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It is a well known fact (or nice exercise) to prove that for a positive integer $$m$$ with prime factorization $$m=\prod_{i=1}^np_i^{a_i}$$, where $$p_1,\ldots,p_n$$ are distinct prime numbers and $$a_1,\ldots,a_n$$ are positive integers, we have $$\sigma(m)=\prod_{i=1}^n\sum_{j=0}^{a_i}p_i^j.$$ In particular this shows that $$\sigma(m)$$ is odd if and only if for every odd prime $$p_i$$ dividing $$m$$ we have $$a_i\equiv0\pmod{2}$$, or equivalently either $$m=k^2$$ or $$m=2k^2$$ for some integer $$k$$. In particular we see that if $$b\equiv3\pmod{4}$$ then $$\sigma(b)$$ is even and hence $$r(b)\equiv r(b-1)\pmod{2}$$.
Here are some small values for $$r(b)$$: $$\begin{array}{r|ccccccccc} b&0&1&2&3&4&5&6&7\\ \hline r(b)&0&0&0&0&0&1&0&2\\ &&&&&&&&\\ b&8&9&10&11&12&13&14&15\\ \hline r(b)&2&2&3&7&2&7&10&8\\ &&&&&&&&\\ b&16&17&18&19&20&21&22&23\\ \hline r(b)&8&15&11&19&16&15&22&32\\ \end{array}$$
• It seems that $r(b)\equiv r(b+1)\pmod{2}\iff\sigma(b+1)\equiv1\pmod{2}$ is false. Take $b=6$. – mathlove Mar 23 at 15:24
• @mathlove I had spotted the same inconsistency. I have removed the error in my argument. I will try to complete this answer soon (likely tomorrow). – Servaes Mar 23 at 15:53 | {
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# If $d$ divides $2n$ and $d$ doesn't divide $n$, then $d$ is even
I have encountered a proof regarding dihedral groups we this fact is used:
If $$d\mid 2n$$ and $$d\nmid n$$, then $$d$$ is even and $${d\over 2}\mid n$$.
I can't seem to understand why this is true. If $$d\nmid n$$, then there are $$q,r \in \mathbb{Z}$$ so that $$0 < r < d$$ and $$n = qd + r$$. On the other hand, $$d\mid 2n$$ means that there is $$m \in \mathbb{Z}$$ so that $$2n = md$$. We need to somehow use these two facts.
Also, my second question is how we can naturally generalize this result?
• Hint: if $d$ was odd, $m$ would be even. – Wojowu Mar 15 at 19:28
• $2n =md$. And $2$ is prime. So $2|m$ or $2|d$. If $2|m$ then $n = \frac m2 d$ and $d|n$. If $2|d$ then $d$ is even. General. If $d|pn$ for a prime $p$ and $d\not \mid n$ then $p|d$. Even more general. If $d|ab$ then $\frac d{\gcd(d,b)}|a$. – fleablood Mar 15 at 21:22
Suppose $$d$$ is odd. Then $$d$$ and $$2$$ are relatively prime and $$d\mid 2n$$, so by Euclid lemma we have $$d\mid n$$.
A contradiction. So $$d$$ must be even.
We could have more general situation.
Say $$d\mid pn$$ for some prime $$p$$ and $$d$$ doesn't divide $$n$$. Then $$p\mid d$$.
The proof goes exactly the same as for $$p=2$$.
It's also possible to see what's going on simply by keeping track of factors of $$2$$; note that this kind of analysis could also be generalized to other prime factors, as has been illustrated in previously posted answers.
Let $$n:= 2^aQ,\ 2n=2^{a+1}Q, d:=2^bP$$ where $$Q$$ represents the product of all of the odd prime factors of $$n$$ and $$P$$ represents the product of all of the odd prime factors of $$d$$. We could write out all of those factors and explicitly show this, but it should be plain that $$d\mid 2n \Rightarrow P\mid Q$$. Note that thus far, the exponents $$a,b$$ might be $$0$$, so we have not assumed that $$d$$ is even.
$$d\mid 2n \Rightarrow b\le a+1$$
$$d\not \mid n \Rightarrow b>a$$ | {
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$$d\mid 2n \Rightarrow b\le a+1$$
$$d\not \mid n \Rightarrow b>a$$
Together, these establish $$b=a+1$$, meaning that $$d$$ has at least one factor of $$2$$ and is even, even if $$a=0$$.
This also illustrates the second point: the exponent of $$2$$ in $$d\over 2$$ is simply $$b-1=a$$. Hence $$\frac{d}{2} \mid n$$
If you know that the remainder $$r$$ in $$n=qd+r$$ with $$0\lt r\lt d$$ is unique, then from $$2n=md$$ we get
$$n=2n-n=md-(qd+r)=(m-q)d-r=(m-q+1)d+(d-r)=q'd+r'$$
with $$0\lt r'=d-r\lt d$$, so that, by uniqueness of the remainder, we have $$r'=r$$, i.e. $$d-r=r$$, hence $$d=2r$$.
$$d| 2n$$ then $$2n=kd;$$
Since $$kd$$ is even, $$2| kd$$.
Euclid's lemma:
1) $$2| k$$ or 2) $$2|d$$.
1) If $$2|k$$ then $$k=2k'.$$
$$2n=2k'd$$;
$$n=k'd$$, i.e. $$d|n$$ , a contradiction.
2) Hence $$2|d$$ , and we are done.
By below $$\ d\mid pn\iff\!\!\!\! \overbrace{d\mid n}^{\large\color{#0a0}{(d,p)}\ =\ \color{#c00}1}\!\!$$ or $$\ \overbrace{{d/\color{#c00}p}\mid n}^{\large\color{#0a0}{ (d,p)}\ =\ \color{#c00} p}\!,\$$ by $$\ \color{#0a0}{(d,p)}\mid \color{#c00}p\,$$ prime.
Lemma $$\,\ d\mid an\iff\smash[t]{\overbrace{ d/\color{#0a0}{(d,a)}\,\mid\, n,\,}\ }$$ where $$\,\ (x,y) := \gcd(x,y)$$
Proof $$\quad\ d\mid an\iff d\mid dn,an,\iff d\mid (dn,an)=(d,a)n\iff d/(d,a)\mid n$$
• Convention: $\ d/p\mid n\$ means $\ d/p\,$ is an integer, so $\,p\mid d\ \$ – Bill Dubuque Mar 15 at 20:23
Intuitively. If $$d|ab$$ and $$d\not \mid b$$ then "some part of $$d$$ must divide $$a$$". So if $$d|2n$$ but $$d\not \mid n$$ then some (non-trivial) part must divide $$2$$ and that part must be $$2$$ so $$d$$ is even.
.....
That's intuition. Let's make a proof.
.....
If $$d|ab$$. Let $$\gcd(d,b) = g$$ and let $$d = d'g$$ and $$b = b'g$$. It's easy to see that $$d'$$ and $$b'$$ are relatively prime: if $$b'$$ and $$d'$$ had any non-trivial factor, $$k$$, in common then $$kg$$ would be a common divisors of $$b$$ and $$d$$ contradicting that $$g$$ is the greatest common divisor. | {
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So $$d=d'g$$ and $$ab = ab'g$$ and $$d'g|ab'g$$ so $$d'|ab'$$. But $$d'$$ and $$b'$$ are relatively prime so they have no factors in common. So $$d'|a$$.
Now if $$d|b$$ then $$d'g|b'g$$ so $$d'|b'$$ but $$d'$$ and $$b'$$ are relatively prime so $$d' = 1$$ and $$d = \gcd(d,b)$$.
Lemma 1: $$d|b \iff \gcd(d,b) = d$$.
If $$d|ab$$ the $$d'|a$$ but if $$d\not \mid b$$ then $$\gcd(d,b)=g \ne d$$ so $$d' = \frac dg > 1$$. $$d'\ne 1$$ and $$d'|a$$.
So
Lemma 2: If $$d|ab$$ but $$d\not \mid b$$ then $$d' = \frac d{\gcd(d,b)} > 1$$ and $$d'|a$$.
So if $$d|2n$$ and $$d\not \mid n$$ then $$\frac d{\gcd(d,n)} > 1$$ and $$\frac d{\gcd(d,n)} |2$$.
So $$\frac d{\gcd(d,n)} = 2$$ and $$d = \frac d{\gcd(d,n)}\gcd(d,n) = 2\gcd(d,n)$$. And $$d$$ is even.
=====
Actually, this may be the best most general Theorem:
Theorem: If $$d|mn$$ then $$\frac d{\gcd(d,n)}|m$$.
I'll leave the proof to you, and I'll leave it to you to figure out how that implies if $$d|2n$$ and $$d\not\mid n$$ then $$d$$ is even. | {
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# Generate covariance matrix from related random variable distributions
Say I have a random variable $$X\sim\mathscr{N}\left(0,\sigma^2\right)$$ and another random variable $$Y=X+\varepsilon$$, where $$\varepsilon\sim\mathscr{N}\left(0,\eta^2\right)$$ is independent of $$X$$. I can calculate $$Cov\left(X,Y\right)=Cov\left(X,X+\varepsilon\right)=Var(X)=\sigma^2$$.
Is there a way to get Mathematica to generate a bivariate normal distribution automatically from the given information? Specifically, rather than working out the covariance matrix myself and then inputting to Mathematica, I would like to be able to do something like (implementing the above example):
\[ScriptCapitalX] = NormalDistribution[0, \[Sigma]]
\[CapitalEpsilon] = NormalDistribution[0, \[Eta]]
\[ScriptCapitalY] = TransformedDistribution[X + \[CurlyEpsilon], {X \[Distributed] \[ScriptCapitalX], \[CurlyEpsilon] \[Distributed] \[CapitalEpsilon]}]
\[Mu] = {Mean[\[ScriptCapitalX]], Mean[\[ScriptCapitalY]]}
\[CapitalSigma] = {{Variance[\[ScriptCapitalX]], Covariance[\[ScriptCapitalX], \[ScriptCapitalY]]}, {Covariance[\[ScriptCapitalX], \[ScriptCapitalY]], Variance[\[ScriptCapitalY]]}}
multi = MultinormalDistribution[\[Mu], \[CapitalSigma]]
want = Expectation[X \[Conditioned] Y = y, {Y, X} \[Distributed] multi]
Obviously, the Covariance function doesn't work that way, but at a minimum I would really like to have Mathematica do the covariance calculation. Also, if there's an easy way to get the mean vector and covariance matrix without explicitly calling Mean, Variance, etc. on individual elements to construct them, that would be even better. I am quite new to the Wolfram language. | {
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• Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. Feb 11 at 10:34
You were almost there:
\[ScriptCapitalX] = NormalDistribution[0, σ]
\[CapitalEpsilon] = NormalDistribution[0, η]
multi = TransformedDistribution[{X, X + ε},
{X \[Distributed] \[ScriptCapitalX], ε \[Distributed] \[CapitalEpsilon]}]
Covariance[multi][[1, 2]]
(* σ^2 *)
Expectation[X \[Conditioned] Y == y, {X, Y} \[Distributed] multi]
(* (y σ^2)/(η^2 + σ^2) *) | {
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# Equivalence of Definitions of Path Component
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## Theorem
The following definitions of the concept of Path Component in the context of Topology are equivalent:
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
### Equivalence Class
Let $\sim$ be the equivalence relation on $T$ defined as:
$x \sim y \iff x$ and $y$ are path-connected.
The equivalence classes of $\sim$ are called the path components of $T$.
If $x \in T$, then the path component of $T$ containing $x$ (that is, the set of points $y \in T$ with $x \sim y$) can be denoted by $\map {\operatorname{PC}_x} T$.
### Union of Path-Connected Sets
The path component of $T$ containing $x$ is defined as:
$\displaystyle \map {\operatorname{PC}_x} T = \bigcup \left\{{A \subseteq S: x \in A \land A}\right.$ is path-connected $\left.\right\}$
### Maximal Path-Connected Set
The path component of $T$ containing $x$ is defined as:
the maximal path-connected set of $T$ that contains $x$.
## Proof
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text { is path-connected in } T}$
Let $C = \bigcup \CC_x$.
### Lemma
$C$ is path-connected in $T$ and $C \in \CC_x$.
$\Box$
Let $C'$ be the equivalence class containing $x$ of the equivalence relation $\sim$ defined by:
$y \sim z$ if and only if $y$ and $z$ are connected in $T$.
### Equivalence Class equals Union of Path-Connected Sets
It needs to be shown that $C = C'$. | {
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### Equivalence Class equals Union of Path-Connected Sets
It needs to be shown that $C = C'$.
$\displaystyle y \in C'$ $\leadstoandfrom$ $\displaystyle x \text{ is path-connected to } y \text{ in } T$ Definition of $\sim$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle \exists B \text{ a connected set of } T, x \in B, y \in B$ Points are Path-Connected iff Contained in Path-Connected Set $\displaystyle$ $\leadstoandfrom$ $\displaystyle \exists B \in \CC_x : y \in B$ Equivalent definition $\displaystyle$ $\leadstoandfrom$ $\displaystyle y \in \bigcup \CC_x$ Definition of Set Union $\displaystyle$ $\leadstoandfrom$ $\displaystyle y \in C$ Definition of $C$
The result follows.
$\Box$
### Union of Path-Connected Sets is Maximal Path-Connected Set
Let $\tilde C$ be any path-connected set such that:
$C \subseteq \tilde C$
Then $x \in \tilde C$.
Hence $\tilde C \in \CC_x$.
$\tilde C \subseteq C$.
Hence $\tilde C = C$.
It follows that $C$ is a maximal path-connected set of $T$ by definition.
$\Box$
### Maximal Path-Connected Set is Union of Path-Connected Sets
Let $\tilde C$ be a maximal path-connected set of $T$ that contains $x$.
By definition:
$\tilde C \in \CC_x$
$\tilde C \subseteq C$
By maximality of $\tilde C$:
$\tilde C = C$
$\blacksquare$ | {
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## Calculus, all content (2017 edition)
### Unit 1: Lesson 16
Limits of piecewise functions
# Worked example: point where a function is continuous
AP.CALC:
LIM‑2 (EU)
,
LIM‑2.A (LO)
,
LIM‑2.A.2 (EK)
Sal finds the limit of a piecewise function at the point between two different cases of the function. In this case, the two one-sided limits are equal, so the limit exists.
## Want to join the conversation?
• at he substitutes x with 3 but the function is defined between 0 < x < 3. In other words it's defined up to three but not three. So isn't that technically wrong?
• in that particular function you mentioned,, it is continuos until x approaches 3.
We needed to find the limit of f(x) of that function as x approached 3. That is only the idea of limit.. It is not defined at x =3, but we can find a value of f(x) so that it would have formed a continuos graph.. So we find the value of f(x) that would have been for x=3. | {
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Hope that was not complicated
• At how we can understand that function is continuous?
• It is not undefined for any positive argument. This means that there are no asymptotes or removable discontinuities, but proving continuity can be done in a variety of ways (for instance, noting that it is differentiable or noting that its inverse is differentiable etc.). Since differentiability is a stronger condition than continuity, all differentiable functions are also continuous over the differentiable interval.
• Can someone please suggest me a video in which all log functions are explained? Because I am only aware of the basic stuff but I guess as we proceed, we need to know the complicated functions of log too!
• What is a piecewise function?
• A piecewise function has different rules in different intervals. For example, look up aat this function:
f(x) = x^2 if x if x<4
= 4 if x<4 or x=4
Between the interval wich goes from negative infinity, it is x^2; and between the interval wich goes from 4 to positive infinity it is always four. | {
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To give a counterexample, g(x)=x^2+1 is not a piecewise function, because it is always equal to x^2+1; without mattering the value of x
• At and , what if the function is non continuous? What numbers do we plug in for x?
• If there is a jump discontinuity, then the limit from the left side and the limit from the right side will not be equal so the overall limit does not exist. You still have to plug in the same x value in both equations and you will get different values so the overall limit does not exist. But if there is a removable discontinuity, both the limit from the left side and the limit from the right side will be equal so the overall limit exists. In any case, you have to plug in the same x value.
• I forgot what a log is :/
• A logarithm is essentially the opposite of the exponential function. What this means is that if a^x = b, the log(base a) b = x.
• Why didn't we find the function value when x =3 to check if that is equal to the limit to satisfy the condition of continuity
• for a limit to be continuous, lim(x tends to c) f(x) = f(c).
IN this case we know that lim(x tends to 3) g(x) = log(9) but we don't know if
g(3)=log(9).
So how can we say that the limit is continuous
• We do know that g(3)=log(9), because the function g is defined at x=3 and we can plug 3 into the function.
g(3) = (4-3)*log(9) = 1*log(9) = log(9)
• I'm a bit confused by the title; this only proves that the limit of g(x) as x -> 3 exists, not that g(x) is necessarily continuous at that point right?
EDIT: nvm guys we used direct sub, so it is indeed continuous
• for a limit to be continuous, lim(x tends to c) f(x) = f(c). so we first see if the limit exists and then we see if it's equal to the function of that point, at this example we are not examining if the function is continuous it is continuous, he is showing us the case.
• How would I explain why g(x) is continuous?
