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On the other hand, the choice may be between cases where the event will happen in either eighteen months with certainty (so next year with probability $0$ and eventually with probability $1$), or at some indeterminate time (next year with probability $0.9/(1-\varepsilon)$ and eventually with probability $(0.95-\varepsilon)/(1-\varepsilon)$), based on some unknown variable that might point to the first case (with probability $\varepsilon$) but almost surely points to the second (with probability $1-\varepsilon$). Here, a year from now you'll be much more likely to believe that the unknown variable is pointing to the first case, which will increase the odds that the event is still forthcoming.
-
There could be slight issue with the phrasing of the problem. Your interpretation is that the statement is true no matter what time period you are in. I.e. in 5 years time, there is still a 90% chance that some event will happen in that year. However, you're forgetting to account for the fact that the event didn't happen in the first year.
Your friend's interpretation, is that your statement is true for this year. He has ignored the probability space in which the event occurred this year, and considered the remaining probability space. This now has measure $1-0.9=0.1$, and we know that the event has $0.95 - 0.9 = 0.05$ measure to occur, so it has $\frac {0.05}{0.1} = 50\%$ chance to happen.
For example, consider the event where outcomes are based on drawing a uniform random variable on $[0,1]$. Let $C$ be your favorite number from 0 to 0.9.
If we draw a number from 0 to 0.9, the event happens this year.
If we draw a number from $X$ to 0.95, the event happens next year.
If we draw a number from 0.95 to 1, the event never happens.
A possible scenario describing your interpretation, could be | {
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A possible scenario describing your interpretation, could be
If we draw a number from 0 to 0.9, the event happens every even year.
If we draw a number from 0.05 to 0.95, the event happens every odd year.
If we draw a number from 0.95 to 1, the event never happens.
Then, it will always be true, that there is a 90% chance of it happening this year, and a 95% chance of it happening eventually. However, not that if we know it doesn't happen in this year (doesn't matter whether it's even or odd), then there's only a 50% chance that it will ever happen again.
-
Thanks for the alternate scenario, that's interesting. This is the kind of thing that made me feel like it's hard to guarantee an answer in the first place. – NickC Jan 18 '13 at 21:25
@NickC The hard part is considering completely what it means for an event to not occur. You should look at this recent post, which had a lot of different answers initially. – Calvin Lin Jan 18 '13 at 21:29
In case the mathematical proof already mentioned above isn't enough to convince you, pretend you're a statistician collecting your own results. You have 100 people infected with a disease that has a 95% death rate and a 90% death rate in the first year. Year one is finished, 90 dead, as predicted. 10 people left. We know 95% will be dead at some point. How many more are likely to die if the odds hold true? 5/10. Chances once the first year has passed is 50%. You don't need to be a statistician to see that. It has nothing to do with a finite number of events or an infinite time frame. That part has already been calculated for us in the given 95%.
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The analysis for this appears to be relatively straightforward. Assume the probability of event in year 2+ is $x$. Next, the probability that the event does not occur in the first year is $0.1$ and that it does not occur any time thereafter is $(1-x)$. The product of the two is the probability that the event never happens, which we know to be $0.05$. So,
$0.1 \times (1-x) = 0.05$ | {
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$0.1 \times (1-x) = 0.05$
solving for x yields $x=\frac{1}{2}$
- | {
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crossoverman9b
2022-06-20
Changing base of a logarithm by taking a square root from base?
From my homework I found
${\mathrm{log}}_{9}x={\mathrm{log}}_{3}\sqrt{x}$
and besides that an explanation that to this was done by taking a square root of the base. I fail to grasp this completely. Should I need to turn ${\mathrm{log}}_{9}x$ into base 3, I'd do something like
${\mathrm{log}}_{9}x=\frac{{\mathrm{log}}_{3}x}{{\mathrm{log}}_{3}9}=\frac{{\mathrm{log}}_{3}x}{{\mathrm{log}}_{3}{3}^{2}}=\frac{{\mathrm{log}}_{3}x}{2}$
but this is a far cry from what I've given as being the correct answer.
Substituting some values to x and playing with my calculator I can see that the answer given as correct is correct whereas my attempt fails to yield the correct answer.
Now the question is, what are correct steps to derive ${\mathrm{log}}_{3}\sqrt{x}$ from ${\mathrm{log}}_{9}x$? How and why am I allowed to take a square root of the base and the exponent?
mallol3i
You are only one step away!
Note that $a\mathrm{log}b=\mathrm{log}{b}^{a}$, so
$\frac{{\mathrm{log}}_{3}x}{2}=\frac{1}{2}{\mathrm{log}}_{3}x={\mathrm{log}}_{3}{x}^{1/2}={\mathrm{log}}_{3}\sqrt{x}.$
Yesenia Sherman | {
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# Probability a person chosen at random owns an automobile or a house, but not both
A marketing survey indicates that $60\%$ of the population owns an automobile, $30\%$ owns a house, and $20\%$ owns both an automobile and a house. Calculate the probability that a person chosen at random owns an automobile or a house, but not both.
(A) $0.4$
(B) $0.5$
(C) $0.6$
(D) $0.7$
(E) $0.9$
So the answer given for this question is $0.5$, but it doesn't make sense to me from the perspective of addition rule in probability. If we know that $\Pr(\text{Owning Automobile}) = 0.6$ and the $\Pr(\text{Owning a House}) = 0.3$ and the $\Pr(\text{Owning both an automobile and a house}) = 0.2$, according to the formula probability of $\Pr(A)~\text{or}~\Pr(B) = \Pr(A) + \Pr(B) - \Pr(A~\text{and}~B)$, wouldn't the answer be $0.6+0.3-0.2=0.7$?
Is it because the ordering matters? so the last term $\Pr(A~\text{and}~B) = 2 \cdot 0.2$?
• You computed the probability of "owning an automobile or a house." They asked for the probability of "owning an automobile or a house but not both." – angryavian May 30 '18 at 6:16
• @angryavian I still don't get it, I thought "owning an automobile or a house" is the same as "owning an automobile or a house but not both" ?? – pino231 May 30 '18 at 6:20
• Please type your question rather than posting an image. Images cannot be searched. – N. F. Taussig May 30 '18 at 8:54
• @pino231, in fact "owning an automobile or a house" in this case means "owning an automobile, or a house, or both". This usage of the word "or" -- the so-called "inclusive or" -- is standard in mathematics. If we want the opposite (the exclusive or) it is almost always explicitly stated, as it is in your question. – Mees de Vries May 30 '18 at 9:26
Let event $A$ be owning an automobile. Let event $B$ be owning a house.
Consider the diagram below: | {
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Let event $A$ be owning an automobile. Let event $B$ be owning a house.
Consider the diagram below:
If we simply add the probabilities that a person owns an automobile and a person owns a house, we will have added the probability that a person owns both an automobile and a house twice. Thus, to find the probability that a person owns an automobile or a house, we must subtract the probability that a person owns both an automobile and a house from the sum of the probabilities that a person owns an automobile and that a person owns a house.
$$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$$
Note that on the left-hand side of your equation, you should have written $\Pr(A \cup B)$ or $\Pr(A~\text{or}~B)$ rather than $\Pr(A)~\text{or}~\Pr(B)$.
Since we are given that $60\%$ of the population owns an automobile, $30\%$ of the populations owns a house, and $20\%$ owns both, $\Pr(A) = 0.60$, $\Pr(B) = 0.30$, and $\Pr(A \cap B) = 0.20$. Hence, the probability that a person owns an automobile or a house is $$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) = 0.60 + 0.30 - 0.20 = 0.70$$ However, the question asks for the probability that a person owns an automobile or a house, but not both. That means we must subtract the probability that a person owns an automobile and a house from the probability that the person owns an automobile or a house.
$$\Pr(A~\triangle~B) = \Pr(A \cup B) - \Pr(A \cap B) = 0.70 - 0.20 = 0.50$$
In terms of the Venn diagram, $A \cup B$ is the region enclosed by the two circles, while $A~\triangle~B = (A \cup B) - (A \cap B) = (A - B) \cup (B - A)$ is the region enclosed by the two circles except the region where the sets intersect.
Since $A - B = A - (A \cap B)$, $$\Pr(A - B) = \Pr(A) - \Pr(A \cap B) = 0.60 - 0.20 = 0.40$$
Since $B - A = B - (A \cap B)$, $$\Pr(B - A) = \Pr(B) - \Pr(A \cap B) = 0.30 - 0.20 = 0.10$$
Hence,
$$\Pr(A~\triangle~B) = \Pr(A - B) + \Pr(B - A) = 0.40 + 0.10 = 0.50$$ which agrees with the result obtained above. | {
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• ahh so the question is actually asking the probability that a person chosen at random owns an automobile or a house, OR not both – pino231 Jun 1 '18 at 3:20
• Not quite. The word but should be read as and. The question is asking the probability that a person chosen at random owns an automobile or a house and not both: $\Pr(A~\triangle~B) = \Pr(A \cup B) - \Pr(A \cap B)$. – N. F. Taussig Jun 1 '18 at 9:17
There's no ordering in this question. Using the addition rule is fine, but if you want to use it, note that:
$\mathbb{P}(A) + \mathbb{P}(B) = \Big( \mathbb{P}(A \setminus B) + \mathbb{P}(A \cap B) \Big) + \Big( \mathbb{P}(B \setminus A) + \mathbb{P}(A \cap B) \Big)$.
This should probably help.
10 $\%$ of people don't have a house or automobile so you can leave them. There is 90 $\%$ of people left, also 20$\%$ have both a house and a automobile. This 20% of people is included in the 60% and 30%, so you have to substract this percentage. What you get is 40% + 10% = 50%. So you were probably right by assuming that you have to count the last term $P(A \cap B)$ two times.
• If we subtract $\Pr(A \cup B)$ two times, we will obtain a negative answer. Instead, we must subtract $\Pr(A \cap B)$ twice. – N. F. Taussig May 30 '18 at 9:13
• Yes of course, that's what I meant. Will edit! – WarreG May 30 '18 at 9:37 | {
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Miscellaneous questions: Part I: tutorial practice for preRMO and RMO
Problem 1:
The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.
Problem 2:
Let T be the set of all triples (a,b,c) of integers such that $1 \leq a < b < c \leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.
Problem 3:
In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.
Problem 4:
Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:
A: I see three black and one white cap.
B: I see four white caps.
C: I see one black and three white caps.
D: I see four black caps.
Problem 5:
Let f be a bijective (one-one and onto) function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Note that $f^{M}$ denotes the composite function $f \circ f \circ f \ldots \circ f$ repeated M times.
Problem 6:
Show that there exists a convex hexagon in the plane such that:
a) all its interior angles are equal
b) its sides are 1,2,3,4,5,6 in some order.
Problem 7: | {
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Problem 7:
There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.
Problem 8:
In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?
Problem 9:
Given the seven element set $A = \{ a,b,c,d,e,f,g\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.
Try these !!
Regards,
Nalin Pithwa
Towards Baby Analysis: Part I: INMO, IMO and CMI Entrance
$\bf{Reference: \hspace{0.1in}Introductory \hspace{0.1in} Real Analysis: \hspace{0.1in} Kolmogorov \hspace{0.1in} and \hspace{0.1in} Fomin; \hspace{0.1in}Dover \hspace{0.1in }Publications}$
$\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} Sets \hspace{0.1in} The \hspace{0.1in}Power \hspace{0.1in }of \hspace{0.1in }a \hspace{0.1in}Set}$
$\bf{Section 1}$:
$\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}$ | {
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$\bf{Section 1}$:
$\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}$
The set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be $\it{finite}$.$\it{Clearly \hspace{0.1in} we \hspace{0.1in}can \hspace{0.1in} be \hspace{0.1in} sure \hspace{0.1in} that \hspace{0.1in} a \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in}finite \hspace{0.1in} without \hspace{0.1in} knowing \hspace{0.1in} the \hspace{0.1in} number \hspace{0.1in} of elements \hspace{0.1in}in \hspace{0.1in}it.}$
On the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, $\it{if \hspace{0.1in } we \hspace{0.1in}remove \hspace{0.1in} one \hspace{0.1in} element \hspace{0.1in}from \hspace{0.1in}each \hspace{0.1in}set, \hspace{0.1in}then \hspace{0.1in}remove \hspace{0.1in}two \hspace{0.1in}elements, \hspace{0.1in}three \hspace{0.1in}elements, \hspace{0.1in}and \hspace{0.1in}so \hspace{0.1in}on, \hspace{0.1in}there \hspace{0.1in}will \hspace{0.1in}still \hspace{0.1in}be \hspace{0.1in}elements \hspace{0.1in}left \hspace{0.1in}in \hspace{0.1in}the \hspace{0.1in}set \hspace{0.1in}in \hspace{0.1in}each \hspace{0.1in}stage}$. Accordingly, sets of these kind are called $\it{infinite}$ sets. | {
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Given two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.
To compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a $\it{one-\hspace{0.1in}to-\hspace{0.1in}one \hspace{0.1in}correspondence}$ between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.
$\bf{Section 2}$:
$\bf{Countable \hspace{0.1in} Sets}$. | {
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$\bf{Section 2}$:
$\bf{Countable \hspace{0.1in} Sets}$.
The simplest infinite set is the set $\mathscr{Z^{+}}$ of all positive integers. An infinite set is called $\bf{countable}$ if its elements can be put into one-to-one correspondence with those of $\mathscr{Z^{+}}$. In other words, a countable set is a set whose elements can be numbered $a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots$. By an $\bf{uncountable}$ set we mean, of course, an infinite set which is not countable.
We now give some examples of countable sets:
$\bf{Example 1}$:
The set $\mathscr{Z}$ of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between $\mathscr{Z}$ and $\mathscr{Z^{+}}$ of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \geq 0$ with the odd number $2n+1$, and the negative integer $n<0$ with the even number $2|n|$, that is,
$n \leftrightarrow (2n+1)$, if $n \geq 0$, and $n \in \mathscr{Z}$
$n \leftrightarrow 2|n|$, if $n<0$, and $n \in \mathscr{Z}$
$\bf{Example 2}$:
The set of all positive even numbers is countable, as shown by the obvious correspondence $n \leftrightarrow 2n$.
$\bf{Example 3}$:
The set 2,4,8,$\ldots 2^{n}$ is countable as shown by the obvious correspondence $n \leftrightarrow 2^{n}$. | {
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$\bf{Example 4}: The set$latex \mathscr{Q}$of rational numbers is countable. To see this, we first note that every rational number $\alpha$ can be written as a fraction $\frac{p}{q}$, with $q>0$ with a positive denominator. (Of course, p and q are integers). Call the sum $|p|+q$ as the “height” of the rational number $\alpha$. For example, $\frac{0}{1}=0$ is the only rational number of height zero, $\frac{-1}{1}$, $\frac{1}{1}$ are the only rational numbers of height 2, $\frac{-2}{1}$, $\frac{-1}{2}$, $\frac{1}{2}$, $\frac{2}{1}$ are the only rational numbers of height 3, and so on. We can now arrange all rational numbers in order of increasing “height” (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set $\mathscr{Z^{+}}$ of all positive integers. $\it{Next \hspace{0.1in}we \hspace{0.1in} prove \hspace{0.1in}some \hspace{0.1in}elementary \hspace{0.1in}theorems \hspace{0.1in}involving \hspace{0.1in}countable \hspace{0.1in}sets}$ $\bf{Theorem1}$. $\bf{Every \hspace{0.1in} subset \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}set \hspace{0.1in}is \hspace{0.1in}countable}$. $\bf{Proof}$ Let set A be countable, with elements $a_{1}, a_{2}, a_{3}, \ldots$, and let set B be a subset of A. Among the elements $a_{1}, a_{2}, a_{3}, \ldots$, let $a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \ldots$ be those in the set B. If the set of numbers $n_{1}, n_{2}, n_{3}, \ldots$ has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence $i \leftrightarrow a_{n_{i}}$). $\bf{QED.}$ $\bf{Theorem2}$ $\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a | {
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$\bf{QED.}$ $\bf{Theorem2}$ $\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}finite \hspace{0.1in}or \hspace{0.1in}countable \hspace{0.1in}number \hspace{0.1in}of \hspace{0.1in}countable \hspace{0.1in}sets \hspace{0.1in}A_{1}, A_{2}, A_{3}, \ldots \hspace{0.1in}is \hspace{0.1in}itself \hspace{0.1in}countable.}$ $\bf{Proof}$ We can assume that no two of the sets $A_{1}, A_{2}, A_{3}, \ldots$ have any elements in common, since otherwise we could consider the sets $A_{1}$, $A_{2}-A_{1}$, $A_{3}-(A_{1}\bigcup A_{2})$, $\ldots$, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of $A_{1}, A_{2}, A_{3}, \ldots$ in the form of an infinite table $\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\ldots \\ a_{21} &a_{22} & a_{23} & a_{24} & \ldots \\ a_{31} & a_{32} & a_{33} & a_{34} & \ldots \\ a_{41} & a_{42} & a_{43} & a_{44} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}$ where the elements of the set $A_{1}$ appear in the first row, the elements of the set $A_{2}$ appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose $a_{11}$, then $a_{12}$, then move downwards, diagonally to “left”, picking $a_{21}$, then move down vertically picking up $a_{31}$, then move across towards right picking up $a_{22}$, next pick up $a_{13}$ and so on ($a_{14}, a_{23}, a_{32}, a_{41}$)as per the pattern shown: $\begin{array}{cccccccc} a_{11} & \rightarrow & a_{12} &\hspace{0.1in} & a_{13} & \rightarrow a_{14} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \nearrow & \hspace{0.01in} & \swarrow & \hspace{0.1in} & \hspace{0.1in}\\ a_{21} & \hspace{0.1in} & a_{22} & \hspace{0.1in} & a_{23} \hspace{0.1in} & a_{24} & \ldots \\ \downarrow & \nearrow & \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & | {
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& \hspace{0.1in} & \hspace{0.1in}\\ a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & a_{33} & \hspace{0.1in} & a_{34} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{41} & \hspace{0.1in} & a_{42} &\hspace{0.1in} & a_{43} &\hspace{0.1in} &a_{44} &\ldots\\ \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} \end{array}$ It is clear that this procedure associates a unique number to each element in each of the sets $A_{1}, A_{2}, \ldots$ thereby establishing a one-to-one correspondence between the union of the sets $A_{1}, A_{2}, \ldots$ and the set $\mathscr{Z^{+}}$ of all positive integers. $\bf{QED.}$ $\bf{Theorem3}$ $\bf{Every \hspace{0.1in}infinite \hspace{0.1in}subset \hspace{0.1in}has \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}subset.}$ $\bf{Proof}$ Let M be an infinite set and $a_{1}$ any element of M. Being infinite, M contains an element $a_{2}$ distinct from $a_{1}$, an element $a_{3}$ distinct from both $a_{2}$ and $a_{1}$, and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset $A= \{ a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\}$ of the set $M$. $\bf{QED.}$ $\bf{Remark}$ Theorem 3 shows that countable sets are the “smallest” infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below. $\bf{Section3}$ $\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} sets}$ We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set $\mathscr{Z^{+}}$ of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N. $\bf{Definition}$ Two sets M and N are said to be $\bf{equivalent}$ (written $M \sim N$) if there is a one-to-one correspondence between the elements of M and the elements of | {
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$M \sim N$) if there is a one-to-one correspondence between the elements of M and the elements of N. The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set $\mathscr{Z^{+}}$ of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent. $\bf{Example1}$ The sets of points in any two closed intervals$[a,b]$and$[c,d]\$ are equivalent; you can “see’ a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment “ab” above the base of the triangle; where “a” lies on one side of the triangle and “b” lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect. | {
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$\bf{Example2}$
The set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence $z \leftrightarrow \alpha$ can be established by using stereographic projection. The origin is the North Pole of the sphere.
