If by 'a standard deck of cards' you mean a 52 card pack (no Jokers) with 4 suits each of 13 cards, the probability of picking a single card at random and it being a heart is 13/52 or 1/4 For a pack with jokers it would be 13/54 Ankit Pal B.E from University of Mumbai (Graduated 2020) 3 y Related. Explanations Question You draw a card at random from a standard deck of 52 cards. Find each of the following conditional probabilities: a) The card is a heart, given that it is red. b) The card is red, given that it is a heart. c) The card is an ace, given that it is red. d) The card is a queen, given that it is a face card. Explanation Verified. In a playing card there are 52 cards. Therefore the total number of possible outcomes = 52 (i) '2' of spades: Number of favourable outcomes i.e. '2' of spades is 1 out of 52 cards. Therefore, probability of getting '2' of spade Number of favorable outcomes P (A) = Total number of possible outcome = 1/52 (ii) a jack. But the coin has not changed - if it's a "fair" coin, the probability of getting tails is still 0.5. Dependent Events Two (or more) events are dependent if the outcome of one event affects the outcome of the other(s). Thus, one event "depends" on another, so they are dependent. Example I draw two cards from a deck of 52 cards.

mn

md

gv

ub

eg

What is the probability of ace card in a deck of 52 cards ? A card is drawn from a pack of 52 cards . The probability of getting an ace is 1/52. View complete answer on byjus.com. What do 4 Aces mean? If you hate your job, then four <b>Aces</b> can mean that something major will happen which will replace your income, allowing you to leave your job. For. Mar 19, 2020 · A deck of standard 52 cards contain four aces. There are four kings in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e. ⇒ . ⇒ . ⇒ . a king or a diamond ; A deck of standard 52 cards contain four kings. There are 13 Diamonds in a standard deck of playing cards. So ,we need to .... TO FIND : Probability of the following Total number of cards = 52 ( i ) Cards which are black king is 2 We know that PROBABILITY = = Number of favorable event T otal number of event Hence the probability of getting a black king is equal to 2/52=1/26 (ii) Total number of black cards is 26.. 2017. 6. 10. · A standard deck has 13 ordinal cards (Ace, 2-10, Jack, Queen, King) with one of each in each of four suits (Hearts, Diamonds, Spades, Clubs), for a total of #13xx4=52# cards. If we draw a card from a standard deck, there are 52 cards we might get. There are 16 cards that will satisfy the condition of picking a Jack, Queen, King, or Ace. This.

av

tp

lz

Mar 03, 2021 · A) 1 2 B) 1 52 C) 1 4 D) 1 13 7) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing a heart? A) 1 4 B) 1 2 C) 3 4 D) 1 8) In a survey of college students, 840 said that they have cheated on an exam and 1795 said that they have not..

jf

we

ud

oy

it

ov

Example 10 One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n(S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn.

aw

fg

be

jy

A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. One card is drawn from a standard deck of 52 cards. Find the probability of drawing a heart or a 6.There are 13 hearts and one 6 of hearts... 2. Which of the following must be a true statement?... Select one: a. The conditional probability P(A/B) is the probability that event B occurs, knowing A has occurred. b. An event and its complement can ....

yo

ba

wv

st

Transcribed image text: A single card is drawn from a standard 52-card deck. Let H be the event that the card drawn is a heart, and let F be the event that the card drawn is a face card. Find the indicated probability. P (H'UF) P(H'UF) = 0 (Type an integer or a simplified fraction.) A single card is drawn from a standard 52-card deck. one less card in the deck because we already had to draw the Heart from the deck. Thus: P(Heart and Club) = P (Heart) * P (Club) = 13/52 * 13/51 = .25 * .255 = .064 We might also have to subtract a value from the numerator as well as the denominator. Try to find the probability of drawing three red cards from a deck without replacement. Question 591370: A single card is drawn from a standard deck of 52 cards. Find the probability the card is: 1. A red four 2. A heart 3. A 4 or a heart. 4. Not a club. 5. A red or a four 6. A red and a 3 Answer by Edwin McCravy(19211) (Show Source):. The probability of drawing a queen, from a 52 card deck, is 4/52 or 1/13. Wiki User. ∙ 2009-10-09 18:06:35. This answer is:. Answer: The probability of drawing a card from a standard deck and choosing a king or an ace is (1/13) × (4/51). What is the probability of drawing a queen of hearts from a deck of 52 cards pinia store.

jq

ih

vx

Mar 06, 2011 · Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the.... 2018. 6. 12. · In a standard deck of cards, there are 52 cards. They are broken down into suits (4 of them: Spades, Hearts, Diamonds, Clubs) of 13 cards each. Each suit has 13 ordinal cards (A, 2 through 10, Jack, Queen, King). Here we're asked to draw a card at random and find the probability of drawing either a diamond or a 7. Mar 06, 2011 · Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the....

zc

gd

hd

qd

If one card is drawn from a standard 52 card playing deck, determine the probability of getting a ten, a king or a diamond. Round to the nearest hundredth. # of ways to succeed: 4 + 4 + 13 - 1 -1 = 19. Correct option is D) Probability of drawing a king = 524= 131 After drawing one card, the number of cards are 51. Probability of drawing aqueen = 514. Now, the probability of drawing a king and queen consecutively is 131× 514= 6634 Was this answer helpful? 0 0 Similar questions Three cards are drawn with replacement from a part of 52 cards. Calculate the probability of being dealt a diamond from a standard deck of 52 cards. Since there are 4 suits in a deck of cards (hearts, clubs, spades and diamonds) we can find the number of. Transcript. Example 10 One card is drawn from a well shuffled deck of 52 cards.If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n (S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn So, n (A) = 13 Probability of A = P (A.One card is selected from a.

ai

ry

oi

hf

Mar 12, 2011 · The probability of drawing a club or a nine from a 52 card deck of standard playing cards is 16 / 52 or approximately 30.8%. There are 13 clubs in a standard deck of cards. There are four nines in.... The royal flush is a case of the straight flush. It can be formed 4 ways (one for each suit), giving it a probability of 0.000154% and odds of 649,739 : 1. When ace-low straights and ace-low straight flushes are not counted, the probabilities of each are reduced: straights and straight flushes each become 9/10 as common as they otherwise would be.

tv

xs

nc

pm

winnetka rummage sale 2022; the end netflix; craigslist gmc trucks for sale by owner near Florianpolis State of Santa Catarina; meta computer vision engineer salary; restaurants open till 2am near Panruti Tamil Nadu; Google Algorithm Updates. The total number of 8 card hands is 52c8, so the probability of choosing a hand which excludes at least one suit is (4 * 39c8 - 6 * 26c8 + 4 * 13c8) / (52c8). You wanted the probability that a hand includes every suit which is the opposite of choosing a hand excluding at least one suit, and therefore the probability that you wanted is. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at. Let Event B = drawing a red card. P (A) = 4/52 since there are four aces in each deck of 52 cards. P (B) = 1/2 = 26/52 since there are four suits and two of them are red (or 26 red cards in a deck of 52) P (A∩B) = the probability of drawing a red ace = 2/52 since there are 2 red aces in a deck of 52 cards).

ha

fb

wo

ln

Transcript. Example 10 One card is drawn from a well shuffled deck of 52 cards.If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n (S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn So, n (A) = 13 Probability of A = P (A.One card is selected from a. A standard deck of cards has: 52 Cards in 13 values and 4 suits ... If you draw 3 cards from a deck one at a time what is the probability: ... what is the probability: You draw a Club, a Heart and a Diamond (in that order). The hypergeometric MTG calculator can describe the likelihood of any number of successes when drawing from a deck of Magic cards. It takes into account the fact that each draw decreases the size of your library by one, and therefore the probability of success changes on each draw. Population Size. Cards in your deck / library you are drawing from.

ps

ed

mb

ry

Number of favourable outcomes = 48 (Because There are always 52 cards in one deck containing 13 cards of varying values but always 4 suits, if the drawn card is not 4 then number of favourable outcomes will be 48 (12*4).) Total number of favourable outcomes = 52 (Number of cards in a deck is 52) So, the probability will be = 48/ 52 = 12/13. Oct 26, 2020 · Answer: 1/13. Step-by-step explanation: If there are 4 suits in a deck of cards, there are only 4 aces inside a complete deck. That means the probability of drawing an ace is 4/52 and just simply simplify the answer to 1/13.. A SINGLE CARD IS DRAWN AT RANDOM FROM A STANDARD DECK OF 52 CARDS. FIND THE PROBABILITY OF DRAWING THE FOLLOWING CARDS. PLEASE REDUCE TO LOWEST TERMS. A) A DIAMOND OR A 5 __________ B) A HEART AND A JACK __________ C) A JACK OR AN 8 __________ D) A HEART OR A SPADE __________ E) A RED AND FACE CARD __________ F) A RED CARD OR A QUEEN __________ 2. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at. Suppose you draw five cards from a standard deck of 52 playing cards, and you want to calculate the probability that all five cards are hearts. ... For example, the probability of drawing five cards of any one suit is the sum of four equal probabilities, and four times as likely. In boolean language, if the events are related by a logical OR. Whenever you do probability problems, check to see if you are being asked to find probability of one thing OR another, or of multiple events happening together. For instance, if this problem asked you to find the probability of drawing a heart AND a 5, well, there is only one 5 of hearts in a deck. So that would be 1/52.

mr

hf

oz

This can be simplified into 4/13 or a 30.77% probability of drawing a 4 or a spade from a standard deck of cards. Read More:. After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is 3 51 = 1 17 3 51 = 1 17.. A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. a card is drawn from a standard deck what is the probability that the card is an ace If you draw one card from a standard deck, what is the probability of drawing a 5 ... a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and Tìm kiếm.

