/10$$
You are to randomly select an urn (ie: without bias). This tells you the (marginal) probabilities for selecting each urn. $$\mathsf P(U_1)=1/2\\ \mathsf P(U_2)=1/2$$
So then the probabilities for selecting each urn and drawing a red ball, will be the products.$$\mathsf P(U_1\cap R)= \mathsf P(U_1)~\mathsf P(R\mid U_1)= 5/20\\\mathsf P(U_2\cap R)=\mathsf P(U_2)~\mathsf P(R\mid U_2)=3/20$$
The Law of Total Probability tells us the (marginal) probability for drawing a red ball.$$\mathsf P(R)~{=\mathsf P(U_1\cap R)+\mathsf P(U_2\cap R)\\=\mathsf P(U_1)~\mathsf P(R\mid U_1)+\mathsf P(U_2)~\mathsf P(R\mid U_2)\\=8/20}$$ | {
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Now then, you are to evaluate the probability that urn 1 was selected given that a red ball was drawn. From this we use wither the defiition for conditional probability, or Bayes' Rule.$$\begin{split}\mathsf P(U_1\mid R) &= \dfrac{\mathsf P(U_1\cap R)}{\mathsf P(R)} &\quad&=\dfrac{\mathsf P(U_1)~\mathsf P(R\mid U_1)}{\mathsf P(U_1)~\mathsf P(R\mid U_1)+\mathsf P(U_2)~\mathsf P(R\mid U_2)}\\ &=\dfrac{5}{8} \end{split}$$
Note: This should be anticipated, because 50 from the 80 red balls are in urn 1 and both urns contain an equal total count for balls. | {
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Class Notes (834,986)
Mathematics (1,919)
MATH 136 (168)
Lecture
# Lecture 33.pdf
6 Pages
110 Views
School
Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter | {
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Friday, March 28 − Lecture 33 : Eigenvectors associated to an eigenvalue. Concepts: 1. Define the family of all eigenvectors associated to an eigenvalue. 2. Find all eigenvectors associated to an eigenvalue. 3. Define an eigenspace. 33.1 Theorem − Let λ be a number. The number λ is an eigenvalue ofA if and only if 1 1 the matrix equation Ax = λ x w1th variable x has a non-trivial solution. Proof : λ 1s an eigenvalue ofA ⇔ λ is a 1olution to det(A − λI) = 0 ⇔ det(A − λ I)1= 0 ⇔ A − λ I 1s not invertible ⇔ (A − λ I)1 = 0 has a non-trivial solution ⇔ Ax − λ Ix1= 0 has a non-trivial solution ⇔ Ax = λ Ix 1as a non-trivial solution ⇔ Ax = λ x h1s a non-trivial solution. Different ways of viewing an eigenvalue. The following are equivalent: 1. λ i1 an eigenvalue ofA 2. (A − λ I1x = 0 has a non-trivial solution 3. Ax = λ x 1as a non-trivial solution 4. Null(A − λ I)1≠ {0} has a non-trivial solution 33.1.1 Example − Show that 1 is an eigenvalue of the following matrix A Solution: By the above theorem it suffices to verify that the following homogeneous system(A – (1)I)x = 0. has a non-trivial solution. This coefficient matrix easily row-reduces to Since (A – (1)I)x = 0 has a non-trivial solution then 1 is an eigenvalue of A. Then Ax = x has a non-trivial solution. Similarly, Ax = 2x and Ax = 3x both have non-trivial solutions and so 2 and 3 are eigenvalues of A. 33.1.2 Different ways of viewing an eigenvalue. We have the following different ways of viewing an eigenvalue of A: λ1= eigenvalue of A ⇔ det(A − λ I) = 0 1 ⇔ A − λ I 1s not invertible, ⇔ (A − λ I)1 = 0 has a non-trivial solution. ⇔ Null(A − λ I) 1ontains infinitely many vectors ⇔ Ax = λ x for infinitely many vectors x 1 33.2. Definition − Suppose λ is a1 eigenvalue of n × n matrix A. Then any non-zero vector x in Null(A − λ I) is called an eigenvectorof A corresponding to λ . 1 1 1 33.2.1 Remark – By definition of “eigenvector of a matrix A” an eigenvector always exists in relation to some eigenvalue. So to say that | {
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of a matrix A” an eigenvector always exists in relation to some eigenvalue. So to say that x is an eigenvector of the matrix 1 A invites the following question: “x is 1ssociated to which of the eigenvalues of A?”. An eigenvector of A can only correspond to a single eigenvalue. If x is an 1 eigenvector of A associated to the eigenvalues λ and λ1of A th2n λ = λ . Thi1 is 2 because x c1nnot be the zero vector. To see this consider: The expression “λ 1 1= λ 2 1mplies λ = λ1” si2ce an eigenvector cannot be the zero vector. 33.2.2 Remark – Extending the notion of “eigenvalue” so that it applies to linear mappings. The notions of “eigenvalue” and “eigenvector” of a matrix A can be extended to a linear map (transformation) T : ℝ → ℝ .n n Definition –We say that λ is 1n eigenvalue of the linear map T: ℝ → ℝ if n n T(x) = λ x1 has a non-trivial solution for x. We say thatx 1 is an eigenvector of the linear transformation T corresponding to the eigenvalue λ 1 if x1is a non-trivial solution to T(x) = λ x1 n n - If λ i1 an eigenvalue of T: ℝ → ℝ and x is an eigenvec1or corresponding to λ 1hen, “T maps x to a1scalar multiple λ x of x ”1 1 1 2 2 33.2.2.1 Example – Suppose R : ℝ → ℝ πs the linear transformation which rotates points in ℝ counterclockwise about the origin by an angle of π radians. Suppose we seek an eigenvalue and its associated eigenvectors for this linear map. See that if x is any ordered pair in ℝ , R (x) = −x. Then λ = −1 is an eigenvalue of π 1 2 R π The eigenvectors associated to λ = −1 ar1 all non-zero vectors in ℝ . The set of all eigenvectors corresponding to an eigenvalue – If λ is an eigen1alue of a matrix A, all the eigenvectors associated to λ form a1set or “package”. This set is the nullspace, Null(A − λ I ), of the mat | {