## Video transcript | {
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- [Voiceover] So we have g of x being defined as the log of 3x when zero is less than x is less than three and four minus x times the log of nine when x is greater than or equal to three. So based on this definition of g of x, we want to find the limit of g of x as x approaches three, and once again, this three is right at the interface between these two clauses or these two cases. We go to this first case when x is between zero and three, when it's greater than zero and less than three, and then at three, we hit this case. So in order to find the limit, we want to find the limit from the left hand side which will have us dealing with this situation 'cause if we're less than three we're in this clause, and we also want to find a limit from the right hand side which would put us in this clause right over here, and then if both of those limits exist and if they are the same, then that is going to be the limit of this, so let's do that. So let me first go from the left hand side. So the limit as x approaches three from values less than three, so we're gonna approach from the left of g of x, well, this is equivalent to saying this is the limit as x approaches three from the negative side. When x is less than three, which is what's happening here, we're approaching three from the left, we're in this clause right over here. So we're gonna be operating right over there. That is what g of x is when we are less than three. So log of 3x, and since this function right over here is defined and continuous over the interval we care about, it's defined continuous for all x's greater than zero, we can just substitute three in here to see what it would be approaching. So this would be equal to log of three times three, or logarithm of nine, and once again when people just write log here within writing the base, it's implied that it is 10 right over here. So this is log base 10. That's just a good thing to know that sometimes gets missed a little bit. All right, now let's think about | {
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just a good thing to know that sometimes gets missed a little bit. All right, now let's think about the other case. Let's think about the situation where we are approaching three from the right hand side, from values greater than three. Well, we are now going to be in this scenario right over there, so this is going to be equal to the limit as x approaches three from the positive direction, from the right hand side of, well g of x is in this clause when we are greater than three, so four minus x times log of nine, and this looks like some type of a logarithm expression at first until you realize that log of nine is just a constant, log base 10 of nine is gonna be some number close to one. This expression would actually define a line. For x greater than or equal to three, g of x is just a line even though it looks a little bit complicated. And so this is actually defined for all real numbers, and it's also continuous for any x that you put into it. So to find this limit, to think about what is this expression approaching as we approach three from the positive direction, well we can just evaluate a three. So it's going to be four minus three times log of nine, well that's just one, so that's equal to log base 10 of nine. So the limit from the left equals the limit from the right. They're both log nine, so the answer here is log log of nine, and we are done. | {
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# Math Help - total number of combinations
1. ## total number of combinations
Hey. I have 3 problems I can't get. If anyone can help with how to do any of them, it would be appreciated.
1) To gain access to his account, a customer using an ATM must enter a four-digit code. If repetition of the same four digits is not allowed (for example, 5555) how many possible combinations are there?
2) Over the years, the state of California has used different combinations of letter of the alphabet and digits on its automobile license plates.
a. At one time license plates were issued that consisted of three letters followed by three digits. How many different license plates can be issued under this arrangement?
b. Later on, license plates were issued that consisted of three digits followed by three letters. How many different license plates can be issued under this arrangement?
3) An exam consists of ten true-or-false questions. In how many ways may the exam be completed if a penalty is imposed for each incorrect answer, so that a student may leave questions unanswered?
2. Originally Posted by xojuicy00xo
Hey. I have 3 problems I can't get. If anyone can help with how to do any of them, it would be appreciated.
1) To gain access to his account, a customer using an ATM must enter a four-digit code. If repetition of the same four digits is not allowed (for example, 5555) how many possible combinations are there?
how many combinations could we have if this restriction was not imposed? find that and then subtract the number of four repeated digits there can be (there are only 10: 0,0,0,0; 1,1,1,1; 2,2,2,2; 3,3,3,3; etc) | {
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2) Over the years, the state of California has used different combinations of letter of the alphabet and digits on its automobile license plates.
a. At one time license plates were issued that consisted of three letters followed by three digits. How many different license plates can be issued under this arrangement?
i assume repetitions are allowed, since you didn't say they weren't.
we choose 3 letters and 3 digits. there are 26 letters and 10 digits (including 0).
so for the first letter we have 26 choices.
for each of those choices, we have 26 choices for the 2nd letter
for each of those choices, we have 26 choices for the 3rd letter
for each of those choices, we have 10 choices for the 1st digit
for each of those choices, we have 10 choices for the 2nd digit
for each of those choices, we have 10 choices for the 3rd digit
so in all, we have 26*26*26*10*10*10 choices
b. Later on, license plates were issued that consisted of three digits followed by three letters. How many different license plates can be issued under this arrangement?
do this in the way i did the above
3) An exam consists of ten true-or-false questions. In how many ways may the exam be completed if a penalty is imposed for each incorrect answer, so that a student may leave questions unanswered?
maybe it's just me, or there is something missing from the question. is there a minimum grade we are hoping the student will get? how many penalties is the student allowed? ...
3. Hello, xojuicy00xo!
3) An exam consists of ten true-or-false questions.
In how many ways may the exam be completed
if a penalty is imposed for each incorrect answer,
so that a student may leave questions unanswered?
For each of the ten questions, the student has three choices:
There are: . $3^{10} \:=\:59,049$ ways.
4. Originally Posted by Soroban
Hello, xojuicy00xo!
For each of the ten questions, the student has three choices:
There are: . $3^{10} \:=\:59,049$ ways. | {
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# Calculating the amount of cubes needed to form a sum
Since I've never done any performance programming (Aside from the better choices such as array vs list etc. The real basics.), I should probably read up on it. But I had to start somewhere, so, someone I know - which is a much better programmer than I am - tasked me with making this "assignment":
You are given the total volume m of the building. Being given m can you find the number n of cubes you will have to build?
The parameter of the function findNb (find_nb, find-nb) will be an integer m and you have to return the integer n such as $n^3 + (n-1)^3 + \dots + 1^3 = m$ if such a n exists or -1 if there is no such n.
He gave me this template with it:
using System;
public class Program
{
public static void Main()
{
Console.WriteLine(findNb(4183059834009));
Console.WriteLine(findNb(24723578342962));
Console.WriteLine(findNb(135440716410000));
Console.WriteLine(findNb(40539911473216));
}
public static long findNb(long m)
{
}
}
In which I inserted:
for(long n = 1; n < (m / 3); n++)
{
double vol = 0;
for(long i = 0; i < n; i++)
vol += Math.Pow((n - i), 3);
if(vol > m)
return -1;
if(vol == m)
return n;
}
return -1;
This code works, but it takes incredibly long for the larger numbers, as is my main problem.
What can I do to shorten the time it takes for this code takes to complete?
• Please try to update your title to reflect what it is your code does, not what improvements you want to make. – forsvarir May 29 '16 at 18:36
• Is this any better, @forsvarir? – sxbrentxs May 29 '16 at 18:47
• Yes, the title change and addition of the function spec make your question much more engaging. – forsvarir May 29 '16 at 19:03
First of all, you have a bug:
variable vol is declared as double that causes the condition vol == m hardly to be satisfied.
Next, you don't need the Math.Pow method, you could relace it with just n * n * n.
And your loops look a bit confusing. Why m / 3? Why do you need the inner loop? | {
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The proposed code:
private static long findNb(long m)
{
long n = 1;
long vol = 0;
while (vol < m)
{
vol += n * n * n;
if (vol == m)
return n;
++n;
}
return -1;
}
• > Why m / 3? Why do you need the inner loop? Apologies, that was/is a remainder of me trying to get the code to execute fully on DotNETFiddle. And the loop was more to calculate the volume. But That's not needed anymore. Thanks for this. Could you tell me, though why n * n * n is better over Math.Pow? – sxbrentxs May 29 '16 at 19:23
• @sxbrentxs Math.Pow is intended for floating point math. You need only two integer multiplications, while Math.Pow calculates it as Exp(y * Log(x)) and also you need to cast arguments to double and back. – Dmitry May 29 '16 at 19:33
• Which makes Math.Pow obsolete and slow in this case. Cool! Thank you. – sxbrentxs May 29 '16 at 19:34
When there's math to be done, always check if some formula exists
$$1^3+2^3 + \dots+n^3 = m = (1+2 + \dots+n)^2$$
in conjunction with some other formula (interesting wikipedia article title)
$$\sum^n_{k=1}k = \frac{n(n+1)}{2}$$
you get some equation for the result
$$\sqrt{m} = 1+2 + \dots+n = \sum^n_{k=1}k = \frac{n(n+1)}{2}$$ and maybe some closed formula
\begin{align} \sqrt{m} &= \frac{n(n+1)}{2}\\ 2\sqrt{m} &= n(n+1)\\ 2\sqrt{m} &= n^2+n\\ 0&= n^2+n - 2\sqrt{m} \end{align}
Which has the 2 solutions
$$n_{1,2} = -\frac12 \pm\sqrt{\frac14+2\sqrt{m}}$$
As $n$ should be a positive number, only the first solution makes sense.
Return $-\frac12 +\sqrt{\frac14+2\sqrt{m}}$ if it is an integer or $-1$ otherwise.
The results are:
$$\begin{array}{r|r} m & n\\ \hline 4183059834009 & 2022\\ 24723578342962 & -1\\ 135440716410000 & 4824\\ 40539911473216 & 3568\\ \end{array}$$
24723578342962 is a very close call, because
$$\sum^{3153}_{k=1}k^3=24723578342961$$ | {
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24723578342962 is a very close call, because
$$\sum^{3153}_{k=1}k^3=24723578342961$$
is only 1 off. The floating point result is $n=3153.00000000003$. If you don't want to get false positives due to floating point precision (lack thereof), you can do the check by converting the result to some integer type and see if you get the exact number by doing the calculations with that.
• Holy shit. That's a lot to take in. I think I know what you did and what I have to do. Thank you! – sxbrentxs May 29 '16 at 19:27
• Wow! Very impressed to see an O(1) solution. +1. – ApproachingDarknessFish May 31 '16 at 0:58
For a significant speedup: In the outer loop, you vary n from 1 to some large value, and in the inner loop you add up n values.
Now in the next iteration of the outer loop, using n+1 instead of n, you run the inner loop again, this time adding up n+1 value. But just before that, in the previous loop, you already added up n values, so to get the sum of n+1 values, all you have to do is add value number n+1 to the sum that you already have.
Have a variable n and a variable sum_n. sum_n will be the sum of the first n cubes. Initially n = 0 and sum_n = 0. Inside the outer loop, increase n by 1 then increase sum_n by nnn.
For example, if n goes from 1 to 1000, you add up one value, then two, then three, and so on, and eventually you add up 1000 values. On average about 500. This change means you add only one value in each iteration of the outer loop.
This is an improvement that works for any task where you calculate consecutive sums. In this particular case, you can find a formula to calculate the sum. That isn't helpful by itself, since calculating the formula is likely slower than calculating the sums. However, you can reverse the formula and calculate what n should be, and then you just have to verify this. So you just need a small fixed number of operations, independent of the size of the m that you were given. | {
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# Showing $A \cup B \cup C$ is countable if $A, B$ are countable and $C$ is finite.
How does one go about showing $A \cup B \cup C$ is countable if $A, B$ are countable and $C$ is finite?
I understand most of the confusion for resolving set theory questions online seem to be the definition. For my course we consider the following definitions:
countable: Finite or $A \sim\mathbb{N}$
uncountable: not countable
finite: The empty set or $A \sim J_n$ where $n \in \mathbb{N}$
infinite: not finite
So I'm not sure if I have this right but looking at the above definitions the way I'm looking to approach this is to consider 6 different cases.
1. Where C is the empty set and A, B are both finite sets
2. Where C is the empty set and A is finite and B is countably infinite
3. Where C is the empty set and both A, B are countably infinite
4 - 6. Repeated above but with C being a non-empty finite set
This seems like quite a round about way but intuitively it seems to me like the only way to cover all bases according to the definitions. I'm hoping I might be absolutely wrong on this. Is there a simpler way to prove this? | {
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• You can reduce the time taken by first proving the lemma $A$ and $B$ both countable implies $A\cup B$ countable. By using the lemma twice you show $A\cup B\cup C = (A\cup B)\cup C$ is countable by noting that it is the union of two countable sets since what is in the parenthesis is also a countable set. Further cut down on the number of cases by saying "Without loss of generality suppose $|A|\leq |B|$" before working too hard since any situation where $|B|<|A|$ is proven the same way by a relabeling of the sets. Also, treat empty sets during the same case as finite sets. – JMoravitz Mar 21 '17 at 23:50
• You now need only prove the following three cases: $A$ finite and $B$ finite, $A$ finite and $B$ countably infinite, and $A$ countably infinite and $B$ countably infinite. In any of those cases, try to find a bijection between $A\cup B$ and a subset of $\Bbb N$. – JMoravitz Mar 21 '17 at 23:52
• @JMoravitz Thank you for the detail. This is very helpful – Sithe Mar 22 '17 at 0:05
If $A$ and $B$ are countable, then they can be enumerated, respectively, as:
$a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$
Since $C$ is finite, its elements can be listed as: $c_1, c_2, \ldots, c_n$.
To prove the union of all three is countable, it will suffice to list them in some order $d_1, d_2, d_3, \ldots$, since there is then a natural bijection with $\mathbb{N}$ in which we match $d_k$ with $k \in \mathbb{N}$.
As to the listing for this union, you could list all of $C$, then interweave enumerations of $A$ and $B$:
$c_1, c_2, \ldots, c_n, a_1, b_1, a_2, b_2, a_3, b_3, \ldots$
(You may also specify that no element is listed more than once.)
If either of $A$ or $B$ is finite (since your definition of countable allows for this) then the listing can be tweaked appropriately, e.g., if just $A$ is finite, then list all elements of $C$, then of $A$, and then enumerate $B$. | {
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• We must have been typing in synchrony ... Great answer by the way!! – Bram28 Mar 21 '17 at 23:56
• Thank you. I've never seen countability being proven that way. It's very straight forward and I like it – Sithe Mar 22 '17 at 0:00
• Slight thing to watch out for if the sets aren't disjoint. But just skip them if they do. That's acceptable "tweaking". – fleablood Mar 22 '17 at 0:04
• This is an injection, not necessarily a bijection, because $A\cap B$ might not be empty. – vadim123 Mar 22 '17 at 0:04
• Okay, I missed that. – fleablood Mar 22 '17 at 0:09
Sets are countable if and only if their elements are 'listable'.
Now, $C$ is finite, so say it contains elements $c_1, c_2, ..., c_n$ for some $n$
$A$ is countable, so we can create a list:
$a_1, a_2, ...$
Same for $B$:
$b_1, b_2, ...$
So, I would create the following list:
$c_1, c_2, ..., c_n, a_1, b_1, a_2, b_2, ...$
• This list may have repeated elements. – vadim123 Mar 22 '17 at 0:04
• @vadim123 sure, but they can be removed if we really wanted to define an injection. So as long as every element appears somewhere on the list, we're fine. – Bram28 Mar 22 '17 at 1:06
Let $j_a: A \rightarrow \mathbb N$ be an injection. We know that is possible because $A$ is countable.
Let $j_b: B \rightarrow \mathbb N$ be an injection. We know that is possible because $B$ is countable.
Let $j_c: B \rightarrow \mathbb N_j$ be an bijection. We know that is possible because $C$ is finite.
Let $k:A\cup B\cup C \rightarrow \mathbb N$ via $k(x) = 3*j_a(x)$ if $x \in A$. If $x \in B$ but $x \not \in A$ let $k(x) = 3*j_b(x) + 1$. And if $x \in C$ but $x \not \in A$ and $x \not \in B$ let $k(x) = 3*j_c(x) + 2$.
Show that $k$ is an injection and that shows $A\cup B \cup C$ is countable. | {
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Show that $k$ is an injection and that shows $A\cup B \cup C$ is countable.
Another way to think of it is to start by picking out the first element of $A$ then pick the first element of $B$, then of $C$, then pick the second element, then the third elements and so on. That's a list of all the elements. Since they can be listed one after another they are countable.
• And before vadim points out there'll be many elements of N not mapped into, I'll point out as long as there is an injection into N that is sufficient to prove countabity. It need not be onto. But it can be made onto if it is infinite. But it is not nescessary. – fleablood Mar 22 '17 at 0:07 | {
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# What does it mean when we say most of the points in a hypercube are at the boundary?
If I have a 50 dimensional hypercube. And I define it's boundary by $0<x_j<0.05$ or $0.95<x_j<1$ where $x_j$ is dimension of the hypercube. Then calculating the proportion of points on the boundary of the hypercube will be $0.995$. What does it mean? Does it mean that rest of the space is empty? If $99\%$ of the points are at the boundary then the points inside the cube must not be uniformly distributed?
• No, it means the periphery is more spacious, and the effect is commensurate with the dimensionality. It is somewhat counterintuitive. This phenomenon has consequences on the distribution of the distance between random pairs of nodes that become relevant when you want to cluster or calculate nearest neighbors in high-dimensional spaces.
– Emre
Feb 2, 2018 at 18:25
• Calculate what proportion of the points on a line segment are near its boundary. Then points in a square. Then points in a cube. What can you say about them? Feb 3, 2018 at 11:38
Speaking of '$$99\%$$ of the points in a hypercube' is a bit misleading since a hypercube contains infinitely many points. Let's talk about volume instead.
