# rolle's theorem pdf

Explain why there are at least two times during the flight when the speed of Theorem (Cauchy's Mean Value Theorem): Proof: If , we apply Rolle's Theorem to to get a point such that . Rolle’s Theorem. Rolle's theorem is one of the foundational theorems in differential calculus. Rolle S Theorem. It is a very simple proof and only assumes Rolle’s Theorem. Without looking at your notes, state the Mean Value Theorem … For problems 1 & 2 determine all the number(s) c which satisfy the conclusion of Rolle’s Theorem for the given function and interval. We seek a c in (a,b) with f′(c) = 0. Learn with content. If f a f b '0 then there is at least one number c in (a, b) such that fc . Watch learning videos, swipe through stories, and browse through concepts. Standard version of the theorem. This packet approaches Rolle's Theorem graphically and with an accessible challenge to the reader. The “mean” in mean value theorem refers to the average rate of change of the function. The result follows by applying Rolle’s Theorem to g. ⁄ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. Practice Exercise: Rolle's theorem … }�gdL�c���x�rS�km��V�/���E�p[�ő蕁0��V��Q. The result follows by applying Rolle’s Theorem to g. ¤ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f 0 . Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. %PDF-1.4 3 0 obj Proof. This version of Rolle's theorem is used to prove the mean value theorem, of which Rolle's theorem is indeed a special case. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with Example - 33. Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change ʹ뾻��Ӄ�(�m���� 5�O��D}P�kn4��Wcم�V�t�,�iL��X~m3�=lQ�S���{f2���A���D�H����P�>�;$f=�sF~M��?�o��v8)ѺnC��1�oGIY�ۡ��֍�p=TI���ߎ�w��9#��Q���l��u�N�T{��C�U��=���n2�c�)e�L����� �����κ�9a�v(� ��xA7(��a'b�^3g��5��a,��9uH*�vU��7WZK�1nswe�T��%�n���է�����B}>����-�& 3�c)'�P#:p�8�ʱ� ����;�c�՚8?�J,p�~$�JN����Υ�����P�Q�j>���g�Tp�|(�a2���������1��5Լ�����|0Z v����5Z�b(�a��;�\Z,d,Fr��b�}ҁc=y�n�Gpl&��5�|���(�a��>? Theorem 1.1. (Rolle’s theorem) Let f : [a;b] !R be a continuous function on [a;b], di erentiable on (a;b) and such that f(a) = f(b). �_�8�j&�j6���Na$�n�-5��K�H 3.2 Rolle’s Theorem and the Mean Value Theorem Rolle’s Theorem – Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). For each problem, determine if Rolle's Theorem can be applied. Rolle’s Theorem is a special case of the Mean Value Theorem in which the endpoints are equal. Be sure to show your set up in finding the value(s). 5 0 obj Rolle’s Theorem, like the Theorem on Local Extrema, ends with f′(c) = 0. Section 4-7 : The Mean Value Theorem. Proof: The argument uses mathematical induction. 172 Chapter 3 3.2 Applications of Differentiation Rolle’s Theorem and the Mean Value Theorem Understand and use Rolle’s Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. When n = 0, Taylor’s theorem reduces to the Mean Value Theorem which is itself a consequence of Rolle’s theorem. 13) y = x2 − x − 12 x + 4; [ −3, 4] 14) y = Stories. If it can, find all values of c that satisfy the theorem. That is, we wish to show that f has a horizontal tangent somewhere between a and b. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. Videos. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. EXAMPLE: Determine whether Rolle’s Theorem can be applied to . Rolle’s Theorem and other related mathematical concepts. The Mean Value Theorem is an extension of the Intermediate Value Theorem.. Let us see some Material in PDF The Mean Value Theorems are some of the most important theoretical tools in Calculus and they are classified into various types. Examples: Find the two x-intercepts of the function f and show that f’(x) = 0 at some point between the (Insert graph of f(x) = sin(x) on the interval (0, 2π) On the x-axis, label the origin as a, and then label x = 3π/2 as b.) f x x x ( ) 3 1 on [-1, 0]. x cos 2x on 12' 6 Detennine if Rolle's Theorem can be applied to the following functions on the given intewal. Now an application of Rolle's Theorem to gives , for some . The value of 'c' in Rolle's theorem for the function f (x) = ... Customize assignments and download PDF’s. Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. If it can, find all values of c that satisfy the theorem. If so, find the value(s) guaranteed by the theorem. If f(a) = f(b) = 0 then 9 some s 2 [a;b] s.t. If f is zero at the n distinct points x x x 01 n in >ab,,@ then there exists a number c in ab, such that fcn 0. 13) y = x2 − x − 12 x + 4; [ −3, 4] 14) y = Concepts. Lesson 16 Rolle’s Theorem and Mean Value Theorem ROLLE’S THEOREM This theorem states the geometrically obvious fact that if the graph of a differentiable function intersects the x-axis at two places, a and b there must be at least one place where the tangent line is horizontal. 3.2 Rolle’s Theorem and the Mean Value Theorem Rolle’s Theorem – Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). In case f  ( a ) = f  ( b ) is both the maximum and the minimum, then there is nothing more to say, for then f is a constant function and … Thus, which gives the required equality. Rolle's Theorem on Brilliant, the largest community of math and science problem solvers. It is a special case of, and in fact is equivalent to, the mean value theorem, which in turn is an essential ingredient in the proof of the fundamental theorem of calculus. Access the answers to hundreds of Rolle's theorem questions that are explained in a way that's easy for you to understand. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. Rolle’s Theorem extends this idea to higher order derivatives: Generalized Rolle’s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . Brilliant. In modern mathematics, the proof of Rolle’s theorem is based on two other theorems − the Weierstrass extreme value theorem and Fermat’s theorem. A similar approach can be used to prove Taylor’s theorem. ?�FN���g���a�6��2�1�cXx��;p�=���/C9��}��u�r�s�[��y_v�XO�ѣ/�r�'�P�e��bw����Ů�#�����b�}|~��^���r�>o���W#5��}p~��Z؃��=�z����D����P��b��sy���^&R�=���b�� b���9z�e]�a�����}H{5R���=8^z9C#{HM轎�@7�>��BN�v=GH�*�6�]��Z��ܚ �91�"�������Z�n:�+U�a��A��I�Ȗ�$m�bh���U����I��Oc�����0E2LnU�F��D_;�Tc�~=�Y��|�h�Tf�T����v^��׼>�k�+W����� �l�=�-�IUN۳����W�|׃_�l �˯����Z6>Ɵ�^JS�5e;#��A1��v������M�x�����]*ݺTʮ���״N�X�� �M���m~G��솆�Yoie��c+�C�co�m��ñ���P�������r,�a The result follows by applying Rolle’s Theorem to g. ⁄ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. f0(s) = 0. f is continuous on [a;b] therefore assumes absolute max and min values and by Rolle’s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. Let f(x) be di erentiable on [a;b] and suppose that f(a) = f(b). <> %PDF-1.4 Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. For example, if we have a property of f 0 and we want to see the effect of this property on f , we usually try to apply the mean value theorem. Take Toppr Scholastic Test for Aptitude and Reasoning and by Rolle’s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with This calculus video tutorial provides a basic introduction into rolle's theorem. In these free GATE Study Notes, we will learn about the important Mean Value Theorems like Rolle’s Theorem, Lagrange’s Mean Value Theorem, Cauchy’s Mean Value Theorem and Taylor’s Theorem. %���� Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Determine whether the MVT can be applied to f on the closed interval. If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that ′ =. Since f (x) has infinite zeroes in \begin{align}\left[ {0,\frac{1}{\pi }} \right]\end{align} given by (i), f '(x) will also have an infinite number of zeroes. For example, if we have a property of f0 and we want to see the eﬁect of this property on f, we usually try to apply the mean value theorem. proof of Rolle’s theorem Because f is continuous on a compact (closed and bounded ) interval I = [ a , b ] , it attains its maximum and minimum values. x��]I��G�-ɻ�����/��ƴE�-@r�h�١ �^�Կ��9�ƗY�+e����\Y��/�;Ǎ����_ƿi���ﲀ�����w�sJ����ݏ����3���x���~B�������9���"�~�?�Z����×���co=��i�r����pݎ~��ݿ��˿}����Gfa�4�����Ks�?^���f�4���F��h���?������I�ק?����������K/g{��׽W����+�~�:���[��nvy�5p�I�����q~V�=Wva�ެ=�K�\�F���2�l��� ��|f�O�n9���~�!���}�L��!��a�������}v��?���q�3����/����?����ӻO���V~�[�������+�=1�4�x=�^Śo�Xܳmv� [=�/��w��S�v��Oy���~q1֙�A��x�OT���O��Oǡ�[�_J���3�?�o�+Mq�ٞ3�-AN��x�CD��B��C�N#����j���q;�9�3��s�y��Ӎ���n�Fkf����� X���{z���j^����A���+mLm=w�����ER}��^^��7)j9��İG6����[�v������'�����t!4?���k��0�3�\?h?�~�O�g�A��YRN/��J�������9��1!�C_$�L{��/��ߎq+���|ڶUc+��m��q������#4�GxY�:^밡#��l'a8to��[+�de. Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. The special case of the MVT, when f(a) = f(b) is called Rolle’s Theorem.. By Rolle’s theorem, between any two successive zeroes of f(x) will lie a zero f '(x). Taylor Remainder Theorem. In the case , define by , where is so chosen that , i.e., . Determine whether the MVT can be applied to f on the closed interval. differentiable at x = 3 and so Rolle’s Theorem can not be applied. exact value(s) guaranteed by the theorem. If f a f b '0 then there is at least one number c in (a, b) such that fc . If it cannot, explain why not. For each problem, determine if Rolle's Theorem can be applied. If Rolle’s Theorem can be applied, find all values of c in the open interval (0, -1) such that If Rolle’s Theorem can not be applied, explain why. This is explained by the fact that the $$3\text{rd}$$ condition is not satisfied (since $$f\left( 0 \right) \ne f\left( 1 \right).$$) Figure 5. �K��Y�C��!�OC���ux(�XQ��gP_'�s���Տ_��:��;�A#n!���z:?�{���P?�Ō���]�5Ի�&���j��+�Rjt�!�F=~��sfD�[x�e#̓E�'�ov�Q��'#�Q�qW�˿���O� i�V������ӳ��lGWa�wYD�\ӽ���S�Ng�7=��|���և� �ܼ�=�Չ%,��� EK=IP��bn*_�D�-��'�4����'�=ж�&�t�~L����l3��������h��� ��~kѾ�]Iz���X�-U� VE.D��f;!��q81�̙Ty���KP%�����o��;$�Wh^��%�Ŧn�B1 C�4�UT���fV-�hy��x#8s�!���y�! %�쏢 Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change We can use the Intermediate Value Theorem to show that has at least one real solution: For example, if we have a property of f0 and we want to see the eﬁect of this property on f, we usually try to apply the mean value theorem. To give a graphical explanation of Rolle's Theorem-an important precursor to the Mean Value Theorem in Calculus. The proof of Rolle’s Theorem is a matter of examining cases and applying the Theorem on Local Extrema. It’s basic idea is: given a set of values in a set range, one of those points will equal the average. Calculus 120 Worksheet – The Mean Value Theorem and Rolle’s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( Õ)−( Ô) Õ− Ô =′( . If a functionfis defined on the closed interval [a,b] satisfying the following conditions – i) The function fis continuous on the closed interval [a, b] ii)The function fis differentiable on the open interval (a, b) Then there exists a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a) This theorem is also known as the first mean value theorem or Lagrange’s mean value theorem. If it cannot, explain why not. x��=]��q��+�ͷIv��Y)?ز�r$;6EGvU�"E��;Ӣh��I���n v��K-�+q�b ��n�ݘ�o6b�j#�o.�k}���7W~��0��ӻ�/#���������$����t%�W ��� For the function f shown below, determine we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0.If not, explain why not. We can see its geometric meaning as follows: \Rolle’s theorem" by Harp is licensed under CC BY-SA 2.5 Theorem 1.2. Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). Question 0.1 State and prove Rolles Theorem (Rolles Theorem) Let f be a continuous real valued function de ned on some interval [a;b] & di erentiable on all (a;b). f c ( ) 0 . The Common Sense Explanation. stream After 5.5 hours, the plan arrives at its destination. Get help with your Rolle's theorem homework. This builds to mathematical formality and uses concrete examples. So the Rolle’s theorem fails here. THE TAYLOR REMAINDER THEOREM JAMES KEESLING In this post we give a proof of the Taylor Remainder Theorem. A plane begins its takeoff at 2:00 PM on a 2500 mile flight. Then, there is a point c2(a;b) such that f0(c) = 0. stream Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Calculus 120 Worksheet – The Mean Value Theorem and Rolle’s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( Õ)−( Ô) Õ− Ô =′( . Proof: The argument uses mathematical induction. �wg��+�͍��&Q�ណt�ޮ�Ʋ뚵�#��|��s���=�s^4�wlh��&�#��5A ! 2\�����������M�I����!�G��]�x�x*B�'������U�R� ���I1�����88%M�G[%&���9c� =��W�>���\$�����5i��z�c�ص����r ���0y���Jl?�Qڨ�)\+�B��/l;�t�h>�Ҍ����X�350�EN�CJ7�A�����Yq�}�9�hZ(��u�5�@�� Forthe reader’s convenience, we recall below the statement ofRolle’s Theorem. 20B Mean Value Theorem 2 Mean Value Theorem for Derivatives If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c on (a,b) such that EX 1 Find the number c guaranteed by the MVT for derivatives for Then there is a point a<˘ab,,@ then there exists a number c in ab, such that fcn 0. View Rolles Theorem.pdf from MATH 123 at State University of Semarang. Make now. Examples: Find the two x-intercepts of the function f and show that f’(x) = 0 at some point between the Rolle’s Theorem extends this idea to higher order derivatives: Generalized Rolle’s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. Then . Let us see some Proof of Taylor’s Theorem. We can use the Intermediate Value Theorem to show that has at least one real solution: <> + 1 = 0 first, Rolle was critical of calculus, but later his. Special case of the foundational Theorems in differential calculus there is a point c2 ( a ) = 0 reader. Approach can be applied to f on the given intewal only assumes Rolle s. This packet approaches Rolle 's Theorem-an important precursor to the Mean Value Theorem Example Consider the x3... Explained in a way that 's easy for you to understand State of. S Theorem is a very simple proof and only assumes Rolle ’ s..! 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# Merge sort - maximum comparisons