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More Less | {
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# Defining compact sets with closed covers
This question is a continuation of this. My book says that a metric space is compact if and only if:
$$M=\cup A_{\lambda}\implies M = A_{\lambda1}\cup\cdots\cup A_{\lambda_n}$$
where each $A_{\lambda}$ is open. Then, it says that the definition can be extended to closed sets, since if $A_\lambda$ is a family of open sets in $M$, then the complements $F_\lambda = F-A_\lambda$ forms a family of closed sets in $M$. We have, then: $$M=\cup A_\lambda\iff \cap F_\lambda = \emptyset.$$ So, by the last question, I understood the reason behind this. No problem. The book then proceeds:
A metric space $M$ is compact, $\iff$ all families $(F_\lambda)_{\lambda\in L}$ of closed sets with empty intersection has a finite subfamily with empty intersection: $F_{\lambda_1}\cap \cdots \cap F_{\lambda_n}=\emptyset$.
Then, the book proceeds and says:
A family $(F_\lambda)_{\lambda\in L}$ has the property of finite intersection when for any finite subset $\{\lambda_1,\cdots,\lambda_n \}\subset L$ we have $F_{\lambda_1}\cap\cdots\cap F_{\lambda_n}\neq \emptyset$.
The following condition is necessary and sufficient to tell that a metric space $M$ is compact:
If $(F_\lambda)_{\lambda\in L}$ if a family of closed sets with the property of finite intersection, then $\cap_{\lambda\in L}F_\lambda \neq \emptyset$
Is the last one the negation of the first one? I mean, for $M$ to be compact, either we have:
family of closed sets with empty intersection has finite subfamily with empty intersection
or
family of closed sets with the property of finite intersection (that is, every intersection of finite subfamilies is not empty) does not have a empty intersection.
Is this correct?
I ask this, because the book proceeds for a proof of Dini's theorem like this: | {
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Is this correct?
I ask this, because the book proceeds for a proof of Dini's theorem like this:
An example of a family of closed subsets with the property of finite intersection is a decrescent sequence $F_1\supset F_2\supset \cdots\supset F_n\supset \cdots$ of nonempty closed subsets of $M$. If $M$ is compact, it follows that $\cap_{n=1}^{\infty}F_n\neq \emptyset$. The following demonstration shows these ideas:
Dini's theorem: If a sequence of continuous real functions $f_n:M\to \mathbb{R}$ defined in a compact metric space $M$, converges simply to a function $f:M\to \mathbb{R}$, and if $f_1(x)\le f_2(x)\le \cdots\le f_n(x)\le \cdots$ for all $x\in M$, then the convergence $f_n\to f$ is uniform in $M$.
Demonstration: Given $\epsilon>0$, choose, for each $n\in \mathbb{N}$:
$$F_n=\{x\in M: |f_n(x)-f(x)|\ge \epsilon\}$$
Then, $F_1\supset F_2\supset \cdots \supset F_n\supset \cdots$ and each $F_n$ is closed in $M$. We must prove that there is $n_0\in \mathbb{N}$ such that $F_{n_0}=\emptyset$ (then $n>n_0\implies |f_n(x)-f(x)|<\epsilon$ for all $x\in M$). Well, since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n = \emptyset$ (WHY?). Being $M$ compact, we must have $F_n=\emptyset$ for some $n$
And besides the chain of sets $F_1\supset F_2\supset \cdots \supset F_n\supset \cdots$ having the property of finite intersection (every subfamily has a not empty intersection), the proof concludes that $\cap_{n=1}^{\infty}F_n = \emptyset$
• Your two bolded statements are contrapositive to each other. – Omnomnomnom Jun 24 '16 at 19:53 | {
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• Your two bolded statements are contrapositive to each other. – Omnomnomnom Jun 24 '16 at 19:53
Since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n =\emptyset$
Note that $$\bigcap_{n=1}^\infty F_n = \left\{x \in M : |f_n(x) - f(x)| \geq \epsilon \text{ for all } n\right\}$$ However, if $x$ is such that $f_n(x) \to f(x)$, then (by definition of convergence) $x$ is not an element of this set. So, if $f_n(x) \to f(x)$ for every $x \in M$, then $\bigcap_{n=1}^\infty F_n = \emptyset$.