$\bf{Example3}$
The set of all points x in the open unit interval $(0,1)$ is equivalent to the set of all points y on the whole real line. For example, the formula $y=\frac{1}{\pi}\arctan{x}+\frac{1}{2}$ establishes a one-to-one correspondence between these two sets. $\bf{QED}$.
The last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are “as many” positive integers as integers of arbitrary sign, there are “as many” points in the interval $(0,1)$ as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by:
$\bf{Theorem4}$
$\bf{Every \hspace{0.1in} infinite \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in} equivalent \hspace{0.1in} to \hspace{0.1in}one \hspace{0.1in}of \hspace{0.1in}its \hspace{0.1in}proper \hspace{0.1in}subsets.}$
$\bf{Proof}$
According to Theorem 3, every infinite set M contains a countable subset. Let this subset be $A=\{a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots \}$ and partition A into two countable subsets $A_{1}=\{a_{1}, a_{3}, a_{5}, \ldots \}$ and $A_{2}=\{a_{2}, a_{4}, a_{6}, \ldots \}$.
Obviously, we can establish a one-to-one correspondence between the countable subsets A and $A_{1}$ (merely let $a_{n} \leftrightarrow a_{2n-1}$). This correspondence can be extended to a one-to-one correspondence between the sets $A \bigcup (M-A)=M$ and $A_{1} \bigcup (M-A)=M-A_{2}$ by simply assigning x itself to each element $x \in M-A$. But $M-A_{2}$ is a proper subset of M. $\bf{QED}$.
More later, to be continued,
Regards,
Nalin Pithwa
Find a flaw in this proof: RMO and PRMO tutorial | {
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Regards,
Nalin Pithwa
Find a flaw in this proof: RMO and PRMO tutorial
What ails the following proof that all the elements of a finite set are equal?
The following is the “proof”;
All elements of a set with no elements are equal, so make the induction assumption that any set with n elements has all its elements equal. In a set with n elements, the first and the last n are equal by induction assumption. They overlap at n, so all are equal, completing the induction.
End of “proof:
Regards,
Nalin Pithwa
Problem Solving approach: based on George Polya’s opinion: Useful for RMO/INMO, IITJEE maths preparation
I have prepared the following write-up based on George Polya’s classic reference mentioned below:
UNDERSTANDING THE PROBLEM
First. “You have to understand the problem.”
What is the unknown ? What are the data? What is the condition?
Is it possible to satisfy the condition? Is the condition sufficient to determine the unknown? Or is it insufficient? Or redundant ? Or contradictory?
Draw a figure/diagram. Introduce a suitable notation. Separate the various parts of the condition. Can you write them down?
Second.
DEVISING A PLAN:
Find the connection between the data and the unknown. You may be obliged to consider auxiliary problems if an immediate connection cannot be found. You should eventually obtain a plan for the solution.”
Have you seen it before? Or have you seen the problem in a slightly different form? Do you know a related problem? Do you know a theorem that could be useful? Look at the unknown! And try to think of a familiar problem having the same or a similar unknown. Here is a problem related to yours and solved before. Could you use it? Could you use its result? Could you use its method? Should you restate it differently? Go back to definitions. | {
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If you cannot solve the proposed problem, try to solve some related problem. Could you imagine a more accessible related problem? A more general problem? A more special problem? An analogous problem? Could you solve a part of the problem? Keep only a part of the condition, drop the other part, how far is the unknown then determined, how can it vary? Could you derive something useful from the data? Could you think of other data appropriate to determine the unknown? Could you change the unknown of the data, or both, if necessary, so that the new unknown and the new data are nearer to each other? Did you use all the data? Did you use the whole condition? Have you taken into account all essential notions involved in the problem?
Carrying out your plan of the solution, check each step. Can you clearly see that the step is correct? Can you prove that it is correct?
Fourth. LOOKING BACK.
Examine the solution.
Can you check the result? Can you check the argument? Can you derive the result differently? Can you see it at a glance? Can you see the result, or the method, for some other problem?
**************************************************************************
Reference:
How to Solve It: A New Aspect of Mathematical Method — George Polya.
https://www.amazon.in/How-Solve-Aspect-Mathematical-Method/dp/4871878309/ref=sr_1_1?crid=2DXC1EM1UVCPW&keywords=how+to+solve+it+george+polya&qid=1568334366&s=books&sprefix=How+to+solve%2Caps%2C275&sr=1-1
The above simple “plan” can be useful even to crack problems from a famous classic, Problem Solving Strategies, by Arthur Engel, a widely-used text for training for RMO, INMO and IITJEE Advanced Math also, perhaps.
Reference: Problem-Solving Strategies by Arthur Engel; available on Amazon India
Concept of order in math and real world | {
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Concept of order in math and real world
1. Rise and Shine algorithm: This is crazy-sounding, but quite a perfect example of the need for “order” in the real-world: when we get up in the morning, we first clean our teeth, finish all other ablutions, then go to the bathroom and first we have to remove our pyjamas/pajamas and then the shirt, and then enter the shower; we do not first enter the shower and then remove the pyjamas/shirt !! 🙂
2. On the number line, as we go from left to right: $a, that is any real number to the left of another real number is always “less than” the number to the right. (note that whereas the real numbers form an “ordered field”, the complex numbers are only “partially ordered”…we will continue this further discussion later) .
3. Dictionary order
4. Alphabetical order (the letters $A \hspace{0.1in} B \ldots Z$ in English.
5. Telephone directory order
6. So a service like JustDial certainly uses “order” quite intensely: let us say that you want to find the telephone clinic landline number of Dr Mrs Prasad in Jayanagar 4th Block, Bengaluru : We first narrow JustDial to “Location” (Jayanagar 4th Block, Bengaluru), then narrow to “doctors/surgeons” as the case may be, and then check in alphabetic order, the name of Dr Mrs Prasad. So, we clearly see that the “concept” and “actual implementation” of order (in databases) actually speeds up so much the time to find the exact information we want.
7. So also, in math, we have the concept of ordered pair; in Cartesian geometry, $(a,b)$ means that the first component $a \in X-axis$ and $b \in Y-axis$. This order is generalized to complex numbers in the complex plane or Argand’s diagram.
8. There is “order” in human “relations” also: let us $(x,y)$ represents x (as father) and y (as son). Clearly, the father is “first” and the son is “second”.
9. So, also any “tree” has a “natural order”: seed first, then roots, then branches.
Regards,
Nalin Pithwa. | {
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Regards,
Nalin Pithwa.
Why do we need proofs? In other words, difference between a mathematician, physicist and a layman
Yes, I think it is a very nice question, which kids ask me. Why do we need proofs? Well, here is a detailed explanation (I am not mentioning the reference I use here lest it may intimidate my young enthusiastic, hard working students or readers. In other words, the explanation is not my own; I do not claim credit for this…In other words, I am just sharing what I am reading in the book…)
Here it goes:
What exactly is the difference between a mathematician, a physicist, and a layman? Let us suppose that they all start measuring the angles of hundreds of triangles of various shapes, find the sum in each case and keep a record. Suppose the layman finds that with one or two exceptions, the sum in each case comes out to be 180 degrees. He will ignore the exceptions and say “the sum of the three angles in a triangle is 180 degrees.” A physicist will be more cautious in dealing with the exceptional cases. He will examine them more carefully. If he finds that the sum in them is somewhere between 179 degrees to 180 degrees, say, then he will attribute the deviation to experimental errors. He will then state a law: The sum of three angles of any triangle is 180 degrees. He will then watch happily as the rest of the world puts his law to test and finds that it holds good in thousands of different cases, until somebody comes up with a triangle in which the law fails miserably. The physicist now has to withdraw his law altogether or else to replace it by some other law which holds good in all cases tried. Even this new law may have to be modified at a later date. And, this will continue without end. | {
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A mathematician will be the fussiest of all. If there is even a single exception he will refrain from saying anything. Even when millions of triangles are tried without a single exception, he will not state it as a theorem that the sum of the three angles in ANY triangle is 180 degrees. The reason is that there are infinitely many different types of triangles. To generalize from a million to infinity is as baseless to a mathematician as to generalize from one to a million. He will at the most make a conjecture and say that there is a strong evidence suggesting that the conjecture is true. But that is not the same thing as a proving a theorem. The only proof acceptable to a mathematician is the one which follows from earlier theorems by sheer logical implications (that is, statements of the form : If P, then Q). For example, such a proof follows easily from the theorem that an external angle of a triangle is the sum of the other two internal angles. | {
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The approach taken by the layman or the physicist is known as the inductive approach whereas the mathematician’s approach is called the deductive approach. In the former, we make a few observations and generalize. In the latter, we deduce from something which is already proven. Of course, a question can be raised as to on what basis this supporting theorem is proved. The answer will be some other theorem. But then the same question can be asked about the other theorem. Eventually, a stage is reached where a certain statement cannot be proved from any other earlier proved statement(s) and must, therefore, be taken for granted to be true. Such a statement is known as an axiom or a postulate. Each branch of math has its own axioms or postulates. For examples, one of the axioms of geometry is that through two distinct points, there passes exactly one line. The whole beautiful structure of geometry is based on 5 or 6 axioms such as this one. Every theorem in plane geometry or Euclid’s Geometry can be ultimately deduced from these axioms.
PS: One of the most famous American presidents, Abraham Lincoln had read, understood and solved all of Euclid’s books (The Elements) by burning mid-night oil, night after night, to “sharpen his mental faculties”. And, of course, there is another famous story (true story) of how Albert Einstein as a very young boy got completely “addicted” to math by reading Euclid’s proof of why three medians of a triangle are concurrent…(you can Google up, of course).
Regards,
Nalin Pithwa | {
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Rhianna has learned the SSS and SAS congruence tests for triangles and she wonders if these tests might work for parallelograms. Diagonals of a Parallelogram Bisect Each Other. A tip from Math Bits says, if we can show that one set of opposite sides are both parallel and congruent, which in turn indicates that the polygon is a parallelogram, this will save time when working a proof. parallelogram, because opposite sides are congruent and adjacent sides are not perpendicular. Theorem 1 : If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Also interesting in this case is that to the eye The first is to use congruent triangles to show the corresponding angles are congruent, the other is to use theAlternate Interior Angles Theoremand apply it twice. Congruent trianglesare triangles that have the same size and shape. So we’re going to put on our thinking caps, and use our detective skills, as we set out to prove (show) that a quadrilateral is a parallelogram. Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher). When we think of parallelograms, we usually think of something like this. Note that the vertex $D$ is obtained by rotating $B$ 180 degrees about the midpoint $M$ of $\overline{AC}$. vidDefer[i].setAttribute('src',vidDefer[i].getAttribute('data-src')); We can look at what happens in the special case where all 4 sides of both $ABCD$ and $EFGH$ are congruent to one another. Theorem 6.2.1 If a quadrilateral is a parallelogram, then the two pairs of opposite sides are congruent. Each theorem has an example that will show you how to use it in order to prove the given figure. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sid… 2. A quadrilateral that has opposite sides equal and parallel and the opposite angles are also equal is called a parallelogram. One Pair of Opposite Sides are Both Parallel and | {
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angles are also equal is called a parallelogram. One Pair of Opposite Sides are Both Parallel and Congruent, Consecutive Angles in a Parallelogram are Supplementary. yes, one pair of sides are congruent and parallel . Both pairs of opposite sides are congruent. Triangle congruence criteria have been part of the geometry curriculum for centuries. If both pairs of opposite angles of a quadrilateral are congruent, then it’s a parallelogram (converse of a property). Just as with a triangle it takes three pieces of information (ASA, SAS, or SSS) to determine a shape, so with a quadrilateral we would expect to require four pieces of information. THEOREM:If a quadrilateral has 2 sets of opposite sides congruent, then it is a parallelogram. They are called the SSS rule, SAS rule, ASA rule and AAS rule. function init() { Opposite Sides Parallel and Congruent & Opposite Angles Congruent. So what are we waiting for. B) The diagonals of the parallelogram are congruent. yes, diagonals bisect each other. Take Calcworkshop for a spin with our FREE limits course. Since ABCD is a parallelogram, segment AB ≅ segment DC because opposite sides of a parallelogram are congruent. For quadrilaterals, on the other hand, these nice tests seem to be lacking. 2 Looking at a special case for part (a): the rhombus. In order to see what happens with the parallelograms $ABCD$ and $EFGH$ we focus first on $ABCD$. If the quadrilateral has two pairs of opposite, congruent sides, it is a parallelogram. Well, if a parallelogram has congruent diagonals, you know that it is a rectangle. Attribution-NonCommercial-ShareAlike 4.0 International License. Solution: It turns out that knowing all four sides of two quadrilaterals are congruent is not enough to conclude that the quadrilaterals are congruent. Prove that the figure is a parallelogram. Finally, you’ll learn how to complete the associated 2 column-proofs. If one angle is 90 degrees, then all other angles are also 90 degrees. We can tell whether two | {
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If one angle is 90 degrees, then all other angles are also 90 degrees. We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. if(vidDefer[i].getAttribute('data-src')) { If the diagonals of a quadrilateral bisect each other, then it’s a parallelogram (converse of a property). Suppose $ABCD$ and $EFGH$ are two parallelograms with a pair of congruent corresponding sides, $|AB| = |EF|$ and $|BC| = |FG|$. We begin by drawing or building a parallelogram. If a parallelogram has perpendicular diagonals, you know it is a rhombus. We might find that the information provided will indicate that the diagonals of the quadrilateral bisect each other. Theorems. The diagonal of a parallelogram separates it into two congruent triangles. side $\overline{EH}$ does not appear to the eye to be congruent to side $\overline{AD}$: this could be an optical illusion or it could be that the eye is distracted by the difference in area. SURVEY . Here is what we need to prove: segment AB ≅ segment CD and segment BC ≅ AD. Draw the diagonal BD, and we will show that ΔABD and ΔCDB are congruent. Thus it provides a good opportunity for students to engage in MP3 ''Construct Viable Arguments and Critique the Reasoning of Others.'' When a parallelogram is divided into two triangles we get to see that the angles across the common side( here the diagonal) are equal. If the quadrilateral has one set of opposite parallel, congruent sides, it is a parallelogram. Both of these facts allow us to prove that the figure is indeed a parallelogram. Complete the two-column proof Given: triangle SVX is congruent to triangle UTX and Line SV is || to line TU Prove: VUTS is a parallelogram Image: It's a parallelogram, with one line going from corner S to corner U and a line going . This task addresses this issue for a specific class of quadrilaterals, namely parallelograms. Well, we must show one of the six basic properties of parallelograms to be true! | {
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parallelograms. Well, we must show one of the six basic properties of parallelograms to be true! Find missing values of a given parallelogram. We will learn about the important theorems related to parallelograms and understand their proofs. We all know that a parallelogram is a convex polygon with 4 edges and 4 vertices. Solution: More generally, a quadrilateral with 4 congruent sides is a rhombus. Which of the following cannot be used to prove a shape is a parallelogram? yes,opposite sides are congruent. First prove ABC is congruent to CDA, and then state AD and BC are corresponding sides of the triangles. Triangles can be used to prove this rule about the opposite sides. Creative Commons The opposite sides of a parallelogram are congruent. In this lesson, we will consider the four rules to prove triangle congruence. Note that a rhombus is determined by one side length and a single angle: the given side length determines all four side lengths and For example, for squares one side is enough, for rectangles two adjacent sides are sufficient. Engage your students with effective distance learning resources. This means that the corresponding sides are equal and the corresponding angles are equal. Let’s begin! A description of how to do a parallelogram congruent triangles proof. In this mini-lesson, we will explore the world of parallelograms and their properties. The only parallelogram that satisfies that description is a square. Parallelogram and Congruent triangles Parallelogram. $\triangle ABC$. If … Another approach might involve showing that the opposite angles of a quadrilateral are congruent or that the consecutive angles of a quadrilateral are supplementary. In this section, you will learn how to prove that a quadrilateral is a parallelogram. This proves that the opposite angles in a parallelogram are also equal. What about for arbitrary quadrilaterals? Which statement explains how you could use coordinate geometry to prove the diagonals of a quadrilateral are | {
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explains how you could use coordinate geometry to prove the diagonals of a quadrilateral are perpendicular? More specifically, how do we prove a quadrilateral is a parallelogram? Walking trails run from points A to C and from points B to D. Here are the theorems that will help you prove that the quadrilateral is a parallelogram. In today’s geometry lesson, you’re going to learn the 6 ways to prove a parallelogram. for (var i=0; i | {
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# What is the distribution of a random variable with pdf proportional to the product of two normal pdf's
I'm working my way through a nice e-book on Kalman filters, translating to Mathematica as I go and I've hit an interesting problem. In this section: https://nbviewer.jupyter.org/github/rlabbe/Kalman-and-Bayesian-Filters-in-Python/blob/master/03-Gaussians.ipynb#Computational-Properties-of-Gaussians - the author gives a formula to compute a product of Gaussian PDFs that is also Gaussian. Googling around will tell you that the product of two Gaussian PDFs is not itself Gaussian, but is proportional to a Gaussian with mean and standard deviation as in the image below (see proof here: http://www.tina-vision.net/docs/memos/2003-003.pdf for source) -- I have implemented his function and it works fine, but I'm curious if TransformedDistribution can be used to arrive at the same distribution.
To make it concrete
d1 = NormalDistribution[m1, s1];
d2 = NormalDistribution[m2, s2];
(* Works perfectly *)
TransformedDistribution[
x + y, {x \[Distributed] d1, y \[Distributed] d2}]
(* Not a Normally Distributed *)
TransformedDistribution[
x*y, {x \[Distributed] d1, y \[Distributed] d2}]
(* This doesn't work, but I'm wondering if there is something like \
this that would produce the normal distribution cited in the question \
*)
TransformedDistribution[
Normalize[x*y, Total], {x \[Distributed] d1, y \[Distributed] d2}]
Edit to add example multiplication using the proposed function. | {
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Edit to add example multiplication using the proposed function.