I recently came across a problem where I was to find the maximum comparison operations when applying the merge sort algorithm on an 8 character long string. I tried implementing the 2r^r model however, the number of comparison operations used in a merge sort varies greatly with different input lists.

My question asked for the greatest number of comparison operations for one list. I applied the r2^r explicit definition which gave me 24. But the answer was 17. I can't find much information online or in the book about elementary algorithms and most solutions do not go into such details.

Does anyone know why this might be? I have seen some solutions where;

let 2^r = length of list, r2^r = greatest number of comparison operations.

2^r = 8
r = log(8)/log(2)
r = 3

Therefore, r2^r = 24

But that is not corroborated in my course.

any ideas?

• What distinguishes this "cardinality" of comparison operations from the computational complexity of the merge sort, which in computer science is usually measured by the number of comparison operations performed? How is any computation complexity problem not a "discrete maths question on cardinality" according to your definition? Apr 29 '20 at 2:25
• Perhaps it would help if you showed, step by step, how you arrived at the answer $24$ so people can see how your methods reflect some kind of discrete maths cardinality approach instead of a computer science complexity approach. It would be better if you write the math in math notation; see math.stackexchange.com/help/notation Apr 29 '20 at 2:27
• I distinguished it from a computer science problem as my understanding is that their implementations are different. In my experience, I use merge sort in Java or C++ to combine two lists and sort them in one function. You are right, the complexity of which would determine the worst-case/ greatest number of comparisons. However, the question specified one list of 8 elements which I am not used to. Apr 29 '20 at 3:31
• Thanks, David I just added my method I used to find 24. I also removed the disclaimer. Apr 29 '20 at 3:35
• Complexity theory in computer science involves no Java or C++. It's an abstract topic. But computer science also is a topic on this site, as you can see by searching the [computer-science] tag. Apr 29 '20 at 3:41

Let $$a_1...a_8$$ be the input and let for simplicity let $$f_{i,j}\begin{cases} 1 & \text{if } a_i\leq a_j \\ 0 & \text{if } a_i> a_j \end{cases}$$, i.e. the $$f_{i,j}$$ are the comparison operations.

Let us go through the steps of Mergesort; there are 3 levels or phases corresponding to top-down recursive calls:

1. Level 1 Compute $$M(a_1,a_2) , ... ,M(a_7,a_8)$$
2. Level 2 Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ and merge $$(a_5,a_6)$$ with $$(a_7,a_8)$$
3. Level 3 Merge $$(a_1,a_2,a_3,a_4)$$ with $$(a_5,a_6,a_7,a_8)$$

Let us count the # of $$f_{i,j}$$ at each of the levels

1. Level 1 has four comparisons $$f_{1,2},...,f_{7,8}$$
2. Level 2 has at most 6 comparisons

• Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ takes at most 3 comparisons

• Merge $$(a_1,a_2)$$ with $$(a_3,a_4)$$ takes at most 3 comaprisons

3. Level 3 has at most 7 comparisons $$f_{1,5},...,f_{4,8}$$

• After performing $$f_{i,j}$$ mergesort will then perform $$f_{i,j+1}$$ or $$f_{i+1,j}$$ until it hits $$f_{4,8}$$; the worst computation path could take 7 comparisons

Let us make an educated guess at the worst-case scenario, say $$(7,4,3,6,5,2,1,8)$$

1. Level 1 will spit out $$(4,7),(3,6),(2,5)$$ and $$(1,8)$$ after 4 comparisons
2. Level 2 will spit out $$(3,4,6,7)$$ and $$(1,2,5,8)$$ after 6 comparisons

• $$(3,4,6,7)$$ will cause the comparisons $$f_{1,3},f_{1,4},f_{2,4}$$ to be computed
• $$(1,2,5,8)$$ will cause the comparisons $$f_{5,7},f_{5,8},f_{6,8}$$ to be computed
3. Level 3 will spit out $$(1,2,3,4,5,6,7,8)$$ after 7 comparisons

• The following comparisons will be computed: $$f_{1,5},f_{1,6},f_{1,7},f_{2,7},f_{3,7},f_{3,8},f_{4,8}$$

For a grand total of 17

BTW the arguments and construction given can easily be generalized ... do you see the general pattern ... Good Luck with your mathematical voyages! Bon Voyage!

