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https://mathhelpboards.com/threads/find-a-b-c.8089/ | # find a,b,c
#### Albert
##### Well-known member
$a,b,c \in N$
(1) $1<a<b<c$
(2)$(ab-1)(bc-1)(ca-1) \,\, mod \,\, (abc)=0$
$find :a,b,c$
#### mathworker
##### Active member
Since modulo is "zero" there is no remainder.
$$\displaystyle \frac{(ab-1)(bc-1)(ca-1)}{abc}$$ is not a fraction
$$\displaystyle abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}$$ is not a fraction
So the little terms must sum it up to zero or 1 so,
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0$$(let)
$$\displaystyle \frac{ab+bc+ca-1}{abc}=0$$
$$\displaystyle ab+bc+ca=1$$ which certainly can't be true
If,
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=1$$
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
But as $$\displaystyle a,b,c>1$$ $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ has a maxmum value of 1 so there are no solutions to this one either so I arrived at "there are no such a,b,c"
Did I do something wrong?
#### mathbalarka
##### Well-known member
MHB Math Helper
I am just going to leave this one here : (a, b, c) = (2, 3, 5). (*) Can you see where you went wrong now?
(*) Note that this doesn't mean there are no other. I haven't even revealed half of the solution yet. Keep trying!
#### mathworker
##### Active member
yeah I got it
.....has a maxmum value of 1.....
max value is not 1 $$\displaystyle its \frac{13}{12}$$
- - - Updated - - -
But,I guess (2,3,5) is the only solution....
#### Albert
##### Well-known member
But,I guess (2,3,5) is the only solution....
yes (2,3,5) is the only solution..,but how to get the answer ?
#### mathworker
##### Active member
In,
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
$$\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}$$
$$\displaystyle c=5$$
#### Albert
##### Well-known member
In,
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
$$\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}$$
$$\displaystyle c=5$$
very nice solution | 2020-09-26T22:09:19 | {
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https://math.stackexchange.com/questions/3367836/sum-of-vandermonde-determinant | # Sum of Vandermonde determinant
Given positive integers $$n$$ and $$d$$, where $$d\geq 2$$, I would like to compute the sum $$\displaystyle\sum_{0\leq i_{1} < i_{2} < ... < i_{d}\leq n} \quad\displaystyle\prod_{1 \leq p < q \leq d}\left(i_{q} - i_{p}\right).$$ Since there are $$d\choose 2$$ factors in the product, the sum should return a polynomial in $$n$$ of degree $$d(d+1)/2$$. Ideally I would like to know all coefficients of the polynomial in $$n$$. The leading coefficient (i.e., coefficient of $$n^{d(d+1)/2}$$) is of particular interest.
For example, when $$d=2$$, our sum becomes $$\displaystyle\sum_{i=0}^{n}\displaystyle\sum_{j=i+1}^{n}(j-i) = \frac{1}{6}n(n+1)(n+2)$$, and the leading coefficient (of $$n^3$$) is $$1/6$$.
When $$d=3$$, our sum gives $$\displaystyle\sum_{i=0}^{n}\displaystyle\sum_{j=i+1}^{n}\displaystyle\sum_{k=j+1}^{n}(k-j)(j-i)(k-i) = \frac{1}{180}(n-1)n(n+1)^{2}(n+2)(n+3)$$, and the leading coefficient (of $$n^{6}$$) is $$1/180$$.
For $$d=4$$, WolframAlpha gives $$\displaystyle\sum_{i=0}^{n}\displaystyle\sum_{j=i+1}^{n}\displaystyle\sum_{k=j+1}^{n}\displaystyle\sum_{l=k+1}^{n}(l-k)(l-j)(l-i)(k-j)(k-i)(j-i)\\ = \frac{1}{25200}(n-2)(n-1)n^{2}(n+1)^{2}(n+2)^{2}(n+3)(n+4)$$ and the leading coefficient (of $$n^{10}$$) is $$1/25200$$.
I am not sure if this object is well-known, or has a name. Any references will be great too.
• Extending to $d = 0$ using the empty product gives a sum of $n$, and even with this extension the sequence of ratios $1, \frac{1}{6}, \frac{1}{30}, \frac{1}{140}$ of successive leading coefficients follows the pattern $\frac{n!^2}{(2n+1)!}$. – Travis Willse Sep 27 '19 at 21:43
• And the reciprocals of these ratios, $\frac{(2n+1)!}{n!^2}$, is OEIS A002457 – rogerl Oct 2 '19 at 19:02
• If your primary interest is the leading coefficient, you should be able to replace all the sums by integrals (so that, e.g. the $d=4$ case becomes $$\int_0^1 \int_0^w \int_0^x \int_0^y (w-x)(w-y)(w-z)(x-y)(x-z)(y-z) \, dz \, dy \, dx \, dw=\frac{1}{25200}.$$ Effectively this is the expected product of the pairwise distances of $d$ randomly placed points in the unit interval, and it feels like in this language it should have been studied somewhere (I don't have a reference though). – Kevin P. Costello Oct 2 '19 at 19:25
• @Kevin: I was thinking along similar lines: the Faulhaber's formula gives $\sum_{k=1}^{p}k^{p} = \frac{1}{p+1}n^{p+1} + O(n^{p})$. When we use this in the nested sum, there are many terms contributing to the leading coefficient .... not sure if there is a good way to keep track of them so as to get the leading coefficient as function of $d$. – Abhishek Halder Oct 3 '19 at 0:56
In terms of the leading term coefficient the following identities hold true: $$\begin{eqnarray} c_d&=&\int\limits_{0 \le x_1 \le \cdots x_d \le 1} \prod\limits_{1 \le p < q \le d} (x_p - x_q)\cdot \prod\limits_{p=1}^d dx_p\\ &=&\sum\limits_{\sigma \in \Pi} \mbox{sign(\sigma)} \frac{1}{\prod\limits_{i=1}^d \sum\limits_{j=1}^i \sigma_j} \quad (1)\\ &=& \int\limits_{[0,1]^d} \left(\prod\limits_{p=1}^d x_p^{\binom{p}{2}+p-1} \right) \cdot \prod\limits_{p=1}^d \prod\limits_{q=p+1}^d \left(1-\prod\limits_{\xi=p}^{q-1} x_\xi\right) \cdot \prod\limits_{p=1}^d d x_p \quad (2) \\ &\underbrace{=}_{?}& \prod\limits_{\xi=1}^{d-1} \frac{(\xi!)^2}{(2 \xi+1)!} \end{eqnarray}$$ where in $$(1)$$ we expanded the Vandermonde determinant in a sum over permutations $$\Pi$$ and then integrated term by term and in $$(2)$$ we we used the trick $$1/p = \int\limits_0^1 x^{p-1} dx$$ and the definition of the Vandermonde determinant again. The representations above are readily used to compute the result for $$d \le 9$$. We have:
In[484]:= d =.;
ss = Table[
Total[Signature[#] Product[1/(Total[Take[#, i]]), {i, 1, d}] & /@
Permutations[Range[1, d]]], {d, 1, 9}]
Table[Expand[
Product[x[p]^(Binomial[p, 2] + p - 1), {p, 1,
d}] Product[(1 - Product[x[xi], {xi, p, q - 1}]), {p, 1, d}, {q,
p + 1, d}]] /. x[n_]^p_. :> 1/(p + 1), {d, 1, 9}]
Table[Product[(xi!)^2/(2 xi + 1)!, {xi, 1, d - 1}], {d, 1, 9}]
Update: Let us denote: $$\begin{eqnarray} {\mathcal S}_d^{(n)} := \sum\limits_{0 \le i_1 < i_2 < \cdots < i_d \le n} \prod\limits_{1 \le p < q \le d} (i_q-i_p) \end{eqnarray}$$ for $$n \ge d-1$$. Then my conjecture is the following: $$\begin{eqnarray} {\mathcal S}_d^{(n)} = \left[\prod\limits_{\xi=1}^{d-1} \frac{(\xi!)^2}{(2 \xi+1)!}\right] \cdot \left[\prod\limits_{j=-d+1}^1 (n+j)^{\lceil \frac{j+d-1}{2} \rceil}\right] \cdot (n+2)^{\lfloor \frac{d}{2}\rfloor} \cdot \left[\prod\limits_{j=3}^d (n+j)^{\lceil \frac{d-j+1}{2} \rceil }\right] \end{eqnarray}$$
I have verified this conjecture for $$d \le 6$$ using the code below:
d = 2; Clear[a]; Clear[aa]; i[0] = 0;
aa = Table[a[p], {p, 0, d - 1}];
smnD = Product[i[q] - i[p], {p, 1, d}, {q, p + 1, d}];
subst = First@
Solve[CoefficientList[
smnD - (Sum[Binomial[i[d] - i[d - 1], p] a[p], {p, 0, d - 1}]),
i[d]] == 0, aa] // Simplify;
(*Sum over i[d] done.*)
S = Sum[Binomial[n + 1 - i[d - 1], p + 1] a[p], {p, 0, d - 1}] /.
subst;
upLim = d - 1;
For[count = 1, count <= d - 1, count++,
upLim = upLim + d - count;
aa = Table[a[p], {p, 0, upLim}];
subst =
First@Solve[
CoefficientList[
FunctionExpand@(S - (Sum[
Binomial[i[d - count] - i[d - count - 1], p] a[p], {p, 0,
upLim}])), i[d - count]] == 0, aa] // Simplify;
(*Sum over i[d-count] done.*)
S = Sum[
Binomial[n + 1 - i[d - count - 1], p + 1] a[p], {p, 0,
upLim}] /. subst;
Print["count=", count, "done"];
];
Factor[FunctionExpand@S]
• I think the coefficient of $\mathrm{sign}(\sigma)$, i.e. $\int \prod x_j^{\sigma_j-1} dx$, should be $(\prod_{j=1}^n \sum_{i=1}^j \sigma_i)^{-1}$, not $(\sum \sigma_j)^{-1}$, right? – user125932 Oct 4 '19 at 16:19
• Yes, yes, of course, I am sorry for that. I fixed it. – Przemo Oct 4 '19 at 17:07
• This looks nice, but my impression is that the polynomial factors completely into linear terms, and that probably cannot be proven by an integral approximation, can it? – darij grinberg Oct 5 '19 at 2:58
• @darij grinberg It is indeed quite remarkable that this factorizes so nicely. I have updated my answer and provided a Mathematica code that computes the sum for higher values of $d$. By the way , why are you interested in this very problem ? – Przemo Oct 7 '19 at 11:54
• I care partly because the sum is the sum of all maximal minors of the Vandermonde-like rectangular matrix $\left(i^{j-1}\right)_{1\leq i\leq n,\ 1\leq j\leq d}$, and it appears that there may be a more general determinant identity behind it. Also, the sum can be reinterpreted as counting something like Gelfand-Tsetlin patterns, since each $i_q - i_p$ factor can be viewed as choosing an integer from the interval $\left[i_p,i_q\right)$. These are not quite Gelfand-Tsetlin patterns, because all these integers in a given addend of the sum are independent, but close enough to be interesting to me. – darij grinberg Oct 7 '19 at 13:18
This is a follow up on Przemo's answer in deriving the final expression of the leading coefficient $$c_{d}$$. Notice that $$d! c_{d}$$ equals the Vandermonde determinant integrated over $$[0,1]^{d}$$. In an 1955 paper by De Bruijn (see toward the end of Sec. 9), it is proved that
$$\int_{[0,1]^{d}}\prod_{1\leq i < j \leq d} |x_{i} - x_{j}| \: {\rm{d}}x_{1} ... {\rm{d}}x_{d} = \frac{\{1! \times 2! \times 3! \times ... \times (d-1)!\}^{2} d!}{1!\times 3! \times 5! \times ... \times (2d-1)!}.$$
That proof utilizes the result (also derived in that paper) that integrals of this type are equal to certain Pfaffian form. Equating the above with $$d! c_{d}$$ recovers the expression conjectured by Przemo:
$$c_{d} = \prod\limits_{\xi=1}^{d-1} \frac{(\xi!)^2}{(2 \xi+1)!}.$$ | 2021-02-27T16:08:30 | {
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https://math.stackexchange.com/questions/3543002/prove-that-a-set-is-a-subset-of-another | # Prove that a set is a subset of another
Prove that a set is a subset of another, using predicates and (if needed) quantifiers:
(A $$\cap$$ C) $$\cup$$ (B $$\cap$$ D) $$\subseteq$$ (A $$\cup$$ B) $$\cap$$(C $$\cup$$ D)
Should I start with the whole statement, and rewrite it using predicates and logic until a tautology comes out or am I supposed to somehow come from the left side of the $$\subseteq$$ to the right side?
I tried the first option and only ended up going in circles and the second one I’m not sure how it’s supposed to look like.
Thanks!
• No, do not start with the statement to be proved and derive a tautology. That is considered very sloppy mathematics. It can also lead to errors if your steps aren't reversible. Feb 11 '20 at 18:44
Let $$x \in (A \cap C) \cup ( B \cap D)$$. Then $$x \in A \cap C$$ or $$x \in B \cap D$$.
Suppose $$x \in A \cap C$$, then $$x \in A$$ and $$x \in C$$. Therefore $$x \in A \cup B$$ and $$x \in C \cup D$$, i.e. $$x \in (A \cup B) \cap (C \cup D)$$.
Suppose $$x \in B \cap D$$, then $$x \in B$$ and $$x \in D$$. Therefore $$x \in A \cup B$$ and $$x \in C \cup D$$, i.e. $$x \in (A \cup B) \cap (C \cup D)$$.
It follows that $$(A \cap C) \cup ( B \cap D) \subseteq(A \cup B) \cap (C \cup D)$$.
• So I can understand the explanation, after drawings a Venn diagram for it. So basicaly as a predicate it would look like: x $\in$ A and x $\in$ C $\rightarrow$ $x \in (A \cup B) \cap (C \cup D)$ . But how could I say using logic that the statment is true.
– Lino
Feb 12 '20 at 13:09
• Using logic you can see the answer from @Siddharth Bhat. However, I think that solve your problem using the definition of union and intersection is the easiest way Feb 12 '20 at 13:21
You can write the predicate for the left hand side and the right hand side, then attempt to show one implies the other. The answers are in a spoiler tag so you can try them out yourself :)
• find some predicate $$P$$ which models $$x \in (A \cap C) \cup (B \cap D)$$
$$P \equiv (x \in A \land x \in C) \lor (x \in B \land x \in D)$$
• Now find a predicate $$Q$$ for $$x \in (A \cup B)\cap (C \cap D)$$
$$P \equiv (x \in A \lor x \in B) \land (x \in C \lor x \in D)$$
• For subsets relations $$L \subseteq R$$, the predicate is: $$\forall x \in L \implies x \in R$$. That is, prove that $$P \implies Q$$.
To show that $$P \implies Q$$:
the rule that will come in handy is the distributivity of $$(\lor)$$ over $$(\land)$$: $$(a \land b) \lor (c \land d) = (a \lor c) \land (b \lor d)$$.
• I don’t see how the 2 are equivalent: $(a \land b) \lor (c \land d) = (a \lor c) \land (b \lor d)$ , after trying to come from one side to other I just go in circles.
