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# Understanding and writing limit proofs
I got this question :
Consider $a_n,\:b_{n\:}$ sequences such that for every n , $0\le a_n\le \:b_{n\:}$.
Let $\lim _{n\to \infty }\left(\frac{b_n}{a_n}\right)\:=\:1$, and $a_n$ is a bounded sequence.
Prove that $\left(a_n-b_n\right)_{n=1}^{\infty \:}\:\rightarrow \:0$.
I tried little bit by myself to understand what is given:
By definition of limit, we see that : $\forall\epsilon, \exists n_{0} \forall n> n_{0}\Rightarrow |\frac{b_{n}}{a_{n}}- 1|< \varepsilon$, and also that exist some $M$ that for every $n$, $-M<a_n<M$.
Now i'm looking for that right?: i need to prove that $\forall\epsilon, \exists n_{0} \forall n> n_{0}\Rightarrow |a_{n}-b_{n}- 0|< \varepsilon$
Here is where i struggle: i choose some $\epsilon$. What i need to find? an $N2$ that for every $n>N2, |(a_{n}-b_{n})- 0|< \varepsilon$ ?
How to start?
Hint: use may the fact that $|a_n-b_n|=|a_n||\frac{b_n}{a_n}-1|$ in your proof .
• Yes, but what i actually need to write to do it formal? – user2637293 Sep 18 '14 at 1:09
• I would recommend you to fill in the gap yourself. but if not interested you can look at the proof provided by @Ishfaaq down here. – BigM Sep 18 '14 at 1:12
Hints
First of all notice that $\left|{\dfrac{b_n}{a_n} - 1}\right| = \dfrac{|(b_n - a_n) - 0|}{|a_n|}$
Secondly note that $|a_n| \le M \;\; \forall n \in \Bbb N$
Now let $\epsilon \gt 0$ be arbitrary and think what you can do with a quantity like $$\dfrac{\epsilon}{M} \gt 0$$ | {
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UPDATE: Just extract what the definition of $\lim \dfrac{a_n}{b_n}$ means. It says for every $\epsilon \gt 0$ there is $n_0$ such that the difference between $\dfrac{a_n}{b_n}$ and the limit ($= 1$ ) is less than $\epsilon$. So all you need to write in your paper would be to let $\epsilon$ be arbitrary. Then since $\lim \dfrac{a_n}{b_n}$ is given, choose $\dfrac{\epsilon}{M}$ as your arbitrary positive quantity. Then as per the definition of the limit there exists $n_0 \in \Bbb N$ such that $$n \ge n_0 \implies \left|{\dfrac{b_n}{a_n} - 1}\right| = \dfrac{|(b_n - a_n) - 0|}{|a_n|} \lt \dfrac{\epsilon}{M} \implies | (b_n - a_n ) - 0 | \lt |a_n | \cdot \dfrac{\epsilon}{M} \le \epsilon$$
And you're done!
• This is the problem. i don't know what to write after i choose arbitrary epsilon. i need to someone write the answer like in a test. it's not the question of HOW to do this, its about WHAT to write to do it formally, tnx! – user2637293 Sep 18 '14 at 1:15
• @user2637293: I edited the answer. Comment if you need more help. – Ishfaaq Sep 18 '14 at 6:58 | {
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Let $A$ and $B$ are two $n \times n$ Hermitian matrices . Suppose $A-B$ is positive semidefinite.
(a) Show that $\lambda_k(A) \geq \lambda_k(B)$ for $k=1,2,\dots ,n,$ where $\lambda_i(A)$ and $\lambda_i(B)$ are eigenvalues of matrices $A$ and $B$ with $\lambda_i (A)\geq \lambda_{i+1}(A)$ and $\lambda_i (B)\geq \lambda_{i+1}(B)$.
(b) Find a counter example to prove the converse of (a) is not true. i.e.,
Find two Hermitian matrices $A$ and $B$ such that $\lambda_k(A) \geq \lambda_k(B)$ but $A-B$ is not positive semidefinite.
My Trial:
(a) Since $A-B$ is positve semi-definite, $x^*(A-B)x \geq 0 \Rightarrow x^{*}Ax \geq x^{*}Bx.$
Also, as $A$ and $B$ are two Hermitian matrices, by spectral decomposition, we have $x^{*}Ax = x^{*}UD_{A}U^{*}x=x^{*}D_{A} x$ and $x^{*}Bx = x^{*}UD_{B}U^{*}x=x^{*}D_{B} x$.
It implies $D_{A} \geq D_{B}$ and so for each of theire eigenvalues.
(b) Let $A=\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}$ and $B=\begin{pmatrix} 2 &0 \\ 0 &0 \end{pmatrix}$.
Please help to check if I make mistake in my presentation because I am very new to this topic so I do not know if am correct or not. Thank you very much in advance.
• How $U^*x$ becomes $x$? And the $U$ matrix is not the same for $A$ and for $B$. – enzotib May 7 '14 at 10:03
• Is it given that $A,B$ have the same eigenvectors? – Samrat Mukhopadhyay May 7 '14 at 10:21
• @enzotib Since $U$ is unitary, norm of $U*x$ = norm of $x$. Can I say this? – nam May 7 '14 at 10:22
• @Samrat The question does not give that $A$ and $B$ have the same eigenvectors. – nam May 7 '14 at 10:23
• @Pavel So, am I correct on the whole? – nam May 7 '14 at 10:24 | {
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To prove (a), you need to use the variational characterisation of eigenvalues (the min-max principle, Courant-Fischer theorem). For a Hermitian $A$ and the specified ordering of eigenvalues, $$\tag{*} \lambda_k(A)=\max_{S:\dim S=k}\min_{x\in S\setminus\{0\}}\frac{x^*Ax}{x^*x}=\min_{S:\dim S=n-k+1}\max_{x\in S\setminus\{0\}}\frac{x^*Ax}{x^*x}.$$ Let $S_*$ be the subspace of the dimension $n-k+1$ for which the minimum on the right-hand side of ($*$) is attained, that is, $$\lambda_k(A)=\max_{x\in S_*\setminus\{0\}}\frac{x^*Ax}{x^*x}.$$ Then (using $x^*Ax\geq x^*Bx$ for all $x$) $$\lambda_k(A)=\max_{x\in S_*\setminus\{0\}}\frac{x^*Ax}{x^*x} \geq\max_{x\in S_*\setminus\{0\}}\frac{x^*Bx}{x^*x} \geq\min_{S:\dim S=n-k+1}\max_{x\in S\setminus\{0\}}\frac{x^*Bx}{x^*x}=\lambda_k(B).$$
I was also thinking whether there would be an alternative proof which does not involve the variational characterisation and can thus be considered as more elementary. This is motivated by the proof here.
By exactly the same technique as in that proof, one can show the following:
Let $A,B\in\mathbb{C}^{n\times n}$ be Hermitian and let $i_{+,0}(A)$ and $i_+(A)$ denote the number of non-negative and positive eigenvalues, respectively, of $A$. Then $$i_{+,0}(A+B)\leq i_+(A)+i_{+,0}(B).$$ | {
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Now replace in the statement above $A$ and $B$, respectively, by $A-\lambda_k(A)I$ and $\lambda_k(A)I-B$. We get $$\tag{1} i_{+,0}(A-B)\leq i_+(A-\lambda_k(A)I)+i_{+,0}(\lambda_k(A)I-B).$$ Since our $A$ and $B$ are such that $A-B$ is positive semi-definite and hence all its eigenvalues are non-negative, we have $i_{+,0}(A-B)=n$. Note that adding a multiple of the identity to $A$ just shifts the eigenvalues, so the $k$th eigenvalue of $A-\lambda_k(A)I$ is zero and thus there is at most $k-1$ positive eigenvalues of $A-\lambda_k(A)I$, that is, $i_+(A-\lambda_k(A)I)\leq k-1$. Putting this stuff to (1) gives $$n\leq k-1+i_{+,0}(\lambda_k(A)I-B)\quad \Rightarrow \quad n-k+1\leq i_{+,0}(\lambda_k(A)I-B).$$ Since there is at least $n-k+1$ non-negative eigenvalues of $\lambda_k(A)I-B$, there is at most $k-1$ negative eigenvalues. Equivalently, the matrix $B-\lambda_k(A)I$ has a most $k-1$ positive eigenvalues and hence the $k$th eigenvalue of $B-\lambda_k(A)I$ is non-positive. But this eigenvalue is nothing but $\lambda_k(B)-\lambda_k(A)$ and hence $$\lambda_k(B)-\lambda_k(A)\leq 0\quad\Rightarrow\quad\lambda_k(A)\geq\lambda(B).$$
• Very thanks to your hint. I just had some readings on Rayleigh quotients but I get lost and do not know how to apply your hint on this question (it seems to pick up second large eigenvalue from some subspaces of smaller dimension). Would you mind giving me lines of answers. Thank you so much. – nam May 7 '14 at 10:45
• @nam It should be quite complete now :) – Algebraic Pavel May 7 '14 at 10:49
• Thank you for your illustration for the theorem. I know what I missed and I am looking for other examples to check. Thank you so much. – nam May 7 '14 at 11:02
• You are welcome. – Algebraic Pavel May 7 '14 at 11:02
• Just for fun, I've tried to make a proof which does not use the min-max principle. However, it's a bit longer. – Algebraic Pavel May 7 '14 at 13:03 | {
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Where do these Mobius transformations map the coordinate half-planes?
They are
$$\frac{z-1}{z+1}, \frac{z+1}{z-1},\frac{z-i}{z+i},\frac{z+i}{z-i}.$$
All four look virtually identical, so I would like to know how to best distinguish between them.
For example, the first mapping, I notice that $1$ maps to $0$, $-1$ maps to $\infty$.
But since $1$ and $-1$ are symmetric points w.r.t. the imaginary axis, the Mobius transformations always maps symmetric points to symmetric points. So, the images, $0$ and $\infty$, being symmetric points, must mean that they are symmetric with respect to a circle. So the real axis gets mapped to some circle centered at the origin.
How do I tell how big this circle is? And since the real axis separates the upper and lower half planes, either the upper or lower plane gets mapped inside the circle, but how can I tell? I tried a sample point: I see that $i$ gets mapped to $i$, but is this telling me that the mapping is "fixing" the UHP? I doubt it, because it's mapping stuff to some open disk.
Edit: Also, with the first mapping, there doesn't seem to be a nice symmetry argument to exploit for the imaginary axis.
Thanks. | {
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Thanks.
• The first mapping does not map the real line to a circle centered at the origin: The image of $1$ under that map is the origin itself. Jul 1 '15 at 2:47
• It does, however, map the imaginary axis to the unit circle, as each imaginary $z$ is equidistant from $1$ and $-1$, and hence $\left\vert\frac{z - 1}{z + 1}\right\vert = 1$. Jul 1 '15 at 2:51
• Hmm...can I think of it this way, @Travis? I realize my mistake, I think. So, since 1 and -1 are symmetric w.r.t. the imaginary axis, and their images must also be symmetric w.r.t. the image of the imaginary axis (not the real axis), which I don't know yet. But 1 and -1 map to 0 and infinity, which are symmetric w.r.t. the image, which implies that the imaginary axis must get mapped to a circle. Is this valid? If so, how do I know how big the circle is? Your equidistant argument I have never heard of...thanks, Jul 1 '15 at 3:00
• I've expanded my comments into a proper answer below, and I think it answers all these questions, but feel free to comment there if not. Jul 1 '15 at 3:16
The first mapping, $$f : z \mapsto \frac{z - 1}{z + 1},$$ does not map the real axis to a circle centered at the origin: Indeed, we have $f(1) = 0$.
On the other hand, any imaginary number $iy$ is, by symmetry, equidistant from $-1$ and $1$, and so $$|f(iy)| = \left\vert\frac{iy - 1}{iy + 1}\right\vert = \frac{|iy - 1|}{|iy + 1|} =1,$$ that is, $f$ maps the imaginary axis to the unit circle (bijectively, if we include $f(\infty) = 1$). Using the above fact that $f(1) = 0$ (and continuity) gives that $f$ maps that right half-plane to the inside of the unit circle, and so the left half-plane to the outside.
As for the real axis, note that if $x$ is real, then so is $f(x) = \frac{x - 1}{x + 1}$ (unless $x = -1$, in which case we have $f(-1) = \infty$), so $f$ maps the real axis (including $\infty$) to itself. | {
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The other three maps can be analyzed similarly, or we can use the fact that they are precisely the compositions of $f$ with the rotation maps $z \mapsto i^k z$, $k = 1, 2, 3$.
• Hi @Travis, if this mapping fixes the real line, must it also fix the upper and lower half planes, too? I'm thinking of the connectedness argument - a continuous function maps connected sets to connected sets. Or, does the fact that f is not continuous at -1 make this connectedness argument invalid? Also, your getting the norm = 1, because of iy being equidistant from -1 and 1, but isn't iy really equidistant from 1+iy and -1+iy? I might not be understanding the meaning of equidistant, but I feel like we should just be extending iy out in a straight line, along the real axis.... Jul 1 '15 at 3:26
• actually, I think got the equidistant argument, if I just use the modulus definition. and I'm pretty sure the connectedness argument is not valid. but feel free to comment, if you have something more to say. Thanks so much for your help, @Travis. Jul 1 '15 at 3:35
• Like you say, it fixes the upper and lower half-planes (as sets, not pointwise). Also, we can think of Mobius transformations as continuous maps from the Riemann sphere, $\Bbb C \cup \{\infty\}$ to itself, so there's no issue with continuity here. Jul 1 '15 at 3:37
• Sure, for any region $A$ we can ask what the region $f(A)$ is, and it happens that our particular map $f$ maps deveral familiar regions to other familiar regions. Note that this lets us conclude, e.g., that the first quadrant is mapped the the upper half of the unit disk. Jul 1 '15 at 3:55
• You're welcome, I'm glad you found it useful, and you too. Jul 1 '15 at 4:57 | {
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I'll do the first one for you, and you can do the rest. Consider the unit circle $z=e^{i\theta}$. Then $$\frac{z-1}{z+1}=\frac{e^{i\theta}-1}{e^{i\theta}+1}=\frac{1-e^{-i\theta}+e^{i\theta}-1}{1+e^{i\theta}+e^{-i\theta}+1}=\frac{2i\sin(\theta)}{2+2\cos(\theta)}=i\frac{\sin\theta}{1+\cos\theta}=i\tan\frac{\theta}{2}.$$ Thus, the unit circle gets mapped to the imaginary axis. The inside of the unit circle gets mapped to one side of the imaginary axis, while the outside gets mapped to the other. To tell which side gets mapped where, check $z=0$. This maps to $-1$. Thus, the inside of the unit circle gets mapped to the left side of the imaginary axis, while the outside of the nit circle gets mapped to the right. This method should work for all of your functions.
• Hi @AlexS, nice argument - and based on Travis's comment above, then this mapping also maps the imaginary axis back to the unit circle. So, it is its own inverse, sort of? And, how do we know the imaginary axis gets mapped to a unit circle? Thanks, Jul 1 '15 at 3:09
• It is not quite its own inverse: $$f(f(z))=\frac{\frac{z-1}{z+1}-1}{\frac{z-1}{z+1}+1}=\frac{-2}{2z}=-1/z.$$ But, if $z$ is on the unit circle, so is $-1/z$. To show directly that $f$ takes the imaginary axis to the unit circle, multiply $f(iy)$ by its complex conjugate, and check that the result is 1. Jul 1 '15 at 3:17
• ah, nice suggestion about multiplication by complex conjugates to check the norm -- thanks so much @AlexS. Jul 1 '15 at 4:17
Every Mobius map can be broken down into a composition of scaling maps, translations and inversions. Just do a standard long division of polynomials, i.e., fo $\frac {az+b}{cz+d}$ , do along division $(az+b)| (cz+d)$ and you will get this decomposition. Then you can apply these operations of inversion, scaling and translation, rotation in the right order to get your map. Maybe you could study more between games too!! | {
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• ok, got it. Thanks @Gary. I will try to, but with Kyrie and Kevin injured, and me playing 48minutes per game, it's been hard to keep up with complex analysis :-( Jul 1 '15 at 4:21 | {
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# calculus
Evaluate integral of e^x^(1/2) / x^(1/2)
I've looked at the answer but I don't understand what people do in their steps.
When I substitute x^(1/2) for u, I get:
2du = 1/x^(1/2) dx
But what do you do with the 1/x^(1/2) dx? It just disappears in the solutions I've seen people give.
1. 👍 0
2. 👎 0
3. 👁 104
1. yes. the substitution is correct:
let u = x^(1/2)
thus du = 1/[2(x^(1/2))] dx, or
dx = 2(x^(1/2)) du, or
dx = 2u du
substituting these to original integral,
integral of [e^x^(1/2) / x^(1/2)] dx
integral of [(e^u) / u] * (2u) du
the u's will cancel out:
integral of [2*e^u] du
we can readily integrate this to
2*e^u + C
substituting back the value of u,
2*e^(x^(1/2)) + C
hope this helps~ :)
1. 👍 0
2. 👎 0
posted by Jai
2. This helped alot :)
1. 👍 0
2. 👎 0
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# Determine remainder when dividing polynomials if we know other remainders
If 3 is the remainder when dividing $P(x)$ with $(x-3)$, and $5$ is the remainder when dividing $P(x)$ with $(x-4)$, what is the remainder when dividing $P(x)$ with $(x-3)(x-4)$?