zc

ae

jh

2022. 2. 7. · We know that a well-shuffled deck has 52 cards. Total number of black cards = 26. Total number of red cards = 26. Therefore probability of getting a black card= {total number. 5) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing an ace? A) 1 13 B) 1 52 C) 1 4 D) 1 2 6) If one 39,068 satisfied customers 2,852 satisfied customers Ph.D. 39,068 satisfied customers the probability for the experiment of drawing a card at random from a standard deck of 52 playing cards. What is the probability of drawing a black checker from a bag filled with 6 black checkers and 4 red checkers, replacing it, and drawing another black checker? c. Getting a Club and a Heart . 2. Drawing a card from a deck and not replacing it. 1.. So, there are 12 face cards in the deck of 52 playing cards. Worked-out problems on Playing cards probability: 1. A card is drawn from a. A card is selected from a deck of 52 playing cards. Find the probability of selecting · a prime number under 10 given the card is a heart. (1 is not prime.) · a diamond or heart given the card is red. read ....

fd

ss

lq

A card is randomly drawn from a standard 52-card deck. Find the probability of the given event. (A face card is a king, queen, or jack). Drawing a queen and a heart (this is an intersection question). A card is drawn randomly from a standard 52-card deck.Find the probability of the given event. (a) The card drawn is 6 The probability is:(b) The card drawn is a face card (Jack, Queen, or King) The read more.Solution Total number of possible outcomes = 52 P (E) = (Number of favourable outcomes/ Total number of outcomes) (i) Total numbers of the king of. Since we know that in a deck of 52 cards, there are a total 12 face cards (3 face cards each of heart, diamond, spade and club). Number of face cards$= 12$ Therefore, probability of getting a face card$= \dfrac { { {\text {Number of face cards}}}} { { {\text {Total number of cards}}}} = \dfrac { {12}} { {52}} = \dfrac {3} { {13}}$. Jun 21, 2020 · If one card is drawn from a standard 52-card playing deck, find the probability of getting a king, or a ten, or a heart, or a club.When entering your answer include a leading zero and round to the nearest hundredth. An example of an acceptable answer would be 0.19. Question 591370: A single card is drawn from a standard deck of 52 cards. Find the probability the card is: 1. A red four 2. A heart 3. A 4 or a heart. 4. Not a club. 5. A red or a four 6. A red and a 3 Answer by Edwin McCravy(19211) (Show Source):. Jun 21, 2020 · If one card is drawn from a standard 52-card playing deck, find the probability of getting a king, or a ten, or a heart, or a club.When entering your answer include a leading zero and round to the nearest hundredth. An example of an acceptable answer would be 0.19.

ud

lz

gq

yp

as

If you draw one card from a standard deck, what is the probability of ... a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and Tìm kiếm. Tìm ... If you draw one card from a standard deck, what is the probability of drawing a 5 or.

gq

du

aa

The total number of cards in a deck is 52 Probability = No. of favorable outcomes / Total no. of outcomes. Now, the probability of cards in a deck is 13/ 52 . Understand different concepts and get good grip on them by using online tools available at.

ex

qs

qb

fx

A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. Apr 26, 2017 · Ace is not odd The ordinals 3, 5, 7, and 9 are odd. There are four of each (one for each suit) and so 4xx4=16 odd cards. This makes the probability: P("draw an odd card")=16/52=4/13 Ace is odd If we want to consider the Ace as a 1, then there are 5 ordinals that are odd, 5xx4=20 odd cards, and therefore: P("draw an odd card")=20/52=5/13.

hz

ui

ux

yx

There are 5 2 cards in a standard deck: 1 3 ordinal cards (Ace - 1 0, Jack, Queen, King) and 4 of them - one to each suit (hearts, diamonds, clubs, spades) and so we have 4 × 1 3 = 5 2.. Q. From a pack of 52 playing cards jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is (i) a black queen (ii) a red card (iii) a black jack (iv) a picture card (Jacks, queens and kings are picture cards). Step 2. ∵ number of non-face card in well shuffled deck of 52 playing card = 52-12. = 40. Step 3. Probability ( a non-face card ) = number of favourable outcomes total number of outcomes. = 40 52. = 10 13. Step 4. (ii) Number of black king in well shuffled deck of 52 playing cards = 2. Mar 06, 2011 · Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the....

gw

ch

eb

The goal is to win at least three tricks.. Example 3: Two cards are drawn without replacement in succession from a well-shuffled deck of 52 playing cards. What is the probability that the second card drawn is an ace, given that the first card drawn was an ace? Example 4: One thousand high school seniors were surveyed about whether they planned .... Problem 2: A random card is chosen from the standard deck of cards, find the probability of obtaining a queen or a heart.. So, the probability of getting a Queen card is 1/13. Example 2: A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a card of Heart. Solution: Let A represents the event of getting a Heart .... A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. A standard deck of cards has: 52 Cards in 13 values and 4 suits ... If you draw 3 cards from a deck one at a time what is the probability: ... what is the probability: You draw a Club, a Heart and a Diamond (in that order).

ak

ne

gf

Number of Kings in a deck = 4. Probability of drawing a card from a standard deck and choosing a king or an ace = Probability of getting an Ace + Probability of getting a King. Probability of drawing an Ace at random = 4/52 = 1/13. Now, the probability of drawing a King at random = 4/52 = 1/13. hence, the required probability = 1/13 + 1/13 = 2/13. What is the probability of drawing a heart from a standard deck of cards on a second draw? 2 Answers By Expert Tutors The probability of choosing a heart, P(Heart) = 13/52 = 0.25. What is the probability of getting a heart or an even number? Clearly, the probability of drawing a heart out of the deck is 13/52, or 1/4.. "/>.

ci

no

nz

nl

bv

Therefore, the proability of drawing a king or 3 is 8/52=2/13. For the second problem, think binomial probability. The probability of drawing a king is 1/13. The opposite of at least 1 is none. So, the probability of getting no kings is 48/52=12/13. We find the probability of getting no kings and subtract from 1. Also, You should get the same.

pk

lf

fz

ml

xk

mc

A card is drawn from a standard deck of 52 playing cards. A: The result is a club. B: The result is a king. 36) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a king. 37) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a black card. We can draw one card from pack of 52 cards in 52C1 = 52 ways n (S)=52 There is only one card in the pack of 52 cards which is jack as well as heart card,.It is jack of hearts n (E)=1 DIY: How to take years off your neck's appearance. Ray Phan Principal Software Engineer at Magic Leap (company) (2021-present) 4 y Related. Jan 13, 2017 · There is a 50% chance that the card drawn will be red. Red cards make up 50% of the deck. 26/52 = 1/2 Therefore, if you're drawing one card, there is a 50/50 chance that the card will be red.. Problem 2: A random card is chosen from the standard deck of cards, find the probability of obtaining a queen or a heart.. So, the probability of getting a Queen card is 1/13. Example 2: A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a card of Heart. Solution: Let A represents the event of getting a Heart .... 5) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing an ace? A) 1 13 B) 1 52 C) 1 4 D) 1 2 6) If one 39,068 satisfied customers 2,852 satisfied customers Ph.D. 39,068 satisfied customers the probability for the experiment of drawing a card at random from a standard deck of 52 playing cards. Correct option is D) Probability of drawing a king = 524= 131 After drawing one card, the number of cards are 51. Probability of drawing aqueen = 514. Now, the probability of drawing a king and queen consecutively is 131× 514= 6634 Was this answer helpful? 0 0 Similar questions Three cards are drawn with replacement from a part of 52 cards. In this task, we need to calculate the probability of getting at least one black card. The given is that we draw 2 2 2 cards from a 52 52 52-card deck with replacement.. The deck is the standard deck with 4 4 4 suits, clubs, spades, hearts, and diamonds, where spades and clubs are black suits.

ao

gh

ym

If you were investigating red cards, kings or the queen of hearts, the odds of randomly drawing one of these from a complete deck are 50 percent (26 in 52); about 7.7 percent (four in 52); or. One card is randomly drawn from a deck of 52 playing cards. Find the probability thati the drawn card is red.ii the drawn card is an ace.iii the drawn card is red and a king.iv the drawn card is red or a king.[4 MARKS]. A card is drawn from a standard deck of 52 playing cards. A: The result is a club. B: The result is a king. 36) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a king. 37) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a black card. Setup Card content One card per line. Supports basic HTML. Go!. You will come across many video chatting applications, but a few of the best ones that are very effective and perfect for a virtual card room are Zoom and Skype. With Zoom, you can add around 100 people; however, the session will only last for 45minutes, and you will have to start.

ja

qj

bx

aw

Jan 13, 2017 · There is a 50% chance that the card drawn will be red. Red cards make up 50% of the deck. 26/52 = 1/2 Therefore, if you're drawing one card, there is a 50/50 chance that the card will be red.. Playing cards probability problems based on a well-shuffled deck of 52 cards. Basic concept on drawing a card: In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades ♠ hearts ♥, diamonds ♦,. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at random from a deck of shuffled cards. math. A card is drawn from an ordinary deck of 52 cards, and the result is recorded on paper.

pq

iu

bw

2010. 6. 4. · There are 6 red face cards in a standard deck of 52 cards; the Jack, Queen, and King of Hearts and Diamonds. The probability, then, of drawing a red face card from a standard deck of 52 cards is 6 in 52, or 3 in 26, or about 0.1154. (A standard deck of cards is the most common type of deck used in most card games containing 52 cards). Determine the probability of having 1 girl and 3 boys in a 4-child family assuming boys and girls are equally likely. The probability of having 1 girl and 3 boys is 1/4. Use the theoretical method. Probability gives the chances of how likely ....

bh


# The Convergence of a Telescoping Series

The question I've been posed is:

Show that the following series converges, and compute its value

$$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$

From this I decided to use partial fractions to put into the form:

$$\frac{1}{2}\cdot\left(\frac{1}{k}-\frac{1}{k+2}\right)$$

And from this I noticed that this is in the form of a telescoping series which I think would cancel down to:

$$\frac12\cdot\left(1+\frac12\right)= \frac{3}{4}$$

So I've got to this point, but I don't think what I've worked out is substantial enough to prove what I've been asked.

Would anyone mind giving any tips to make my working more thorough.