• Okay yep, that's a great explanation. I see how they arrived at 17 now. I was quite confused. We have just covered proofs for strong induction, so I think I can induce an explicit formula from your solution that can solve for the greatest number of comparison operations. However, without skipping a beat we are now combining: Probability, propositional logic, matrices and algorithms - so RIP me. But knowing I can count on my math stack exchange community to help me out here and there gives me the confidence to continue strong on my mathematical voyage. Thank you Pedrpan !! Apr 29 '20 at 15:07
• No problem, I am glad that I could be of use to you! Take care! Apr 29 '20 at 20:05


# What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is …

What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is $2^33^3$ where $x, y,z\in \Bbb N$?

What I tried :

At least one of $x, y$ and $z$ should have factor $2^3$ and at least one should have factor $3^3$. I then tried to figure out the possible combinations but couldn't get the correct answer.

We use Inclusion/Exclusion.

First we find the number of (positive) triples in which each entry divides $2^33^3$. At each of $x$, $y$, $z$ we have $(4)(4)$ choices, for a total of $16^3$.

We want to subtract the number of such triples in which each entry divides $2^23^3$. There are $12^3$ such triples. There are also $12^3$ such triples in which each element divides $2^33^2$.

But we have subtracted once too many times the $9^3$ triples in which each entry divides $2^23^2$.

So the total is $16^3-2\cdot 12^3+9^3$.

Consider all candidate triples of the form: $$(2^{a_1}3^{b_1}, 2^{a_2}3^{b_2}, 2^{a_3}3^{b_3})$$ where for each $i \in \{1, 2, 3\}$, we have $a_i, b_i \in \{0, 1, 2, 3\}$.

We define such a candidate triple to be valid if for some $j, k \in \{1, 2, 3\}$, we have $a_j = 3$ and $b_k = 3$. Otherwise, if ($a_j \in \{0, 1, 2\}$ for all $j \in \{1, 2, 3\}$) or ($b_k \in \{0, 1, 2\}$ for all $k \in \{1, 2, 3\}$), then such a candidate triple is considered invalid.

Observe that: \begin{align*} \text{# of valid triples} &= \text{# of candidate triples} - \text{# of invalid triples} \\ &= 4^6 - (3^3 \cdot 4^3 + 4^3 \cdot 3^3 - 3^6) \end{align*}


# S1 Probability Coin Toss

Tags:
1. Feb 25, 2015

### AntSC

Having trouble with certain binomial and geometric distribution questions, which is indicating that my understanding isn't completely there yet. Any help would be greatly appreciated.

1. The problem statement, all variables and given/known data

A bag contains two biased coins: coin A shows Heads with a probability of 0.6, and coin B shows Heads with a probability 0.25. A coin is chosen at random from the bag and tossed three times.
Find the probability that the three tosses of the coin show two Heads and one Tail in any order.

2. Relevant equations

3. The attempt at a solution
Probabilities:
$H_{A}=0.6$ and $T_{A}=0.4$
$H_{B}=0.25$ and $T_{B}=0.75$

Possibilities for 2 heads and one tail in any order:
$$3\left ( H \right )^{2}\left ( T \right )$$

Is this correct so far?
My question is how to incorporate the probability of picking coin A or coin B into the problem?

2. Feb 25, 2015

### Simon Bridge

What is the probability you picked coin A?

3. Feb 25, 2015

### AntSC

A half

4. Feb 25, 2015

### Simon Bridge

So if you picked a coin at random and tossed just once, what is the probability the result is a head?

5. Feb 25, 2015

### AntSC

Ah i see it now.
$$P=\frac{1}{2}3\left ( H_{A} \right )^{2}\left ( T_{A} \right )+\frac{1}{2}3\left ( H_{B} \right )^{2}\left ( T_{B} \right )$$
Is this right?

6. Feb 25, 2015

### Simon Bridge

You can check it with a probability tree if you are unsure.

7. Feb 25, 2015

### AntSC

Sure. I want to start to dispense with the need for visual aids and make sure i can construct the problem without.
Especially when dealing with a larger set of choices, like 52 cards. A tree then won't be so helpful.
Thanks for the dialogue. I think i needed to get it out there to help work it through.
You might see a few more questions from me in future :)

8. Feb 25, 2015

### Ray Vickson

QUOTE="AntSC, post: 5021860, member: 450435"]Ah i see it now.
$$P=\frac{1}{2}3\left ( H_{A} \right )^{2}\left ( T_{A} \right )+\frac{1}{2}3\left ( H_{B} \right )^{2}\left ( T_{B} \right )$$
Is this right?[/QUOTE]

If $E$ is the event "2H, 1T (any order)", does your formula satisfy the basic relationship
$$P(E) = P(E|A) P(A) + P(E|B) P(B) ?$$
If it does, it is OK.

BTW: you might compare this with the scenario where you replace the coin after each toss and then ask about $E$.


# Are all conjugacy classes in $\text{GL}_n(\mathbb R)$ path-connected?

Suppose $$A$$ and $$B$$ are conjugate invertible real $$n \times n$$-matrices. Does there always exist a path from $$A$$ to $$B$$ inside their conjugacy class?

I thought I had an easy proof for odd $$n$$ which goes as follows, but it was incorrect as pointed out in this answer. To show where my misunderstanding arised, here is the wrong argument.

Suppose there exists a real matrix $$P$$ such that $$B = PAP^{-1}$$. By replacing $$P$$ with $$-P$$ if necessary, we can assume that $$\det P > 0$$ (this is what goes wrong in even dimensions, see this question). Then we have that $$P = e^Q$$ for some real matrix $$Q$$ (since the image of the exponential map is the path-component of the identity in $$\text{GL}_n(\mathbb R)$$). But now the path $$t \mapsto e^{tQ}Ae^{-tQ}$$ is a path connecting $$A$$ to $$PAP^{-1} = B$$.

This edit comes a bit late, but the way I see it is a bit different than the other answers so I'll write it anyway.

Again, I use the example from the question that you link to: $$A$$ is the rotation by $$\frac \pi 2$$ in the euclidean plane, and $$B$$ is the rotation by $$-\frac \pi 2$$.

Now any conjugate of $$A$$ by a matrix with positive determinant will correspond to a linear map $$\varphi$$ such that for any non-zero vector $$v$$, the vectors $$v$$ and $$\varphi (v)$$ in this order make a positive basis of the plane.

Conversely, a conjugate of $$A$$ by a matrix with negative determinant (which is the case of $$B$$) will correspond to a linear map $$\varphi$$ such that for any non-zero vector $$v$$, the vectors $$v$$ and $$\varphi (v)$$ in this order make a negative basis of the plane.

A path from $$A$$ to $$B$$ has to cross the set of matrices that have real eigenvalues, and such a matrix cannot be conjugate to $$A$$.

• I like this geometric approach, this is the solution I will not forget. – Levi Nov 4 '19 at 0:50
• @Levi Thanks :) – Arnaud Mortier Nov 4 '19 at 0:51

We can use the counterexample from your other question to answer this one. Let $$A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$. Matrices $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ must have trace $$a+d$$ equal to $$0$$ and determinant $$ad-bc=-a^2-bc$$ equal to $$1$$. This implies that such a matrix cannot have $$b=0$$, since $$-a^2\leq 0$$. Therefore the set of conjugates of $$A$$ is disjoint union of open sets defined by $$b>0$$ and $$b<0$$. Neither of those sets is empty, since one contains $$A$$ and the other contains $$-A$$. Thus the conjugacy class of $$A$$ is disconnected.

• This is much nicer than my argument! (But slightly less detailed, whence I'm leaving mine up.) – darij grinberg Nov 4 '19 at 0:30
• This is an awesome way of seeing this. This particular example also shows that I have been trying to prove the wrong thing when answering the other question. So I am particularly grateful! – Levi Nov 4 '19 at 0:33
• Sorry for unaccepting your answer, but I cannot resist going with coordinate-freeness :) – Levi Nov 4 '19 at 0:50

Other people have given counterexamples, so I would like to demonstrate that counterexamples are somewhat rare. Here is an attempt at a conceptual explanation for both why this is not true in general and also when it is true. First, we note that if we have a path $$P(t)$$ in $$GL_n(\mathbb R)$$ with $$P(0)=P_0$$ and $$P(1)=P_1$$, then $$P(t)AP(t)^{-1}$$ gives us a path inside the conjugacy class of $$A$$. Since $$GL_n(\mathbb R)$$ has two path components (given by sign of determinant), this shows that the conjugacy class of $$A$$ is the union of two path connected subsets (conjugating by things with positive or negative determinant), and it will be path connected if these two subsets intersect.

If $$PAP^{-1}=QAQ^{-1}$$ where $$\det(P)>0$$ and $$\det(Q)<0$$, we have $$A=(P^{-1}Q)A(P^{-1}Q)^{-1}$$, so $$A$$ commutes with a matrix with negative determinant. The converse is also true, so we have the following result.

Lemma: The conjugacy class of $$A$$ is path connected if and only if $$A$$ commutes with some matrix with negative determinant.

This gives several conditions that would ensure conjugacy classes are path connected.