– Lino
Feb 12 '20 at 13:28
Use the definition of a subset.
$$B$$ is a subset of $$A$$ if for all $$x\in B$$, $$x\in A.$$
Now let $$x\in (A\cap C)\cup(B\cap D)$$. Then either $$x\in (A\cap C)$$ or $$x\in (B\cap D)$$
Case I: $$x\in (A\cap C)$$
Then $$x\in A$$ and $$x\in C$$. Thus $$x\in (A\cup B)$$ and $$x\in (C\cup D)$$. Thus $$x\in (A\cup B) \cap (C\cup D)$$.
Case II: $$x\in (B\cap D)$$
Then $$x\in B$$ and $$x\in D$$. Thus $$x\in (A\cup B)$$ and $$x\in (C\cup D)$$. Thus $$x\in (A\cup B) \cap (C\cup D)$$.
So in either case $$x\in (A\cup B) \cap (C\cup D)$$. Therefore $$(A\cap C)\cup(B\cap D) \subseteq (A\cup B) \cap (C\cup D)$$
Below an almost completed proof
• It would be better to write out the proof using MathJax instead of including a picture. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. Feb 11 '20 at 19:24 | 2021-09-18T23:41:24 | {
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https://www.physicsforums.com/threads/open-sets-and-closures.735099/ | # Open sets and closures
1. Jan 27, 2014
### Shaji D R
Suppose A and B are open sets in a topological Hausdorff space X.Suppose A intersection B is an empty set. Can we prove that A intersection with closure of B is also empty? Is "Hausdorff" condition necessary for that?
2. Jan 27, 2014
### pasmith
Given that $A$ and $B$ are disjoint, the only way $A$ can intersect with the closure of $B$ is if there exists $a \in A$ such that $a$ is a limit point of $B$.
But that's impossible: $A$ is an open neighbourhood of $a$ which contains no points in $B$. Hence $a$ is not a limit point of $B$.
This holds whether or not $X$ is Hausdorff.
3. Jan 27, 2014
### economicsnerd
Rephrasing pasmith's argument without reference to individual points...
Let $X$ be any topological space, and suppose $A,B\subseteq X$ are disjoint and $A$ is open. Then:
-$X\setminus A$ is closed because $A$ is open.
-$X\setminus A \supseteq B$ because $A,B$ are disjoint.
-As the closure of $B$, the set $\bar B$ is the smallest closed set that contains $B$.
-In particular, $\bar B \subseteq X \setminus A$.
Rephrasing the last point, $A,\bar B$ are disjoint sets.
4. Jan 28, 2014
### Shaji D R
Thank you very much
5. Feb 4, 2014
### ibdsm
A \cap B empty
If A and B are disjoint, the B is a subset of the complement of A.
If A is open, its complement is closed.
Hence, in this case, the closure of B is contained in the complement of A.
Hence, A and the closure of B are disjoint.
There is no need for the ambient space to be Hausdorff. There is no need for B to be ooen.
6. Feb 5, 2014
### JG89
The space has to be Hausdorff and B has to be open. A = (0,1) and B = [1,0] is a counter example to what you say
7. Feb 5, 2014
### ibdsm
1. [1,0] is the empty set, as there are no real numbers which are both at least 1 and at most 0. Hence your "counter-example" fails.
2. By definition,
(i) a subset of a topological space is closed if and only if it is the complement of an open set
(ii) the closure of a subset, B, of a topological space is the smallest closed subset of the space which contains B.
There is no question of being a Hausdorff space or even a T1 or T0 space for this.
8. Feb 5, 2014
### JG89
Wow, I can't believe I wrote that. I was dead tired and my counter-example I was trying to type out was A = (0,1), B = [1,2]. But even that fails.
My apologies, I'll try not to post when I'm half awake anymore lol.
9. Feb 5, 2014
### ibdsm
I'm glad I'm not the only one to blunder at times! | 2017-12-18T13:07:00 | {
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https://math.stackexchange.com/questions/2884094/trace-of-product-of-two-hermitian-matrices | # Trace of product of two Hermitian matrices
Let $A$ and $B$ be two Hermitian complex matrices. (a) Prove that $\operatorname{tr}(AB)$ is real. (b) Prove that if $A, B$ are positive, then $\operatorname{tr}(AB)>0$.
(a) The trace of Hermitian matrix is a real number, since $a_{ii} = \bar{a}_{ii}$, that also means that all eigenvalues are real. I can't proceed to conclusion that $\operatorname{tr}(AB)$ is real, since $\operatorname{tr}(AB) \neq \operatorname{tr}A\cdot\operatorname{tr}B$ and product of two Hermitian matrices is also Hermitian only if these matrices commute, which is not the case for arbitrary Hermitian matrices.
(b) Am I missing something or the question is indeed so easy? If all entries $a_{ij}>0$ and $b_{ij}>0$, then all $c_{ij}=\sum_{m}a_{im}\cdot b_{mj}>0$ too, finally $\operatorname{tr}(AB)=\sum_{i}c_{ii}>0$.
• Don't they mean positive definite when they just say positive? – Nik Pronko Aug 15 '18 at 20:57
• @NikPronko, maybe there is a typo and there should be positive definite, but the task states just positive. – Hasek Aug 15 '18 at 20:59
• in context of complex matrices, positive usually means positive definite... – user251257 Aug 15 '18 at 21:00
• @user251257, it makes sense since we can't compare complex numbers with zero. I didn't pay attention to this, thanks! – Hasek Aug 15 '18 at 21:02
• As part (b) isn't actually addressed to, a hint: As $A$ is positive definite, it has a positive definite square root, say $R$. That is $A = R^2$. Now, $AB=R(RB)RR^{-1}$ is similar to $RBR$, which is positive definite. – user251257 Aug 15 '18 at 21:13
Since $A^T = \overline{A}$ and using basic properties of trace :
$$tr(AB) = tr((AB)^T) = tr(B^T A^T) = tr(\overline{B} \ \overline{A})= tr(\overline{A} \ \overline{B})$$
$$tr(AB) = tr(\overline{A} \ \overline{B}) = tr(\overline{AB}) = \overline{tr(AB)}$$
Finally $tr(AB) \in \mathbb R$.
You can't just say that because the trace is over the diagonal elements of $(AB)_{ii}$. Instead $\lambda = tr(AB)=\sum A_{ij}B_{ji}$ then $\lambda^* = tr(A^*B^*)=tr(A^TB^T)=\sum A^T_{ij}B^T_{ji} = \sum A_{ji}B_{ji}$ which is clearly $tr(AB)$ under renaming dummy indicies, so $\lambda=\lambda^*$ | 2019-12-13T00:20:18 | {
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https://www.physicsforums.com/threads/find-p-and-c-such-that-a-pcp-1.747142/ | Homework Help: Find P and C such that $A=PCP^{-1}$
1. Apr 5, 2014
Zondrina
1. The problem statement, all variables and given/known data
Find an invertible matrix $P$ and a matrix $C$ of the form $\left( \begin{array}{cc} a & -b \\ b & a \end{array} \right)$ such that $A=PCP^{-1}$ when $A = \left( \begin{array}{cc} 1 & -2 \\ 1 & 3 \end{array} \right)$.
2. Relevant equations
Eigenvalues for A:
$\lambda_1 = 2+i$
$\lambda_2 = 2-i$
Eigenvectors for eigenvalues:
$v_1 = [-1 + i, 1]^t$
$Re(v_1) = [-1, 1]^t$
$Im(v_1) = [1, 0]^t$
$v_2 = [-1 - i, 1]^t$
$Re(v_2) = [-1, 1]^t$
$Im(v_2) = [-1, 0]^t$
A theorem:
Let $A$ be a real 2x2 matrix with a complex eigenvalue $\lambda = a - bi, (b≠0)$ and an associated eigenvector $v$.
Then $A=PCP^{-1}$ where $P = [ Re(v), Im(v) ]$ and $C = \left( \begin{array}{cc} a & -b \\ b & a \end{array} \right)$
3. The attempt at a solution
I've basically solved this, but I had a question.
Does the theorem state that I only care about the eigenvalue $a - bi$? In this case it would be $\lambda_2 = 2-i$. If that's the case, $\lambda_1$ is useless in finding the required matrices.
Granted that $\lambda_1$ and $v_1$ have nothing to do with the problem, the matrices required are:
$C = \left( \begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array} \right)$
$P = \left( \begin{array}{cc} -1 & -1 \\ 1 & 0 \end{array} \right)$
$P^{-1} = \left( \begin{array}{cc} 0 & 1 \\ -1 & -1 \end{array} \right)$
Where I have plugged the magnitudes of $a$ and $b$ into $C$.
My real curiosity lies in why I only care about the eigenvalue $a - bi$?
2. Apr 5, 2014
Staff: Mentor
Complex eigenvalues come in pairs, so if you know one of them, the other is uniquely determined. Also notice that the first components of your eigenvectors are conjugates of one another. Notice also that the real parts of your two eigenvectors are the same, and the imaginary parts are scalar multiples of one another.
3. Apr 5, 2014
Zondrina
This much I have already seen. I was more curious as to why the eigenvalue $a+bi$ is being neglected in the theorem. Is it solely due to the definition of $C$?
4. Apr 5, 2014
Staff: Mentor
I'll bet the theorem could be rewritten so that it uses a + bi instead of a - bi.
5. Apr 5, 2014
Zondrina
Indeed, I conjecture that:
Let $A$ be a real 2x2 matrix with a complex eigenvalue $\lambda = a + bi, (b≠0)$ and an associated eigenvector $v$.
Then $A=PCP^{-1}$ where $P = [ Re(v), Im(v) ]$ and $C = \left( \begin{array}{cc} a & b \\ -b & a \end{array} \right)$
I have verified the application of this theorem. I'll go prove it in a more general setting now. | 2018-07-19T10:10:20 | {
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https://math.stackexchange.com/questions/90290/importance-of-least-upper-bound-property-of-mathbbr | # Importance of Least Upper Bound Property of $\mathbb{R}$
My professor asserts that the Least Upper Bound Property of $\mathbb{R}$ (Completeness Axiom) is the most essential piece in the study of real analysis. He says that almost every theorem in calculus/analysis relies directly upon on this Property.
I know that the Archimedian property of $\mathbb{R}$ directly uses the property for the proof, but I'm trying to think of other major theorems that use the property directly. Do you think he means consequences of the property? Because then the gates are wide open...
Does anyone know of any other theorems that use the properly directly?
Sorry this is more of a general question
• The conclusion of Bolzano's theorem (intermediate value theorem) would be false if $\mathbb{R}$ weren't complete. Now think of all the times you say "since $f$ is a continuous function, negative here and then positive, it must be zero somewhere in the middle"! – Bruno Stonek Dec 10 '11 at 22:08
• @pigishpig: You may want to hold off before accepting an answer for several reasons: (i) You are more likely to attract answers if the question does not have an accepted answer than if it does; I think the particular kind of question you are asking is also one that can attract a lot of good answers that don't overlap. (ii) It's always a good idea to let a question lie for at least a couple of hours before accepting an answer, unless it's something that has a definitive, clear, singular answer. Note you can only accept one question, so you may want to think a bit on which one you'll accept here – Arturo Magidin Dec 10 '11 at 22:46
• @ArturoMagidin thank you for the advice. I'm somewhat new. – pigishpig Dec 10 '11 at 22:48
One way to spot the importance of the Least Upper Bound property is to think about results that work in $\mathbb{R}$ but fail in $\mathbb{Q}$; this, because $\mathbb{Q}$ still has a lot of the algebraic properties of $\mathbb{R}$ (including the Archimedean property), but does not satisfy the Least Upper Bound property. So results that are true in $\mathbb{R}$ but are false in $\mathbb{Q}$ are good candidates for needing the Least Upper Bound property.
Some examples:
1. The Intermediate Value Theorem uses the Least Upper Bound Property of $\mathbb{R}$ (in fact, the IVT is equivalent to the Least Upper Bound Property; see below).
2. The fact that Cauchy sequences converge in $\mathbb{R}$ depends on the Least Upper Bound Property; without it, you can have sequences that are Cauchy but do not converge (as you do with $\mathbb{Q}$. That Cauchy sequences converge is very important in, for example, the definition of integration as limits of Riemann sums.
3. Bounded sequences have convergent subsequences (the Bolzano-Weierstrass Theorem); the proof in $\mathbb{R}$ rests on showing that such a sequence has a monotone subsequence, and the proof that monotone bounded sequences converge uses the Least Upper Bound Property directly.
The equivalence of the IVT and the Least Upper Bound Property:
IVT. Let $f\colon [a,b]\to\mathbb{R}$ be a continuous function. If $f(a)\lt 0$ and $f(b)\gt 0$, then there exists $c\in(a,b)$ such that $f(c)=0$.
Least Upper Bound Property. Let $S\subseteq \mathbb{R}$ be a nonempty set that is bounded above. Then $S$ has a least upper bound bound in $\mathbb{R}$.
Theorem. IVT is equivalent to the Least Upper Bound Property.
Proof. Assume the Least Upper Bound Property, and let $f\colon [a,b]\to\mathbb{R}$ be continuous. let $S=\{x\in [a,b]\mid f(x)\lt 0\}$. Then $S$ is bounded above by $b$, and is nonempty, since $a\in S$. Let $c$ be the least upper bound for $S$. Note that $a\lt c\lt b$.
If $f(c)\neq 0$, then there exists $\delta\gt 0$ such that for all $x$, $|x-c|\lt\delta$, $|f(x)-f(c)|\lt \frac{|f(c)|}{2}$. In particular, $f(x)$ and $f(c)$ have the same sign.
But: if $f(c)\gt 0$, then this means that for all $\delta\gt 0$ there exists $s\in S$ such that $c-\delta \lt s\leq c$, so $f(s)\lt 0$, hence $|f(s)-f(c)| = f(c)+|f(s)|\gt \frac{f(c)}{2}$, a contradiction. In particular, $c\lt b$.
And if $f(c)\lt 0$, then there exists $x\in [a,b]$ with $x\lt c+\delta$ (since $c$ cannot be $b$), and then $f(x)\lt 0$, so $x\in S$, contradicting the fact that $c$ is an upper bound for $S$.
Therefore, we cannot have $f(c)\neq 0$, so $f(c)=0$, with $a\lt c\lt b$, and we are done.
(The following is an argument suggested to me some years ago by Lee Rudolph on sci.math) Conversely, assume that we have the Intermediate Value Theorem. Let $S$ be a nonempty subset of $\mathbb{R}$ that is bounded above; let $Y$ be the set of upper bounds of $S$, and define $f$ as follows: $$f(x) = \left\{\begin{array}{ll} 1 & x\in Y\\ -1 & x\notin Y \end{array}\right.$$ Note that there are points where $f$ takes value $-1$, and points where $f$ takes value $1$, but there are no point where $f$ takes the value $0$. That means, by the Intermediate Value Theorem, that $f$ is not continuous. Therefore, there is a point $z$ where $f$ is not continuous.