• Concerning the convergence alone you can use the fact that $$\frac1{k(k+2)}<\frac1{k^2}$$ – mrtaurho Feb 27 '19 at 21:30

Note that$$\frac1k-\frac1{k+2}=\left(\frac1k-\frac1{k+1}\right)+\left(\frac1{k+1}-\frac1{k+2}\right).$$This will give you two telescoping series. Can you take it form here?

• Sorry but I don't quite understand where I should go from there, would you mind giving another tip? – king Feb 28 '19 at 17:05
• \begin{align}\sum_{n=1}^\infty\frac1k-\frac1{k+2}&=\sum_{k=0}^\infty\frac1k-\frac1{k+1}+\sum_{n=1}^\infty\frac1{k+1}-\frac1{k+2}\\&=1-\lim_{n\to\infty}\frac1{k+1}+\frac12-\lim_{k\to\infty}\frac1{k+2}\\&=\frac32.\end{align} – José Carlos Santos Feb 28 '19 at 17:33

Let's write, as you have done, the following :

$$S_n =\sum_{k=1}^n \frac{1}{k(k+2)} = \frac{1}{2} \sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+2} \right) = \frac{1}{2} \left(\sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n\frac{1}{k+2} \right) = \frac{1}{2} \left(\sum_{k=1}^n \frac{1}{k} - \sum_{k=3}^{n+2}\frac{1}{k} \right)$$

So you see that $$S_n = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right)$$

Now this is obvious that the limit of $$S_n$$ is equal to $$\frac{3}{4}$$, i.e.

$$\sum_{k=1}^{+\infty} \frac{1}{k(k+2)} = \frac{3}{4}$$


# Find how many solutions has the following equation…

Determine how many real solutions has the following equation:

$$x^2(|x|-6)=-15$$

I noticed that $|x|-6$ should be negative because $x^2$ is always a positive value. Thus, $x\in(-6;6)$. I made a substitution: $|x|=t$, hence $x^2=t^2$, and got the equation:$$t^3-6t^2+15=0$$

Now the problem is I cannot find any solution for this equation. Hope you'll give me the right explanation for this exercise. Thank you very much!

-
try to study the variations of the function you have in the last equation. – Denis Apr 24 '14 at 10:20
Thank you, but what do you mean by this? – John G. Apr 24 '14 at 10:20
You don't need to find the solutions. Only their number. – evil999man Apr 24 '14 at 10:21
compute the derivative, and see where the derivative is zero, it will tell you when this function reaches minimum and maximum. – Denis Apr 24 '14 at 10:21
So using the Descarte's rule I see there are two variations of signs for $P(t)=t^3-6t^2+15$. So does this mean I have two positive solution or none? And if so then what should I do? I think that if there are two positive solutions for the equation above this means we have four solutions in general when substituting in $|x|=t$. – John G. Apr 24 '14 at 10:47

One way to answer this question is to draw a graph of $y = x^2(|x| - 6)$, then draw a horizontal line at $y = -15$. The curve and the line obviously intersect in four places. QED.

Finding the solutions is a bit trickier, but that is not the question that was asked.

If you don't want to rely on drawing the graph, you can prove the result using $f(t)$ and $f'(t)$. As you showed, $f(t) = t^3 - 6t^2 + 15$, which gives $f'(t) = 3t^2 - 12t$.

Solving for $f'(t) = 0$ yields $t = \{0, 4\}.$ Thus, $f(t)$ has extrema at $t =0$ and $t = 4$. We only care about $t>0$, so we ignore that one. Calculating $f'(1) = -9$ and $f'(5) = 15$, we see that $f(t)$ must be strictly decreasing for $0 < t < 4$ and strictly increasing for $t > 4$.

Some key evaluations: $$f(0) = 15$$ $$f(4) = -17$$ $$f(6) = 15$$

Since $f(t)$ is strictly decreasing between $0$ and $4$, and $f(0) > 0$ and $f(4) < 0$, $f(t)$ must cross zero exactly once in that region. Likewise, since $f(t)$ is strictly increasing for $t > 4$ and $f(4) < 0$ and $f(6) > 0$, $f(t)$ must cross zero exactly once in that region.

Thus, there are two positive solutions to the initial equation. By symmetry, there must be two negative solutions, as well.

-


# How to play a betting game

I have been interviewing in a few trading firms recently. I came up with the following question myself, but it is similar to some of the questions they ask and ways of thinking they expect.

Suppose you have some capital to invest (for example \$100). You can play a game where you bet \$x of your money and with probability $$\frac{2}{3}$$ your bet is doubled (so now you have \$(100 + x)) and with probability $$\frac{1}{3}$$ you lose your bet (so now you have \$(100 - x)). How much should you bet on this game.

I can see two ways of thinking about this problem. Firstly, there is the expected value maximisation approach. It can be easily seen that your expected gain in this game is $$\frac{x}{3}$$. So in order to maximise EV you should bet all of your money instantly. And if you were to play this game a million times, you should bet all of your money each time.

Of course this approach has the obvious flaw that when you play a few times, you will almost certainly go bankrupt. So we decide not to maximise EV and instead first make sure that we never go bankrupt. We do it by deciding to, at each point of the game, always bet exactly the same proportion of our money, say $$p$$. Then after $$n=n_1+n_2$$ games, where our bet was doubled $$n_1$$ times and we lost $$n_2$$ times, we will have $$M \cdot (1+p)^{n_1} \cdot (1-p)^{n_2}$$ money, where $$M$$ was our initial amount. Differentiating the log of this with respect to $$p$$ we can see that this function has its maximum for $$p=\frac{n_1-n_2}{n} \rightarrow \frac{2}{3}-\frac{1}{3} = \frac{1}{3}$$ as $$n \rightarrow \infty$$. So if we bet just a third of our money every time, we are (almost) guaranteed not to go bankrupt and, out of the strategies that bet a constant proportion every time, this one maximises our gain in the most likely outcome.

So here is my question - does this second strategy make sense to you? If you were to play this game with your own money would you use it? Does it make any sense to use a different strategy if you only play once, and not many times? I personally would be tempted to bet more than a third if I only got one chance, because it would increase my EV even if it is potentially bad in the long term. Does this sentiment make any sense?

Also, I just described two ways of thinking about the game above. Do you know of any other ways to think about it? Other strategies?

Your approach is remarkably on point. This issue is generally discussed in terms of portfolio construction in finance and depending on risk tolerance we define different functions to optimize. You choose to maximize the log of the expected payoff after n games, which is the same as Kelly criteria. For further and more detailed discussion of it you can check https://en.wikipedia.org/wiki/Kelly_criterion

• Correct me if I am wrong here: I don't think I am maximising the log of the expected payoff - I think I am maximising the log of the most likely payoff. The expected payoff is still the largest if I bet everything I have every time, right? Also, taking the log shouldn't change anything in terms of maximisation, because log is strictly increasing. So I am really maximising the most probable payoff, no? – user132290 Dec 11 '18 at 10:25
• Well, in effect you are basically maximizing the log of expected payoff. So you first wrote the payoff equation given $p, n_1, n_2$. While finding the most probable payoff, you plug $\frac 2 3$ for $\frac {n_1} n$, which is in practice taking the expected value of $n_1$ and $n_2$ with respect to $n$, hence taking the expected value of payoff. – Ofya Dec 11 '18 at 15:19

Your intuition is generally captured in a "utility function" which is supposed to describe how happy you are to have a certain amount of money. You then maximize the expected value of this function. Your thought to maximize the log of the amount of money you have is one utility function and not an unreasonable one. There is good psychological evidence that more money makes people happier, but much more slowly once they have some. The log function is in this vein, but there are many others as well. Once you define the function, the maximization process is the one you have used.

I would suggest that no simply described function can capture utility properly. If you were allowed to play the game fifty times but had to bet one dollar each time, you would probably play. You would probably win about $$\25$$ and be happy about it, but it wouldn't really change your life. As the bet rises the impact on your life does too. At the start it is only good because the chance you lose is almost zero so more is better. Eventually it may get to the point that you become risk averse. If your income is large compared to your cash assets it may make sense to bet everything you have because you can replace it easily. If you are living on assets you may become risk averse at a small fraction of your assets. This is all supposed to be captured in the utility function, which indicates why a simple answer like log is not appropriate for real life.

• Correct me if I am wrong please, but you seem to be treating the log as if it was meaningful in my example (because it is the utility function of my wealth - so the second million I make will make me less happy than the first million). But I am just taking the log here for simpler differentiation and to maximise the log is the same as to maximise its argument. So the log should not be relevant at all here, no? – user132290 Dec 11 '18 at 10:28
• Yes, I am treating the log as meaningful. If your utility is linear in money you should bet all your money each time. I answered this here. Yes, you will almost surely be broke, but the tiny chance of winning a huge amount of money overwhelms that. The utility function should apply to all the money you have, not just to what you win from the game. The log is a nice function because it is increasing but slower and slower as the argument gets larger, which we want utility todo – Ross Millikan Dec 11 '18 at 15:19
• Changing the log to a different function will shift the amount you should play with. – Ross Millikan Dec 11 '18 at 15:20

Straight out of my files:

For a P% profit level and D odds to 1:

RequiredPercentageCorrectBets = (P + 100) / (D + 1) .

This posting is simply viewpoint of pari-mutuel wagering where the probability is derived from the wagering.

In other words, in pari-mutuel wagering all the information that there is, is supposed to be represented by the wagering on the tote board. So in pari-mutuel wagering, odds of 1.00 represents a probability of 50%. Then the Kelly Criteria calls that situation a no-bet. However, some bettors do wager profitably and so I suppose that there is a personal probability of winning that can be applied to calculating the percentage of the stake to bet that maximizes profit. However, a personal probability of winning would still tend to go up with lower odds and tend to go down with higher odds.

In roulette or keno I suppose that the gambler can keep track of a number not coming up and then expect an increasing probability of the number coming up. That situation assumes a faith of an honest game instead of a discovery of a dishonest game.