The volume of a hypercube is the product of its side lengths. For the 50-dimensional unit hypercube we get $$\text{Total volume} = \underbrace{1 \times 1 \times \dots \times 1}_{50 \text{ times}} = 1^{50} = 1.$$ | {
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Now let us exclude the boundaries of the hypercube and look at the 'interior' (I put this in quotation marks because the mathematical term interior has a very different meaning). We only keep the points $$x = (x_1, x_2, \dots, x_{50})$$ that satisfy $$0.05 < x_1 < 0.95 \,\text{ and }\, 0.05 < x_2 < 0.95 \,\text{ and }\, \dots \,\text{ and }\, 0.05 < x_{50} < 0.95.$$ What is the volume of this 'interior'? Well, the 'interior' is again a hypercube, and the length of each side is $$0.9$$ ($$=0.95 - 0.05$$ ... it helps to imagine this in two and three dimensions). So the volume is $$\text{Interior volume} = \underbrace{0.9 \times 0.9 \times \dots \times 0.9}_{50 \text{ times}} = 0.9^{50} \approx 0.005.$$ Conclude that the volume of the 'boundary' (defined as the unit hypercube without the 'interior') is $$1 - 0.9^{50} \approx 0.995.$$
This shows that $$99.5\%$$ of the volume of a 50-dimensional hypercube is concentrated on its 'boundary'.
Follow-up: ignatius raised an interesting question on how this is connected to probability. Here is an example.
Say you came up with a (machine learning) model that predicts housing prices based on 50 input parameters. All 50 input parameters are independent and uniformly distributed between $$0$$ and $$1$$.
Let us say that your model works very well if none of the input parameters is extreme: As long as every input parameter stays between $$0.05$$ and $$0.95$$, your model predicts the housing price almost perfectly. But if one or more input parameters are extreme (smaller than $$0.05$$ or larger than $$0.95$$), the predictions of your model are absolutely terrible.
Any given input parameter is extreme with a probability of only $$10\%$$. So clearly this is a good model, right? No! The probability that at least one of the $$50$$ parameters is extreme is $$1 - 0.9^{50} \approx 0.995.$$ So in $$99.5\%$$ of the cases, your model's prediction is terrible. | {
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Rule of thumb: In high dimensions, extreme observations are the rule and not the exception.
• Worth using the OP's quote "Does it mean that rest of the space is empty?" and answering: No, it means that the rest of the space is relatively small . . . Or similar in your own words . . . Feb 2, 2018 at 15:41
• Really nice explanation of the term "curse of dimensionality" Dec 19, 2018 at 12:04
• Wondering if the following is correct: taking this example, if a set of features are evenly distributed along [0,1] in each of the 50 dimensions, the (99.5% -0.5%) = 99% of the the volume (hypercube feature space) captures only the 10% values of each feature Dec 19, 2018 at 12:16
• "Any given input parameter is extreme with a probability of only 5%." I think this probability is 10%. Feb 1, 2020 at 15:04
• @Rodvi: You are right of course, thanks! Fixed it. Feb 3, 2020 at 10:39
You can see the pattern clearly even in lower dimensions.
1st dimension. Take a line of length 10 and a boundary of 1. The length of the boundary is 2 and the interior 8, 1:4 ratio.
2nd dimension. Take a square of side 10, and boundary 1 again. The area of the boundary is 36, the interior 64, 9:16 ratio.
3rd dimension. Same length and boundary. The volume of the boundary is 488, the interior is 512, 61:64 - already the boundary occupies almost as much space as the interior.
4th dimension, now the boundary is 5904 and the interior 4096 - the boundary is now larger.
Even for smaller and smaller boundary lengths, as the dimension increases the boundary volume will always overtake the interior.
The best way to "understand" it (though it is IMHO impossible for a human) is to compare the volumes of a n-dimensional ball and a n-dimensional cube. With the growth of n (dimensionality) all the volume of the ball "leaks out" and concentrates in the corners of the cube. This is a useful general principle to remember in the coding theory and its applications. | {
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The best textbook explanation of it is in the Richard W. Hamming's book "Coding and Information Theory" (3.6 Geometric Approach, p 44).
The short article in Wikipedia will give you a brief summary of the same if you keep in mind that the volume of a n-dimensional unit cube is always 1^n.
I hope it will help. | {
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# How can I find the dimension of an eigenspace?
I have the following square matrix
$$A = \begin{bmatrix} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 &-1 \end{bmatrix}$$
I found the eigenvalues:
• $2$ with algebraic and geometric multiplicity $1$ and eigenvector $(1,2,7/3)$.
• $-1$ with algebraic multiplicity $2$ and geometric multiplicity $1$; one eigenvector is $(0,0,1)$.
Thus, matrix $A$ is not diagonizable. My questions are:
1. How can I find the Jordan normal form?
2. How I can find the dimension of the eigenspace of eigenvalue $-1$?
3. In Sagemath, how can I find the dimension of the eigenspace of eigenvalue $-1$?
• Thanks for edit Rodrigo – user539638 Mar 17 '18 at 14:52
• Haven't you answered the question already with "... geometric multiplicity $1$"? – Rodrigo de Azevedo Mar 17 '18 at 14:59
• I am not sure for the answer.I want the command in sagemath for the dimension of eigenspace. – user539638 Mar 17 '18 at 15:08
• Compute the rank of $A+I_3$. Problem solved. – Rodrigo de Azevedo Mar 17 '18 at 15:15
Define the matrix:
sage: a = matrix(ZZ, 3, [2, 0, 0, 6, -1, 0, 1, 3, -1])
and then type a.jor<TAB> and then a.eig<TAB>, where <TAB> means hit the TAB key. This will show you the methods that can be applied to a that start with jor and with eig.
Then, once you found the method a.jordan_form, read its documentation by typing a.jordan_form? followed by TAB or ENTER.
You will find that you can call a.jordan_form() to get the Jordan form, or a.jordan_form(transformation=True) to also get the transformation matrix.
sage: j, p = a.jordan_form(transformation=True)
sage: j
[ 2| 0 0]
[--+-----]
[ 0|-1 1]
[ 0| 0 -1]
sage: p
[ 1 0 0]
[ 2 0 1]
[7/3 3 0]
Here is an exploration of the eigenvalues, eigenspaces, eigenmatrix, eigenvectors. | {
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Here is an exploration of the eigenvalues, eigenspaces, eigenmatrix, eigenvectors.
sage: a.eigenvalues()
[2, -1, -1]
sage: a.eigenspaces_right()
[
(2, Vector space of degree 3 and dimension 1 over Rational Field
User basis matrix:
[ 1 2 7/3]),
(-1, Vector space of degree 3 and dimension 1 over Rational Field
User basis matrix:
[0 0 1])
]
sage: a.eigenmatrix_right()
(
[ 2 0 0] [ 1 0 0]
[ 0 -1 0] [ 2 0 0]
[ 0 0 -1], [7/3 1 0]
)
sage: j, p
(
[ 2| 0 0]
[--+-----] [ 1 0 0]
[ 0|-1 1] [ 2 0 1]
[ 0| 0 -1], [7/3 3 0]
)
sage: a.eigenvectors_right()
[(2, [
(1, 2, 7/3)
], 1), (-1, [
(0, 0, 1)
], 2)]
Most Jordan Normal Form questions, in integers, intended to be done by hand, can be settled with the minimal polynomial. The characteristic polynomial is $\lambda^3 - 3 \lambda - 2 = (\lambda -2)(\lambda + 1)^2.$ the minimal polynomial is the same, which you can confirm by checking that $A^2 - A - 2 I \neq 0.$ Each linear factor of the characteristic polynomial must appear in the minimal polynomial, which exponent at least one, so the quadratic shown is the only possible alternative as minimal.
Next, $$A+I = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 0 \\ 1 & 3 & 0 \end{array} \right)$$ with genuine eigenvector $t(0,0,1)^T$ with convenient multiplier $t$ if desired.
$$(A+I)^2 = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 18 & 0 & 0 \\ 21 & 0 & 0 \end{array} \right)$$
The description I like is that we now take $w$ with $(A+I)w \neq 0$ and $(A+I)^2 w = 0.$ I choose $$w = \left( \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right)$$ This $w$ will be the right hand column of $P$ in $P^{-1}A P = J.$ The middle column is $$v = (A+I)w,$$ so that $v \neq 0$ but $(A+I)v = (A+I)^2 w = 0$ and $v$ is a genuine eigenvector. You already had the $2$ eigenvector, I take a multiple to give integers. i like integers. | {
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$$P = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right)$$ with $$P^{-1} = \frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right)$$
leading to $$\frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 & -1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right)$$
It is the reverse direction $PJP^{-1} = A$ that allows us to evaluate functions of $A$ such as $e^{At},$
$$\frac{1}{9} \left( \begin{array}{rrr} 3 & 0 & 0 \\ 6 & 0 & 1 \\ 7 & 3 & 0 \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ -7 & 0 & 3 \\ -18 & 9 & 0 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 6 & -1 & 0 \\ 1 & 3 & -1 \end{array} \right)$$ | {
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# addition of decimals word problems with solutions | {
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$4,522.08. On Sunday, he worked 5.5 hours. A collection of mathematics problems with an answer and solution to each problem. Each page has a random set of 6 problems. View All . We can perform all arithmetic operations on fractions by expressing them as a decimal. So he's starting with$4,522.08. See how rounding can help in finding an approximate solution. Search . She put 0.35kg of chocolate on … Example 1: If 58 out of 100 students in a school are boys, then write a decimal for the part of the school that consists of boys. Word problems on mixed fractrions. e.g. Will you draw a picture? 2. Choose a word problem card. Print them. Quiz & Worksheet Goals. Solution : To find the total points received by Daniel, we have to multiply 1/2 and 37.5 ... Decimal word problems. Category: Decimals Add/Subtract Decimals Estimate Sums and Differences . The cashier gave her $1.46 in change from a$50 Word problems on fractions. DECIMAL ADDITION; Decimal Addition worksheets here contains 1-digit , 2-digit, 3-digit, 4-digit and 5-digit numbers with decimal values varying from 1 decimal place to 3 decimal places. * You receive 4 full worksheets of 20 word problems involving addition and subtraction of decimals. Explain the mistakes you see. 48 differentiated decimal addition and subtraction questions set over three worksheets. Remind your child to fill in the tens and hundredths places for the first number, so it should look like this: 8.00 - 3.41. This is a fun activity for students during or after your decimal unit. Students will move around the room to find the next problem matched up with the previous problem. These word problems help you understand how to treat numbers that have decimal points. Welcome to the math word problems worksheets page at Math-Drills.com! If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Practice solving word problems by adding or subtracting decimal numbers. Addition and Subtraction | {
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Practice solving word problems by adding or subtracting decimal numbers. Addition and Subtraction Vocabulary and Language. fraction: decimal: 0.58: Answer: 0.58. 10 word problem scavenger hunt using addition, subtraction, multiplication, and division of decimals. 2) Melissa purchased $39.46 in groceries at a store. Consider the following situations. 2. Improve your math knowledge with free questions in "Add and subtract decimals: word problems" and thousands of other math skills. What do you know? Adding and Subtracting Decimals Home Play Multiplayer Unit Challenge Decimal Addition Overview Adding decimals … Yes, it IS all right here: decimal place value, comparing and ordering, addition, subtraction, multiplication, exponents, division, order of operations, converting from fractions to decimals, and decimal word problems are all included in this unit. Create an account! It's also a great spiral act how many in total, altogether, combined, more than, difference, how many are needed. Addition word problems Addition word problems arise in any situations where there is a gain or an increase of something as a result of combining one or more numbers. How much will I earn after 5 days? 3. 4th through 6th Grades. 1. These addition word problems worksheet will produce 2 digits problems with missing addends, with ten problems per worksheet. Problems for 3rd Grade . Grade 5 Decimals Word Problems Name: _____ Class: _____ Question 1 I earn 5.50$ per hour. Addition and subtraction word problems. Example 2: A computer processes information in nanoseconds. These decimal addition problems were not solved correctly. So decimal notations are used to represent an exact value or the amount or the quantities less than zero. Video transcript. You may select between regrouping and non-regrouping type of problems. Leo has $4,522.08 in his bank account. Worksheet: Fifth Grade Estimate- Decimal Sums and Differences - II. Decimal Word Problems Addition (Linda Alexander-Buckley) DOC; 3dp | {
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Decimal Sums and Differences - II. Decimal Word Problems Addition (Linda Alexander-Buckley) DOC; 3dp Decimals Game (JandJ) DOC; Adding VAT (Sue Vaughan) PDF; Rounding Decimals Word Problems (Alan Bartlett) PDF; Making 1, 10 & 100 (3dp) (Mrs. Bangee) DOC; Writing "Mixed Up" Decimals (Alison Wild) DOC; Swamp Monster Decimals (Bernadette Gill) Decimal Fractions (to 3 digits) PDF; Making 1 & 10 (2dp) (Mrs. Bangee) DOC; SATs … * Each problem involves either addition or subtraction or BOTH operations together in one problem. Round three decimals to whole numbers and then estimate the sums. New User? Adding and Subtracting to Tenths with Bar Models. And then he deposits, or he adds, another$875.50. Write $$\frac{127}{1000}$$ as decimal number. Add, subtract and multiply decimals step-by-step. View PDF. Learning Zone Standards Sign up Sign In Username or email: Password: Forgot your username or password? Find the Mistakes (Subtraction) Look very carefully at these subtraction (tenths) problems and search for errors in the solutions. Reread and visualize the problem. Next lesson. Telling time 1 Telling time 2 Telling time 3 Reading pictographs. Problem #1: John has 800 dollars in his checking account. Think of addition as combining parts to form a whole. Word Problem With Multiplication of Two Decimals - Daniel earns $8.80 per hour at his part-time job. Plan how to solve the problem. Multiplying decimals. How much is left in his account? However, you could also cut the pages into strips to assign one problem at a time. View PDF. Each Page also has a speed and accuracy guide, to help students see how fast and how accurately they should be doing thses problems. Search form. A set of word problems linked to decimals. Time and work word problems. Solving Decimal Word Problems. Then solve to find the correct sum. Home > Numbers & Operations > Decimal > Decimal Addition DECIMAL ADDITION Decimal Addition worksheets here contains 1-digit , 2-digit, 3-digit, 4-digit and 5-digit | {
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ADDITION Decimal Addition worksheets here contains 1-digit , 2-digit, 3-digit, 4-digit and 5-digit numbers with decimal values varying from 1 decimal place to 3 decimal places. Sample Decimal Problems and Answers Addition and Subtraction. An exclusive set of worksheets are here for children. Word problem worksheets: Adding and subtracting decimals. Given problem situations, the student will use addition, subtraction, multiplication, and division to solve problems involving positive and negative fractions and decimals. She made 10 chocolate cakes. Practical Problems Involving Decimals Reporting Category Computation and Estimation Topic Solving practical problems involving decimals Primary SOL 6.7 The student will solve single-step and multistep practical problems involving addition, subtraction, multiplication, and division of decimals. DECIMALS WORD PROBLEMS ADDITION AND SUBTRACTION * NO FLUFF - JUST 20 PURE WORD PROBLEMS! To add decimals, follow these steps: Write down the numbers, one under the other, with the decimal points lined up; Put in zeros so the numbers have the same length (see below for why that is OK) Then add, using column addition, remembering to put the decimal point in the answer; Example: Add 1.452 to 1.3. What do you need to find out? Solve problems with two, three or more decimals in one expression. You'll find addition word problems, subtraction word problems, multiplication word problems and division word problems, all starting with simple easy-to-solve questions that build up to more complex skills necessary for many standardized tests. If Supriya used 21.19 liters how much water is left in the barrel. One step equation word problems. All Decimal Operations with Word Problems 1) Ellen wanted to buy the following items: A DVD player for$49.95 A DVD holder for $19.95 Personal stereo for$21.95 Does Ellen have enough money to buy all three items if she has $90. Each set includes four different worksheets with answer keys. Question 2 Jenny bought | {
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she has $90. Each set includes four different worksheets with answer keys. Question 2 Jenny bought 4.35kg of chocolate. Filtered HTML. 3rd through 5th Grades. As they progress, you'll also find a mix of operations that require students to figure out which type of story problem they need to solve. Word Problems: Decimals Materials: Word Problems: Decimals cards _____ 1. Analysis: We can write a fraction and a decimal for the part of the school that consists of boys. Being able to solve each type of problem described above requires students to master the vocabulary of addition and subtraction. 1-8 Math » . Example 1: A barrel has 56.32 liters capacity. Description: This packet heps students practice solving word problems that require subtraction with decimals. Each page includes 3 problems, with space for kids to write out their thinking and solution. Line up the decimal points: 1. Two and three-digit subtraction Subtraction with borrowing. Quick links to download / preview the worksheets listed below : 1 Digit 1 Decimal place, 1… Read More » I work 8 hours per day. What operation will you use? Linear inequalities word problems. He deposits another$875.50 and then withdraws $300 in cash. Adding decimals is easy when you keep your work neat. Improve your skills with free problems in 'Adding decimals in word problems' and thousands of other practice lessons. Read the problem. Below are three versions of our grade 4 math worksheet with word problems involving the addition and subtraction of simple one-digit decimals. Problem 21 This set of 27 decimal word problems covers all the different structures, which kids can solve using multiplication and division. On this page, you will find Math word and story problems worksheets with single- and multi-step solutions on a variety of math topics including addition, multiplication, subtraction, division and other math topics. Get this Worksheet. Materials Newspaper and/or magazine ads Shopping list What were his total earnings for | {
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Worksheet. Materials Newspaper and/or magazine ads Shopping list What were his total earnings for the day? 8 - 3.41 The answer for this problem is 4.59. Practice: Adding & subtracting decimals word problems. If you're seeing this message, it means we're having trouble loading external resources on our website. Let's write that down. Let’s practice some word problems on decimal fractions by using various arithmetical operations. Worksheets > Math > Grade 4 > Word Problems > Addition & subtraction of decimals. This calculator uses addition, subtraction, multiplication or division for calculations on positive or negative decimal numbers, integers, real numbers and whole numbers. Addition And Subtraction On Decimals Word Problems - Displaying top 8 worksheets found for this concept.. These addition worksheets provide practice adding tenths and adding hundredths. These word problems worksheets are appropriate for 3rd Grade, 4th Grade, and 5th Grade. Word problems on ages. Addition with decimals is an important skill in financial and scientific applications, even though the basic addition skills are fairly easy. Web page addresses and e-mail addresses turn into links automatically. Counting involving multiplying Multiplying 1-digit numbers Multiplying by multiples of 10. Word problems on sets and venn diagrams. Ratio and proportion word problems. Apply rounding skills to solve word problems requiring estimation, addition, and subtraction. Decimals » G.4. So he's going to add$875.50. In groceries at a store.kastatic.org and *.kasandbox.org are unblocked, multiplication, and 5th Grade with... Per worksheet, combined, more than, difference, how many in total, altogether, combined, than..., combined, more than, difference, how many in total, altogether, combined, more than difference! To find the total points received by Daniel, we have to multiply and! Total, altogether, combined, more than, difference, how many are needed or! Fractions by using various arithmetical | {
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page includes 3,. Have to multiply 1/2 and 37.5... decimal word problems '' and thousands of other practice.!, which kids can solve using multiplication and division is 4.59 provide practice adding tenths and hundredths! Decimal Sums and Differences easy when you keep your work neat when you keep your work neat one.! > Grade 4 > word problems '' and thousands of other math skills: _____ Class: _____ Question I! Are needed Forgot your Username or Password for kids to write out their thinking and to... In Add and subtract decimals: word problems covers all the addition of decimals word problems with solutions structures, which kids can using! Grade 5 decimals word problems involving the addition and subtraction * NO FLUFF - JUST 20 word. To each problem involves either addition or subtraction or BOTH operations together in one problem amount... Used 21.19 liters how much water is left in the solutions versions of our Grade 4 word! Problems ' and thousands of other practice lessons subtracting decimal numbers previous.! Subtraction ) Look very carefully at these subtraction ( tenths ) problems and search for errors the! That consists of boys then withdraws $300 in cash 27 decimal word problems involving addition and of... Students during or after your decimal unit one problem with free problems in decimals... Solve problems with two, three or more decimals in one expression the that! Packet heps students practice solving word problems involving the addition and subtraction these worksheets...: John has 800 dollars in his checking account practice solving word problems on decimal fractions expressing. # 1: John has 800 dollars in his checking account the vocabulary of addition as combining parts form... Worksheet with word problems: decimals Add/Subtract decimals Estimate Sums and Differences - II at subtraction! To treat numbers that have decimal points in 'Adding decimals in word problems on decimal fractions using. Problems > addition & subtraction of simple | {
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decimals in word problems on decimal fractions using. Problems > addition & subtraction of simple one-digit decimals the previous problem Estimate Sums Differences. Subtraction or BOTH operations together in one problem will produce 2 digits problems with missing addends, space... Of problems FLUFF - JUST 20 PURE word problems Name: _____ Question 1 I earn$... Forgot your Username or Password select between regrouping and non-regrouping type of problems during or after your unit... Treat numbers that have decimal points page addresses and e-mail addresses turn into links automatically Melissa purchased \$ 39.46 groceries... Time 1 Telling time 2 Telling time 2 Telling time 2 Telling time 2 Telling time 3 Reading pictographs or... A decimal for the part of the school that consists of boys multiplication and division packet heps students solving! Is an important skill in financial and scientific applications, even though the basic addition are. Can write a fraction and a decimal a computer processes information in.... The quantities less than zero for 3rd Grade, 4th Grade, 4th,. The domains *.kastatic.org and *.kasandbox.org are unblocked understand how to numbers! Problems '' and thousands of other practice lessons or BOTH operations together in one.... Collection of mathematics problems with two, three or more decimals in word problems decimal! Worksheets > math > Grade 4 math worksheet with word problems '' and of! This set of worksheets are here for children decimal notations are used to represent an exact value the. Example 1 addition of decimals word problems with solutions John has 800 dollars in his checking account problems by or. Consists of boys by multiples of 10 Zone Standards Sign up Sign Username! To form a whole web filter, please make sure that the domains * and. 3.41 the answer for this problem is 4.59 as combining parts to form whole... | {
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Tato stránka používá Akismet k omezení spamu. Podívejte se, jak vaše data z komentářů zpracováváme.. | {
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# How can I find the integral of this function using trig substitution?