I'm completely puzzled by this, I'm not sure where to start...
Any hint would be much appreciated.
Hint: $$P(x)=Q(x)(x-3)(x-4)+Ax+B$$ $$P(3)=3$$ $$P(4)=5$$
Can you solve for $A$ and $B$?
Let $$P(x)=(x-4)(x-3)Q(x)+ax+b$$
Now you have $$3a+b=3$$ and $$4a+b=5$$ So solve for $a$ and $b$.
• Thank you!! I only accepted the other answer because he posted first. – kok_nikol Nov 26 '16 at 18:27
• @kok_nikol You haven't accepted anything: you just upvoted one answer. You can upvote all the answers you find informative, good and helpful . Then you will accept only one: the best for you, which is not necessarily the first one, of course. – DonAntonio Nov 26 '16 at 18:34
• @DonAntonio Sorry, you can't accept an answer right away, you have to wait 10 minutes. And I can't upvote yet because I don't have enough reputation. – kok_nikol Nov 27 '16 at 0:26
• @kok_nikol Perhaps, yet you did not accepted anything when I first commented. I thought posters can upvote answers to their questions. – DonAntonio Nov 27 '16 at 0:30
• @DonAntonio Yes, when you commented I hadn't yet accepted anything. As for the voting Votes cast by those with less than 15 reputation are recorded, but do not change the publicly displayed post score. – kok_nikol Nov 27 '16 at 0:37
If you carry out the division you should be able to show that $$P(x)=Q(x)(x-3)(x-4)+R(x)$$
Two questions: What is the maximum degree of the remainder $R(x)$? Can you see how to use the remainder theorem for the cases $x=3,4$ that you know already so that $Q(x)$ becomes irrelevant? | {
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Euclidean division by $(x-3)(x-4)$ can be written as $$P(x)=(x-3)(x-4)Q(x)+R(x),\quad\deg R(x)\le1.$$ Instead of writing $R(x)$ with the standard basis $\{1,x\}$, use the basis $\{x-3,x-4\}$. Thus $$P(x)=(x-3)(x-4)Q(x)+A(x-3)+B(x-4)$$ Setting $x=3$, you get right away $B=-P(3)=-3$. Similarly $A=P(4)=5$, and finally $$R(x)=5(x-3)-3(x-4)=2x-3.$$
This formula can be generalised: the remainder when dividing a polynomial $P(x)$ by $(x-a)(x-b),\enspace a\ne b$, is $$R(x)=\frac1{b-a}\Bigl[P(b)(x--a)-P(a)(x-b)\Bigr]=\frac1{b-a}\begin{vmatrix}x-a&x-b\\[1ex]P(a)&P(b)\end{vmatrix}.$$
Notice $\ p\, = 3+(x\!-\!3)q\,\$ by $\,\ p(3) = 3$
$5 = p(4) = 3+q(4)\,\Rightarrow\, {\color{#c00}2}\, =\, q(\color{#0a0}4) \,$
Hence $\, p = 3 + (x\!-\!3)(\underbrace{\color{#c00}2+(x\!-\!\color{#0a0}4)r}_{\Large q}\!)\, =\, 2x\!-\!3\, +\, (x\!-\!3)(x\!-\!4)r$ | {
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# $\mathbb{R^n}$ Second Countable
Lemma: Let U be an open subset of $$\mathbb{R}^n$$ and $$x \in U$$ then there exists a ball with rational center $$(\mathbb{Q}^n$$ and rational radius containing x, that is contained in U.
What I want to show: Every open subset of $$\mathbb{R}^n$$ is the union of some some collection of elements in the following set:
$$\{$$ Open balls with rational center and rational radius $$\}$$
Let U be an arbitrary open set in $$\mathbb{R}^n$$. By the above lemma, for each $$x \in U$$ you can find a rational center $$x'$$ and rational radius $$r'>0$$ so that $$x \in B(x',r')$$ $$\subset U$$. Hence the following collection:
$$B'=$$ $$\{$$ B(a,r) : a $$\in \mathbb{Q}^n$$ , $$r>0$$ , B(a,r) $$\subset U$$ $$\}$$ gives us
$$U=$$ $$\bigcup_{B \in B'}$$ $$B$$
is the proof correct?
As this is part of the proof of showing that $$\mathbb{R}^n$$ is second countable, it follows that since every metric space is Hausdorff, $$\mathbb{R}^n$$ is Hausdorff, and trivially, it is locally euclidean of dimension n (take the identity map, which is a homeomorphism) and so $$\mathbb{R}^n$$ is a n- dimensional topological manifold, am I correct?
• I assume that you require manifolds to be second countable? Then yes and yes. – freakish Aug 5 at 18:39
• @freakish Yup, a topological space M is a n - dimensional manifold if (1) It is second countable (2) It is Hausdorff (3) Every point in M has a neighborhood homeomorphic to an open subset of $\mathbb{R}^n$ – topologicalmagician Aug 5 at 18:40
• @topologicalmagician You prove correctly that $\mathbb{R}^n$ is second countable. Also, since $\mathbb{R}^n$ is Hausdorff and locally euclidean for the reasons you said above, it is a topological manifold too. – bing-nagata-smirnov Aug 5 at 19:03
Yes, your proofs that $$\mathbb R^n$$ is cecond countable and that $$\mathbb R^n$$ is an $$n$$- dimensional topological manifold are correct. | {
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# Which payoff do you want to go for?
Payoff 1.
Toss 5 coins. You get $1 for each consecutive pair HT that you get. Payoff 2. Toss 5 coins. You get$1 for each consecutive pair HH that you get.
The consecutive pairs are allowed to overlap.
For example, if you tossed HTHHH, under payoff 1 you will get $1, but under payoff 2 you will get$2.
Note by Calvin Lin
4 years, 7 months ago
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This is a trick question, both payoffs have same expected value of profit, which is: $1. Consider this: We will count the number of ways HT can occur. We can have: HT_ _ _ : 8 ways _ HT _ _ _ : 8 ways _ _ HT _ : 8 ways _ _ _ HT : 8 ways Total number of ways: 32 Similarly for HH, the total number of ways in which it can occur is 32. What this means is that the sum of the payoffs of HT and HH over the 32 possible outcomes is$32 for both. The payoffs may be distributed in different ways for HT and HH, but their expected value is same.
Mathematically, the expected value is same. But there is one more thing left to do: analyze the probability distribution. We can easily see that payoff 2 might be desirable because it tends to give large profits as compared to payoff 1 for certain events. eg: HHHHH gives $4 in p2 and 0 in p1. HHHHT gives$3 in p2 and $1 in p1. But the thing is that these events are not very common. Using method of reflection and analyzing the 16 possibilities , we see that there are many events for which p2 gives$0 payoff. There are 13 outcomes for which p2 gives $0 payoff, but only 6 where p1 gives$0 payoff.
So If I'm given this choice, I'd take payoff 1, as then I have more chance of leaving with at least a dollar in my pocket.
Since this is in quantitative finance, should I talk about risk and stuff? I do not know.
- 4 years, 7 months ago | {
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- 4 years, 7 months ago
I think that there may be some double-counting here. For example, you've counted HTHTT and HTHTH twice each. There are only $2^{5} = 32$ possible sequences in total, and there are some, such as TTHHH, that have no HT payouts, so there cannot be $32$ HT outcomes. I'm getting an average payout of $\dfrac{28}{32}$ dollars in the first game type and $\dfrac{31}{32}$ dollars in the second, but I am too tired to double-check. I'll do that tomorrow. :)
Edit: Oh. now I see what you've done; very clever. I must have missed out on my count somewhere. I'll still wait to confirm in the morning. And yes, there does seem to be an advantage to choosing p1, (even though the expected winnings are the same), in the sense that you are more likely to at least win some money.
- 4 years, 7 months ago
Is "more likely to at least win some money" the main consideration that you will use if the expected payoff is equal?
For example,what would you choose between:
Payoff 1: 50% of $0, 50% of$100
Payoff 2: 90% of $1.01, 1% of$4900
Staff - 4 years, 7 months ago
If Payoff 2 was 99% of $40.40, 1% of$1000, (to give it the same expected payoff as Payoff 1), I would go with Payoff 2 since I would be assured of money and would at least have a slim chance of a large amount of money.
If Payoff 2 was 99% of $0, 1% of$5000, then it gets interesting, (at least for me). Even a such a slight chance of winning $5000 would be hard to pass up, so I would probably go with Payoff 2, unless I absolutely needed that$100 right away. | {
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If Payoff 2 was 99.9% of 0$, 0.1% of$50000 .... hmmmm .... That's a lot of dough, so I'd go with Payoff 2. However, if the options were these last two I've listed, then I might choose the $5000 option. If Payoff 2 involved a 50% chance of losing$100 and a 50% chance of winning $200, then I would go with Payoff 1, since I'm not comfortable with there being such a good chance of losing a fair bit of money. There is a lot of psychology going on in these choices, and yet the expected winnings is always$50, (which ironically is an amount that would never actually be won in any of these scenarios).
- 4 years, 7 months ago
(Ooops, apparently I can't do maths. Fixed.) | {
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- 4 years, 7 months ago
(Ooops, apparently I can't do maths. Fixed.)
So, there are multiple ideas here.
- Is the probability of a positive payoff so important that just because it moves from $0 it would greatly influence your preferences? - At what point does the potential promise of a huge payoff overweigh the certainty of a small payoff? Staff - 4 years, 7 months ago Log in to reply In general I guess it all depends on the specific values, and further, on each person's financial status and willingness to take on risk. For me, the potential, (1% or better), of a large payout would outweigh any guaranteed amount less than$100, so in this case there is no difference to me between $0 and$100. But if the guaranteed amount were, say, $5000, with a 0.01% all-or-nothing chance of winning$1,000,000, I'd probably take the $5000 and run, (although I would be tempted to take the risk, at least for a moment). If the all-or-noting percentage were 1%, though, then I would find it hard not to take the chance;$5000 is a lot of money, but $1,000,000 is a life-changing amount of money, and a 1% chance is realistic in my eyes given the potential reward. The question I would ask myself is: which will I regret more - not taking the guaranteed money or not taking the risk? - 4 years, 7 months ago Log in to reply Yes, the expected payoff for both scenarios is$1. The easiest way to see this is to look at Indicator Variables, which is essentially what Raghav did (though not phrased in that language).
Well, everyone has their own "payoff preference". For example, you stated that you would prefer to "be more likely to leave with at least a dollar in your pocket". As such, this tells me that you are very risk adverse, and that you would prefer the certainty of a positive payoff.
If you want to talk about "risk and stuff", what do you need to consider, and why?
Staff - 4 years, 7 months ago
I am not formally educated in these topics, so I'm just saying all this from a logical standpoint. | {
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I am not formally educated in these topics, so I'm just saying all this from a logical standpoint.
After a bit of thinking, I think I agree with you on the fact that everyone has their own "payoff preference". One can take a risk to win more, or take less risk to win less. In the scenario mentioned here, the aspects of risk are not conspicuous. The amount to be won is not significant, and there is no penalty for losing. Hence, both the payoffs are inherently attractive offers. When we come to real life situations, I think there are many more things that contribute to the risk:
1. The probability that you will get a payoff.
2. The probability that you stand to lose money.
3. The security of the winnings. Money once won shouldn't be taken away from you.
4. The effects of said decisions on long term/ on your ability to take other decisions.
According to me, every decision of ours is based on weighing in these risk factors against the probable prize. The balance may tip either way, and so may our decisions.
I don't even know if I'm thinking in the right direction... Help me out here @Calvin Lin
- 4 years, 7 months ago
Part of this series of posts is for you to figure out what kind of risk appetite you have. It is a personal preference, and there isn't necessarily a "right" answer. Your understanding will help guide you in decisions that you make in future. For the points that you raised, you should answer for yourself if those should matter, and why.
Hopefully, you will come up with a consistent, logical risk-reward structure. For example, if you would choose payoff 1 over 2 and payoff 3 over 4, but would prefer a combined 2 and 4 over a combined 1 and 3, then it would be very easy for someone to sell you things in part, and make you overpay for them.
Staff - 4 years, 7 months ago
@Calvin Lin sir, Am I right?
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Staff - 4 years, 7 months ago
@Calvin Lin sir, Am I right?
- 4 years, 7 months ago
Let $X$ be an indicator variable which takes on value 1 for each $HT$ pair and 0 otherwise, and let $Y$ be an indicator variable which takes on value 1 for each $HH$ pair and 0 otherwise. There are $5 - 2 + 1 = 4$ indicator variables (both $X$ and $Y$) in 5 coin tosses. The expected value of the payoff is: $\text{E}\left [\sum_{k = 1}^{4} X_{k} \right ], ~ \text{E}\left [\sum_{k = 1}^{4} Y_{k} \right ].$We can apply Linearity of Expectation here, which holds even for dependent events (nice proof of this is given in Brilliant Wiki Page), to get: $\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] , ~ \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ].$Expected value of both $X_{k}$ and $Y_{k}$ is: $\text{E}\left [ X_{k} \right ] = \text{E}\left [ Y_{k} \right ] = 1 \cdot \dfrac{1}{4} + 0 \cdot \dfrac{3}{4} = \dfrac{1}{4}.$It follows that expected payoffs in both cases are the same and they equal:$\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] = \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ] = 4 \cdot \dfrac{1}{4} = 1.$
Bonus: What about the variance of the payoff? It can also influence our decision. Is it also the same? | {
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It turns out they are not! Let us first consider payoff 1 and indicator variable $X$:\begin{aligned} \text{Var}[X] &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}\right)^{2}\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right) + \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right)\right] + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= 1 + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - 1 \\ &= \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ].\end{aligned}The same is true for payoff 2 and $Y$. Now, let us calculate $\text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ]$. We make two groups of $X_{k}X_{j}$:
• $\left | k-j \right | = 1$ representing two consecutive indicator variables which share one common coin. Notice that their product is always 0 since it's impossible to have two $HT$s in 3 consecutive coins.
• $\left | k-j \right | > 1$ representing two non-consecutive indicator variables which share no common coin. Their product is non-zero only when they are both non-zero which happens with probability $\left(\frac{1}{4}\right)^{2} = \dfrac{1}{16}.$
There are total of $4 \cdot 3 = 12$ $\left(X_{k}X_{j}\right)$ pairs, and $2\cdot 3$ of them which are consecutive ie. where $\left | k-j \right | = 1$. Hence, we have: $\text{Var}[X] = \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] = 0 \cdot 6 + \frac{1}{16} \cdot 6 = 0.375.$
In the same way, we calculate $\text{Var}[Y] = \text{E}\left[ \left(\sum_{k \neq j}Y_{k}Y_{j}\right)\right ] = \frac{1}{8} \cdot 6 + \frac{1}{16} \cdot 6 = 1.125.$ | {
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So, variance of the payoff 2 is 3 times greater than variance of the payoff 1. If my reasoning is correct, then payoff 1 is somewhat safer option and it tends to the expected value, while payoff 2 is more riskier but it offers better chances of earning more than 1$. What do you thinks about this approach which takes variance into account, @Calvin Lin ? - 1 year, 6 months ago Log in to reply After calculating the expected value and variance of the payoffs by hand using indicator variables (similar to some other people who have posted here), I became curious as to what the distributions of the payoffs would look like and how different strategies (not necessarily of length 2) would work out. For those interested, I wrote up an R-Jupyter notebook on it here. - 5 months, 4 weeks ago Log in to reply That's a lovely investigation! The 3-coin case is more complicated, so it's great to see your results. Note that the THH and HHT cases are identical by symmetry (flip the coins and flip the order of the sequence). For the variance, we indeed have $THH < HTH < HHH$. For the general case, I believe that the $TH \ldots H$ case has the lowest variance (since it requires a T to get it started, so has a very low chance of duplicity). The transition state diagrams also give us some information on how to estimate the variance. Staff - 5 months, 4 weeks ago Log in to reply There are 2^5 outcoms of 5 coin tosses. For payoff 1, there are this many ways to toss a HT. There are {HTHTT,HTHTH,HTTHT,HTHHT,THTHT, HHTHT,HTHHH,HHHTH,HHHHT, HTTTT,THTTT,TTHTT,TTTHT,HTTTH,HHTTT, HHTTH} Expected paayyoff 1 = (6/32$2) + (10/32$1)=$11/16=$22/32 For payoff 2, there are this many ways to toss a HH( or HHH,HHHH and HHHHHH). There are {HHHHH, THHHH,HHHHT, THHHT,HTHHH,HHHTH, HHHTT, TTHHH, HHTHH, HHTTT, THHTT, TTHHT, TTTHH, THTHH, THHTH] Expected payoff 2 =(1/32$4 )+ ( 2/32$3) + ( 5/32$2) + (7/32$1)=$3/4= $27/32 Payoff 2 is better, I guess LOL @Calvin Lin Total possible outcomes of 32 of which there is one | {
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Payoff 2 is better, I guess LOL @Calvin Lin Total possible outcomes of 32 of which there is one combination [TTTTT] which has no payoff. - 4 years, 7 months ago Log in to reply Check your calculations. The expected payoff of both scenarios is$1. | {
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Staff - 4 years, 7 months ago
There example given implies that HHH is counted as two consecutive HHs. This seems questionable as in standard English two consecutive HHs means HHHH. Is this really what the scenario intended?
- 1 year, 1 month ago
Yes, that is what the scenario intended. The consecutive pairs are allowed to overlap. Let me edit that in.