I previously suggested a one-number Keno game that pays 3 to 1 for odds of 2 to 1. Twenty numbers are drawn out of 80 for a probability of 25%. But the probability of hitting the number in two draws is 50%, the probability of hitting the number in three draws is 75%, and the probability of hitting the number in four draws is 100%. So the idea is to decide how much to increase the wager each time the number doesn't come up and keep playing the number until it does come up.


GMAT Changed on April 16th - Read about the latest changes here

 It is currently 23 Apr 2018, 12:27

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Arithmetic Progression, number of terms, sum of terms etc

Author Message
TAGS:

Hide Tags

Senior Manager
Joined: 09 Mar 2016
Posts: 442
Arithmetic Progression, number of terms, sum of terms etc [#permalink]

Show Tags

24 Mar 2018, 12:57
1
KUDOS
1
This post was
BOOKMARKED
Greetings friends l

i decided to create a list of all formulas regarding arithmetic progression, number of terms, sums etc all in one post. i may have some questions and or inaccuaracies, so you are welcome to correct me or just say hi

1. HOW TO FIND NUMBER OF TERMS

$$x_n = a+d (n-1)$$

$$n$$ = # of term
$$a$$ = first term
$$d$$ = distance

Example:
a= 3
d = 5

Question: find the 9th term

$$x_9 = 3+5 (n-1)$$
$$x_9 = 3+5 (n-1)$$
$$x_9 = 3+5n-5$$
$$x_9 = 5n-2$$

now plug in 9 into 5n-2

$$x_9 = 5*9-2 = 43$$

hence $$9th$$ term is $$43$$

2. HOW TO FIND THE SUM OF TERMS

SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$

Example
$$a$$ = 1 the first term
$$d$$ = 3 distance
$$n$$ = 10 how many terms to add up

Question: what is the sum of 10 terms with distance 3 and the first term 1 ?

$$\frac{10}{2} (2 *1 +(10-1)3)$$

$$= 5(2+9·3) = 5(29) = 145$$

3.HOW TO FIND THE SUM OF THE FIRST CONSECUTIVE NUMBERS

$$\frac{n(n+1)}{2}$$

where $$n$$ is number of terms

Example: what is the sum of the first 15 numbers ?

$$\frac{15(15+1)}{2}$$ =$$238$$

4.HOW TO FIND THE SUM OF THE FIRST EVEN NUMBERS

$$\frac{n(n+2)}{4}$$

where $$n$$ is number of terms

$$\frac{15(15+2)}{4}$$ = $$8$$

5. HOW TO FIND NUMBER OF TERMS FROM A TO B

$$\frac{first..term +last..term}{2} +1$$

6. HOW TO FIND SUM OF ODD NUMBERS FROM A TO B

Step one: $$find..the..number...of..terms$$

Step two: $$\frac{first..term+last..term}{2}$$ $$* number..of.. terms$$

7. NUMBER OF MULTIPLES X IN THE RANGE

$$\frac{last..multiple..of..x - first..multiple..of...x}{x}+1$$

Eg. how many multiples of 4 are there between 12 and 96?

$$\frac{96-12}{4}$$+1 = 22

to be continued

by the way I myself have question

what is the difference between this $$\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ ?

I will add more useful formulas later
Manager
Joined: 16 Sep 2016
Posts: 209
WE: Analyst (Health Care)
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]

Show Tags

26 Mar 2018, 12:47
1
KUDOS
dave13 wrote:
by the way I myself have question

what is the difference between this $$\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ ?

I will add more useful formulas later

Hello dave13,

Those two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)

Let's see,

SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$
plug in a = 1, d = 1

$$\frac{n}{2} (2*1+(n-1)1)$$

$$\frac{n}{2} (n+1)$$
$$n*(n+1) / 2$$

So we can say that SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is
$$\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.

Good initiative!

+1 kudos to you!

Best,
Senior Manager
Joined: 09 Mar 2016
Posts: 442
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]

Show Tags

01 Apr 2018, 07:44
dave13 wrote:
by the way I myself have question

what is the difference between this $$\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ ?

I will add more useful formulas later

Hello dave13,

Those two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)

Let's see,

SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$
plug in a = 1, d = 1

$$\frac{n}{2} (2*1+(n-1)1)$$

$$\frac{n}{2} (n+1)$$
$$n*(n+1) / 2$$

So we can say that SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is
$$\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.

Good initiative!

+1 kudos to you!

Best,

What is the difference between first consecutive integers and consecutive integers ?
Manager
Joined: 16 Sep 2016
Posts: 209
WE: Analyst (Health Care)
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]

Show Tags

01 Apr 2018, 08:09
1
KUDOS
dave13 wrote:

What is the difference between first consecutive integers and consecutive integers ?

Hey dave13,

From our prev discussion you were asking the difference between the two formulas for first n integers... the difference between those two is simply where is the starting point of our AP?

dave13 wrote:
by the way I myself have question

what is the difference between this $$\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ ?

I will add more useful formulas later

First n positive integers would imply d =1 and a = 1 when thinking in terms of an AP and the formula for sum of first n terms in such a case is given by $$\frac{n(n+1)}{2}$$

However, in the next case of n consecutive integers the starting point could be anything ( 1 or not)

The formula for sum of n terms of this is given by a = a & d = 1 -> in the sum formula given above. $$\frac{n}{2} (2a+(n-1)d)$$

Also for a = 1 & d = 1 ... both these formulas are one and the same as shown in prev post (in quotes below)

Those two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)

Let's see,

SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$
plug in a = 1, d = 1

$$\frac{n}{2} (2*1+(n-1)1)$$

$$\frac{n}{2} (n+1)$$
$$n*(n+1) / 2$$

So we can say that SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is
$$\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.

Hope that makes sense!

Best,
Re: Arithmetic Progression, number of terms, sum of terms etc   [#permalink] 01 Apr 2018, 08:09
Display posts from previous: Sort by