$$\int_0^1x^3\sqrt{1 - x^2}dx$$
I need to find the integral of this function using trigonometric substitution.
Using triangles, I found that $x = \sin\theta$, and $dx = \cos\theta d\theta$; so I have
$$\int_0^{\pi/2}\sin^3\theta\sqrt{1 - \sin^2\theta}\cos\theta d\theta$$ then, using identities:
$$\int_0^{\pi/2}\sin^3\theta\sqrt{\cos^2\theta}\cos\theta d\theta$$
$$\int_0^{\pi/2}\sin^3\theta\cos^2\theta d\theta$$
After this point, I don't know where to go. My teacher posted solutions, but I don't quite understand it.
Does anyone know how this can be solved using trig substitution? The answer is $2/15$.
Much thanks,
Zolani13
-
If you really want to learn how to integrate that last expression watch tinyurl.com/6m4u5ye and use all the answer suggestions. – Kirthi Raman Mar 14 '12 at 22:08
Factor out a $\sin\theta$ from the $\sin^3\theta$. Write the remaining $\sin^2\theta$ as $1-\cos^2\theta$. Now let $u=\cos\theta$.
$$\sin^3\theta\, \cos^2\theta \,d\theta= (1- {\cos^2\theta} )\cos^2\theta\sin\theta \,d\theta = (\cos^2\theta- {\cos^4\theta} ) \sin\theta \,d\theta\ \ \buildrel{u=\cos\theta}\over = \ \ -(u^2-u^4)\,du$$
-
I have it!! Much thanks ^_^ – Zolani13 Mar 15 '12 at 0:10
1) if $x = \sin \theta$ then you write $dx = \cos \theta d\theta$. (You missed the $d\theta$).
2) Since you are integrating in the range $0$, $\pi/2$, $\sqrt{\cos^2 \theta} = \cos \theta$. You should probably mention that in your work.
Now you are at the integral
$$\int_{0}^{\pi/2} \sin^3 \theta \cos ^2 \theta d \theta$$
Normally when you have to integrate something like $\sin^{2n+1}\theta \cos^m \theta$, the substitution $t = \cos \theta$ usually works, because the $dt$ swallows one of the $\sin \theta$, and then you can use $\sin^{2n} \theta = (1 - \cos^2 \theta)^n$
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-
From here, convert $\sin^3(\theta)\cos^2(\theta) = \sin(\theta)(1 - \cos^2(\theta))\cos^2(\theta) = \sin(\theta)\cos^2(\theta) - \sin(\theta)\cos^4(\theta)$. Now, solve both integrals using a u-substitution: $u = \cos(\theta), du = -\sin(\theta) d\theta$. Can you finish from here?
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# Linear Algebra, Cramers Rule - Solving for unknowns in a matrix
Hello everyone, I have a linear algebra question regarding Cramer's rule.
## Homework Statement
Using Cramer's rule, solve for x' and y' in terms of x and y.
$$\begin{cases} x = x' cos \theta - y' sin \theta\\ y = x' sin \theta + y'cos \theta \end{cases}$$
2. Homework Equations
##sin^2 \theta + cos^2 \theta = 1 ##
## The Attempt at a Solution
I need a matrix to start off, so I form a matrix based on the right-hand side of x and y. I'm assuming that x' and y' are just alternative ways of writing ##x_1## and ##x_2##.
$$Let A = \begin{bmatrix} cos \theta & -sin \theta\\ sin \theta & cos\theta \end{bmatrix}$$
I form two more matrices, ##A_1## and ##A_2##.
$$Let A_1 = \begin{bmatrix} x & -sin \theta\\ y & cos\theta \end{bmatrix}$$
$$Let A_2 = \begin{bmatrix} cos \theta & x\\ sin \theta & y \end{bmatrix}$$
I then find ##det(A)##. I get ##cos^2 \theta + sin^2 \theta ## which is ##1##.
##det(A_1) = x cos \theta + y sin \theta##
##det(A_2) = y cos \theta - x sin \theta##
Lastly, I need to find the value of ##x'## and ##y'##, using Cramer's Rule.
$$x' = \frac{det(A_1)}{det A} = \frac{x cos \theta + y sin \theta}{1}\\ y' = \frac{det(A_2)}{det A} = \frac{y cos \theta - x sin \theta}{1}$$
Can anyone tell me if I'm on the right track for this problem?
Last edited:
## Answers and Replies
Related Precalculus Mathematics Homework Help News on Phys.org
Ray Vickson
Homework Helper
Dearly Missed
Hello everyone, I have a linear algebra question regarding Cramer's rule.
## Homework Statement
Using Cramer's rule, solve for x' and y' in terms of x and y.
$$\begin{cases} x = x' cos \theta - y' sin \theta\\ y = x' sin \theta + y'cos \theta \end{cases}$$
2. Homework Equations
##sin^2 \theta + cos^2 \theta = 1 ##
## The Attempt at a Solution | {
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2. Homework Equations
##sin^2 \theta + cos^2 \theta = 1 ##
## The Attempt at a Solution
I need a matrix to start off, so I form a matrix based on the right-hand side of x and y. I'm assuming that x' and y' are just alternative ways of writing ##x_1## and ##x_2##.
$$Let A = \begin{bmatrix} cos \theta & -sin \theta\\ sin \theta & cos\theta \end{bmatrix}$$
I form two more matrices, ##A_1## and ##A_2##.
$$Let A_1 = \begin{bmatrix} x & -sin \theta\\ y & cos\theta \end{bmatrix}$$
$$Let A_2 = \begin{bmatrix} cos \theta & x\\ sin \theta & y \end{bmatrix}$$
I then find ##det(A)##. I get ##cos^2 \theta + sin^2 \theta ## which is ##1##.
##det(A_1) = x cos \theta + y sin \theta##
##det(A_2) = y cos \theta - x sin \theta##
Lastly, I need to find the value of ##x'## and ##y'##, using Cramer's Rule.
$$x' = \frac{det(A_1)}{det A} = \frac{x cos \theta + y sin \theta}{1}\\ y' = \frac{det(A_2)}{det A} = \frac{y cos \theta - x sin \theta}{1}$$
Can anyone tell me if I'm on the right track for this problem?
A good working rule to learn is: check it for yourself, by substituting your proposed solution in the original equations to see if it all works out.
A good working rule to learn is: check it for yourself, by substituting your proposed solution in the original equations to see if it all works out.
Ray, I completely forgot about being able to check questions like these. I want to sincerely thank you for reminding me of this rule.
HallsofIvy
Homework Helper
By the way, the original equations,
$x= x'cos(\theta)- y'sin(\theta)$
$y= x'sin(\theta)+ y'cos(\theta)$
give the (x, y) coordinates of point (x', y') after a rotation through angle $\theta$, clockwise, around the origin. Going from (x, y) back to (x', y') is just a rotation in the opposite direction. So
$x'= xcos(-\theta)- y sin(-\theta)= xcos(\theta)+ ysin(\theta)$,
$y'= xsin(-\theta)+ ycos(-\theta)= -x sin(\theta)+ y cos(\theta)$. | {
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# How do I count the number of increasing functions from $\{1, 2, 7\}$ to $\{1, 2, 3, 4, 5, 6, 7, 8\}$?
I can't seem to figure it out, one way I thought about it is:
Let's take $f(1) = 1$:
• $f(2) = 1$ (because it's not strictly increasing) so for $f(7)$ I have $8$ possibilities
• $f(2) = 2$ so $f(7)$ now has $7$ possibilities
• $f(2) = 3$ so $f(7)$ now has $6$ possibilities
.
.
and so on.
So for $f(1) = 1$ we have $8+7+6+5+...+1$ functions (this can be expressed as the formula $\frac{(n+1)n}{2}$)
So for $f(1) = 2$:
• $f(2) = 2$ results in $7$ possibilities of $f(7)$
.
.
and so on and it result in $7+6+5+..+1$
so by my reasoning I would say that the number of increasing functions is equal to: $S_8 + S_7 + S_6 + ... + S_1$ (where $S_n = \frac{(n+1)n}{2}$ )
But I'm pretty sure that's not right but I don't know how to think about it.
• Consider the derivative instead. $g(n)=f(x_{n+1})-f(x_{n})$, where $x_1=1, x_2=2, x_3=7$. Find how many solutions does this inequality has $g(1)+g(2)\leq 8$. Turn the inequality into an equality $g(1)+g(2)+Y=8$ with $g(1),g(2),y\geq 0$. – Bettybel Jul 12 '17 at 13:11
• You say you're pretty sure your intuition is wrong. Why do you say this? – abiessu Jul 12 '17 at 13:11
• Are you sure that "increasing" does not mean "strictly increasing" here? – Did Jul 12 '17 at 13:12
• Your reasoning for counting the number of non-decreasing functions seems spot-on. If that answer (120) is not correct, then you must be looking for strictly increasing functions, of which there are $\binom83$. – G Tony Jacobs Jul 12 '17 at 13:55
First, to count the number of strictly increasing functions, note that any combination of $3$ numbers from the set $\{1,2,\ldots, 8\}$ gives such a function and all such functions correspond to a combination. So there are $8 \choose 3$ strictly increasing functions. | {
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Second, count the number of functions $f(x)$ where two numbers go to the same image and one is different. Either $f(1)=f(2) \neq f(7)$ or $f(1)\neq f(2) =f(7)$. These two cases are identical, so we count the first one and multiply by $2$. By the same reasoning above, there are $2{8 \choose 2}$ of these functions.
Third, count the number of functions with $f(1)=f(2)=f(3)$. There are $8$ of these.
Final tally: $${8 \choose 3} +2 {8 \choose 2} +8 = 120,$$
The function $g$ given by $g(1)=f(1)$, $g(2)=f(2)+1$, $g(7)=f(7)+2$ is a strictly increasing map to $\{1,2,\ldots,10\}$ for every increasing $f$ and vice versa. The number of such $g$ is simly the number of ways to pick three distinct elements of a ten-element set, so $10\choose 3$.
• really slick! :) – muzzlator Jul 12 '17 at 14:02
A small variation of the theme. Note, that in order to determine the number of increasing functions the elements of the domain $\{1,2,7\}$ are not relevant. Just the number of elements $|\{1,2,7\}|=3$ of the domain is essential.
We can reformulate the problem: Find the number of triples $(x_1,x_2,x_3)$ of positive integers with \begin{align*} 1\leq x_1\leq x_2\leq x_3\leq 8\tag{1} \end{align*}
This can be solved using stars and bars, here with $n=3$ and $k=8$ giving \begin{align*} \binom{n+k-1}{n}=\binom{10}{3}=120 \end{align*} | {
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# Integer midpoint
Two points $$(c, d, e),(x, y, z) \in \mathbb{R}^3$$, we say that the midpoint of these two points is the point with coordinates $$\left(\frac{c+x}{2} , \frac{d+y}{2} , \frac{e+z}{2}\right)$$
Take any set $$S$$ of nine points from $$\mathbb{R}^3$$ with integer coordinates.
Prove that there must be at least one pair of points in $$S$$ whose midpoint also has integer coordinates.
I've tried to do an example with set $$S=\{(2,2,2),(1,8,2),(3,4,5),(5,2,2),(4,2,9),\\(2,1,4),(6,8,2),(0,0,0),(5,2,3)\}$$
So taking $$2$$ points, $$(3,4,5)$$ and $$(5,2,3)$$ so $$(3+5)/2=4$$, $$(4+2)/2=3$$, $$(5+3)/2 = 4$$, which are integers. I'm wanting this argument to hold in general and I'm finding it tricky to prove this does anyone have suggestions would be grateful!
• These are 3-tuples. You can look at each individual coordinate as being either even (e) or odd (o). So for example, point 1 could be (e,e,e) and point 2 could be (o,o,e) and so on.... Look at all possible cases (how many possible cases?) and then ask yourself, when would two points give integer midpoints using the formula you provided? Finally, think to yourself, if I have 9 points....... since its a discrete math class, you probably learned about a particular "principle"... – Eleven-Eleven Oct 8 '18 at 12:21
• pigeon hole principle is probably onto the right track ? – sphynx888 Oct 8 '18 at 12:32
• By the way, a big +1 for doing an example and seeing what happened. A good next step might have been to try to build a counterexample, and seeing where you got stuck --- that might have given further insight. – John Hughes Oct 8 '18 at 12:47
Just to reiterate what I said earlier, if we list the possible coordinate parities of the $$3$$-tuples, we have
$$\{(e,e,e),(e,e,o),(e,o,e),(e,o,o),(o,e,e),(o,e,o),(o,o,e),(o,o,o)\}$$
So there are $$8$$ different parity $$3$$-tuples possible. Also note the addition of even and odd integers;
$$e+e=e,$$ $$o+o=e,$$ $$e+o=o$$ | {
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$$e+e=e,$$ $$o+o=e,$$ $$e+o=o$$
So in order for your midpoint to be an integer, we need to have the points being added to have coordinates of the same parity. Therefore, since we have 9 points, assuming our first 8 points are all of different parity combinations as above, the 9th must be one of these 8 possibilities. If you add up the two points that have the same coordinate parity structure, this will yield an even number, which is divisible by 2. For example, if both ordered triples have coordinate parity structure $$(e,e,o)$$, then by the midpoint formula,
$$Mid((e,e,o),(e,e,o))=\left(\frac{e+e}{2}, \frac{e+e}{2},\frac{o+o}{2}\right)=\left(\frac{2e}{2}, \frac{2e}{2},\frac{2o}{2}\right)=(e,e,o)$$
And so by the Pigeonhole principle, there is at least 1 midpoint that contains integer coordinates. And as Robert Z noted in his solutions, this can be extended to $$n$$-tuples as well.