• ok, now, in the end, it says that being $M$ compact, we must have $F_n=\emptyset$ for some $n$. Why? And which one of the bolded definitions it's using? I ask because in the beggining, it says that the chain of descending closed sets has the property of finite intersection, that is, any subfamily of them has a not empty intersection. How is it possible to conclude that the intersection IS empty? – Guerlando OCs Jun 24 '16 at 20:14 | {
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# verifying solution to first order differential equation
#### find_the_fun
##### Active member
Verify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.
$$\displaystyle (y-x)y'=y-x+8$$ where $$\displaystyle y=x+4\sqrt{x+2}$$
So the derivative is $$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$
and the LHS becomes $$\displaystyle (y-x)(1+\frac{2}{\sqrt{x+2}})= y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=$$
$$\displaystyle x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}$$
and the RHS becomes $$\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$
which isn't equal but the answer key seems to think it is because it gives an interval of definition.
#### MarkFL
Staff member
I agree that if:
$$\displaystyle y=x+4\sqrt{x+2}$$
then:
$$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$
Now, let's look at the left side of the ODE:
$$\displaystyle (y-x)y'=\left(x+4\sqrt{x+2}-x \right)\left(1+\frac{2}{\sqrt{x+2}} \right)=4\sqrt{x+2}+8$$
And next, let's look at the right side:
$$\displaystyle y-x+8=x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$
This then shows that:
$$\displaystyle (y-x)y'=y-x+8$$
#### Prove It
##### Well-known member
MHB Math Helper
Verify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.
$$\displaystyle (y-x)y'=y-x+8$$ where $$\displaystyle y=x+4\sqrt{x+2}$$
So the derivative is $$\displaystyle y'=1+\frac{2}{\sqrt{x+2}}$$
and the LHS becomes $$\displaystyle (y-x)(1+\frac{2}{\sqrt{x+2}})= y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=$$
$$\displaystyle x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}$$ | {
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and the RHS becomes $$\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$
which isn't equal but the answer key seems to think it is because it gives an interval of definition.
\displaystyle \begin{align*} \left( y - x \right) \, \frac{dy}{dx} &= y - x + 8 \end{align*}
If \displaystyle \begin{align*} y = x + 4\sqrt{x+2} \end{align*} then \displaystyle \begin{align*} \frac{dy}{dx} = 1 + \frac{2}{\sqrt{x + 2}} \end{align*} and so substituting into the DE we have:
\displaystyle \begin{align*} LHS &= \left( y - x \right) \, \frac{dy}{dx} \\ &= \left( x + 4\sqrt{x + 2} - x \right) \left( 1 + \frac{2}{\sqrt{x + 2}} \right) \\ &= x + \frac{2x}{\sqrt{x + 2}} + 4\sqrt{x + 2} + 8 - x - \frac{2x}{\sqrt{x + 2}} \\ &= \left( x + 4\sqrt{x + 2} \right) - x + 8 \\ &= y - x + 8 \\ &= RHS \end{align*}
#### Prove It
##### Well-known member
MHB Math Helper
Another possibility is brute force, actually solving the DE:
\displaystyle \begin{align*} \left( y - x \right) \, \frac{dy}{dx} &= y - x + 8 \\ \frac{dy}{dx} &= \frac{y - x + 8}{y - x} \end{align*}
Make the substitution \displaystyle \begin{align*} u = y - x \implies \frac{du}{dx} = \frac{dy}{dx} - 1 \implies \frac{dy}{dx} = \frac{du}{dx} + 1 \end{align*} and the DE becomes
\displaystyle \begin{align*} \frac{dy}{dx} &= \frac{y - x + 8}{y - x} \\ \frac{du}{dx} + 1 &= \frac{u + 8}{u} \\ \frac{du}{dx} &= \frac{u + 8}{u} - 1 \\ \frac{du}{dx} &= \frac{8}{u} \\ u\,\frac{du}{dx} &= 8 \\ \int{ u\,\frac{du}{dx}\,dx} &= \int{8\,dx} \\ \int{u\,du} &= 8x + C_1 \\ \frac{1}{2}u^2 + C_2 &= 8x + C_1 \\ \frac{1}{2}u^2 &= 8x + C_1 - C_2 \\ u^2 &= 16x + C \textrm{ where } C = 2C_1 - 2C_2 \\ \left( y - x \right) ^2 &= 16x + C \\ y - x &= \sqrt{16x + C} \\ y &= x + \sqrt{16x + C} \end{align*}
Of course, your given function \displaystyle \begin{align*} y = x + 4\sqrt{x + 2} \end{align*} is the case where \displaystyle \begin{align*} C = 32 \end{align*}, thus your given function is a solution to the DE. | {
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# Evaluate the triple integral.
Evaluate the triple integral of $x=y^2$ over the region bounded by $z=x$, $z=0$ and $x=1$ My order of integration was $dx\:dy\:dz$.