• If $$n$$ is odd, since then $$\det(-I_n)=-1$$
• If $$\det(A)<0$$
• If $$\mathbb R^n$$ as a direct sum of two $$A$$-invariant subspaces where one summand is odd dimensional, (e.g, if the Jordan normal form of $$A$$ has a block of odd size with a real eigenvalue).
• If $$\mathbb R^{n}$$ is the direct sum of two $$A$$-invariant subspaces, and the restriction of $$A$$ to either subspace has negative determinant.

One can check that Jordan blocks only commute with matrices that are upper triangular and have only a single eigenvalue, and if $$n$$ is even and that eigenvalue is real, such a matrix could not have negative determinant. So these give counterexamples.

I suspect one could give a nice characterization of all counterexamples in terms of JNF. However, I have not worked out the details.

No.

Here is a counterexample: Let $$n=2$$, $$A=\left( \begin{array} [c]{cc} 0 & -1\\ 1 & 0 \end{array} \right)$$ and $$B=-A=\left( \begin{array} [c]{cc} 0 & 1\\ -1 & 0 \end{array} \right)$$. Then, $$A$$ and $$B$$ are conjugate, since $$B=C^{-1}AC$$ for the invertible matrix $$C=\left( \begin{array} [c]{cc} 1 & 0\\ 0 & -1 \end{array} \right)$$. Thus, there exists a conjugacy class $$\mathcal{C}$$ in $$\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$ that contains both $$A$$ and $$B$$. However, there exists no path from $$A$$ to $$B$$ in $$\mathcal{C}$$. Why not?

Probably there is a nice conceptual reason [EDIT: yes, and @Wojowu explains it in his answer], but you can just as well brute-force it: I claim that every matrix in $$\mathcal{C}$$ has a nonzero $$\left( 1,2\right)$$-th entry. To see this, just notice that any arbitrary element of $$\mathcal{C}$$ has the form \begin{align} \left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) ^{-1}A\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) =\left( \begin{array} [c]{cc} -\dfrac{ab+cd}{ad-bc} & -\dfrac{b^{2}+d^{2}}{ad-bc}\\ \dfrac{a^{2}+c^{2}}{ad-bc} & \dfrac{ab+cd}{ad-bc} \end{array} \right) \end{align} for some $$\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) \in\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$, and thus its $$\left( 1,2\right)$$-th entry $$-\dfrac{b^{2}+d^{2}}{ad-bc}$$ is nonzero (because $$b^{2}+d^{2}$$ can only be $$0$$ if both $$b$$ and $$d$$ are $$0$$, but then $$\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right)$$ cannot belong to $$\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$).

Thus, every matrix in $$\mathcal{C}$$ has a nonzero $$\left( 1,2\right)$$-th entry. But if there was a path from $$A$$ to $$B$$ in $$\mathcal{C}$$, then some point on this path would be a matrix in $$\mathcal{C}$$ with zero $$\left( 1,2\right)$$-th entry (since the $$\left( 1,2\right)$$-th entries of $$A$$ and $$B$$ have opposite signs). Thus, there cannot be a path from $$A$$ to $$B$$ in $$\mathcal{C}$$.

• There is also a geometric way to see it if you're interested. – Arnaud Mortier Nov 4 '19 at 0:48
• @ArnaudMortier I'm interested. – Abhimanyu Pallavi Sudhir Nov 29 '19 at 13:22
• @AbhimanyuPallaviSudhir Thanks! You will find it in my answer which is currently the accepted one I think. – Arnaud Mortier Nov 29 '19 at 13:47
• Ah I missed that, thanks. – Abhimanyu Pallavi Sudhir Nov 29 '19 at 15:04

Notice that the assertion "since the image of the exponential map is the path-component of the identity in $$GLn(R)$$" is false.

Firstly, the path-component of the identity in $$GLn(R)$$ is the set of matrices with $$>0$$ determinant. Thus (inside this component) there is a path linking your $$P$$ and $$I$$ (in fact, there is nothing to prove).

Secondly, $$diag(-1,-2)$$ has $$>0$$ determinant and is not in the image of the real matrices by the exponential map.

EDIT.

Let $$SC(A)$$ be the conjugacy class of $$A\in M_n(\mathbb{R})$$.

*In dimension $$2$$, $$SC(A)$$ is non-connected in 2 cases

a. The eigenvalues of $$A$$ are non-zero conjugate complex; then $$SC(A)$$ is homeomorphic to a hyperboloid of two sheets.

b. $$A$$ is non-zero and non-diagonalizable; then $$SC(A)$$ is homeomorphic to a conical surface with the apex cut off.

*In dimension $$4$$, we consider the matrices $$A_1=diag(U,U)$$ where $$U=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ and $$A_2=diag(V,V)$$ where $$V=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$. Then, using the Aaron's test, we can prove that $$SC(A_1),SC(A_2)$$ are non-connected (it's easy for $$A_1$$ and more difficult for $$A_2$$).

*Assume that we randomly choose $$A\in M_n(\mathbb{R})$$ -the $$(a_{i,j})$$ follow iid normal laws- We deduce that follows

$$\textbf{Proposition}.$$ When $$n\rightarrow +\infty$$, the probability that $$SC(A)$$ is connected tends to $$1$$.

$$\textbf{Proof}$$. A random matrix $$A$$ has distinct complex eigenvalues with probabiity $$1$$. Then, up to a real change of basis, $$A=diag(a_1I_2+b_1V,\cdots,a_pI_2+b_pV,\lambda_1,\cdots,\lambda_q)$$, where $$2p+q=n$$, the $$(b_i)$$ are non-zero and the $$(\lambda_j)$$ are real distinct.

Thus $$SC(A)$$ is connected iff $$q\not=0$$ (Aaron's test). When $$n$$ tends to $$+\infty$$, the mean of the number of real zeroes of a polynomial of degree $$n$$ is in $$\Omega(\sqrt{n})$$; we can deduce that the probability that $$A$$ has, at least, one real eigenvalue tends to $$1$$ when $$n$$ tends to $$+\infty$$. $$\square$$

• Sorry for not reacting to this sooner. I have now finally managed to convince myself that what you say is true. Thanks for pointing it out! – Levi Nov 29 '19 at 13:09
• This statement also struck me when I first read it but then I focused on answering the question and forgot about it. Thanks for this. – Arnaud Mortier Nov 29 '19 at 13:47
• @Levi . More precisely, the real matrices that can be written in the form $e^Q$ are the squares of real matrices. – loup blanc Nov 29 '19 at 14:02
• @Arnaud Mortier . Thanks Arnaud. I wrote this because I was surprised that no one would react. – loup blanc Nov 29 '19 at 14:05


# Iterative derivatives of log

## Homework Statement:

Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =
87 178 291 200.

## Relevant Equations:

n=k
n=k+1
Derivative of lnx is 1/x.
I have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
BvU
Homework Helper
2019 Award
I don't know where to start
A little googling perhaps ?

PeroK
Homework Helper
Gold Member
Homework Statement: Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =
87 178 291 200.
Homework Equations: n=k
n=k+1
Derivative of lnx is 1/x.

I have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?
You can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?

FactChecker
Gold Member
The general outline of a basic inductive proof is in two steps:
1) Prove it is true for N=1
2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.

There is another variation where in step #2 you assume that it is true for N##\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.

A little googling perhaps ?
I tried to google it, but I still did not understand.

You can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?
Induction is one of the things I struggle with in math, that why it was not obvious to me. Thanks for the tips by the way.

The general outline of a basic inductive proof is in two steps:
1) Prove it is true for N=1
2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.

There is another variation where in step #2 you assume that it is true for N##\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.
Thank you for the tips and tricks. But how do I prove for n when I don't have a general formula? Do I have to make a general formula for the derivative of lnx?

BvU
Homework Helper
2019 Award
general formula for the derivative of lnx?
For the nth derivative

For the nth derivative
Yes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.

FactChecker
Gold Member
Do you know what the first derivative of ##ln(x)## and of ##x^{-n}## are? That should be all you need.

PS. When you are stuck by an intimidating problem, do what you can do. You might find out that you can do a lot more than you anticipated.

HallsofIvy
Homework Helper
The first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, etc.
Did you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is $$f^{(n)}(x)= 0=(-1)^{n+1}(n-1)! x^{-n}$$. It remains to use induction to prove that.
When n= 1, the derivative is $$\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.
Now assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$

Kolika28
PeroK
Homework Helper
Gold Member
Yes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.
If you want to find a derivative, you differentiate the function! If you want to find the second derivative, you differentiate the first derivative. And so on.

The first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, etc.
Did you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is $$f^{(n)}(x)= 0=(-1)^{n+1}(n-1)! x^{-n}$$. It remains to use induction to prove that.
When n= 1, the derivative is $$\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.
Now assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$
Thank you so much!!! I finally understand it. I really appreciate your help! :)


Probability of filling all urns

We have $K$ urns and $N$ balls. where $N\geq K$.

For each ball we uniformly select one of the urns and place the ball in it.

Question 1: What is the probability that all the urns will now have at least one ball in them? Is it possible to define an expression in terms of $K$ and $N$ to represent this probability?

I've implemented a simulation, here are some of the results for the following $K$-urns and $N$-Balls:

\begin{align*} Pr(2,3) &\sim 75.0\% \\ Pr(5,10) &\sim 52.2\% \\ Pr(10,20) &\sim 21.5\% \\ Pr(20,40) &\sim 3.6\% \\ Pr(50,100) &\sim 0.0177\%\\ \end{align*}

Question 2: Is there an expression that can describe this probability if we wanted a fill percentage.

eg: What is the probability that only 60% of urns have at least one ball in them?

eg: What is the probability that at least 75% of urns have at least one ball in them?