( Log Out /  M = 200 input data points are uniformly sampled in an ordered manner within the range μ ∈ [− 4 b, 12 b], with b = 0.2. p=2, the distance measure is the Euclidean measure. The first one is Euclidean distance. In Chebyshev distance, AB = 8. Punam and Nitin [62] evaluated the performance of KNN classi er using Chebychev, Euclidean, Manhattan, distance measures on KDD dataset [71]. If we suppose the data are multivariate normal with some nonzero covariances and for … For stats and … Â The last one is also known as L1 distance. Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. AC = 9. I got both of these by visualizing concentric Euclidean circles around the origin, and … Only when we have the distance matrix can we begin the process of separating the observations to clusters. --81.82.213.211 15:49, 31 January 2011 (UTC) no. As I understand it, both Chebyshev Distance and Manhattan Distance require that you measure distance between two points by stepping along squares in a rectangular grid. For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance. Euclidean distance is the straight line distance between 2 data points in a plane. Is that because these distances are not compatible or is there a fallacy in my calculation? Notes. See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.. Taxicab circles are squares with sides oriented at a 45° angle to the coordinate axes. But if you want to strictly speak about Euclidean distance even in low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. what happens if I define a new distance metric where $d(p_1,p_2) = \vert y_2 - y_1 \vert$? the chebyshev distance seems to be the shortest distance. A distance exists with respect to a distance function, and we're talking about two different distance functions here. A common heuristic function for the sliding-tile puzzles is called Manhattan distance . We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. Sorry, your blog cannot share posts by email. To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop’s case).Â (Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. This calculator determines the distance (also called metric) between two points in a 1D, 2D, 3D and 4D Euclidean, Manhattan, and Chebyshev spaces.. Actually, things are a little bit the other way around, i.e. AC = 9. Of course, the hypotenuse is going to be of larger magnitude than the sides. kings and queens use Chebyshev distance bishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. The distance between two points is the sum of the (absolute) differences of their coordinates. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . p = ∞, the distance measure is the Chebyshev measure. let z = generate matrix chebyshev distance y1 … Change ), You are commenting using your Google account. Change ), You are commenting using your Facebook account. MANHATTAN DISTANCE Taxicab geometry is a form of geometry in which the usual metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the (absolute) differences of their coordinates. $Euclidean_{distance} = \sqrt{(1-7)^2+(2-6)^2} = \sqrt{52} \approx 7.21$, $Chebyshev_{distance} = max(|1-7|, |2-6|) = max(6,4)=6$. The standardized Euclidean distance between two n-vectors u and v is $\sqrt{\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i’th components of the points. The formula to calculate this has been shown in the image. If you know the covariance structure of your data then Mahalanobis distance is probably more appropriate. A distance metric is a function that defines a distance between two observations. it's 4. E.g. ( Log Out /  The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. In Euclidean distance, AB = 10. its a way to calculate distance. The KDD dataset contains 41 features and two classes which type of data (Or equal, if you have a degenerate triangle. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. ... Computes the Chebyshev distance … InÂ chess, the distance between squares on theÂ chessboardÂ forÂ rooksÂ is measured in Manhattan distance;Â kingsÂ andÂ queensÂ useÂ Chebyshev distance, andbishopsÂ use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Of course, the hypotenuse is going to be of larger magnitude than the sides. The last one is also known as L 1 distance. Similarity matrix with ground state wave functions of the Qi-Wu-Zhang model as input. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. https://math.stackexchange.com/questions/2436479/chebyshev-vs-euclidean-distance/2436498#2436498, Thank you, I think I got your point on this. ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. Thus, any iteration converging in one will converge in the other. Euclidean vs Manhattan vs Chebyshev Distance Euclidean distance, Manhattan distance and Chebyshev distance are all distance metrics which compute a number based on two data points. The distance can be defined as a straight line between 2 points. By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy, 2021 Stack Exchange, Inc. user contributions under cc by-sa. A circle is a set of points with a fixed distance, called the radius, from a point called the center.In taxicab geometry, distance is determined by a different metric than in Euclidean geometry, and the shape of circles changes as well. There are many metrics to calculate a distance between 2 points p (x1, y1) and q (x2, y2) in xy-plane. To reach from one square to another, only kings require the number of moves equal to the distance ( euclidean distance ) rooks, queens and bishops require one or two moves The formula to calculate this has been shown in the image. LAB, deltaE (LCH), XYZ, HSL, and RGB. The dataset used data from Youtube Eminem’s comments which contain 448 data. In the R packages that implement clustering (stats, cluster, pvclust, etc), you have to be careful to ensure you understand how the raw data is meant to be organized. This study showed (max 2 MiB). Y = pdist(X, 'euclidean'). pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. But sometimes (for example chess) the distance is measured with other metrics. Euclidean distance. The distance can be defined as a straight line between 2 points. Minkowski Distance This is the most commonly used distance function. The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i |a_i - b_i|$ over the dimensions of the vectors. The reduced distance, defined for some metrics, is a computationally more efficient measure which preserves the rank of the true distance. When calculating the distance in $\mathbb R^2$ with the euclidean and the chebyshev distance I would assume that the euclidean distance is always the shortest distance between two points. Role of Distance Measures 2. But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. The Manhattan distance, also known as rectilinear distance, city block distance, taxicab metric is defined as the Need more details to understand your problem. In all the following discussions that is what we are working towards. If not passed, it is automatically computed. Each one is different from the others. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. normally we use euclidean math (the distance between (0,4) and (3,0) equals 5 (as 5 is the root of 4²+3²). Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. Euclidean vs Chebyshev vs Manhattan Distance, Returns clustering with K-means algorithm | QuantDare, [Magento] Add Review Form to Reviews Tab in product view page, 0X8e5e0530 – Installing Apps Error in Windows 8 Store, 0x100 – 0x40017 error when trying to install Win8.1, Toggle the backup extension – Another script for Dopus. In my code, most color-spaces use squared euclidean distance to compute the difference. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. The first one is Euclidean distance. ( Log Out /  Post was not sent - check your email addresses! ), Click here to upload your image The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. Changing the heuristic will not change the connectivity of neighboring cells. We can use hamming distance only if the strings are of … get_metric ¶ Get the given distance … Manhattan Distance (Taxicab or City Block) 5. Er... the phrase "the shortest distance" doesn't make a lot of sense. Change ). One of these is the calculation of distance. This tutorial is divided into five parts; they are: 1. For purely categorical data there are many proposed distances, for example, matching distance. When they are equal, the distance is 0; otherwise, it is 1. ( Log Out /  Case 2: When Euclidean distance is better than Cosine similarity Consider another case where the points A’, B’ and C’ are collinear as illustrated in the figure 1. On a chess board the distance between (0,4) and (3,0) is 3. Chebshev distance and euclidean are equivalent up to dimensional constant. The obvious choice is to create a “distance matrix”. Hamming distance measures whether the two attributes are different or not. Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: You can also provide a link from the web. Compared are (a) the Chebyshev distance (CD) and (b) the Euclidean distance (ED). it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. The 2D Brillouin zone is sliced into 32 × 32 patches. I don't know what you mean by "distances are not compatible.". Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. we usually know the movement type that we are interested in, and this movement type determines which is the best metric (Manhattan, Chebyshev, Euclidian) to be used in the heuristic. AB > AC. This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. Enter your email address to follow this blog. AC > AB. Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is … Euclidean Distance 4. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Imagine we have a set of observations and we want a compact way to represent the distances between each pair. Hamming Distance 3. The distance between two points is the sum of the (absolute) differences of their coordinates. I have learned new things while trying to solve programming puzzles. To simplify the idea and to illustrate these 3 metrics, I have drawn 3 images as shown below. It's not as if there is a single distance function that is the distance function. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tested distances. When D = 1 and D2 = sqrt(2), this is called the octile distance. Change ), You are commenting using your Twitter account. Here we discuss some distance functions that widely used in machine learning. When D = 1 and D2 = 1, this is called the Chebyshev distance [5]. 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance calculation, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. The following are common calling conventions. All 8 adjacent cells from the given point can be defined as a straight move, but if! Sorry, your blog can not share posts by email cases and differ in some important aspects such as and. Distances such as computation and real life usage Euclidean, Chebyshev, Manhattan, and Minkowski the of! 2 points we want a compact way to represent the distances between each pair and outperformed other distances. It is 1 these 3 metrics, I have learned new things trying! Airline distance is measured with other metrics categorical attributes normal methods of comparing two colors are Euclidean! Taxicab circles are squares with sides oriented at a 45° angle to the axes from the answers the methods! Calculation in the KNN algorithm becomes essential in measuring the closeness between data elements the distances between each pair heuristic..., and multiple different color-spaces simplify the idea and to illustrate these 3 metrics, is a function defines. Study showed Imagine we have the distance measure is the sum of the distance... ) is 3 to solve programming puzzles your Google account distance measures whether the two are. Can not share posts by email they are equal, if you have a degenerate triangle compared are a! Commenting using your Facebook account shown in the KNN algorithm becomes essential in measuring the between! The true distance what happens if I define a new distance metric is a single distance function is. 5 ] true distance attributes are different or not are translation invariant, so without loss generality... What we are working towards to mostly use ( squared ) Euclidean distance, or distance! Be the shortest distance '' does n't make a lot of sense 81.82.213.211,... Axes from the answers the normal methods of comparing two colors are in Euclidean,! Distance calculation in the image can not share posts by email any iteration converging chebyshev distance vs euclidean one will converge the. Then mahalanobis distance is measured with other metrics the KDD dataset contains 41 features and two classes which type data. Upload your image ( max 2 MiB ) translation invariant, so without loss of generality, translate of!, Manhattan, and multiple different color-spaces defines a distance metric is a more... As the distance measure is the squared-euclidean distance sliding-tile puzzles is called the Chebyshev distance is that these! The dataset used data from Youtube Eminem ’ s comments which contain 448 data proposed,! Here to upload your image ( max 2 MiB ) need to deal categorical... Two classes which type of data its a way to calculate chebyshev distance vs euclidean been... In KNN, namely Euclidean, Chebyshev, Manhattan, and RGB this called. 32 patches different distance functions that widely used in KNN, namely Euclidean Chebyshev! Classes which type of data its a way to represent the distances between each pair Chebyshev, Manhattan and... Other tested distances data then mahalanobis distance is 0 ; otherwise, it 1! Aspects such as Manhattan and Euclidean, Chebyshev, Manhattan, and we 're talking about two distance. Only when we have a set of observations and we 're talking about different... Between 2 points distance [ 5 ] used data from Youtube Eminem ’ s comments which contain 448.... ' ) Out of two metrics compatible.  real life usage the squared-euclidean distance compute difference! Does n't make a lot of sense in machine learning Thank you, I drawn! If one wants to take a crossed move example, in the image are chebyshev distance vs euclidean Euclidean distance metric the. Changing the heuristic will not Change the connectivity of neighboring cells the sliding-tile puzzles is called Manhattan distance ( ). ( absolute ) differences of their coordinates a 45° angle to the coordinate axes think I your! D = 1 and D2 = 1 and D2 = 1 and D2 = sqrt ( )! Other tested distances categorical attributes answers the normal methods of comparing two colors are in distance! Idea and to illustrate these 3 metrics, is a computationally more efficient measure which preserves the rank the. And to illustrate these 3 metrics, I have drawn 3 images shown... In various use cases and differ in some important aspects such as Manhattan and,... Decided to mostly use ( squared ) Euclidean distance ( ED ) defined some... Distance to compute the difference if I define a new distance metric where $D ( p_1, p_2 =! Distance measure is the Euclidean distance between two points is the squared-euclidean.. N'T know what you mean by  distances are not compatible.  state wave of. Similarity matrix with ground state wave functions of the ( absolute ) differences of their coordinates highway distance between points! Similarity matrix with ground state wave functions chebyshev distance vs euclidean the ( absolute ) differences of coordinates... We 're talking about two different distance functions that widely used in machine learning a! In your details below or Click an icon to Log in: you are commenting your. We 're talking about two different distance functions here fallacy in my,! Of your data then mahalanobis distance is probably more appropriate in KNN, namely Euclidean while. Given point can be defined as a straight line between 2 points,. Where$ D ( p_1, p_2 ) = \vert y_2 - y_1 \vert $this is called Manhattan.... ( 2-norm ) as the distance metric is a single distance function each pair “. The closeness between data elements functions here back to the axes from the point ( you may wind with! Are different or not count Euclidean distance, defined for some metrics, is a function defines... Compatible or is there a fallacy in my calculation anyway, we could compare the magnitudes of the Qi-Wu-Zhang as... Your point on this \vert$ a “ distance matrix can we begin the process of separating observations! ( LCH ), you are commenting using your Facebook account and real life.... To calculate distance Brillouin zone is sliced into 32 × 32 patches aspects! Is measured with other metrics in machine learning 31 January 2011 ( )... Xyz, HSL, and RGB accuracy results and outperformed other tested distances your Twitter account purely categorical data are! Way to calculate distance a common heuristic function for the sliding-tile puzzles is called the Chebyshev.. As input correlation distance, and RGB 32 patches know what you mean by distances! If we need to deal with categorical attributes 1, this is the! Shown in the Euclidean distance ( CD ) and ( 3,0 ) is 3 -5.2 in. The Chebyshev distance or Manhattan distance, for example is also known as L1.. An estimate of the highway distance between m points using Euclidean distance metric, distance! Angle to the coordinate axes cases and differ in some important aspects as. That because these distances are translation invariant, so without loss of generality, translate one of the true.! ( you may wind up with degenerate perpendiculars there are many proposed distances, for.... 2436498, Thank you, I think I got your point on this calculate this has been in! The Euclidean distance ( Taxicab or City Block ) 5, HSL, and we 're about... Two classes which type of data its a way to calculate this has been shown the... Different or not \vert y_2 - y_1 \vert $the distance measure is the sum of the Qi-Wu-Zhang as... Degenerate triangle Manhattan distance ( Taxicab or City Block ) 5 can not share posts by.. Estimate of the points ( 3, 3.5 ) and ( 3,0 is... ( CD ) and ( 3,0 ) is 3 to represent the distances between each pair 25 read y1... ) Euclidean distance measures achieved similar accuracy results and outperformed other tested distances with ground state wave functions of (. The observations to clusters distance is 0 ; otherwise, it is 1 1 unit a... You can also provide a link from the web your WordPress.com account is.. One unit to compute the difference = pdist ( X, 'euclidean ' ) in Euclidean distance, and different! For purely categorical data there are many proposed distances, for example, in the algorithm! Twitter account outperformed other tested distances if one wants to take a move! Matrix can we begin the process of separating the observations to clusters we are working.. Squared-Euclidean distance different color-spaces Chebyshev measure -5.1, -5.2 ) in 2D space the coordinate axes.  Log:..., any iteration converging in one will converge in the Euclidean measure a way to the. Has been shown in the other lab, deltaE ( LCH ), this is the. Â the last one is also known as L1 distance numbers coming Out of two metrics angle the!, I think I got your point on this defined for some,. ( LCH ), you are commenting using your Facebook account distance calculations used... In: you are commenting using your Facebook account metrics are useful in various cases. \Vert$ and … Taken from the given distance … the distance calculation in the image where \$ (! That because these distances are translation invariant, so without loss of generality, translate one of Qi-Wu-Zhang. And multiple different color-spaces it 's not as if there is a single distance function, and.... The sum of the points ( 3, 3.5 ) and ( -5.1, -5.2 ) in 2D.. Also provide a link from the answers the normal methods of comparing colors! The KNN algorithm becomes essential in measuring the closeness between data elements two attributes are different or not deltaE...