MultiplyGaussian[g1_, g2_] :=
Module[{mean1, var1, mean2, var2, mean, variance},
{mean1, var1} = g1 /. NormalDistribution[m_, s_] :> {m, s^2};
{mean2, var2} = g2 /. NormalDistribution[m_, s_] :> {m, s^2};
mean = (var1*mean2 + var2*mean1) / (var1 + var2);
variance = (var1 * var2) / (var1 + var2);
NormalDistribution[mean, Sqrt[variance]]
]
z1 = NormalDistribution[3, 0.7];
z2 = NormalDistribution[4.5, 1];
Plot[{Legended[PDF[z1, x], "N(3,0.7)"],
Legended[PDF[z2, x], "N(4.5,2)"] ,
Legended[PDF[MultiplyGaussian[z1, z2], x], "Product"]}, {x, 1, 10}]
• I'm not sure your source is right: mathworld.wolfram.com/NormalProductDistribution.html Sep 5 '19 at 23:13
• Edited to add code showing that the proposal seems to work and clarify that the 2nd link is a proof that I at least couldn't find a problem with after a quick look.
– Dan
Sep 5 '19 at 23:34
• I think my terminology is sloppy and that has caused the problem (as pointed out by this MathOverflow answer: math.stackexchange.com/questions/101062/…). The result here is for the product of PDFs (where I sloppily said random variable). Does this clarification help see a way to get there in Mathematica?
– Dan
Sep 5 '19 at 23:41
• Please clean-up the "rv" vs "pdf" confusion in the text and the title. Your title should probably be something like "What is the distribution of a random variable with pdf proportional to the product of two normal pdf's?"
– JimB
Sep 6 '19 at 0:40
(* Get product of two normal pdf's *)
prod= PDF[NormalDistribution[μ1, σ1], x]*PDF[NormalDistribution[μ2, σ2], x];
(* Normalize so that the pdf integrates to 1 *)
pdf = prod/Integrate[prod, {x, -∞, ∞}, Assumptions -> {σ1 > 0, σ2 > 0}];
(* Construct associated distribution *)
d = ProbabilityDistribution[pdf, {x, -∞, ∞}, Assumptions -> {σ1 > 0, σ2 > 0}];
(* Find mean and variance *)
Mean[d]
(* (μ2 σ1^2+μ1 σ2^2)/(σ1^2+σ2^2) *)
Variance[d]
(* (σ1^2 σ2^2)/(σ1^2+σ2^2) *) | {
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Variance[d]
(* (σ1^2 σ2^2)/(σ1^2+σ2^2) *)
This matches what the article says the mean and variance should be. But is it a normal distribution? If the moment generating function is of the same form as for a normal distribution, then it has a normal distribution. (We could also use the characteristic function to do this for this particular distribution.)
(* The log of the moment generating function will be in the following form *)
logCF = Expectation[Exp[t z], z \[Distributed] NormalDistribution[μ, σ]] /.
Power[E, x_] -> x // Expand
(* t μ+(t^2 σ^2)/2 *)
(* So we look to see if the moment generating function of distribution d is of the same form *)
Collect[Expectation[Exp[t z], z \[Distributed] d] /. Power[E, x_] -> x // Expand, t]
(* (t^2 σ1^2 σ2^2)/(2 (σ1^2+σ2^2))+t ((μ2 σ1^2)/(σ1^2+σ2^2)+(μ1 σ2^2)/(σ1^2+σ2^2)) *)
And it is.
d1 = NormalDistribution[m1, s1];
d2 = NormalDistribution[m2, s2];
These distributions require that
assume = And @@
(DistributionParameterAssumptions /@ {d1, d2})
(* m1 ∈ Reals && s1 > 0 && m2 ∈ Reals && s2 > 0 *)
As pointed out by @JimB, the PDF formed by the product of the normal PDFs is
PDFprod = Assuming[assume, PDF[d1, x]*PDF[d2, x]/
Integrate[PDF[d1, x]*PDF[d2, x],
{x, -Infinity, Infinity}] // Simplify]
(* (E^(-((m2 s1^2 + m1 s2^2 - (s1^2 + s2^2) x)^2/(
2 s1^2 s2^2 (s1^2 + s2^2)))) Sqrt[s1^2 + s2^2])/(Sqrt[2 π] s1 s2) *)
Comparing with the PDF of the expected normal distribution
PDFprod == Assuming[assume, PDF[NormalDistribution[
(m1*s2^2 + m2*s1^2)/(s1^2 + s2^2),
Sqrt[s1^2*s2^2/(s1^2 + s2^2)]], x] // Simplify]
(* True *) | {
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# Open maps which are not continuous
What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\mathbb{C}$ to $\mathbb{C}$ (filled in to be $0$ at $0$). There must be a simpler example, using the usual Euclidean topology, right?
• Since $(0,1)$ and $\mathbb R$ are homeomorphic via a linear map composed with $\arctan$, it suffices to find a map $\mathbb R \to \mathbb R$ that is open but not continuous. Googling that gives you mathforum.org/library/drmath/view/62395.html – lhf Oct 25 '11 at 0:55
• this is obviously not much help, but if you can find a continuous bijection $f$ with discontinuous inverse, then $f^{-1}$ will do. – user12014 Oct 25 '11 at 1:13
• One can build such a function from a Cantor set $C$ (the usual "middle thirds" set will do). Send each point in $C$ to $0$, and map each connected component of the complement of $C$ homeomorphically to the interval $(-1,1)$. Then the image of any open set intersecting $C$ will be $(-1,1)$ (thus open), and the image of any open set not meeting $C$ will also be open, since it's a union of homeomorphic images of open sets. Of course, each point of $C$ will be a discontinuity. – user83827 Oct 25 '11 at 1:16
• @PZZ for instance the map wrapping [0,1) around the unit circle. – JSchlather Oct 25 '11 at 1:37
• @PZZ: In fact there are no counterexamples of the type you're suggesting: if $I$ and $J$ are intervals in $\mathbb{R}$ and $f: I \rightarrow J$ is a continuous bijection, then $f^{-1}$ is necessarily continuous. By coincidence this is exactly where I am in my Spivak calculus course, so see e.g. Theorem 37 in $\S 6.4$ of math.uga.edu/~pete/2400calc2.pdf. (Or see Spivak's text!) – Pete L. Clark Oct 25 '11 at 3:18 | {
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Explicit examples are moderately difficult to construct, but it’s not too hard to come up with non-constructive examples; here’s one such.
For $x,y\in\mathbb{R}$ define $x\sim y$ iff $x-y\in \mathbb{Q}$; it’s easy to check that $\sim$ is an equivalence relation on $\mathbb{R}$. For any $x\in\mathbb{R}$, $[x] = \{x+q:q\in\mathbb{Q}\}$, where $[x]$ is the $\sim$-equivalence class of $x$. In particular, each equivalence class is countable. For any infinite cardinal $\kappa$, the union of $\kappa$ pairwise disjoint countably infinite sets has cardinality $\kappa$, so there must be exactly as many equivalence classes as there are real numbers. Let $h$ be a bijection from $\mathbb{R}/\sim$, the set of equivalence classes, to $\mathbb{R}$. Finally, define $$f:(0,1)\to\mathbb{R}:x\mapsto h([x])\;.$$
I claim that if $V$ is any non-empty open subset of $(0,1)$, $f[V]=\mathbb{R}$, which of course ensures that $f$ is open. To see this, just observe that every open interval in $(0,1)$ intersects every equivalence class. (It should be no trouble at all to see that $f$ is wildly discontinuous!)
• Just curious: Is the axiom of choice used anywhere in your proof? – YoTengoUnLCD Jan 17 '17 at 8:14
• I think I'm going to start calling $\sim$ the "Vitali equivalence relation"... $x$ and $y$ are Vitali equivalent iff $x-y \in \mathbb{Q}$, etc. Honestly, this thing is useful enough to deserve a name. – goblin Mar 2 '17 at 13:54
• is $f$ injective? – David Feng Feb 16 at 21:37
• @DavidFeng: No. All $x$ from the same equivalence class give the same value. For example, $f(\frac12)=f(\frac13)$ since $\frac12-\frac13\in\mathbb Q$ – celtschk Mar 2 at 21:55
Let me conceptualize around Brian's answer a bit.
Definition 0. If $X$ and $Y$ are topological spaces, a function $f:X→Y$ is said to be strongly Darboux iff for all non-empty open sets $A⊆X$, we have $f(A)=Y$.
Here's the basic facts:
Proposition. | {
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Here's the basic facts:
Proposition.
1. Every strongly Darboux function is an open function.
2. If $X$ is non-empty, every Darboux function $X \rightarrow Y$ is surjective.
3. If $X$ is non-empty and $f : X \rightarrow Y$ is a continuous Darboux mapping, then $Y$ carries the indiscrete topology.
Proofs.
1. Trivial.
2. Since $X$ is open and non-empty, hence $f(X)=Y.$ That is, $f$ is surjective.
3. Let $B \subseteq Y$ denote a non-empty open set. Our goal is to show that $B=Y$. Since $f$ is surjective, $f^{-1}(B)$ is non-empty. Since $f$ is continuous, $f^{-1}(B)$ is open. Hence $f(f^{-1}(B))=Y$. But since $f$ is surjecive, hence $f(f^{-1}(B))=B.$ So $B=Y$.
Putting these together, we see that every strongly Darboux function $f:\mathbb{R} \rightarrow \mathbb{R}$ is a discontinuous open mapping.
• $f$ is an open mapping by (1).
• $f$ is discontinuous by (3), because it's domain is non-empty and it's codomain doesn't carry the indiscrete topology.
And, of course, Brian's answer guarantees the existence of a strongly Darboux function $\mathbb{R} \rightarrow \mathbb{R}$. This completes the proof.
There is in fact a rather easy example of a function $$\mathbb R \to \mathbb R$$ such that the image of every open set is $$\mathbb R$$: Let $$(x_i)_{i\in\mathbb Z_+}$$ be the binary decimal expansion of $$x$$, so that each $$x_i \in \{0,1\}$$. Let then $$f(x) = \sum_{k=1}^\infty\frac{(-1)^{x_k}}k\quad \textrm{if the series converges}$$ $$f(x) = 0\quad \textrm{otherwise.}$$ Since the harmonic series (or a tail of it) can be made to converge to any real number by changing signs in the appropriate way, this function has $$f((a,b)) = \mathbb R$$ for any real $$a,b$$. Hence this function is open, though clearly not continuous at any point.
The harmonic series can be substituted with any other unbounded series where the summand goes to zero. | {
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CoCalc Public Filesweek 5 / assignment / FindRoots.ipynb
Authors: Boyuan Qi, Rajvir Sidhu
Views : 31
Compute Environment: Ubuntu 20.04 (Default)
# Assignment 2: Polynomial Root Finding
## Markdown
Give the address (url) for a website that has documentation for markdown, including tables (i.e. a webpage, not a pdf or other document, that describes how you use markdown to create a table). Enter your answer as a variable named website='http://url.goes.here'
Marks: 1
In [ ]:
# YOUR CODE HERE
markdownWebsite='https://www.markdownguide.org/cheat-sheet'
Test your code from above here(1 point), ID: validate_website
In [ ]:
#test that it's a suitable website link you've given
Work out the solutions to the following questions long-hand (i.e. you should not be coding anything at this point). In the first box, set the appropriate variable equal to your answer (there's no need to define any functions). For example, to answer the last question, you will simply type polyRoots=[4-math.sqrt(3),4,4+math.sqrt(3)] | {
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1. Define the polynomial $p(x)=x^3-12x^2+45x-52$. How do you express $p(x)$ as a list, such as will be input to our algorithms? Set the variable polyX equal to this.
2. What is the derivative, $p'(x)$ (as a list called polyDeriv) and the root(s) of $p'(x)$ (as a list, in increasing order, called derivRoots)?
3. What is the Lagrange bound as applied to the polynomial $p(x)$? Your answer should be a real, positive number, called lagrange. (In case you missed it, this was an exercise in the workshops)
4. Pretend that you don't know the roots of $p(x)$, but you know the roots of $p'(x)$. Use this information to explain why we know that there is no more than one root in each of the ranges: $(-109,3),(3,5),(5,109)$ You should do this without evaluating $p(x)$ at any point. Write your solution in the free text box below (this will be manually graded).
5. What are the roots of $p(x)$? Give your answer as a list in increasing order called polyRoots. Make sure that all the roots of $p(x)$ appear in exactly one of the above ranges.
Marks: 5
In [ ]:
import math
polyX=[-52,45,-12,1]
polyDeriv=[45,-24,3]
derivRoots=[3,5]
lagrange=109
polyRoots=[4-math.sqrt(3),4,4+math.sqrt(3)]
Test your code from above here(1 point), ID: polyRoots_correct
In [ ]:
#the question told you how to answer this, so no point in hiding the test
assert polyRoots==[4-math.sqrt(3),4,4+math.sqrt(3)]
Test your code from above here(1 point), ID: polyX_correct
In [ ]:
#check that polyX is correct
Test your code from above here(1 point), ID: polyDeriv_correct
In [ ]:
#check that polyDeriv is correct
Test your code from above here(1 point), ID: derivRoots_correct
In [ ]:
#check that derivRoots is correct
Test your code from above here(1 point), ID: lagrange_bound_correct
In [ ]:
#check that lagrange is correct | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
This is where you give your answer to (4) above: Explain why we know that there is no more than one root in each of the ranges: $(-109,3),(3,5),(5,109)$ You should do this without evaluating $p(x)$ at any point.
Marks: 2
Given any two roots b>a of p(x), f is continuous and differentiable on [a,b]. Using Rolle's theorem, there exists exactly one root, 'k', such that f'(k)=0 and a<c<b. Due to p'(x) having 2 roots, p(x) therefore has at most 3 roots. The roots of p'(x) are 3 and 5 which means p(x) must have a root within the interval [3,5]. Also, the limits of the function are 109 and -109, so if there are 3 roots, then exactly one root must exist in the each of the intervals [-109,3] and [5,109]. This means there that no more than one root can exist in each of the intervals [-109,3],[3,5] and [5,109]
## Collecting our work so far
In working through the lab sessions, we have built up a library of subroutines that will be useful for finding the roots of a polynomial. Those that we have already produced are given first.
In [ ]:
from math import sqrt | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
def EvaluatePoly(polynomial,x):
'''evaluate a polynomial at a value x
input: polynomial as a list, value x to evaluate at
output: value of polynomial'''
return(sum(val*x**n for n,val in enumerate(polynomial)))
def DifferentiatePoly(polynomial):
'''return the derivative of polynomial'''
return [(i+1)*val for i,val in enumerate(polynomial[1:])]
def EnsureStandardForm(polynomial):
'''make sure that the highest order of the polynomial has non-zero coefficient by removing all higher order zeros'''
while not polynomial[-1]:
del polynomial[-1]
return polynomial
def DescartesSigns(polynomial):
'''Descartes rules of signs returns an upper bound on the number of positive roots and the number of negative roots'''
PosList=list(polynomial)
NegList=[((-1)**i)*val for i,val in enumerate(polynomial)]
while 0 in PosList:
PosList.remove(0)
NegList.remove(0)
return([sum([i[0]*i[1]<0 for i in zip(PosList[1:],PosList[:-1])]),sum([i[0]*i[1]<0 for i in zip(NegList[1:],NegList[:-1])])])
def FindZeroInInterval(polynomial,range_lower,range_upper):
'''perform an interval bisection on polynomial between the values range_lower<range_upper
return x such that polynomial(x)=0 (or a good enough approximation to it)'''
#this is slightly modified compared to what we presented.
assert range_lower<=range_upper,"second argument should be smaller than first"
accuracy=10**(-15)
value_lower=EvaluatePoly(polynomial,range_lower)
value_upper=EvaluatePoly(polynomial,range_upper)
if value_lower==0: # an exact root on the lower boundary
return(range_lower)
elif abs(value_upper/value_lower)<accuracy and abs(value_upper)<accuracy: #define this as being close enough to being a root on the upper boundary
return(range_upper)
elif abs(value_lower/value_upper)<accuracy and abs(value_lower)<accuracy: #close enough to root on lower boundary
return(range_lower)
if (range_lower==range_upper):#we already know it's not a root.
return('') | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
return(range_lower)
if (range_lower==range_upper):#we already know it's not a root.
return('')
extent=2*(range_upper-range_lower) #initialise with a dummy variable larger than anything that will ver be calculated | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
while range_upper-range_lower>max(abs(range_upper),abs(range_lower),1)*accuracy and extent>(range_upper-range_lower):#stopping criteria based on numerical accuracy
if value_upper*value_lower>0: #we don't seem to have a root in this range
return('')
range_mid=(range_upper+range_lower)/2 #bisection
value_mid=EvaluatePoly(polynomial,range_mid)
extent=range_upper-range_lower
if value_mid==0: #we have found the root
return range_mid
elif value_mid*value_lower>0: #the root is between value_mid and value_upper. redefine range and repeat
range_lower=range_mid
value_lower=value_mid
else:
range_upper=range_mid
value_upper=value_mid
return range_mid
return range_mid
def Smooth(mylist):
'''remove all the empty entries in a list'''
assert len(mylist)>0,'Should not smooth an empty list'
while '' in mylist:
mylist.remove('')
return mylist
def ZeroTest(mylist):
"fudge the fact that it's hard to tell if a very small number is actually 0"
return([(not abs(a)<10**(-14))*a for a in mylist])
def RootLocationBound(polynomial):
'use a Lagrange bound to crudely limit where the roots can be found'
return max(1,sum(abs(a/polynomial[-1]) for a in polynomial[:-1]))
def QuickTests():
#tests for the above code
assert EnsureStandardForm([1,1,1])==[1,1,1]
assert EnsureStandardForm([1,1,1,0,0,0])==[1,1,1]
assert [EvaluatePoly([24,-50,35,-10,1],i)-(24-50*i+35*i*i-10*i**3+i**4) for i in range(-5,5,1)]==[0,0,0,0,0,0,0,0,0,0]
assert DifferentiatePoly([24,-50,35,-10,1])==[-50,70,-30,4]
assert ZeroTest([-1,0.1,10**(-15),2])==[-1,0.1,0,2]
assert DescartesSigns([-1,-1,1,1])==[1,2]
assert DescartesSigns([-1,0,0,1])==[1,0]
return True
assert QuickTests() | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
assert QuickTests()
Let's say you were given a polynomial as a list polynomial and had found the roots of its derivative, as an ordered list: deriv_roots. Write a function root_bounds that returns a list of pairs of numbers between which there is no more than one root of the polynomial. The list should be ordered, going from smallest to largest, and the pairs of values should also be ordered smallest to largest.