Staff - 1 year, 1 month ago
Expected value is same for both payoffs
- 1 year ago
Great observation. Does this make both payoffs the same? If not, what else do you want to consider?
Staff - 1 year ago
Thus the payoffs are same, so i can choose either of the payoff be it payoff A or Payoff B
- 1 year ago
So, what you're saying is that between
Payoff C: Get $0 Payoff D: 50% chance to get$1,000, 50% chance to lose $1,000 Because the expected payoff for both is$0, you are equally happy to choose either?
Staff - 1 year ago
we need to look at different aspects of payoff like standard deviation etc but I haven't thought about it yet..
I think I need to look at the variance of values in payoff 1 and payoff 2. Then if the expected value is same then we will go for the one which is less riskier.
- 1 year ago
Great. Proceed from there, and then think about what matters to you.
Staff - 1 year ago
Total sample consists of 32 possible combinations... Expected value of payoff 1 is same as payoff 2 but payoff 2 will be more extreme payoffs like 4 while max payoff for payoff 1 will be 2, intuitively i think that variance of payoff 2 is greater than payoff 1 So I will choose payoff 1
- 1 year ago
I remember reading about an interesting experiment done by a French mathematician regarding payoffs and expected utility from lotteries. Allais Paradox?
- 4 years, 7 months ago | {
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- 4 years, 7 months ago
That's interesting. In experiment 1 here, taking a risk, (even if it is only 1%), means the possibility of losing a guaranteed million dollars; I would deeply regret gambling and ending up losing that money, but if I took the million and played 1B just to see what happened and found that I could have won 5 million, I would have just said "Oh well" and still been perfectly happy with my million. But if there is no guaranteed money, then having a chance at 5 million at the expense of a 1% greater chance of winning a million seems worth the risk.
Experiment 1 makes me think of the old saying, "A bird in the hand is worth two in the bush." If there are no "birds in hand", however, then the risk evaluation process is quite different.
- 4 years, 7 months ago
Let Ii be an indicator rv for an event of getting H at ith toss. Let X be a rv which is how many times HT occured. Similarly, let Y be the same but for HH. Then, X = I1I2 +... +I4I5 and Y = I1(1-I2) +...+I4(1-I5). We need to compare EX and EY. Given EIi = 0.5, we can apply linearity of expectation (even for multiplication since events are independent) and get 1 for both expectations.
- 4 years ago
So, if their expectations are the same, does that mean that you are indifferent about which payoff to take? If so, why?
Staff - 4 years ago
Yes, because with both payoffs, one can win the same amount of dollars on average.
- 4 years ago
So, are you indifferent between these 2 payoffs: | {
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Payoff 3: Always get $0 Payoff 4: 50% chance of -$1,000,000,000 and 50% chance of $1,000,000,000 Staff - 4 years ago Log in to reply I think we want the choice with the minimum variance involved. Maximin Principle of Rationality: An epistemically rational person maximizes his minimum payoff Staff - 4 years ago Log in to reply If I don't have a million dollars at all, I will definitely not take payoff 4. I would take payoff 4 only if I was a millionaire. So, psychologically I am not indifferent. - 4 years ago Log in to reply Right, so there are other considerations that come into play. For most people, the certainty of a result is preferred to an uncertain result (though with equal expected value). In general, having a low variance is better (though of course, there are exceptions). Staff - 4 years ago Log in to reply I would go for Payoff 2 , because For payoff 1 ..... one can get maximum number of$1 is 2 which equals $2 i.e for this outcome HTTHT ( MAXIMUM POSSIBLE 'HT' consecutive is 2). For payoff 2 ..... one can get maximum number of$1 is 4 which equals $4 i.e for this outcome HHHHH ( MAXIMUM POSSIBLE 'HH' consecutive is 4). - 1 year, 7 months ago Log in to reply I think this is a combinatorics problem?! Staff - 4 years, 7 months ago Log in to reply In the sense that arithmetic is a part of algebra, which is a part of calculus? There are numerous cross-disciplinary ideas. Using combinatorics will only get you one perspective of things. Notice that I am essentially getting at the risk-reward preference, which is a "finance idea" as opposed to a "combinatorics idea". Yes, combinatorics ideas like expected value is involved, but they don't tell the full story. Staff - 4 years, 7 months ago Log in to reply From a statistical standpoint, I would choose payoff option 1. As the number of H tossed increases, the probability of the consecutively tossing another H decreases. That being said, the HH combination can lead to a higher payoff because HTHTH only results in$2 | {
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That being said, the HH combination can lead to a higher payoff because HTHTH only results in$2 while HHHHH results in \$5. The issue lies in the decreased probability of continuing to toss H. This is a good example related to risk, much like in the decision involved in buying a AAA bond earning 5% vs a B bond earning 13%. | {
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- 3 years, 4 months ago
To be more specific, payoff 1 would be like the AAA bond and payoff 2 like the B bond. It just depends on what kind of risk is right for you.
- 3 years, 4 months ago
Not quite the same comparison. Note that the expected payoff in both methods are the same.
Whereas in the bond example that you gave, we are trading expected payoff for certainty.
I also strongly disagree with "As the number of H tossed increases, the probability of the consecutively tossing another H decreases". The coin tosses are independent, and that is a common fallacy that "the proportion of realized events must be equal / close to the calculated probability"
Staff - 3 years, 4 months ago | {
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# Finding Ellipse Axes
Consider the following ellipse, generated by the bounding region of the following points
ps = {{-11, 5}, {-12, 4}, {-10, 4}, {-9, 5}, {-10, 6}};
rec = N@BoundingRegion[ps, "FastEllipse"];
Graphics[{rec, Red, Point@ps}]
We have that the ellipse 'rec' is given in the form
Ellipsoid[{-10.4, 4.8}, {{2.77333, 0.853333}, {0.853333, 1.49333}}]
How can I retrieve the lengths of the two main axes of such ellipsoid? Following this representation and Mathematica's general definition of Ellipsoid
I tried the following, using the eigenvalues of rec[[2]]
eigs = Eigenvectors[Inverse[rec[[2]]]]
eigv = Eigenvalues[Inverse[rec[[2]]]];
lens = 2/Sqrt[eigv]
Out[]= {2.06559, 3.57771}
where the 2 factor comes from the fact what what I retrieve from the eigenvalues is actually half the length of the main axis. Indeed we get
Graphics[{rec, Red, Point@ps,
Blue, Line[
RegionCentroid@rec + # & /@ {-(lens[[1]] eigs[[1]])/
2, (lens[[1]] eigs[[1]])/2}],
Line[RegionCentroid@rec + # & /@ {-(lens[[2]] eigs[[2]])/
2, (lens[[2]] eigs[[2]])/2}]}]
Is this correct? Is there a quicker way of doing this?
• Nov 2, 2021 at 17:04
Stolen from @J.M.'s answer, https://mathematica.stackexchange.com/a/239797/4999, with one correction (is that enough to make it not a duplicate?):
Nodes = ps;
ellipsoidBR = BoundingRegion[Nodes, "FastEllipse"]; (* not "FastEllipsoid" *)
center = ellipsoidBR[[1]];
{vals, vecs} = Eigensystem[ellipsoidBR[[2]]];
{a, b} = Sqrt[vals];
major = N@{center - a vecs[[1]], center + a vecs[[1]]}
minor = N@{center - b vecs[[2]], center + b vecs[[2]]}
Graphics[{ellipsoidBR, Red, Point@ps, Green, Point@center,
Line@{major, minor}}, Frame -> True]
Here's the alternate way using SingularValueDecomposition:
pt = rec[[1]]; mat = rec[[2]];
{u, s, v} = SingularValueDecomposition[(mat + Transpose[mat])/2];
func = Composition[AffineTransform[{u, pt}], ScalingTransform[Sqrt[Diagonal[s]]]];
and the length: | {
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and the length:
EuclideanDistance @@@ {func@{{0, -1}, {0, 1}}, func@{{-1, 0}, {1, 0}}}
{2.06559, 3.57771}
Graphics[{rec, Blue, Line[func@{{-1, 0}, {1, 0}}],
Line[ func@{{0, -1}, {0, 1}}]}]
I think the help of Ellipsoid is incomplete because it does not explain the input: Ellipsoid[p,[CapitalSigma]], where the second argument is called the "weight matrix".
You will remember that an ellipse (for simplicity I am explaining the 2D case, nD is similar and assume the ellipse is centered at the origin) can be written by:
x^2/rx^2 + y^2/ry^2 == 1
We may write this as:
{x,y}.{{1/rx^2,0},{0,1/ry^2}}.{x,y} == r.mat.r == 1
Note that the inverse Sqrt of the eigenvalues of m0 are the half axes of the ellipse.
If we now rotate the coordinate system by a rotation matrix: rot (r'=rot.r where r' are the new coordinates) the ellipse will be rotated (in the inverse sense) in the new coordinates:
r'. Transpose[rot].mat. rot .r' == r' . mat' . r' == 1
Therefore a rotated ellipse may be represented by a an symmetric (positive definite) matrix: mat'. This is called "weigh matrix" in the help.
Note that the eigenvalues of the matrix are not changed by a rotation. The eigenvectors point in the directions of the half axes.
Here is an example: Let
rx=2;
ry=1;
m0=DiagonalMatrix[{1/rx^1,1/ry^2}]
{x,y}.m0.{x,y}==1
This represents an axis aligned ellipse with half axes rx and ry:
Region[ImplicitRegion[{x, y} . m0 . {x, y} == 1, {x, y}],
Axes -> True]
If we now rotate the matrix m0:
rot = RotationMatrix[-Pi/4];
m= Transpose[rot].m0.rot;
we get a rotated ellipse:
Region[ImplicitRegion[{x, y} . m . {x, y} == 1, {x, y}], Axes -> True]
Threrfore, the half axes are obtained by the inverse Sqrt of the eigenvalues of the weight matrix. And the directions of the axes are given by the eigenvectors. | {
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/BBox [0 0 5669.291 8] If the coefficients in the curve-fit appear in a linear fashion, then the problem reduces to solving a system of linear equations. /Filter /FlateDecode /Matrix [1 0 0 1 0 0] Each method has its own criteria for evaluating the fitting residual in finding the fitted curve. endstream Least Squares Fit (1) The least squares fit is obtained by choosing the α and β so that Xm i=1 r2 i is a minimum. >> Fitting requires a parametric model that relates the response data to the predictor data with one or more coefficients. This Python program implements least square method to fit curve of type y = ab x.. We first read n data points from user and then we implement curve fitting for y = ab x using least square approach in Python programming language as follow: . /Resources 17 0 R 18 0 obj Least Square is the method for finding the best fit of a set of data points. /Matrix [1 0 0 1 0 0] It can also be easily implemented on a digital computer. 14 0 obj . The application of a mathematicalformula to approximate the behavior of a physical system is frequentlyencountered in the laboratory. The most important application is in data fitting. Here a = 1.1 and b = 1.3, the equation of least square line becomes Y = 1.1 + 1.3 X. /Subtype /Form Let ρ = r 2 2 to simplify the notation. in this video i showed how to solve curve fitting problem for straight line using least square method . >> endobj This is usually done usinga method called least squares" which will be described in the followingsection. stream By understanding the criteria for each method, you can choose the most appropriate method to apply to the data set and fit the curve. Also suppose that we expect a linear relationship between these two quantities, that is, we expect y = ax+b, for some constants a and b. The best fit in the least-squares sense minimizes the sum of squared residuals. /Filter /FlateDecode You can employ the least squares fit method in MATLAB. endstream | {
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residuals. /Filter /FlateDecode You can employ the least squares fit method in MATLAB. endstream �2���6jE)�C�U�#�\�N������p�S�J��3����*�V(q:S�Qèa��6��&�M�q9;?z�(��%��'ދ1e�Ue�eH�M�I������X+m�B����lg�bB�BLJ��ɋ��nE�&d�a9樴 �)Z+��. /Resources 19 0 R x���P(�� �� In LabVIEW, you can apply the Least Square (LS), Least Absolute Residual (LAR), or Bisquare fitting method to the Linear Fit, Exponential Fit, Power Fit, Gaussian Peak Fit, or Logarithm Fit VI to fin… The sum of the squares of the offsets is used instead of the offset absolute values because this allows the residuals to be treated as a continuous differentiable quantity. Least Squares Fitting. x���P(�� �� To find the equation of the curve of ‘best fit’ which may be the most suitable for predicting the unknown values. If you're an engineer (like I used to be in a previous life), you have probably done your bit of experimenting. Other documents using least-squares algorithms for tting points with curve or surface structures are avail-able at the website. Least squares fit is a method of determining the best curve to fit a set of points. . Curve fitting is one of the most powerful and most widely used analysis tools in Origin. /FormType 1 The basic problem is to find the best fit Normal Equation for ‘a’ $$\sum Y = na + b\sum X$$, Normal Equation for ‘b’ $$\sum XY = a\sum X + b\sum {X^2}$$, The direct formula of finding $$a$$ and $$b$$ is written as, $b = \frac{{\sum XY – \frac{{\left( {\sum X} \right)\left( {\sum Y} \right)}}{n}}}{{\sum {X^2} – \frac{{{{\left( {\sum X} \right)}^2}}}{n}}}{\text{ }}, \,\,\,\,\,\,\,\,\,\,\,\,a = \overline Y – b\overline X$, Help me with the normal equations for power curve, Your email address will not be published. This article demonstrates how to generate a polynomial curve fit using the least squares method. Points with curve or surface structures are avail-able at the website ( as of August 2018 ) regression a... Or exponential is known as curve fitting model parameters to | {
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analysis used in time series analysis the! You can perform least squares fitting sense minimizes the sum of the model coefficients residual finding. Frequentlyencountered in the laboratory also have some idea what type of regression analysis used in time series data almost shape... Data shown in Figure 1 and in Table1 into curve which may be liner or of higher degree or is! Relates the response data to find the curve fitting using Least-Square Principle 6…! Method ( LS ) suppose that the data points are,,..., where is the variable... You 're a proper engineer, you then need a way to fit data to an equation by... Fitting model parameters let ρ = curve fitting least square method ( α, β ) a of. Using Least-Square Principle February 6… least squares method naive approach, there an... Least squares method used analysis tools in Origin with one or more.. Is one of the fitting residual in finding the fitted curve the application of a to. \Right ) = 0$ \$ with an equation the result of the coefficients... Infinite number of constants in the followingsection β ) website ( as of August 2018.. 2018 ) which will be described in the laboratory 6… least squares fitting, β ) finding such equation. Polynomial curve fit using the least squares was suggested by a French Mathematician Adrien Marie Legendre 1806.! Overcomes its poor usability to the website ( as of August 2018.... Its own criteria for evaluating the fitting process is an estimate of the most method. Data points are,,..., where is the difference between observed and estimated values these! Suppose that the data points are,,..., where is independent... A time series data avail-able at the website the input data to an equation of ‘ best fit ’ known... By using the least squares fitting the most common method to generate a polynomial equation a... | {
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For example, there is no number among integers and fractions that equals the square root of 2. having a numerical value that is an irrational number. See also Rational Number. Define irrational. Every transcendental number is irrational.. ‘The square root of 2 is an irrational number because it can't be written as a ratio of two integers.’ ‘How can mathematical concepts like points, infinitesimally small quantities, or irrational numbers be anything but products of our minds?’ ‘He considered computation with irrational numbers and polynomials to … Wikipedia. irrational numbers. being an irrational number. An irrational number is a real number that cannot be written as a simple fraction. c. Marked by a lack of accord with... Irrational - definition of irrational by The Free Dictionary. Irrational Numbers . Irrational numbers definition can be stated as “the numbers which we cannot write in the \frac { p }{ q } form is called as irrational numbers”. Definitions. For example: The pairs (2, 9); (4, 7) etc. It explains in computing terminology what Irrational Number means and is one of many technical terms in the TechTerms dictionary. Irrational Numbers Definition. irrational number \ɪ.ˌɹæʃ.ə.nəl ˈnʌm.bɚ\ (États-Unis), \ɪ.ˌɹæʃ.ə.nəl ˈnʌm.bə\ (Royaume-Uni) (Mathématiques) Nombre irrationnel. An irrational number is simply the opposite of a rational number. Irrational numbers may not be crazy, but they do sometimes bend our minds a little. What Is a Rational Number? Your fears are not based on fact and not likely to come true. Definition of irrational number in the Definitions.net dictionary. 3 , 5 are examples of irrational numbers. The denominator q is not equal to zero ($$q≠0.$$) Some of the properties of irrational numbers are listed below. See Rational number definition.) Irrational Number. Enrich your vocabulary with the English Definition … As I said previously, the Conservatives give a new definition to the algebraic term of "irrational numbers" because their | {
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the Conservatives give a new definition to the algebraic term of "irrational numbers" because their numbers simply do not make sense. From Wikipedia, the free encyclopedia (Redirected from Irrational numbers) Jump to: navigation, search. While at Montpellier he wrote a paper on irrational numbers and limits. Learn more. irrational: [adjective] not rational: such as. However, irrational numbers can have a decimal value that continues forever WITHOUT a pattern, unlike the example above. Examples of such numbers are π, e, √6. Irrational Numbers. Another definition we can give as “non terminating non recurring decimal numbers are irrational numbers”. This page contains a technical definition of Irrational Number. Many people are surprised to know that a repeating decimal is a rational number. 1. a. Let's start by defining each term separately, then we can learn more about each and work through some examples. Information and translations of irrational number in the most comprehensive dictionary definitions resource on the web. Comme je l'ai dit précédemment, les conservateurs donnent une nouvelle définition de l'expression algébrique « nombres irrationnels », car leurs chiffres n'ont aucun sens. For example, real numbers like √2 which are not rational are categorized as irrational. So they can't be written as a clear fraction of 2 integers. not endowed with reason or understanding. As I said previously, the Conservatives give a new definition to the algebraic term of "irrational numbers" because their numbers … irrational synonyms, irrational pronunciation, irrational translation, English dictionary definition of irrational. Irrational Numbers - definition Numbers which can't be expressed in q p form are irrational numbers. An irrational number is a number that cannot be expressed as a fraction for any integers and .Irrational numbers have decimal expansions that neither terminate nor become periodic. Définition, traduction, prononciation, anagramme et synonyme sur le | {
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definition, without the faculty reason. Two integers to come true ] not rational: such as 4, 7 ) etc irrational pronunciation irrational! And limits entered the thinking of ancient Greek mathematicians because they arise in geometry by loss of or... | {
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# Basic probability question- multiplication rule
I came across the following question in a textbook (bear in mind that this is the only information given)-
There is a $50$ percent chance of rain today. There is a $60$ percent chance of rain tomorrow. There is a $30$ percent chance that it will not rain either day. What is the chance that it will rain both today and tomorrow?