• Note: As an example, if we had 3 balls and 2 urns, there would be 8 ways we could place the balls in the urns, however there are only 6 ways in which there is at least one ball in each urn. – J Mkdjion May 13 '17 at 7:35

Question 1 can be done with inclusion-exclusion.

The probability of a specific urn being empty is $\big(1-\frac1K\big)^N$, because to avoid putting a ball in this urn, you have to choose one of the other urns at each step. Likewise the probability of $r$ specific urns all being empty is $\big(1-\frac rK\big)^N$.

Now the probability of at least one urn being empty is $$\binom K1\Big(1-\frac1K\Big)^N-\binom K2\Big(1-\frac2K\Big)^N+\cdots+(-1)^{r+1}\binom Kr\Big(1-\frac rK\Big)^N+\cdots+(-1)^{K}\binom K{K-1}\Big(1-\frac {K-1}K\Big)^N,$$ so to get the probability that no urns are empty, subtract this from $1$.

• I attempted to evaluate your expression, but am not able to get results which are similar to the ones obtained using a simulation. any ideas on what could be amiss? – J Mkdjion May 13 '17 at 19:21
• I don't know. What values of $n$ and $k$ did you try, and what did the simulation give? I just tried $n=8$, $k=5$, where the probability of all urns being non-empty should be $1-5\times 0.8^8+10\times 0.6^8-10\times 0.4^8+5\times 0.2^8\approx 0.323$. A quick simulation had this happening in $32113$ of $100000$ trials. – Especially Lime May 13 '17 at 21:58
• is the probability defined as the following correct? : $$1 - \sum_{i=1}^{K-1} \left( (-1)^{i+1} \begin{pmatrix} K \\ i \end{pmatrix} \left( 1 - \frac{i}{K} \right)^{N} \right)$$ – J Mkdjion May 14 '17 at 0:01
• Yes, that's correct – Especially Lime May 14 '17 at 7:46

Total number of ways to fill the urns without any restriction is $\binom{N+K-1}{K}$.

Question 1 is same as the number of tuples of nonnegative integer solutions to $$x_1+\cdots+x_K=N\\s.t.~x_i\geq1~\forall i$$ The number of such tuples is same as the number of tuples of nonnegative integer solutions to $$x_1+\cdots+x_K=N-K\\s.t.~x_i\geq0~\forall i$$ which is $$\binom{N-1}{K}$$ Thus the probability that all the urns will be non-empty is $$\frac{\binom{N-1}{K}}{\binom{N+K-1}{K}}$$

Question 2: Suppose more than fraction $p,~0\leq p\leq(1-1/K),$ of the urns contains at least one ball. Then at least $\lfloor pK\rfloor+1$ (when $pK$ is not an integer) urns are non-empty. It can be any $\lfloor pK\rfloor+1$ urns from the $K$ urns. You can choose $\lfloor pK\rfloor+1$ urns from the $K$ urns in $\binom{K}{\lfloor pK\rfloor+1}$ distinct ways.

Now the number of tuples of nonnegative integer solutions to $$x_1+\cdots+x_K=N-K\\s.t.~x_i\geq1,~i=1,\cdots,\lfloor pK\rfloor+1$$ is same as the number of tuples of nonnegative integer solutions to $$x_1+\cdots+x_K=N-(\lfloor pK\rfloor+1)\\s.t.~x_i\geq0~\forall i$$ which is $$\binom{N-(\lfloor pK\rfloor+1)+K-1}{K}$$

So the number of ways so that more than fraction $p$ of the urns are nonempty, is $$\binom{K}{\lfloor pK\rfloor+1}\binom{N-\lfloor pK\rfloor+K-1}{K}$$

If $pK$ is an integer then the solution is $$\binom{K}{pK}\binom{N- pK+K-1}{K}$$ Thus the probability that more than fraction $p$ of the $K$ urns will be non-empty is $$\frac{\binom{K}{\lfloor pK\rfloor+1}\binom{N-\lfloor pK\rfloor+K-1}{K}}{\binom{N+K-1}{K}}~\text{if}~ pK ~\text{is not an integer}$$ and $$\frac{\binom{K}{pK}\binom{N- pK+K-1}{K}}{\binom{N+K-1}{K}}~\text{if}~ pK ~\text{is an integer}$$

• I don't quiet see how your answers apply. In both questions a probability is required (a value between 0 and 1), also please take into account the scenario were all the balls are placed into one urn or the scenario where all the balls bar one are placed into the same one urn and the remaining ball is in another urn - how is that $\begin{pmatrix} N-1\\ k \end{pmatrix}$ – J Mkdjion May 13 '17 at 6:38
• These are the number of ways to fill the urns in the ways specified in the above questions. If you want the probability, just divide these numbers by the total number of ways to fill the urns which is $$\binom{N+K-1}{K}$$ – Abishanka Saha May 13 '17 at 6:40
• edited for your help – Abishanka Saha May 13 '17 at 6:46
• This is incorrect because the things you are counting are not equally likely. – Especially Lime May 13 '17 at 7:32
• If you randomly place two balls in two bins, there are $3$ possible arrangements: one in each, both in the first, or both in the second. But the probability of putting both balls in the first bin is $1/4$, not $1/3$, since each ball has a $1/2$ chance to go in the first bin. – Especially Lime May 13 '17 at 7:48


# The expected time to sort $n$ elements is bounded below

Prove that the expected time to sort $n$ elements is bounded below by $cn \log n$ for some constant $c$.

Could you give me some hints how I could do that?

• Is this for a specific algorithm? – Robert Israel Dec 8 '14 at 0:47
• @RobertIsrael No, it is not for a specific algorithm. It is for every comparison sort. – Mary Star Dec 8 '14 at 0:50

## 2 Answers

It's false as stated. For instance, radix sort is $O(n)$ (but see @mlo105's answer as well). But if you're talking about comparison-based sorting, it's true.

One proof (sketch): consider all possible inputs to the algorithm and draw an execution tree, where each internal node represents a comparison and its two out-edges are the two branches that could be taken as a result of that comparison, depending on the input data. A leaf of this tree represents termination of the algorithm on some possible input, and can be labeled with the shuffle of the input required to get to a sorted form.

Because there are $n!$ possible outputs (corresponding to the $n!$ ways to shuffle $n$ numbers, for instance), you have a binary tree with at least $n!$ leaves. It must therefore have depth at least $log_2 (n!)$, which is $O(n \log_2 n)$, by, say, Stirling's approximation.

Oh...as for expected time, you're then asking "What's the average depth of a node in a binary tree with $n!$ leaves?"

Well, I can't answer that, but I can show it's larger than the min depth $k = \lceil n \log_2 n \rceil$ of such a tree.

For any leaf $A$ whose depth is greater than $k$, take some leaf $L$ whose depth is LESS than $k$, replace it with a node whose left child is $L$ and whose right child is $A$. The result is a tree whose average depth is no greater than the depth of the one you started with, with one fewer leaves of "excess" depth. Repeat this process until all leaves are at depth no more than $k$. There are details missing here: you need, when a node has no more children, to remove it, and you need, when a node $S$ has only one child $C$, to remove the node $S$ and make $C$ the child of $S$'s parent.

But the idea is clear enough -- just move stuff up, without increasing the average path-length to the root. [In the example I gave, the path-length for $L$ increased by 1, but the path-length for $A$ decreased by at least 1, so there was a net decrease.]

• While this reasoning is sound, it deals with worst, not with the expected (= average) number of comparisons. – Peter Košinár Dec 8 '14 at 0:57
• Now it does. :) – John Hughes Dec 8 '14 at 0:58
• Stirling's approximation is a bigger hammer than you need to show that $\log_2(n!) > cn$. It suffices to observe that $n!$, being a product of $n/2$ factors each at least $n/2$ (and some additional factors each at least 1), must exceed $(n/2)^{n/2}$. – MJD Dec 8 '14 at 1:56
• Good point. I was trying to be terse, but was unnecessarily so. – John Hughes Dec 8 '14 at 3:58

Radix Sort is an interesting one. In practice, it performs in linear time $O(kn)$. In theory, it is possible to argue $k = log(n)$, where $n$ is the number of elements. Suppose we are considering an array of $n$ consecutive elements. Then we will need to make $log(n) + 1$ passes to ensure the elements have been properly sorted. If we consider $1, ..., 100$, we need $log_{10}(100)$ passes. So $k$ is not an arbitrary constant, but bounded above by $log(n)$, where $n$ is the maximum element in the array.

Binary sort is one that does perform in $\Theta(n)$. Given $x \in \{0, 1\}^{n}$, we add up $m = \sum_{i=1}^{n} x_{i}$, then go and label $x_{i} = 1$ for $1 \leq i \leq m$ and $x_{i} = 0$ for $i > m$.

We can write binary sort as a parallel algorithm using threshold functions: $x_{sort}(x) = (\tau_{1}(x), \tau_{2}(x), \tau_{3}(x), ..., \tau_{n}(x))$, where $\tau_{k}(x)$ denotes the threshold-k function (at least $k$ bits in $x$).

To show the result, I agree that Stirling's approximation isn't necessary. We can use the trick that $log(n!) \leq log(n^{n})$ for $n \in \mathbb{N}$. Then by rule of logs, $log(n^{n}) = n log(n)$.