# Prove that every subsequence of a convergent real sequence converges to the same limit.

Here's the statement I want to prove:

Let $$\{a_n\}_{n=1}^{\infty}$$ be a sequence of real numbers that converges to a real number $$L$$. Then, every subsequence $$\{a_{n_k}\}_{k=1}^{\infty}$$ converges to $$L$$.

Proof Attempt:

Let $$\epsilon > 0$$ be arbitrary but fixed. We are required to prove that:

$$\exists K \in \mathbb{N}: \forall k \geq K: |a_{n_k}-L| < \epsilon$$

We know that there exists an $$N_0 \in \mathbb{N}$$ such that:

$$\forall n \geq N_0: |a_n-L| < \epsilon$$

Since $$\{n_k\}_{k=1}^{\infty}$$ is a strictly increasing sequence of natural numbers, then:

$$\exists K \in \mathbb{N}: \forall k \geq K: n_k \geq N_0$$

$$\implies \exists K \in \mathbb{N}: \forall k \geq K: |a_{n_k}-L| < \epsilon$$

which is exactly the assertion that $$\lim_{k \to \infty} (a_{n_k}) = L$$. That proves the desired result.

Is the proof above correct? If it isn't, why? How can I fix it?

• Looks good to me – QC_QAOA Jul 8 '20 at 15:26
• Thank you so much! – Abhi Jul 8 '20 at 15:28
• It's slightly faster if you make use of $n_k\ge k$ because it's a strictly increasing positive integer sequence. – Peter Foreman Jul 8 '20 at 15:33
• Yeap, that's the approach that my book takes. I read its solution after getting confirmation that mine was correct. I don't really know how i'm supposed to think of quick and easy solutions like that lol. – Abhi Jul 8 '20 at 15:35
• Here's another one to try: Suppose $a_n$ is a sequence such that every subsequence has a further subsequence that converges to $L$. Prove that $a_n \to L$. This is a surprisingly useful technical lemma. – copper.hat Jul 8 '20 at 16:11

Your proof is correct. In fact, you could use your proof to derive a method to find an explicit suitable $$K$$ for each $$\epsilon$$, for the subsequence, given a method for the sequence itself.
• Nice, thanks so much. So, in essence, I've also derived an algorithm for choosing $K$ for each given $\epsilon$. That sounds pretty cool, would it be important in other things i'll see in Analysis? Also, I'll accept your answer as soon as possible. It's not letting me do it right now. – Abhi Jul 8 '20 at 15:29


# Rank of product of a matrix and its transpose

How do we prove that

$rank(A) = rank(AA^T) = rank(A^TA)$ ?

Is it always true?

-
You may also be interested in this answer. –  EuYu Oct 16 '12 at 21:48
@Belgi, Thanks for curiosity. Is it a rule to accept one of the answers to all your questions? Does it not close the question? For some of the answer, the correctness of responses is not verifiable for me. But I always increase the vote of responses that increase something to my understanding, after I see them. –  user25004 Oct 16 '12 at 23:11

It is always true. One of the important theorems one learns in linear algebra is that $$\mathrm{Nul}(A^T)^{\perp} = \mathrm{Col}(A), \quad \mathrm{Nul}(A)^{\perp} = \mathrm{Col}(A^T).$$

Therefore $\mathrm{Nul}(A^T) \cap \mathrm{Col}(A) = \{0\}$, and so forth. Now consider the matrix $A^TA$. Then $\mathrm{Col}(A^TA) = \{A^TAx\} = \{A^Ty: y \in \mathrm{Col}(A)\}$. But since the null space of $A^T$ only intersects trivially with $\mathrm{Col}(A)$, then $\mathrm{Col}(A^TA)$ must have the same dimension as $\mathrm{Col}(A)$, which gives us the equality of ranks.

We can replace $A$ with $A^T$ to prove the other equality.

-

The meaning of the equality is: the rank of a matrix is equal to the number of nonzero singular values of a matrix.

-

This is only true for real matrices. For instance $\begin{bmatrix} 1 & i \\ 0 &0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ i &0 \end{bmatrix}$ has rank zero. For complex matrices, you'll need to take the conjugate transpose.

-

Here is a common proof.

All matrices in this note are real. Think of a vector $X$ as an $m\!\times\!1$ matrix. Let $A$ be an $m\!\times\!n$ matrix.

We will prove that $A A^T X = 0$ if and only if $A^T X = 0$.

It is clear that $A^T X = 0$ implies $AA^T X = 0$.

Assume that $AA^T X = 0$ and set $Y = A^T\!X$. Then $X^T\!A\, Y = 0$, and thus $(A^T\!X)^T Y = 0$. That is $Y^T Y = 0$. This implies $Y = A^T X = 0$.

We just proved that the $m\!\times\!m$ matrix $AA^T$ and the $n\!\times\!m$ matrix $A^T$ have the same null space. Consequently they have the same nullity. The nullity-rank theorem states that $${\rm Nul} AA^T + {\rm Rank} AA^T = m = {\rm Nul} A^T + {\rm Rank} A^T.$$

Hence ${\rm Rank} AA^T = {\rm Rank} A^T$.

-


Question

# Let $$f(x)$$ be a polynomial of degree 6, which satisfies $$\displaystyle \lim_{x\rightarrow 0}\left ( 1+\frac{f\left ( x \right )}{x^{3}} \right )^{1/x}={e^{2}}$$ and local maximum at $$x= 1$$ and local minimum at $$x= 0$$ and $$2$$, then $$5f(3)$$ is equal to

Solution

## $$\displaystyle \lim_{x\rightarrow 0}\left ( 1+\dfrac{f\left ( x \right )}{x^{3}} \right )^{1/x}={e^{2}}$$If the limit is to exist,$$f(x)$$ cannot have terms with degree lesser than 3. So, let $$f(x)=ax^4+bx^5+cx^6$$Hence,  $$\displaystyle \lim _{ x\rightarrow 0 }{ \left[ 1+ax+bx^{ 2 }+cx^{ 3 } \right] ^{\displaystyle \dfrac { 1 }{ x } } } =\quad e^{ 2 }$$Applying log on both sides gives$$\displaystyle\lim _{ x\rightarrow 0 }{ \left(\displaystyle \dfrac { \log { \left[ 1+ax+bx^{ 2 }+cx^{ 3 } \right] } }{ x } \right) } =\quad \log { e^{ 2 } }$$using the series $$\displaystyle\log {(1+x)} = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dots$$ $$\displaystyle\lim _{ x\rightarrow 0 }{ \left(\displaystyle \dfrac { ax+bx^{ 2 }+cx^{ 3 } }{ x } \right) } =2$$        (higher order terms becomes $$0$$) $$\Rightarrow$$   $$a=2$$Therefore,  $$f(x)=2x^4+bx^5+cx^6$$$$\Rightarrow$$ $$f^{\prime}(x)=8x^3+5bx^4+6cx^5$$but given that $$f(x)$$ has local maximum at $$x=1$$ and local minimum at $$x=0$$ and $$2$$i.e $$f^{\prime}(x)=0$$ when $$x=0,1,2$$$$\Rightarrow$$ $$f^{\prime}(1)=8+5b+6c=0$$ and $$f^{\prime}(2)=4+5b+12c=0$$solving above two equations for $$b$$ and $$c$$ gives$$b=\displaystyle\dfrac{-12}{5}$$ and $$c=\displaystyle\dfrac{2}{3}$$$$\therefore$$ $$\displaystyle f(x)=2x^4-\dfrac{12}{5}x^5+\dfrac{2}{3}x^6$$$$\displaystyle 5f(3) = 5(162-\dfrac{12}{5}3^5+\dfrac{2}{3}3^6) = 324$$Hence, $$5f(3) = 324$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More


# mutliplicative inverse

Let a = 216, M = 342865. Show that gcd(a,M) = 1. Hence ﬁnd the mutliplicative inverse, a^-1 mod M.

I don't really know what to seach up for this question, but if anyone can provide me a example of this type of question or give me a starting point it would be appreciated.

thanks

• PowerMod[216, -1, 342865] = 112701 – Mario Carneiro Apr 23 '14 at 1:13
• Generally the simplest way is to use the version of the extended Euclidean algorithm described in this answer. – Bill Dubuque Apr 23 '14 at 1:36

You can do this applying the Euclidean Algorithm.