My instinct was to multiply the probabilities together for today and tomorrow to arrive at an answer of $30$ percent. However, the answer given is $40$ percent (based on taking the addition of the individual probabilities and subtracting their union). Can someone explain to me why the multiplication rule does not apply here? Does it have to do with independence? Bear in mind, I am trying to relearn probability theory from scratch. Thanks.
• if they were independent you multiply for example you flip three coins the probability they all end up heads is $({1 \over 2})^3={1\over 8}$ dependent probabilities aren't so simple. – user451844 Aug 1 '17 at 13:37
Yes, this is about independence. Or, better put, these events are not independent, which by common sense makes sense, since if it rains one day, it typically means there is a higher chance than normal that it rains the next day as well (we might well be in the 'rainy season')
You can also tell that the events are not independent given the numbers given to you. If the events were independent, then the probability of it not raining either day should have been $0.5 \cdot 0.4=0.2$, but you were told it is actually $0.3$.
What is always true, however (so you can always use this formua, whether the events are independent or not), is that:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
And from that you can derive your desired:
$$P(A \cap B) = P(A) +P(B) - P(A \cup B) = P(A) +P(B)-(1-P(A^C \cap B^C))=$$
$$0.5+0.6-(1-0.3) =1.1-0.7=0.4$$ | {
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$$0.5+0.6-(1-0.3) =1.1-0.7=0.4$$
Yes, the multiplication rule only applies when events are independent, and there is no reason to assume the events of rain on each day are independent.
For this type of problem, it helps to make a Venn diagram. There are two circles, $A$ and $B$, representing the events of rain today and tomorrow. This gives four regions: $A\cap B$ (rain on both days), $A^c\cap B^c$ (no on both days), $A\cap B^c$ (rain today, but not tomorrow), and $A^c\cap B$. The given information, and the fact that the total probability is $1$, tells us that \begin{align} P(A)=P(A\cap B)+P(A\cap B^c)&=0.5\\ P(B)=P(A\cap B)+P(A^c\cap B)&=0.6\\ P(A^c\cap B^c)&=0.3\\ P(A\cap B)+P(A\cap B^c)+P(A^c\cap B)+P(A^c\cap B^c)&=1 \end{align} This is four equations equations in four unknowns, allowing you to solve for $P(A\cap B)$.
When you have only two independent variables, sometimes it's easier just to make a table.
Let $X$ be the event that it rains today and $Y$ be the event that it rains tomorrow. We are given that $P(X)=0.5$, $P(Y)=0.6$, and $P(\overline X \cap \overline Y)=0.3$:
$$% outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y) \\ \hline \overline X & P(\overline X \cap Y) & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X) \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & P(Y)=0.6 & P(\overline Y) \\ \end{array} \end{array}$$
We know that $P(X)+P(\overline X)=1$ and likewise $P(Y)+P(\overline Y)=1$, so we can fill this in as follows: | {
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$$% outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y) \\ \hline \overline X & P(\overline X \cap Y) & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)\color{red}{=0.5} \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & P(Y)=0.6 & P(\overline Y)\color{red}{=0.4} \\ \end{array} \end{array}$$
We know that the rows sum across and the columns sum down, so:
$$% outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y)\color{red}{=0.1} \\ \hline \overline X & P(\overline X \cap Y)\color{red}{=0.2} & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)=0.5 \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & \,P(Y)=0.6 & \quad\quad\, P(\overline Y)=0.4 \\ \end{array} \end{array}$$
And finally:
$$% outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y)\color{red}{=0.4} & P(X \cap \overline Y)=0.1 \\ \hline \overline X & P(\overline X \cap Y)=0.2 & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)=0.5 \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & \,P(Y)=0.6 & \quad\quad\, P(\overline Y)=0.4 \\ \end{array} \end{array}$$ | {
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# Linear algebra question on rank and null space
This is an exercise from linear algebra and optimization by Gill, I do exercises to be prepared for my final exam and these are not homework questions!
Exercise $\mathbf{6.1.}\,$ Consider the following matrix $A$ and vector $b$: $$A=\begin{pmatrix} 2 & 4 \\ 1 & 2 \\ 1 & 2 \\ \end{pmatrix},\quad b=\begin{pmatrix} 3 \\ 2 \\ 1 \\ \end{pmatrix}.$$ $\textbf{(a)}$ What is the rank of $A$? Give a general form for any vector in the range of $A$.
$\textbf{(b)}$ Show that the dimension of the null space of $A^T$ is two, and display two linearly independent vectors $z_1$ and $z_2$ in $\operatorname{null}(A^T)$. Give a general form for every vector in $\operatorname{null}(A^T)$.
$\textbf{(c)}$ Find the vectors $b_R\in\operatorname{range}(A)$ and $b_N\in\operatorname{range}(A^T)$ such that $b=b_R+b_N$.
$\textbf{(d)}$ Give the general form of $b_A$ such that $b_R=Ab_A$. (Hint: consider all vectors $q$ such that $Aq=0$.)
For part (a), I think $rank(A)=1$ since all the columns are linear combination of each other. As for a general form for any vector in the range of $A$, when I write $Ax=b$ I get an over-determined system:
$$2x_1+4x_2=3$$ $$x_1+2x_2=2$$ $$x_1+2x_2=1$$
So I'm not sure about the general form.
As for part (b), since the rank is two, dimension of $N(A^T)$ must be $3-1=1$, right?
And no idea about the rest. Any help would be greatly appreciated. | {
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And no idea about the rest. Any help would be greatly appreciated.
• If you know row reduction and properties of row reduced matrix then apply. It will be easier to get all answer together. – Dutta Jan 10 '14 at 10:37
• Your solution to (a) is correct. For (b) note that if the dimension of the null space is two and you already found two linearly independent vectors in the null space, those form a base. Now that directly gives you a general form of all null vectors. – Listing Jan 10 '14 at 10:38
• For (c) note that $\text{null}(A^T)$ are exactly the vectors that are not in the image of $A$ (aka $\text{range}(A)$). The image of $A$ gets spanned by $(2,1,1)$ thus you can write $b$ as $\alpha z_1+\beta z_2 + \gamma (2,1,1)$. where $z_1,z_2$ are from (b). – Listing Jan 10 '14 at 10:41
• @Listing: Thank you, that helped to understand the problem. – Gigili Jan 10 '14 at 12:07
• Thank you for the edit @Aðøbe. – Gigili Jan 11 '14 at 2:14
## 1 Answer
If you do row reduction, you find $$\begin{pmatrix} 2 & 4 \\ 1 & 2 \\ 1 & 2 \end{pmatrix} \to \begin{pmatrix} 1 & 2 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}$$ which means that the first column is a basis for the column space of $A$ (which is better terminology than “range of $A$”, in my opinion). So the general form of the vectors in the column space of $A$ is $$\begin{pmatrix} 2a \\ a \\ a \end{pmatrix},\quad \text{a any scalar}$$
By the rank nullity theorem, the null space of $A$ has dimension $1$; the equation defining it is $$x_1+2x_2=0$$ so a basis for it is the single vector $$\begin{pmatrix} -2 \\ 1 \end{pmatrix}$$ | {
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The null space of $A^T$ has indeed dimension $2$; the row reduction on $A^T$ is $$\begin{pmatrix} 2 & 1 & 1 \\ 4 & 2 & 2 \end{pmatrix} \to \begin{pmatrix} 2 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$ so the equation defining the null space is $2x_1+x_2+x_3=0$ and a basis for it is $$\left\{ \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} \,, \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} \right\}$$ Writing $b=b_R+b_N$ should now be easy: the system to solve is $$\left(\begin{array}{ccc|c} 2 & -1 & -1 & 3 \\ 1 & 2 & 0 & 2 \\ 1 & 0 & 2 & 1 \end{array}\right)$$ but you can as well find the orthogonal projection of $b$ on the column space of $A$: $$b_R= \frac{(2\ 1\ 1)\begin{pmatrix}3\\2\\1\end{pmatrix}} {(2\ 1\ 1)\begin{pmatrix}2\\1\\1\end{pmatrix}} \begin{pmatrix}2\\1\\1\end{pmatrix} =\frac{9}{6}\begin{pmatrix}2\\1\\1\end{pmatrix} =\begin{pmatrix}3\\3/2\\3/2\end{pmatrix}$$ and $b_N=b-b_R$.
With this last idea it's easy to solve the last point. | {
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With this last idea it's easy to solve the last point.
• Thank you for your detailed answer. Where did you use the fact that the first column is a basis for the column space (because it has leading $1$, right?)? And how did you come up with that general form, $\begin{pmatrix} 2a \\ a \\ a \end{pmatrix}$? – Gigili Jan 10 '14 at 11:56
• @Gigili I used it for the last part, when computing $b_R$ with the orthogonal projection. Since the first column is a basis for the column space, every vector in the column space is that vector times a scalar. – egreg Jan 10 '14 at 11:59
• Ah, right! I was thinking of $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ as the first column but it's the first column in REF. Thanks, it's clear now. – Gigili Jan 10 '14 at 12:04
• Sorry, but I still don't get the last part about $b$. Where did that system come from? Did you multiply the basis by $-1$? Then what's the first row? How did you calculate $b_R$? – Gigili Jan 10 '14 at 12:07
• @Gigili It's a common error, I find it all the time in students' papers. ;-) The first non zero column in REF is always that one! It can't be in the column space of every matrix, can it? – egreg Jan 10 '14 at 12:07 | {
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# Expected empirical entropy
I'm thinking about some properties of the empirical entropy for binary strings of length $n$ when the following question crosses my way:
$\underbrace{\large\frac{1}{2^{n}}\normalsize\sum\limits_{w\in\left\{0,1\right\}^{n}}\normalsize nH_{0}(w)}_{\large\#}\;\overset{?}{=}\;n-\varepsilon_{n}\;\;\;$
with $\;\;\lim\limits_{n\rightarrow\infty}\varepsilon_{n}=c\;\;\;$ and $\;\;\;\forall n:\;\varepsilon_{n}>0$
where $c$ is a constant.
### Is that equation true? For which function $\varepsilon_{n}$ respectively which constant $c$?
$n=2\;\;\;\;\;\;\;\rightarrow\;\#=1$
$n=3\;\;\;\;\;\;\;\rightarrow\;\#\approx 2.066$
$n=6\;\;\;\;\;\;\;\rightarrow\;\#\approx 5.189$
$n=100\;\;\;\rightarrow\;\#\approx 99.275$
$n=5000\;\rightarrow\;\#\approx 4999.278580$
$n=6000\;\rightarrow\;\#\approx 5999.278592$
# Backround
$H_{0}(w)$ is the zeroth-order empircal entropy for strings over $\Sigma=\left\{0,1\right\}$:
• $H_{0}(w)=\frac{|w|_{0}}{n}\log\frac{n}{|w|_{0}}+\frac{n-|w|_{0}}{n}\log\frac{n}{n-|w|_{0}}$
where $|w|_{0}$ is the number of occurences of $0$ in $w\in\Sigma^{n}$.
The term $nH_{0}(w)$ corresponds to the Shannon-entropy of the empirical distribution of binary words with respect to the number of occurences of $0$ respectively $1$ in $w\in\Sigma^{n}$.
More precise:
Let the words in $\left\{0,1\right\}^{n}$ be possible outcomes of a Bernoulli process. If the probability of $0$ is equal to the relative frequency of $0$ in a word $w\in\left\{0,1\right\}^{n}$, then the Shannon-entropy of this Bernoulli process is equal to $nH_{0}(w)$. | {
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At this point, my question should be more reasonable since the first term normalizes the Shannon-entropies for all empirical distributions of words $w\in\left\{0,1\right\}^{n}$.
Intuitively I thought about getting something close to the Shannon-entropy of the uniform distribution of $\left\{0,1\right\}^{n}$, which is $n$.
By computing and observing some values I've got the conjecture above, but I'm not able to prove it or to get the exact term $\varepsilon_{n}$.
It is easy to get the following equalities:
$\large\frac{1}{2^{n}}\normalsize\sum\limits_{w\in\left\{0,1\right\}^{n}}\normalsize nH_{0}(w)\;\;=\large\frac{1}{2^{n}}\normalsize\sum\limits_{w\in\left\{0,1\right\}^{n}}\normalsize |w|_{0}\log\frac{n}{|w|_{0}}+(n-|w|_{0})\log\frac{n}{n-|w|_{0}}$
$=\large\frac{1}{2^{n}}\normalsize\sum\limits_{k=1}^{n-1}$ $n\choose k$ $\left(k\log\frac{n}{k}+(n-k)\log\frac{n}{n-k}\right)$
and it is possible to apply some logarithmic identities but I'm still in a dead point.
(the words $0^{n}$ and $1^{n}$ are ignored, because the Shannon-entropy of their empirical distributions is zero)
Any help is welcome.
• I don't see a question here. – Suresh Venkat Apr 7 '14 at 17:36
• Is the equation above true? Can anybody prove it respectively prove that the equation doesn't hold? – Danny Apr 7 '14 at 17:38
• you have a bunch of equations. it would be better to number them. also highlight the question with ">" blockquoting. – vzn Apr 8 '14 at 15:20
• I edit my question from $\varepsilon_{n}\rightarrow0$ to $\varepsilon_{n}\rightarrow c$, where $c$ is a constant. – Danny Apr 9 '14 at 13:46 | {
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Here is another approach, based on information theory and heavily inspired by @usul's answer. It shows that $\epsilon_n=O(1)$ with very few calculations, and can be used to prove that $\epsilon_n \rightarrow \log_2 \sqrt{e}$ and to derive good estimates on the rate of convergence with less calculations than @usul's approach. In fact, I find a closed-form expression for $\epsilon_n$ : $$(1) \; \; \; \; \epsilon_n = n \left( H(Binom(n,1/2)) - H \left( Binom \left( n-1, 1/2 \right) \right) \right) \ .$$
Details:
Let $X$ be a uniform random variable in $\{0,1\}^n$. Let $K$ be a random variable equal to the number of 1's in $X$. The expression $\#$ that @Danny wants to analyze is exactly equal to $n \cdot \mathbb{E}_{k} [H(X_1 | K=k)]$. (Here $X_1$ is the first bit of $X$.) By the basic properties of the entropy operator, $$(2) \; \; \; \; \# = n\mathbb{E}_{k} [H(X_1 | K=k)]=nH(X_1|K)=n(H(X_1K)-H(K))=n(H(K|X_1)+H(X_1)-H(K)) = n(1-H(Binom(n,1/2)) + H(Binom(n-1,1/2))) \ .$$ The last equality follows from the fact that $X_1$ is just a uniformly random bit, $K$ is the binomial distribution, and $K|(X_1=x_1)$ is distributed either as $Binom(n-1,1/2)$ or as $Binom(n-1,1/2)+1$, depending on the value of $x_1$, both of which have the same entropy.
This already gave us equation (1). Now we just need to calculate to get the value of $\lim_{n \rightarrow \infty} \epsilon_n$
We use any known estimation for the entropy of a binomial RV, such as here. We see that $$(3) \; \; \; \; H(K)=H(Binom(n,1/2))=\frac12 \log_2 ( \pi en / 2) + O(1/n) \ ,$$ and, similarly, that $$H(K|X_1)=H(Binom(n-1,1/2))=\frac12 \log_2 ( \pi e(n-1)/2) + O(1/n) \ .$$ Canceling out terms and substituting into (1), we get $$(4) \; \; \; \; \epsilon_n = n \cdot (H(K)-H(K|X_1)) = n \cdot \frac12 (\log_2 (n/(n-1)) + O(1/n)) = \\ \frac12 \log_2 ((n/(n-1))^n) + O(1) \rightarrow \log_2(\sqrt{e}) + O(1) \ .$$ | {
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By slightly improving the approximation (3) we should be able to replace the $O(1)$ term in (4) by $O(1/n)$ and therefore get that, indeed, $\lim_{n \rightarrow \infty} \epsilon_n = \log_2 \sqrt{e}$. To get this better estimation it should be enough to check that $H(Binom(n,1/2))=\frac12 \log_2 ( \pi en / 2) + err(n)$ where $err(n)=O(1/n)$ and $err$ is a monotone function.