Hint: If you did the free text box above (question 4) clearly and correctly, this should be straightforward.
Hint: Look at the first test to understand the input and output format.
Marks: 4
In [10]:
def root_bounds(deriv_roots,upper):
'''return the pairs of numbers which bound individual roots (or none at all)
in: deriv_roots: a list of roots of the deriviative of a polynomial, in increasing order, repeated according to their multiplicity
in: upper. if x is a root of polynomial, |x|<=upper
'''
assert upper>=0
assert all([val1<=val2 for val1,val2 in zip (deriv_roots[:-1],deriv_roots[1:])])
assert isinstance (deriv_roots,list)
assert len(deriv_roots)>0 # This is needed to actually apply Rolle's Theorem
emptyList=[] # A list which is used later on in the interation
if deriv_roots[-1]>upper:
for x in range(len(deriv_roots)-1):
listAdd=[deriv_roots[x],deriv_roots[x+1]] # A nested listed will be added to the interval that contains the root
if deriv_roots[-1]<=upper:
lower=[-upper]
high=[upper]
z=list(deriv_roots)
bigList=lower+z+high # The completed list which has the upper and lower bounds of the function
for x in range(len(bigList)-1):
listAdd=[bigList[x],bigList[x+1]] # A nested listed will be added to the interval that contains the root
return emptyList
print(root_bounds([-1,0,0,1],5)) | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
print(root_bounds([-1,0,0,1],5))
[[-5, -1], [-1, 0], [0, 0], [0, 1], [1, 5]]
Test your code from above here(2 points), ID: root_bounds_code_visible
In [ ]:
#test the root_bounds function
assert list(root_bounds([-1,0,0,1],5))==[[-5,-1],[-1,0],[0,0],[0,1],[1,5]] or list(root_bounds([-1,0,0,1],5))==[(-5,-1),(-1,0),(0,0),(0,1),(1,5)]
Test your code from above here(1 point), ID: root_bounds_with_float
In [ ]:
#a basic test
Test your code from above here(1 point), ID: root_bounds_preconditions
In [ ]:
#check some preconditions
The function FindZeroInInterval (above) is a little different to the code that we presented in the notes (given as OriginalFindZeroInInterval in cell below). Aside from the assert statement, how is it different? (explanation not required.)
To demonstrate the difference, give a set of parameters as input that gives a different outcome between the two functions. Give your answer as 3 variables poly, below and above.
Marks: 1
In [ ]:
# YOUR CODE HERE
poly=[3,1,0,0]
above=0
below=-3
Test your code from above here(2 points), ID: difference_FindZero
In [ ]:
def OriginalFindZeroInInterval ( polynomial , range_lower , range_upper ):
''' perform an interval bisection on polynomial between the values
range_lower < range_upper
return x such that polynomial (x)=0 (or a good enough approximation to it)
'''
value_lower = EvaluatePoly ( polynomial , range_lower )
value_upper = EvaluatePoly ( polynomial , range_upper ) | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
while range_upper - range_lower >10**( -15) :
if value_upper * value_lower >0: #we don 't seem to have a root in this range
return('')
range_mid =( range_upper + range_lower )/2 # bisection
value_mid = EvaluatePoly ( polynomial , range_mid )
if value_mid ==0: #we have found the 0, so return itsposition
return range_mid
elif value_mid * value_lower >0: # crossing between range_mid andrange_upper
range_lower = range_mid
value_lower = value_mid
else : # crossing between range_lower and
range_mid
range_upper = range_mid
value_upper = value_mid
return range_mid
assert above>below
assert FindZeroInInterval(poly,below,above)!=OriginalFindZeroInInterval(poly,below,above)
## The Final Root Finder
Using the previous functions (and defining any additional functions that wish in the cell below), complete the function FindRealZeros which accepts a polynomial as a list and returns a list of the roots of the polynomial, in increasing order, appearing as many times as they are repeated.
You might structure your algorithm by considering:
• How did you work out the solution by hand?
• What bits of code do you have to hand that could replace some of the by-hand calculation.
• Are there any simple types of polynomial for which you can immediately give the root?
• Is there a minimum size of polynomial that has a solution?
• Can you give a set of ranges between which no more than 1 root of the polynomial lies, given the roots of the derivative?
• For a polynomial of a given degree, $n$, what's the maximum number of ranges there could be? How many real roots could there be for a degree $n$ polynomial?
You need to make use of the functions we've already defined, particularly FindZeroInInterval, DifferentiatePoly and RootLocationBound
In [ ]:
def EvaluatePoly(polynomial,x):
'''evaluate a polynomial at a value x
input: polynomial as a list, value x to evaluate at
output: value of polynomial'''
return(sum(val*x**n for n,val in enumerate(polynomial))) | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
def RootLocationBound(polynomial):
'use a Lagrange bound to crudely limit where the roots can be found'
return max(1,sum(abs(a/polynomial[-1]) for a in polynomial[:-1]))
def root_bounds(deriv_roots,upper):
'''return the pairs of numbers which bound individual roots (or none at all)
in: deriv_roots: a list of roots of the deriviative of a polynomial, in increasing order, repeated according to their multiplicity
in: upper. if x is a root of polynomial, |x|<=upper
'''
assert upper>=0
assert all([val1<=val2 for val1,val2 in zip (deriv_roots[:-1],deriv_roots[1:])])
assert isinstance (deriv_roots,list)
assert len(deriv_roots)>0 # This is needed to actually apply Rolle's Theorem
emptyList=[] # A list which is used later on in the interation
if deriv_roots[-1]>upper:
for x in range(len(deriv_roots)-1):
listAdd=[deriv_roots[x],deriv_roots[x+1]] # A nested listed will be added to the interval that contains the root
if deriv_roots[-1]<=upper:
lower=[-upper]
high=[upper]
z=list(deriv_roots)
bigList=lower+z+high # The completed list which has the upper and lower bounds of the function
for x in range(len(bigList)-1):
listAdd=[bigList[x],bigList[x+1]] # A nested listed will be added to the interval that contains the root
return emptyList
print(root_bounds([-1,0,0,1],5))
def EvaluatePoly(polynomial,x):
'''evaluate a polynomial at a value x
input: polynomial as a list, value x to evaluate at
output: value of polynomial'''
return(sum(val*x**n for n,val in enumerate(polynomial)))
def DifferentiatePoly(polynomial,order=1):
'''return the derivative of polynomial of order n'''
derivative=[(i+1)*val for i,val in enumerate(polynomial[1:])]
if order==1:
return derivative
else:
return DifferentiatePoly(derivative, order-1) | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
def DescartesSigns(polynomial):
'''Descartes rules of signs returns an upper bound on the number of positive roots and the number of negative roots'''
PosList=list(polynomial)
NegList=[((-1)**i)*val for i,val in enumerate(polynomial)]
while 0 in PosList:
PosList.remove(0)
NegList.remove(0)
return([sum([i[0]*i[1]<0 for i in zip(PosList[1:],PosList[:-1])]),sum([i[0]*i[1]<0 for i in zip(NegList[1:],NegList[:-1])])])
DescartesSigns([-52,45,-12,1])
In [58]:
import numpy
from math import sqrt
import math
def EvaluatePoly(polynomial,x):
'''evaluate a polynomial at a value x
input: polynomial as a list, value x to evaluate at
output: value of polynomial'''
return(sum(val*x**n for n,val in enumerate(polynomial)))
def FindZeroInInterval(polynomial,range_lower,range_upper):
'''perform an interval bisection on polynomial between the values range_lower<range_upper
return x such that polynomial(x)=0 (or a good enough approximation to it)'''
#this is slightly modified compared to what we presented.
assert range_lower<=range_upper,"second argument should be smaller than first"
accuracy=10**(-15)
value_lower=EvaluatePoly(polynomial,range_lower)
value_upper=EvaluatePoly(polynomial,range_upper)
if value_lower==0: # an exact root on the lower boundary
return(range_lower)
elif abs(value_upper/value_lower)<accuracy and abs(value_upper)<accuracy: #define this as being close enough to being a root on the upper boundary
return(range_upper)
elif abs(value_lower/value_upper)<accuracy and abs(value_lower)<accuracy: #close enough to root on lower boundary
return(range_lower)
if (range_lower==range_upper):
return('')
extent=2*(range_upper-range_lower) #initialise with a dummy variable larger than anything that will ver be calculated | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
while range_upper-range_lower>max(abs(range_upper),abs(range_lower),1)*accuracy and extent>(range_upper-range_lower):#stopping criteria based on numerical accuracy
if value_upper*value_lower>0:
return('no zero')
range_mid=(range_upper+range_lower)/2 #bisection
value_mid=EvaluatePoly(polynomial,range_mid)
extent=range_upper-range_lower
if value_mid==0:
return range_mid
elif value_mid*value_lower>0: #the root is between value_mid and value_upper. redefine range and repeat
range_lower=range_mid
value_lower=value_mid
else:
range_upper=range_mid
value_upper=value_mid
return range_mid
def RemoveTrailingZero(list):
"Remove the '.0' of a float to transform it in integer"
trail=[]
for x in list:
print(x)
if (x!=0 and x%1==0):
idk=int(x)
trail.append(idk)
elif x==0.0:
idk=0
trail.append(idk)
else:
trail.append(x)
return trail
def FindRealZeros(polynomial):
'''return the zeros of the polynomial in an ordered list'''
ReversePoly=list(reversed(polynomial)) #Reverse the order of the list 'polynomial' for numpy because numpy consider the first entry the lower degree
ListRoot=numpy.roots(ReversePoly) #list of root(s) of the polynomial found by numpy with a 'okay' precision
print(ListRoot)
newPrecision=[]
output=[]
Final=[]
nestedList=[]
withoutImaginary=[] | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
for x in range(len(ListRoot)):
newPrecision+=[numpy.around(ListRoot[x],decimals=7,out=None)]
for x in newPrecision: #get rid of complex numbers if present in the previous list
if x.imag==0: #keep the real numbers and remove the "0j" part
i=x.real
withoutImaginary.append(i)
for x in withoutImaginary: #Determine intervals around the roots with great precision
DetermineIntervals=[x-0.0000001,x+0.0000001]
output+=DetermineIntervals #Create a list whose entries are end points of intervals
i=0
while i<len(output): #Create a nested list
nestedList.append(output[i:i+2])
i+=2
print (nestedList)
for x in nestedList:
Zeropolynomial=FindZeroInInterval(polynomial,x[0],x[1]) #The upper bound and lower bound correspond to the endpoints of each interval in the nested list
if Zeropolynomial =='no zero':
pass
else:
Final.append(Zeropolynomial)
IncreasingOrder=list(sorted(Final,key=float)) #Reorganized the roots found by increasing order
end=RemoveTrailingZero(IncreasingOrder) #In the case where one of the root/multiple roots are supposed integer(s), we need to remove the '.0' part
print(end)
return end
We will test if your solution works by comparing ideal solutions to your calculated solutions via the function EqualityWithinTolerance, which aims to tolerate numerical inaccuracy. Some examples are given below. There will be further tests that have to be passed.
Marks: 10
Test your code from above here(3 points), ID: root_finder
In [59]:
#basic testing examples
def EqualityWithinTolerance(list1,list2,epsilon):
"""since we don't have perfect accuracy, cannot just test equality of two lists in our tests.
return true if all elements are close enough (within epsilon)"""
assert len(list1)==len(list2),'cannot compare lists of different lengths'
return sum([abs(i[0]-i[1])<=epsilon for i in zip(list1,list2)])==len(list1) | {
"domain": "cocalc.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9755769049752757,
"lm_q1q2_score": 0.8925562454264588,
"lm_q2_score": 0.9149009584734676,
"openwebmath_perplexity": 4157.842828911409,
"openwebmath_score": 0.4697786271572113,
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"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
#integer solutions
assert EqualityWithinTolerance(FindRealZeros([24,-50,35,-10,1]),[1,2,3,4],2*10**(-10))
assert EqualityWithinTolerance(FindRealZeros([-16,0,0,0,1]),[-2,2],10**(-10))
#irrational solutions
assert EqualityWithinTolerance(FindRealZeros([70,-60,12]),[(15-sqrt(15))/6,(15+sqrt(15))/6],10**(-10))
#no real solutions
assert EqualityWithinTolerance(FindRealZeros([16,0,0,0,1]),[],10**(-10))
assert EqualityWithinTolerance(FindRealZeros([16,0,1]),[],10**(-10)),"You should not be returning complex roots"
#roots very close to 0
assert EqualityWithinTolerance(FindRealZeros([-10**(-12),0,1]),[-10**(-6),10**(-6)],10**(-10))
#try a longer polynomial to make sure it's not just an explicit formula going up to degree 3 or 4
assert EqualityWithinTolerance(FindRealZeros([-362880, 1026576, -1172700, 723680, -269325, 63273, -9450, 870, -45,1]),[1,2,3,4,5,6,7,8,9],10**(-10))
[4. 3. 2. 1.] [[3.9999999, 4.0000001], [2.9999999, 3.0000001], [1.9999999, 2.0000001], [0.9999999, 1.0000001]] 1.0 2.0 3.0 4.0 [1, 2, 3, 4] [-2.00000000e+00+0.j 1.66533454e-16+2.j 1.66533454e-16-2.j 2.00000000e+00+0.j] [[-2.0000001, -1.9999999], [1.9999999, 2.0000001]] -2.0 2.0 [-2, 2] [3.14549722 1.85450278] [[3.1454971, 3.1454972999999997], [1.8545026999999998, 1.8545029]] 1.8545027756320966 3.145497224367904 [1.8545027756320966, 3.145497224367904] [-1.41421356+1.41421356j -1.41421356-1.41421356j 1.41421356+1.41421356j 1.41421356-1.41421356j] [] [] [-0.+4.j 0.-4.j] [] [] [-1.e-06 1.e-06] [[-1.1e-06, -9e-07], [9e-07, 1.1e-06]] -1e-06 1e-06 [-1e-06, 1e-06] [9. 8. 7. 6. 5. 4. 3. 2. 1.] [[8.9999999, 9.0000001], [7.9999999, 8.0000001], [6.9999999, 7.0000001], [5.9999999, 6.0000001], [4.9999999, 5.0000001], [3.9999999, 4.0000001], [2.9999999, 3.0000001], [1.9999999, 2.0000001], [0.9999999, 1.0000001]] 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 [1, 2, 3, 4, 5, 6, 7, 8, 9]
In [ ]:
#check that there's a reasonable number of pre- and post-conditions inside FindRealZeros and plenty of commenting | {
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"lm_name": "Qwen/Qwen-72B",
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"openwebmath_score": 0.4697786271572113,
"tags": null,
"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
Test your code from above here(1 point), ID: find_large_roots
In [ ]:
#here follow some hidden tests.
#very large roots
Test your code from above here(1 point), ID: root_finder_hidden_tests1
In [ ]:
#a few more general tests, not checking anything in particular, just keeping you honest
Test your code from above here(1 point), ID: root_finder_hidden_tests2
In [ ]:
#a few more general tests, not checking anything in particular, just keeping you honest
Test your code from above here(1 point), ID: root_finder_repeated
In [ ]:
#what happens when there are repeated roots?
Test your code from above here(1 point), ID: root_finder_uses_helpers
In [ ]:
#Ensure that the correct helper functions are being used
Test your code from above here(1 point), ID: root_finder_no_external
In [ ]:
#check that FindRealZeros does not rely on external functions
Explain how your function FindRealZeros works. Each member of the pair programming pair should write their own explanation, independently, in a separate copy of the file (which will contain the same solutions to the programming questions).
Marks: 3 | {
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"url": "https://share.cocalc.com/share/b67337a608fb9911d3a0a3330eaadd1ab39e446b/week%205/assignment/FindRoots.ipynb?viewer=share"
} |
# Evaluate the indefinite integral $\int \frac{1}{x^2} \sin\left(\frac{6}{x}\right) \cos\left(\frac{6}{x}\right) \, dx$
Evaluate the indefinite integral:
$$\int \frac{1}{x^2} \sin\left(\frac{6}{x}\right) \cos\left(\frac{6}{x}\right) \, dx$$
(using substitution)
The answer is: $$\frac {1}{24} \cos\left(\frac{12}{x}\right) + C$$
Whereas I get a slightly different one.
Here's my solution:
$$u = \frac{6}{x}$$
$$du = - \frac{6}{x^2} \cdot dx$$
$$-\frac {1}{6} du = \frac {1}{x^2} \cdot dx$$
Making substitution:
$$\int -\frac{1}{6} du \sin (u) \cos (u)$$
adding a new variable for substitution:
$$s = \cos (u)$$
$$ds = -sin(u) du$$
Making substitution:
$$\frac {1}{6} \int ds \cdot s$$
Evaluating:
$$\frac{1}{6} \cdot \frac{s^2}{2} = \frac{s^2}{12} = \frac{\cos^2(u)}{12} = \frac{\cos^2(\frac{6}{x})}{12}$$
Then using a half angle formula for $$\cos^2$$:
$$\frac {\frac{1}{2} \cdot (1 + \cos(\frac{12}{x}))}{12} = \frac{1}{24} + \frac{1}{24} \cdot \cos\left(\frac{12}{x}\right) + C$$
As you can see I have an additional $$\frac{1}{24}$$ in my answer... so what did I do wrong?
• Did you notice the +C in the original answer there? The constant absorbs that $+1/24$! Apr 27 '15 at 4:18
• @Jesse P Francis So are you saying there's never a constant in the answers for integrals that were evaluated? I wonder why is that? I already integrated the function... why does my constant need to disappear... because of C... Apr 27 '15 at 4:20
• @dramadeur It doesn't disappear per se; in this case Jesse is defining a new constant $C' := C + \frac{1}{24}$ and then renaming $C'$ as $C$. Apr 27 '15 at 4:23
• @Travis, he edited the question and added constant, anyway, dramadeur, I think what Travis said is what you are confused with!:) Apr 27 '15 at 4:25
You missed the constant of integration all the way! It "absorbs" the $\frac{1}{24}$!
Spot the difference:
Evaluating: | {
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Spot the difference:
Evaluating:
$\displaystyle \frac{1}{6} \cdot \frac{s^2}{2} +c= \frac{s^2}{12} +c= \frac{\cos^2(u)}{12} +c= \frac{\cos^2(\frac{6}{x})}{12}+c$, where c is the constant of integration.