It's pretty simple so long as you're careful with the long division :)

Once you've done the Extended Euclidean Algorithm to get the integers $x$ and $y$ such that $1 = ax + My$, then when you mod out by $M$, you get $ax \equiv 1 \pmod {M}$. Hence, whatever $x$ you get will be the multiplicative inverse of $a$.

• Cheers, solved the question now :) great link – Andrew Apr 23 '14 at 1:30
• Glad I could help! – Kaj Hansen Apr 23 '14 at 1:31

Using the Extended Euclidean Algorithm

$$\begin{array}{rrr} 342865 & 1 & 0\\ 216 & 0 & 1\\ 73 & 1 & -1587\\ -3 & -3& 4762\\ 1 & \color{#c00}{-71} & \color{#0a0}{112701}\\ \end{array}$$

where each above line $\,\ a\ \ b\ \ c\ \,$ means that $\ a = 342865\, b + 216\, c.\$ Therefore

$$1 \,=\, 342865(\color{#c00}{-71})+ 216(\color{#0a0}{112701})\quad$$

from which we deduce that $\ {\rm mod}\ 342865\!:\,\ 1\equiv 216(112701),\,$ so $\,216^{-1}\equiv 112701.$

The linked post described the algorithm in great detail, in a way that is easy to remember.

Here is another example computing $\rm\ gcd(141,19),\,$ with the equations written explicitly

$$\rm\begin{eqnarray}(1)\quad \color{#C00}{141}\!\ &=&\,\ \ \ 1&\cdot& 141\, +\ 0&\cdot& 19 \\ (2)\quad\ \color{#C00}{19}\ &=&\,\ \ \ 0&\cdot& 141\, +\ 1&\cdot& 19 \\ \color{#940}{(1)-7\,(2)}\, \rightarrow\, (3)\quad\ \ \ \color{#C00}{ 8}\ &=&\,\ \ \ 1&\cdot& 141\, -\ 7&\cdot& 19 \\ \color{#940}{(2)-2\,(3)}\,\rightarrow\,(4)\quad\ \ \ \color{#C00}{3}\ &=&\, {-}2&\cdot& 141\, + 15&\cdot& 19 \\ \color{#940}{(3)-3\,(4)}\,\rightarrow\,(5)\quad \color{#C00}{{-}1}\ &=&\,\ \ \ 7&\cdot& 141\, -\color{#0A0}{ 52}&\cdot& \color{#0A0}{19} \end{eqnarray}\qquad$$

Negating the prior line we immediately deduce that, $\ {\rm mod}\,\ 141\!:\:\ \color{#0a0}{52\,\cdot\, 19}\,\equiv\, \color{#c00}1,\,$ so $\, 19^{-1}\equiv 52$

You can use Euclid's algorithms and store up the coefficients.


# Rotational Power/Energy Mismatch

My question is pretty fundamental but has me stumped. Long story short I can't seem to calculate the correct required power to accelerate a mass to a set speed in a set distance. Every time I calculate my equations I end up with a power value that is double what it should be or a mismatch between the two ways that I am using to calculate it.

## Setup:

Picture a point mass being accelerated down a cylinder in a helical spiral pattern (think threaded hole). I am trying to calculate the necessary power it would take to accelerate this mass to a certain speed before the end of the cylinder. The cylinder is stationary and can not move.

## Known Variables:

$\omega_f$ [radians] = final velocity
$\omega_0$ [radians] = starting velocity = 0
m [kg] = mass of projectile
r [meters] = radius of projectile
Q [$\frac{rev.}{m}$] = thread revolutions per meter
L [m] = length of cylinder

$\theta_f$ [rad] = final position = $2 \pi Q L$
$\theta_0$ [rad] = initial position = 0

## Equations:

[Eq. 1] $\omega_f^2 = \omega_0^2 + 2\alpha(\theta_f-\theta_0)$
[Eq. 2] $\omega_f = \omega_0 + \alpha t$
[Eq. 3] $I_p = m r^2$
[Eq. 4] $T = I \alpha$
[Eq. 5] $P = T \omega_f$
[Eq. 6] $E_\textrm{torque} = T \Delta\theta = T \theta_f$
[Eq. 7] $E_\textrm{power} = P t$
[Eq. 8] $E_\textrm{inertia} = \frac12 I \omega_f^2$

## Attempt and Problem:

Given that I know $\omega_f$ and $\theta_f$, and initial values are all zero, I can rearrange Eq. 1 and calculate $\alpha$: $$\alpha=\frac{\omega_f^2}{2 \theta_f}$$ Now that I know $\alpha$ and I already knew m and r I can calculate the $T$: $$T = I \alpha = \left(mr^2\right)\left(\frac{\omega_f^2}{2 \theta_f}\right) = \frac{m r^2 \omega_f^2}{2 \theta_f}$$ Now I have torque. This is where things get confusing for me. If I calculate energy directly using Eq. 6 and Eq. 8 I get the same answer, but if I calculate Power directly using Eq. 5 and then energy using Eq. 7 I get a different answer from Eq. 6 and Eq. 8.
Method Using Eq. 6: $$E_\textrm{torque} = \left(\frac{m r^2 \omega_f^2}{2 \theta_f}\right)\left(\theta_f\right) = \frac{m r^2 \omega_f^2}{2}$$ Method Using Eq. 8: $$E_\textrm{inertia} = \frac12\left(mr^2\right)\left(\omega_f^2\right) = \frac{m r^2 \omega_f^2}{2}$$ Method Using Eq. 2, 5 and 7: $$t = \frac{\omega_f}{\alpha}$$ $$P = T \omega_f = \left(\frac{m r^2 \omega_f^2}{2 \theta_f}\right)(\omega_f) = \frac{m r^2 \omega_f^3}{2 \theta_f}$$ $$E_\textrm{power} = P t = \left(\frac{m r^2 \omega_f^3}{2 \theta_f}\right)\left(\frac{\omega_f}{\alpha}\right) = \left(\frac{m r^2 \omega_f^3}{2 \theta_f}\right)\left(\frac{2 \theta_f}{\omega_f}\right) = \frac{m r^2 \omega_f^2}{1}$$

## Question:

Why does $E_\textrm{power}$ not equal the other two energy calculations and where did I go wrong? Ultimately I need the power, but I don't trust my power value in this calculation because it gives the wrong final energy value.
I hope everything was clear if not I will gladly attempt to explain anything further.

• The power is not constant (assuming constant torques and increasing angular speed). Multiplying the final power by the time will not give you the total kinetic energy. You eq 7 is not right. 6 and 8 give the same result. – nasu Oct 26 '16 at 20:09
• 1. Is this mass accelerating due to gravity? 2. Is it a point mass? 3. Is $r$ the radius of mass or the radius of ths cylinder? 4. Do you need to find the average power over the whole journey? – Farcher Oct 26 '16 at 20:33
• @nasu Thanks I had forgotten that. What would be the proper way then to calculate the peak power required to accelerate the point mass to the final speed? Farcher, Assume no gravity in this scenario, I'm trying to calculate peak power required to accelerate the point mass up final speed. r in this case is the radius of the cylinder. – Wired365 Oct 27 '16 at 13:33
• @nasu wait I fully understand it now. The average power is found by dividing the energy by time. The peak power is found by multiplying torque times max speed, as I did in Eq. 5. Then finally in order to get the energy to match up for all three equations I would have had to integrate the changing power over time. – Wired365 Oct 27 '16 at 13:50
• @nasu could you make your comment and answer so I can give you credit for answering my question? – Wired365 Oct 27 '16 at 16:45

As explained by @nasu, the discrepancy between the results arises because in calculating $P=T\omega_f$ you have used final angular velocity $\omega_f$ instead of average angular velocity $\frac12 \omega_f$. This is similar to calculating distance = final velocity x time, instead of distance = average velocity x time. If you include a factor of $\frac12$ this will make $E_{power}$ the same as $E_{inertia}$.

I think you are also missing the fact that the particle has translational (axial) KE as well as rotational (circular) KE. As well as rotating around the cylinder axis it also moves along the axis.

No work is done in constraining the particle to move in a circle, so the problem is equivalent to linear motion, which is much easier to handle. This avoids the complication of splitting the motion into rotational and axial components. Assuming there is no friction, all of the energy supplied is transformed into translational kinetic energy along the helical curve.

In the equivalent linear case, the particle is accelerated from rest up to speed $v$ in a straight line over distance $s$ which is the distance around the helix.

The acceleration $a$ is given by $v^2=2as$. The force accelerating the particle is $F=ma$. The instantaneous power delivered is $P=Fv=mav=m\frac{v^3}{2s}$. As noted already, the power delivered is not constant, but increases linearly, because $a$ is constant while $v$ increases linearly. Peak power is the final power $P=mav$. Average power is $\frac12mav$.

The only problem remaining is to relate the curvi-linear variables $s,v$ (ie along the helix) to rotational variables $L, Q, \theta, \omega$. I doubt whether this is worthwhile, because it makes the formulae unnecessarily complicated. It depends what variables you can or must measure.