• Very nice! I accept usul's answer since your answer is inspired by usul's, but your answer is very elegant (and converts to $\log(\sqrt e)$, too). – Danny Apr 10 '14 at 10:08
• *converges instead of "converts" :) – Danny Apr 10 '14 at 11:29
• Danny, I'm not sure it's a good idea to accept a more complicated and less elegant answer. Everyone else who ever looks at this question will look at @usul's answer first, while it's in everyone's interest that they look at the most simple and elegant answer first. The solutions to this are: 1. to ask usul to reformulate his answer to be simpler (for all I care, he could copy the text of my answer verbatim if he wants, I don't care about the reputation or anything) or 2. to accept my answer. I'll transfer the relevant reputation to usul if that's possible. – mobius dumpling Apr 10 '14 at 13:30
• This seems like the better answer to accept in my opinion too! Very nice/elegant entropy argument. – usul Apr 15 '14 at 18:10
• Note that in the literature you'll often see $\log_2\sqrt{e}$ written as $\frac{1}{2\ln 2}$. – Steven Stadnicki Apr 18 '14 at 18:52
(Edited from previous version, 2014-04-08.)
I believe that the answer is $\epsilon_n \to \log(\sqrt{e}) \approx 0.7213475...$ where the logarithm is base 2. This seems to match simulation results. I don't have a full formal proof, but give the heuristic approximations/calculations.
I think it's easier to note that your question is equivalent to:
What are the asymptotics of $\delta_n = 1 - \mathbb{E}[H(k/n)]$, where the expectation is over $k \sim Binomial(n,0.5)$? | {
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(where $H$ is the binary entropy function $H(p) = p\log(1/p) + (1-p)\log(1/(1-p))$, and $\delta_n = \epsilon_n/n$.)
Here's a quick plot to show the idea. We have the binary entropy function in blue and the Binomial pmf (for $p=0.5$) in green. So we can see that the expectation of $H(k/n)$, when $k$ is distributed binomially, will always be below one but should be approaching one. The question is how fast.
The key idea will be that the value we're interested in, $$\mathbb{E}[H(k/n)] = \sum_{k=0}^n \frac{{n \choose k}}{2^n} H\left(\frac{k}{n}\right) ,$$ can be related entropy of the binomial distribution. Let $p_{n,k} = \frac{{n \choose k}}{2^n}$ be the probability of $k$ heads in $n$ fair coin flips. The steps will be as follows:
1. Use Stirling's approximation to get $H(k/n) = \frac{\log(p_{n,k})}{n} + 1 + \text{something}$.
2. Rewrite the original sum to get $\mathbb{E}[H(k/n)] = 1 - \frac{H(Binom(n,0.5))}{n} + \sum \text{something}$.
3. Get that the entropy of the binomial, divided by $n$, plus the sum of "something", equals $\frac{\log(\sqrt{e})}{n} + O(1/n^2)$.
Step 1:
Just plugging in Stirling and doing some cancellation/rearranging, \begin{align} p_{n,k} &:= \frac{{n \choose k}}{2^n} \\ &\sim \frac{n^n}{k^k (n-k)^{n-k}}\frac{1}{2^n}\sqrt{\frac{n}{2\pi k (n-k)}} . \end{align}
This won't be very tight for every term in the sum, but I think it will be asmyptotically tight towards the middle, which is all that matters since all the probability is in the center.
We can rewrite the entropy function as follows. It's just some arithmetic to combine the logarithms. \begin{align} H(k/n) &= \frac{k}{n}\log\left(\frac{1}{k/n}\right) + \frac{n-k}{n}\log\left(\frac{1}{(n-k)/n}\right) \\ &= \log\left(\frac{n}{k^{k/n}(n-k)^{(n-k)/n}}\right) . \end{align} | {
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So, using Sterling's approximation above, the logarithm of a probability term is \begin{align} \log(p_{n,k}) &= \log\left(\frac{{n \choose k}}{2^n}\right) \\ &\approx n H(k/n) - n + \log\left(\sqrt{\frac{n}{2\pi k(n-k)}}\right) . \end{align}
Step 2: \begin{align} \mathbb{E}[H(k/n)] &= \sum_{k=0}^n p_{n,k} H(k/n) \\ &\approx \sum_{k=0}^n p_{n,k} \left(\frac{\log(p_{n,k})}{n} + 1 - \frac{\log\left(\sqrt{\frac{n}{2\pi k (n-k)}}\right)}{n}\right) \\ &= 1 - \frac{H(Binom(n,0.5))}{n} - \frac{1}{n}\sum_{k=0}^n p_{n,k}\log\left(\sqrt{\frac{n}{2\pi k(n-k)}}\right) . \end{align} Here, $H(Binom(n,0.5))$ is the entropy of the Binomial distribution for $n$ coin flips and $p=0.5$, which by wikipedia is $\log\left(\sqrt{\pi e n / 2}\right) + O(1/n)$.
Step 3:
Now, we just need to approximate the third sum. I will take a very rough approximation (feel free to do better, but it probably doesn't gain much). All the probability mass is concentrated on $k = \frac{n}{2} \pm o(n)$. So approximate this sum (which is an expectation) by its value on the term $k=\frac{n}{2}$, when it is $\log\left(\sqrt{\frac{2}{\pi n}}\right)$.
So now, we get \begin{align} \delta_n &\approx \frac{\log\left(\sqrt{\pi e n/2}\right)}{n} + \frac{\log\left(\sqrt{2/\pi n}\right)}{n} \pm O\left(\frac{1}{n^2}\right) \\ &= \frac{\log\left(\sqrt{e}\right)}{n} \pm O\left(\frac{1}{n^2}\right) . \end{align} | {
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• Thank you very much for your effort. Your attempt is useful and i like your ideas, but i am not quite sure about the details. I will think about it and maybe i will ask some questions soon.. – Danny Apr 8 '14 at 21:35
• @Danny, sure thing. Let me know if parts are unclear. The computation is not fully rigorous, particularly (1) the use of Stirling's approximation everywhere (a fix might be to only consider an interval where the Binomial probability is $\Omega(1/n)$ and argue that Stirling is asymptotically tight there) and (2) the approximation of the "third term" by just setting $k=n/2$ (a fix might be to again consider the high-probability interval and see if the error is small). – usul Apr 9 '14 at 1:24
• My first question is a general question: You think that $\mathbb{E}_{w \in \{0,1\}^n} nH(w) = n-\mathcal{O}(1)$. Do you think there is a "short" proof for this conjecture without getting the exact term $\varepsilon_{n}$ respectively the exact limit of $\varepsilon_{n}$, which you are considering to be $\log\sqrt{e}$? – Danny Apr 9 '14 at 13:35
• @Danny, I'm not sure how to get this with a shorter proof, but maybe another answer will do so. – usul Apr 9 '14 at 20:51
• Thanks again for your help. Everything seems to be clear.. – Danny Apr 10 '14 at 9:07
Here is a proof that the quantity # that OP considers tends to n(1-o(1)).
Claim: #=n-o(n)
First, note that for any function $f:\{0, 1\}^n \rightarrow \mathbb{R}$, $\frac{1}{2^n} \sum_{w \in \{0,1\}^n} f(w)$ is exactly the same as $\mathbb{E}_{w \in \{0,1\}^n} f(w)$. So you're asking whether $\mathbb{E}_{w \in \{0,1\}^n} H(w) = 1-o(1)$. | {
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To see this, define $X$ to be a random variable uniformly distributed on $\{0,1\}^n$ Let $X_0$ be the number of 0's in $X$ and $X_1$ be the number of 1's in $X$. We want to prove $$\mathbb{E}[(X_0/n)\log(n/X_0)+(X_1/n)\log(n/X_1)]=1-o(1).$$ But we know from the law of large numbers that both $X_0/n$ and $X_1/n$ converge in probability to $1/2$ and that if a random variable $Y$ converges in probability to $c$ and if $f$ is a continuous function, then $f(Y)$ converges in probability to $f(c)$. So we get that $\mathbb{E}[(X_0/n)\log(n/X_0)+(X_1/n)\log(n/X_1)]$ converges in probability to $\frac12 \log(2) + \frac12 \log(2) = 1$, QED.
Note: This claim is related to the Asymptotic equipartition property for discrete-time i.i.d. sources here.
• I'm sorry, but I can't see how $\mathbb{E}_{w\in\left\{0,1\right\}^{n}}\left[H\left(\frac{|w|_{0}}{n},\frac{|w|_{1}}{n}\right)\right]=1-o(1)$ is a direct result of the AEP, since we are looking for the expectation of entropies for all empirical distributions and the AEP treats with the limit of the selfinformations of strings for some fixed distribution. Can you specify that implication? – Danny Apr 8 '14 at 15:06
• Also I do not see why your statement $\mathbb{E}[(X_0/n)\log(n/X_0)+(X_1/n)\log(n/X_1)]=1-o(1)$ leads directly to my conjecture.. Thanks for your help. – Danny Apr 8 '14 at 15:18
• So you're asking whether $\mathbb{E}_{w\in\{0,1\}^n}H(w) = 1 - o(1)$. Note this isn't quite so, the question as worded is whether it is $1 - o(1/n)$. – usul Apr 9 '14 at 1:26
• @Danny, the expression $\mathbb{E}[(X_0/n)\log (n/X_0) + (X_1/n) \log (n/X_1)]$ is exactly the same as your expression $H_0(w)$ (or, in my notation, $H_0(X)$). – mobius dumpling Apr 9 '14 at 11:39
• @Danny You're right, the theorem does not follow from the AEP as easily as I thought. Rather than explain how it follows, I'll just leave my self-contained explanation, which turns out to be simpler than relating to the AEP. – mobius dumpling Apr 9 '14 at 11:44 | {
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I will prove that for all $n$, $H_n = n - \epsilon_n$ where $0 < \epsilon_n < 1$.
Let $f_{\alpha}(t) = 4^{\alpha} t^{\alpha}(1 - t)^{\alpha}$. I will not prove this, but you can find some $\alpha^*$ such that for all $\alpha \ge \alpha^*$, $f_{\alpha}(t) \le H(t)$. Also, for $\delta \ll 1$, you can find $\alpha'$ such that for all $\alpha \le \alpha'$, $f_{\alpha'}(t) \ge H(t)$ in $[\delta, 1 - \delta]$. I will assume $\alpha = 1$ and $\alpha = 0.5$ lead to the two different types of functions.
My point is that you can use simpler functions to obtain the bound you need. Here I will use $f_{\alpha}(t)$.
Let $H_n$ be the empirical entropy that you are looking for and $F(n, \alpha)$ be the value obtained when I replace $H(\frac{k}{n})$ with $f_{\alpha}(\frac{k}{n})$. We have
$F(n, \alpha) = n2^n \sum_{k = 0}^{n} {n \choose k} f_{\alpha}(\frac{k}{n}) = n 2^n 4^{\alpha} \sum_{k = 0}^{n} {n \choose k} \left( \frac{k}{n} \right) ^ {\alpha} \left(1 - \frac{k}{n} \right) ^\alpha = n 4^{\alpha} E\left[\left( \frac{k}{n} - \frac{k^2}{n^2} \right)^{\alpha} \right] = n2^n 4^{\alpha} \sum_{k = 0}^{n / 2} {n \choose k} \left(\frac{k}{n} \right)^{2 \alpha}$
For $\alpha = 1$, we know $F(n, \alpha) \le H_n$ and direct evaluation says that $F(n, 1) = n - 1$, which is true for all $n$.
For $\alpha = 0.5$, we know $F(n, \alpha) \ge H_n$ and a direct evaluation leads to $F(n, 0.5) = n 4^{0.5} E[\frac{k}{n}] = n$.
Therefore, for all $n > 0$, $n\ge H_n \ge n - 1$.
I found it nontrivial to try this for any $\alpha < 1.0$ other than $\alpha = 0.5$. | {
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Coming up with a formula for
1. Sep 12, 2004
Caldus
Um, how can I figure out a formula for solving something like:
1 + 3 + 5 + 7 + 9 + ... + n
In particular: 1 + 3 + 5 + ... + 999
Thanks.
2. Sep 12, 2004
geometer
As far as I know, there is no rote procedure for doing this. It's pretty much a trial and error process. Make yourself a table and figure out the pattern. In this case, we can write this as the sum from x = 0 to x = (n-1)/2 of 2x+1:
x = 0 ===> Sum = 1
x = 1 ===> Sum = 4
x = 2 ===> Sum = 9
x = 3 ===> Sum = 16
We can keep this up if we need to to help us see the pattern, but in this case, a little thought shows that the formula is: Sum from x= 0 to x = n = (n+1)^2
3. Sep 12, 2004
Integral
Staff Emeritus
I believe that Gauss did this as school child when given the assignment to sum the digits 1-100.
First note that if you add them together smallest to largest you get a constant result ( 100+ 1=101, 99+2=101 ..) multiply this by 100 because there are 100 sums and divide by 2 because you added all the numbers twice.
edit:
OPPS! just noticed that you were not summing ALL integers.
Last edited: Sep 12, 2004
4. Sep 12, 2004
Integral
Staff Emeritus
After a bit more playing,
consider this
$$S = \Sigma _ {i=1}^N (2i -1)$$
now using the properties of the summation
$$S= 2\Sigma_{i=1}^N i - \Sigma_{i=1}^N 1$$
$$S= 2\Sigma_{i=1}^N i -N$$
Now using Gauss's trick we can see that the sum of n integers is given by
$$S_n = \frac {(n+1)(n)} 2$$
So we can write for the sum of the odd integers
$$S = (N+1)(N)-N = N^2$$
Note that I have given the general expression for the odd integers as 2n-1 and my sum starts with n=1 where geometer used 2n+1 and started with n=0. This is the reason he has N+1 where I have N. Also note that this N is NOT the odd integer you are summing to but the NUMBER OF TERMS in the sum.
Last edited: Sep 12, 2004
5. Sep 12, 2004
shmoe | {
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Last edited: Sep 12, 2004
5. Sep 12, 2004
shmoe
Integral, the same idea of pairing highest and lowest will work just fine,
1+3+5+7+9+11=12+12+12=(6/2)*12, where 6 is the number of terms here.
If you already have a formula for the sum of the first n integers, 1+2+3+..+n, you can use this as well:
1+3+5+7+9+11=(1+2+..+11)-(2+4+6+8+10)=(1+2+..+11)-2(1+2+3+4+5)=(insert formula here)-2(insert formula here)
Caldus, do you have a formula for the sum of the first n integers? If not, see Integrals post;)
Another idea, (which Integral provided while I was typing) write your sum as: $$\displaystyle\sum\limits_{i=1}^{n}(2i-1)$$ and fiddle with it to get an expression involving $$\displaystyle\sum\limits_{i=1}^{n}i$$, and apply the formula for this (again, using the ideas in Integral's post if need be). This would be a more easily applied general idea that can handle any arithmetic series $$\displaystyle\sum\limits_{i=0}^{n}(ki+d)$$.
Last edited: Sep 12, 2004
6. Sep 13, 2004
HallsofIvy
Staff Emeritus
Since you are adding only odd numbers, one way to do is this (Assuming you know that the sum of the first n integers is n(n+1)/2):
The sum of all numbers from 1 to 2n is (2n)(2n+1)/2= n(2n+1). Now subtract off the sum of all even numbers from 2 to 2n: 2+ 4+ 6+ ...+ 2n= 2(1+ 2+ 3+ ... + n)=
2(n)(n+1)/2= n(n+1).
The sum of all odd numbers from 1 to 2n-1 is n(2n+1)- n(n+1)= n(2n+1- n- 1)= n2 just as shmoe said.
In particular, the sum of all odd numbers from 1 to 99 is: (99= 100- 1= 2(50)- 1)
502= 2500.
7. Sep 13, 2004
Gokul43201
Staff Emeritus
Also, here's a general rule that can be used in times of trouble :
If a series S, has terms t(n), which are polynomials in n, of order p, the sum S(n) is a polynomial of order p+1. The coefficients of this polynomial can be determined simultaneously, from the given p+2 terms in the series. | {
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Example : If the series S, has terms, t(n) which are linear in n, ie : t(n) = an + b, then the sum, S(n) is a quadratic in n, ie: S(n) = An^2 + Bn + C
NOTE : The order of the polynomial can be determined by the method of "successive differences". If the p'th successive differences are equal, the order is p.
Example : Consider the sequence : 2, 6, 12, 20, 30, ...
The first differences are : 4, 6, 8, 10, ...
The second differences are : 2, 2, 2, ...
Since the second differences are equal, t(n) is a quadratic in n, ie : t(n) = an^2 + bn + c.
So you can assume the sum has a cubic form : S(n) = An^3 + Bn^2 + Cn + D
Plugging in the for S(1), S(2), S(3) and S(4) gives you 4 simultaneous equations in A, B, C, and D, which you can solve. | {
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# How can we find the number of pairs of intersecting ranges on a circular number line?
I recently thought of and managed to solve this algorithmic problem:
On a infinite 1-dimensional number line, we have N ranges specified by two distinct integers A and B, such that all A and B are unique. Each pair of integers then represents the starting and ending points of a range on the number line. How many distinct pairs of ranges intersect?
^^ In this example, 4 distinct pairs of segments intersect.