I want to calculate the volume of this surface. I solved it for $dz\:dy\:dx$ and it was:
$$V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx$$
And for $dz\:dx\:dy$ would be this:
$$V=\int_{-1}^{1}\int_{y^2}^{1}\int_{0}^{x}dz\:dx\:dy$$
I tried to solve it and the result is this:
$$V=\int_{0}^{1}\int_{-\sqrt{x}}^{\sqrt{x}}\int_{z}^{1}dx\:dy\:dz + \int_{0}^{1}\int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{y^2}^{1}dx\:dy\:dz$$
But i think its wrong please advice me the best solution .
I wanted to post the shape of this surface in 3-dimensional region but I couldn't because I am new user.
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What are the ▒ supposed to be? To get a proper integral sign with limits, enclose \int_0^1 in dollar signs to get $\int_0^1$ – Ross Millikan Aug 5 '12 at 21:18
## 2 Answers
Integrating in three dimensions will give you a volume, not the area of a surface. Your region is not well defined in the first line-it is a triangle in the $xz$ plane but there is no restriction in the $y$ direction. If you want the region to be bounded by $x=y^2$ then your integral is correct, $V=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}\int_{0}^{x}\:dz\:dy\:dx=\int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}x\:dy\:dx=\int_0^12x\sqrt{x}\:dx=\frac 45 x^{\frac 52}|_0^1=\frac 45$. This is the triple integral of $1$ over that volume. | {
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yes it would be 4/5 for dzdydx & dzdxdy but for the dxdydz with that formula that i wrote its not the same so its wrong i don't know how to fix it – shahin Aug 6 '12 at 0:37
for $dx \:dy \:dz$, the lower $x$ limit has to be the maximum of $y^2$ and $z$, you don't add the two integrals together. In your last line, you can't have $x$ in the limits of the $y$ integral as you have already integrated over $x$. In the second integral on that line, I don't know where the $y$ limits came from-we don't have any $\frac 12$'s around. – Ross Millikan Aug 6 '12 at 0:50
yes i think it would be $$\int_0^1\int_{-1}^{1 }\int_1^{y^2}dxdydz$$ – shahin Aug 6 '12 at 1:02
but its still wrong – shahin Aug 6 '12 at 1:28
(to follow up on where this was left in August 2012):
Here is a graph of the region in question, made just incomplete enough to allow the interior to be viewed. The surface $\ x = y^2 \$ is a parabolic cylinder extending indefinitely in the $\ z-$ directions; the surface $\ z = x \$ is an oblique plane; and $\ z = 0 \$ is, of course, the $\ xy-$ plane. The volume is then a sort of wedge with a tilted "roof" and a parabolic "wall".
The integrals for the integration orders $\ dz \ dy \ dx \$ , which is sort of the "natural" order many people would use, and $\ dz \ dx \ dy \$ are shown in the other posts. One should not be too quick to arbitrarily re-arrange the order of integration in multiple integrals for a variety of reasons, sometimes because of the geometric configuration of the integration region, sometimes because of the integrands one would be left to grapple with. | {
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In this problem, the amount of symmetry of the region permits us to choose alternative orders of integration without producing any "crisis". For the order $\ dx \ dy \ dz \$ , we are able to make use of the fact that one of the boundary surfaces is $\ z \ = \ x \$ . Since we want to "herd" the integration toward working in the variable $\ z \$ , we can carry out the integration in $\ x \$ from $\ 0 \$ to $\ z \$ . For the integration in $\ y \$ , we can therefore also "replace" $\ x \$ by $\ z \$ to express the relevant portion of the parabolic surface as $\ z \ = \ y^2 \$ . The integration limits on $\ z \$ become $\ 0 \$ to $\ 1 \$ , as they were for $\ x \$ .
We can now write the "re-ordered" integration as
$$\int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ \ \text{or} \ \ \ 2 \ \int_0^1 \int_0^{\sqrt{z}} \int_0^z \ \ dx \ dy \ dz \ \ ,$$
by exploiting the symmetry of the region about the $\ xz-$ plane. Since this new integral simply looks like a "relabeled" version of the first integral in the original post, using the order $\ dz \ dy \ dx \$ , this will plainly give the same result for the volume of $\ \frac{4}{5} \$ .
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# distance formula: simplifying sqrt(148)
#### marshall1432
##### Banned
i have a small question...regarding the distance formula...
(-7, 1) (5,3)
using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)
How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...
I don't need it in a decimal
#### Mrspi
##### Senior Member
Re: distance formula
marshall1432 said:
i have a small question...regarding the distance formula...
(-7, 1) (5,3)
using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)
How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...
I don't need it in a decimal
You need to see if you can simplify sqrt(148)
Does 148 have any perfect square factors? If so, you can remove the square root of any perfect square factor from under the radical sign.
Here's an example:
Simplify sqrt(60)
Note that 60 = 4*15, and that 4 is a perfect square. So,
sqrt(60) = sqrt(4*15), or sqrt(4)*sqrt(15)
Now, you know that sqrt(4) is 2, so sqrt(4)*sqrt(15) can be written as 2 sqrt(15).
And sqrt(60 = 2 sqrt(15).
Since 15 does not have any perfect square factors, this is the simplest form for sqrt(60).
Try the same process on sqrt(148).
#### Subhotosh Khan
##### Super Moderator
Staff member
Re: distance formula
marshall1432 said:
i have a small question...regarding the distance formula...