If there exists $s\in S$ such that $z\lt s$, then letting $\delta=s-z\gt 0$ we have that $(z-\delta,z+\delta)\cap Y = \emptyset$, and therefore $f$ is constant on this interval, hence continuous. So no such $s$ can exist, and therefore $z$ is an upper bound for $S$.
If there exists $y\in Y$ such that $y\lt z$, then letting $\delta=z-y\gt 0$ we have that $(z-\delta,z+\delta)\subseteq Y$, and so $f$ is constant on this interval, hence continuous. This contradiction tells us that no such $y$ can exist either.
So $z$ is in $Y$ and is a lower bound for $Y$, hence $z$ is the least element of $Y$. Thus, $z$ is the least upper bound of $S$, and $S$ has a least upper bound, as desired. QED
The importance is not so much because of the fact that the least upper bound property itself can be used directly to prove major thoerems. As Gerry pointed out, the important point is that it says that the field of real numbers is complete, which is how it is distinguished from the field of rational numbers, for example. Consequences of completeness, however it is formulated, are what is behind all of the analysis that would be impossible with $\mathbb Q$. Compactness of closed and bounded intervals, connectedness of intervals, convergence of bounded monotone sequences, convergence of Cauchy sequences, nonemptiness of intersections of nested sequences of closed bounded intervals, and existence of limit points of bounded infinite subsets, are some of the important properties in analysis that $\mathbb R$ has and $\mathbb Q$ doesn't, and all are consequences of some formulation of completeness. (Contrast this with the Archimedean property, which also holds for $\mathbb Q$.)
One could take other axioms as a starting point for analysis. For example, completeness for metric spaces is formulated in terms of Cauchy sequences, mentioned in Arturo's and Thomas's answers, and that can also be done for $\mathbb R$. There is one difference, however, which is that convergence of Cauchy sequences is not enough to guarantee the Archimedean property, so the latter would have to be taken as an additional axiom. Arturo's answer shows that the intermediate value property for continuous functions on intervals could also be taken as the defining axiom. I have seen at least one analysis textbook, Körner's excellent A companion to analysis, that makes convergence of bounded monotone sequences the defining axiom of the real numbers instead of the least upper bound property; the equivalence of the two is the topic of a recent question here.
I think that just the fact that we define almost everything in analysis in terms of limits in enough to see that this is fundamental ; since one way to define limits is to say that the inferior limit is equal to the superior limit, we need those somehow implicitly to speak of limits, and thus need the completeness axiom.
But more explicitly, you can think about the theorem that states that monotone increasing bounded above sequences converges : what you do is consider $$\alpha = \sup_{n \in \mathbb N} \{ x_n \}$$ which exists by the completeness axiom. Then you use $\delta$'s and $\varepsilon$'s to show that $\alpha$ is actually the limit.
Another example would be the definition of $n^{\text{th}}$ roots in analysis, which are defined in terms of supremums. Basically everything you know that is defined in terms of supremums uses that axiom implicity.
Hope that helps,
Just imagine trying to do analysis/calculus if all you had was the rational numbers, and you'll see how important completeness is.
The fact that every bounded non-decreasing sequence has a limit is a direct result of this theorem.
The fact that every bounded sequence has a convergent subsequence is a relatively direct consequence.
From there, you can get that any Cauchy sequence is convergent.
There are two essential ways to define the reals from the rationals.
The first is to define the reals as Cauchy sequences of rational numbers upto some equivalence relation. This definition assures directly the completeness of the reals.
The second uses something called Dedekind cuts on the rationals. This technique assures directly the Least Upper Bound property of the reals.
Both definitions are equivalent, so it is a matter of emotion to say which property is more primary. I suppose that the least upper bound property is easier to state in predicate logic as an axiom of the reals, so, from a logician's point of view, it might be preferable. | 2021-02-28T06:20:48 | {
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https://math.stackexchange.com/questions/1645371/is-it-true-that-mathbbq-sqrt2-cap-mathbbqi-mathbbq | # Is it true that $\mathbb{Q}(\sqrt{2}) \cap \mathbb{Q}(i) = \mathbb{Q}$?
Is it true that $\mathbb{Q}(\sqrt{2}) \cap \mathbb{Q}(i) = \mathbb{Q}$?
I know that \begin{align*} \mathbb{Q}(\sqrt{2}) &= \{a+b\sqrt{2} \mid a,b \in \mathbb{Q}\}, \\ \mathbb{Q}(i) &= \{a+bi \mid a,b \in \mathbb{Q}\} \end{align*}
I tend to believe it is, since every element in $\mathbb{Q}(\sqrt{2})$ belongs to $\mathbb{Q}(i)$ iff $b=0$. Also, an element in $\mathbb{Q}(i)$ belongs to $\mathbb{Q}(\sqrt{2})$ iff $b = 0$.
Is that enough for a formal proof?
• In general it is risky to ask questions about intersections, because it is heavily dependent on the meaning. For some rings, list $\mathbb Q[\sqrt[3]{2}]$, there ar emultiple embeddings into $\mathbb C$, so whether the rings intersects another field extension of $\mathbb Q$ is ambiguous. For quadratic extensions such as these, though, the image of the embeddings are unique (even if the embedding functions are not.) Feb 8 '16 at 1:08
• Where did this question come from? Homework! Feb 8 '16 at 1:08
• you know $Q \subset Q(√2)$ and $Q\subset Q(i)$ then $Q\subset Q(√2)\cap Q(i)$ then to see $=$ you have to show "give $a,b\in Q$ $(b\noteq 0)$ that $a+b√2 \notin Q(i)$ and so... The answer it is true
– Elll
Feb 8 '16 at 1:09
• @MhenniBenghorbal I found this question in an old exam while studying for my exam (Linear Algebra). Feb 8 '16 at 1:13
Alternatively, show that $$\Bbb Q\subseteq\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr),$$ which I leave to you. Then, suppose $z\in\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr).$ Since $z\in\Bbb Q\bigl(\sqrt2\bigr),$ then $z\in\Bbb R.$ From there, we can use the fact that $z\in\Bbb Q(i)$ to readily show that $z\in\Bbb Q,$ completing the proof.
yes it is true $Q\sqrt2$ and $Q(i)$ are extension of degree $2$ of $Q$ so the degree of $Q(i)\cap Q(\sqrt2)$ is either 1 or $2$, if it is 2 it implies that $Q(i)=Q(\sqrt2)$ thus $\sqrt2=a+bi, a,b\in Q$, by writing $(\sqrt2-a)^2=-b^2$, you obtain $a=0$ unless $\sqrt2\in Q$ a fact which is not true. If $a=0$, $\sqrt2=bi$ thus the $2=-b^2$ this not true also. | 2022-01-16T20:14:30 | {
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http://stacks.math.columbia.edu/tag/01CV | # The Stacks Project
## Tag: 01CV
This tag has label modules-definition-gamma-star and it points to
The corresponding content:
Definition 16.21.4. Let $(X, \mathcal{O}_X)$ be a ringed space. Assume that all stalks $\mathcal{O}_{X, x}$ are local rings. Given an invertible sheaf $\mathcal{L}$ on $X$ we define the associated graded ring to be $$\Gamma_*(X, \mathcal{L}) = \bigoplus\nolimits_{n \geq 0} \Gamma(X, \mathcal{L}^{\otimes n})$$ Given a sheaf of $\mathcal{O}_X$-modules $\mathcal{F}$ we set $$\Gamma_*(X, \mathcal{L}, \mathcal{F}) = \bigoplus\nolimits_{n \in \mathbf{Z}} \Gamma(X, \mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes n})$$ which we think of as a graded $\Gamma_*(X, \mathcal{L})$-module.
\begin{definition}
\label{definition-gamma-star}
Let $(X, \mathcal{O}_X)$ be a ringed space.
Assume that all stalks $\mathcal{O}_{X, x}$ are local rings.
Given an invertible sheaf $\mathcal{L}$ on $X$ we define
the {\it associated graded ring} to be
$$\Gamma_*(X, \mathcal{L}) = \bigoplus\nolimits_{n \geq 0} \Gamma(X, \mathcal{L}^{\otimes n})$$
Given a sheaf of $\mathcal{O}_X$-modules $\mathcal{F}$ we set
$$\Gamma_*(X, \mathcal{L}, \mathcal{F}) = \bigoplus\nolimits_{n \in \mathbf{Z}} \Gamma(X, \mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes n})$$
which we think of as a graded $\Gamma_*(X, \mathcal{L})$-module.
\end{definition}
To cite this tag (see How to reference tags), use:
\cite[\href{http://stacks.math.columbia.edu/tag/01CV}{Tag 01CV}]{stacks-project}
## Comments (4)
Comment #11 by Pieter Belmans on July 22, 2012 at 9:17 am UTC
In the notes following this definition (there's no associated tag) there is the typo "allthough".
Comment #12 by Pieter Belmans on July 22, 2012 at 9:21 am UTC
I am not sure about your stance on using _* versus _\bullet, but given that complexes of cohomology in tag 07AS are written in the latter and less ambiguous way, it might be beneficial to do this here (and elsewhere) too?
Comment #13 by Pieter Belmans on July 22, 2012 at 9:23 am UTC
Missed this earlier on: there is the typo "compatibilties" in the last sentence of the remarks following this definition.
Comment #20 by Johan on July 22, 2012 at 10:58 pm UTC
@#11 and #13. Fixed. Thanks. @#12: Yes, hmm, I think of the notation $\Gamma_*$ as standard for the direct sum of global sections of the powers of an invertible sheaf. This isn't a complex!
## Add a comment on tag 01CV
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In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).
In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this is tag 0321 you just have to write 0321. This captcha seems more appropriate than the usual illegible gibberish, right? | 2013-05-25T18:21:28 | {
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https://math.stackexchange.com/questions/3024896/seeking-methods-to-solve-i-int-0-frac-pi2-frac-arctan-left-sin/3024913 | # Seeking Methods to solve $I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$
I was wondering what methods people knew of to solve the following definite integral? I have found a method using Feynman's Trick (see below) but am curious as to whether there are other Feynman's Tricks and/or Methods that can be used to solve it:
$$I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$$
My method:
Let
$$I(t) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(t\sin(x)\right)}{\sin(x)}\:dx$$
Thus,
\begin{align} I'(t) &= \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\left(t^2\sin^2(x) + 1\right)\sin(x)}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{t^2\sin^2(x) + 1}\:dx \\ &= \left[\frac{1}{\sqrt{t^2 + 1}} \arctan\left(\sqrt{t^2 + 1}\tan(x) \right)\right]_{0}^{\frac{\pi}{2}} = \sqrt{t^2 + 1}\frac{\pi}{2} \end{align}
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t) + C$$
Now
$$I(0) = C = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(0\cdot\sin(x)\right)}{\sin(x)}\:dx = 0$$
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t)$$
And finally,
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$
• Accepting an answer will take your question off of the "unanswered questions" page, and you probably don't want to do that until you've seen a wider variety of answers. – Franklin Pezzuti Dyer Dec 4 '18 at 0:09
• @Frpzzd - Thanks again for your info. I have removed the "answered" status for the moment. – user150203 Dec 4 '18 at 0:17
• I know I am a bit late in the game, but I just found an answer which I'm pretty proud of :) – clathratus Jan 12 '19 at 23:22
\begin{align} \int_0^{\pi/2}\frac{\arctan \sin(x)}{\sin(x)}dx &=\int_0^{\pi/2}\frac{1}{\sin(x)}\sum_{n=0}^\infty \frac{(-1)^n \sin^{2n+1}(x)}{2n+1}dx\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^{\pi/2}\sin^{2n}(x)dx\\ &=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2n+1}\cdot \frac{(2n-1)!!}{(2n)!!}\\ &=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2^{2n-1}(2n+1)}\cdot \binom{2n-1}{n} \\ &=\frac{\pi}{2}+\frac{\pi}{2}\cdot (\sinh^{-1}(1)-1) \\ &=\frac{\pi}{2}\ln(1+\sqrt{2}) \\ \end{align}
Using the following relation: $$\frac{\arctan x}{x}=\int_0^1 \frac{dy}{1+(xy)^2} \Rightarrow \color{red}{\frac{\arctan(\sin x)}{\sin x}=\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}$$ We can rewrite the original integral as: $$I = \color{blue}{\int_{0}^{\frac{\pi}{2}}} \color{red}{\frac{\arctan\left(\sin x\right)}{\sin x}}\color{blue}{dx}=\color{blue}{\int_0^\frac{\pi}{2}}\color{red}{\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}\color{blue}{dx}=\color{red}{\int_0^1} \color{blue}{\int_0^\frac{\pi}{2}}\color{purple}{\frac{1}{1+(\sin^2 x )y^2}}\color{blue}{dx}\color{red}{dy}$$ $$=\int_0^1 \left(\frac{\arctan\left(\sqrt{1+y^2}\cdot\tan(x)\right) }{\sqrt{1+y^2}} \bigg|_0^\frac{\pi}{2}\right) dy=\frac{\pi}{2}\int_0^1 \frac{dy}{\sqrt{1+y^2}}=\frac{\pi}{2}\ln\left(1+\sqrt 2\right)$$
• Absolutely love this method. With respect to your method of taking the integrand and converting to an integral - do you know of any texts/examples that explain the process in detail? Thanks again for the post. – user150203 Dec 5 '18 at 11:09
• @DavidG More or less, it's similar to "Feynmann Trick". We learn many tricks "along the way". For example, in MSE. – Felix Marin Dec 5 '18 at 16:13
$$I = \int_{0}^{1}\frac{\arctan x}{x\sqrt{1-x^2}}\,dx =\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1}\frac{x^{2n}}{\sqrt{1-x^2}}\,dx=\frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}\cdot\frac{\binom{2n}{n}}{4^n}$$ is a fairly simple hypergeometric series, namely $$\frac{\pi}{2}\cdot\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2};-1\right)$$. Since $$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n,\qquad \arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n} x^{2n+1}$$ we clearly have $$I=\frac{\pi}{2}\,\text{arcsin} \color{red}{\text{h}}(1) = \color{red}{\frac{\pi}{2}\log(1+\sqrt{2})}$$.
By enforcing the substitution $$x\mapsto\frac{1-x}{1+x}$$ (involution) and exploiting the Maclaurin series of $$\frac{1}{x}\left(\frac{\pi}{4}-\arctan(1-x)\right)$$ I got the mildly interesting acceleration formula
$$\frac{\pi}{2}\log(1+\sqrt{2})=\small{\sum_{k\geq 0}(-1)^k\left[\frac{2^{6k}}{(4k+1)(8k+1)\binom{8k}{4k}}+\frac{2^{6k+2}}{(4k+2)(8k+3)\binom{8k+2}{4k+1}}+\frac{2^{6k+3}}{(4k+3)(8k+5)\binom{8k+4}{4k+2}}\right]}.$$ In this case we have that a $$\phantom{}_2 F_1(\ldots,-1)$$ decomposes as a linear combination of three $$\phantom{}_6 F_5(\ldots,-1/4)$$.