Hint. Read carefully Eleven-Eleven's comment (or take a look at Pigeon-Hole Principle and 2d grid?) and, by using the Pigeonhole Principle, show the following more general statement: in any set of $$2^n+1$$ points in $$\mathbb{Z}^n$$ there is at least one pair that has the midpoint in $$\mathbb{Z}^n$$. | {
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# Lebesgue measure-preserving differentiable function
Let $\lambda$ denote Lebesgue measure and let $f: [0,1] \rightarrow [0,1]$ be a differentiable function such that for every Lebesgue measurable set $A \subset [0,1]$ one has $\lambda(f^{-1}(A)) = \lambda (A)$. Prove that either $f(x) = x$ or $f(x) = 1- x$.
I will appreciate a hint or a solution that doesn't use ergodic theory as this is an old qual problem in measure theory. $f$ is bounded and continuous and so it is integrable and hence the hypothesis of Lebesgue's Differentiation Theorem are satisfied, but I haven't seen a way to use this or if it is even applicable.
• To understand why the "differentiable" hypothesis is there: examine the function $f(x) = |2x-1|$ on $[0,1]$. – GEdgar Apr 23 '16 at 18:15
First note $f$ is onto: Otherwise the range of $f$ is a proper subinterval $[a,b],$ which implies $\lambda(f^{-1}([a,b])) = \lambda([0,1]) = 1 > b-a = \lambda([a,b]),$ contradiction.
Suppose $f'(x) = 0$ for some $x\in [0,1].$ Then $f(B(x,r))\subset B(f(x),r/4)$ for small $r>0.$ ($B(y,r)$ denotes the open ball with center $y$ and radius $r$ relative to $[0,1].$) Thus for such $r,$ $f^{-1}(B(f(x),r/4))$ contains $B(x,r).$ But $B(f(x),r/4)$ has measure no more than $2\cdot (r/4) = r/2,$ while $B(x,r)$ has measure at least $r.$ That's a contradiction. Hence $f'(x) \ne 0$ for all $x\in [0,1].$
By Darboux, $f'>0$ on $[0,1]$ or $f'<0$ on $[0,1].$ Let's assume the first case holds. Then $f$ is strictly increasing, hence a bijection on $[0,1].$ Thus if $A$ is an interval, $\lambda(A) = \lambda(f^{-1}(f(A)))= \lambda(f(A)).$ Since $f(0) = 0,$ this gives $x = \lambda([0,x]) = \lambda(f([0,x])) = f(x) - f(0) = f(x)$ for all $x\in [0,1].$
• This is a much better proof! – Giovanni Apr 23 '16 at 16:52 | {
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• This is a much better proof! – Giovanni Apr 23 '16 at 16:52
The following argument should work under the additional assumption that $f'$ is integrable. I don't know if this is something that you need to assume or if it is a limitation of my proof. It is worth mentioning that I need integrability of the derivative for two reasons: first to apply Lebesgue differentiation theorem for $f'$, and second to get that $f$ is absolutely continuous, which I need to apply the Fundamental Theorem of Calculus.
Let $\mu(\cdot) = \lambda(f^{-1}(\cdot))$. Then $\mu = \lambda$ and hence for a.e. $x \in (0,1)$ we have
$$1 = \frac{d\lambda}{d\mu}(x) = \lim_{r \to 0}\frac{\lambda(B_r(x))}{\mu(B_r(x))} = \lim_{r \to 0}\frac{\int_{B_r(x)}d\lambda}{\mu(B_r(x))} = \lim_{r \to 0}\frac{1}{\lambda(f^{-1}(B_r(x)))}\int_{f^{-1}(B_r(x))}|f'|d\lambda,$$ where in the last equality we used the change of coordinates $z = f(y)$. Notice that $\lambda(f^{-1}(B_r(x))) = \lambda(B_r(x)) \to 0$, hence by Lebesgue differentiation theorem, for every $y \in f^{-1}(x)$ we have that $|f'(y)| = 1.$ Moreover we have that for $y \in (0,1)$, $y \in f^{-1}(f(y))$, thus proving that $|f'| = 1$ a.e. in $[0,1]$.
To conclude the argument, we only need to show that $f'$ cannot change sign. Indeed, by the FTC, we would have that $$f(1) = f(0) + \int_0^1f'(y) = f(0) \pm 1,$$ showing that either $f(0) = 0$ and $f' = 1$ or $f(0) = 1$ and $f' = -1$.
To prove that the derivative is constant a.e., I'll show that $f$ is surjective. By continuity, $f([0,1])$ is a compact subset of $[0,1]$. Suppose by contradiction that $f$ is not surjective, then $\lambda(f[0,1]) < 1$. To find a contradiction it is enough to notice that $$1 = \lambda([0,1]) = \lambda(f^{-1}(f([0,1]))) = \lambda(f([0,1])) < 1.$$
In response to Ramiro's comment: Everywhere differentiable + integrable derivative implies absolute continuity, i.e. the FTC holds. Then I can prove what you asked as follows. | {
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Necessarily, $f(0)$ is either $0$ or $1$. To see this notice that if there is $x \in (0,1)$ such that $f(x) \in \{0,1\}$ then let $y$ be the point such that $f(y) = \{0,1\} \setminus \{f(x)\}$. Assume without loss that $x< y$, then $$1 = |f(x) - f(y)| \le \int_x^y|f'| = y - x < 1.$$ Then assume without loss that $f(0) = 0$. Reasoning as above shows that $f(1) = 1$. Again by the FTC, $$1 = f(1) - f(0) = \int_0^1f' = \lambda(\{f' = 1\}) - \lambda(\{f' = -1\}).$$ Finally, if $\lambda(\{f' = -1\}) > 0$ we get a contradiction, hence $f' = 1$ a.e.
• Why proving that $f$ is surjective implies that the derivative does not change sign? – Ramiro Apr 23 '16 at 5:12
• @Ramiro: I have edited with a proof, let me know if it is convincing :) – Giovanni Apr 23 '16 at 5:41
• Thanks for writing down the details. Please note that your statement "To prove that the derivative does not change sign I'll show that f is surjective." is NOT accurate. The fact that f is surjective implies only that $\lambda({f\neq 1})=0$ or $\lambda({f\neq -1})=0$. This is all you need, but it is weaker than claiming "the derivative does not change sign". (In fact, in principle, in a set of measure zero we may even have $|f'|\neq 1$). – Ramiro Apr 23 '16 at 12:48
• Thank you for your solution. The problem mentioned that we cannot assume the derivative is continuous. I left that out because we can't in general assume things not given to us and which we cannot deduce easily as parts of the properties of the object given to us. Can we remove the assumption of integrability of the derivative or say why it suffices to consider this case only? – akech Apr 23 '16 at 13:34
• If you allow me a suugestion, I suggest that instead of writing "To prove that the derivative does not change sign I'll show that f is surjective", it would be better to write something like "To prove that the derivative is constant a.e., I'll show that f is surjective". – Ramiro Apr 23 '16 at 16:57 | {
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# Does L'Hôpital's work the other way?
Hello fellows,
As referred in Wikipedia (see the specified criteria there), L'Hôpital's rule says,
$$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$
As
$$\lim_{x\to c}\frac{f'(x)}{g'(x)}= \lim_{x\to c}\frac{\int f'(x)\ dx}{\int g'(x)\ dx}$$
Just out of curiosity, can you integrate instead of taking a derivative? Does
$$\lim_{x\to c}\frac{f(x)}{g(x)}= \lim_{x\to c}\frac{\int f(x)\ dx}{\int g(x)\ dx}$$
work? (given the specifications in Wikipedia only the other way around: the function must be integrable by some method, etc.) When? Would it have any practical use? I hope this doesn't sound stupid, it just occurred to me, and I can't find the answer myself.
## Edit
Take 2 functions $f$ and $g$. When is
$$\lim_{x\to c}\frac{f(x)}{g(x)}= \lim_{x\to c}\frac{\int_x^c f(a)\ da}{\int_x^c g(a)\ da}$$
true?
Not saying that it always works, however, it sometimes may help. Sometimes one can apply l'Hôpital's even when an indefinite form isn't reached. Maybe this only works on exceptional cases.
Most functions are simplified by taking their derivative, but it may happen by integration as well (say $\int \frac1{x^2}\ dx=-\frac1x+C$, that is simpler). In a few of those cases, integrating functions of both nominator and denominator may simplify.
What do those (hypothetical) functions have to make it work? And even in those cases, is is ever useful? How? Why/why not? | {
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• You cannot use L'Hospital the other way to evaluate a limit. You can try an example, as I did and it just won't work. Realizing that taking a derivative of num and denom really means that you are linearizing and so that the ratio of the slopes (i.e. derivatives) will give you the answer to the limit, makes sense. But when you are anti deriving, what does that geometrically mean to the limit? I am not aware that "anti-L'Hospitaling" does anything... – imranfat Nov 22 '13 at 21:14
• @JMCF125: This is a good question. Perhaps you should clarify it by adding bounds to the integrals. Specifically, how about writing $\int_{c}^{x}\; f(t)\; dt$ instead of $\int f(x) dx$? Same for $g(x)$. I point out that neither answer addressed this question. – Chris K Nov 22 '13 at 21:18
• @ChrisK, good, then the constants are eliminated! Forgot about them... I'll add that ASAP- – JMCF125 Nov 22 '13 at 21:33
• @JMCP125 : based on Umberto P.'s answer, the answer seems to be "yes". The next question is, is this ever useful, like the "regular" L'Hopital's Rule itself is? – Stefan Smith Nov 23 '13 at 2:28
• @miracle173, I never said l'Hôpital's implied this. And I didn't say the integrals had to be real anti-derivatives; in fact, in the edit, I added ChrisK's suggestion, which clears that up. – JMCF125 Nov 24 '13 at 8:52
I recently came across a situation where it was useful to go through exactly this process, so (although I'm certainly late to the party) here's an application of L'Hôpital's rule in reverse: | {
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We have a list of distinct real numbers $\{x_0,\dots, x_n\}$. We define the $(n+1)$th nodal polynomial as $$\omega_{n+1}(x) = (x-x_0)(x-x_1)\cdots(x-x_n)$$ Similarly, the $n$th nodal polynomial is $$\omega_n(x) = (x-x_0)\cdots (x-x_{n-1})$$ Now, suppose we wanted to calculate $\omega_{n+1}'(x_i)/\omega_{n}'(x_i)$ when $0 \leq i \leq n-1$. Now, we could calculate $\omega_{n}'(x_i)$ and $\omega_{n+1}'(x_i)$ explicitly and go through some tedious algebra, or we could note that because these derivatives are non-zero, we have $$\frac{\omega_{n+1}'(x_i)}{\omega_{n}'(x_i)} = \lim_{x\to x_i} \frac{\omega_{n+1}'(x)}{\omega_{n}'(x)} = \lim_{x\to x_i} \frac{\omega_{n+1}(x)}{\omega_{n}(x)} = \lim_{x\to x_i} (x-x_{n+1}) = x_i-x_{n}$$ It is important that both $\omega_{n+1}$ and $\omega_n$ are zero at $x_i$, so that in applying L'Hôpital's rule, we intentionally produce an indeterminate form. It should be clear though that doing so allowed us to cancel factors and thus (perhaps surprisingly) saved us some work in the end.
So would this method have practical use? It certainly did for me!
PS: If anyone is wondering, this was a handy step in proving a recursive formula involving Newton's divided differences. | {
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• Very interesting, +1! I certainly didn't expect it to have been successfully used before. But why didn't you make the question you are answering when you found it worked? – JMCF125 Nov 27 '13 at 11:01
• Ah, after having reread your question, I think the technique you're asking about is a little different. In my case, I went from $f'/g'$ (where both $f'$ and $g'$ were non-zero) to $f/g$ (where both $f$ and $g$ were zero). I didn't use the reverse L'Hôpital to resolve an indeterminate form; I used it to make one. – Omnomnomnom Nov 27 '13 at 15:38
• However, when you apply the function $\omega_n(x)$ and cancel out $\Pi_{i=0}^n (x-x_i)$ (sorry, couldn't resist but to use big product notation :)), the "indeterminacy" disappears, it only appears as an intermediate step. Call $\omega'$ as $\alpha$ and $$\lim_{x\to x_i} \frac{\alpha_{n+1}(x)}{\alpha_n(x)} = \lim_{x\to x_i} \left(\lim_{c\to x}\frac{\int\alpha_{n+1}(z)\ dz}{\lim_{c\to x}\int\alpha_{n}(z)\ dz}\right)=x_i-x_{n}$$. Right? (didn't check it thoroughly, there's no comment preview) – JMCF125 Nov 27 '13 at 21:07
• Exactly! So, you could say that the trick in using L'Hopital's rule backwards is choosing the antiderivatives of $\alpha_{n+1}$ and $\alpha_n$ that give you the indeterminate fraction. I'm not sure what the ruole of $c$ is in your statement, though – Omnomnomnom Nov 27 '13 at 21:16
• About the $c$, it's to eliminate the constants, see the problem Arkamis referred and the answer of Kaz, which has a related idea. The point is to make the integral so small it would be alike a derivative, only as a different expression. (BTW, the second $c\to x$ limit was not suppose to be there, I pasted it twice) – JMCF125 Nov 27 '13 at 21:55 | {
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With L'Hospital's rule your limit must be of the form $\dfrac 00$, so your antiderivatives must take the value $0$ at $c$. In this case you have $$\lim_{x \to c} \frac{ \int_c^x f(t) \, dt}{\int_c^x g(t) \, dt} = \lim_{x \to c} \frac{f(x)}{g(x)}$$ provided $g$ satisfies the usual hypothesis that $g(x) \not= 0$ in a deleted neighborhood of $c$.
• Good, interpreted the question in the "right" way. – André Nicolas Nov 22 '13 at 21:26
• Or $\frac{\infty}{\infty}$. – marty cohen Nov 23 '13 at 2:56
• Very important, this only works if you know that $\lim \frac{f(x)}{g(x)}$ exists. Even if the first limit exists, you cannot conclude that the second one does. – N. S. Nov 23 '13 at 5:58
• @N.S., neither can you conclude the division of the derivatives exists, even when the division of the original functions does, as in the original l'Hôpital's. – JMCF125 Nov 23 '13 at 13:48
• @JMCF125 True, but the goal in L'H is: starting with a limit, you replace by a different limit. In L'H you replace the functions by the Derivatives, not the derivatives by the functions. In classical L'H, you check $\lim\frac{f'}{g'}$ and if this limit exists you are done. Here you check $\lim \frac{\int f}{\int g}$ and if the limit exists, you cannot say anything. Note that if you go the other direction, you just apply classical L'H – N. S. Nov 23 '13 at 15:40
No it does not, for example consider the limit $\lim_{x\to 0}\frac{\sin x}{x}$, L'Hôpital's gives you 1, but reverse L'Hôpital's goes to $-\infty$ (without introducing extra constants). | {
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• +1 for being a simple, easy counterexample, although I assume that you're taking the integral of sin(x) to be -cos(x) - but that implies either a constant or a definite interval on the integral. – fluffy Nov 23 '13 at 4:33
• @fluffy : Yes, exactly, thank you – Arjang Nov 23 '13 at 4:43
• The conditions under which it works may be more restricted... – JMCF125 Nov 23 '13 at 17:20
• If we want to use reverse L'Hopital, it is our 'responsibility' to ensure that the newly integrated function is also of an indeterminate form, and that can be tackled easily using the integration constant. Here, instead of -cosx, one should use 1-cosx (in the numerator) for the limit to exist, and hence evaluate it to be 1. – arya_stark May 8 '18 at 8:08
Recall that the integral of a function requires the inclusion of an arbitrary constant. So, if $f(x) = x$, then $\int f(x)\ dx = \frac12 x^2 + C$.
Then, assuming your rule holds, then $$\lim_{x\to 0} \frac{x}{x} = \lim_{x \to 0} \frac{\int x\ dx}{\int x\ dx} = \lim_{x \to 0} \frac{\frac12 x^2 + C_f}{\frac12 x^2 + C_g} = \frac{C_f}{C_g}.$$
This means you can get literally any value you wish. | {
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This means you can get literally any value you wish.