(-7, 1) (5,3)
using the distance formula.....sqrt(x2-x1)+(y2-y1)...I was able to determine that the answer is sqrt(148)
How do I go about getting the final answer or is this...I am unsure, but I think it stays this way...
I don't need it in a decimal
This is good enough in my view.
However, you can reduce it one step further by factorizing148 and taking some factor/s out-of-the radical sign. That is
$$\displaystyle \sqrt{50} \,=\, \sqrt{5^2\cdot2} \, = \,5\sqrt{2}$$
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$$\displaystyle \sqrt{50} \,=\, \sqrt{5^2\cdot2} \, = \,5\sqrt{2}$$
#### marshall1432
##### Banned
Re: distance formula
well there is a perfect square being sqrt 4 * sqrt 37
so would you consider the answer to be 2 sqrt 37?
#### Subhotosh Khan
##### Super Moderator
Staff member
Re: distance formula
marshall1432 said:
well there is a perfect square being sqrt 4 * sqrt 37
so would you consider the answer to be 2 sqrt 37<<<<< Correct
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# Probability of dying from smallpox?
A family of four is infected with Variola major. There is a fatality rate of 30%. Calculate the probability that...
Here are my attempts,
The probability that nobody dies, $$0.7^4\cdot100\%=24.01\%$$
The probability that everybody dies, $$0.3^4\cdot100\%=0.81\%$$
The probability that at least one person dies is equal to $$(1-3^4)\cdot100\%=99.19\%$$ or the probability that 1 person dies + 2 people + 3 people + 4 people
The probability that one person dies, two people die, three people die are what I'm having trouble with. For instance, for the probability 1 person dies: $0.3\cdot\binom{4}{1}\cdot n$
I know I have to multiple it by some value n, but I just can't figure out the logic behind what I need to do. I figure I have to find the probability that 1 person dies and multiply that by the probability that the other 3 people live.
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The probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier. It is about $0.76$, or if you prefer percent (I don't) it is about $76\%$. – André Nicolas Aug 26 '14 at 4:57
What about exactly one person dying, exactly two people dying, exactly three people dying? Sorry, I must have worded my question poorly, but that's what I meant. I thought about this after watching an episode of House, M.D., and my curiosity sparked. – Jason Aug 26 '14 at 5:00
Andre is pointing out that your third answer is incorrect. You found the complement probability to everyone dying, but you need the complement probability to no one dying. – alex.jordan Aug 26 '14 at 5:01
Ah yes, I see how I was wrong. Thank you for pointing that out. – Jason Aug 26 '14 at 5:04
But hasnt smallpox been eradicated ? – Rene Schipperus Aug 26 '14 at 5:05
The probability at least $1$ person dies is $1$ minus the probability nobody does, which you computed earlier. | {
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Now let us compute the probability exactly one person dies. Call the people A, B, C, D. The event exactly one person dies can happen in $4$ ways: (i) A dies and the rest survive; (ii) A survives, B dies, C and D survive; (iii); (iv).
If we assume independence, the probability of (i) is $(0.3)(0.7)(0.7)(0.7)$. The probability of (ii) is $(0.7)(0.3)(0.7)(0.7)$, the same. We also get the same answer for (iii) and (iv). So the probability exactly $1$ dies is $(4)(0.3)(0.7)^3$.
Now let us find the probability exactly $2$ die. Again, there are several ways this can happen. How many ways? As many as there are ways to choose the $2$ unlucky people. There are $6$ ways to do that. Each of them has probability $(0.3)^2(0.7)^2$. So the probability exactly $2$ die is $(6)(0.3)^2(0.7)^2$.
The probability exactly $3$ die is calculated in a similar way. There are $4$ "patterns."
In general, if the probability of "success" (in this case death) is $p$, and the experiment is repeated independently $k$ times, then the probability of exactly $k$ successes is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ For details, please see Wikipedia, Binomial distribution. There are many other web sources. The Khan Academy stuff is good.
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For choosing 2 people to die, why do we count each person 3 times? i.e., if people are named ABCD, we can choose AB to die, AC to die, AD to die; then we also add BC, BD and CD. Sure, this is six. But we're counting each person three times. Doesn't this cause problems? – Jason Aug 26 '14 at 6:25
The two people to die could be A and B, A and C, A and D, B and C, B and D, or C and D. Then for example the probability A and C die and the other two survive is $(0.3)(0.7)(0,3)(0.7)$, that is, $(0.3)^2(0.7)^2$. Same for the other five. No problems, there is no double-counting involved. – André Nicolas Aug 26 '14 at 6:31 | {
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The probability that say, exactly one person dies is $(0.3)(0.7)^3\binom{4}{1}$. The $0.3$ is the chance that the first person will die, and then the $(0.7)^3$ is the chance that the second, third, and fourth survive. The $\binom{4}{1}$ counts for the other "combinations" where exactly one person dies: maybe it was the first person, maybe it was the second, maybe it was the third, and maybe it was the fourth.