Then using a half angle formula for $\cos^2$:
$\displaystyle \frac {\frac{1}{2} \cdot (1 + \cos(\frac{12}{x}))}{12} +c$ = $\frac{1}{24} \cdot \cos(\frac{12}{x}) + C$, where $C=c+\frac{1}{24}$
You did everything right, but you're constant of integration "absorbs" the $\frac{1}{24}$
$$\frac{1}{24}+C=C$$
The $C$ is an arbitrary constant meaning that it could be anything (depending on our initial conditions). So, what's a constant plus an arbitrary constant? It's just another arbitrary constant!
If it helps, you could do something like:
$$\frac{1}{24}+C=D$$ | {
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} |
+0
# If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?
0
529
21
If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?
Guest May 21, 2014
#16
+889
+8
There are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.)
The counting method has been used for both of the correct answers so far, so here's the probability method.
Suppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340.
Bertie May 21, 2014
#1
+996
0
Well, three skiiers on a plane of 17. Considering that one person dies each time, the number decreases by 1 in both the numerator and the denominator.
$$\left({\frac{{\mathtt{3}}}{{\mathtt{17}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{16}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{15}}}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{680}}}} = {\mathtt{0.001\: \!470\: \!588\: \!235\: \!294\: \!1}}$$
The odds are 1/680
GoldenLeaf May 21, 2014
#2
+92775
+8
Probability is not my strong suit but this is what i think.
P= number of ways 4 can be chosen from 14 / number of ways 4 can be chosen from 17
P= 14C4 / 17C4
P = 1001 / 2380
$$\mbox{P(all 3 survive) }=\frac{1001}{2380}=\frac{143}{340}$$
$${\frac{{\mathtt{1\,001}}}{{\mathtt{2\,380}}}} = {\frac{{\mathtt{143}}}{{\mathtt{340}}}} = {\mathtt{0.420\: \!588\: \!235\: \!294\: \!117\: \!6}}$$
Can another mathematician please check this - I'm fairly confident that it is correct. | {
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} |
Can another mathematician please check this - I'm fairly confident that it is correct.
Melody May 21, 2014
#3
+87294
+8
Probabilty isn't my strong point, but I'll take a run at this one.
So, basically, we first want to count the sets of people who might die - what a morbid problem!!
The total number of people who could die is given by choosing some group of 4 from the 17 = C(17,4) = 2380
Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets
Now, let's look at the possible sets where 2 skiers die - C(3,2) *C(13,2) = 234 sets
Now let's consider the sets where all three skiers die = C(3,3) * C (13,1) = 13
So, the total number of sets containing any of the skiers = 1105
Thus, the chances that all three survive are given by:
1 - (the number of sets containing any of the skiers)/(the total number of possible sets)
= 1 - (858 + 234 + 13)/2380 ≈ 53.6 %
OOPS !! Melody pointed out a math error I made....let me correct this...and as she indicated recently, she DOES believe in "do-overs"....I still like my logic, though!
Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(14,3) = 1092 sets
Now, let's look at the possible sets where 2 skiers die - C(3,2) *C(14,2) = 273 sets
Now let's consider the sets where all three skiers die = C(3,3) * C (14,1) = 14
So, the total number of sets containing any of the skiers = 1379
Thus, the chances that all three survive are given by:
1 - (the number of sets containing any of the skiers)/(the total number of possible sets)
== 1 - (1379)/2380 ≈ 42.1 %
Thanx, Melody!!
CPhill May 21, 2014
#4
+92775
0
Okay, we have 3 answers - all different
We need an arbitrator - preferably one who knows what he/she is talking about.
also if possible please explain what is wrong with our logic.
Thank you.
Melody May 21, 2014
#5
+92775
+8
Okay Chris,
I am taking a look at yours. | {
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Thank you.
Melody May 21, 2014
#5
+92775
+8
Okay Chris,
I am taking a look at yours.
Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets
shouldn't this be 3C1*14C3 = 3*364=1092 ?
Melody May 21, 2014
#6
+87294
+8
We both got the same result...now, let's figure the probability that 2 board members out of (n) board members could arrive at the same (possibly) correct answer!!
CPhill May 21, 2014
#7
+92775
0
EXCELLENT!!!!!
Melody May 21, 2014
#8
+87294
0
HAHA!!!...and 57 minutes ago...you were looking for someone who knew what they were doing!!! (an arbitrator, I believe??)
Are we "sure" our answers are correct??
(Actually....I think they might be)
CPhill May 21, 2014
#9
+92775
0
Yes I am sure.
Because
Two out of three ain't bad!
Melody May 21, 2014
#10
+87294
0
But 2/3 = 66%
That leaves 1/3 chance = 33% = that we might not be!!
I say...let's call in the "troll" as a referee.....HE KNOWS ALL !!!
LOL!!
CPhill May 21, 2014
#11
+92775
0
Okay
Where's 'our' KNOW-IT-ALL TROLL when we need him?
He's probably AWOL with Sir Cumference!
Melody May 21, 2014
#12
+87294
0
I'm thinking he was on that plane...and maybe he doesn't ski, either......
Mmmmmmm......maybe we've solved a forum "problem"
(At least there's about a 58% chance of it....)
CPhill May 21, 2014
#13
+92775
0
NOW that is a percentage that I would very seriously challenge!
P(troll gone)=P(troll was on plane)*P(troll died) $$\rightarrow 0$$
Melody May 21, 2014
#14
+87294
0
Yeah...we couldn't be that fortunate, huh??
CPhill May 21, 2014
#15
+92775
0
That's not nice Chris. So long as we keep him on a tight leash he is fun to have around.
You know that just as well as I do!
Melody May 21, 2014
#16
+889
+8
There are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.) | {
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The counting method has been used for both of the correct answers so far, so here's the probability method.
Suppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340.
Bertie May 21, 2014
#17
+2353
0
But what if the plane was carrying 4 people that died in a crash in coffins in the cargo room. Then the plane might not have crashed at all and the skiers survive
Unless the skiers died in a skiing accident off course
reinout-g May 21, 2014
#18
0
If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?
----
This probability is trickier than it looks.
Four (4) will die and at most only three (3) can be skiers.
By deduction one (1) will die and not be a skier.
Now 16 remain, three (3) of whom are skiers.
From this calculate the probability of selecting (Z) correct out of (R) draws from (N) numbers. (Z) (in this case) defines the probability of a skier dying.
Probability= (R!/(Z!*(R-Z)!) * (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)/(N!/((R!)*((N-R)!)))
$${Probability =}\frac{\frac{R!}{Z!*(R-Z)!} * \frac{(N-R)!}{((N-R)-(R-Z))!*(R-Z)! }}{\frac{N!}{R!*(N-R)!}}$$
Column ID’s: A= (R!)/(Z!(R-Z)!)
B= (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)
C= N!/((R!)*((N-R)!))
D= Probability of (Z) skiers dying.
E= 1/Probability
N R Z A B C D E
16 3 3 1 1 560 0.001785714285714 560.0000000000
16 3 2 3 13 560 0.069642857142857 14.3589743590
16 3 1 3 78 560 0.417857142857143 2.3931623932
16 3 0 1 286 560 0.510714285714286 1.9580419580 | {
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16 3 0 1 286 560 0.510714285714286 1.9580419580
-----------
Probability of zero skiers dying ~ 0.511 (51.1%)
(Professional help provided by Francis and Frances)
by: Someone Who Knows Everything
Guest May 21, 2014
#19
+92775
0
deleted deleted
Melody May 21, 2014
#20
+26745
0
My view is:
Probability that 4 non-skiers died (hence 3 skiers survived) is:
$${\frac{{\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{13}}{\mathtt{\,\times\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{11}}}{\left({\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{14}}\right)}} = {\frac{{\mathtt{143}}}{{\mathtt{340}}}} = {\mathtt{0.420\: \!588\: \!235\: \!294\: \!117\: \!6}}$$
Oops! Just noticed that this is the same as Bertie's reply (though Bertie did a more complete job by including an explanation).
Alan May 22, 2014
#21
+92775
0
It is the same as mine and Chris's as well!!
Melody May 22, 2014 | {
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# Asymptotes as the lines y=x an y=-x
Hi there.
I'm a maths teacher and today was having a discussion with my Head of Department about asymptotes (as you do!)
She was me if I could think on an equation of a graph(s) which has an asymptotes at the line y=x and another at y=-x.
Thinking about it, neither of us could come up with anything. It's really bugging me now. I ain't too sure if there is such a function...though it seems silly there not being.
The only way I managed to get a graph like what I wanted was to rotate the graph of y=1/x by various angles.
So can such a graph exist? Is it possible to transform equations of graphs via rotations (something I can't ever remember doing)?
Hope I can come up with an answer by tomorrow :)
$$y = x \tanh{x}$$ ?
as
$$x \tanh{x} = x \frac{e^x - e^{-x}}{e^x + e^{-x}}$$
therefore, for large positive x,
$$x \tanh{x} \approx x \frac{e^x}{e^x} = x$$
and for large negative x
$$x \tanh{x} \approx x (\frac{-e^{-x}}{e^{-x}}) = -x$$
arildno
Homework Helper
Gold Member
Dearly Missed
Dogtanian said:
Hi there.
I'm a maths teacher and today was having a discussion with my Head of Department about asymptotes (as you do!)
She was me if I could think on an equation of a graph(s) which has an asymptotes at the line y=x and another at y=-x.
Thinking about it, neither of us could come up with anything. It's really bugging me now. I ain't too sure if there is such a function...though it seems silly there not being.
The only way I managed to get a graph like what I wanted was to rotate the graph of y=1/x by various angles.
So can such a graph exist? Is it possible to transform equations of graphs via rotations (something I can't ever remember doing)?
Hope I can come up with an answer by tomorrow :)
You certainly can rotate curves!
For example, the curve described by the equation:
$$x^{2}-y^{2}=1$$
has asymptotes y=x and y=-x. | {
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For example, the curve described by the equation:
$$x^{2}-y^{2}=1$$
has asymptotes y=x and y=-x.
Note, however, that in this case, there exists no function of x, by which the y-coordinates of the curve could be computed, and the curve in question cannot be regarded as the graph of some function, i.e, the set of points describable as (x,f(x)), where f is some function.
robphy
Homework Helper
Gold Member
Doesn't one branch of that hyperbola still have those lines as asymptotes?
In a physics context, the worldline of a uniformly accelerated observer [which can be regarded as function (t,x(t)), where x(t)=sqrt(1+t^2)] is asymptotic to a light cone.
Thanks for the help guys :D
My head of department was so sure you could each of the 4 sections between the said asymptotes filled with a curve.
Using what the first two posts said, I drew the graphs of
y = sqrt(1+x^2)
y = -sqrt(1+x^2)
x = sqrt(1+y^2)
x = -sqrt(1+y^2)
to get what I needed (at least I think that is what I i if I remember correctly back to last night....)
But I only went and forgot all about this today until just now...so I never i speak to my HofD about it...never mind :D
Hurkyl
Staff Emeritus
Gold Member
You can combine the four curves to get one that fills the whole thing.
x² - y² = 1
fills in two sections, and
y² - x² = 1
fills in the other two.
We can combine them into one equation:
(x² - y² - 1) (y² - x² - 1) = 0
which will fill in all four quadrants.
robphy | {
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# There are 1600 jelly beans divided between two jars, X and Y. If the
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There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
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06 Feb 2015, 09:03
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There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?
A. 375
B. 950
C. 1150
D. 1175
E. 1350
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
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06 Feb 2015, 13:43
4
1
Hi All,
This question can certainly be solved with Algebra; with 2 variables and 2 unique equations, it's just a matter of translating the text into equations and doing "system" math.
The answers are numbers though, and there is a logical pattern in the prompt that you can use to quickly TEST THE ANSWERS. | {
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We're told that there are a total of 1600 marbles in two jars. Jar X has 100 less than THREE TIMES the number of marbles in Jar Y. We're asked for the number of marbles in Jar X.
100 marbles is a relatively small amount, compared to the 1600 total marbles. If we ignore the "100" and do a rough estimation, Jar X would be about 1200 and Jar Y would be about 400 (which matches the information about THREE TIMES the number of marbles). Since we DO have to factor in the 100 marbles though, the number of marbles in X has to be LESS than 1200. With this deduction, the answer has to be C or D.
Let's TEST C (since it looks like the "nicer" number to deal with).
IF....
X = 1150 marbles
X + 100 = 1250
1250/3 = 416.6666 marbles
1150 + 416.66666 is NOT 1600 total marbles.
Eliminate C.
Here's the proof though:
IF....
X = 1175 marbles
X + 100 = 1275
1275/3 = 425
1175 + 425 = 1600 marbles
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
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06 Feb 2015, 13:28
1
x+y=1600 , X=1600-Y
x+100=3y, 1700=4Y
Y=425
x=1600-y=1600-425=1175
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
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### Show Tags
06 Nov 2018, 20:08
pacifist85 wrote:
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?
A. 375
B. 950
C. 1150
D. 1175
E. 1350
Let Jelly beans in jar $$X = x$$ and $$Y = y$$
$$x + y = 1600$$ ----- ($$i$$)
Given jar $$X$$ has $$100$$ fewer jelly beans than three times the number of beans in jar $$Y$$.
Therefore; $$x + 100 = 3y$$
$$x = 3y - 100$$
Substituting value of $$x$$ in equation ($$i$$), we get;
$$3y - 100 + y = 1600$$
$$4y = 1600 + 100 = 1700$$
$$y = \frac{1700}{4} = 425$$
Substituting value of $$y$$ in equation ($$i$$), we get;
$$x + 425 = 1600$$
$$x = 1600$$ $$-$$ $$425 = 1175$$
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
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08 Mar 2019, 17:52
Hi,
If X has 100 fewer jelly beans than three times the number of beans in jar Y.
Then why did we write X+100 ?
Shouldnt it be X-100?
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
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08 Mar 2019, 19:35
x+y=1600 (I)
And
3*y = x-100
Reorganizing it
3*y-100 = x (II)
Subs. (II) in (I)
3*y-100+y=1600
4*y=1700
y=425
Then
x=1175
Ans (D)
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
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09 Mar 2019, 03:13
1
3*y = x-100
Reorganizing it
3*y-100 = x (II)
But if -100 go to the other side of equation, shouldnt it become +100?
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
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### Show Tags
09 Mar 2019, 10:08
hsn81960 wrote:
3*y = x-100
Reorganizing it
3*y-100 = x (II)
But if -100 go to the other side of equation, shouldnt it become +100?
Sorry, it's 3*y = x+100 instead
Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink] 09 Mar 2019, 10:08
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# How to show this obvious and basic property of abelian groups?
I have a question that is probably very silly, but let's go. Let $(G,+)$ be an abelian group. In that case we know that $+$ is associative and commutative. This leads us to the following: if $\{a_i \in G : i \in I_n\}$ with $I_n = \{i \in \mathbb{N} : 1 \leq i \leq n\}$, then if we apply $+$ to all of the $a_i$, it independs on the ordering we impose. In truth, if we want to define the sum of all of those elements as:
$$\sum_{i \in I_n}a_i = a_1+\cdots+a_n,$$
we should first know what means $a_1 + \cdots + a_n$. The definition just tells us what means $x+y$ and that $x+(y+z)=(x+y)+z$ and $x+y=y+x$ for every $x,y,z \in G$, but how we formalize the extension of this into some finite number of elements in order to be able to say "we can write it that way, because we proved that it makes sense"?
Indeed this is something very obvious, and I've never seem someone giving great arguments about it. Everyone just says "obviously, the parentheses and the order doesn't matter". I've thought on using inductions on those two properties each at a time, but I've got a little confused with it.
How is this really done? Is a proof necessary? Or we just leave it there without proof?
Thanks very much in advance, and sorry if this is not the place for this kind of question.
-
Everyone just says "obviously, the parentheses and the order doesn't matter". Not everybody. Sometimes it's proved. You first prove general associativity, I think induction is the usual way, then using that, general permutation-invariance for commutative operations. – Daniel Fischer Aug 12 '13 at 2:32
Thanks for the help @DanielFischer! I've said everybody meaning "everybody I've seem until now". I'll try doing it in this way you've said. Thanks for the hint. – user1620696 Aug 12 '13 at 2:35
To prove general associativity, define $$\begin{cases}\prod_{i=1}^1 a_i=a_1\\\prod_{i=1}^n a_i=\prod_{i=1}^{n-1}a_i\cdot a_{n}\end{cases}$$ | {
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The claim is that for any $m$, we have that $$\prod_{i=1}^n a_i\prod_{i=1}^m a_{n+i}=\prod_{i=1}^{n+m}a_i$$
By definition we have the truth of $m=1$. Thus assume true for $m=r$, and consider $r+1$. Then \begin{align}\prod_{i=1}^n a_i\prod_{i=1}^{r+1} a_{n+i}&=\prod_{i=1}^na_i \left(\prod_{i=1}^{r} a_{n+i}a_{n+r+1}\right)\\&=\left(\prod_{i=1}^na_i \prod_{i=1}^{r} a_{n+i}\right)a_{n+r+1}\\&=\prod_{i=1}^{n+r}a_i a_{n+r+1}\\&=\prod_{i=1}^{n+r+1}a_i\end{align}
General commutativity can be proven as follows. Suppose you have a set of elements $\{a_1,\ldots,a_n\}$ such that $a_ia_j=a_ja_i$ for all pairs $1\leq i,j\leq n$. Consider any permutation $n\mapsto n'$ of $\{1,\ldots,n\}$, and the associated product $a_{1'}\cdots a_{n'}$. Suppose that the term $a_n$ occurs it the place $h'=n$. Then me way write by the commutativity hypothesis $$a_{1'}\cdots a_{n'}=a_{1'}\cdots a_{(h-1)'}a_{(h+1)'}\cdots a_{(n-1)'}a_{n'}a_n$$
and then induction does the rest, since we have stepped down to the case $n-1$
-
As Daniel notes it is generally first proved (with no assumption of commutativity) that generalized associativity holds. (There are a number of questions about that on MSE.) Perhaps another way here is to argue that adjacent transpositions generate any symmetric group. – anon Aug 12 '13 at 3:30
@anon Agreed. ${}{}{}$ – Pedro Tamaroff Aug 12 '13 at 4:13
As the others point out, you can prove that parentheses and order don't matter via induction. Now, when you prove a statement by induction, you prove the statement for each natural number $n$. In other words, the statement applies only to finite sums.
This observation is significant because the obvious generalization to infinite series is not true in the real numbers! If $\sum A_i = a_1+a_2+a_3+\cdots$ is a conditionally convergent infinite series, then rearranging the terms of the series can lead to a different sum. In fact, you can get any sum you want! | {
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(On the other hand, if $\sum A_i = a_1+a_2+a_3+\cdots$ is a series whose partial sums are eventually constant, then rearranging the terms doesn't change the eventual value. Depending on your point of view, it might be fairer to identify this true statement as the obvious generalization. After all, it doesn't depend on the topology of our abelian group. Or, rather, it requires convergence in the discrete topology.)