• What if we forget the constants? Wait, don't answer yet, I'll edit my question first. – JMCF125 Nov 22 '13 at 21:32
• @JMCF125 Note that it really makes no sense to talk of "forgetting the constants". For example, the following three functions are antiderivatives of $2\sin(x)\cos(x)$: $\sin^2(x)$, $-\cos^2(x)$, $-\cos(2x)/2$. Which one is the "correct" one to use if we are to "forget constants"? – Andrés E. Caicedo Nov 22 '13 at 21:59
• @AndresCaicedo, what if we use a definite integral $\displaystyle\int_0^x x\ dx=\frac{x^2}2$, as referred in my edit. That way, wouldn't the result be the same? – JMCF125 Nov 24 '13 at 8:57
• Why use $0$ as the lower bound. Also, it should be $\int_0^x t\ dt$, as it makes no sense to have the variable of integration be the same as a bound. In any case, a lower bound of $c$ makes some sense if $x \to c$, since, as others have said, one of the ways this question makes sense is if $f, g \to 0$ at $x=c$. But it elides the case of $f,g \to \infty$. – Emily Nov 24 '13 at 16:13
• @Arkamis, «(...) as it makes no sense to have the variable of integration be the same as a bound.», sorry, distraction. :S I meant for the top limit to be an $a$. I used the lower limit as $0$ so that in your example, the integral would be $\int^a_0 x\ dx=\frac{a^2}2+C-\frac{0^2}2-C=\frac{a^2}2$. Isn't it right like this? – JMCF125 Nov 27 '13 at 10:57
If the limit of the ratio of those functions exists, I suspect that it may be possible for a certain definite integral to have the same limit. Something like:
$$\lim_{x\to \infty}\frac{\int_x^{x+1}f(x)\ dx}{\int_x^{x+1}g(x)\ dx}$$
The idea is that we grow $x$, and compare the ratios of some definite integrals in the neighborhood of x, of equal width and position. | {
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For instance, if we imagine the special case that the functions separately converge to constant values like 4 and 3, then if we take $x$ far enough, then we are basically just dividing 4 by 3. A definite integral of width 1 of a function that converges to 4 has a value that is approximately four, near a domain value that is large enough.
But this is like a derivative in disguise. If we divide a definite integral by the interval length, and then shrink the interval, we are basically doing derivation to obtain the original function that was integrated. Except we aren't shrinking the interval, since we don't have to: the fixed width 1 becomes smaller and smaller in relation to the value of $x$, so it is a "quasi infinitesimal", so to speak.
The real problem with this, even if it works, is that integrals don't seem to offer any advantage. Firstly, even if the function has the properties of being integrable, it may not be symbolically integrable, or it may be hard to integrate. Integration produces something more complex than the integrand.
The advantage of L'Hôpital's Rule is that we can reduce the power of the functions. If they are polynomials, they lose a degree in the differentiation, which can be helpful, and gives us a basis for instantly gauging the limit of the ratios of polynomials of equal degree simply by looking at the ratios of their highest degree coefficients. And then certain common functions at least don't grow any additional hair under differentiation, like sine, cosine, e to the x.
• Interesting, +1. But what about those exceptional functions whose derivative is more complicated? Wouldn't the integral ease calculation? – JMCF125 Nov 23 '13 at 13:55
L'Hospital's Rule is really a compact way of looking at ratio of the Taylor's polynomial for the numerator and denominator functions (w/o remainder). This polynomial is an exact fit to the function and its derivatives at the indicated point and close in a small neighborhood thereof. | {
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When the limit is 0/0 that means that both Taylor's polynomials have 0 constant terms. So then you look at the x terms (which represent the derivatives). If you can pick out the limit from there, good -- if they are both zero you go on to the $x^2$ terms (if any).
So L'Hospital's Rule isn't some vague thing that was plucked out of the air and happens to use derivatives. Any method of evaluating limits ought to be based on a clear mathematical derivation. You can use integrals or gamma functions or field theory or anything that pleases you, as long as you show you have a sound basis for evaluating the limit that way.
• Can you help me finding such a basis? – JMCF125 Nov 28 '13 at 12:26
• @ Betty Mock What is exact permissible basis to modify the vanishing numerator and denominator ? – Narasimham Nov 20 '14 at 20:11
• Expand the top and bottom in a Taylor's polynomial. If the first term of both polynomials is zero, then you look at the second term. If both of those are zero, you go on to the third term, etc. Sooner or later you get a term which is not zero, and that is the one which shows you the limit. – Betty Mock Nov 22 '14 at 2:46
$$\lim_{ x \to 0} \frac{ sin \,x \,}{\,x}= \lim_{ x \to 0} \frac{\int sin \,x \,\ dx}{\int x\ dx} = \lim_{x \to 0} \frac{\ (- cos \, x)}{( x^2/2)-1} =\lim_{x \to 0} \frac{\ (-cos \, x)+1}{( x^2/2)} = 1$$
Constants should be appropriately chosen from function/ power series expansion of the integrands and distributed between numerator and denominator maintaining the balance.
The statement is vague but I am sure it can be properly worded/stated to include even second, third integrands in the numerator and denominator.
More simply:
When $\frac yx =const = m$ as L'Hospital limit, it is =$\frac{dy}{dx}$ by Quotient Rule.
$m = \frac {dy}{ dx} = \frac {y \, \pm C_1 }{ x} =\frac {y \, }{ x \pm C_2}.$ | {
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# Improve speed for calculating a recursive sequence
I want to calculate following recursive sequence:
$\alpha_{0}=0,\\ \cos(\alpha_{i})=\cos(\alpha_{i-1})\cdot\cos(\beta_{i})+\sin(\alpha_{i-1})\cdot\sin(\beta_{i})\cdot\cos(\gamma_{i}).$
In mathematica 8.0.1 I try to do this with:
endRecursion = 20;
beta = Import["beta.txt", "List"];
beta = beta[[1;;endRecursion]];
gamma = RandomReal[{0,2*Pi}, endRecursion];
alpha[0]:= 0;
alpha[i_]:=
ArcCos[Cos[alpha[i-1]]*Cos[beta[[i]]]+
Sin[alpha[i-1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]];
result=Table[alpha[i], {i, 1, endRecursion}];
The values for beta are imported because they are generated by another formula which is not important for my question here. Values for beta can be found here.
The problem is that the calculation of the result is very slowly. It takes about 14s for just 20 iterations! And I need about 300 iterations (times ~10000, because I have several lists of beta values).
I recognized that the calculation is much faster if I skip the Cos[alpha[i-1]] in the first or the Sin[alpha[i-1]] in the 2nd summand. Then it takes just ~ 0.01 s for 20 iterations. But unfortunately this falsifies my result. I cannot understand why it is so much faster if there is just one time alpha[i-1] on the right side of the equation. Am I doing something wrong? Is there any way to increase the speed of this calculation? I would be glad about any help! | {
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-
Have you tried caching? alpha[i_] := alpha[i] = ArcCos[Cos[alpha[i-1]]*Cos[beta[[i]]] + Sin[alpha[i-1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]] – J. M. May 26 '12 at 12:44
I am such a *** idiot! Thank you so much for this comment! I totally forgot the caching although I used it some time ago (5 years, perhaps too long time ago). Unfortunately, I did not see this trick when I searched for a solution, otherwise I would not have asked this question... Thank you anyway! You saved my weekend! How can I accept your comment as answer? – partial81 May 26 '12 at 12:58
@J.M. hmm, in my amazement that nobody had answered this, I wrote my answer without seeing your comment... Hope you were not writing up one also! – acl May 26 '12 at 13:07
You can accept acl's answer. :) I didn't answer because I can't test at the moment. – J. M. May 26 '12 at 13:07
Ok, if you do not mind! Thanks again! – partial81 May 26 '12 at 13:12
You can speed it up by "memoization", i.e., remembering previous values of alpha[i]:
endRecursion = 20;
beta = Import["beta.txt", "List"];
beta = beta[[1 ;; endRecursion]];
gamma = RandomReal[{0, 2*Pi}, endRecursion];
alpha[0] := 0;
alpha[i_] :=
alpha[i] =
ArcCos[Cos[alpha[i - 1]]*Cos[beta[[i]]] +
Sin[alpha[i - 1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]];
result = Table[alpha[i], {i, 1, endRecursion}];
-
Thanks acl and @J.M. for answering my question so quickly! – partial81 May 26 '12 at 13:14
There was a mistake: you were importing strings (you didn't convert to numbers after StringSplit.) I fixed it. – Szabolcs May 26 '12 at 13:15
@Szabolcs yes, I had just realised it and our edits almost overlapped... thanks – acl May 26 '12 at 13:16
Okay, I'll remove the comments then. – Szabolcs May 26 '12 at 13:20
Although you can always implement recursion by hand the way you are trying to do, there are also some specialized functions that are designed to make your life a little easier. In particular, there is FoldList. | {
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This approach is also slightly faster than the manual recursion approach (assuming you start from a clean slate):
endRecursion = 20;
beta = Import["beta.txt", "List"];
beta = beta[[1 ;; endRecursion]];
gamma = RandomReal[{0, 2*Pi}, endRecursion];
FoldList[ArcCos[Cos[#1]*#2[[1]] + Sin[#1]*#2[[2]]] &, 0,
Transpose[{Cos[beta], Sin[beta] Cos[gamma]}]]
After initializing the lists as in your example, I need only a single statement to generate the list.
The folding function is what appears in the original recursive formulation, but it is written as a pure function with two formal arguments #1 and #2.
The first argument, #1, refers to your alpha and its starting value is passed to FoldList as the second argument (0). The second argument in the pure function (#2) refers to the elements of the list Transpose[{Cos[beta], Sin[beta] Cos[gamma]}] in which I collected the pre-existing lists beta and gamma (but in a form that is already pre-processed to avoid having to evaluate cosines and sines individually in the recursion). | {
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-
With this particular formulation, one could even simplify the FoldList[] expression as FoldList[ArcCos[{Cos[#1], Sin[#1]}.#2] &, 0, Transpose[{Cos[beta], Sin[beta] Cos[gamma]}]]. One might even consider using the Weierstrass substitution here, since the beta values are all $< \pi/2$... – J. M. May 26 '12 at 18:47
@J.M. That's correct - the dot product may give a tiny additional speed boost but I didn't check that. – Jens May 27 '12 at 0:50
Thanks @Jens for this additional answer. FoldList is something I have never used so far – that will change now (although the additional speed boost is just small). – partial81 May 27 '12 at 11:24
@ Jens, Let’s say: alpha[i_]:=alpha[i]=ArcCos[Cos[alpha[i - 1]]*Cos[beta[[i+1]]]+Sin[alpha[i - 1]]*Sin[beta[[i]]]*Cos[gamma[[i]]]]. How do I have to adapt your solution with FoldList for this case? I really have problems to see a solution because now I would need in my iteration the i+1 element of beta. I would be happy if you could give me a hint again! – partial81 May 31 '12 at 13:10
@partial81 The term you changed is the first one in Transpose[{Cos[beta], Sin[beta] Cos[gamma]}], and the change is simply a shift in the index. The fastest way this can be implemented is to use RotateLeft: Transpose[{RotateLeft[Cos[beta]], Sin[beta] Cos[gamma]}] – Jens May 31 '12 at 15:39
The function runs slowly because each iteration calls alpha twice the number of times of the previous iteration. Thus, $n$ iterations invoke alpha $2^n$ times -- exponential run time! Fortunately, the two nested calls to alpha are identical. If we change alpha to call itself only once instead of twice, the algorithm runs in linear time while still retaining simple recursive form:
endRecursion = 20;
beta = Import["beta.txt", "List"];
beta = beta[[1;;endRecursion]];
gamma = RandomReal[{0,2*Pi}, endRecursion];
alpha[0] = 0;
alpha[i_]:=
Module[{a = alpha[i-1], b = beta[[i]], g = gamma[[i]]}
, ArcCos[Cos[a]*Cos[b]+Sin[a]*Sin[b]*Cos[g]]
] | {
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result = Table[alpha[i], {i, 1, endRecursion}];
Even so, I agree with @Jens that functional iteration is frequently preferred over recursion in order to avoid hitting the $RecursionLimit (or$IterationLimit).
-
Thanks @WReach for answering the other part of my question. Now I understand why it was so slowly at the beginning. And it is nice that you give a link to the useful functional iterations. – partial81 May 27 '12 at 20:53 | {
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# Finding the Exact Value
How would one go about finding the exact value of $\theta$ in the following:$\sqrt{3}\tan \theta -1 =0$? I am unsure of how to begin this question. Any help would be appreciated!
-
are you familiar with $\arctan$ function? – Alex Jun 18 '13 at 16:15
I am not familiar with that function – ComradeYakov Jun 18 '13 at 16:16
@Tim I meant the first, sorry for the confusion. It isn't cubed – ComradeYakov Jun 18 '13 at 16:17
Hint: $\sqrt{3}\tan \theta -1 =0\implies \sqrt{3}\tan \theta =1\implies \tan \theta=1/\sqrt{3}$
Recall from SOH CAH TOA, that $\tan \theta=\text{opposite}/\text{adjacent}$, so in this case $1$ is the opposite side and $\sqrt{3}$ is the adjacent side.
The above is a special triangle. Since you know that $\tan \theta=1/\sqrt{3}$, it follows that the angle $\theta=\pi/6$. However since $\tan \theta$ is a trigonometric periodic function, that value will come up again, and again. (Just like $\sin0=\sin2\pi).$
$\tan \theta$ has a period of $\pi.$
So to be fully accurate, $\theta=\pi/6 +n\pi, n \in \mathbb{Z}$
-
Thank you! So theta= pi/4(+-)pi(n), 5pi/4(+-) is not a solution? – ComradeYakov Jun 18 '13 at 16:31
theta= pi/4(+-)pi(n), 5pi/4(+-) is not a solution. – Sujaan Kunalan Jun 18 '13 at 16:35
Ok, thanks for your help! – ComradeYakov Jun 18 '13 at 16:37
You are welcome! – Sujaan Kunalan Jun 18 '13 at 16:38
$2\cos^{2} \theta+\cos \theta-1=0 \implies 2\cos^{2}+\cos \theta =1 \implies \cos \theta (2\cos \theta+1)=1\implies \cos \theta =1 \implies \theta =2n\pi, n\in \mathbb{Z}$ However, $\cos \theta(2\cos \theta +1)=1$ also implies that $2\cos \theta+1=1 \implies 2\cos \theta =0 \implies \cos \theta =0 \implies \theta = (2n-1)/\pi, n \in \mathbb{Z}$ – Sujaan Kunalan Jun 18 '13 at 17:07
If $\sqrt{3}\tan \theta - 1= 0$ then $\sqrt{3}\tan\theta = 1$ and $\tan\theta = 1/\sqrt{3}$. Hence $\theta =\arctan(1/\sqrt{3}) = \frac{\pi}{6}$. | {
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Notice that the graph $y = \tan\theta$ repeats every $\pi$-radians. Hence, all solutions are given by
$$\theta = \frac{\pi}{6} + \pi n$$
where $n$ is any whole number.
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# Does the square of the minimum of two correlated Normal variables have a chi-squared distribution?
If the random variables $X$ and $Y$ are normally distributed with mean $\mu = 0$ and standard deviation $\sigma = 1$, then define the random variable $Z = \min(X, Y )$. The problem is to prove that $Z^2$ follows a gamma distribution with parameters $(\alpha = 1/2, \lambda= 1/2)$ (i.e. a chi-squared distribution).
I have: $$P(Z^2 \le z) = P((\min\{X,Y\})^2 \le z) =P\left(-\sqrt{z}\le \min(X,Y)\le \sqrt{z}\right) \\ = P\left(-\sqrt{z}\le X\le \sqrt{z}, -\sqrt{z}\le Y\le \sqrt{z}\right)$$ But I am not sure where to go from here, since the random variables aren't assumed to be independent. I would appreciate any hints!
• Your logic at the last step seems to go astray. If the minimum of (X,Y) is within $\pm \sqrt z$ (as you have in your second-last line), that doesn't make both of them in that interval, as you have in your last line. You only need one of the two of them in there. – Glen_b -Reinstate Monica Nov 28 '16 at 3:57
• @glen_b What then would be the interval on the other one? – Mjt Nov 28 '16 at 14:08
• That's not a productive way to approach writing the event. But first you might want to ponder whether anything has been left out of your framing of the question (which is possibly why whuber is able to offer a counterexample). – Glen_b -Reinstate Monica Nov 28 '16 at 15:55
The result is not true. | {
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The result is not true.
As a counterexample, let $(X,Y)$ have standard Normal margins with a Clayton copula, as illustrated at https://stats.stackexchange.com/a/30205. Generating 10,000 independent realizations of this bivariate distribution, as shown in the lower left of the figure, produces 10,000 realizations of $Z^2$ that clearly do not follow a $\Gamma(1/2,1/2)$ distribution (in a Chi-squared test of fit, $\chi^2=121, p \lt 10^{-16}$). The qq plots in the top of the figure confirm that the marginals look standard Normal while the qq plot at the bottom right indicates the upper tail of $Z^2$ is too short.
The result can be proven under the assumption that the distribution of $(X,Y)$ is centrally symmetric: that is, when it is invariant upon simultaneously negating both $X$ and $Y$. This includes all bivariate Normals (with mean $(0,0)$, of course).
The key idea is that for any $z \ge 0$, the event $Z^2 \le z^2$ is the difference of the events $X \ge -z\cup Y \ge -z$ and $X \ge z \cup Y \ge z$. (The first is where the minimum is no less than $-z$, while the second will rule out where the minimum exceeds $z$.) These events in turn can be broken down as follows:
$$\Pr(Z^2 \le z^2) = \Pr(X\ge -z) - \Pr(Y \le -z) + \Pr(X,Y\lt -z) - \Pr(X,Y\gt z).$$
The central symmetry assumption assures the last two probabilities cancel. The first two probabilities are given by the standard Normal CDF $\Phi$, yielding
$$\Pr(Z^2 \le z^2) =1 - 2\Phi(-z).$$
That exhibits $Z$ as a half-normal distribution, whence its square will have the same distribution as the square of a standard Normal, which by definition is a $\chi^2(1)$ distribution.