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Let $N$ be the number of people who die. Then we have a binomial distributution. $$N\sim\mathcal{Bin}(n, p), n=4, p=0.3$$
Hence the probability that exactly $k$ people die is: $$\mathsf P(N=k)= {n\choose k}\cdot p^k \cdot (1-p)^{n-k} = \frac{4! \cdot 0.3^k \cdot 0.7^{4-k}}{k! \cdot (4-k)!}$$
The binomial coefficient counts the ways to select the required number of people. The exponentials on the probability and its complement measure the probability of a each of those people either dying or not, as selected.
\begin{align} \mathsf P(N=0) & = 1 \cdot (0.7)^4 & = 24.01\% \\ \mathsf P(N=1) & = 4 \cdot (0.3) \cdot (0.7)^3 & = 41.16\% \\ \mathsf P(N=2) & = 6 \cdot (0.3)^2 \cdot (0.7)^2 & = 26.46\% \\ \mathsf P(N=3) & = 4 \cdot (0.3)^3 \cdot (0.7) & = 7.56 \% \\ \mathsf P(N=4) & = 1 \cdot (0.3)^4 & = 0.81 \% \\ \end{align}
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# permutation with repetition allowed | {
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When the order doesmatter it is a Permutation. The formula is written: n r. where, n is number of things to choose from; r is number of things we choose of n; repetition is allowed; order matters; Permutation without Repetition In this post, we will see how to find all lexicographic permutations of a string where repetition of characters is allowed. 216. Another definition of permutation is the number of such arrangements that are possible. Permutations with Repetition Permutation when repetition is allowed. When a permutation can repeat, we just need to raise n to the power of however many objects from n we are choosing, so . For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition. Male or Female ? "With repetition" means that repetition is allowed. First position can have N choices The second position can have ( N-1 ) choices. Permutation without Repetition: This method is used when we are asked to reduce 1 from the previous term for each time. To improve this 'Permutation with repetition Calculator', please fill in questionnaire. There are basically two types of permutation: Repetition is Allowed: It could be “333”. A permutation is an arrangement of objects, without repetition, and order being important. Repeating of characters of the string is allowed. Permutations are items arranged in a given order meaning […] List permutations with repetition and how many to choose from. Like combinations, there are two types of permutations: permutations with repetition, and permutations without repetition. They have sometimes been referred to as permutations with repetition, although they are not permutations in general. Permutations. The number of permutations of ‘n’ things taken ‘r’ at a time is denoted by n P r It is defined as, n P r A permutation with repetition of n chosen elements is also known as an "n-tuple". | {
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defined as, n P r A permutation with repetition of n chosen elements is also known as an "n-tuple". Total […] No Repetition: for example the first three people in a running race. That was an $$r$$-permutation of $$n$$ items with repetition allowed. Permutations without Repetition In this case, we have to reduce the number of available choices each time. All the different arrangements of the letters A, B, C. All the different arrangements of the letters A, A, B A permutation is an ordering of a set of objects. Print all permutations with repetition of characters, Given a string of length n, print all permutation of the given string. In some cases, repetition of the same element is allowed in the permutation. Please update your bookmarks accordingly. (Repetition allowed, order matters) Ex: how many 3 litter words can be created, if Repetition is allowed? We should print them in lexicographic order. D. 320. For this case, n and k happens to be the same. Permutation formulas. Permutation without Repetition: for example the first three people in a running race. 1. There is a subset of permutations that takes into account that there are double objects or repetitions in a permutation problem. C. 120. When additional restrictions are imposed, the situation is transformed into a problem about permutations with restrictions. With permutations, every little detail matters. These are the easiest to calculate. This is a permutation with repetition. It could be "333". Permutation with Repetition. You can’t be first and second. When the number of object is “n,” and we have “r” to be the selection of object, then; Choosing an object can be in n different ways (each time). For example, what order could 16 pool balls be in? Noel asks: Is there a way where i can predict all possible outcomes in excel in the below example. Permutations with repetition. Permutation with repetition occurs when a set has r different objects, and there are n choices every time. Repetition of characters is | {
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when a set has r different objects, and there are n choices every time. Repetition of characters is allowed. Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. Permutations without repetition A permutation is an arrangement, or listing, of objects in which the order is important. For the given input string, print all the possible permutations. Compare the permutations of the letters A,B,C with those of the same number of letters, 3, but with one repeated letter $$\rightarrow$$ A, A, B. After choosing, say, number "14" we can't choose it again. Permutations: There are basically two types of permutation: Repetition is Allowed: such as the lock above. Technically, there's no such thing as a permutation with repetition. n different things taking r at a time without repetition - definition The number of permutations of n different things, taking r at a time without repetition is denoted by n P r . We can actually answer this with just the product rule: $$10^5$$. It has following lexicographic permutations with repetition of characters - AAA, AAB, AAC, ABA, ABB, ABC, … 7.1.5 When repetition of objects is allowed The number of permutations of n things taken all at a time, when repetion of objects is allowed is nn. Permutations with repetition. OR 125. 1. Ordered arrangements of n elements of a set S, where repetition is allowed, are called n-tuples. There are two types of permutations: Repetition is Allowed: For the number lock example provided above, it could be “2-2-2”. For example, locks allow you to pick the same number for more than one position, e.g. Post, we have 26 choices of letter: $26^ { }! S, where repetition is allowed, is nr from the previous term for each time repetition. Will be equal to, '' with repetition by treating the elements as an ordered set and. K happens to be the same arrangements that are identical opens with 1221,. More than one position \ ( n\ ) items | {
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permutations in general to find all lexicographic permutations of a of... From a zero-based index to the nth permutation allowed, are called n-tuples C Program k=combination length.... An ordered combination a way where i can predict all possible outcomes in excel in the example... Are items arranged in a running race the nth permutation of characters is allowed instance, be 333 have positions. Permutations without repetition, and language arts that there are basically two types of permutation is an arrangement a! Transformed into a problem about permutations with repetition allowed, is nr ] List with.: repetition is allowed: we can select 4 multiple times if we want letters go. Digits from 1 to 5 letter: $26^ { 4 }$ possibilities the possible.. Math, science, and permutations without repetition a permutation is an arrangement objects! Set S, where repetition is allowed: such as the lock above n=number of elements k=combination!, in each of the words are allowed items with repetition of the different which... | {
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# Conditional Probability: probability of having a girl.