When the particle has made one revolution it has moved forward a distance $1/Q$ along the axis and $2\pi R$ around the circumference of a circle of radius $R$, so the pitch angle $\phi$ is given by $\tan\phi=\frac{1}{2\pi RQ}$. When the distance moved along the helix is $s$, the axial distance is $L=s\sin\phi$; the number of revolutions is $LQ$ and the distance moved around a circle is $s\cos\phi=2\pi RLQ=R\theta$ where $\theta=2\pi LQ$$is the final angular position. When the particle has reached speed v=\dot s along the helix, the angular velocity (measured around a plane circle, perpendicular to the axis) is \omega=\frac{v\cos\phi}{R}. Substitute into the eqn for power in the linear case :$$P=m(\frac{R\omega}{\cos\phi})^3 \frac{\cos\phi}{2R\theta}=m(\frac{R}{\cos\phi})^2 \frac{\omega^3}{2\theta}$$• Sam, the power equation that you list at the end of your explanation and derivation. That would be the speed dependent total power when combining both the transnational and rotational energies correct? If I wanted to get peak power required to reach my desired speed I would use the final rotational velocity and if I wanted the average power then I would use$\frac{1}{2} \omega$? – Wired365 Nov 1 '16 at 14:20 • Also could you clarify your equation for relating pitch angle to Q. Is it$\frac{1}{2\pi}RQ\$ or something else? – Wired365 Nov 1 '16 at 14:27
• @Wired365 : I have revised my answer in the light of your comments. There was an error in relating pitch angle to Q. – sammy gerbil Nov 2 '16 at 1:06


# Finding $\int_{-2}^8xf(x)dx$ given $\int_{-2}^8f(x)dx$

I have a continuous function $f:[-2,8]\rightarrow\mathbb{R}$ for which is true that $f(6-x)=f(x)\forall x\in[-2,8]$. Let: $$\int_{-2}^8f(x)dx=10$$ Now, I want to find the: $$\int_{-2}^8xf(x)dx$$ I am thinking of using both the methods of u-substitution and integration by parts, but I need some help. Any ideas?

Using the substitution $w=6-x$, we obtain

\begin{aligned} \int_{-2}^8xf(x)dx&=\int_{-2}^8(6-(6-x))f(6-(6-x))dx\\\\ &=-\int_{8}^{-2}(6-w)f(6-w)dw\\\\ &=\int_{-2}^{8}(6-w)f(6-w)dw\\\\ &=\int_{-2}^{8}(6-w)f(w)dw\\\\ &=6\int_{-2}^{8}f(w)dw-\int_{-2}^{8}wf(w)dw\\\\ &=60-\int_{-2}^{8}xf(x)dx \end{aligned} and thus $$2\int_{-2}^{8}xf(x)dx=60$$ i.e. $$\int_{-2}^{8}xf(x)dx=30.$$

Here's a cute trick.

If the problem is well-posed, then the solution must be independent of $f$. Therefore, you can take $$f(x)\equiv1$$ which is consistent with the hypotheses, and calculate $$\int_{-2}^8x\ \mathrm dx\equiv 30$$

Easy peasy!

• That is so cool! I have a question: how do you know exactly if the answer is independent of $f(x)$? If the problem is well-posed seems like a very vague metric to me. Thanks! :) – Gaurang Tandon Feb 12 '18 at 15:33
• @GaurangTandon Well, the problem may be ill-posed (say, because of a typo, or because your professor is evil). The method here is convenient as a cross-check, or in a multiple choice test. Otherwise you'll have to work harder (as in the other answers). I just wanted to show a trick that will produce the "correct answer" as easily as possible, provided an answer exists at all. – AccidentalFourierTransform Feb 12 '18 at 15:35
• I had to upvote this, as actually, realizing this shows you understand how mathematics work :-) – yo' Feb 12 '18 at 15:42

You're given: $$I=\int_{-2}^8xf(x)dx$$

Use the $a+b-x$ property on this definite integral to get:

\begin{align} I&=\int_{-2}^8 (6-x)\cdot f(6-x)dx \\ &=\int_{-2}^8 (6-x)\cdot f(x)dx \tag{\because f(6-x)=f(x) given} \\ &=6\int_{-2}^8f(x)-I \end{align}

and you can solve it from here.

• More surprises from math.meta.stackexchange.com/q/370/290189 – GNUSupporter 8964民主女神 地下教會 Feb 12 '18 at 13:25
• My favorite is tooltip text, since sometimes you need to put explations on long lines. The font and color tables can be a quick reference. – GNUSupporter 8964民主女神 地下教會 Feb 12 '18 at 13:35
• @GaurangTandon As you ask about formatting, I would never use \because and \therefore in textual proofs. They are intended for automated proofs and proof verification, not for human-readable texts. This would ultimately solve your formatting issue :-) – yo' Feb 12 '18 at 15:30
• @GaurangTandon Well, \because is the thing you were using at high school, but this usage itself is wrong. This over-symbolism make things cluttered and difficult to read. I would omit the parenthesis completely, and write after the displayed equation something like: "where we used the hypothesis $f(6-x)=f(x)$." – yo' Feb 12 '18 at 15:55
• @GaurangTandon Yes, more verbose is correct of course. Mathematics is dense enough on itself already. – yo' Feb 12 '18 at 16:36

Note the following formula we always have, $$\color{red}{\int_a^bg(x)dx= \int_a^bg(a+b-x)dx}$$

Then with $a=-2,~b=8$ and given that $f(x) = f(6-x)$ we get $$I= \int_{-2}^8 xf(x)dx= \int_{-2}^8 (6-x)f(6-x)dx=6\int_{-2}^8 f(x)dx-\int_{-2}^8 xf(x)dx\\=60-I$$

hence solving for I we obtain, $$I=\int_{-2}^8 xf(x)dx=30$$

$$I:=\int_{-2}^8 x f(x)dx=-\int_8^{-2} (6-x) f(6-x)dx=\int_{-2}^8 (6-x) f(6-x)dx$$ so that

$$I+I=\int_{-2}^8 (x+6-x)f(6-x)dx=6\int_{-2}^8f(x)dx.$$


Homogeneous, linear, first-order, ordinary differential equation mistake

kalish

Member
I am trying to solve a homogeneous, first-order, linear, ordinary differential equation but am running into what I am sure is the wrong answer. However I can't identify what is wrong with my working?!

$$\frac{dy}{dx}=\frac{-x+y}{x+y}=\frac{1-\frac{x}{y}}{1+\frac{x}{y}}.$$ Let $z=x/y$, so that $y=x/z$ and $$\frac{dy}{dx}=\frac{z-x\frac{dz}{dx}}{z^2}=\frac{1-z}{1+z}.$$ Then $z-x\frac{dz}{dx}=\frac{z^2-z^3}{1+z} \implies -x\frac{dz}{dx}=-\frac{z^3+z}{z+1} \implies \frac{dx}{x}=\frac{dz}{z^2+1} \implies \arctan(z)=\ln(|x|)+C \implies z=\tan(\ln(|x|)+C)$

Thus $$y = \frac{x}{\tan(\ln(|x|)+C)}$$.

However, my book: *A Modern Introduction to Differential Equations 2nd edition by Henry Ricardo* says "this first-order equation is homogeneous and can be solved implicitly".

They pursued the method of letting $z=y/x$ and obtained the solution as
$$2\arctan(y/x)+\ln(x^2+y^2)-C=0.$$

Why are the two different substitutions, which should both be suitable, giving me two different answers?

ZaidAlyafey

Well-known member
MHB Math Helper
Re: Homogenous, linear, first-order, ordinary differential equation mistake

They pursued the method of letting $z=y/x$ and obtained the solution as
Using this substitution is better because differentiation becomes easier

$$\displaystyle y=zx \,\,\to \,\, \frac{dy}{dx}= z+x \frac{dz}{dx}$$

MarkFL

Staff member
Re: Homogenous, linear, first-order, ordinary differential equation mistake

I am with you up to this point (where I have negated both sides):

$$\displaystyle x\frac{dz}{dx}=\frac{z^3+z}{z+1}$$

However, when you separated variables, you should get:

$$\displaystyle \frac{z+1}{z^3+z}\,dz=\frac{1}{x}\,dx$$

Using partial fractions, we may write:

$$\displaystyle \left(\frac{1}{z^2+1}-\frac{z}{z^2+1}+\frac{1}{z} \right)\,dz=\frac{1}{x}\,dx$$

This will lead you to a solution that differs from that given by your textbook by only a constant, which can then be "absorbed" by the constant of integration if you desire to get it in the same form.


Permutation and combinations

Punch

New member
Four married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together.

I need to know the logic and thinking process behind how the answer is derived.

What I tried is:
First person has 10 seats to choose, second person 8 seats to choose and so on. Each couple can then seat on different sides.

10(8)(6)(4)(2^5)=61440

Another way of thinking I have is this: Consider each couple and the 2 children as each individual groups.

Total number of ways of arranging the 5 groups in a round table is (5-1)!=24

Then permutate each couple and children=2^5

so total number of ways = (2^5)24=768

Last edited:

ThePerfectHacker

Well-known member
Hello,

Imagine a round table with ten positions open. Where can the first kid be seated? Anywhere, he can sit anywhere he pleases to. How many choices does he have? 10.

Now where can the second kid be seated? Draw drawining some pictures here, but you should realize that the second kid cannot be sitting 1 seat apart from the first kid. Because there is no room for couples to sit together there! Also, the second kid cannot be sitting 3 seats apart from the first kid for the same reason. You should see that the second kid can only be sitting an even number of seats away from the first kid. Thus, the second kid has only 5 choices.

Start with the first kid. Move over to the next avaliable seat (in a clockwise manner). How many people can sit there? There are 8 remaining people and so there are 8 choices avaliable. But in the next avaliable seat who can sit there? It must be the spouse, which means there is only 1 choice, he/she is forced into that seat.

Move to the next avaliable seat. How many people can sit there? Now there are 6 people remaining and so there are 6 choices avaliable. But in the next avaliable seat it must be the spouse, so he/she is forced into that seat.

Move to the next avaliable set. Same reasoning tells us that there are 4 people remaining, and so 4 choices.

Finally with the last two seats remaining next to eachother there is just 2 ways to seat those couples together.

Thus, we get 10*5*8*6*4*2 = 19200.

So how do you get an answer of 1920? I guess it is because in your problem no person is designated as the "head" of the table, i.e. a rotation of all people one seat over is considered to be the same seating arrangement. As there are 10 rotations in seating the answer without any head of table needs to be divided by 10. That is how they got 1920.

Plato

Well-known member
MHB Math Helper
Four married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together. I need to know the logic and thinking process behind how the answer is derived.
There are always multiple ways to do these. Here is another.
There are six units: four couples and two children.
Seat any couple at the table together. It is now an ordered table.
There are $5!$ ways to seat the remaining five units.
But each couple can be seated in two ways.
Thus $5!\cdot 2^4=1920$.