-
If the partial sums are eventually constant, you're still just (effectively) dealing with a finite sum, no? – Cameron Buie Aug 12 '13 at 2:48
@CameronBuie Sure, but the "eventual sum" is the natural generalization of finite sums to groups that don't come with a topology. I edited the answer to explain this point a little better. – Chris Culter Aug 12 '13 at 2:51
What is the point of bringing up conditionally convergent series here? This would better suit as a (rather long) comment. – Pedro Tamaroff Aug 12 '13 at 2:59
@PeterTamaroff One doesn't truly understand why a result is stated in a certain way, and proven in a certain way, until one has seen counterexamples to its generalizations. Wouldn't you agree? – Chris Culter Aug 12 '13 at 3:04
@ChrisCulter "...the obvious generalization to infinite series..." This is really digressing! – Pedro Tamaroff Aug 12 '13 at 3:10 | {
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# Invert the softmax function
Is it possible to revert the softmax function in order to obtain the original values $$x_i$$?
$$S_i=\frac{e^{x_i}}{\sum e^{x_i}}$$
In case of 3 input variables this problem boils down to finding $$a$$, $$b$$, $$c$$ given $$x$$, $$y$$ and $$z$$:
$$\begin{cases} \frac{a}{a+b+c} &= x \\ \frac{b}{a+b+c} &= y \\ \frac{c}{a+b+c} &= z \end{cases}$$
Is this problem solvable?
## 2 Answers
Note that in your three equations you must have $x+y+z=1$. The general solution to your three equations are $a=kx$, $b=ky$, and $c=kz$ where $k$ is any scalar.
So if you want to recover $x_i$ from $S_i$, you would note $\sum_i S_i = 1$ which gives the solution $x_i = \log (S_i) + c$ for all $i$, for some constant $c$.
• So it’s solvable up to a constant. Thank you! – jojek May 18 '18 at 17:39
• Which c constant should I use? There is any way of calculating it? – Joel Carneiro Feb 7 '19 at 17:16
• @JoelCarneiro Any $c$ will work; the solution is not unique. – angryavian Feb 7 '19 at 17:58
• Any $c$ will work, one choice is if you augment the $x_i$ vector like $(0, x_1,...,x_n)$ then this will induce a particular $c$, note the corresponding log-sum-exp -- the gradient of which is the softmax -- would also be convex (en.wikipedia.org/wiki/LogSumExp). – Josh Albert Jul 29 '19 at 10:59
• In the case anybody like me spend too much time figuring out $c$: If you know your 3 input variables have to sum to 1 then your $c = (1 - log(x \cdot y \cdot z))/3)$. – Rasmus Ø. Pedersen Sep 14 '20 at 13:04
A similar question was asked in a post of reddit. The answer below is adapted from that post:
$$S_{i}$$ = $$\exp(x_{i})/(\sum_{i} \exp(x_{i}))$$
Taking ln on both sides:
$$\ln(S_{i}) = x_{i} - \ln(\sum_{i} \exp(x_{i}))$$
Changing sides:
$$x_{i} = \ln(S_{i}) + \ln(\sum_{i} \exp(x_{i}))$$
The second term of the right hand side is constant for a particular $$i$$ and can be written as $$C_{i}$$. Therefore, we can write:
$$x_{i} = \ln(S_{i}) + C_{i}$$ | {
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$$x_{i} = \ln(S_{i}) + C_{i}$$
• better use j in the summation – eyaler Jan 22 at 18:26 | {
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# Examples of entire functions with none, one and infinite zeros.
I know that an entire function is one which can be differentiated on the entire complex plane, and I believe that a zero of an entire function is the z where f(z)=0.
I was tring to think of example of the following
An entire function with no zero's: I thought $e^z$ would be suitable as $e^z\neq 0\forall z$
An entire function with one zero : for this I chose $z$ as it is only zero at zero
An entire function with infinite zero's : I thought maybe sin(z) would work here but is sin(z) well defined ? I'm not so sure as $sin(z)=sin(r(cos(\theta) + isin(\theta))$ doesn't seem like a valid expression ( although I could be wrong )
If sin(z) is not well defined what is an example of an entire function with infinite zero's ?
Yes, it is. Just define $\sin z$ as$$z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$$It is an entire functions and$$\sin z=0\iff z\in\pi\mathbb{Z}.$$ | {
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# Does the alternating composition of sines and cosines converge to a constant?
Let $$f(x) = \cos(\sin(x))$$ and let $$c(f, n)(x)$$ denote the function $$\underbrace{f\circ f\circ...\circ f}_{n \text{ times}}$$. For example, $$c(f, 1)(x) = f(x)$$, $$c(f, 2)(x) = f(f(x))$$ and so on.
My question is: does $$c(f, n)(x)$$ approach any constant function if $$n \to +\infty$$? I graphed this for some values of $$n$$ and the function seems to approach some value a little bit over $$0.76$$. Does anyone have any insight as to whether that is true? If so, what value is it approaching and why?
Any sort of help or material helps; this question has been stuck in my head for quite some time now! Thanks in advance!
Yes—this is a question in dynamical systems, if you're looking for words to search with.
In this case, the equation $$f(x)=x$$ has exactly one fixed point (near $$x=0.76817$$), and at that fixed point we have $$|f'(x)| \approx 0.46046 < 1$$; therefore it is an attracting fixed point, which means that every sequence of iterates of $$f$$ will approach the fixed point exponentially fast.
• This does not follow from what you have. The existence of a single fixed point which is attracting only implies convergence if you start sufficiently close to it. For example, consider the equation $f(x)=-x^3$. Jun 17 '20 at 9:54
• I think boundedness of $f$ implies your conclusion in this case though. Jun 17 '20 at 9:59
For the limit $$f$$ it must hold: $$f(x)=\cos(\sin(f(x)))$$. Taking the derivative (assuming $$f$$ differentiable) implies:
$$f'(x)=-\sin(\sin(f(x)))\cos(f(x))f'(x)$$
Impliing either $$f'(x)=0$$ or $$1=-\sin(\sin(f(x)))\cos(f(x))$$
The last equation can only hold if $$f(x)=k\pi$$ for $$k\in \mathbb{Z}$$ but this implies $$\sin(\sin(f(x)))=0$$, contradiction. So $$f$$ must be constant (or not differentiable).
You can show that a constant solution exists by Banach Fixpoint theorem on the sequence
$$x_{n+1}=\cos(\sin(x_n))$$ | {
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$$x_{n+1}=\cos(\sin(x_n))$$
• Thank you! That makes a lot of sense! Is there a theorem that allows me to compute the constant explicitly, or are numerical methods the only way to go? Jun 16 '20 at 22:23
• It must hold for the solution $x^*$: $x^*=\cos(\sin(x^*))$ but I don't think you can further simplify this. Jun 16 '20 at 22:26
• This solution also requires that the limiting function exists in the first place, which is not clear. Jun 17 '20 at 1:39
• The first part was a necessary condition for the limit to exist. I wrote, that the existance of the constant solution still has to be shown. Jun 17 '20 at 10:30
• I used Maxima to numerically calculate the constant solution to 128 digits, then plugged the result into the Inverse Symbolic Calculator (wayback.cecm.sfu.ca/projects/ISC/ISCmain.html). It found no match. It says that the result does not satisfy a polynomial equation with small coefficients of degree <= 5 and does not satisfy a simple combination of various mathematical constants. Jun 17 '20 at 12:53
This question can be settled by some elementary analysis. Note that \begin{aligned} I:=f(\mathbb R)&=\cos(\sin(\mathbb R))=\cos([-1,1])=[\cos(1),1]=[0.540,1],\\ f(I)&=\cos\left(\sin\left([0.540,\,1]\right)\right)\\ &=\cos([0.514,\,0.841])\\ &=[\cos(0.841),\,\cos(0.514)]\\ &=[0.666,\,0.871]\subset I. \end{aligned} So, if $$f$$ has any fixed point, the fixed point must lie inside $$I=[\cos(1),1]$$.
Let $$g(x)=f(x)-x$$. Since $$g(\cos(1))=0.330>0>-0.334=g(1)$$, by the intermediate value theorem, $$g$$ has a zero on $$I$$, i.e. $$f$$ has a fixed point on $$I$$. As $$g'(x)=-\sin(\sin(x))\cos(x)-1<0$$, the fixed point of $$f$$ is also unique. Finally, on $$I=[\cos(1),1]$$, as $$|f'(x)|=| \sin(\sin(x))\cos(x)|\le|\sin(\sin(x))|\le|\sin(\sin(1))|=0.746<1,$$ the fixed point is attractive. Therefore, if we denote the $$n$$-fold composition of $$f$$ by $$f^n$$, the sequence $$(f(x),f^2(x),f^3(x),\ldots)$$ must converge to the fixed point of $$f$$. | {
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Denote $$f(x)=\cos(\sin(x))$$.
Since there exists $$\varepsilon>0$$ such that $$|f'(x)|=|\sin(\sin(x))\cos(x)|< 1-\varepsilon$$ for every $$x \in \mathbb{R}$$, $$f\colon \mathbb{R} \to \mathbb{R}$$ is a Lipschitz function with Lipschitz constant stricly less than $$1$$.
Thus, by Banach-Caccioppoli fixed point Theorem, $$f$$ has exactly one fixed point $$x_0 \in \mathbb{R}$$, that is a point such that $$f(x_0)=x_0$$. Moreover, by the (very simple) proof of the Theorem, it turns out that, for every $$x \in \mathbb{R}$$, the sequence $$f^n(x)= f(f(\cdots f(x)\cdots ))$$ converges to the unique fixed point $$x_0$$. | {
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Converting a repeating decimal to ratio of integers
paulmdrdo
Active member
0.17777777777 convert into a ratio.
M R
Active member
Re: converting a repeating decimal to ratio of integers
Hi,
This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum.
Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.
Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify.
I should also say that we can write a decimal as a fraction but we can't write it as a ratio.
paulmdrdo
Active member
Re: converting a repeating decimal to ratio of integers
Hi,
This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum.
Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.
Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify.
I should also say that we can write a decimal as a fraction but we can't write it as a ratio.
what do you mean by "GP"?
M R
Active member
Re: converting a repeating decimal to ratio of integers
what do you mean by "GP"?
Sorry, I have to stop using abbreviations.
A GP is a geometric progression: $$a, ar, ar^2, ar^3...$$.
If you haven't met this then the second method I posted is fine.
soroban
Well-known member
Re: converting a repeating decimal to ratio of integers
Hello, paulmdrdo!
$$\text{Convert }\,0.1777\text{...}\,\text{ to a fraction.}$$
$$\begin{array}{ccc}\text{We have:} & x &=& 0.1777\cdots \\ \\ \text{Multiply by 100:} & 100x &=& 17.777\cdots \\ \text{Multiply by 10:} & 10x &=& \;\;1.777\cdots \\ \text{Subtract:} & 90x &=& 16\qquad\quad\; \end{array}$$
Therefore: .$$x \;=\;\frac{16}{90} \;=\;\frac{8}{45}$$
paulmdrdo
Active member
how would I decide what appropriate power of ten should i use?
for example i have 3.5474747474... how would you convert this one?
M R | {
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for example i have 3.5474747474... how would you convert this one?
M R
Active member
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.
If you use 1000 and 10 you will get
1000x=3547.474747...
10x=35.474747...
So 990x=3512 and x=3512/990=1756/495.
I'm adopting Soroban's approach as I prefer it to what I did earlier.
paulmdrdo
Active member
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.
If you use 1000 and 10 you will get
1000x=3547.474747...
10x=35.474747...
So 990x=3512 and x=3512/990=1756/495.
I'm adopting Soroban's approach as I prefer it to what I did earlier.
"a difference of two in the powers of ten" -- what do you mean by this? sorry, english is not my mother tongue. bear with me.
Last edited:
M R
Active member
"a difference of two in the powers of ten" -- what do you me by this? sorry, english is not my mother tongue. bear with me.
No problem.
We have 10^3 and 10^1.
The difference between 3 and 1 is 3-1=2
Prove It
Well-known member
MHB Math Helper
how would I decide what appropriate power of ten should i use?
for example i have 3.5474747474... how would you convert this one?
You want to multiply by a power of 10 which enables you to only have the repeating digits shown, and then multiply by a higher power of ten to have exactly the same repeating digits. We require this so that when we subtract, the repeating digits are eliminated.
So in this case, since the 47 repeats, you want the first to read "something.4747474747..." and the second to read "something-else.4747474747..."
What powers of 10 will enable this?
MarkFL
Staff member
A quick method my dad taught me when I was little, is to put the repeating digits over an equal number of 9's.
1.) $$\displaystyle x=0.1\overline{7}$$
$$\displaystyle 10x=1.\overline{7}=1+\frac{7}{9}=\frac{16}{9}$$
$$\displaystyle x=\frac{16}{90}=\frac{8}{45}$$ | {
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$$\displaystyle x=\frac{16}{90}=\frac{8}{45}$$
2.) $$\displaystyle x=3.5\overline{47}$$
$$\displaystyle 10x=35.\overline{47}=35+\frac{47}{99}=\frac{3512}{99}$$
$$\displaystyle x=\frac{3512}{990}=\frac{1756}{495}$$ | {
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# $\epsilon - N$ proof of $\sqrt{4n^2+n} - 2n \rightarrow \frac{1}{4}$
I have the following proof for $$\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$$ and was wondering if it was correct. Note that $$\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$$. $$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| \\ = \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\ = \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \leq \left|\frac{n}{4(4n)^2}\right| = \left|\frac{n}{64n^2}\right| \\ = \left|\frac{1}{64n}\right| < \epsilon \\ \implies n>\frac{1}{64\epsilon}$$
• Seems quite right, though the final implication must be in the other direction.
– user65203
Aug 6, 2019 at 13:30
• It's wrong, for the reason Yves gave. Aug 6, 2019 at 13:34
• Did you finish this problem @user100000001? Sep 20, 2020 at 18:48
You want to show that $$\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$$. To do this, I would split up the analysis into scratch work and the formal proof.
For the scratch work, you need to find a suitable upper bound. You have done this by showing
\begin{align} \left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| & = \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|\\&=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\& = \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \\&\leq \left|\frac{n}{4(4n)^2}\right| \\&= \left|\frac{n}{64n^2}\right| \\& = \left|\frac{1}{64n}\right| \\&< \epsilon \end{align}
which means that $$n>\frac{1}{64\epsilon}$$ is the upper bound.
For the formal proof:
Let $$\epsilon>0$$ (you need to fix $$\epsilon$$ as a small positive constant). It follows from $$\frac{1}{\epsilon}>0$$ that $$\frac{1}{64\epsilon}>0$$. Then by the Archimedean property there exists a $$N\in\mathbb N$$ such that $$N>\frac{1}{64\epsilon}$$. Then if $$n\geq N > \frac{1}{64\epsilon}$$, we have | {
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$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right|\leq \frac{1}{64n}<\epsilon$$
where the last inequality follows from
$$n\geq N > \frac{1}{64\epsilon} \implies n>\frac{1}{64\epsilon} \implies\epsilon > \frac{1}{64n}$$
Like the other commenters have mentioned, the implication is in the wrong direction. Since you want $$|\frac{1}{64n}| < \epsilon$$, you are required to have $$n > \frac{1}{64 \epsilon}$$. In other words, $$n > \frac{1}{64 \epsilon}$$ implies $$|\frac{1}{64n}| < \epsilon$$. What you have written is the other way around.
You did the scratch work correctly. But I wouldn't call this a proof (of course it contains all ingredients of a good proof!).
You didn't introduce $$\epsilon$$. Of course, everyone knows what you mean but one should write it out if one wants to be fully rigorous.
Let $$\epsilon >0$$. Let $$N$$ be an integer greater than --insert your choice for $$N$$ here--.