Slightly different from @Frpzzd's answer $$I=\int_0^{\pi/2}\frac{\arctan\sin x}{\sin x}\mathrm dx$$ Recall that $$\arctan x=\sum_{n\geq0}(-1)^n\frac{x^{2n+1}}{2n+1},\qquad |x|\leq1$$ And since $$\forall x\in\Bbb R ,\ \ |\sin x|\leq1$$, we have that $$\arctan\sin x=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\sin(x)^{2n+1},\qquad \forall x\in\Bbb R$$ So we have that $$I=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\int_0^{\pi/2}\sin(x)^{2n}\mathrm dx$$ I leave it as a challenge to you to prove that $$\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ So $$I=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\frac{\Gamma(\frac{2n+1}2)\Gamma(\frac{1}2)}{2\Gamma(\frac{2n}2+1)}$$ $$I=\frac{\sqrt\pi}2\sum_{n\geq0}\frac{(-1)^n}{2n+1}\frac{\Gamma(n+\frac{1}2)}{\Gamma(n+1)}$$ Then recall that $$\frac{d}{dx}\operatorname{arcsinh}x=(1+x^2)^{-1/2}$$. This function has the hypergeometric representation $$\frac{d}{dx}\operatorname{arcsinh}x=\,_1\mathrm{F}_0[1/2;;-x^2]$$ $$\frac{d}{dx}\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^n(1/2)_n}{n!}x^{2n}$$ Thus $$\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^n(1/2)_n}{n!}\frac{x^{2n+1}}{2n+1}$$ then recalling that $$(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$$, we have $$\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^nx^{2n+1}}{2n+1}\frac{\Gamma(n+\frac12)}{\Gamma(\frac12)\Gamma(n+1)}$$ $$\sqrt{\pi}\,\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^nx^{2n+1}}{2n+1}\frac{\Gamma(n+\frac12)}{\Gamma(n+1)}$$ And (drum roll please)... $$I=\frac{\pi}2\operatorname{arcsinh}1$$ $$I=\frac{\pi}2\log(1+\sqrt2)$$
Extra: proving the hypergeometric identity
We start by finding the Taylor Series representation for $$x^\alpha$$ about $$x=1$$. Here $$\mathrm{D}^n$$ represents differentiating $$n$$ times wrt $$x$$.
It is easily shown that $$\mathrm{D}^nx^\alpha=p(\alpha,n)x^{\alpha-n}$$ Where $$p(\alpha,n)=\prod_{k=1}^{n}(\alpha-k+1)$$ is the falling factorial. Hence $$\mathrm{D}_{x=1}^nx^\alpha=p(\alpha,n)$$ So $$x^{\alpha}=\sum_{n\geq0}\frac{p(\alpha,n)}{n!}(x-1)^n$$ $$(1+x)^{\alpha}=\sum_{n\geq0}\frac{p(\alpha,n)}{n!}x^n$$ Then using the identity $$p(\alpha,n)=(-1)^n(-\alpha)_n$$ with $$(x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$$, we have that $$(1+x)^\alpha=\,_1\mathrm{F}_0[-\alpha;;-x]$$ $$(1+x^2)^{-1/2}=\,_1\mathrm{F}_0[1/2;;-x^2]$$ As desired.
• Nice solution. Especially enjoy the hypergeometric identity component. I need to spend more time focussing on these functions! I've been living in the land of Polygamma Functions lately :-) – user150203 Jan 12 '19 at 23:41
• Thank you :) I'm finishing my stay in polylogarithm land and working my way towards polygamma land. – clathratus Jan 12 '19 at 23:44 | 2021-04-17T23:59:43 | {
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https://math.stackexchange.com/questions/22348/how-to-add-and-subtract-values-from-an-average | # How to add and subtract values from an average?
Say I have 100 numbers that are averaged:
number of values = 100
total sum of values = 2000
mean = 2000 / 100 => 20
If I want to add a value and find out the new average:
total sum of values = 2000 + 100
mean = 2100 / 101 => 20.79
If I want to subtract a value and find out the new average:
total sum of values = 2100 - 100
mean = 2000 / 100 => 20
It seems to work, but is the above correct?
Is this the proper way to add/subtract values from a average without having to re-sum all the 100 numbers first?
I know that's an old thread but I had the same problem. I want to add a value to an existing average without calculate it back to the total sum.
to add an value to an exisitng average we only must know for how much values the average is calculated: $$average_{new} = average_{old} + \frac{ value_{new} - average_{old}}{size_{new}}$$
• I may be a smart guy, but the accepted answer above was incomprehensible to me. This answer works perfectly, and I can understand it. Thanks! Nov 13, 2014 at 16:37
• @501 - not implemented thank you! Jan 13, 2015 at 14:34
• This answer was exactly what I needed. Elegant!
– visc
Nov 16, 2016 at 23:02
• -1 Careful this formula works only for adding 1 new value! Please work out the correct equation using this answer and you'll see! math.stackexchange.com/a/1153800/503549 Nov 16, 2017 at 4:25
• Is it possible to calculate median in a similar way? Mar 9, 2020 at 22:14
$$s=\frac{a_1+...+a_n}{n}$$.
If you want the average of $$a_1,...,a_n$$ and $$a_{n+1}$$, then $$s'=\frac{a_1+...+a_n+a_{n+1}}{n+1}=\frac{ns+a_{n+1}}{n+1} = \frac{(n+1)s+a_{n+1}}{n+1} - \frac{s}{n+1} = s + \frac{a_{n+1}-s}{n+1}$$
If you want the average of $$a_1,...,a_{n-1}$$ then $$s''=\frac{a_1+...+a_{n-1}}{n-1}=\frac{ns-a_n}{n-1}= \frac{(n-1)s-a_n}{n-1} + \frac{s}{n-1}=s+\frac{s-a_n}{n-1}$$.
• And $ns$ is the old sum, as you can derive from the first equation. Feb 16, 2011 at 13:21
• I'm tempted to downvote as ns is not explained in the answer Aug 22, 2014 at 20:32
• At first I commented that @501's answer was clearer but realized I didn't fully understand why the question worked. This answer, while denser, actually explains the result.
– jds
Dec 10, 2014 at 23:05
• @Celeritas $ns = n \cdot s$, $n$ and $s$ are defined quantities in the post. Jan 13, 2015 at 15:13
• This is the working answer. @501's answer doesn't work when increasing n by more than 1. Nov 16, 2017 at 4:26
To put it programmatically, and since the question was about how to both add and subtract:
average = average + ((value - average) / nValues)
Subtract a value:
average = (average * nValues - value) / (nValues - 1)
• Oh thank god there's another programmer on here. Those math equations make my head hurt.
– Matt
Jan 5, 2019 at 6:38
• @MattD: see also this answer on StackOverflow for replacing a number in a running average. Jan 13, 2020 at 5:03
• Shouldn't the first formula end with / nValues + 1) rather than / nValues)? Jun 18, 2021 at 19:56
Another formula can be
$$\text{NewAvg} = \frac{( \text{OldAvg} \cdot \text{OldSize} ) + \text{NewValue} } { \text{NewSize}}$$
• I was looking for this answer. It's intuitive because newAvg = (sum + newValue) / newSize (by definition of average). And also by definition sum = avg * size. Mar 3 at 6:57
• And also easy for multiple values: newAvg = (sum1 + sum2) / (size1 + size2). Or in the other format newAvg = (avg1 * size1 + avg2 * size2) / (size1 + size2). Mar 3 at 7:00 | 2022-05-21T13:31:13 | {
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http://math.stackexchange.com/questions/801745/is-it-possible-to-sort-an-array-of-10-elements-with-20-comparisons-of-two-elemen | Is it possible to sort an array of 10 elements with 20 comparisons of two elements?
I think it's not, because there are $10!$ permutations of $10$ elements, and $10! \gt 2^{20}$.
Can anyone point me how to prove it rigorously?
-
It depends on the sorting algorithm used. – npisinp May 19 '14 at 16:50
The best comparison sorting algorithms use $O(n \log n)$ operations in the worst case scenario (and even simple comparison sorts like insertion sort use $O(n^2)$ in the worst case). Of course, these are asymptotic statements, but you don't need to compare every element to every other element to sort them (since $a < b$ and $b<c$ tells you $a<c$, for example). For example, using the bound on wikipedia for merge sort, we see worst case upper bound 25 operations. The general case is treated in Knuth's TAOCP, Vol 3, Section 5.3.1. – Batman May 19 '14 at 16:54
This should probably be re-worded as "is it possible to sort every array of $10$ elements with $20$ comparisons of two elements?" – NotNotLogical May 19 '14 at 16:56
@NotNotLogical Well, that should be understood. For every array there exists an algorithm that sorts it without any comparisons. – Hagen von Eitzen May 19 '14 at 16:59
Hmm I did it with 0 comparisons a long time ago. – Joshua May 19 '14 at 21:16
Construct a tree whose interior nodes say which pairwise comparisons are performed: the root node says which two elements are compared first, and then one proceeds to one of its two children depending on the result of that comparison. Each interior node of the tree has exactly two children. Each leaf is labeled with the instructions for swapping items in the list in order based on the results of the preceding comparisons. Any comparison sort can be put into this form, and a comparison sort which does at most 20 comparisons can be put into the form of a tree with depth 20 and therefore only $2^{20}=1048576$ leaves.
For a list with 10 elements there are $10! = 3628800$ possible input lists and therefore, for any given comparison sort, two such lists must arrive at the same leaf, and the sort therefore uses the same sequence of swaps to order the items in both lists. But this must leave at least one of them out of order.
This all assumes that the 20 items in the input list are all distinct. But you can assume that they are because a comparison sort which could work for any list of 20 elements must as a special case work for any list of 20 distinct elements.
It is instructive to consider the case of radix sort, for which this analysis does not apply.
-
You are right and give the correct reason. Already from this simple aspect it follows that at least $22$ comparisons are needed - whether that is possible is then not quite as simple to decide. The flow of the algorithm (including which entries get swapped and which get compared next) is completely determined by the results of the comparisons. By the pigeon-hole principle, two of the $10!$ possible permutations of $\{1,2,\ldots,10\}$ produce the same of $2^{20}$ possible sequences of comparison results, hence produce the same flow and the same swaps - which cannot sort both.
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http://givelife.ph/yygjmih/srlbmqa.php?gzrorzaen=geometric-sequence-examples | Geometric sequence examples
### Geometric sequence examples
Seq. Geometric sequences are used in several branches of applied mathematics to engineering, sciences, computer sciences, biology, finance Problems and exercises involving geometric sequences, along with detailed solutions and answers, are presented. The formula is then used to find another term of the sequence. For example, the sequence 2, 6, 18, 54, is a geometric progression with In a Geometric Sequence each term is found by multiplying the previous term And, yes, it is easier to just add them in this example, as there are only 4 terms. Geometric Sequences and Sums Sequence. One less than n, or n − 1, is the exponent on r. n is our term number and we plug the term number into the function to find the valueSo this is a geometric series with common ratio r = –2. It is usually denoted by r . This sequence can also be defined recursively, by the formula a_ {1} =1 \quad \text {, and} \quad a_ {n} = 3a_ {n-1} \text { for } n\geq 2. Examples of Geometric Series that could be encountered in the “real world” include:Infinite Geometric Series. Recursively, this can be expressed as: . 20=0. 4, 2. iCoachMath is a one stop shop for all Math queries. What is geometric series ? Geometric series is a series in which ratio of two successive terms is always constant. maxresdefault. In the first part of the race the runner runs 1/2 of the track. Use geometric sequences and series to model real-life quantities, such as monthly bills for cellular telephone service in Example 6. The explicit form can be expressed as: So to find the 5th element, for example,Using Recursive Formulas for Geometric Sequences. 1. In his math book they define a series as a func Sequences and Series Series Geometric series T he sum of an infinite geometric sequence, infinite geometric series The sum of an infinite converging geometric series, examples Converting recurring decimals (infinite decimals) to fraction: T he sum of an infinite geometric sequence, infinite geometric series This unit introduces sequences and series, and gives some simple examples of each. 01) Month 5 Research Sources a = (100) (1. Geometric Sequence Calculator Find indices, sums and common ratio of a geometric sequence step-by-step Geometric sequence Before talking about geometric sequence, in math, a sequence is a set of numbers that follow a pattern. Now that we know how these geometric guys work, we won't have to do the table A geometric series sum_(k)a_k is a series for which the ratio of each two consecutive terms a_(k+1)/a_k is a constant function of the summation index k. 8 , r = −5 16) a 1 = 1, r = 2 Given the first term and the common ratio of a geometric sequence find the recursive formula and the three terms in the sequence after the last one given. 5 Section 6. General Term: Geometric Sequence - is a sequence of terms that have a common _____ between them. Explain your decisions. 8+ Sample Geometric Sequence Examples If you are a math teacher or student, you will definitely require lots of geometric sequence examples. Example 3: Find the sum of the first 8 terms of the geometric series if …iCoachMath. The sequence 9,3,1,1/3,… = is a geometric sequence with common ratio 1/3. r = 2. In mathematics, a geometric series is a series with a constant ratio between successive terms. 911 sequences word problems. Edgar is getting better at math. 2)*(1+0. S 8 = 1 ( 1 − 2 8 ) 1 − 2 = 255. Arithmetic series. . We will just need to decide which form is the correct form. Plugging those values into the general form of the geometric sequence (as done in Example 2) we find that the general term for the denominator is a n Arithmetic and Geometric Sequence Examples Name_____ ©e I2J0y1_5D nKHuOtka[ fSioLfmthwQakr_eZ vLxLICC. Growth Rates Example. Examples and calculation steps for the geometric mean. A sequence is a list of numbers or objects, called terms, in a certain order. Here the ratio of any two terms is 1/2 , and the series terms values get increased by factor of 1/2. In variables, it looks like In variables, it looks like Geometric Examples. Example 1. Now let’s have a look at some examples where we can use all this! Pay it Forward. It is one of the most commonly used tests for determining the convergence or divergence of series. arithmetic sequence, the difference between one term and the next is always the same. Fortunately, geometric series are also the easiest type of series to analyze. A recovering heart attack patient is told to get on a regular walking program. Arith. Sum of the lengths of the sides intersecting in one of the edges is 13 cm. g. com/bitesize/guides/zy6vcj6/revision/4Example 1. The following are examples of sequences: Exploration. A geometric sequence is a sequence of numbers that follows a pattern were Example. 80^3 \times 0. Example: Find a1 for the sequence described by a leads to the next term and the explicit form for an Geometric sequence, and use the explicit formula to find Geometric sequence Summing a Geometric Sequence. Lets take a example. If his scores continued to increase at the same rate, what will be his score on his 9th quiz? Show all work. Tattoos of interlinked geometric shapes glowed on his arms before fading. If they are geometric, state r. Geometric sequence Before talking about geometric sequence, in math, a sequence is a set of numbers that follow a pattern. The behaviour of the geometric sequence { qn } n=0,1,2,3, Powers. Hence I can conclude that this is the answer to the given example. Show the class some examples of objects or pictures that have a repeated pattern. A geometric sequence is a sequence of numbers in which each term is a fixed multiple of the previous term. 2. An arithmetic sequence has a common difference, or a constant difference between each term. 5 . Geometric series. 