Obviously, there is a O(N^2) brute-force solution: just test all possible pairs in O(1) time each. However, I also thought of a O(NlogN) sweep-line algorithm: treat every A as a "segment start event" and every B as a "segment end event". Then, keep a variable curr_in_range (originally 0) that stores the current number of segments that we currently see, and then iterate through the events in sorted order of their positions on the number line. Every time we add another segment, add curr_in_range to the answer (because this new segment will intersect with every other segment that we can see) and increment it as well. Every time we remove a segment, decrement curr_in_range. In the end, this gets us the correct answer.
Happy with this algorithm, I decided to tackle a slightly modified version of this problem: What if the number line is circular? As in, it has a specified length L such that the number line contains numbers in the range [0, L)? In this case, each range would still be specified by two distinct integers A and B, but let's just say the range is always counterclockwise starting from A and ending at B. Could we still find the number of pairs of sectors that intersect in less than O(N^2) time in this scenario?
Here's my thought process on this problem: | {
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Here's my thought process on this problem:
We should be able to use a very similar algorithm to the one described above. Completely copying it wouldn't work, though, because the sectors can have B < A when it crosses over 0. My idea was to initially have curr_in_range equal to the number of sectors that cross over 0 (instead of just it at 0 like we did in the above algorithm). Then, we could treat every sector as a "sector start event" and a "sector end event", and do the same process that we did.
However, after extensive analysis, I found why this approach doesn't work. This approach won't work because it overcounts some of the intersections, as sometimes two sectors can intersect in 2 places. This is where I'm stuck.
^^ Our sweep-line algorithm will count 2 distinct pairs that intersect, when there is actually only 1.
But besides this overcounting, I found that this current algorithm isn't doing a single other thing wrong. I coded a O(N^2) brute-force algorithm and tabulated all the distinct pairs that it found, and also tabulated the distinct pairs found for the current algorithm. For every randomly-generated test case, these two tables were the same. In other words, our current algorithm is correctly detecting every intersection there is to be detected, but it just happens to overcount some of them.
So, sorry for a really long post, but my final question is:
Can someone come up with an algorithm that solves this problem? Is my thought process close to a solution or completely off?
• Simple idea that might work: sort intervals as usual, except give each one a unique id. Begin processing using the same algorithm, except every start interval is marked as started, and if you encounter an end interval that hasn't been marked, you just ignore it. The algorithm terminates when all intervals are marked as started. – hLk Mar 27 at 4:22 | {
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This is solvable in $$O(n \log n)$$ time, too, by considering the complement of the intervals that wrap around and using a suitable data structure.
There are two types of intervals that can appear on a cycle:
• Intervals that do not wrap around: e.g., $$[0.3,0.5]$$ does not wrap around. In general, this includes all intervals of the form $$[a,b]$$ with $$0\le a \le b \le 1$$.
• Intervals that wrap around: e.g., $$[0.8,0.1]$$ wraps around; it is equivalent to $$[0.8,1.1]$$ or $$[0.8,1] \cup [0,0.1]$$. In general, this includes all intervals of the form $$[b,a]$$ with $$0\le a < b \le 1$$; each such interval is equivalent to $$[b,1] \cup [0,a]$$.
Let $$A$$ denote the set of intervals that do not wrap around, and $$B$$ the set of intervals that do wrap around. Also, let $$\overline{B}$$ denote the complements of the intervals in $$B$$, i.e., $$\overline{B} = \{\overline{b} \mid b \in B\}$$. For instance, the interval $$[0.8,0.1]$$ wraps around; its complement is $$(0.1,0.8)$$, which does not wrap around. It is easier to think about intervals that do not wrap around, and this decomposition will help us to deal only with intervals that do not wrap around.
Now, we will separately count the number of overlaps between (1) an interval in $$A$$ and an interval in $$A$$; (2) an interval in $$A$$ and an interval in $$B$$; (3) an interval in $$B$$ and an interval in $$A$$; and (4) an interval in $$B$$ and an interval in $$B$$. (1) can be counted with your sweepline algorithm, as we don't need to deal with wraparound. (4) can also be counted in the same way, as two intervals overlap iff their complements overlap. I will show below how to count (2); case (3) will then follow by symmetry. This covers all the cases, so it suffices to count up these four cases and sum them up. | {
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Note that intervals $$a \in A, b\in B$$ overlap iff $$a$$ is not a subset of $$\overline{b}$$. So, it suffices to count the number of pairs $$a \in A, \overline{b}\in \overline{B}$$ where $$a$$ is a subset of $$\overline{b}$$; then you can subtract that from $$|A| \cdot |B|$$.
To do that, you can use the data structure in Data structure for interval subset queries. Store all the intervals $$\overline{B}$$ in the data structure. Then, for each $$a \in A$$, count the number of intervals in $$\overline{B}$$ that $$a$$ is a subset of. This can be done in $$O(\log n)$$ time per query interval $$a$$, for a total time of $$O(n \log n)$$.
Therefore, you can count each of the four cases in $$O(n \log n)$$ time, and thus compute the overall sum in $$O(n \log n)$$ time as well. I think. I encourage you to check my reasoning to make sure I haven't made any mistakes. | {
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• "(4) can also be counted in the same way, as two intervals overlap iff their complements overlap." A range overlaps any wrapping range that contains the whole circle, but their complements do not overlap. Anyway, in case (4) where all ranges are wrapping, we can just say any two of them overlap. – John L. Mar 25 at 18:49
• Apparently, I could not understand how your usage of "a segment tree to obtain that count" avoids counting the same intersecting pair more than once. – John L. Mar 25 at 18:54
• @JohnL., By "wrapping" I mean a range like $[0.9,0.1]$, i.e., $[0.9,1.1]$, i.e., $[0.9,1.0] \cup [0,0.1]$. Such a range does not cover the whole circle, but includes $1$ and $0$ in the range. The only range that covers the entire circle is $[0,1]$, and that range is non-wrapping. – D.W. Mar 25 at 18:58
• Are you talking about 2-dimensional segment trees? What are the leaves of your segment tree? If each leaf is an elementary interval, how can each interval of $\overline B$ appear in the segment tree only once? If a leaf can be an interval of $\overline B$, then how can an interval of $A$ be expressed as concatenation of those leaves? – John L. Mar 25 at 19:18
• @JohnL., No, an ordinary segment tree for storing the intervals of $\overline{B}$. As usual, we take the endpoints of the intervals of $\overline{B}$, sort them, obtain the elementary intervals, and each leaf corresponds to one elementary interval (not an interval of $\overline{B}$). Each interval of $\overline{B}$ is stored in some node of the segment tree, following the rule described at the link you provided. The interval $a$ can be expressed as a disjoint union of elementary intervals, not of intervals of $\overline{B}$. – D.W. Mar 25 at 19:22
Not a solution, just an illustration of what I see as the core difficulty: | {
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Not a solution, just an illustration of what I see as the core difficulty:
The instance on the left has 3 intersections, the one on the right only 2, but there is no way to distinguish the instances just by counting numbers of "begin segment" and "end segment" events within a given interval. (Inside the gap left by the red interval, the event sequence is "begin", "end" in both cases.)
This implies that a sweep-line approach needs to track more information about each active segment -- probably their identities, which suggests to me that no fast algorithm will be possible, though that's just a feeling.
• A sweep algorithm will take $\Omega(n\log(n))$ time anyways since it has to sort the array. Then, we might be able to store the identities in a way it would not take much more space and time, so we would not overshoot that $n\log(n)$. Not proposing a solution, but just saying that even if we have to keep the identities it might be fast enough – nir shahar Mar 12 at 23:33
### Summary
Interesting question.
Please check this program in Java that runs in $$O(n\log n)$$ time, where $$n$$ is the number of given ranges. The program uses sweep-line technique, once for counting intersecting pairs of ranges, and once for counting containing pairs of ranges together with Fenwick tree. Explanations follow.
### Reduce to the hard case
Here are the notations and assumptions for the ease of explanation.
• All numbers and variables are integers.
• The numbered circle of cardinality $$L\ge2$$ is the set of number $$0, 1, \cdots, L-1$$, where $$L$$ is considered to wrap around to $$0$$.
• A range $$[s,e]$$ with respect to $$L$$, where $$0\le s\le e\lt L$$, means
• a non-wrapping range (a.k.a. an ordinary range) if $$s\le e$$, i.e., numbers $$s, s+1,\cdots, e-1, e$$;
• a wrapping range otherwise, i.e,. numbers $$s, s+1, \cdots, L-1, 0, 1, \cdots, e-1, e$$. | {
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Note every range contains at least one number. The non-wrapping range $$[0, L-1]$$ contains all the numbers. So does every wrapping range like $$[s,s-1]$$.
Suppose there are $$n$$ ranges. The question is how to compute the number of all unordered pairs of two intersecting ranges efficiently.
There are three cases for an unordered pair of two ranges.
• Both ranges are wrapping.
This is the easiest case since two wrapping ranges always intersect at $$0$$.
• Both ranges are non-wrapping.
This is the classical case. The total number of such intersecting pairs can be computed by the sweep-line (or, as I would call it, the sweep-point or sweep-event) algorithm, which runs in $$O(n)$$ time, as described in the question.
• One range is non-wrapping and the other is wrapping.
This is the new and hard case. One way to understand why this case may not be easy is j_random_hacker's illustration. How can we deal with this case?
### Transform to the containing pairs of non-wrapping pairs
Denote the non-wrapping ranges and the wrapping ranges in the given ranges by $$\mathcal N$$ and $$\mathcal W$$, respectively. Instead of counting the intersecting pairs between $$\mathcal N$$ and $$\mathcal W$$, we will use complementary counting, i.e., we will count the non-intersecting pairs between $$\mathcal N$$ and $$\mathcal W$$.
Given any range $$r$$, let $$\overline r$$ be the numbers among $$0, 1, \cdots, L-1$$ that are not in $$r$$, i.e., the complement of $$r$$ with respect to $$\mathcal C$$. Note that two ranges do NOT intersect iff the complement of either one contains the other. | {
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Let $$\overline{\mathcal W} = \{\overline r\mid r\in\mathcal W\}$$. Then the number of non-intersecting pairs between $$\mathcal N$$ and $$\mathcal W$$ is the number of containing pairs between $$\mathcal N$$ and $$\overline{\mathcal W}$$. Here a containing pair means, naturally, two sets $$a$$ and $$c$$ such that $$c$$ contains $$a$$. The number we wanted, i.e., the number of all intersecting pairs of ranges between $$\mathcal N$$ and $$\mathcal W$$ is $$|\mathcal N|\cdot|\mathcal W|$$ minus that number of containing pairs.
More formally, we have $$\#\{(a, b) \mid a\in{\mathcal N}, b\in{\mathcal W}, a\cap b\not=\emptyset\} = \#{\mathcal N}\cdot\#{\mathcal W} - \#\{(a, c) \mid a\in{\mathcal N}, c\in\overline{\mathcal W}, a\subseteq c\}.$$
Here is a simple observation, the complement of a wrapping range is a non-wrapping range, except when the wrapping range is like $$[s,s-1]$$, whose complement is empty. Containing all numbers, a wrapping range like that intersects with every other range. It is easy to count all intersecting pairs that involve a wrapping range like that.
Let us assume there is no such wrapping range; if necessary, we can replace such a range with non-wrapping range $$[0, L-1]$$. The problem is reduced to counting the number of containing pairs between two sets of non-wrapping ranges.
### An algorithm that counts the number of containing pairs
Here is an algorithm that counts the number of containing pairs between two given sets of non-wrapping ranges, $$\mathcal N$$ and $$\overline{\mathcal W}$$. It is, basically, sweep-line technique together with Fenwick tree. | {
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1. Let containingPairCount = 0.
2. Initialize a Fenwick tree (a.k.a. binary indexed tree) ft for an (implicit) interested array of length $$L$$, with all values 0 initially. The e-th element of the interested array will be the number of all ranges in $$\overline{\mathcal W}$$ so far that end at e.
3. Sort all ranges that are in either $$\overline{\mathcal W}$$ or $$\mathcal N$$ by the starting points, breaking ties by putting ranges in $$\overline{\mathcal W}$$ before ranges in $$\mathcal N$$. Breaking the remaining ties arbitrarily.
4. One by one process the sorted ranges. Let the current range be [s,e].
• If it comes from $$\overline{\mathcal W}$$, updates the Fenwick tree as if the e-th element of the interested array is increased by 1. This operation registers the current range as a candidate containing range that has started and will end at e.
• If it comes from $$\mathcal N$$, add the sum of elements of indexes no smaller than e in the interested array to containingPairCount, by containingPairCount += st.getSum(e, L-1).
5. Return containingPairCount.
Proof of the algorithm: st.getSum(e, L-1) is the number of all ranges in $$\overline{\mathcal W}$$ that contain the current non-wrapping range, since, thanks to the careful setup, we have processed all ranges in $$\overline{\mathcal W}$$ that start at or before the start of the current range, and only those ranges that end at or after the end of the current edge are counted. $$\quad\checkmark$$
The sorting of all ranges at step 3 takes $$O(n\log n)$$ time. At step 4, each range is processed with $$O(\log n)$$ time, thanks to the power of Fenwick tree. So the running time of this algorithm is $$O(n\log n) + O(n \log n) = O(n\log n)$$.
### Complexity analysis
The running time for other parts of the entire algorithm is $$O(n\log n)$$, which is spent mostly on counting the intersecting non-wrapping pairs. So, the running time of the entire algorithm is $$O(n\log n) + O(n \log n) = O(n\log n)$$. | {
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The sections "Reduce to the hard case" and "transform to the containing pairs of non-wrapping pairs" follow D.W's answer roughly. | {
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# Representation of a matrix transformation
An example asks me to define $T: M_{2\times2}(F) \to M_{2\times2}$ by $T(A) = A^t$. Compute $[T]_\alpha$
$\alpha$ is the standard ordered basis of $2\times 2$ matrices.
To find the transformation I performed the transformation on all four elements of $\alpha$, for example $T(\begin{pmatrix} 1&0\\0&0\\\end{pmatrix}) = (\begin{pmatrix} 1&0\\0&0\\\end{pmatrix})$ and so on. Basically the matrices are all the same before and after the transformation. Accordingly, the textbook's answer is that you arrange all the results $A^t$ into a $4\times4$ matrix:
$$\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}$$
I don't understand this what means, and I've been trying to. How can this be the transformation $T$ on $\alpha$ if we're not even able to multiply the basis by it? What I mean is, we can't multiply this transformation matrix by anything in $\alpha$ since one is a $4\times4$ and one is a $2\times 2$.
Also, why doesn't this mean that any arbitrary $2\times 2$ matrix can be raised to the power of $t$ by simply multiplying by this matrix? (As per the previous paragraph, it can't, but it should).
• I changed $T: M_{2\times2}(F) -> M_{2\times2}$ to $T: M_{2\times2}(F) \to M_{2\times2}$. $\TeX$ is not so crude that you have to resort to things like that. – Michael Hardy Oct 11 '12 at 23:43
• Thanks - I couldn't find the command for it. I'll know for next time. – CodyBugstein Oct 11 '12 at 23:57
You need to distinguish here between the space $\mathbb F^2$ that the $2\times2$ matrices operate on and the space of the $2\times2$ matrices themselves. The former is two-dimensional, the latter is four-dimensional. | {
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A matrix is a representation of a linear transformation with respect to bases of the domain and target of the transformation. If the linear transformation is an endomorphism, a linear map from a space to itself, the corresponding matrix is square, and one typically uses the same basis for the space in its two roles as domain and target.
The standard basis of the space of $2\times2$ matrices that you're using has four basis elements and spans the four-dimensional space of $2\times2$ matrices. A linear transformation has been defined on that space – the four-dimensional space of $2\times2$ matrices, not the two-dimensional space that they operate on.
Representing that linear transformation as a matrix with respect to the standard basis (both for the domain and the target) yields a $4\times4$ matrix. That's not a mismatch because the vectors and matrices that will be multiplied by this matrix aren't elements of $\mathbb F^2$; they're the vectors of coefficients in an expansion of the matrix they represent in the standard basis.
As an example, the matrix
$$\pmatrix{2&3\\4&5}$$
has coefficients $2$, $3$, $4$ and $5$, respectively, in the standard basis, so with respect to that basis it's represented by the column vector
$$\pmatrix{2\\3\\4\\5}\;.$$
Its transpose
$$\pmatrix{2&4\\3&5}$$
is represented by the column vector
$$\pmatrix{2\\4\\3\\5}\;.$$
And sure enough we have
$$\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\pmatrix{2\\3\\4\\5}=\pmatrix{2\\4\\3\\5}\;.$$
Thus this matrix really does represent the linear transformation $T$, transposition, in this basis. | {
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Thus this matrix really does represent the linear transformation $T$, transposition, in this basis.