## Homework Statement
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?
## The Attempt at a Solution
I think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.
But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.
Which is right?
1/2
it doesn't matter
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My vote's 1/3. This is a different question to 'My eldest child of two is a girl, what is the probability that they are both girls', which is 1/2. In your case, the piece of information provided is not 'ordered' but, instead, excludes the BB possibility.
Using Bayes theorem, the probability of event A, given event B, (with events independent) is the probability of A intersect B divided by probability of B.
So, the probability of the kid being a girl, given that there is already a boy, is the probability of a girl intersected with a boy divided by the probability of a boy.
To find A intersect B for independent events, just multiply the probability of each. (1/2)(1/2). So, (1/4) divided by (1/2).
It's the probability that it's a girl, given that at least one is a girl - nothing involving a boy. Does that change things?
I think you're right. The outcome of one boy and one girl should be twice as likely as the other 2 outcomes, since this is a http://en.wikipedia.org/wiki/Binomial_distribution" [Broken] with n=2 and p=1/2. | {
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More formally, let $p = 1/2$ be the probability a girl is born and on our universe U define the random variables X = number of girls, Y = number of boys. Define the events $C = \{X = 2 \}$ and $D = \{X \geq 1 \}$. Certainly $C \subseteq D$ so $C \cap D = C$. Since $p = 1/2$ and the two births are independent events, then $\mathbf{P}(C) = (1/2)^2 = 1/4$. The probability of the complement of D, $\mathbf{P}(U \setminus D)$ is the same as $\mathbf{P}\{X=0\}$, which is the probability of 2 boys. Similarly, since $1-p = 1/2$ and the two births are independent events, then $\mathbf{P}(U \setminus D) = (1/2)^2 = 1/4$. So $\mathbf{P}(D) = 1 - \mathbf{P}(U \setminus D) = 3/4$. Finally
$$\mathbf{P}(C|D) = \frac{\mathbf{P}(C \cap D)}{\mathbf{P}(D)} = \frac{\mathbf{P}(C)}{\mathbf{P}(D)} = \frac{1/4}{3/4} = 1/3 \, .$$
I think your proof using the reduced sample space is correct as well--and shorter, too!
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diazona
Homework Helper
Yep, 1/3. For a less technical explanation, if it helps anyone:
Imagine surveying a large number of families with 2 children. In half of them, the older child will be a boy. Within that half, half again (or a quarter of the total) will have the younger child be a boy, and the rest (another quarter of the total) will have the younger child be a girl. The same goes for the remaining half of the families whose older child is a girl. So you have four possibilities: BB, BG, GB, GG (in order of age), each representing a quarter of the total population.
Now, the statement "given that at least one of them is a girl" excludes one of those possibilities, BB. There are three equally likely possibilities left (BG, GB, GG), of which exactly one has both children being girls. So the probability is 1/3.
Imagine surveying a large number of families with 2 children...
Yes, that was the the means I convinced myself of my analysis. | {
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Yes, that was the the means I convinced myself of my analysis.
I think the comment about Bayes has actually taken the conclusion the wrong way. What we learn in the statement 'there is a girl' is a piece of information that teaches us to exclude 1/4 of the population set of 2-child families.
@AN; You have concluded Bayes says 'To have a boy is twice as likely as to have a girl'. Therefore probability of two girls is 1/3 and a boy and a girl is 2/3, because 2/3 is twice 1/3!!! (Because we've excluded the BB option from the P=1 outcome.)
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Oops, I was talking about girls and boys. That doesn't change the answer though. The Bayes theorem is sound. The answer is 1/2.
But what's the question? '1/2' is not the answer to the op, but is the answer to 'what is the probability of GG compared with not GG' (conditional on 'there is a girl').
Ray Vickson
Homework Helper
Dearly Missed
## Homework Statement
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?
## The Attempt at a Solution
I think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.
But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.
Which is right?
This is the same problem as: we toss a (fair) coin twice and get at least one head; what is the probability we get another head as well? The results LOOK a bit paradoxical when written out as follows: events are C = {two heads}, A = {first toss is heads}, B = {second toss is heads}. We have P{C|A} = P{C|B} = 1/2, but P{C|A or B} = 1/3. That is your problem in a nutshell. This looks paradoxical because the statement that we get at least one head is the same as saying that A or B occur, and each of these separately implies a 1/2 probability of C; however, when taken together they imply a 1/3 probability of C!