625 + 0. A sequence made by multiplying by the same value each time. It's because it is a different kind of a sequence - a geometric progression . An infinite geometric sequence is a geometric sequence Problems. Then give a recursive definition and a closed formula for the number of dots in the $$n$$th pattern. For example, the sequence 2, 10, 50, 250, 1250, 6250, 31250, 156250, 781250 is a geometric progression with the common ratio being 5. The geometric mean is used to tackle continuous data series which the arithmetic mean is unable to accurately reflect. to recognize, write, and use geometric sequences. In finer terms, the sequence in which we multiply or divide a fixed, non-zero number, each time infinitely, then the progression is said to be geometric. We have three numbers in an arithmetic progression, and another three numbers in a geometric progression. Formula 3: This form of the formula is used when the number of terms ( n), the first term ( a 1), and the common ratio ( r) are known. What is the 51st term? Box 4 apply their knowledge of geometric sequences to everyday life situations apply the relevant formula in both theoretical and relevant applications calculate the value of a the first term, r the common ratio and T Improve your math knowledge with free questions in "Geometric sequences" and thousands of other math skills. A sequence of numbers {an} is called a geometric sequence if the quotient of successive terms is a constant, called the common ratio. Thus, arithmetic sequences always graph as points along a line. Finding the nth term of an arithmetic sequence. If it's right they get the next questions if it&'s wrong they go back and check it. Geometric Sequences. Then this sequence is a geometric sequence. The variable n indicates the number term in the sequence that the equation is evaluating. An arithmetic sequence is a sequence with the difference between two consecutive terms constant. Be careful here. A geometric sequence (geometric progression) is defined as a sequence in which the quotient of any two consecutive terms is a constant. Since a geometric sequence is a sequence, you find the terms exactly the same way that you do a sequence. 3 Arithmetic and Geometric Sequences Worksheet Determine if the sequence is arithmetic. (A fascinating object for number theorists. geometric_sequences_1. Example 3: The first term of an geometric progression is 1, and the common ratio is 5 determine how many terms must be added together to give a sum of 3906. 68 1, 4, 16, 64, Is this sequence below arithmetic or geometric? How do you know? Write an equation for the sequence. Convergent, divergent, oscillating and alternating sequences, examples, exercises and problems with solutions. A geometric sequence is a sequence where the next term is found by multiplying the previous term by a number. Geometric Sequences: A Formula for the' n - …EXAMPLE 1 Finding the nth term Write a formula for the nth term of the geometric sequence 6, 2, 2 3 , 2 9 , . Example Find the nth term of the geometric sequence: 2, 2. Solution. png Geometric Sequences and Series -. This constant is called the Common Difference. General Term: 1. To find the sum of a finite geometric series, use the formula, S n = a 1 ( 1 − r n ) 1 − r , r ≠ 1 , where n is the number of terms, a 1 is the first term and r is the common ratio . 3. g. Arithmetic progression is a sequence of numbers such that the difference between the consecutive terms in a constant. (b) Show that if a sequence has the property above, it must be a geo-metric sequence. m. , moving from term to term) give rise to equal changes in the output (determined by the common difference). In general singular decisions can be anything - but typically arithmetic. (a) Show that every term of a geometric sequence with non-negative terms, except the first term and the last term (in case of a finite sequence), is the geometric average of the preceding term and the following term. ARITHMETIC AND GEOMETRIC SEQUENCE WORD PROBLEMS PRACTICE. com › Calculus › Sequences › ExamplesExample 1. Notation used in this video is relatively advanced. A geometric sequence has a1 = 1 and r = 2. Use the information you've gathered and the general rule of a geometric sequence to create an equation with one variable, n. Thus, geometric sequences always graph as points along the graph of an exponential function. Since the common ratio has value between -1 and 1, we know the series will converge to some value. This sequence can also be defined recursively, by the formula a_ {1} =1 \quad \text {, and} \quad a_ {n} = 3a_ {n-1} \text { for } n\geq 2. Geometric sequences are formed by choosing a starting value and generating each subsequent value by multiplying the previous value by some constant called the geometric ratio. In 2013, the number of students in a small school is 284. Multiplying numbers together hides patterns. Get smarter on Socratic. This means that it can be put into the form of a geometric series. The formula is broken down into a 1 which is the first term of the sequence, r being the common ratio, and n being the number of the term to find. Unlike arithmetic sequences, these sequences progress by multiplication. Find the number of cubes in the next three figures. c) Find the value of the 13 th term (as a fraction). Example 1 : Find the first five terms and the common ratio of the geometric sequence . Find the first four terms of the sequence. 8 , r = −5 16) a 1 = 1, r = 2 Given the first term and the common ratio of a geometric sequence find the recursive formula and the three terms in the sequence after the last one given. We dealt a little bit with geometric series in the last section; Example 1 …Illustrated definition of Geometric Sequence: A sequence made by multiplying by the same value each time. So once you know the common ratio in a geometric sequence you can write the recursive form for that sequence. 8+ Sample Geometric Sequence Examples If you are a math teacher or student, you will definitely require lots of geometric sequence examples. Thus I can say that the second, third and fourth terms and the general term of the geometric sequence are and respectively. We start with alternating sequence and return to it again at the end, we briefly cover arithmetic sequences, but the most important type is the geometric sequence. 14. For example: 2, 6, 18, 54, . The nth term of a geometric sequence is , where is the first term and is the common ratio. This series of slides introduce the idea of exponential decay. For example, the series + + + + ⋯ is geometric, because each successive term can be obtained by multiplying the previous term by 1/2. The patient is told to walk a distance of 5 km the first week, 8 km the second week, 11 km the third week and so on for a period of 10 weeks. For example, the series is geometric, because each successive term can be obtained by multiplying the previous term by 1/2. A geometric series is the indicated sum of the terms of a geometric sequence. In contrast, a geometric sequence is one where each term equals the one before it multiplied by a certain value. Example 1: {2,6,18,54,162,486,1458,}Apr 8, 2010Nov 8, 2013A geometric sequence is a sequence of numbers that follows a pattern were Example. In the plenary, the class are challenged to apply finding the nth term of a geometric sequence to compound percentage changes. & Geo. In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Hence the geometric sequence will be . Using Explicit Formulas for Geometric Sequences. This video shows how derive the formula to find the 'n-th' term of a geometric sequence by considering an example. Example 2: Find the common ratio if the fourth term in geometric series is \frac {4} {3} and the seventh term is \frac {64} {243}. ), all seperated by + 1 . sequence an ordered list of numbers series the sum of the terms of a sequence term a specific number in a sequence arithmetic sequence a sequence of numbers where the Geometric Sequences …An arithmetic sequence is a sequence with the difference between two consecutive terms constant. Geometric Sequence. jpg geometric sequence examples. This is a sequence of numbers. In formal terms, a complex sequence is a function whose domain is the positive integers and whose range is a subset of the complex numbers. Another formula for the sum of a geometric sequence is. a geometric sequence is a sequence that satisfies a n = a n-1 r where r is the common ratio. Definitions emphasize the parallel fea-tures, which examples will clarify. However, items are multiplied, not added. All final solutions MUST use the formula. . Draw the next term if this represents a geometric sequence. Geometric Sequences Worksheet Determine whether each of the following sequences is arithmetic, geometric, or neither. Geometric Series Example doe. Well, it’s an old topic from high school. Arithmetic and Geometric Sequence Examples Name_____ ©e I2J0y1_5D nKHuOtka[ fSioLfmthwQakr_eZ vLxLICC. We have detailed definitions, easy to comprehend examples and video tutorials to help understand complex mathematical concepts. (c) Generalize the statement in (a) by proving that the kth term is the geometric average of the (k−i)th term and the (k+i)th term for all i such that these terms exist. Since n ! > n, our intuition suggests that the sequence { n !} n=1,2,3, Exponential. Step (2) The given series starts the summation at , so we shift the index of summation by one: Our sum is now in the form of a geometric series with a = 1, r = -2/3. Learn more about Geometric sequences and see some examples. i V nAQlele crLiJghBtosJ SrFexsJeYrKvmegdV. There are also other sequences like arithmetic sequence , harmonic sequence and so on. and . The difference is called the common difference . The value of equals. The second differences are the same. Now, continue multiplying each product by the common ratio (3 in my example) and writing the result down… over, and over, and over: By following this process, you have created a “Geometric Sequence”, a sequence of numbers in which the ratio of every two successive terms is the same. Concept 16 Arithmetic & Geometric Sequences Concept 16: Arithmetic & Geometric Sequences Assessment (Level 4 Example Question Level 3 Example Question Level 2 Example Question Write an equation for this geometric sequence and find the 10th term of the sequence. An arithmetic-geometric progression (AGP) is a progression in which each term can be represented as the product of the terms of an arithmetic progressions (AP) and a geometric progressions (GP). We dealt a little bit with geometric series in the last section; Example 1 showed that n 1 1 2n 1, This series doesn’t really look like a geometric series. The formulas applied by this geometric sequence calculator are detailed below while the following conventions are assumed: - the first number of the geometric progression is a; - the step/common ratio is r;Infinite Geometric Series Definition of an Infinite Geometric Series. LIMITS OF RECURSIVE SEQUENCES 3 Two simple examples of recursive definitions are for arithmetic sequences and geomet- ric sequences. In mathematics, a sequence is usually meant to be a progression of numbers with a clear starting point. LIMITS OF RECURSIVE SEQUENCES 3 Two simple examples of recursive definitions are for arithmetic sequences and geomet- ric sequences. The sequence <1,2,4,8,16,… = is a geometric sequence with common ratio 2, since each term is obtained from the preceding one by doubling. It is estimated that the student population will increase by 4% each year. …is a sequence of numbers where the ratio of consecutive terms is constant. The value r is called the common ratio. 2, 6, 18, 54, … This is an increasing geometric sequence with a common ratio of 3. Each term of a geometric sequence is the geometric mean of the terms preceding and following it. iCoachMath. It’s supposed that q≠0 and q≠1. Learning Zone Standards Sign up Sign In Username or email: Finding the Sum of a Geometric Series Solving Word Problems Using Geometric Series Infinite Geometric SeriesA sequence is a set of numbers determined as either arithmetic, geometric, or neither. Geometric Sequences and Series -. However, if we write each term as its prime factorization, then we can see what's going on. Example. Example 3: Find the sum of the first 8 terms of the geometric series if a 1 = 1 and r = 2 . We l In barely a passable British accent, as a class we explore geometric sequences. edu/stat414/node/77Example (continued) Solution. 9. Calculating the Infinite Geometric Series Example Suppose that a runner begins on a one mile track. This constant is called the Common Ratio. Adding the corresponding terms of the two series, we get $$120 , 116 , 130$$. Now I’ll give some examples of geometric sequences. exampleChapter 13 - Sequences and Series Section 13. jpeg geometric sequence examples. Let's say that your starting point is #2#, and the common ratio is #3#. Real life problem with geometric sequences. A sequence is a list of numbers in which each number depends on the one before it. The equality given in Example 4. Homework problems on geometric sequences often ask us to find the nth term of a sequence using a formula. Given the first term and the common ratio of a geometric sequence find the first five terms and the explicit formula. On his first quiz he scored 57 points, then he scores 61 and 65 on his next two quizzes. In the above sequence, n = 3 when evaluating 6/3, the third term in the series. 1) a n = 40 - 5n 2) a n = 176 - 200n Determine if the sequence is arithmetic. In a geometric sequence, a term is determined by multiplying the previous term by the rate, explains to MathIsFun. However, the recursive formula can become difficult to work with if …Then this sequence is a geometric sequence. 1)*(1-0. 2 Geometric sequences (EMCDR) Geometric sequence. 2 covering the Arithmetic and Geometric Sequences. For example, the sequence 1, 3, 9, 27, 81 is a geometric sequence. where r is the common ratio between successive terms. For the series: 5 + 2. Suppose there are 100 insects in the first generation. i V nAQlele crLiJghBtosJ SrFexsJeYrKvmegdV. Rather than write a recursive formula, we can write an explicit formula. Complete the quiz and then go on to the related lesson entitled Geometric Sequences: Formula & Examples to learn more about: Defining geometric sequences Exploring ratios in geometric sequenceswhere a 1 is the first term of the sequence and r is the common ratio which is equal to 4 in the above example. Geometric series are among the simplest examples of infinite series with finite sums, MATH 1090 Sec. geometric sequence example. The graph of the sequence 4, 7, 10, 13,Improve your math knowledge with free questions in "Geometric sequences" and thousands of other math skills. To solve real-life problems, such as finding the number of tennis matches played in Exs. If we multiply, it is a geometric sequence. Ask students to find the patterns. a sequence (such as 1, 1/2, 1/4) in which the ratio of a term to its predecessor is always the same —called also geometrical progression,…A geometric sequence is an ordered list of numbers in which each term is the product of the previous term and a fixed, non-zero multiplier called the common factor. Geometric Sequences. e. Obviously, our arithmetic sequence calculator is not able to analyze any other type of sequence. 14 illustrates an important point when evaluating a geometric series whose beginning index is other than zero. SERIES AND SEQUENCES. Sal introduces geometric sequences and gives a few examples. The denominator is a geometric sequence with a 1 = 2 and r = 2. a; ar; ar2; ar3 ¢¢¢: We note that the ratio between any two consecutive terms of each of the above sequences is always the same. The geometric mean is used in finance to calculate average growth rates and is referred to as the compounded annual growth rate. 35. The ratio of two successive terms is always the same . The sequence is quadratic and will contain an term. ) The first term of the sequence is a = –6. We can obtain the common ratio by dividing any term after the first by the term preceding it. Arithmetic and Geometric Sequences and Series and geometric sequences and series to solve real-world problems, including writing the first n terms, finding the nth term, and evaluating summation formulas. (a) How many will there be in the fifth generation? Note: The population can be written as a geometric sequence with a1 as the first- A sequence is called a geometric sequence if the ratio of any term to its previous term is a constant. If the sequence has a definite number of terms, the simple formula for the sum is. Sal introduces geometric sequences and gives a few examples. The geometric mean of the growth rate is calculated as ((1+0. 1, 0, 3, 0, 5, 0, 7, Arithmetic and Geometric Sequences. 