• Nice point about radix-sort, etc. – John Hughes Dec 9 '14 at 0:07


Given a finite collection of disjoint subsets of $I$ must every ultrafilter on $I$ contain exactly one?

The title pretty much contains the question, but here's some elaboration:

The following is one of the first results one encounters while learning about Ultrafilters.

Fact: If $\mathfrak{U}$ is an ultrafilter on an index set $I$, and $X\subset I$, then exactly one of $X$ and $I\backslash X$ is in $\mathfrak{U}$.

Question: Can this be generalized to a collection $X_{1}, ... X_{n}$ of disjoint subsets of $I$ for any $n\geq 1$? That is, if $\mathfrak{U}$ is an ultrafilter on $I$, then there is exactly one value of $j\in\{1,...,n\}$ such that $X_{j}\in \mathfrak{U}$?

I haven't been able to find it in the literature anywhere or prove it myself (though I thought I did briefly) so I'm beginning to suspect it's false in general.

Second Question: If the answer to the first question is false, is it true if $\mathfrak{U}$ is an ultrafilter on $\mathbb{N}$ containing the order filter?

• Yes, and in fact one can define ultrafilters this way. See, for example, Proposition 1.5 in arxiv.org/abs/1209.3606. – Qiaochu Yuan May 25 '15 at 23:09
• @QiaochuYuan: You need the $X_i$ to be a partition. This is not what the OP asked (but it may be what they meant). – Michael Albanese May 25 '15 at 23:16
• A partition is what I intended to start with, but forgot to mention. Thanks to all for the great answers. – roo May 25 '15 at 23:59
• I was not sure whether to tag this (elementary-set-theory) or (set-theory). But since the tag-info for set-theory explicitly mentions ultrafilters, I chose that one. – Martin Sleziak Jul 11 '16 at 8:02
• – Martin Sleziak Jul 15 '16 at 16:01

If $X_1, \dots, X_n$ is a collection of disjoint sets with $\bigcup_{i=1}^nX_i = I$, then $X_k \in \mathfrak{U}$ for precisely one $k$.

Note that for each $k$, either $X_k \in \mathfrak{U}$ or $\bigcup_{i \neq k}X_i \in \mathfrak{U}$. If $X_k \not\in \mathfrak{U}$ for all $k$, then $\bigcup_{i\neq k}X_i \in \mathfrak{U}$ for every $k$, and so the intersection of such sets would also belong to $\mathfrak{U}$. This is a contradiction as the intersection is empty, i.e. $\bigcap_{k=1}^n\bigcup_{i\neq k}X_i = \emptyset$. Therefore, there exists $k \in \{1, \dots, n\}$ such that $X_k \in \mathfrak{U}$. If there were $l \in \{1, \dots, n\}$, $l \neq k$, with $X_l \in \mathfrak{U}$, then $X_l\cap X_k \in \mathfrak{U}$, but this intersection is empty, so no such $l$ exits. Therefore, $k$ is unique.

Note, you need the sets $X_i$ to cover $I$. If they don't cover $I$, choose $m \in I\setminus\bigcup_{i=1}^nX_i$ and consider the principal ultrafilter $\mathfrak{U}_m$. As $m \not\in X_i$ for all $i$, $X_i \not\in \mathfrak{U}_m$ for all $i$.

Let $X_1,X_2,\dots, X_n$ be a finite collection of sets (not necessarily pairwise disjoint) such that the union $X_1\cup X_2\cup \cdots\cup X_n$ is in the ultrafilter. Then at least one of the $X_i$ is in the ultrafilter. If the $X_i$ are pairwise disjoint, then exactly one of the $X_i$ is in the ultrafilter. The proof is straightforward, by induction.

Added: From comments, it became clear that the OP knew the standard proof that if $D$ is an ultrafilter on the index set $I$, then $X$ or $X^c$ is in $D$. The fact that if the union $X_1\cup X_2$ is in $D$, then $X_1$ or $X_2$ is in $D$ follows. For let $O$ be the "rest" of $I$. If $X_1$ is not in $D$, then its complement $X_2\cup O$ is in $D$. Also, we know that $X_1\cup X_2$ is in $D$. Now note that $X_2=(X_2\cup O)\cap(X_1\cup X_2)$, so $X_2\in D$.

Remark: I think it helps the intuition to think of an ultrafilter as defining a two-valued "measure" $\mu$ on the collection of subsets of $I$, where $\mu(X)=1$ if $X$ is in the ultrafilter, and $\mu(X)=0$ otherwise. The "measure" is almost always not a real measure, since it is ordinarily not countably additive. But it is finitely additive. The finite additivity makes the answer to your question clear. If the $X_i$ all had measure $0$, their union would have measure $0$.

• It is the intuition you describe that led me to guess that the result might be true. I could not come up with a proof, however. Thanks for your additional information! – roo May 26 '15 at 1:12
• To see how easy the induction is, let's do it for $3$ sets. Let $X_1\cup X_2\cup X_3$ be in the ultrafilter $D$. Then $(X_1\cup (X_2\cup X_3))\in D$. By the $n=2$ case, either $X_1\in D$, and we are finished, or $(X_2\cup X_3)\in D$, in which case again by the $n=2$ case we are finished. – André Nicolas May 26 '15 at 1:41
• I agree that the induction part of the proof is indeed easy. But you are using the following lemma, which I think is the tricky part: If $X_{1}\cup X_{2}\in \mathfrak{U}$, with $X_{1}\cap X_{2} = \phi$, then exactly one of $X_{1}$ or $X_{2}$ is in $\mathfrak{U}$. This is slightly stronger than the fact I mentioned, and cannot be proved in quite the same way (using a brief maximality argument). The lemma clearly follows immediately from Michael Albanese's argument above, but I do not see an easy way to get it directly. – roo May 26 '15 at 1:55
• That it cannot be both is obvious, for part of the usual definition of a filter $D$ is that $\emptyset\not\in D$. For the proof that at least one is, it depends on the definition of ultrafilter. If we define it as a maximal filter, one shows that if neither $A$ nor $A^c$ is in $D$, then $A$ can be added to $D$, along with all intersections of elements of $D$ with $A$, and their supersets, to form a larger filter $D'$. – André Nicolas May 26 '15 at 2:06
• Is it the part $O$ "outside" $X_1\cup X_2$ that bothers you? If $X_1$ is not in $D$, then its complement $X_2\cup O$ is in $D$, and by assumption $X_1\cup X_2$ is in $D$, so the intersection of $X_2\cup O$ and $X_1\cup X_2$ is in $D$, that is, $X_2$ is in $D$. – André Nicolas May 26 '15 at 3:54

It is true, granted the sets form a partition of $I$. Otherwise it is easy to come by counterexamples, simple consider a few singletons and a free ultrafilter.

The proof is simple by induction.


# Hexagonal rooks

Suppose you have a triangular chessboard of size $n$, whose "squares" are ordered triples $(x,y,z)$ of nonnegative integers that add up to $n$. A rook can move to any other point that agrees with it in one coordinate -- for example, if you are on $(3,1,4)$ then you can move to $(2,2,4)$ or to $(6,1,1)$, but not to $(4,3,1)$.

What is the maximum number of mutually non-attacking rooks that can be placed on this chessboard?

More generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?

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Can you prove what the maximal number of mutually non-attacking bishops is on an ordinary n×n chessboard? –  Zsbán Ambrus Jul 30 '12 at 21:52
The bishop question is easy - it's 2n-2. Put bishops along two opposite edges and remove two at adjacent corners. This is maximal since there are only 2n-1 different North-East diagonals, and two of these are opposite corners so only 2n-2 can be filled. –  Carl Jul 30 '12 at 22:27
The graph is regular, of degree $2n$. If you put a rook on $(n,0,0)$, the problem reduces to finding the maximum number of non-attacking rooks on the board of order $n-2$. I think the graph has enough symmetry that no matter where you put the first rook you reduce the problem to that of order $n-2$, but I'm not sure. If the graph does have that symmetry, then by induction you get more-or-less $n/2$ rooks ($(n+2)/2$ if $n$ is even, $(n+1)/2$ if $n$ is odd). –  Gerry Myerson Jul 30 '12 at 23:01
That's a nice idea, but I don't think it works. If you remove (2,1,0) from $G_3$, you get a path of length $3$, not a cycle. –  David Speyer Jul 30 '12 at 23:23
One of the coordinates must have an average value of no more than $n/3$ among all the rooks. The maximal number of distinct nonnegative integers whose avergae is $n/3$ is $2n/3+1$. This is a better bound. –  Will Sawin Jul 31 '12 at 1:00

And here's another one "Putting Dots in Triangles"

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Good references. The first paper (Nivasch-Lev, Mathematics Magazine 2005) ends up giving another proof of the $\lfloor 2n/3 \rfloor + 1$ bound, and the same construction to attain this bound. –  Noam D. Elkies Jul 31 '12 at 15:23
Cristi, thanks! I didn't know about either of these references. –  Jeremy Martin Jul 31 '12 at 19:35
You're welcome, Jeremy. –  Cristi Stoica Jul 31 '12 at 20:21

Nice question!