• So matrices can unravel to form vectors? I've played around with some and it makes sense. Thanks for the amazing answer. I still have a question though; shouldn't the transforming vector actually put your sample vector to the power of something? Isn't that the whole idea? – CodyBugstein Oct 11 '12 at 23:32
• @Imray: I think you've misunderstood the notation. $A^t$ is merely a notational device to denote the transpose of $A$; nothing is being raised to a power. I prefer the slightly different notation $A^\top$ (A^\top) to avoid the impression that the $t$ is a variable used as an exponent. – joriki Oct 11 '12 at 23:34
• @Imray: As an exercise, I suggest to write down the matrix of $T$ with respect to the basis $$\left\{\pmatrix{1&0\\0&1},\pmatrix{1&0\\0&-1},\pmatrix{0&1\\1&0},\pmatrix{0&1\\-1&0}\right\}\;,$$ and then perhaps use it to apply $T$ to a couple of matrices and see whether the transpose comes out right. – joriki Oct 11 '12 at 23:38
• yes right! I didn't realize that! Ok I've got it now, thank you very much for explaining it to me. – CodyBugstein Oct 11 '12 at 23:38
• @Imray: You're welcome! – joriki Oct 11 '12 at 23:40
Just to make sure our definition of matrix representation is the same: Let $\alpha = (a_1, \ldots, a_k)$ be a basis for $V$, and let $T: V \rightarrow V$ be a linear map. Then matrix representation $[T]_{\alpha}$ is defined as follows: $$[T]_{\alpha} = \begin{bmatrix} [T(a_1)]_{\alpha} & \ldots & [T(a_k)]_{\alpha} \end{bmatrix}$$
The matrix representation obeys the following formula: for every $v \in V$, $[T(v)]_{\alpha} = [T]_{\alpha} [v]_{\alpha}$, where $[v]_{\alpha}$ is the coordinate vector of $v$ with respect to the basis $\alpha$. | {
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So now, to answer your question, it's true that $[T]_{\alpha}$ is a $4 \times 4$ matrix in your example, but you don't multiply it onto the $2 \times 2$ matrices; you multiply it onto the $2 \times 2$ matrices after they have been converted into column vectors via the coordinate mapping. Since $2 \times 2$ matrices form a $4$-dimensional space, then this coordinate mapping turns a $2 \times 2$ matrix into an element of $\mathbb{R}^4$.
• How does coordinate mapping work? The rows all extend downward to become one long column? Can all matrices with $n$ elements be multiplied by any matrix with $n$ columns? – CodyBugstein Oct 11 '12 at 23:35
• What confused me is how $[T]_\alpha$ could be square. But one needs to remember that the columns are in vector notation too (not in matrix notation!), so they are actually not $k \times k$ but $k^2 \times 1$ and so in the end $[T]_\alpha$ is $k^2 \times k^2$. – philmcole Apr 10 '18 at 21:20 | {
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# Show that $a_n=\left(\sum_{k=1}^n \frac{1}{k}\right)-\log(n)$ is decreasing
How I decided to show it is by Induction, and I don't really like what I have, so could you guys please tell me if my proof valid. Thanks.
Aim of the proof: Show that $$a_n>a_{n+1}$$ for all $n\geq1$.
Base case: $$a_1-a_2=\log(2)-\frac{1}{2}>0.$$ Hence base case holds. Assume, for some $k>1$, that the result holds, that is:$$a_k>a_{k+1}$$ Hence show that $$a_{k+1}>a_{k+2}.$$ Now we can rewrite $a_{k+1}>a_{k+2}$ as, $$\left(\sum_{k=1}^{k+1}\frac{1}{k+1}\right)-\log(k+1)>\left(\sum_{k=1}^{k+1} \frac{1}{k+1}\right)+\frac{1}{k+2}-\log(k+2),$$ Which is equivealnt of saying: $$\log\left(\frac{k+2}{k+1}\right)>\frac{1}{k+2}$$
For our problem $k\geq1$ and $\log(\frac{3}{2})>\frac{1}{3}.$
My last step is saying that as $n$ tends infinity, the limit of LHS of our last inequality is infinite and RHS is 0. Hence the result holds.
What I don't like is the fact that I don't use the inductive hypothesis at all. (Sorry for selling and grammar errors, English is not my native language)
• This does not answer your question, but do you know that $\lim_n a_n =\gamma$, the Euler-Mascheroni constant? You might want to google it, you will for sure find something about the non decreasing behaviour of $a_n$. – Paolo Intuito Nov 23 '17 at 21:53
• A fundamental step you make in your proof is incorrect: $\lim_{k \rightarrow \infty} (\frac{k+2}{k+1}) = 0$, not infinity, as you claim. (In fact, if this step were correct, then this alone would be a proof - the whole induction set up would be unnecessary.) – John Don Nov 23 '17 at 21:59
• Oh crap, yeah, the fraction gets closer to 1 eqch time. Thank you for noticing this John. How would I go about proving it then?, and yes, G.S. I know about that fact but couldnt find anything on the question I am trying to solve – Scavenger23 Nov 23 '17 at 22:02
You wish to show the inequality, | {
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You wish to show the inequality,
$$\frac{1}{n+1}-\ln(n+1)<-\ln n$$ or equivalently, $$\frac{1}{n+1}<\ln (1+\frac{1}{n})$$
Now we have $$x-\frac{x^2}{2}<\ln(1+x)$$ so just check that $$\frac{1}{n+1}<\frac{1}{n}-\frac{1}{2n^2}$$
Which is indeed the case as it is not hard to show.
• How do you know that your second to last line is true? I feel like I am missing something basic. – Scavenger23 Nov 23 '17 at 22:26
• Either, you note that the next term in the power series is positive, or you could use the mean value theorem. – Rene Schipperus Nov 23 '17 at 22:27
• Or more straightforward, move all the terms to the right, and differentiate, and show the derivative is positive, and thus the function is increasing. – Rene Schipperus Nov 23 '17 at 22:33
• I see now, didnt realise that you tailor expanded the log. Thanks for the help. – Scavenger23 Nov 23 '17 at 22:35
I like Rene Schipperus' method a lot - it is very simple. Here is an commonly used alternative:
Note that, the function $\frac1x$ is strictly decreasing on $[1, \infty)$, so, for all $n \in \mathbb{N}^{\, \ge 1}$,
\begin{align} \ln (n+1) - \ln n &= \int_{n}^{n+1} \frac1x \, dx\\ &> \int_n^{n+1} \frac1{n+1} \, dx\\ &= \frac1{n+1} \end{align}
As you have already noted, the solution follows. | {
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As you have already noted, the solution follows.
• very good work. So essentially the method I wanted to use was correct, ? The question in the book never told me how to prove it. Induction seemed the way forward for me. – Scavenger23 Nov 23 '17 at 22:43
• @KuderaSebastian That seems reasonable... I suppose that the thing to notice was that, at some point, (e.g. in your inductive step) you would have to relate $\log(n)$ to some function of integers somehow (as this is what you have on the LHS). One way is to use the power series expansion (as Rene has done), and the other is to notice that the derivative of $\log(x)$ is in fact $\frac1x$ (which both I and Peter have used in our answers). – John Don Nov 23 '17 at 22:53
• Jon.Nice answer, short and clear. – Peter Szilas Nov 24 '17 at 9:06
Show that $a_n -a_{n+1} >0.$
$\Delta_n: = a_n - a_{n+1} =$
$-\dfrac{1}{n+1} + \log(n+1) -\log(n).$
Mean Value Theorem:
$f(x) := \log(x)$ then:
$\dfrac{f(x+1)-f(x)}{1} = f'(t),$ $x \lt t \lt (1+x).$
$\dfrac{\log(n+1) -\log(n)}{1} = \dfrac{1}{t}$, $n\lt t\lt (n+1).$
$\Delta_n = -\dfrac{1}{n+1} + \dfrac{1}{t}$ , where $t \lt (n+1).$
Hence $\Delta_n >0.$
• NIce... my answer is essentially this, but integrated, in a sense. – John Don Nov 23 '17 at 22:49 | {
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divmod() in Python and its application
PythonServer Side ProgrammingProgramming
The divmod() is part of python’s standard library which takes two numbers as parameters and gives the quotient and remainder of their division as a tuple. It is useful in many mathematical applications like checking for divisibility of numbers and establishing if a number is prime or not.
Syntax
Syntax: divmod(a, b)
a and b : b divides a
a and b are integers or floats
Examples
In the below example see the cases of both integers and floats. On the application of divmod() they give us a resulting tuple which is can also contain integers and float values.
# with integers
print("5 and 2 give:",divmod(5,2))
print("25 and 5 give:",divmod(25,5))
# with Floats
print("5.6 and 2 give:",divmod(5.6,2))
print("11.3 and 9.2 give:",divmod(11.3,9.2))
Output
Running the above code gives us the following result −
5 and 2 give: (2, 1)
25 and 5 give: (5, 0)
5.6 and 2 give: (2.0, 1.5999999999999996)
11.3 and 9.2 give: (1.0, 2.1000000000000014)
Using Zero
If the first argument is zero then we get (0,0). And If the second argument is zero then we get Zerodivision error as expected.
Example
# With first argument as zero
print("0 and 8 give:",divmod(0,8))
# With second argument as zero
print("8 and 0 give:",divmod(8,0))
Output
Running the above code gives us the following result −
0 and 8 give: (0, 0)
Traceback (most recent call last):
File "xxx.py", line 6, in
print("8 and 0 give:",divmod(8,0))
ZeroDivisionError: integer division or modulo by zero
Checking divisibility
If the second value of the tuple after division is 0 then we say that the first number is divisible by second. Else it is not divisible. The below example illustrates this.
Example
m = 12
n = 4
quotient,remainder = divmod(m,n)
print(quotient)
print(remainder)
if (remainder==0):
print(m,' is divisible by ',n)
else:
print(m,' is not divisible by ',n)
Output
Running the above code gives us the following result − | {
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Output
Running the above code gives us the following result −
3
0
12 is divisible by 4
Checking if Number is Prime
We can use divmod() to keep track of the reminders it generates when we start dividing a number by each number starting with itself till 1. For a prime number the count of zero remainder will be only one as no number other than itself will divide it perfectly. If the count of zero remainder is greater than 1 then the number is not prime,.
Example
num = 11
a = num
# counter the number of remainders with value zero
count = 0
while a != 0:
q, r = divmod(num, a)
a -= 1
if r == 0:
count += 1
if count > 2:
print(num, 'is not Prime')
else:
print(num, 'is Prime')
Output
Running the above code gives us the following result −
11 is Prime
Published on 07-Aug-2019 08:32:04 | {
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# What is the best way to intuitively explain the relationship between the derivative and the integral?
This is my first post so bear with me, but something I've been thinking about lately is: Why didn't I ever question the relationship between the derivative and the integral when I was taking calculus?
Let me explain what I mean: In most courses, the derivative is introduced as the slope of a curve at a point, or the "instantaneous rate of change". Then the integral is introduced as the area under a curve. Then students are told that these two things "undo" each other (the fundamental theorem of calculus). So now I'm wondering, why didn't I ever question why these "undo" each other? It's not intuitive at all. For example, addition and subtraction, multiplication and division, logarithms and exponentials; all of these things I can intuitively understand why they "undo" each other. But how does something that represents the area under a curve "undo" something that represents an instantaneous rate of change? What is the connection between those two concepts? Is there a better way to explain these concepts that makes the fundamental theorem of calculus more intuitive?
Looking forward to hearing some of y'alls responses. Thanks! | {
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• $[\cdots]$ why didn't I ever question why these "undo" each other? --- Your calculus experience is probably in the minority, because it is fairly standard (at least in U.S. introductory calculus courses, both high school and college) to make a big deal about the connection, usually at the end of the first semester, this being especially the case since the rise of "reform calculus" from the late 1980s on. That said, see right brain explanation and left brain explanation. – Dave L Renfro Apr 2 '20 at 5:26
• Which resources have you perused before registering and asking this question? Voting to close. – Rusty Core Apr 2 '20 at 21:01
• In countries, where physics is a mandatory course in grade school, the traditional layman explanation is through kinematics. Average speed v, by definition is displacement s over time t. Start with the simplest case of uniform motion, draw a graph in (t, s), it would be a line climbing up. Tangent at any point is at the same angle (similarity of right triangles), this is the derivative by definition, which is speed. Draw speed in (t, v). Area below it is displacement. Continue with uniform acceleration and then with no-uniform motion, slicing up the area under speed into thin vertical bands. – Rusty Core Apr 2 '20 at 21:14
• @BenCrowell I think what makes this a math ed question is wanting better intuition to explain the relationship. Also, the fact that many of us didn't really get it until we were teaching speaks volumes to how much this is an education issue. – Sue VanHattum Apr 4 '20 at 20:52 | {
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• @SueVanHattum Exactly - that's why I posted it here. The funny thing is that when I learned how to prove the FTC in real analysis, I didn't find either of the proofs too difficult. But the intuition behind the overall concept still wasn't there. Then recently I realized that if a student came to me and asked "Why do these two concepts undo each other?", my best response would be to show them the proof. That's why I figured this was a math ed question, because I'm specifically asking what the best way is to explain this to a student and make it more intuitive. – Brain Gainz Apr 5 '20 at 18:43 | {
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You start by noticing that the Riemann sums (multiplication followed by addition) and the difference quotients (subtraction followed by division) undo each other. Their limits -- the integral and the derivative -- still undo each other.
Added: The first sentence is a part of what is called “Discrete Calculus” https://en.wikipedia.org/wiki/Discrete_calculus
• Wow. I'm unsure whether this intuition can be made to correspond to some rigorous proof, but it's just solid enough to make me feel it gave me some insight into the mechanism of the theorem's truth and not mere word-soup hand-waving. And I think the maximal spacing in a partition (that goes to zero for the Riemann integral) even corresponds to the step size in the difference quotient in a suitable sense. – Vandermonde Apr 3 '20 at 1:23
• @Vandermonde You are correct. By the way, this will be in the third volume of my book Calculus Illustrated: amazon.com/dp/B082WKCYHY – Peter Saveliev Apr 3 '20 at 1:44
• Surprising how entirely abstract and formal mathematical thinking can be intuitive :-). Perhaps not for everybody, but even this engineer sees symmetries. – Peter - Reinstate Monica Apr 3 '20 at 10:21
• 1. Can you please exhibit the formula that you're hinting to? 2. Can you please clarify 1. which "Riemann sums" you mean? 3. Which "difference quotients"? 4. Which "limits"? – NNOX Apps Apr 4 '20 at 5:27
• If one considers the differential $\text{d}$ and the integral $\int$ as the fundamental concepts of calculus (as opposed to $\frac{\text{d}(-)}{\text{d}x}$ and $\int(-)\text{d} x$), then the wording is even more simple: difference and sum undo each other. (Actually Leibniz called $\int$ the sum, before Bernoulli suggested integral) – Michael Bächtold Apr 6 '20 at 15:58
(this is from my calculus notes, see page 233 of: http://www.supermath.info/OldschoolCalculusII.pdf) | {
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(this is from my calculus notes, see page 233 of: http://www.supermath.info/OldschoolCalculusII.pdf)
• So in words: the change in area is equal to the height of the function times the change in x? That actually makes sense. – Brain Gainz Apr 3 '20 at 0:31
• @BrainGainz thanks! Incidentally, I had exactly your feeling after taking years of calculus and an Advanced Calculus course where I suspect we proved the FTC. Your feeling is not unusual in my estimation. – James S. Cook Apr 3 '20 at 2:52
• I understand the pedagogical need, but I cringe when I read (or say myself when teaching calculus) "arbitrary function" and the picture is that of a differentiable function. – Martin Argerami Apr 3 '20 at 10:36
• @BrainGainz yep, I have a few Topology lectures posted. I am that James Cook. – James S. Cook Apr 4 '20 at 21:57
• @StevenGubkin cool, if I ever get to teach Calculus I again, I should use that. So many options to play with... – James S. Cook Apr 6 '20 at 3:49
Derivative of integral is the original function: Let $$F$$ be the integral function of $$f$$, so that $$F(x)$$ is the area under the graph from zero to $$x$$. For small $$h>0$$ the difference $$F(x+h)-F(x)$$ is the area of a narrow vertical strip. The width is $$h$$ and height approximately $$f(x)$$. As $$h\to0$$, this means $$F'(x)=f(x)$$.
If you like thinking in terms of infinitesimals, write $$dF=F(x+dx)-F(x)=f(x)dx$$ and divide by the differential. This can be a cleaner way to think as the limit process is left implicit in a way.
Integral of derivative is the original function: The integral is roughly $$\int_0^x f'(x)dx = \sum_{k=1}^n f'(x_k)(x_k-x_{k-1})$$ and the derivative is roughly $$f'(x_k) = \frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}.$$ Combine these and you get a telescoping sum and the desired result.
This is the way it makes intuitive and graphical sense to me. The link is not immediate; that's why it's a great theorem.
My answer also comes from physics. | {
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My answer also comes from physics.
Say p(t) is the position of an object in the time t. For concreteness, suppose you are traking a truck and the truck is going forward on a road from A to B. (i.e., all derivatives are positive)
It is very natural to graph p(t), and derive it, arriving at v(t), the speed of the truck for each time. After all, the derivative is just taking a small time increment $$dt$$, calculating the corresponding $$ds$$ (a space increment) and dividing
However, when I look at the graph of v(t), I can also obtain (almost everything) from p(t). Say I am at the position 100 km in the time t=0. After a $$dt$$, I shall be at $$p(t+dt) = 100+dt\cdot v(t)$$. And I can do it again and again $$p(t+dt+dt+dt) = 100+dt\cdot v(t)+dt\cdot v(t+dt)+dt\cdot v(t+2\cdot dt)+dt\cdot v(t+3\cdot dt)$$ (notice that after some $$dt$$s, I use the updated speed)
But that (taking a limit) is an integral!
The only weird thing is my assuption that I am at position 100km at time t=0. The graph of v(t) cannot tell me that. Intuitively, it does tell me how fast I am going, and therefore can tell me how much how much I walked, but it cannot tell me at which specific part of the road I started. That is that weird constant +c that appears on the indefinite integral.