This demonstration can be reversed to show $Z^2$ has a $\chi^2(1)$ distribution if and only if $\Pr(X,Y\le -z) = \Pr(X,Y\ge z)$ for all $z\ge 0$.
Here is the R code that produced the figures.
library(copula)
n <- 1e4
set.seed(17)
xy <- qnorm(rCopula(n, claytonCopula(1)))
colnames(xy) <- c("X", "Y")
z2 <- pmin(xy[,1], xy[,2])^2 | {
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cutpoints <- c(0:10, Inf)
z2.obs <- table(cut(z2, cutpoints))
z2.exp <- diff(pgamma(cutpoints, 1/2, 1/2))
rbind(Observed=z2.obs, Expected=z2.exp * length(z2))
chisq.test(z2.obs, p=z2.exp)
par(mfrow=c(2,2))
qqnorm(xy[,1], ylab="X"); abline(c(0,1), col="Red", lwd=2)
qqnorm(xy[,2], ylab="Y"); abline(c(0,1), col="Red", lwd=2)
plot(xy, pch=19, cex=0.75, col="#00000003", main="Data")
qqplot(qgamma(seq(0, 1, length.out=length(z)), 1/2, 1/2), z2,
xlab="Theoretical Quantiles", ylab="Z2",
main="Gamma(1/2,1/2) Q-Q Plot")
abline(c(0,1), col="Red", lwd=2)
Let $M=min(X,Y)^2$, $$P(M<m) = P(M<m,X<Y) + P(M<m,X>Y) ~~~~~~~~~~~\\ = P(M<m|X<Y)P(X<Y) + P(M<m|X>Y)P(X>Y) \\ = P(X^2<m)P(X<Y) + P(Y^2<m)P(X>Y) ~~~~~~~~~~~~~~~~~~~~~~\\ = \frac{1}{2}P(X^2<m) + \frac{1}{2}P(Y^2<m) \quad \quad \quad\quad\quad\quad\quad\quad\quad\quad~ \\ =P(X^2<m) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ~~~$$ | {
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+0
# Direct
0
330
6
the time needed to paint a fence varies directly with the length if the fence and indirectly with the number of painters. if it takes 5 hours to paint 200 feet of fence with 3 painters, how long will it take 5 painters to paint 500 feet of fence?
Guest May 26, 2015
#3
+91510
+10
ok we might as well all answer this question!
This is how I was trained to do it
$$\\T\alpha \frac{L}{N} \\\\ T=k* \frac{L}{N}\qquad but\\\\ 5=k* \frac{200}{3}\qquad so\\\\ k=\frac{3}{40}\qquad so\\\\ T=\frac{3L}{40N}\\\\ When N=5 and L=500\\\\ T=\frac{3*500}{40*5}\\\\ T=7.5 \;\;hours$$
Melody May 26, 2015
Sort:
#1
+26412
+10
The rate at which fence painting is done is 200/(3*5) feet per painter-hour.
In order to paint 500 feet we need 500*3*5/200 painter-hours
With 5 painters the time taken is 500*3*5/(200*5) hours = 7.5 hours
.
Alan May 26, 2015
#2
+81090
+10
If 3 painters paint 200 ft of fence in 5 hours, in 1 hour, they must paint 40 ft.......so each painter paints 40/3 ft per hour.
Then, 5 painters can paint 200/3 ft every hr........so........500/ (200/3) = 1500/200 = 7.5 hrs.
CPhill May 26, 2015
#3
+91510
+10
ok we might as well all answer this question!
This is how I was trained to do it
$$\\T\alpha \frac{L}{N} \\\\ T=k* \frac{L}{N}\qquad but\\\\ 5=k* \frac{200}{3}\qquad so\\\\ k=\frac{3}{40}\qquad so\\\\ T=\frac{3L}{40N}\\\\ When N=5 and L=500\\\\ T=\frac{3*500}{40*5}\\\\ T=7.5 \;\;hours$$
Melody May 26, 2015
#4
+81090
0
Which can we do faster.......paint that fence or answer this question?????
{ I say its a toss up..... }
CPhill May 26, 2015
#5
+72
+5
Depends on how much You can take 60 seconds on writing this while 70 painters paint a small fence 3*3 fence
xXmathXx May 26, 2015
#6
+72
0
Hello CPhill
xXmathXx May 26, 2015
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xXmathXx May 26, 2015
#6
+72
0
Hello CPhill
xXmathXx May 26, 2015
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# Minimum steps on a numberline
Given an infinite numberline, you start at zero. On every i'th move you can either move i places to the right, or i places to the left. How, in general, would you calculate the minimum number of moves to get to a target point x? For example:
if x = 9:
move 1: starting at zero, move to 1
move 2: starting at 1, move to 3
move 3: starting at 3, move to 0
move 4: starting at 0, move to 4
move 5: starting at 4, move to 5
• This seems like another mathbook-type problem to me. – Mithical Aug 25 '16 at 11:47
• @Mithrandir I think so. VTC. – IAmInPLS Aug 25 '16 at 11:50
• I say "don't close" -- unless it is mandatory to have a real-life story enclosing the real puzzle to be solved. – Rosie F Aug 26 '16 at 3:15
• I want to go to school with these VTCers; clearly their textbooks have much more interesting problems than mine. – ffao Aug 26 '16 at 3:28
• Shouldn't the last line end with "move to 9"? – celtschk Aug 26 '16 at 9:23
First,
you need to find $n$, for which $S_n=1+2+\dots+n=\frac{n(n+1)}2\ge x$
Then,
find a subset of the numbers in range $[1, n]$ which sum to $\frac{S_n-x}2$. This can always be done if the parity of $S_n$ is the same as the parity for $x$. This means, that if $m$ is the lowest number for which $S_m\ge x$, then at least one of $S_m$, $S_{m+1}$ or $S_{m+2}$ will do the work.
With the help of this
adding the rest, and subtracting the above will give you the desired solution
First you calculate the smallest $n$ with $\frac{1}{2}n(n+1) \geq x$
if $\frac{1}{2}n(n+1)-x$ is odd then increase $n$ by $1$ if $n$ is even or increase $n$ by $2$ if $n$ is odd.
Now $\frac{1}{2}n(n+1)-x$ is even
Now you can put a minus sign before some numbers that have the sum $\frac{1}{2} (\frac{1}{2}n(n+1)-x)$ and you are done.
## Example:
$x=23$
The smallest $n$ with $\frac{1}{2}n(n+1) \geq x$ is $n=7$
$\frac{1}{2}n(n+1)-x = 28-23 = 5$ is odd and $n$ is also odd so we have to increase $n$ by $2$.
That means $n=9$ | {
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That means $n=9$
$\frac{1}{2}(\frac{1}{2}n(n+1)-x) = \frac{1}{2}(45-23) = 11$
So we have to put a minus in front of numbers with the sum $11$. For example $2$ and $9$.
Result: 1-2+3+4+5+6+7+8-9
It comes out to be a better puzzle than it was voted to close for!
One thing is for sure, if the given number $x$ is a triangular number, the answer is its position in the sequence 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66
which is given by
$$\left\lfloor\dfrac{\sqrt{1+8x}-1}{2}\right\rfloor$$
Also:
here is my code, which gives a very interesting pattern,
output: For x=1, ans : 1 For x=2, ans : 3 For x=3, ans : 2 For x=4, ans : 3 For x=5, ans : 5 For x=6, ans : 3 For x=7, ans : 5 For x=8, ans : 4 For x=9, ans : 5 For x=10, ans : 4 For x=11, ans : 5 For x=12, ans : 7 For x=13, ans : 5 For x=14, ans : 7 For x=15, ans : 5 For x=16, ans : 7 For x=17, ans : 6 For x=18, ans : 7 For x=19, ans : 6 For x=20, ans : 7 For x=21, ans : 6 For x=22, ans : 7 For x=23, ans : 9 For x=24, ans : 7 For x=25, ans : 9 For x=26, ans : 7
which is :
$1,3,2,3,5,3,5,4,5,4,5,7,5,7,5,7,6,7,6,7,6,7,9,7,9,7,9$,... OEIS/A140358
check the values up to $1000$ here. see this final code
#include<bits/stdc++.h>
using namespace std;
int i,n,t;
int main()
{ int x,ans;
for(cin>>t;t;t--)
{ cin>>x;
n=floor((sqrt(1+8*x)-1)/2);
//cout<<n<<endl;
i=(n*(n+1))/2;
//cout<<x<<", sum:"<<i<<", N:"<<n<<endl;
if(n%2) //n is odd
{
if((x-i)%2) ans=n+2;
else ans=n+1;
}
else
{
if((x-i)%2) ans=n+1;
else ans=n+3;
}
cout<<">! For x="<<x<<", ans : "<<ans<<endl;
}
} | {
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"tags": null,
"url": "https://puzzling.stackexchange.com/questions/41436/minimum-steps-on-a-numberline"
} |
The addition problem above shows four of the 24 different in : GMAT Problem Solving (PS)
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# The addition problem above shows four of the 24 different in
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1,234
1,243
1,324
.....
....
+4,321
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?
A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660
[Reveal] Spoiler: OA
Last edited by Bunuel on 06 Nov 2012, 02:17, edited 1 time in total.
Renamed the topic and edited the question.
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02 Nov 2010, 23:45
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student26 wrote:
1,234
1,243
1,324
.....
....
+4,321 | {
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"url": "http://gmatclub.com/forum/the-addition-problem-above-shows-four-of-the-24-different-in-104166.html?kudos=1"
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The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?
A.24,000
B.26,664
C.40,440
D.60,000
E.66,660
Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.
The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660
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### Show Tags
08 Feb 2011, 05:26
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1,2,3,4 can be arranged in 4! = 24 ways
The units place of all the integers will have six 1's, six 2's, six 3's and six 4's
Likewise,
The tens place of all the integers will have six 1's, six 2's, six 3's and six 4's
The hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's
The thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's
Addition always start from right(UNITS) to left(THOUSANDS);
Units place addition; 6(1+2+3+4) = 60.
Unit place of the result: 0
carried over to tens place: 6
Tens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66
Tens place of the result: 6
carried over to hunderes place: 6
Hundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66
Hundreds place of the result: 6
carried over to thousands place: 6
Thousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66
Thousands place of the result: 6
carried over to ten thousands place: 6
Ten thousands place of the result: 0+6(Carried over from thousands place) = 6
Result: 66660
Ans: "E"
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Ans: "E"
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### Show Tags
08 Feb 2011, 05:48
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Merging similar topics.
Formulas for such kind of problems (just in case):
1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
Similar questions:
nice-question-and-a-good-way-to-solve-103523.html
can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html
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hellscream wrote:
Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make.
How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed?
You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit)
1111
1112
1121
... and so on till 4444
By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4.
Same for 10s, 100s and 1000s place. | {
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Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)
= (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4)
= 1111*64*10 = 711040
or use the formula given by Bunuel above:
Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times)
=$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above)
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09 Nov 2010, 05:24
Thanks for the great explaination shrouded1!
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08 Feb 2011, 04:09
the addition problem below shows four of the 24 different integers that can be formed by using each of the digits 1,2,3, and 4 exactly once in each integer.what is the sum of these 24 integers?
1,234
1,243
1,324
.......
.......
+4,321
a) 24,000
b) 26,664
c) 40,440
d) 60,000
e) 66,660
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### Show Tags
22 Jan 2013, 09:44
Bunuel wrote:
Merging similar topics.
Formulas for such kind of problems (just in case):
1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$. | {
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sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html
Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
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The addition problem above shows four of the 24 different intege [#permalink]
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21 May 2013, 06:35
This can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives
1234 +4321 = 5555
1243 +3421 = 5555
we may see that there are 4!/2 pairings we can make, giving us
5555(12) = 66660
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Re: The addition problem above shows four of the 24 different in [#permalink]
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28 Sep 2014, 09:41
For those who could not memorize the formular, you can guess the answer in 30 secs:
Since we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something
So,
6x1(thousand something) = 6 (thousand something)
6x2(thousand something) = 12 (thousand something)
6x3(thousand something) = 18 (thousand something)
6x4(thousand something) = 24 (thousand something)
Add them all 6+12 +18 + 24 = 60 (thousand something)
----> E
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Re: The addition problem above shows four of the 24 different in [#permalink]
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15 Aug 2016, 13:38
each term is repeating 6 times..(total 24/4=4)
now at unit...each term will repeat 6 times... (1x6+ 2x6 + 3x6 + 4x6 = 60) , so unit digit is "0" and 6 remaining.
repeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6. | {
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repeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6.
only E has 60 as last two digits.
Re: The addition problem above shows four of the 24 different in [#permalink] 15 Aug 2016, 13:38
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# The addition problem above shows four of the 24 different in
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# Conditional Probabilty
• October 27th 2007, 06:10 AM
Revilo
Conditional Probabilty
First little piggy is 25% likely to go to market if second little piggy goes. Second little piggy is 50% likely to go to if first little piggy goes. Sadly at least one of them must go to market. How likely are they both to go?
So far i have:
Let A = First little piggy goes to market
B = Second little piggy goes to market
P(A given B) = 0.25
P(B given A) = 0.5
P(A union B) = P(A) + P(B) - P(A intersection B)
Is the P(A union B) = 1 as atleast one must go, so this is always true?
Does anyone have any ideas, i'm really stuck.
• October 28th 2007, 01:11 AM
kalagota
Quote:
Originally Posted by Revilo
First little piggy is 25% likely to go to market if second little piggy goes. Second little piggy is 50% likely to go to if first little piggy goes. Sadly at least one of them must go to market. How likely are they both to go?
So far i have:
Let A = First little piggy goes to market
B = Second little piggy goes to market
P(A given B) = 0.25
P(B given A) = 0.5
P(A union B) = P(A) + P(B) - P(A intersection B)
Is the P(A union B) = 1 as atleast one must go, so this is always true?
Does anyone have any ideas, i'm really stuck.
yeah, it is correct..
but, i think, you should not look for the union, instead for the intersection..
note that P(A|B)= P(A int B) / P(B) [also, P(B|A) = P(A int B) / P(A) ]; this theorem should help you find the answer.
• October 28th 2007, 08:51 PM
Soroban
Hello, Revilo!
Quote:
First little piggy is 25% likely to go to market if second little piggy goes.
Second little piggy is 50% likely to go to if first little piggy goes.
Sadly at least one of them must go to market.
How likely are they both to go?
All your ideas are good ones . . .
Bayes' Theorem says: . $P(A|B) \:=\:\frac{P(A \wedge B)}{P(B)}$
We are given: | {
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Bayes' Theorem says: . $P(A|B) \:=\:\frac{P(A \wedge B)}{P(B)}$
We are given:
. . $P(1^{st}|2^{nd}) \:=\:\frac{1}{4}\quad\Rightarrow\quad \frac{P(1^{st} \wedge 2^{nd})}{P(2^{nd})}\: = \: \frac{1}{4}\quad\Rightarrow\quad P(1^{st} \wedge 2^{nd}) \:=\:\frac{1}{4}\!\cdot\!P(2^{nd})$ .[1]
. . $P(2^{nd}|1^{st}) \:=\:\frac{1}{2}\quad\Rightarrow\quad \frac{P(2^{nd} \wedge 1^{st})}{P(1^{st})} \:=\:\frac{1}{2}\quad\Rightarrow\quad P(1^{st} \wedge 2^{nd}) \:=\:\frac{1}{2}\!\cdot\!P(1^{st})$ . [2]
Equate [1] and [2]: . $\frac{1}{4}\cdot P(2^{nd}) \:=\:\frac{1}{2}\cdot P(1^{st}) \quad\Rightarrow\quad P(2^{nd}) \:=\:2\cdot P(1^{st})$ . [3]
Since at least one of them goes to the market: . $P(1^{st} \vee 2^{nd}) \:=\:1$
We have: . $P(1^{st} \vee 2^{nd}) \;=\;P(1^{st}) + P(2^{nd}) - P(1^{st} \wedge 2^{nd}) \;=\;1$ . [4]
. . [3] $\; P(2^{nd}) \: = \: 2\cdot P(1^{st})$
. . [2] $\; P(1^{st} \wedge 2^{nd}) \: = \: \frac{1}{2}\cdot P(1^{st})
$
Substitute into [4]: . $P(1^{st}) + 2\cdot P(1^{st}) -\frac{1}{2}\cdot P(1^{st}) \;=\;1\quad\Rightarrow\quad \frac{5}{2}\cdot P(1^{st}) \:=\:1$
Therefore: . $P(1^{st}) \:= \:\frac{2}{5}\qquad P(2^{nd}) \: = \: \frac{4}{5} \qquad \boxed{P(1^{st} \wedge 2^{nd}) \: = \: \frac{1}{5}}$ | {
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# The base of a logarithm
What exactly is the base of a logarithm ? and how should it be understood ? I used to think it was the base of a "normal" exponent e.g. the $2$ in $2^{75}$ would be the base in logarithmic form, but the change of base formula can accept ANY base, and when finding the number of digits in $2^{75}$, you use the common log:
$$2^{75}$$
$$\log_{10}(2^{75})$$
$$75\log_{10}(2)$$
$$75(0.301)+1=23 \textrm{ digits}$$
I understand that these formulas work, I just can't wrap my head around why they work the way they do, and the heart of my issue is how I should understand the base.
I did ponder that maybe a base of 10 represents a decimal system, and a base 2 would represent a binary system, but I haven't found any validation for that. But if that were the case, then would a base 16 represent a hexadecimal system ? and how would that work considering we use letters in addition to numbers ?