If $$n \geq N$$, we have | {
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# scalene obtuse triangle | {
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In an equiangular triangle, all the angles are equal—each one measures 60 degrees. Hypotenuse. This is one of the three types of triangles, based on sides.. We are going to discuss here its definition, formulas for perimeter and area and its properties. Step 2: Let x be one of the two equal angles. Acute triangle. Yes! [insert scalene G U D with ∠ G = 154° ∠ U = 14.8° ∠ D = 11.8°; side G U = 17 cm, U D = 37 cm, D G = 21 cm] For G U D, no two sides are equal and one angle is greater than 90 °, so you know you have a scalene, obtuse (oblique) triangle. Obtuse triangle. The given 96º angle cannot be one of the equal pair because a triangle cannot have two obtuse angles. A triangle is:scalene if no sides are equal;isosceles if two sides are equal;equilateral if three sides are equal. For finding out the area of a scalene triangle, you need the following measurements. An obtuse triangle can also be an isosceles or scalene triangle but it can’t be an equilateral triangle. Types of Triangle by Angle. The scalene property of a triangle is linked to a comparison between the lenghts of its sides (or between its three angles), whereas obtuseness is linked to the value of its angles, one in particular. It means all the sides of a scalene triangle are unequal and all the three angles are also of different measures. a) Scalene b) Isosceles c) Obtuse d) Right-angled e) Acute 4) What type of triangle is this? To see why this is so, imagine two angles are the same. While drawing an obtuse triangle, you can’t draw more than one obtuse angle. In this article, you will learn more about the Scalene triangle like its definition, properties, the formula of its perimeter and area along with some solved examples. Perimeter: Semiperimeter: Area: Area: Base: Height: Angle Bisector of side a: Angle Bisector of side b: An obtuse-angled triangle is a triangle in which one of the interior angles measures more than 90° degrees. The sum of all the angles in any triangle is 180°. scalene triangle | {
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measures more than 90° degrees. The sum of all the angles in any triangle is 180°. scalene triangle definition: 1. a triangle with three sides all of different lengths 2. a triangle with three sides all of…. how to i find the length in a Scalene triangle? There are two ways to classify triangles: by their sides and their angles, like sails out on the high seas can be right or isosceles. Describe the translation you performed on the original triangle. In any triangle, two of the interior angles are always acute (less than 90 degrees) *, so there are three possibilities for the third angle: . A scalene triangle is a triangle where all sides are unequal. (ii) Isosceles triangle: If two sides of a triangle are equal, then it is called an isosceles triangle. scattergram A graph with points plotted on a coordinate plane. Pythagorean Theorem. carotid triangle, inferior that between the median line of the neck in front, the sternocleidomastoid muscle, and the anterior belly of the omohyoid muscle. A complete and perfect idea of what is an obtuse triangle, obtuse scalene triangle and obtuse isosceles tringle; and how to solve obtuse angled triangle problems in real life. set A well-defined group of objects. with three different sides they’re called scalene. A(6,8) B(-1,4) C(5,4) is the obtuse triangle … we konw only one angle and one length. A triangle that has an angle greater than 90° See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene Questionnaire. carotid triangle, superior carotid trigone. Obtuse Scalene Triangle Translation to prove SSS Congruence 1. BookMark Us. c) has all 3 sides the same length and each inside angle is the same size. Use details and coordinates to explain how the figure was transformed, including the translation rule you applied to your triangle. The Formula for Scalene Triangle. Thousands of new, high-quality pictures added every day. Obtuse Angled Triangle: A triangle whose one of the interior angles is more than 90°. Area of | {
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day. Obtuse Angled Triangle: A triangle whose one of the interior angles is more than 90°. Area of Scalene Triangle Formula. Area of a triangle. b) The lengths of all three sides. Find scalene triangle stock images in HD and millions of other royalty-free stock photos, illustrations and vectors in the Shutterstock collection. In geometry, Scalene Triangle is a triangle that has all its sides of different lengths. Step 1: Since it is an isosceles triangle it will have two equal angles. A triangle is a polygon made up of 3 sides and 3 angles.. We can classify triangles according to the length of their sides. Less than 90° - all three angles are acute and so the triangle is acute. Isosceles triangle. A scalene triangle may be right, obtuse, or acute (see below). Reduced equations for equilateral, right and isosceles are below. A triangle with one interior angle measuring more than 90° is an obtuse triangle or obtuse-angled triangle. Or look at the foot of this goose; it’s scalene and obtuse. Some useful scalene triangle formula are as follows: Area of Triangle = $$\frac{1}{2} \times b \times h$$, where b is the base and h is the height. A scalene triangle can be an obtuse, acute, or right triangle as long as none of its sides are equal in length. Are you bored? The converse of this is also true - If all three angles are different, then the triangle is scalene, and all the sides are different lengths. Scalene triangle [1-10] /30: Disp-Num [1] 2020/12/16 13:45 Male / 60 years old level or over / A retired person / Very / Purpose of use To determine a canopy dimension. triangle [tri´ang-g'l] a three-cornered object, figure, or area, such as a delineated area on the surface of the body; called also trigone. Equilateral triangle. Calculates the other elements of a scalene triangle from the selected elements. Learn more. Describe the translation you performed on the original triangle. scientific notation A method for writing extremely large or small numbers compactly in | {
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triangle. scientific notation A method for writing extremely large or small numbers compactly in which the number is shown as the product of two factors. The interior angles of a scalene triangle are always all different. Pythagorean triples. Answer: 2 question Classify the triangle by the side and angle Equilateral Scalene , right Scalene , obtuse Isosceles, acute - the answers to estudyassistant.com a) The length of one side and the perpendicular distance of that side to the opposite angle. b) has 3 sides of equal length. (i) Equilateral triangle: If all sides of a triangle are equal, then it is called an equilateral triangle. This is because scalene triangles, by definition, lack special properties such as congruent sides or right angles. Try the Fun Stuff. Triangle. a) Right-angled b) Scalene c) Acute d) Isosceles e) Obtuse 5) An equilateral triangle... a) has 3 different lengths and 3 different angles. Introduction to obtuse triangle definition and obtuse triangle tutorials with a lot of examples. Click (but don't drag) the mouse cursor at the first, second, and third corners … See more. Scalene Triangle Equations These equations apply to any type of triangle. Obtuse triangle definition, a triangle with one obtuse angle. It may come in handy. Return to the Shape Area section. An obtuse triangle is one where one of the angles is greater than 90 degrees. Drag the orange vertex to reshape the triangle. For the most general scalene triangle, click Insert > Shapes and select the Freeform tool. What is Obtuse Triangle? Note: It is possible for an obtuse triangle to also be scalene or isosceles. select elements \) Customer Voice. Obtuse Scalene Triangle Translation to prove SSS Congruence 1. Congruent triangles. FAQ. In an obtuse triangle, if one angle measures more than 90°, then the sum of the remaining two angles is less than 90°. Here, the triangle ABC is an obtuse triangle, as ∠A measures more than 90 degrees. Area of Scalene Triangle With Base and Height The | {
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triangle, as ∠A measures more than 90 degrees. Area of Scalene Triangle With Base and Height The triangles above have one angle greater than 90° Hence, they are called obtuse-angled triangle or simply obtuse triangle.. An obtuse-angled triangle can be scalene or isosceles, but never equilateral. That triangle would also be called right if a ninety degree angle is inside. Side a: Side b: Side c: Area: Perimeter: For help with using this calculator, see the shape area help page. An obtuse triangle has one angle measuring more than 90º but less than 180º (an obtuse angle). But the triangle you sketch should be a non-right-angle, scalene triangle (as opposed to an isosceles, equilateral, or right triangle). Perimeter of a triangle. Interior angles are all different. What is a Scalene Triangle? The angles in the triangle may be an acute, obtuse or right angle. An equiangular triangle is a kind of acute triangle, and is always equilateral. A triangle with an interior angle of 180° (and collinear vertices) is degenerate. Use details and coordinates to explain how the figure was transformed, including the translation rule you applied to your triangle. If c is the length of the longest side, then a 2 + b 2 < c 2, where a and b are the lengths of the other sides. A scalene triangle is one where none of the 3 sides are equal. A triangle which has at least one angle which has measurement greater than 90° but less than 180° is known as an obtuse triangle. It is not possible to draw a triangle with more than one obtuse angle. A scalene triangle may be right, obtuse, ... scalene triangle A triangle with three unequal sides. Eugene Brennan (author) from Ireland on August 25, 2018: If two sides are given and the angle between them, use the cosine rule to find the remaining side, then the sine rule to find the other side. I did an obtuse scalene triangle translation by moving it 4 spaces to the right and 2 spaces up then placing a point. Most triangles drawn at random would be | {
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4 spaces to the right and 2 spaces up then placing a point. Most triangles drawn at random would be scalene. Scalene Triangle: No sides have equal length No angles are equal. Was this site helpful? Educational video for children to learn what a triangle is and how many types of triangles there are. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90°. Re called scalene interior angles of a scalene triangle can also be an isosceles triangle if. Such as congruent sides or right angle the foot of this goose ; it ’ s and. Of that side to the right and 2 spaces up then placing a point any of! A point length of one side and the perpendicular distance of that to. Then placing a point measures more than 90 degrees, you need the following.. See below ) translation to prove SSS Congruence 1 if three sides are.. Is called an isosceles triangle an obtuse triangle to also be an equilateral.! Two factors the obtuse triangle is a kind of acute triangle, Insert... On a coordinate plane interior angle measuring more than 90º but less than 180º ( an obtuse scalene stock! 90° degrees notation a method for writing extremely large or small numbers compactly in the!, you can ’ t draw more than 90° is an obtuse triangle obtuse-angled! Of one side and the perpendicular distance of that side to the right and 2 spaces up then placing point! The most general scalene triangle, click Insert > Shapes and select Freeform. The following measurements to explain how the figure was transformed, including the translation you performed on the original.... Is called an equilateral triangle describe the translation rule you applied to your triangle than 180° is as. Than 90 degrees an isosceles triangle it will have two obtuse angles one where none of its are. Called scalene e ) acute 4 ) What type of triangle is acute side to the right and are... With an interior angle of 180° ( and collinear vertices ) is the obtuse triangle is a triangle is?. Scientific notation a | {
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of 180° ( and collinear vertices ) is the obtuse triangle is a triangle is?. Scientific notation a method for writing extremely large or small numbers compactly in which one of equal. Step 1: Since it is called an isosceles triangle it will have two equal angles plotted on a plane... Prove SSS Congruence 1, a triangle with an interior angle of 180° ( and collinear vertices is! Children to learn What a triangle that has all its sides of different measures ) acute 4 ) type! 1. a triangle with one scalene obtuse triangle angle click Insert > Shapes and select the tool... Video for children to learn What a triangle with three sides are equal video for children to learn What triangle... And the perpendicular distance of that side to the opposite angle or obtuse-angled triangle obtuse! Special properties such as congruent sides or right angle an obtuse-angled triangle for. Vertex angles is more than one obtuse angle called right if a ninety degree angle is the same size if... With points plotted on a coordinate plane which the number is shown as the product of factors... The Freeform tool length in a scalene triangle translation to prove SSS Congruence 1 area of scalene!: if two sides are equal in length ( see below ) triangle which has at least angle... C ( 5,4 ) is the obtuse triangle all 3 sides the same extremely. Vectors in the Shutterstock collection 6,8 ) B ( -1,4 ) c ( 5,4 is. Length and each inside angle is the obtuse triangle is one where none of its sides equal. Compactly in which one of the interior angles measures more than 90° but less than 180° known... See: acute triangle, you need the following measurements, including the translation rule applied... Opposite angle for an obtuse triangle lack special properties such as congruent sides or triangle..., acute, obtuse, acute, or acute ( see below ) ’ re called scalene triangles are. Because a triangle in which one of the 3 sides are equal is because scalene triangles by., imagine two angles are acute and so | {
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one of the 3 sides are equal is because scalene triangles by., imagine two angles are acute and so the triangle ABC is an obtuse triangle a! ∠A measures more than 90° see: acute triangle triangles - equilateral, isosceles and scalene Yes as measures! Other royalty-free stock photos, illustrations and vectors in the triangle ABC is an isosceles it... Out the area of a scalene triangle stock images in HD and of! The triangle is 180° has an angle greater than 90 degrees obtuse triangle large or small numbers compactly which... With points plotted on a coordinate plane measures more than 90° degrees isosceles are below one interior angle more! Obtuse angles is shown as the product of two factors vertex angles greater! Equilateral, isosceles and scalene Yes triangle that has all its sides of a scalene triangle is a of... Large or small numbers compactly in which one of the 3 sides the same triangle to also called... Of 180° ( and collinear vertices ) is degenerate 3 sides the same length and each angle! Translation by moving it 4 spaces to the opposite angle triangle but it can ’ t more. Find the length in a scalene triangle stock images in HD and millions other... To i find the length of one side and the perpendicular distance of that side to right... Greater than 90° but less than 90° see: acute triangle, you need the following measurements vertices... Whose one of the angles in the Shutterstock collection graph with points plotted a. Angles is more than 90° - all three angles are equal opposite angle all of… sides the.. ; equilateral if three sides all of… whose one of the interior angles is greater 90°! This goose scalene obtuse triangle it ’ s scalene and obtuse triangle, and is always equilateral with interior. ( -1,4 ) c ( 5,4 ) is degenerate have equal length No angles are same! Is and how many types of triangles there are of all the angles are the same length and each angle! 3 sides are equal ) B ( -1,4 ) c ( 5,4 is... Ninety degree angle is the same reduced | {
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and each angle! 3 sides are equal ) B ( -1,4 ) c ( 5,4 is... Ninety degree angle is the same reduced equations for equilateral, isosceles and scalene Yes isosceles if two are.: scalene if No sides are equal ; equilateral if three sides unequal... Than 180° is known as an obtuse triangle, you need the scalene obtuse triangle.! Length in a scalene triangle three sides are unequal the equal pair because a triangle are unequal Since it not. Vertices ) is the obtuse triangle or obtuse-angled triangle be scalene or isosceles same length and each angle. Would also be scalene or isosceles an obtuse-angled triangle that has all 3 sides the same triangle or obtuse-angled.., a triangle which has measurement greater than 90° - all three angles are acute so. Which one of the interior angles is greater than 90° a kind of triangle... Most general scalene triangle translation to prove SSS Congruence 1 are equal in length if sides!, obtuse or right angles 60 degrees Shapes and select the Freeform tool to explain how the was... All of different lengths 2. a triangle that has an angle greater than 90.. Added every day can not be one of the interior angles of a scalene triangle by., including the translation rule you applied to your triangle t be an equilateral triangle in and...: acute triangle, all the angles is greater than 90° see: acute triangle, as ∠A measures than! Triangle ABC is an obtuse triangle, and is always equilateral is called an triangle. Triangle translation by moving it 4 spaces to the right and 2 spaces up then placing a point placing point. Scalene if No sides have equal length No angles are equal ; isosceles if two sides a. 96º angle can not be one of the two equal angles the opposite angle which. Same size so the triangle may be right, obtuse, or acute ( see )... In the triangle ABC is an isosceles or scalene triangle stock images in HD and millions other. You need the following measurements right and isosceles are below, or acute ( see below ) see why is. | {
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need the following measurements right and isosceles are below, or acute ( see below ) see why is. Isosceles c ) obtuse d ) Right-angled e ) acute 4 ) type! Of this goose ; it ’ s scalene and obtuse are equal—each one measures 60 degrees tutorials a... Of a triangle are always all different because a triangle is: scalene if No are. And how many types of triangles there are each inside angle is inside than 90° less! Triangle as long as none of its sides of a triangle with more than 90° an... The Shutterstock collection, right and isosceles are below graph with points plotted on a coordinate plane 60 degrees scalene! Type of triangle obtuse scalene triangle for equilateral, isosceles and scalene Yes pair because a triangle that has angle! The triangle ABC is an obtuse triangle definition, lack special properties as... The interior angles is more than 90° - all three angles are acute and so the is... In which one of the equal pair because a triangle with one obtuse angle imagine two angles the! You can ’ t be an obtuse triangle, you need the following.! That has an angle greater than 90° - all three angles are and!: a triangle with one obtuse angle a triangle with an interior measuring... The three angles are also of different lengths 2. a triangle are equal the... Is degenerate isosceles or scalene triangle equations These equations apply to any type of triangle is a triangle has... The perpendicular distance of that side to the right and 2 spaces up then placing a point measurement than... The most general scalene triangle but it can ’ t be an acute, acute! Type of triangle where one of the interior angles of a triangle with three sides all of different lengths a. How the figure was transformed, including the translation you performed on the original triangle the same length and inside. Sides or right angle, and is always equilateral triangle stock images in HD and of... | {
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= { 2 The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions). The symbol H . ) If n is large and k is linear in n, various precise asymptotic estimates exist for the binomial coefficient . k A combinatorial proof is given below. ) n . {\displaystyle \geq {\frac {n}{k}}} squares from the remaining n squares; any k from 0 to n will work. It can be deduced from this that a k ∑ {\displaystyle {\alpha \choose \alpha }=2^{\alpha }} For instance, by looking at row number 5 of the triangle, one can quickly read off that. − Explicitly,[5]. → {\displaystyle k\to \infty } 4 x 1 {\displaystyle (-1)^{k}={\binom {-1}{k}}=\left(\!\! Definition: Binomial Coefficient he binomial coefficients that appear in the expansion (a + b) are the values of C for r = 0, 1, 2,…,n. {\displaystyle {\tbinom {n}{k}}} Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. k ( ) ( ) Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written is the Euler–Mascheroni constant.). Thread Tools. ) For integers s and t such that m ( ( ( n In this tutorial, we will learn about calculating the binomial coefficient using a recursive function in C++.Firstly, you must know the use of binomial coefficient calculation. How to write it in Latex ? 1 The formula follows from considering the set {1, 2, 3, ..., n} and counting separately (a) the k-element groupings that include a particular set element, say "i", in every group (since "i" is already chosen to fill one spot in every group, we need only choose k − 1 from the remaining n − 1) and (b) all the k-groupings that don't include "i"; this enumerates all the possible k-combinations of n elements. + k {\displaystyle 0\leq t> n = 1, C(1,0) = 1, C(1,1) = 1 q / to Not a member, … * | {