RGV | {
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RGV
I like Serena
Homework Helper
Hi everyone!
Here's my view on the matter.
There is a difference between saying:
"given the first (or youngest) kid is a girl" (chance 2/4)
and "given that at least one kid is a girl" (chance 3/4).
In the first case the chance would be 1/2 as Arcana claims.
We can also write this as:
$$P(\text{2 girls | the youngest kid is a girl}) = {\text{number of favorable outcomes} \over \text{number of possible outcomes}} = {1 \over 2}$$
In the second case it is:
$$P(\text{2 girls | at least 1 girl}) = {\text{number of favorable outcomes} \over \text{number of possible outcomes}} = {1 \over 3}$$
So I believe the reasoning of pearapple is sound for this problem and the chance is 1/3.
vela
Staff Emeritus
Homework Helper
But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.
Since the others have explained how to get the correct answer, I'll point out what mistake your friend is making:
You have four possible outcomes: BB, BG, GB, and GG. BG and GB are distinct outcomes. The order matters when you're enumerating outcomes.
When you combine them to form events, then the order may or may not matter depending on how the event is defined. The event "exactly 1 girl" would be {BG, GB}. Whether the girl comes first or last doesn't really matter.
Note that even though the order doesn't matter in determining what outcomes are included in an event, you still have to add up the probabilities for each outcome to get the probability of the event. That is, P({BG, GB}) = P({BG})+P({GB}) = 1/2. It's incorrect to say: order doesn't matter, so BG=GB and P({BG, GB})=P({BG})=1/4. | {
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The event "at least one girl" is {BG, GB, GG}, which corresponds to a probability of 3/4. If you incorrectly reduce this to {BG, GG}, as your friend has done, you get a probability of 1/2. If you take the probability GG, which is 1/4, and divide it by 3/4, you get the correct answer of 1/3. If instead you divide it by the incorrect probability of 1/2, you get 1/2, the answer your friend got.
BruceW
Homework Helper
Its amazing that even such a simple example has an answer that is not immediately obvious.
The answer to this question is ambiguous because the question itself is ambiguous.
I believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.
So you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.
If you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB
BruceW
Homework Helper
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?
Is different to:
I have two kids. I know one of them, which is a girl. What is the probability that they are both girls?
I suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true. | {
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I guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.
HallsofIvy
Homework Helper
Pretty much what others are saying: there are two different questions that can be confused here:
1) A person has two children, one of which is a girl. What is the probability that the other child is a girl?
2) A person has two children, the older of which (or the younger or the one we happen to know- anything that distinguishes one child from another) is a girl. What is the probability that the other child is a girl?
With two children, and assuming that "boy" and "girl" for any given child are equally likely, There are four equally likely possibilities: BB, BG, GB, GG where, of course, "B" means "boy", "G" means "girl" and the order depends upon which was born first, or was the one we knew, etc.
With problem (1) above, knowing that one of the children is a girl removes "BB" from the list, leaving "BG", "GB", and "GG". In only one of those is the other child also a girl. Probability of two girls, 1/3.
With problem (2) above, knowing that the older child (or younger child or whichever is distinguished in some way) we remove both "BB" and "BG". We are left with "GB" and "GG" of which 1 has both girls. Probability of two girls, 1/2.
I like Serena
Homework Helper
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?
Is different to:
I have two kids. I know one of them, which is a girl. What is the probability that they are both girls?
I suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true.
I guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.
Uhh :uhh:. Aren't those the same? | {
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Uhh :uhh:. Aren't those the same?
The answer to this question is ambiguous because the question itself is ambiguous.
I believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.
So you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.
If you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB
Uhh :uhh:. Aren't those the same?
Is there anything that you don't understand specifically about the above explanation?
I like Serena
Homework Helper
Is there anything that you don't understand specifically about the above explanation?
It seems to me that BruceW's explanation is not correct.
It seems to me that BruceW's explanation is not correct.
I feel similar about his explanation, not that it is incorrect, just maybe incomplete.
vela
Staff Emeritus
Homework Helper
I think his point might have been clearer if he had changed the question in the second example to "What's the probability the other one is a girl?" to make it clear he's picked one child and that she is a girl and now we're only wondering about the other one. | {
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I don't think the question as originally stated (the one in the original post) is ambiguous. The mistake occurs because of a misconception. It boils down to the fact that the phrase at least one is a girl doesn't mean you can single one girl out and then worry about the rest.
Last edited:
I like Serena
Homework Helper
BruceW might have said: "I have two kids. I know the youngest of them, which is a girl. What is the probability that they are both girls?"
Then it would be right.
Now it's not.
BruceW
Homework Helper
The age doesn't matter. I agree with HallsofIvy's post. As long as there is some way to distinguish between the two children, then there are 2 very similar questions. (But the questions are different, and the answer is different).
And the question asked was the first one, but pearapple's friend mistook it to mean the second one. (I am guessing that is where his mistake came from).
I like Serena
Homework Helper
Right. In retrospect I concur. | {
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