2: Compound Interest; Geometric Sequences De nitions: If $P is invested at an interest rate of r per year, compounded annually, the future A sequence is a list of numbers, geometric shapes or other objects, that follow a specific pattern. We begin with two basic examples. org right now: https://www. An arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. ) Example. Author: slcmath@pcViews: 1. n is our term number and we plug the term number into the function to find the value Arithmetic and Geometric Sequences 17+ Amazing Examples! An Arithmetic Sequence is such that each term is obtained by adding a constant to the preceding term. 3, 1, a in the above examples) is called the initial term , which Formulas for the nth terms of arithmetic and geometric sequences For an arithmetic sequence, a formula for thenth term of the sequence is a n 5 a 1 ~n 2 1!d. A Sequence is a set of things (usually numbers) that are in order. Example. Geometric Sequences and Series 1) No 2) a) The common ratio is 6 b) The common ratio is − 1 2 3) a) The missing terms are 144, 24, 4 b) The missing terms are 7 4, 7 8, 7 16 4) The 10th term is 1310720 and the n th term is 5 × 4 n − 1 5) The first term is 4 3 and the 10th term is 26244 a) Show that the sequence is geometric. GEOMETRIC PROGRESSION Examples The following are called geometric progressions: 1. Determine the volume of the cuboid. Calculating the last term using the general form listed above, the term is: There are easier ways to generate the 100th term of a geometric sequence than listing all 99 terms before it. ARITHMETIC AND GEOMETRIC SEQUENCES. example The ratios that appear in the above examples are called the common ratio of the geometric progression. Vocabulary Example: 1, 3, 5, 7, d=2 10, 20, 30, d=10 10, 5, 0, 5,. This ratio is called the common ratio. Instructor: Dr. Here are a few examples of geometric sequences. Example 4. Algebra > Sequences and Series > Geometric Sequences. An arithmetic sequence is a list of numbers with a definite pattern. Each term (except the first term) is found by multiplying the previous term by 2. See also. A geometric progression (GP), also called a geometric sequence, is a sequence of numbers which differ from each other by a common ratio. 1; ¡1=3; 1=9; ¡1=27; ¢¢¢. The situation can be modeled by a geometric sequence with an initial term of 284. The difference is called the common difference. So r 2 6 1 3 . The first term is 12 and the ratio between terms is 1 3, so 12 4 4 3 4 9 4 27 first term 1 …Both arithmetic and geometric sequences begin with an arbitrary first term, and the sequences are generated by regularly adding the same number (thecom-mon difference in an arithmetic sequence) or multiplying by the same number (the common ratio in a geometric sequence). The explicit formula for a geometric sequence is of the form a n = a 1 r-1 , where r is the common ratio. Although n a is just one type of expression, we will actually split it into two cases. Sequences and Series Terms. Each term is the product of the common ratio and the previous term. 9+ Geometric Sequence Examples – DOC, Excel, PDF Geometric sequence can be defined by a series where a fixed amount is multiplied to reach at each of the number of the series, starting from the first. For example, the sequence 2, 4, 8, 16, 32 does not have a common difference. Thus A simple example is the geometric series for a = 1 and r = 1/2, or 1 + 1/2 + 1/4 + 1/8 +⋯, which converges to a sum of 2 (or 1 if the first term is excluded). 3, 6, 12, 24, 48, … Write an equation for this arithmetic sequence and find the geometric sequence examples. A recursive formula for a sequence tells you the value of the n th term as a function of . In a Geometric Sequence each term is found by multiplying the previous term by a constant. Full Answer. The sides of the cuboid make a geometric progression. The value of the $$n^{th}$$ term of the arithmetic sequence, $$a_n$$ is computed by using the following formula: $a_n = a_1 + (n-1)d$ This means that in order to get the next element in the sequence we add $$d$$, to the previous one. The best videos and questions to learn about Sums of Geometric Sequences. Remark 4. N ∞ 1, n. Geometric Series and the Test for Divergence - Part 1 - Duration: 9:57. Whereas, in a Geometric Sequence each term is obtained by multiply a constant to the preceding term. Definition and Basic Examples of Arithmetic Sequence. Example 1: {2,6,18,54,162,486,1458,} is a geometric sequence where each term is 3 times the previous term. A geometric sequence is a sequence of numbers in which each new term (except for the first term) is calculated by multiplying the previous term by a constant value called the constant ratio ($$r$$). p. In this section we will cover basic examples of sequences and check on their boundedness and monotonicity. Solution: The sides lengths are a = 1cm, b = 3cm, c = 9cm. A geometric sequence has a constant ratio r between consecutive terms. More formally, a geometric sequence may be defined recursively by: . 2,4,6,8,10…. This ratio is called the . Without explaining why, have students look at several examples of infinite geometric 911 sequences word problems. The sum of a finite geometric sequence (the value of a geometric series) can be found according to a simple formula. The geometric shapes changed as they circled her neck rather than stuck to a pattern; she assumed it was some kind of writing. Example: 2, 4, 8, 16, 32, 64, 128, 256, (each number is 2 times the number before it)The common difference for an arithmetic sequence is the value being added between terms, and is represented by the variable d. (I can also tell that this must be a geometric series because of the form given for each term: as the index increases, each term will be multiplied by an additional factor of –2. Concept 16 Arithmetic & Geometric Sequences Concept 16: Arithmetic & Geometric Sequences Assessment (Level 4 Example Question Level 3 Example Question Level 2 Example Question Write an equation for this geometric sequence and find the 10th term of the sequence. However, if we write each term as its prime factorization, then we can see what's going on. Solutions for the assessment 7. Author: yaymathViews: 35KGeometric sequences calculator that shows stepshttps://www. To get the next term, we multiply the term we're on by r = 2. Our Math dictionary is both extensive and exhaustive. My son (7th grade) had this homework problem last night: Come up with a function for a non arithmetic, non geometric series starting with 4 and 8. a sequence (such as 1, 1/2, 1/4) in which the ratio of a term to its predecessor is always the same —called also geometrical progression,… 3. Section 2. Let $P$ be the student population and $n$ be the number of years after 2013. When they have an answer they bring it to be checked. The explicit formula is also sometimes called the closed form. 20, 1− p = 0. Thus an+1 an=q or an+1=qan for all terms of the sequence. EX: Example 7: Tell whether each sequence is arithmetic, geometric, or neither. To find the desired probability, we need to find P ( X = 4), which can be determined readily using the p. The Geometric Sequence Concept. An Geometric Sequence describes something that is periodically growing in an exponential fashion (by the same percentage each time), and a Geometric Series describes the sum of those periodic values. 456 and then find the 10th term. Sequences whose rule is the multiplication of a constant are called geometric sequences, similar to arithmetic sequences that follow a rule of addition. Primary SOL . 70 and 71. c) Find the value of the 15 th term. Are the following sequences arithmetic, geometric, or neither? If they are arithmetic, state the value of d. Therefore, 10 + (−7) = 3. Geometric Sequences: A Formula for the nth Term. Example of a geometric sequence. Example (continued) A representative from the National Football League's Marketing Division randomly selects people on a random street in Arithmetic-Geometric Progression (AGP): This is a sequence in which each term consists of the product of an arithmetic progression and a geometric progression. Geometric Series. Do you want to start something new, that no other people have? Try the geometric sequence example …Now, remember, and Arithmetic Sequence is one where each term is found by adding a common value to each term and a Geometric Sequence is found by multiplying a fixed number to each term. The sequence is able to be found by using the geometric sequence formula. Page 5 of 5 . More over geometric sequence example has viewed by 134 visitors. (2) The definitions allow us to recognize both arithmetic and geometric sequences. arithmetic sequence, the difference …A sequence of numbers $$\left\{ {{a_n}} \right\}$$ is called a geometric sequence if the quotient of successive terms is a constant, called the common ratio. notebook April 25, 2014. Here is the first term and is the common ratio in the sequence. Consider. What makes a sequence geometric is a common relationship In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. If we add a number to get from one element to the next, we call it an arithmetic sequence. 80, and x = 4: P(X=4)=0. That's our total number of terms. 12, which is known as the ratio test. The sequence <1,2,4,8,16,… = is a geometric sequence with common ratio 2, since each term is obtained from the preceding one by doubling. 444 5. Geometric Series. Finite geometric series applications Video transcript - [Voiceover] We're asked to find the sum of the first 50 terms of this series, and you might immediately recognize it is a geometric series. If the first term of an arithmetic sequence is a1 and the common difference is d, then the nth term of the sequence is given by:Geometric Sentence Examples Its edges were gilded with gold marking a lazy geometric design across the marble. Examples of arithmetic and geometric sequences and series in daily life. The following figure gives the formula for the nth term of a geometric sequence. A geometric sequence is a sequence with the ratio between two consecutive terms constant. geometric sequence examples, geometric sequence example and solution, geometric sequence examples with solutions pdf, geometric sequence examples real life, geometric sequence example problems, geometric sequence example problems in real life, geometric sequence examples with answer, geometric sequence examples with formula, geometric sequence examples pdf A geometric sequence is a sequence in which each term is found by multiplying the preceding term by the same value. org/math/precalculus/seq_inductionAuthor: Khan AcademyViews: 711KSequences - Edexcel - Revision 4 - GCSE Maths - BBC Bitesizehttps://www. Sum of Arithmetic Geometric Sequence In mathematics, an arithmetico–geometric sequence is the result of the term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression. psu. Convergent and Divergent Sequences Convergent Sequences An example of a geometric sequence with starting number 3, common factor 2 and eight terms is 3, 6, 12, 24, 48, 96, 192, 384. Here are the all important examples on Geometric Series. The geometric sequence can be rewritten as where is the amount Infinite Geometric Sequences. OK, so I have to admit that this is sort of a play on words since each element in a sequence is called a term, and we’ll talk about the terms (meaning words) that are used with sequences and series, and the notation. Geometric sequences contain a pattern where a fixed amount is multiplied from one term to the next ( common ratio r ) after the first term Geometric sequence examples :Geometric progression definition is - a sequence (such as 1, 1/2, 1/4) in which the ratio of a term to its predecessor is always the same —called also geometrical progression, geometric sequence. geometric sequence examplesIn mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Complete the quiz and then go on to the related lesson entitled Geometric Sequences: Formula & Examples to learn more about: Defining geometric sequences Exploring ratios in geometric sequences To write the explicit or closed form of a geometric sequence, we use a n is the nth term of the sequence. This isn't a very useful form for seeing patterns. Find the common ratio in …geometric sequences What is an arithmetic sequence? What is a geometric sequence? How do we find the nth term of an arithmetic or geometric sequence? We call a list of numbers written down in succession a sequence; for example, the numbers drawn in a lottery: 12,22,5,6,16,43,A geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio, r. We call each number in the sequence a term. Explain that patterns (sequences) can occur with a list of numbers. 2)*(1+0. for some q> 0. an Dan1 Cd or an an1 Dd: The common difference, d, is analogous to the slope of a line. In an . If you're going on to Calculus, these are going to be important! Remember that with arithmetic sequences we added something each time. We learned that a geometric series has the form Definition The series Example Suppose that there is a geometric sequence with r = 2 and a 1 = 1 then the infinite geometric series isExample 4. 17) a 1 = −4, r = 6 18) a 1 Arithmetic Sequence - is a sequence of terms that have a common _____ between them. 15) a 1 = 0. We are going to use the computers to learn about sequences and to create our own sequences. In order to master the techniques explained here it is vital that you undertake plenty of practice Free Interior Design Geometric Brochure. Geometric Sequences Examples What's the common ratio for the sequence ? List the first four terms and the 10th term of a geometric sequence with a first The recursive formula for a geometric sequence is written in the form For example, when writing the general explicit formula, n is the variable and does not Geometric Sequences and Sums Sequence. Geometric-Sequence-Example-Problems. For any interior design firm, it is important to have that one document where your prospective clients can have a look at all your designs. 1024 There is about a 10% chance that the marketing representative would have to select 4 peopleThe sequence is able to be found by using the geometric sequence formula. Examples: 1. Example 3: What is the geometric mean of 1/2, 1/4, 1/5, 9/72 and 7/4? First, multiply the numbers together and then take the 5th root: (1/2*1/4*1/5*9/72*7/4) (1/5) = 0. -1-Given the explicit formula for an arithmetic sequence find the first five terms. science. Definition of Geometric Sequence. Sometimes the terms of a geometric sequence get so large that you may need to express the terms in scientific notation rounded to the nearest tenth. Geometric sequence example. There are exceptions of course like the ball bouncing is geometric even though it is singular because of coefficient of restitution. So for example, and this isn't even a geometric series, if I just said 1, 2, 3, 4, 5. Note that after the first term, the next term is obtained by multiplying the preceding element by 3. a1 = 1. You can do it by yourself, too - it's not that hard! Look at the first example sequence: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. geometric sequence. Now the second, third and fourth terms of geometric sequence will be . Without explaining why, have students look at several examples of infinite geometric series, including common ratios that are Arithmetic and Geometric Sequences Reporting Category Number and Number Sense Topic Describing arithmetic and geometric sequences . This tells us that the sequence is geometric with ratio 3, and initial term 1, so we get that the sequence is given by a_ {n} = 3^ {n-1}. If the first term of an arithmetic sequence is a1 and the common difference is d, then the nth term of the sequence is given by:The sequence is able to be found by using the geometric sequence formula. Loading Unsubscribe from slcmath@pc? Cancel Unsubscribe. In order to work with these application problems you need to make sure you have a basic understanding of arithmetic sequences, arithmetic series, geometric sequences, and geometric series. In the second part of the race the runner runs halfThe geometric series is used in the proof of Theorem 4. Example: 2, 4, 8, 16, 32, 64, 128, 256, (each number is 2 times the number before it) A sequence is a set of numbers, called terms, arranged in some particular order. The geometric sequence has its sequence formation: To findGeometric sequence example convergent geometric series the sum of an infinite geometric recursive formulas for geometric sequences practice khan academy solved given the first three terms of a geometric sequence fi geometric sequence formula examples video lesson transcript ppt section 57 arithmetic and geometric sequences powerpoint 9 Menu Algebra 2 / Sequences and series / Geometric sequences and series A geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio, r. Aug 30, 2016 · Unlike arithmetic sequences, these sequences progress by multiplication. is an arithmetic sequence with the common difference 2. Evaluate . Geometric progression definition is - a sequence (such as 1, 1/2, 1/4) in which the ratio of a term to its predecessor is always the same —called also geometrical progression, geometric sequence. D. If the first term of an arithmetic sequence is a 1 and the common difference is d, then the nth term of the sequence is given by: "Common Core F. Find the sum of the first six terms of the sequence: 27, –9, 3, –1, … Geometric with r = –1/3 and a first term of 27 so sum = € 271−− 1 3 #6$ % & ’ ( # $% & ’ ( 1−− 1 3 #$ % & ’ ( =40. Example 1: { 2 , 6 , 18 , 54 , 162 , 486 , 1458 , is a geometric sequence where each term is 3 times the previous term. Write the first five terms of a geometric sequence in which a1=2 and r=3. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence. A geometric series is simply the sum of a geometric sequence, n 0 arn. Geometric series formula is given by Here a will be the first term and r is the common ratio for all the terms, n is number of terms. 04. This isn't a very useful form for seeing patterns. org//geometric-sequences-calculator. AY . Find the common ratio in each of the following geometric sequences. An example would be 3, 6, 12, 24, 48, … Each term is equal to the prior one multiplied by 2. Thus making both of these sequences easy to use, and allowing us to generate a formula that will enable us to find the sum in just a few simple steps. The more general case of the ratio a rational function of the summation index k produces a series called a hypergeometric series. I think it is really important students appreciate the practical, real life aspect of geometric sequences and compound interest links really nicely with this topic. Examples: 1) A bank offers an account with interest rate 10 % The numerator is the same arithmetic sequence that we have encountered in Examples 1 & 4 that has a general term of a n = 3n - 1. Plugging into the summation formula, I …Arithmetic and Geometric Sequences 17+ Amazing Examples! An Arithmetic Sequence is such that each term is obtained by adding a constant to the preceding term. For both of these sequences, u 1 is the first term, and n is the number of terms in the sequence. Students work in paris to answer questions. Example: Determine which of the following sequences are geometric. Sequence goes back through sub-Mycenaean wares to simpler geometric and plain black and grey fabrics. , moving from term to term) cause the output to be successively multiplied by a constant (determined by the common ratio). (A) f(x) = 3x (B) f(x) = x + 3 (C) f(x) = 2x + 6 (D) f(x) = 3x + 4 What is the missing term in the sequence below?Growth Rates Example. We can use this formula to find the first and second set of number. Definition The series is called the infinite geometric series . with a fixed first term and common ratio . In mathematics, the geometric sequence is a collection of numbers in which each term of the progression is a constant multiple of the previous term. Explains the n-th term IntroExamples Arith. com. 1 Sequences and Series. 3. Another way to look at it is that we are multiplying each term by ½ to get the next term in the sequence. Examples of arithmetic and geometric sequences and series in daily life. 1)-4, 1, 6, 11, … 2) 2 Solutions for the assessment 7. Geometric Series Examples Geometric sequences have this same special property: equal changes in the input (e. For example 32/64 = ½ and 2/4 = ½. If it is, find the common difference. 2) Student/Teacher Actions (what students and teachers should be doing to facilitate learning) 1. Introduces arithmetic and geometric sequences, and demonstrates how to solve basic exercises. The geometric series is used in the proof of Theorem 4. Use the information you've gathered and the general rule of a geometric sequence to create an equation with one variable, n. The Sum of the First n Terms of a Geometric Sequence 4:57 Crank out the common ratio, first term, and last term of the sequence. In a geometric sequence, the ratio between consecutive terms is always the same. 14. Improve your math knowledge with free questions in "Geometric sequences" and thousands of other math skills. This is a full lesson on determining whether a graph represents an Arithmetic or Geometric Sequence. A geometric sequence is given by a starting number, and a common ratio. 5 times as large as the previous generation. Consider a sequence of trials, where each trial has only two possible outcomes (designated failure and success). The first differences are not the same, so work out the second differences. Our arithmetic sequence calculator can also find the sum of the sequence (called the arithmetic series) for you. Have a look!! Geometric sequence. A geometric sequence is a sequence in which the ratio of any term to the previous term is constant. Plugging those values into the general form of the geometric sequence (as done in Example 2) we find that the general term for the denominator is a n 974 Chapter 10 Sequences, Induction, and Probability Thus, the formula for the term is a n =a 1 r n–1. A recursive formula allows us to find any term of a geometric sequence by using the previous term. The formula is broken down into a 1 which is the first term of the sequence, r being the common ratio, and n …May 30, 2014 · Geometric Sequences - Examples slcmath@pc. ARITHMETIC AND GEOMETRIC SEQUENCE WORD PROBLEMS PRACTICE. 2 Geometric sequences (EMCDR) Geometric sequence. Examples of Geometric Series that could be encountered in the “real world” include:For example, the sequence 2, 10, 50, 250, 1250, 6250, 31250, 156250, 781250 is a geometric progression with the common ratio being 5. Also notice that the ratio of any term and its preceding term is ½. bbc. Geometric Series and the Test for Divergence - Part 1 - …Geometric series formula is given by Here a will be the first term and r is the common ratio for all the terms, n is number of terms. For example, the sequence 2, 10, 50, 250, 1250, 6250, 31250, 156250, 781250 is a geometric progression with the common ratio being 5. Compute 12 4 4 3 4 9 4 27 . A geometric sequence is a sequence of numbers in which the ratio between consecutive terms is constant. Geometric Sequences This lesson will work with arithmetic sequences, their recursive and explicit formulas and finding terms in a sequence. LE. Alternatively, the difference between consecutive terms is always the same. 88, 3. We could sum all of …This series doesn’t really look like a geometric series. 3125 , the first term is given by a1 = 5 and the common ratio is r = 0. Author: Khan AcademyViews: 221KGeometric Sequences Examples - Shmoopwww. Series If you try to add up all the terms of a sequence, you get an object called a series. To enable students recognise a geometric sequence (geometric progression) • To enable students apply their knowledge of geometric sequences to everyday applications If»students»are»unable»to»suggest»examples»of» geometric»sequences»direct»them»to»examples» 4. 4. notebook April 25, 2014 IF Checking: p. In this case, 2 is called the common ratio of the sequence. A geometric sequence is a sequence of numbers that follows a pattern were the next term is found by multiplying by a constant called the common ratio, r. In the explicit formula for a geometric sequence, the variable r represents the common ratio for the sequence. Engaging math & science practice! Improve your skills with free problems in 'Solving Word Problems Using Geometric Series' and thousands of other practice lessons. 3; 6; 12; 24; ¢¢¢. Find k given that the following Example: An insect population is growing in such a way that each new generation is 1. ; An arithmetic sequence is a sequence with the difference between two consecutive terms constant. 3,6,12,24,48,96, ARITHMETIC AND GEOMETRIC SEQUENCE WORD PROBLEMS PRACTICE. 3))^(1/3)ARITHMETIC AND GEOMETRIC SEQUENCE WORD PROBLEM EXAMPLES. 444 5. Now, lets apply this:For example: 2, 6, 18, 54, . If you're seeing this message, it means we're having trouble loading external resources on our website. com. Example 4 : The average person’s monthly salary in a certain town jumped from $2,500 to$5,000 over the course of ten years. 5. Examples of geometric sequences. khanacademy. A geometric sequence is a sequence of numbers in which each new term (except for the first term) is calculated by multiplying the previous term by a constant value called the constant ratio ($$r$$). Their terms alternate from upper to lower or vice versa. For example, the sequence $$2, 4, 8, 16, \dots$$ is a geometric sequence with common ratio $$2$$. The sequence we saw in the previous paragraph is an example of what's called an arithmetic sequence: each term is obtained by adding a fixed number to the previous term. A sequence is a set of numbers determined as either arithmetic, geometric, or neither. Examples of How to Apply the Concept of Arithmetic Sequence. So this sequence, which is not a geometric sequence, we can still define it explicitly. The student population will be 104% of the prior year, so the common ratio is 1. Arithmetic-Geometric Progression (AGP): This is a sequence in which each term consists of the product of an arithmetic progression and a geometric progression. Geometric Series ExamplesLIMITS OF RECURSIVE SEQUENCES 3 Two simple examples of recursive definitions are for arithmetic sequences and geomet- ric sequences. Example: 1/2,1/4,1/8,1/16,. Example: 2, 4, 8, 16, 32, 64, 128, 256, (eachA geometric sequence is a sequence of numbers where each term after the first term is found by multiplying the previous one by a fixed non-zero number, called the common ratio. The constant ratio is called the common ratio and represented by 'r'. 3, 6, 12, 24, 48, … Write an equation for this arithmetic sequence and find theOne of the most important typesof infinite series are geometric series. Resume, geometric sequence example was posted by readthis. for example in (1. Arithmetic and Geometric Sequences. The sum of the first n terms of a geometric sequence is given byA geometric series is the sum of the terms in a geometric sequence. They can connect these values to the values of the output from the function( 3^-1, 3^-2, 3^-3, etc) (MP7). How can we use arithmetic and geometric sequences to model real-world Example 1 Testing for an arithmetic sequence a Is the sequence 20,17,14,11,8, EXAMPLE 1 common ratio geometric sequence, GOAL 1 Write rules for geometric sequences and find sums of geometric series. Part 4: Geometric Series Important examples. The first term (e. nth General Term of a Geometric Sequence Geometric Sequences Worksheet Determine whether each of the following sequences is arithmetic, geometric, or neither. Sequences and series, whether they be arithmetic or geometric, have may applications to situations you may not think of as being related to sequences or series. Find the sum of the first six terms of the sequence: 27, –9, 3, –1, … Geometric with r = –1/3 and a first term of 27 so sum = € 271−− 1 3 #6 $% & ’ ( #$ % & ’ ( 1−− 1 3 # \$ % & ’ ( =40. Finite Geometric Series. A geometric sequence goes from one term to the next by always multiplying (or dividing) by the same value. One example of a geometric series, where r=2 is 4, 8, 16, 32, 64, 128, 256 If the rate is less than 1, but greater than zero, the number grows smaller with each term, as in 1, 1/2, 1/4, 1/8, 1/16, 1/32… where r=1/2. A sequence made by multiplying by the same value each time. Geometric Sequences and Series 1) No 2) a) The common ratio is 6 b) The common ratio is − 1 2 3) a) The missing terms are 144, 24, 4 b) The missing terms are 7 4, 7 8, 7 16 4) The 10th term is 1310720 and the n th term is 5 × 4 n − 1 5) The first term is 4 3 and the 10th term is 26244 Algebra 2 CCSS Lessons and Practice is a free site for students (and teachers) studying a second year of high school algebra. Looking at this definition I can say that arithmetic progression can applied in real life by analyzing a certain pattern that we . virginia. Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms. Oscillating sequences are not convergent or divergent. We must now compute its sum. Jo Steig. So 1, 2, 4, 8, 16, is geometric, because each step multiplies by two; and 81, 27, 9, 3, 1, , is geometric, because each step divides by 3. Suppose I have a sequence like . For example, in the sequence 2/1, 4/2, 6/3, the common ration is 2. Plugging those values into the general form of the geometric sequence (as done in Example 2) we find that the general term for the denominator is a n = 2 (2) n-1 = 2 n. The individual items in the sequence are called terms , and represented by variables like x n . A geometric sequence is a sequence of numbers in which the ratio between consecutive terms is constant. The Achilles paradox is an example of the difficulty that ancient Greek mathematicians had with the idea that an infinite series could produce a …Geometric progression definition is - a sequence (such as 1, 1/2, 1/4) in which the ratio of a term to its predecessor is always the same —called also geometrical progression, geometric sequence. So I’ll not go into much detail. Evaluate: a. To find the n-th term, I can just plug into the formula an = ar(n – 1): To find the value of the tenth term, I can plug n = 10 into the n-th term formula and simplify: Then my answer is:Crank out the common ratio, first term, and last term of the sequence. khanacademy. Solution We can obtain the common ratio by dividing any term after the first by the term The geometric sequence is sometimes called the geometric progression or GP, for short. The formulas applied by this geometric sequence calculator are detailed below while the following conventions are assumed: - the first number of the geometric progression is a; - the step/common ratio is r;An Geometric Sequence describes something that is periodically growing in an exponential fashion (by the same percentage each time), and a Geometric Series describes the sum of those periodic values. Indeed we can observe that every term in the sequence is found by multiplying the previous term by 3. 5KGeometric Examples | STAT 414 / 415https://newonlinecourses. EXAMPLE 1 Finding the nth term Write a formula for the nth term of the geometric sequence 6, 2, 2 3, 2 9, . Examples: d = the common difference . As a function of q, this is the Riemann zeta function ζ(q). phpExample problems that can be solved with this calculator. The lesson includes three examples for the teacher to use and a two page worksheet for students. org/math/precalculus/seq_induction CHAPTER 1. Geometric Mean Formula for Investments Geometric Mean = [Product of (1 + Rn)] ^ (1/n) -1 Where: Rn = growth rate for year N Using the same example as we did for the arithmetic mean, the geometric mean calculation equals: Identify the Sequence This is a geometric sequence since there is a common ratio between each term . Consider a stock that grows by 10% in year one, declines by 20% in year two and then grows by 30% in year three. 911 sequences word problems. f. q j=1. 01) Month 4 a = 100 Month 2 a = 100 n-1 Why this scenario? 1 I chose this scenario because this is something I deal with in my day to day life. shmoop. An introduction to geometric sequences Practice this lesson yourself on KhanAcademy. " Geometric sequences contain a pattern where a fixed amount is multiplied from one term to the next ( common ratio r ) after the first term Geometric sequence examples : Geometric Sequences / Progressions. 2 Arithmetic and Geometric Sequences Investigate! 18 For the patterns of dots below, draw the next pattern in the sequence. Each term after the first term is ½ of the preceding term. However, notice that both parts of the series term are numbers raised to a power. 7. When writing the general expression for a geometric sequence, you will not actually find a value for this. 1. 