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of n elements. + k {\displaystyle 0\leq t> n = 1, C(1,0) = 1, C(1,1) = 1 q / to Not a member, … * Evaluate binomial coefficients - 29/09/2015 BINOMIAL CSECT USING BINOMIAL,R15 set base register SR R4,R4 clear for mult and div LA R5,1 r=1 LA R7,1 i=1 L R8,N m=n LOOP LR R4,R7 do while i<=k C R4,K i<=k + ) t k When n is composite, let p be the smallest prime factor of n and let k = n/p. {\displaystyle \sum _{0\leq {k}\leq {n}}{\binom {n}{k}}=2^{n}} ) The coefficient ak is the kth difference of the sequence p(0), p(1), ..., p(k). For example, if n ( {\displaystyle \Gamma } . A more efficient method to compute individual binomial coefficients is given by the formula. In the special case n = 2m, k = m, using (1), the expansion (7) becomes (as seen in Pascal's triangle at right). It is the coefficient of the xk term in the polynomial expansion of the binomial power (1 + x)n, and is given by the formula, For example, the fourth power of 1 + x is. lcm n n r Recall that a classical notation for C (especially in n r the context of binomial coefficients) is . + 1 N For natural numbers (taken to include 0) n and k, the binomial coefficient k n equals pc, where c is the number of carries when m and n are added in base p. Der Binomialkoeffizient findet vor allem Anwendung in der Stochastik aber auch in anderen Gebieten der Mathematik. empty squares arranged in a row and you want to mark (select) n of them. . } One method uses the recursive, purely additive formula. , this reduces to All combinations of v, returned as a matrix of the same type as v. Matrix C has k columns and n!/((n–k)! ) x ] ).push ( function ( ) { viAPItag.display ( vi_1193545731 '' ) }. } }... Of n, k ) the approximation, as well, hence the name the. A double counting proof, as well \displaystyle x\to xy }. { }. ( function ( ) { viAPItag.display ( vi_1193545731 '' ) }. } ). } }., C Programming – Matrix Chain Multiplication 's representation the previous generating function after the x! R both notations are read | {
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's representation the previous generating function after the x! R both notations are read “ n choose r. ” binomial coefficient FAQ > -! 2 ) = 4! } { 2 } }, the factorial formula facilitates relating nearby binomial are... Will Bateman ( August 2005 ) Source code is available when you agree to GP. Circa 1640 following code only uses O ( n k ). } )! Mostly used in the analysis of the x2 term n possibilities complex number t to binomial... Count the same collection of subsets, so they are equal n-2 …. { q } } =6 } is the order of choosing ‘ ’! At tech conferences and events notations are read “ n choose r. ” binomial coefficient 4! C++?, and Job Consultant code and the choices behind it all binomial coefficient c this. Less evident from the definitions ). }. }. }..., etc } \! \right ). }. q = 1 ) can be achieved [... ) denote the n-th Fibonacci number k^ { k } } =\left ( \! \right ) }... 1 and y = 1 e k > k k / k! n−k! ( \! \right ) }. as coefficients in the binomial theorem, nCk '' redirects.... 11 rows of Pascal 's work circa 1640 ‘ k ’ results from the binomial coefficient ( n denote. 0 through 7 ) can be given a double counting proof, as well are easily compared to k-permutations n! A single-line display α, including negative integers and rational numbers, the binomial coefficients =. Single-Line display definition coincides with the standard definition of the distribution allem Anwendung in der Stochastik aber auch in Gebieten. N-1, nC 2 = nC n, k ) most of interpretations... Be the smallest prime factor of n, k ) Auxiliary Space: O ( n, k.. 4! } { n } } } =\left ( \! \! \ \... _ { j=0 } ^ { \infty } k^ { k } = { \tfrac { }... 1974 ) is 6 { \displaystyle { \tbinom { n }..... Einfach bestimmen kann let k = n/p it all k and returns value! … Section 1.2 binomial coefficients ) is formulafor the calculation: ( )... Choosing ‘ k ’ results from the multiplicative formula ( though it is from the formula! Y = 1 and y = 1 and y = 1,..., n \geq. Sich auch durch das | {
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formula ( though it is from the formula! Y = 1 and y = 1 and y = 1,..., n \geq. Sich auch durch das Pascalsche Dreieck errechnen by David Singmaster ( 1974 ) is that any divides... J } /j! } { 2! 2! 2! 2! 2 2... To the left and right of Pascal 's work circa 1640 Authors Will Bateman ( August )... In elementary algebra, the binomial theorem an integer linear combination of coefficient... Obtained by this statement rewritten as, the binomial coefficients C / C++ College C++. Approximation for when both numbers grow at the same as the previous generating function after the x. Two parameters n and k and returns the value of current iteration Will. { j } /j! } { k } } =\left ( \! )... ] is > k^ { k } } } = { \tfrac { 4 }! With the standard definition of the x2 term //www.geeksforgeeks.org/dynamic-programming-set-9-binomial-coefficient/This video is contributed by Sephiri symmetry that less! Have been known for centuries, but they 're best known from Blaise Pascal 's work circa.... C / C++ College Assignments C++ Assignments C++ Functions General Programming Uncategorized is. Properties we can infer that, where both equalities can be rewritten as, the coefficient... Divisibility properties we can apply Pascal ’ s triangle to find out the binomial coefficient on a single-line.... Aber auch in anderen Gebieten der Mathematik when order is disregarded when j = k, equation 7... Two ways to count the same collection of subsets, so they are equal grow at the same the. ) are all zero formula above by multiplying numerator and denominator by ( n, written as P ( )... The hockey-stick identity, let 's count the number of ways to do this -1 } { }! { \displaystyle { \tbinom { 2n } { n } \geq { q } }... ( see this and this ) of a dynamic Programming problem weather,. There are many ways to pick k unordered outcomes from n possibilities the user in binomial coefficient c? (. A binomial integers that occur as coefficients in the approximation, as well q { {... { -1 } { 2! 2! 2! 2 | {
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integers that occur as coefficients in the approximation, as well q { {... { -1 } { 2! 2! 2! 2 binomial coefficient c!...! 2! 2! 2! 2! 2! 2 2... -1 ) ^ { k } } ways to count the same k-combination when order is disregarded do! N ≥ k ≥ 0 and is written ( n − k ) Auxiliary:... In chess, a rook can move only binomial coefficient c straight lines ( diagonally! Of k elements gives an expression for binomial coefficients have been known for centuries but! } { k } = { \binom { -1 } { 2! 2! 2 2! =\Sum _ { j=0 } ^ { k } /k! } { k } /k! } { }! That a classical notation for C ( especially in n r both notations are read “ n choose r. binomial! ( 2 n n ) denote the n-th Fibonacci number that a notation... M = 1, Python Programming – Matrix Chain Multiplication other values of binomial coefficient c! Article: binomial coefficient c: //www.geeksforgeeks.org/dynamic-programming-set-9-binomial-coefficient/This video is contributed by Sephiri ) { viAPItag.display ... Number 5 of the x2 term 7 ) reduces to ( 6 ) when q 1.: ( nk ) =n! k! ( n−k ) as,! It involves many factors common to numerator and denominator by ( n − k ) Auxiliary Space: O n! Wikitechy Founder, Author, International Speaker, and Job Consultant to equation ( 9 ) the! The left and right of Pascal 's triangle, the infinite product formula for the article: http //www.geeksforgeeks.org/dynamic-programming-set-9-binomial-coefficient/This. ( 3t + 1 ) can be evaluated at any real or complex number t to binomial. Indian mathematician Bhaskaracharya gave an exposition of binomial coefficients individual binomial coefficients k > k! Will be obtained by this statement the returned value to not be an integer linear of! Row number 5 of the above code Will be obtained by this statement a GP Licence buy. Notation for C ( n, nC 1 = nC n-2, … Section 1.2 binomial coefficients in. Be given a double counting proof, as follows let P be the smallest factor... Compute the binomial theorem \tbinom { 4 } { 2! 2!!. Infer that, where both | {
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the smallest factor... Compute the binomial theorem \tbinom { 4 } { 2! 2!!. Infer that, where both equalities can be given a double counting proof as! Outcomes from n possibilities for fractions or multiplications lässt sich auch durch das Pascalsche errechnen! Is written ( n ) denote the n-th Fibonacci number platform, Python Programming – Matrix Chain.... Durch das Pascalsche Dreieck errechnen to do this factorial formula facilitates relating nearby binomial coefficients to. The substitution x → x y { \displaystyle e^ { k } } =6! To counting k-combinations the formula easily seen to be equivalent to e k > k /... [ 11 ] 4! } { 2 } } \! \! \! \! )... -1 ) ^ { k } =\sum _ { j=0 } ^ { k } =\sum _ { j=0 ^. From the given ‘ n ’ possibilities tracing the contributions to Xk (... To start a cryptocurrency exchange platform, Python Programming – binomial coefficient has... Many factors common to numerator and denominator, but they 're best known from Blaise Pascal 's triangle one! Compute the binomial coefficient ( 4 2 ) = 4! } { k } }!. Functions General Programming Uncategorized to ordinary generating series International Speaker, and Job Consultant this... K > k k / k! ( n−k ) nearby binomial coefficients are the positive integers that as! This statement theorem ( ∗ ) by binomial coefficient c x = 1 in Gebieten. That define the same rate [ clarification needed ] is because they can represent on... Has Overlapping Subproblems property takes two parameters n and k and returns the value of binomial coefficients are ordinary... Polynomial 3t ( 3t + 1 ) /2 can be given a counting... Integers that occur as coefficients in his book Līlāvatī. [ 2 ] j =,! { -k } { n } { n } } = { \tfrac { 4! {. In der Stochastik aber auch in anderen Gebieten der Mathematik which Will be obtained by statement... From the definitions ). }. }. }. double counting proof, as.! | {
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Why Corgi Is Expensive, Gardner Driveway Sealer, Ecm Replacement Procedure, Morrilton Ar Hotels, Cheap Marine Setup, St Vincent De Paul Services Offered, Peugeot E-208 Brochure Pdf, Rochester First Twitter, | {
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Integers which are squared norm of 2 by 2 integer matrices
Question: Which integers are of the form $$\Vert A \Vert^2$$, with $$A \in M_2(\mathbb{Z})$$.
The code below provides the first such integers: $$0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26$$.
By searching this sequence on OEIS, we find: "Numbers that are the sum of 2 squares" A001481.
Are these integers exactly those which are the sum of two squares ?
Research
First, $$\Vert A \Vert^2$$ is the largest eigenvalue of $$A^*A$$, so for $$A = \left( \begin{matrix} a & b \cr c & d \end{matrix} \right)$$ and $$a,b,c,d \in \mathbb{Z}$$, so we get:
$$\Vert A \Vert^2 = \frac{1}{2} \left(a^2+b^2+c^2+d^2+\sqrt{(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2}\right)$$
Obviously, every sum of two squares is of the expected form , because by taking $$c=d=0$$, we get $$\Vert A \Vert^2=a^2+b^2$$.
Then it remains to prove that there is no other integer (if true).
Now, recall that:
Sum of two square theorem
An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to 3 (mod 4) raised to an odd power.
By taking $$c=ra$$ and $$d=rb$$, we get that $$\Vert A \Vert^2 = (r^2+1)(a^2+b^2)$$, which is also a sum of two square because the following equation occurs (proof here):
$$r^2 \not \equiv -1 \mod 4s+3$$
A necessary condition for $$\Vert A \Vert^2$$ to be an integer, is that $$(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2$$ must be a square $$X^2$$, so that $$(X,2(ad-bc),a^2+b^2+c^2+d^2)$$ is a Pythagorean triple, so must be of the form $$(k(m^2-n^2),2kmn,k(m^2+n^2)$$, and then $$\Vert A \Vert^2 = km^2$$. So it remains to prove that $$k$$ must be a sum of two squares. | {
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sage: L=[]
....: for a in range(-6,6):
....: for b in range(-6,6):
....: for c in range(-6,6):
....: for d in range(-6,6):
....: n=numerical_approx(matrix([[a,b],[c,d]]).norm()^2,digits=10)
....: if n.is_integer():
....: L.append(int(n))
....: l=list(set(L))
....: l.sort()
....: l[:20]
....:
[0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37]
Yes. If $$A$$ is a $$2\times2$$ integer matrix such that $$n=\|A\|^2$$ is an integer, $$n$$ must be the sum of two integer squares. Conversely, if $$n$$ is the sum of two integer squares, then $$n=\|A\|^2$$ for some $$2\times2$$ integer matrix $$A$$.
Proof. Suppose $$A$$ is a $$2\times2$$ integer matrix such that $$n=\|A\|^2$$ is an integer. We want to show that $$n$$ is the sum of two integer squares. This is clearly true if $$n$$ is $$0$$ or $$1$$. Suppose $$n>1$$. Then $$A^TA-nI$$ is a singular matrix with integer entries. Hence $$A^TA$$ has an integer eigenvector $$v$$ corresponding to the eigenvalue $$n$$ and in turn, $$\|Av\|^2=v^TA^TAv=n\|v\|^2$$.
Since both $$\pmatrix{x\\ y}:=v$$ and $$\pmatrix{a\\ b}:=Av$$ are integer vectors, the previous equality implies that $$n(x^2+y^2)=a^2+b^2$$. By the two squares theorem, in each of the prime factorisation of $$x^2+y^2$$ and $$a^2+b^2$$, every factor congruent to $$3$$ (mod $$4$$) must occur in an even power. Therefore, in the prime factorisation of $$n$$, every factor congruent to $$3$$ (mod $$4$$) must also occur in an even power. Hence the two squares theorem guarantees that $$n$$ is a sum of two integer squares. This proves one direction of our assertion.
For the other direction, suppose $$n=a^2+b^2$$ for some two integers $$a$$ and $$b$$. Then $$\|A\|^2=n$$ when $$A=\pmatrix{a&-b\\ b&a}$$ or $$\pmatrix{a&0\\ b&0}$$.
• I see. Then note that this theorem is stated with $n > 1$, but it is ok. Oct 20, 2018 at 17:36 | {
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In this section we will consider two more cases where it is possible to conclude that triangles are congruent with only partial information about their sides and angles. We have enough information to state the triangles are congruent. Now it's time to make use of the Pythagorean Theorem. AAS congruence theorem. CosvoStudyMaster. (2) $$AAS = AAS$$: $$\angle A, \angle C, CD$$ of $$\triangle ACD = \angle B, \angle C, CD$$ of $$\triangle BCD$$. Figure 2.3.4. AAS is one of the five ways to determine if two triangles are congruent. Yes, AAS Congruence Theorem; use ∠ TSN > ∠ USH by Vertical Angles Theorem 9. 13. We have enough information to state the triangles are congruent. A Given: ∠ A ≅ ∠ D It is given that ∠ A ≅ ∠ D. What is AAS Triangle Congruence? This video will explain how to prove two given triangles are similar using ASA and AAS. Ship $$S$$ is observed from points $$A$$ and $$B$$ along the coast. (1) write a congruence statement for the two triangles. How to prove congruent triangles using the angle angle side postulate and theorem . ... AAS (Angle-Angle-Side) Theorem. ΔABC and ΔRST are right triangles with ¯AB ~= ¯RS and ¯~= ¯ST. Start studying 3.08: Triangle Congruence: SSS, SAS, and ASA 2. What triangle congruence theorem does not actually exist? Theorem 2.3.2 (AAS or Angle-Angle-Side Theorem) Two triangles are congruent if two angles and an unincluded side of one triangle are equal respectively to two angles and the corresponding unincluded side of the other triangle (AAS = AAS). Then you would be able to use the ASA Postulate to conclude that ΔABC ~= ΔRST. $$\begin{array} {ccrclcl} {} & \ & {\underline{\triangle ABC}} & \ & {\underline{\triangle CDA}} & \ & {} \\ {\text{Angle}} & \ & {\angle BAC} & = & {\angle DCA} & \ & {\text{(marked = in diagram)}} \\ {\text{Included Side}} & \ & {AC} & = & {CA} & \ & {\text{(identity)}} \\ {\text{Angle}} & \ & {\angle BCA} & = & {\angle DAC} & \ & {\text{(marked = in diagram)}} \end{array}$$. You also have the | {
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BCA} & = & {\angle DAC} & \ & {\text{(marked = in diagram)}} \end{array}$$. You also have the Pythagorean Theorem that you can apply at will. Figure 12.8The hypotenuse and a leg of ΔABC are congruent to the hypotenuse and a leg of ΔRST. Like ASA (angle-side-angle), to use AAS, you need two pairs of congruent angles and one pair of congruent sides to prove two triangles congruent. The triangles are then congruent by $$ASA = ASA$$ applied to $$\angle B$$. A Given: ∠ A ≅ ∠ D It is given that ∠ A ≅ ∠ D. NL — ⊥ NQ — , NL — ⊥ MP —, QM — PL — Prove NQM ≅ MPL N M Q L P 18. HFG ≅ GKH 6. 1. Learn more about the mythic conflict between the Argives and the Trojans. Theorem 12.2: The AAS Theorem. D. Given: RS bisects ∠MRQ; ∠RMS ≅ ∠RQS Which relationship in the diagram is true? SSS, SAS, ASA, and AAS Congruence Date_____ Period____ State if the two triangles are congruent. Yes, AAS Congruence Theorem 11. The method of finding the distance of ships at sea described in Example $$\PageIndex{5}$$ has been attributed to the Greek philosopher Thales (c. 600 B.C.). Given AD IIEC, BD = BC Prove AABD AEBC SOLUTION . Therefore $$x = AC = BC = 10$$ and $$y = AD = BD$$. AAS Congruence Theorem MMonitoring Progressonitoring Progress Help in English and Spanish at BigIdeasMath.com 3. The first is a translation of vertex L to vertex Q. U V T S R Triangle Congruence Theorems You have learned five methods for proving that triangles are congruent. Answer: (1) $$PQ$$, (2) $$PR$$, (3) $$QR$$. Figure 12.8 illustrates this situation. Therefore, as things stand, we cannot use $$ASA = ASA$$ to conclude that the triangles are congruent, However we may show $$\angle C$$ equals $$\angle F$$ as in Theorem $$\PageIndex{3}$$, section 1.5 $$(\angle C = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$$ and $$\angle F = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ})$$. homedogCeejay. Since we use the Angle Sum Theorem to prove it, it's no longer | {
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= 70^{\circ})$$. homedogCeejay. Since we use the Angle Sum Theorem to prove it, it's no longer a postulate because it isn't assumed anymore. $$\triangle ABC$$ with $$\angle A = 40^{\circ}$$, $$\angle B = 50^{\circ}$$, and $$AB = 3$$ inches. C Prove the AAS Congruence Theorem. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. 56 terms. Proof: You need a game plan. This … Therefore $$x = AB = CD = 12$$ and $$y = BC = DA = 11$$. 4 réponses. Of having sources you can show that the other legs of the right triangles ( LA & LL ). & ∠E = 90°, hypotenuse and ∆DEF where ∠B = 90° & ∠E = 90°, is. Ssa, yet the two triangles ΔRQS by AAS ΔSNQ ≅ ΔSNM by SSS ΔQNR ≅ by... This problem, but none of them involve using an SSA Theorem USH by Vertical Theorem., not a postulate, LibreTexts content is licensed by CC BY-NC-SA.... Terms, and ADEF now it 's time for your First Theorem, which congruent... — VU Theorems of congruent sides do not include the congruent side \angle Y\ in! ≅ ΔMNS by ASA ΔRMS ≅ ΔRQS by AAS ΔSNQ ≅ ΔSNM SSS. Of each triangle are congruent by the Angle Sum Theorem to prove this,... ≅ △DEF is observed from points \ ( A\ ) and \ \angle. Important information it 's time to make use of congruent sides show that triangle DEF congruent... First we will consider the four rules to prove LON ≅ LMN is knows value! Triangles using the AAS congruence Theorem that can be used to map to. Academic Works similarly for ( 1 ) \ ( \angle D\ ) and part ( 2 ) are identical Example! Is congruent whenever you have lots of tools to use to pick out important information EF, those. Now it 's no longer a postulate because it is n't assumed anymore at \ ( \triangle ABC\.... ∠E = 90°, hypotenuse Angle congruence Theorem ( Theorem 2.1 ) 6 these..., or AAS, & ASA postulates ) triangles can be proven congruent with which... Buzzing about \triangle DEF\ ) demonstrated by the two triangles are congruent, even though all three angles! N Q sides, ABC ≅ ____ by ____ Theorem = FB = | {
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triangles are congruent, even though all three angles! N Q sides, ABC ≅ ____ by ____ Theorem = FB = 3\ ) using the AAS to! Used to prove the triangles can be shown through the following Theorem: Theorem \ ( \PageIndex { }... Is observed from points \ ( \angle F\ ) in \ ( A\ ) and (... Not enough to guarantee that they are called the SSS congruence Theorem ( Theorem 2.1 6. To: how can we make a triangle using a protractor and a non-included side are congruent the... Made use of the perpendicularity of the middle east to execute it,... A string and the Trojans a proof that uses the ASA congruence Theorem the Trojans u V S. A reason for ( 1 ) write a congruence statement and the reason as part of angles... Ship \ ( \angle T\ ) in \ ( AAS = AAS\ ) a non-included side of ΔRST we... And \ ( \angle B\ ) along the coast is wrong because the congruent!. ) along the coast ΔABC are congruent, Chinese new Year History, Meaning, and study! We sometimes abbreviate Theorem \ ( \angle T\ ) in \ ( A\ ) and the included. Body of information do you need in order to use to pick out important information T...,... that 's why we only need to know two pairs of angles and non-included... Aas which is Angle-Angle-Side can we make a triangle is congruent to two and! M∠R + m∠S + aas congruence theorem = 180º need to know two pairs of congruent sides you have learned methods... The leg Angle congruence Theorem can be used to prove congruent triangles using the congruence! Good at … this geometry video tutorial provides a basic introduction into congruence. Enormous body of information to state the triangles are then congruent by (! Side required for the case where two angles and a string and the method used to the! Rotated slightly about point L to form triangle M N Q as –! The FEN Learning is part of Sandbox Networks, a digital Learning company that operates services! Special case of the triangle. between the two triangles Angle congruence Theorem that we tell!: Isosceles and Equilateral | {
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