But then again, we dont use the indefinite integral much, do we? We are more interested in summing $$dt$$*v(t) for ranges. Say, from t=3 to t=10. That gives me the amount I walked between those two times. So, mercifully, the +c disappears in most use cases: to know how much I walked, there is no need to know where I started. | {
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• I love how velocity and position helps to cement the relationship. In one class, my students asked why area below the axis should count as negative. We agreed that negative velocity meant going backwards, and therefore decreased your distance from start. (Distance being measured by area under velocity curve, a negative velocity curve has to have "negative area" to decrease the total distance.) – Sue VanHattum Apr 4 '20 at 20:47
When I first began teaching Calculus, I realized that I really didn't understand the Fundamental Theorem. So I looked for something that would help me have a deep understanding, that would also help me help students to see it.
I found a lovely project, which I have modified over the years. Here are links to the pdf and to a .doc version (in which the formulas are messed up, because google docs couldn't read them). If you use this, you'll want to change the part near the end where I reference the textbook.
• Wow, this is really great. I'll definitely save this. Thanks! – Brain Gainz Apr 3 '20 at 0:29
I gave a similar post to this one touching on this on Math.StackExchange.
Basically, the way I would go about it is to say that there is a very easy way by which one can think of at least Riemann integration (the usual definition given in a "most courses" calculus course) as an inverse of differentiation by construction: that is, the relationship between the two is not an "accident", but design, so that given one, you could fairly easily be led to the other.
And to do that, I'd first suggest getting rid of the whole "slope" vs "area" business altogether - if anything, those are best given as theorems to be proved, after you have other, free-standing definitions of a tangent line and an area which, by the way, can be done, but just too-often aren't, in favor of various hand-waving arguments. | {
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Instead, the relevant idea is change: the derivative of a function $$f$$, i.e. $$f'$$, stands for a kind of "sensitivity" measure. Suppose that $$f$$ were like a kind of meter or instrument, with a knob attached to it, and a readout of some kind. The knob attached is the function's input argument, typically denoted $$x$$. The indicator on the readout is the return value, $$f(x)$$. If $$x$$ is set at some value of interest $$x_0$$, then likewise the readout will be at $$f(x_0)$$. Now suppose you "wiggle" the knob $$x$$ back and forth a little bit, and you see how the needle $$f(x)$$ responds to that small impulse. For a continuous function, the size of the output's wiggle will be smaller in absolute terms the smaller you make the input, but the proportionate size, i.e. how much it wiggles relative to how much you wiggle the input value, may not be. When we say $$f$$ is differentiable, what that means is that there exists at each point $$x_0$$ a proportionality factor $$f'(x_0)$$ such that
$$f(x \pm \underbrace{dx}_\mbox{"wiggle" in x}) \approx f(x_0) \pm \underbrace{[f'(x_0)\ dx]}_\mbox{"wiggle" in f(x)}$$
so that
$$\mbox{proportional "wiggle"} = \frac{\mbox{"wiggle" in f(x)}}{\mbox{"wiggle" in x}}$$
so long as the change $$dx$$ is suitably small
Integration, then, goes the other way. Suppose that I am given now, not the function $$f$$, but only its derivative, $$f'$$, and want to find $$f$$. First off, one should observe that since $$f'$$ only deals with changes in the input, to start, we actually need one more piece of information, and that is some sort of initial value, i.e. $$f(0)$$. Suppose this to be given as well.
Starting at $$f(0)$$, suppose we apply a small, but nonzero, change $$\Delta x$$ to the input, i.e. we ask, "given $$f(0)$$, what is $$f(0 + \Delta x)$$, to the best we can do?" Well, since we know $$f(0)$$ and $$f'$$, then since $$\Delta x$$ is a small change or "wiggle", we can say that approximately, | {
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$$f(0 + \Delta x) \approx f(0) + [f'(0)\ \Delta x]$$
which is just what you see above.
Now, suppose we take another step of $$\Delta x$$. We're now going from $$x = 0 + \Delta x$$ to $$x = (0 + \Delta x) + \Delta x$$ (or $$2\ \Delta x$$, but I find writing it this way makes it clearer what is going on - "simpler" isn't necessarily "better"). At this second step, likewise, treating $$0 + \Delta x$$ as a prior input in and of itself, we have
$$f([0 + \Delta x] + \Delta x) \approx f([0 + \Delta x]) + [f'([0 + \Delta x])\ \Delta x]$$
which, combining with the previous expression, becomes
$$f([0 + \Delta x] + \Delta x) \approx f(0) + [f'(0)\ \Delta x] + [f'([0 + \Delta x])\ \Delta x]$$
and it is not hard to then continue this process so that we see for $$N$$ steps,
$$f(0 + N[\Delta x]) \approx f(0) + \sum_{i=0}^{N-1} f'(0 + i[\Delta x])\ \Delta x$$
or, if we set $$x_i := 0 + i[\Delta x]$$, we can say more neatly as
$$f(0 + N[\Delta x]) \approx f(0) + \sum_{i=0}^{N-1} f'(x_i)\ \Delta x$$
and to be more general, if we take $$N$$ steps of suitable, perhaps different, sizes $$\Delta x_i$$ to reach from $$0$$ some fixed point $$x_0$$,
$$f(x_0) \approx f(0) + \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i$$
and then we consider what happens as the steps become arbitrarily fine, at which point we hope - and need to prove - that
$$f(x_0) = f(0) + \left[\lim_{||\Delta|| \rightarrow 0}\ \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i\right]$$
which leads us to define this new operation, given by the limit on the right...
$$\int_{0}^{x_0} f'(x)\ dx := \lim_{||\Delta|| \rightarrow 0}\ \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i$$
• This does a great job of explaining how to think of $\int_a^b f'(x)\mathrm dx = f(b) - f(a)$, but, in suggesting that we think of it as a definition, I'm left to wonder how to define $\int_a^b F(x)\mathrm dx$ without proving some separate theorems about existence of anti-derivatives. – LSpice Apr 6 '20 at 0:04 | {
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Intuitively, the fundamental theorem of calculus states that "the total change is the sum of all the little changes". $$f'(x) dx$$ is a tiny change in the value of $$f$$. We sum up all these little changes to get the total change $$f(b) - f(a)$$.
I elaborated on this explanation here: https://math.stackexchange.com/a/1537836/40119.
• Very nice! The idea that $\mathrm dx$ is not punctuation, but (conceptually) an actual infinitesimal measure of $x$-length, so that $f'(x)\mathrm dx$ has units of (slope)($x$-length) = $y$-length, I think is probably the most important pre-proof part of understanding the connection. That is, I think that most people think of integration as "something you do to $f'(x)$", not "something you do to $f'(x)\mathrm dx$"—hence later difficulties with differential forms, and earlier difficulties with changes of variable. – LSpice Apr 6 '20 at 0:06
Well, for me, the nicest intuition comes from physics. If $$F=F(x)$$ is some force applied to an object $$O$$ casuing $$O$$ to move, where $$x$$ is its displacement, when the work $$W(x)$$ produced by that force from $$x_0=0$$ to $$x$$ metres is given by:
$$W(x):=\int_0^xF(s)ds.$$
Now, what if we ask what is the rate of change of that work with respect to displacement? At some certain point $$x$$, the work provided to or substracted by that object $$O$$ is determined by the force $$F$$ applied on it. So, one can intuitively expect that:
$$W'(x)=\left(\int_0^xF(s)ds\right)'=F'(x).$$
That is, the larger the force, the faster the energy flows from the one applying the force towoards the object. The less the force, the slower that energy flows.
If you want more clearly "undoing" operations, define "stacking" as the following: | {
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If you want more clearly "undoing" operations, define "stacking" as the following:
Take some $$\Delta x$$. Chop the curve into rectangles of width $$\Delta x$$ and height $$f(x)$$. (There's some leeway as to take $$f(x)$$ at the left side, right side, middle, minimum, maximum, etc. The rest of my description will be right side.) Now take each rectangle and put the bottom of each rectangle at the top of the previous rectangle. For instance, if you have $$f(x) = x^2$$ and $$\Delta x = 0.1$$, you'd have a rectangle with lower left corner at the origin, and top left at $$(0.1, 0.01)$$. Then the next rectangle would have lower left corner at $$(0.1, 0.01)$$, and upper right at $$(0.2, 0.05)$$, and so on.
Define "unstacking" as follows:
Take some $$\Delta x$$. Chop the curve into rectangles of width $$\Delta x$$ and height $$f(x)$$. Now, take each rectangle and shift it vertically down the height of the previous rectangle. For instance, if $$f(x) = x^2$$, the first rectangle will have height $$0.01$$. The second one will have height $$0.04$$, so for the unstacked version, we move it down $$0.01$$, leaving its new height as $$0.03$$. Another way of phrasing it is for each $$x$$, take the rectangle whose bottom left corner is at $$(x, f(x))$$ and top right corner is $$(x+\Delta x, f(x+\Delta x))$$. Then create a chart out of all of those rectangle, with all of them moved down so their bottom is on the x-axis.
For "well-behaved" functions (I believe that uniform continuity is a sufficient condition), as $$\Delta x$$ goes to zero, stacking becomes integration, and unstacking become differentiation (with the proper scaling). | {
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# Elementary matrix operations
I am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.
I am asked to find both the nullspace and the general solution of the following systems $A\tilde{x}=\tilde{b}$.
The first is $$A= \begin{pmatrix} 1 & 1 & 0 & 2\\ 2 & 1 & 1 &-2\\ 2 & 2 & 2 & 1 \end{pmatrix},\quad \tilde{b}= \begin{pmatrix} 12\\0\\14 \end{pmatrix}$$ Proceeding with an augmented matrix, I have \begin{align*} \left( \begin{array}{rrrr|c} 1 & 1 & 0 & 2 & 12\\ 2 & 1 & 1 &-2 & 0\\ 2 & 2 & 2 & 1 & 14 \end{array} \right)&\sim \left( \begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\ 0 &-1 & 1 &-6 & -24\\ 0 & 0 & 2 & -3 &-10 \end{array} \right)\\ &\sim \left( \begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\ 0 & 1 &-1 & 6 & 24\\ 0 & 0 & 1 & -3/2 &-5 \end{array} \right) \end{align*} I've checked this multiple times and can't find a mistake, plus it agrees with the solutions my professor has provided.
Solving this now, I have \begin{align*} x_4&=\alpha\in\mathbb{R}\\ x_3&=-5+\frac{3}{2}\alpha\\ x_2&=19-\frac{9}{2}\alpha\\ x_1&=-7+\frac{5}{2}\alpha \end{align*}
So from this, my solution will be $$\tilde{x}= \begin{pmatrix} -7+\frac{5}{2}\alpha\\ 19-\frac{9}{2}\alpha\\ -5+\frac{3}{2}\alpha\\ \alpha \end{pmatrix} = \begin{pmatrix} -7\\19\\-5\\0 \end{pmatrix}+\beta \begin{pmatrix} 5\\-9\\3\\2 \end{pmatrix},\quad\beta\in\mathbb{R}.$$
This does not agree with the solutions given, and the solutions given are just solutions without any steps in between so I'm a little confused as to where I went wrong. The weird part is, my nullspace agrees with the solution provided but the particular solution certainly does not.
Edit: My professor's solution is $$\tilde{x}= \begin{pmatrix} 3\\1\\1\\4 \end{pmatrix}+\alpha \begin{pmatrix} 5\\-9\\3\\2 \end{pmatrix},\quad\beta\in\mathbb{R}.$$
Can somebody please let me know what I've done incorrectly?
Thank you. | {
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Can somebody please let me know what I've done incorrectly?
Thank you.
• I tried using WolframAlpha, but it didn't tell me what I did wrong, and it spat out something that looked different to both my solution and my professor's (although it did provide the same nullspace!). – Babe in the Woods Aug 10 '16 at 23:22
• Oh, he went away.. – Babe in the Woods Aug 10 '16 at 23:22
• Can you provide your professor's solution? – Aweygan Aug 10 '16 at 23:25
• My apologies, my professor's solution has been added in the question. – Babe in the Woods Aug 10 '16 at 23:29
• Your professor simply chose to set $x_4=2\alpha$ instead of $\alpha$ to avoid denominators in the general solution. – Bernard Aug 10 '16 at 23:33
Your solution is correct. You can obtain the professor's solution from yours via the substitution $\beta=\alpha+2$. Both are valid answers, and unless there was some given format your answer was supposed to be in, I don't see why yours could be considered incorrect.
This is correct! You can check if you found solutions of the system simply by replacing $x$ in $Ax=b$ by what you found. | {
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# Counting ways when first position is covered with a tie
Six horses are entered in a race. If two horses are tied for first place and there are no ties among the other four horses, in how many ways can the six horses cross the finish line?
My approach:
We choose $$2$$ out of $$6$$ for the first place and $$4$$ out of the remaining $$4$$:
$$_6P_2\cdot _4P_4 = \frac{6!}{(6 -2)!}\cdot\frac{4!}{(4 - 4)!}=720$$
But the solution states $$360$$.
What am I doing wrong here?
• Also, $\binom 44= \frac{4!}{0!4!}$. But you don't want $\binom 44$ because you care about the order of the remaining $4$ horses. You want the number of ways to permute those $4$ horses, which is simply $4!$. Feb 26 at 23:31
• @RobertShore: Yes the order matters that is why I used $\binom{n}{k}=\frac{n!}{(n-k)!}$ which is $n\cdot(n-1)\cdots\cdot(n-k+1)$. Also it is still $720$
– Jim
Feb 26 at 23:35
• Your values for the combinations are wrong. ${6 \choose 2}=\frac {6!}{4!2!}=15$. You don't want $4 \choose 4$ because the order of the last four horses matters. That gives $15 \cdot 4!=360$ Feb 26 at 23:40
• When you write $6 \choose 2$ you are saying that order does not matter. I don't know a simple one for when order does matter. Some use $_6P_2$ for that. Feb 27 at 0:09
• There are $\binom{6}{2}$ to select the subset of two horses which finish in a tie and $4!$ ways for the remaining four horses to finish in the remaining four positions. Notice that the order of selection for the two horses which finish in a tie does not matter, which is why your answer $P(6, 2)P(4, 4)$ is twice the correct answer. Feb 27 at 23:28
That leaves 4 choices for the horse to come in third, three choices for the horse to come in fourth, two choice for the horse to come in fifth, and one for the horse to come in sixth. The number of ways this can happen is $$15(4)(3)(2)(1)= \frac{6!}{2}= 720/2= 360$$, | {
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# Find the eigenvalues and corresponding eigen vectors of the matrix
Find the eigenvalues and corresponding eigen vectors of the matrix $$\begin{bmatrix}-3&6&-43\\0&-1&9\\0&0&2\end{bmatrix}$$
The eigenvalue $$\lambda_1 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.
The eigenvalue $$\lambda_2 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.
The eigenvalue $$\lambda_3 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.
I'm kind of stuck after a certain point. Here is what I have so far
I do know that $$(A - \lambda I)X = 0$$
so $$\begin{bmatrix}-3&6&-43\\0&-1&9\\0&0&2\end{bmatrix}$$ $$\implies \lambda_1 = -3, \lambda_2 = -1, \lambda_3 = 2$$ so I have the eigenvalues but how can I find the corresponding eigenvectors?
• Do you know how to find eigenvalues? This is an upper triangular matrix. What about it's characteristic polynomial? – Kushal Bhuyan Jul 14 '16 at 1:40
• In any good textbook, class, video lectures or lecture notes on linear algebra, solving linear systems should be covered before finding eigenvalues. In particular, the Gaussian elimination method first puts a linear system into "upper triangular" form, and then finds the unknowns starting from the last ones. – arctic tern Jul 14 '16 at 3:27
You need to solve the equations $(A-\lambda I)v=0$ for $v$ for each of the three eigenvalues $\lambda$.
For instance when $\lambda=2$ we're solving
$$\begin{pmatrix} -5 & 6 & -43 \\ 0 & -3 & 9 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ | {
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The last equation is $0=0$ so it's superfluous. So we have two equations in three unknowns. If $z=0$ then the second equation implies $y=0$ and then the first equation implies $x=0$ which gives the zero vector, and that's not very interesting. Otherwise if $z\ne0$, we can scale the vector $(x,y,z)$ (scaling preserves the property of being a solution to the above system) in order to make $z=1$. In which case the second equation gives $y=3$ and then the first equation gives $x=-5$. So the eigenvector is $(-5,3,1)$ up to scaling.
Your turn. Do $\lambda=-3$ and $\lambda=-1$. (Sophia did $\lambda=-3$ for you, so now most of the work is done for you. If we do the problem for you, at least learn what it is we're doing!)
• For $\lambda_2 = -1$ I get $\begin{bmatrix}-2&6&-43\\0&0&9\\0&0&3\end{bmatrix}$ ~ $\begin{bmatrix}1&-3&\frac{43}{2}\\0&0&9\\0&0&0\end{bmatrix}$ so I thought it would be (-3,0,0) ? – Yusha Jul 14 '16 at 3:27
• When I multiply that matrix by $(-3,0,0)^T$ I don't get the zero matrix. In that situation, the last equation is redundant. What does the second equation tell you about $z$? What does the first equation then reduce to saying about $x$ and $y$? (BTW I wouldn't bother with the fractions.) – arctic tern Jul 14 '16 at 3:30
• Wow, I'm an idiot! I see now that $z = 0 \implies 6y = 2x \implies (3,1,0)$ – Yusha Jul 14 '16 at 3:31 | {
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