• How many digits in $10^k$? and what is value of $\log_{10}(10^{k})$ ? Try searching for pattern. Feb 1 '18 at 17:45
• "I did ponder that maybe a base of 10 represents a decimal system, and a base 2 would represent a binary system, but I haven't found any validation for that." Really? I would have thought those were the definitions. Feb 1 '18 at 18:00
I think you have the idea.
If $y = 2^{75}$ then $\log_2 y = 75$
One way to use bases is to find the number of digits that that number would have in that system. If $32<y<64$ then $5<\log_2 y < 6$
You can take logarithms in any base, and convert between bases.
$\log_a x = \frac {\log_b x}{\log_b a}$
For example, $\log_{10} 2^{75} = 75 \log_{10} 2$ as you have above.
But you could also say $\log_{10} 2^{75} = \frac {\log_2 2^{75}}{\log_2 10}$
Which implies $\log_2 10 = \frac 1{\log_{10} 2}$ | {
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Which implies $\log_2 10 = \frac 1{\log_{10} 2}$
You're right about the number of digits of a number in a specific base but why? To prove it let $(x)_b$ denote the number x in base $b$. If $b=10$ it is the same natural and convenient base we often use. Obviously if any number is between two consecutive powers of $10$, for example $1253$ is between $10^3$ and $10^4$ and it has 4 digits. In fact for any number $x$ if it belongs to $(10^n,10^{n+1}-1)$ for some n then $n+1$ is the number of digits of $x$. Then we can find the number of digits as following:$$10^n\le x<10^{n+1}\\n\le \log_{10}x<n+1\\n=\lfloor \log_{10}x\rfloor+1$$the same argument is true for any arbitrary base $b$, so the number of digits in base $b$ is $$dig_{b}(x)=\lfloor \log_{b}x\rfloor+1$$
The base of a logarithm is the base of the corresponding exponential function. Take $10^x$ for example. The base of this exponential function is $10$. The corresponding logarithm is the standard log, $\log_{10}{x}$, or simply $\log{x}$. The logarithm is the inverse of the exponential function; for example, if $x = 2$, then $10^2 = 100$. The inverse therefore (the $\log{x}$) will get us the exponent, or 2. Therefore, $\log_{10}(100) = 2.$
The base of the logarithm in this case is the $10$ from $10^x$.
More generally, if you had an exponential function $b^x$ then the corresponding logarithm would be $\log_b{x}$ in which the base is $b$.
$\log_{16} 2^{75} = \frac {75}{4} = 18.75$ so there are $18 + 1 = 19$ digits in hex.
And indeed $2^{75} = 8*2^{72} = 8*16^{18}$ so $2^{75}$ in hexadecimal is $8000000000000000000$.
You seem to understand the base of the logorithms perfectly so far as I can tell.
## Decimals
As for the number of digits of a number in the decimal system. consider the following:
Let $x$ be any real number - for our convenience, assume it is positive. Then, obviously, there exists exactly a natural number $n\in\mathbb{N}$ such that: $$10^n\leq x<10^{n+1}$$ | {
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Now, if a number is between $10^n$ and $10^{n+1}$, as above, it means that it has exactly than $n+1$ digits. Imagine $x=9468$, for instance. In this case, $n=3$, since: $$10^3=1000\leq9468<10,000=10^4$$ So, we would like to express $n$ in terms of $x$, so as to have a formula that, given $x$, returns $n+1$. Since $n$ is appearing as an exponent of $10$, we think that taking $\log_{10}$ on every side is a way to "get $n$ out of the exponent". So, we have that: \begin{align*} &10^n\leq x<10^{n+1}\\ \Leftrightarrow&\log_{10}10^n\leq\log_{10}x<\log_{10}10^{n+1}\\ \Leftrightarrow&n\leq\log_{10}x<n+1 \end{align*} So, $n=\lfloor\log_{10}x\rfloor$ which means that the number of digits of $x$ is exactly: $$n+1=\lfloor\log_{1}x\rfloor+1$$
## Logarithms
Now, as for the logarithms, let us remind the definition:
Given $x,b>0$, $\log_bx$ is the - only - number that has the following property: $$b^{\log_bx}=x$$ So, the logarithm of $x$ with base $b$ is the only number which can be the exponent of $b$ such that $b$ to that exponent is equal to $x$.
So, given the symbol $\log_bx$, $b$ being the logarithms base means that $b$ is the number to which we may set $\log_bx$ as an exponent to get $x$. You can imagine the following game that helped me understand logarithms:
## The Log-Game
Imagine two friends, Alice and Bob - hehe, these are probably the only friends we, mathematicians, have ever heard of - have a secret code to exchange messages. So, every message they want to send is being turned into a - probably very large - number. So, in our case, Alice and Bob are exchanging numbers in order to communicate.
But, in order to make it more difficult for other people to read their messages and find out the code, they use the following trick:
If $x$ is the message Alice wants to send to Bob, then, instead of this Alice sends Bob another number, which is $\log_b x$, for some base $b$ that they have pre-decided. | {
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So, Bob receives a number $y$ that is not the message he wants, and, probably, has no meaning at all. How will he find the message? As mentioned above, Alice and Bob have pre-decided a base for the logarithms they are exchanging, so, what Bob has to do is to calculate: $$b^y=b^{\log_bx}=x$$ since, by the definition of logarithm, the only way to find $x$ given $y=\log_bx$ is to set it as an exponent to $b$. Every other, not knowing the exact value of $b$ cannot find that message.
## Over-decimal Systems
We have heard about decimal system, binary system etc. But, the most usual case is that we use the usual symbols $0,1,2,\dots,9$ for such systems. What about hexadecimal system? There, we need $16$ different - distinct - symbols for our elementary digits. So, we introduce $6$ new symbols: $A,B,C,D,E,F$. These are just symbols, as the previous ones; $0,1,2,\dots,9$. There is no matter about that.
To make it a little more clear, we have decided that: $$0+1=1,\ 1+1=2,\ 2+1=3,\dots,8+1=9$$ Moreover, we have decided that $$9+1=10,\ 99+1=100$$ Or, to be more precise, the nature of our enumerating system - that the position of a digit is important to its evaluation - is a property that makes it feasible to use only some - finite in number - symbols and not infinitely many or strange combinations - see, for instance, the roman digits.
We have been used to using these symbols to represent numbers, which has driven us to the wrong conclusion that these symbols are the numbers they represent. Well, no symbol is any number, it just represents one, probably different with respect to our system.
For instance, in binary system: $$101$$ represents what in the decimal system we would write as: $$5$$ or what in the ternary (3) system we would write as: $$12$$ or what in the hexadecimal system we would write as: $$5$$ | {
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So, if we want to use some enumeration system with more that $10$ elementary digits, we have to introduce some new symbols for these digits and we also have to define the basic operations with them. So, in hexadecimal system we deicide that: $$9+1=A,\ A+1=B,\ B+1=D,\dots,E+1=F$$ and all the other properties we know.
So, to find how many elements a number $x$ has in hexadecimal, we simply note that there exists exactly on $n$ such that: $$16^n\leq x<16^{n+1}$$ and, with the same thoughts as above, we take $\log_16$ to every side, so we have: $$n\leq\log_{16}x<n+1$$ and, we have, finally that number $x$ in the hexadecimal system has: $$\lfloor\log_{16}x\rfloor+1$$ digits. | {
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# Prove that $4x^2-8xy+5y^2\geq0$ - is this a valid proof?
I need to prove that $4x^2-8xy+5y^2\geq0$ holds for every real numbers $x, y$.
First I start with another inequality, i.e. $4x^2-8xy+4y^2\geq0$, which clearly holds as it can be factorized into $(2x-2y)^2\geq0$. Now we can add $y^2$ (which is always nonnegative) to the left side, thus obtaining $4x^2-8xy+5y^2\geq0$, which is always true - we've just added a nonnegative expression to another, so the sum is still greater or equal to zero.
Is my proof correct? I had it on my final math exam and would like to be sure.
• Looks good to me. – simonzack May 5 '15 at 13:23
• +1. An example of converse not being true is the polynomial $$x^6+y^2z^4+y^4z^2-3x^2y^2z^2$$ – Leg May 5 '15 at 14:46
$4x^2-8xy+5y^2 = (2x-2y)^2+y^2 \ge 0$.
Using this sum-of-squares technique you also get that equality occurs exactly when $2x-2y = 0$ and $y = 0$, or equivalently when $(x,y) = (0,0)$. | {
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1. ## integral with cotx
Hi,
Could someone please give me a hint on how to do this problem? I know that I have to use a formula to do it, but it looks like I need to simplify it.
Int. cotx/sq. 1+2sinx
I have these formulas:
int. cotudu=ln|secu|+c
int. cot^udu=-1/n-1cot^n-1u-int. cot^n-2udu
Thank you
2. $\displaystyle\int\frac{\cot x}{\sqrt{1+2\sin x}}dx=\int\frac{\cos x}{\sin x\sqrt{1+2\sin x}}dx$
Substitute $\sin x=t\Rightarrow\cos xdx=dt\Rightarrow\int\displaystyle\frac{1}{t\sqrt{ 1+2t}}dt$.
Substitute $\sqrt{1+2t}=u\Rightarrow 2t=u^2-1\Rightarrow dt=udu$
Then $\displaystyle\int\frac{u}{\frac{u^2-1}{2}\cdot u}du=2\int\frac{1}{u^2-1}du=\ln \left|\frac{u-1}{u+1}\right|+C$
Now back substitute.
3. Hello, chocolatelover!
I found a method . . . but it took two substitutions.
I'm certain someone will streamline it for us.
$\int \frac{\cot x}{\sqrt{1+2\sin x}}\,dx$
Let $u \,=\,\sin x\quad\Rightarrow\quad x \,=\,\arcsin u\quad\Rightarrow\quad dx \,=\,\frac{du}{\sqrt{1-u^2}}$
. . and: . $\cot x \,=\,\frac{\sqrt{1-u^2}}{u}$
Substitute: . $\int\frac{\frac{\sqrt{1-u^2}}{u}}{\sqrt{1+2u}}\,\frac{du}{\sqrt{1-u^2}} \;= \;\int\frac{du}{u\sqrt{1+2u}}$
Let $v \,=\,\sqrt{1+2u}\quad\Rightarrow\quad u \,=\,\frac{v^2-1}{2}\quad\Rightarrow\quad du \,=\,v\,dv$
Substitute: . $\int\frac{v\,dv}{\left(\frac{v^2-1}{2}\right)\cdot v} \;=\;2\int\frac{dv}{v^2-1} \;= \;\ln\left|\frac{v-1}{v+1}\right| + C$
Back-substitute: . $\ln\left|\frac{\sqrt{1+2u} - 1}{\sqrt{1+2u} + 1}\right| + C$
Back-substitute: . $\boxed{\ln\left|\frac{\sqrt{1+2\sin x} - 1}{\sqrt{1+2\sin x} + 1}\right| + C}$
Ha! . . . red_dog beat me to it!
4. Mine just take one.
Set
$u=\sqrt{1+2\sin x}\implies du=\frac{\cos x}{\sqrt{1+2\sin x}}\,dx$, the intregral becomes to
$\int\frac{\cot x}{\sqrt{1+2\sin x}}\,dx=2\int\frac1{u^2-1}\,du$
Cheers,
K. | {
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# Some part of a 50% solution of acid was replaced with an
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Some part of a 50% solution of acid was replaced with an [#permalink]
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Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$
M25Q30
Originally posted by chrissy28 on 12 Aug 2008, 14:27.
Last edited by ENGRTOMBA2018 on 19 Aug 2015, 12:12, edited 2 times in total.
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12 Aug 2008, 15:10
chrissy28 wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5 | {
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(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5
Original Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1-x) of the original. So volume of acid left is (1-x)50.
in the new solution.. you added x of 30% solution. i.e 30x acid
(1-x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40%
[(1-x)50 + 30x] / 100 = 40/100
solve to get x=1/2
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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12 Aug 2008, 15:16
The reason that 1/2 is correct is this:
What would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).
The result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.
Here we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.
chrissy28 wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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16 Jun 2013, 16:55
2
2
If 50% solution becomes 40% solution, then 40% solution must be equally weighted between 50% solution and 30% solution - i.e. $$\frac{1}{2}$$ of 50% solution and $$\frac{1}{2}$$ of 30% solution. If the solution must be equally weighted, then 30% solution must replace $$\frac{1}{2}$$ of 50% solution (assuming original solution was 50% solution of acid).
Just another way of saying the same thing. Cheers
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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16 Jun 2013, 23:07
1
3
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$
Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.
Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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17 Jun 2013, 21:57
Hi Bunuel, What is wrong in this equation? Kindly guide me.
Let total solution =100
Replacement = x (say)
equation:
(100-x)*50% + X*30% = 40%
Solving this equation i am getting x= 248
I looked every way possible but could not figure out the problem in the equation....
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Re: Some part of a 50% solution of acid was replaced with an [#permalink] | {
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17 Jun 2013, 23:14
atalpanditgmat wrote:
Hi Bunuel, What is wrong in this equation? Kindly guide me.
Let total solution =100
Replacement = x (say)
equation:
(100-x)*50% + X*30% = 40%
Solving this equation i am getting x= 248
I looked every way possible but could not figure out the problem in the equation....
(100-x)*0.5+0.3x=100*0.4 --> x=50 --> 50/100=1/2.
Hope it's clear.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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18 Jun 2013, 00:03
1
atalpanditgmat wrote:
Hi Bunuel, What is wrong in this equation? Kindly guide me.
Let total solution =100
Replacement = x (say)
equation:
(100-x)*50% + X*30% = 40% Solving this equation i am getting x= 248
I looked every way possible but could not figure out the problem in the equation....
Try using below equation.
50% of (100-x) + 30% of x = 40% of 100
x=50litres.
So 50 of 100 litres got replaced would amount to 1/2 in ratio.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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18 Jun 2013, 00:33
Ahhhh!! I got it....Thank you both of you guys....I hope silly mistakes won't trouble me on the Real Test...
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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10 Jul 2014, 02:09
2
Did using allegation method
Refer diagram below
Bunuel, kindly update the OA.
Attachments | {
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Refer diagram below
Bunuel, kindly update the OA.
Attachments
all.png [ 3.61 KiB | Viewed 5115 times ]
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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20 Jul 2015, 06:11
3
chrissy28 wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$
M25Q30
Responding to a pm:
It is worded to look like a replacement question but essentially, it is a simple mixtures problem only.
You have some 50% solution of acid and you remove some of it. Now to that 50% solution, you are adding 30% solution to get 40% solution. The ratio in which you mix the 50% solution and the 30% solution is given by w1/w2 = (50 - 40)/(40 - 30) = 1/1
Basically, you mix equal parts of 50% solution and 30% solution.
So initially you must have had 2 parts of 50% solution out of which you removed 1 part and replaced with 1 part of 30% solution. So you replaced half of the original solution.
The first question of this post discusses exactly this concept: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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19 Aug 2015, 11:41
Can you please explain to me how you got the answer to this problem using the allegation methos? | {
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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19 Aug 2015, 11:44
tinnyshenoy wrote:
Can you please explain to me how you got the answer to this problem using the allegation methos?
Thanks
Look at: some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html#p1381700
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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19 Aug 2015, 12:02
I cannot understand how they divided 10/20 to get 10? Where did 20 come from?
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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19 Aug 2015, 12:22
tinnyshenoy wrote:
I cannot understand how they divided 10/20 to get 10? Where did 20 come from?
It is from the alligation method
As shown in the post above, per the alligation method, the distrbution will be:
50 10 (=40-30) parts
40
30 10 (=50-40) parts
Thus, you see that you need 10 parts out of 20 (=(10+10)) parts of 50% solution, giving you 10/20 = 1/2 or 50% as your answer.
Hope this helps.
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
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09 Apr 2016, 00:38
chrissy28 wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced? | {
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A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$
M25Q30
w1 is the weight 50% acid solution which has concentration C1 = 50%
w2 is the weight of 20% acid solution which has concentration C1 = 30%
Concentration of mixture (average) = 40%
using weihted average formula w1/w2 = (c2 - avg) / (Avg-c1)
w1/w2 = (30-40)/(40-50)
w1/w2 = 1/1
w1 is 1 part and w2 is 1 part in a total 2 part of solution.
question is What part of the original solution was replaced?
w2 is what was replaced, hence 1 out of 2 i.e. 1/2 part was replaced.
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Example What is the minimum number of terms of the series one needs to be sure to be within of the true sum? However, it holds also in the sense of Riemann integral provided the (k+1)th derivative of f is continuous on the closed interval [a,x]. This will present you with another menu in which you can select the specific page you wish to download pdfs for. Sometimes you'll see something like N comma a to say it's an Nth degree approximation centered at a. Source
The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of Lebesgue integration theory for the full generality. Solution Here are the first few derivatives and the evaluations. Note that, for each j = 0,1,...,k−1, f ( j ) ( a ) = P ( j ) ( a ) {\displaystyle f^{(j)}(a)=P^{(j)}(a)} . Your cache administrator is webmaster. https://www.khanacademy.org/math/calculus-home/series-calc/taylor-series-calc/v/error-or-remainder-of-a-taylor-polynomial-approximation
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Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. These estimates imply that the complex Taylor series T f ( z ) = ∑ k = 0 ∞ f ( k ) ( c ) k ! ( z − In the mean time you can sometimes get the pages to show larger versions of the equations if you flip your phone into landscape mode. In this example we pretend that we only know the following properties of the exponential function: ( ∗ ) e 0 = 1 , d d x e x = e | {
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