For the maximum number of pairwise non-defending rooks, Will Sawin proved an upper bound of $(2n/3) + 1$ in his comment to the original question. This bound is attained, at least to within $O(1)$, by two rows of $n/3 - O(1)$ rooks each, starting from around $(2n/3,n/3,0)$ and $(n/3,2n/3,1)$ and proceeding by steps of $(-1,-1,2)$ until reaching the $y=0$ or $x=0$ edge of the triangle. This construction generalizes Sawin's five-Rook placement for $n=6$.

On further thought, it seems we actually achieve $\lfloor (2n/3) + 1 \rfloor$ exactly for all $n$. Here's how it works for $n=12$ and $n=15$, with $(2n/3)+1 = 9$ and $11$ respectively:

                                            .
. .
. . .
.                            . . . .
. .                          . . . . .
. . .                        R . . . . .
. . . .                      . . . . . . R
R . . . .                    . R . . . . . .
. . . . . R                  . . . . . . . R .
. R . . . . .                . . R . . . . . . .
. . . . . . R .              . . . . . . . . R . .
. . R . . . . . .            . . . R . . . . . . . .
. . . . . . . R . .          . . . . . . . . . R . . .
. . . R . . . . . . .        . . . . R . . . . . . . . .
. . . . . . . . R . . .      . . . . . . . . . . R . . . .
. . . . R . . . . . . . .    . . . . . R . . . . . . . . . .

Starting from such a solution with $n=3m$, we can add an empty row to get an optimal solution for $n=3m+1$, and remove an edge (and the Rook it contains) to get an optimal solution for $n=3m-1$. So this should solve the problem for all $n$.

More generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?

I don't remember reading about this graph before. Experimentally (for $3 \leq n \leq 16$) its adjacency matrix has all eigenvalues integral, the smallest being $-3$ with huge multiplicity $n-1\choose 2$; more precisely:

Conjecture. For $n \geq 3$ the eigenvalues of the adjacency matrix are: a simple eigenvalue at the graph degree $2n$; a $n-1\choose 2$-fold eigenvalue at $-3$; and a triple eigenvalue at each integer $\lambda \in [-2,n-2]$, except that $\mu := \lfloor n/2 \rfloor - 2$ is omitted, and $\mu - (-1)^n$ has multiplicity only $2$.

This is probably not too hard to show. For example, the $\lambda = -3$ eigenvectors constitute the codimension-$3n$ space of functions whose sum over each of the $3(n+1)$ Rook lines vanishes. [Added later: in the comment Jeremy Martin reports that he and Jennifer Wagner already made and proved the same conjecture.]

Given that the minimal eigenvalue is $-3$, it follows by a standard argument in "spectral graph theory" that the maximal cocliques have size at most $3(n+1)(n+2)/(4n+6) = 3n/4 + O(1)$. But that's asymptotically worse than $2n/3 + O(1)$, though it's still good enough to prove the optimality of Will Sawin's cocliques of size $5$ for $n=6$ and of size $7$ for $n=9$.

Here's some gp code to play with this graph and its spectrum:

{
R(n)=
l = [];
for(a=0,n,for(b=0,n-a,l=concat(l,[[a,b,n-a-b]])));
matrix(#l,#l,i,j,vecmin(abs(l[i]-l[j]))==0) - 1
}

running "R($n$)" puts a list of the vertices in "l" and returns the adjacency matrix with the corresponding labeling. So for instance

matkerint(R(7)-2)~
matkerint(R(8)-1)~

returns matrices whose rows are nice generators of the $2$-dimensional eigenspaces of the $n=7$ and $n=8$ graphs.

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Noam, you've scooped me! :-) Jennifer Wagner and I have recently proved this conjecture by giving an explicit basis of eigenvectors. Writeup forthcoming. (Even more generally, we conjecture that the "simplicial rook graph" --- put a vertex at each lattice point in the $n$th dilate of the standard simplex in $\mathbb{R}^d$; edges are pairs of vertices at Hamming distance 2 -- has integer eigenvalues.) Corollary: the independence number of the triangular rook graph is at most $3(n+2)(n+1)/(2(2n+3))$. But, indeed, this bond is not tight. Back to the independence question: it appears tha –  Jeremy Martin Jul 31 '12 at 4:14
Seems like your comment was cut off by the 600-character limit. Anyway, it's not really a "scoop" since I only conjectured it (though I see that the lower bound of $-3$ on the spectrum is not hard, and likewise for your generalization to higher dimension). –  Noam D. Elkies Jul 31 '12 at 4:34
I was going to ask about the independence number about the higher-dimensional simplicial rook graph. The computational evidence I have suggests that the least eigenvalue is $\min(-n,-\binom{d}{2})$. E.g., for $d=4$, $n\geq 6$, this would imply that the independence number $\alpha(n)$ is at most $a(n)=\lfloor(n+1)(n+3)/3\rfloor$. This is not a tight bound (e.g., $a(6)=21$, $\alpha(6)=16$) and I would guess that it is not even asymptotically tight. –  Jeremy Martin Aug 1 '12 at 14:36
The eigenvalue bound (which you probably mean to be $\max(-n,-{d\choose 2})$, not $\min$) can be proved in the same way, by writing the adjacency matrix as the sum of $d\choose 2$ adjacency matrices (one for each direction) each with minimal eigenvalue $-1$. –  Noam D. Elkies Aug 1 '12 at 14:47

For n=6 you can fit 5 rooks

(0,2,4) (4,0,2) (1,4,1) (3,3,0) (2,1,3)

For n=9 you can fit 7 rooks

(0,3,6) (6,0,3) (2,6,1) (4,5,0) (3,1,5) (5,2,2) (1,4,4)

-
A visualization aid, $n=10$:


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# A polynomial of degree ≤ 2 ? what does this mean.

1. Apr 5, 2013

### mahrap

A polynomial of degree ≤ 2 ? what does this mean.

Would it just be

a + bt + c t^2 = f(t)

Or

at^2 + bt + c = f(t)

Is there even a difference between the two equations considering the fact that a,b, and c are unknown?

2. Apr 5, 2013

### Mentallic

There's no difference in whether you had

$$f(t)=at^2+bt+c$$

or

$$f(t)=ct^2+bt+a$$

or

$$f(t)=xt^2+yt+z$$

But the first is customary, and the last is using letters that usually denote variables as opposed to constants, so unless you have a good reason to otherwise deviate from the first, just stick with that.

Also, with any polynomial of degree n, the leading coefficient (coefficient of tn) must be non-zero, else the polynomial will no longer be degree n. Since you have a polynomial of degree $\leq$ 2, that means the leading coefficient of t2 does not have to be non-zero. You could even have all coefficients equal to 0 and thus simply have f(t)=0.

3. Apr 5, 2013

### mahrap

So what is the difference between a polynomial with degree = 2 and a polynomial with degree ≤ 2 or in general what is the difference between a polynomial with degree = 2 vs a polynomial with degree ≤ n ?

4. Apr 5, 2013

### Staff: Mentor

f(t) = at2 + bt + c, a 2nd-degree polynomial, also called a quadratic polynomial.
Degree ≤ 2 would also include 1st degree polynomials, such as g(t) = at + b, or zero-degree polynomials, such as h(t) = a.
I assume you mean degree = n vs. degree ≤ n. An nth degree polynomial has to have a term in which the variable has an exponent of n. A polynomial of degree ≤ n includes lower-degree polynomials.

5. Apr 5, 2013

### mahrap

Yes sorry for the typo. I meant to say degree = n. The reason I started this thread was with regards to a problem in my linear algebra class where the problem states:

Find all polynomials f(t) of degree ≤ 2 whose graphs run through the points (1,3) and (2,6) , such that f`(1) = 1 .

When I started to solve the problem I used the form f(t) = a + bt + ct^2 for my polynomial and after solving the matrices I got c = 2. However when I checked the solutions in the back of the book they had a = 2 which makes sense because they used f(t) = at^2 + bt + c . So what confused me was, how is one suppose to know which form to use to get the right a or c even though both essentially yield the same variables considering 2 is the value of the variable in front of the t^2 term. In general you mentioned a polynomial of degree ≤ n includes lower-degree polynomials. According to that statement, how would I know to use f(t) = at^2 + bt + c or f(t) = at + c when setting up my system of equations?

6. Apr 5, 2013

### Mentallic

Since the polynomial is of degree $\leq$ 2, that means the degree can at most be 2, so you should have $f(t)=at^2+bt+c$ which also ensures that even if the polynomial is just degree 1, then a=0. If you used $f(t)=at+b$ or $f(t)=a$ then you're assuming the equation must be of that form, and you'll soon find that there is no possible solution to the question if you begin with the assumption that the polynomials are of degree $\leq$ 1, or equivalently, $f(t)=at+b$.

7. Apr 6, 2013

### HallsofIvy

Staff Emeritus
If f(t) is a "polynomial of degree 2 or less" it can be written in the form $at^2+ bt+ c$ where a, b, and c can be any numbers.

If f(t) is a "polynomial of degree 2" it can be written in the form $at^2+ bt+ c$ where a, b, and c can be any numbers- except that a cannot be 0.


# Finding the definite integral of a function that contains an absolute value

The integral in question is this:

$\int_{-2\pi}^{2\pi}xe^{-|x|}$

My attempt:

Since there is a modulus, we split it up into cases. I'm not really sure which cases to split it into, do I just separately integrate these two functions?

$\int_{-2\pi}^{2\pi}xe^{-x}$

$\int_{-2\pi}^{2\pi}xe^{x}$

Or do I split it into these two? $\int_{0}^{2\pi}xe^{-x}$ $\int_{-2\pi}^{0}xe^{x}$

I am leaning towards the second split (splitting the bounds of the integral), which seems better.

The question is: What does it mean by 'splitting it into cases', and why does it work? Another side question I have is how to differentiate a function that has a modulus somewhere inside it.

• Let $f\colon [-2\pi, 2\pi]\to \mathbb R, x\mapsto xe^{-|x|}$. You should know by now that $\displaystyle \int \limits_{-2\pi}^{2\pi}f=\int \limits_{-2\pi}^{0}f+\int \limits_{0}^{2\pi}f$ and also $\forall x\in [-2\pi, 0]\left(f(x)=xe^{-(-x)}\right)$. – Git Gud Sep 15 '14 at 10:29
• @GitGud What if there is a modulus and the bounds of the integral are both positive? – robertmartin8 Sep 15 '14 at 10:30
• @ surelyyourjoking. Then $x$ would only ever be a positive value (or zero) and so the modulus can be replaced simply by $x$ alone since $\mid x\mid\geq 0$ for all $x$. – Pixel Sep 15 '14 at 10:50
• ah I see, so the modulus is meaningless in an indefinite integral? – robertmartin8 Sep 15 '14 at 11:24
• No, the modulus is not meaningless for indefinite integrals. When considering definite integrals as your question does we have bounds on the integral, which tells us the domain of the integrand to consider. Because we know the domain, we also know the sign (positive/negative) of a given $x$ value. Hence we can then split the integral into positive/negative parts to evaluate it. Notice also that an indefinite integral can be written as a definite integral since $$\int f(x)dx = \int_\lambda^x f(t)dt,$$ where the "lower bound" $\lambda$ gives a constant of integration. – Pixel Sep 15 '14 at 12:00

$$I=\int_{-2\pi}^{2\pi} xe^{-\mid x\mid}dx.$$

Loosely speaking, you can think about the definite integral as the area bounded by the function $xe^{-\mid x\mid}$ and the $x$-axis, as the variable $x$ moves from $x=-2\pi$ to $x=0$ then from $x=0$ through to $x=2\pi$. So, intuitively it's not too much of a step to see that $$I=\int_{-2\pi}^{0} xe^{-\mid x\mid}dx+\int_{0}^{2\pi} xe^{-\mid x\mid}dx.$$ Notice in the left integral the $x$ values are only ever negative or zero, and in the right integral the $x$ values are only ever positive or zero, so we can rewrite the whole expression $$I=\int_{-2\pi}^{0} xe^{x}dx+\int_{0}^{2\pi} xe^{-x}dx,$$ since $-|x|=x$ for $x\leq 0$ and $-|x|=-x$ for $x\geq 0$. You can now evaluate the integrals separately to obtain the correct result. Hope this helps.

$$\int_{-2\pi}^{2\pi}xe^{-\left|x\right|}dx=\int_{-2\pi}^{0}xe^{-\left|x\right|}dx+\int_{0}^{2\pi}xe^{-\left|x\right|}dx=\int_{-2\pi}^{0}xe^{x}dx+\int_{0}^{2\pi}xe^{-x}dx$$

This is a split of cases killing the annoying modulus.

• Alright, I understand it for this example now. But in general? when the bounds of the integral are both positive? – robertmartin8 Sep 15 '14 at 10:35
• @ surelyyourjoking. Then $x$ would only ever be a positive value (or zero) and so the modulus can be replaced simply by $x$ alone since $∣x∣≥0$ for all $x$. – Pixel Sep 15 '14 at 10:58

If you have a function $f: [a,b] \to \Bbb R$ defined by parts, as in: $$f(x) = \begin{cases} f_1(x), \mbox{if } a \leq x \leq c \\ f_2(x), \mbox{if } c < x \leq b\end{cases}$$ then: $$\int_a^b f(x) \ \mathrm{d}x = \int_a^c f_1(x) \ \mathrm{d}x + \int_c^bf_2(x) \ \mathrm{d}x.$$

In your case, $f(x) = xe^{-|x|}$, so $c = 0$ and $f_1(x) = xe^{x}$ and $f_2(x) = xe^{-x}$, using the definition of absolute value. Remember the interpretation of the integral for a positive function: the integral is the area, so the sum of the areas is the sum of integrals.

• so this works even when the bounds of the integral are both positive? – robertmartin8 Sep 15 '14 at 10:34
• Sure, why not? You have to pay attention where to split the integral. Try splitting $$\int_{-2}^3 |x - 1|e^x \ \mathrm{d}x$$ to see if you can do it (you don't need to solve it, that's not the point here) – Ivo Terek Sep 15 '14 at 10:36

$|x|=x$ for x>0

and = -x for x<0

and 0 for x=0

since x is positive and negative in the limits $-2\pi$ to $2\pi$ you have to split it into two parts and then evaluate your integral.

So in this function you are splitting into cases for |x| where x>0 and x<0.

this is done because |x| is defined that way in its domain.since you get two different functions for different intervals in the domain, you have to consider two different limits. imagine a function

f(x)=0 for x<0

and 1 for x>0

and we are evaluating an integral $\int{xf(x)}dx$ with limits -1 to 1.

so because of the definition of f(x) you have to split the integral into two.

here its just two integrals sometimes you have to split into many more.

• I understand that you have to split it up into cases, I just dont get technically how or why this is done – robertmartin8 Sep 15 '14 at 10:33


# Are the following sets exactly the same: $\mathbb{N}$, $\mathbb{N_{-\{0\}}}$, $\mathbb{N^+}$ and $\mathbb{Z^+}$?

At university, my Maths Analysis lecturer said that this is still debated nowadays. For example, a definition of an rational number is $\cfrac{p}{q}$ where $p\in \mathbb{Z}$ and $q\in \mathbb{N}$ (or should it be $q\in \mathbb{N^+}$ or the other two in the title?).

I know that $\mid \mathbb{N}|= \mid \mathbb{N^+}|$ since I can find a bijection by defining the map: $f: \mathbb{N^+} \mapsto \mathbb{N}$ by $f(x)=x-1$.

But the problem is that this raises the question as to whether $0 \in \mathbb{N}$ or $0 \in \mathbb{Z^+}$. This is similar to asking whether $0$ is positive?

So which set should I use to define the denominator $q$ of a rational number and why? The choices are $\mathbb{N}$, $\mathbb{N_{-\{0\}}}$, $\mathbb{N^+}$ and $\mathbb{Z^+}$.

I acknowledge that this question is subject somewhat to opinion, but I value the opinions of users of this site. So all I am asking is which set of the above would you choose $q$ to belong to for the definition of the rational number?

• In class-related work you should use what your lecturer uses. Elsewhere, you can use what you like and is most convenient for what you want to do. Maybe you work closely with other people that already have a standard. If the distinction is important for what you do, you should mention whether or not $0 \in \mathbb{N}$. Dec 8, 2015 at 23:21
• (1) We don't need a tag for "mappings" because those are in fact functions. (2) This question is about notation, and not at all about functions, so the tag (and the functions tags) is entirely irrelevant. Dec 10, 2015 at 10:38
• @AsafKaragila Okay, understood; thanks for letting me know. Dec 10, 2015 at 10:41

There are two different sets of relevance here; each has several notations in use.

The set $\{0,1,2,3,\ldots\}$ is considered fundamental in areas such as logic, set theory and computer science. It can be unambiguously written $\mathbb N_0$. Set theorists often refer to it as $\omega$, deftly avoiding the notational trouble surrounding the $\mathbb N$s.

The set $\{1,2,3,4,\ldots\}$ is often needed in most of the rest of mathematics, where people find it more natural to start counting at $1$. It can be unambigously written $\mathbb N_+$ or $\mathbb Z_+$ (the location of the plus sign can vary).

Either of these sets is often notated just $\mathbb N$ -- it is up to the reader who encounters this to know (or guess) which convention the author is following, in the cases where the difference matters. It is considered polite for an author to state which convention he follows before he uses the naked $\mathbb N$, but this is not always done in practice.

In English it is unambiguous that $0$ is not "positive". However other languages may follow other conventions; in French the number $0$ counts as both "positif" and "négatif" and one has to speak about "strictement positif" if one needs to exclude $0$. (This is not a mathematical difference; the concepts $>0$ and $\ge 0$ both exist independently of which words we use about them, and the two languages simply chose different concepts to have a short word for).

• Nice answer, thanks very much. So in your opinion, which of the four would you select, I know it's subject to opinion and therefore arbitrary, but could you tell me which one you would use? Dec 10, 2015 at 7:59
• @BLAZE: Personally? I just use $\mathbb N$ for whatever of the sets I'm speaking about (usually, but not always, the one that contains $0$) unless there seems to be a particular risk that the reader will miss my point if he misunderstands me. In that case I write $\mathbb N_0$ or $\mathbb N_+$. (Except forsuch set-theory contexts where it is conventional to use $\omega$). I'm not holding this up as a shining example to follow, though. Dec 10, 2015 at 15:44

It usually depends on the course and the lecturer.

$0 \in \Bbb N$ vs $0 \not \in \Bbb N$ depends on the definition of $\Bbb N$. The first one is usually found in most logic courses, while on say, analysis, this is not as common.

Whenever I see $\Bbb Z^+$ I'll think they refer to $\{1,2,3,...\}$, same for $\Bbb N^+$(this last one is quite uncommon).