17) a 1 = −4, r = 6 18) a 1 Geometric sequence Summing a Geometric Sequence. In the following series, the numerators are in AP and the denominators are in GP: Sequences & Series - General terms. Example problems that can be solved with this calculator. The geometric sequence is sometimes called the geometric progression or GP, for short. Each number of the sequence is given by multipling the previous one for the common ratio. This algebra lesson explains geometric sequences. It's not a geometric sequence, but it is a sequence. In this lesson, it is assumed that you know what an arithmetic sequence is and can find a common difference. us. Example 2: Find the common ratio if the fourth term in geometric series is \frac {4} {3} and the seventh term is \frac {64} {243}. For example, the sequence 2, 6, 18, 54, is a geometric progression with Introduces arithmetic and geometric sequences, and demonstrates how to solve basic exercises. -1-Given the explicit formula for an arithmetic sequence find the first five terms. For example: 1, 2, 4, 8, 16, 32, is a geometric sequence because each term is twice the previous term. An arithmetic sequence is a sequence of numbers in which each term is given by adding a fixed value to the previous term. For example, suppose the common ratio is …Geometric Sequences - nth Term. In variables, it looks like In variables, it looks like Calculating the Infinite Geometric Series Example Suppose that a runner begins on a one mile track. geometric sequence examples the common ratio r = or . of a geometric random variable with p = 0. A geometric sequence can be defined recursively by the formulas a 1 = c, a n+1 = ra n, where c is a constant and r is the common ratio. We can put this expression in the form of a geometric series: Explore Solution 4. A formula for the n th term of the sequence is. If so, give the value of the common ratio, r. Geometric sequence example convergent geometric series the sum of an infinite geometric recursive formulas for geometric sequences practice khan academy solved given the first three terms of a geometric sequence fi geometric sequence formula examples video lesson transcript ppt section 57 arithmetic and geometric sequences powerpoint 9 Introduction. a = (a ) (1. The geometric mean is similar to the arithmetic mean. Well, our website offers hundreds of free examples of geometric sequences. Example 4: Full Answer. Given the first term and the common ratio of a geometric sequence find the first five terms and the explicit formula. EXAMPLE 1 common ratio geometric sequence, GOAL 1 Write rules for geometric sequences and find sums of geometric series. A geometric sequence is a sequence where the next term is found by multiplying the previous term by a number. an = 131,072. One of the most important typesof infinite series are geometric series. A Geometric Series results when a geometric sequence is expressed as a sum. The formulas applied by this geometric sequence calculator are detailed below while the following conventions are assumed: - the first number of the geometric progression is a; - the step/common ratio is r; EX: A sequence in which the ratio of any term to the term before it is a constant is a geometric sequence. 14 illustrates an important point when evaluating a geometric series whose beginning index is other than zero. 6. It is found by taking any term in the sequence and dividing it by its preceding term. A great Brochure design is an ultimate rescue in these situations. 1024 There is about a 10% chance that the marketing representative would have to select 4 peopleab is called their geometric average. Sequences and Series Worked Examples. ¡1=3 1 =. To get the next term, we multiply the term we're on by r = 2. 1)-4, 1, 6, 11, … 2) 2 The numerator is the same arithmetic sequence that we have encountered in Examples 1 & 4 that has a general term of a n = 3n - 1. In this case, multiplying the previous term in the sequence by gives the next term . What is the 51st term? Box 4 Geometric series formula is given by Here a will be the first term and r is the common ratio for all the terms, n is number of terms. Apr 08, 2010 · An introduction to geometric sequences Practice this lesson yourself on KhanAcademy. is a geometric sequence with r = 3. org right now: https://www. 25 + 0. Example 3: Find the next three terms in . In this lesson, students learn to work flexibly with explicit and recursive expressions of a geometric sequence. is an arithmetic sequence with the common difference 2. the common ratio r = or . We could say that its set or it's the sequence a sub n from n equals 1 to infinity with a sub n being equal to, let's see the fourth one is essentially 4 factorial times a. The geometric sequence is sometimes called the geometric progression or GP, for short. ARITHMETIC AND GEOMETRIC SEQUENCE WORD PROBLEM EXAMPLES. mathportal. CHAPTER 1. If a formula is provided, terms of the sequence are calculated by substituting #n=0,1,2,3,# into the formula. Example 7: Solving Application Problems with Geometric Sequences. 6=3 = 12=6 = 24=12 = ¢¢¢ = 2: 2. It also explores particular types of sequence known as arithmetic progressions (APs) and geometric progressions (GPs), and the corresponding series. Finite Geometric Series. Definition. gov These arithmetic sequence examples are been designed and crafted with sophisticated looks and style that makes them suitable for regular usage by pupils, researchers and even teachers in this domain. 5 + 1. Work out the nth term of the sequence 2, 5, 10, 17, 26, Work out the first differences between the terms. How to find Arithmetic and Geometric Series 13 Surefire Examples! Facebook Tweet Pin Shares 118 As we’ve already seen, using Summation Notation, also called Series Notation, enables us to add up the terms of a sequence by creating Partial Sums. Geometric Sequences Examples What's the common ratio for the sequence ? List the first four terms and the 10th term of a geometric sequence with a first The recursive formula for a geometric sequence is written in the form For example, when writing the general explicit formula, n is the variable and does not Example: 1, 2, 4, 8, 16, 32, 64, 128, 256, This sequence has a factor of 2 between each number. Consider the following geometric sequence: To find the ratio , we have to divide on term by its previous term: . That's our total number of terms. For a geometric series to be convergent, its common ratio must be between -1 and +1, which it is, and so our infinite series is convergent. (1) For a geometric sequence, a formula for thenth term of the sequence is a n 5 a · rn21. Important examples Geometric sequence. The sides of the cuboid make a geometric progression. Improve your skills with free problems in 'Solving Word Problems Using Geometric Series' and thousands of other practice lessons. Concept 16 Arithmetic & Geometric Sequences Practice #3 Practice #4 Draw the next term if this represents an arithmetic sequence. Quiz & Worksheet - Geometric Sequences Quiz; Print Geometric Sequence: Formula & Examples Worksheet 1. Geometric Series ExamplesThe sequence is able to be found by using the geometric sequence formula. 1 Arithmetic and Geometric Sequences Definitions: (yes, that's right, this is important, know these!) A sequence is a set of numbers, called terms, arranged in some particular order. Don’t assume that if the terms in the sequence are all negative numbers, it is a decreasing sequence. Geometric Sequences - nth Term Examples, solutions, videos, worksheets, games and activities to help Algebra II students learn about how to find the nth term of a geometric sequence. pattern, sequence (earlier grades) geometric sequence, arithmetic sequence, common ratio, common difference (7. (Optional) Provide students with the formulas for working with arithmetic and geometric sequences and series, but give them neither explanations of when to use the formulas nor the meanings of the symbols. Arithmetic sequences have this same special property: equal changes in the input (e. Geometric Sequence. For convenience, at times we use …Example (continued) Solution. common ratio (r). 1)*(1-0. For example, -2, 1, 4, 7, 10, is an arithmetic sequence because each term is three more than the previous term. The proof is similar to the one used for real series, and we leave it for you to do. Slide 4 and 5. This means that the first number of the sequence, #a_0#, is 2. Factorial. ) 1,2,3,4,5,6,7 are all seperated by + 1 ~> Arithmetic sequence), is the geometric average of the preceding term and the following term. Geometric Series Examples. A geometric series is the sum of the terms in a geometric sequence. May 30, 2014 · Geometric Sequences - Examples slcmath@pc. To write the explicit or closed form of a geometric sequence, we use Introduction to the geometric distribution. An easy way to remember this theorem is geometric series first term 1 ratio between terms . b) Find the equation for the general term. Remember, it is decreasing whenever the common difference is negative. )So working backwards from 1 you would go to 1/3, 1/9, 1/27, etc. So clearly this is a geometric sequence with common ratio r = 2, and the first term is a = . A geometric sequence is a sequence in which each term is found by multiplying the preceding term by the same value. Example of a geometric progression calculation. 3))^(1/3) Engaging math & science practice! Improve your skills with free problems in 'Solving Word Problems Using Geometric Series' and thousands of other practice lessons. Consider the sequence {12, -6, 3, 2 3 −,…} a) Show that the sequence is geometric. Let’s take an example of a geometric progression having first number a= 2, r = 3 for which we try to figure out which is the 10 th number in the sequence: ■ The 10 th value of the sequence (a 10) is 39,366 ■ Sample of the first ten numbers in the geometric sequence: 2; 6; 18; 54; 162; 486; 1,458; 4,374; 13,122; 39,366. An arithmetic sequence (arithmetic progression) is defined as a sequence of numbers with a constant difference between each consecutive term. The next one, #a_1#, will be #2 \times 3=6# | 2019-03-22T23:13:44 | {
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https://math.stackexchange.com/questions/1981186/solve-the-system-of-inequalities | # Solve the system of inequalities
Solve the system: $$|3x+2|\geq4|x-1|$$
$$\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$$
So for $|3x+2|\geq4|x-1|$ I got the solution $x=6$ and $x=2/7$ and Wolframalpha agrees with me. I'm having troubles writing the final solution for $\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$.
$1)$ $x^{2}+x-2\leq0$ and $-2x^{2}+3x+2>0$
$x^{2}+x-2\leq0$ $\Rightarrow x\in [-2,1]$
$-2x^{2}+3x+2>0$ $\Rightarrow x\in (-1/2,2)$
Now I need the intersection of those two sets, which is $(-1/2,1]$. From the first inequality I got $x=6$ and $x=2/7$ so the final solution is $x=2/7$.
$2)$ $x^{2}+x-2\geq 0$ and $-2x^{2}+3x+2<0$
$x^{2}+x-2\geq 0$ $\Rightarrow x\in(-\infty, -2]\cup[1,\infty]$
$-2x^{2}+3x+2<0$ $\Rightarrow x\in(-\infty, -1/2)\cup(2,\infty)$
Now I'm having troubles finding the intersection of those sets.
• Formatting tip: open interval endpoints are () and not <>. – Parcly Taxel Oct 23 '16 at 10:44
• Just one more question, how would you solve this by not studying the regions, in other words, how would you find the intersection of $((-\infty, -2]\cup[1,\infty]$) and ($(-\infty, -1/2)\cup(2,\infty)$)? – lmc Oct 23 '16 at 13:16
If $x\ge 1$ we have $$|3x+2|\geq4|x-1|\iff 3x+2\geq 4x-4 \iff x\le 6.$$ So $[1,6]$ is solution of the first inequality. Now, if $-2/3\le x\le 1$ we have $$|3x+2|\geq4|x-1|\iff 3x+2\geq 4-4x \iff x\ge 2/7.$$ So $[2/7,1]$ is solution of the system. Finally, if $x\le -2/3$ then $$|3x+2|\geq4|x-1|\iff -3x-2\geq 4-4x \iff x\ge 6,$$ which is impossible. That is, the solution set of the first inequality is $[2/7,6].$
(The idea is to solve $3x+2=0$ and $x-1=0$ and study the inequality on each region you obtain.)
Now, we will work with the second inequality. Write it as
$$\dfrac{(x-1)(x+2)}{2(x-2)(x+1/2)}\ge 0.$$ Study the sign on $(-\infty,-2),$ $(-2,-1/2),$ $(-1/2,1),$ $(1/2,1)$ and $(1,\infty).$ You should obtain that the set solution is $(-\infty,-2]\cup (-1/2,1]\cup (2,\infty).$
(The idea is to solve $x^2+x-2=0$ and $2+3x-2x^2=0$ and study the inequality on each region you obtain. Note that we can't divide by $0.$ So $x\ne-1/2$ and $x\ne 2.$)
Finally, one gets the intersection to obtain $(2/7,1]\cup (2,6].$
• Thanks, I appreciate your help! – lmc Oct 23 '16 at 13:08
• @mfl: I think you have to edit your answer, dear friend, discarding the point $x=1$. Or am I wrong?. Regards. – Piquito Oct 23 '16 at 13:22
• @Piquito Why do you want to discard $x=1?$ It is $|3\cdot 1+2|=5\ge 0=4 |1-1|$ and $\dfrac{1^2+1-2}{2+3\cdot 1 -2 \cdot 1^2}=\dfrac{0}{3}=0\le 0.$ – mfl Oct 23 '16 at 16:18
Consider $$f(x)=\frac{x^{2}+x-2}{2+3x-2x^{2}}=\frac{(x-1)(x+2)}{-(2x+1)(x-2)}$$ It follows $$f(x)=\frac{-(x-1)(x+2)}{(2x+1)(x-2)}\le0\iff\frac{(x-1)(x+2)}{(2x+1)(x-2)}\ge0$$ One has at once the solution set for $f(x)\le 0$ is $$S_1=(-\infty,-2]\cup(-\frac 12,1]\cup(2,\infty)$$ Now consider $$g(x)=\left|\frac{3x+2}{x-1}\right|=\left|\frac{3+\frac 2x}{1-\frac 1x}\right|$$ it follows $$\begin{cases}\lim_{x\to\pm\infty}=3\\\lim_{x\to-1}g(x)=\infty\\g\text { decreasing on } (-\infty,-\frac 23)\cup(-1,\infty)\\g\text { increasing on } (-\frac 23,1)\end{cases}$$ Furthermore $$\left|\frac{3x+2}{x-1}\right|=4\iff\begin{cases}3x+2=4(x-1)\iff x=6\\3x+2=-4(x-1)\iff x=\frac 27\end{cases}$$ Hence the solution set for $g(x)\ge 4$ is $$S_2=[\frac 27,1)\cup(1,6]$$ Thus our solution is $$S_1\cap S_2=\left((-\infty,-2]\cup(-\frac 12,1]\cup(2,\infty)\right)\cap\left([\frac 27,1)\cup(1,6]\right)=\color{red}{(\frac 27,1)\cup (2,6]}$$ | 2020-02-25T03:44:38 | {
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https://math.stackexchange.com/questions/2800782/finding-the-probability-that-an-ace-is-found-in-every-pile-when-a-deck-of-cards/2800791 | # Finding the probability that an ace is found in every pile when a deck of cards is split into 4
I'm trying to answer this question and you are supposed to use the multiplication rule to solve it:
A deck of 52 playing cards is randomly divided into four piles of 13 cards each. Compute the probability that each pile has exactly 1 ace.
I started off by defining following 4 events: $A_{1}, A_{2}, A_{3}$ and $A_{4}$ where $A_{i}$ denotes the event that exactly one ace is found in the i$^{th}$ pile - so to find the probability I need to find the probability of the intersection of all these events which is where I can use the multiplication rule.
The multiplication rule says that $$\mathbb{P}(A_{1} \cap A_{2} \cap A_{3} \cap A_{4})= \mathbb{P}(A_{1}) \mathbb{P}(A_{2}|A_{1}) \mathbb{P}(A_{3}|A_{2} \cap A_{1}) \mathbb{P}(A_{4}|A_{3} \cap A_{2} \cap A_{1})$$
to find each of the probabilities on the RHS I looked compared the possible combinations allowed for each situation:
The number of possible card combinations such that $A_{1}$ holds is $\binom{48}{12}$ and the total number of possible card combinations for the first pile is $\binom{52}{13}$. It then follows that $$\mathbb{P}(A_{1})= \frac{\binom{48}{12}}{\binom{52}{13}}=\frac{1406}{4165}$$
When moving on to the second pile, it follows that we now have 39 cards remaining so in order for the pile 2 to have exactly one ace, it leads to $\binom{36}{12}$ possible combinations out of the $\binom{39}{13}$ total number of combinations and so we get that $$\mathbb{P}(A_{2}|A_{1}) = \frac{\binom{36}{12}} {\binom{39}{13}} =\frac{225}{703}$$.
Continuing on in this way I got that
$$\mathbb{P}(A_{3}|A_{2} \cap A_{1}) = \frac{\binom{24}{12}}{\binom{26}{13}}=\frac{13}{50}$$
and by the way I have defined my events, it means that $$\mathbb{P}(A_{4}|A_{3} \cap A_{2} \cap A_{1}) = 1$$
so by the multiplication rule I get that $$\mathbb{P}(A_{1} \cap A_{2} \cap A_{3} \cap A_{4}) = \frac{1406}{4165} \frac{225}{703} \frac{13}{50} \approx 0.0281$$
However the answer I am given says it should be $\approx 0.105$. Can anyone help me to see where I have gone wrong? Would it perhaps be that defining the events differently lead to different probabilities? Thanks!
• In the first your denominator should be $52 \choose 13$. You have missed multiplying by the number of aces available at each stage, which gives a factor $24$ – Ross Millikan May 29 '18 at 16:47
• Thanks for pointing that out! i've changed it now – BigWig May 29 '18 at 17:44
First, a small typo:
$$\mathbb{P}(A_{1})= \frac{\binom{48}{12}}{\binom{48}{12}}=\frac{1406}{4165}$$
I assume you meant:
$$\mathbb{P}(A_{1})= \frac{\binom{48}{12}}{\binom{52}{13}}$$
However, when I calculate that, I get
$$\frac{\binom{48}{12}}{\binom{52}{13}} = \frac{9139}{83300}$$
More importantly, since there are $4$ aces to choose from for pile $1$, it really should be:
$$\mathbb{P}(A_{1})= \frac{4 \cdot \binom{48}{12}}{\binom{52}{13}}=\frac{9139}{20825}$$
Likewise, something went wrong with your calculation here:
$$\mathbb{P}(A_{2}|A_{1}) = \frac{\binom{36}{12}} {\binom{39}{13}} =\frac{225}{703}$$
When I calculate that, I get
$$\frac{\binom{36}{12}} {\binom{39}{13}} = \frac{325}{2109}$$
But again, more importantly, since there are $3$ aces left to choose from, it really should be:
$$\mathbb{P}(A_{2}|A_{1})=\frac{3\cdot \binom{36}{12}} {\binom{39}{13}} =\frac{325}{703}$$
And likewise, since there are $2$ aces left for pile $3$, it should be:
$$\mathbb{P}(A_{3}|A_{2} \cap A_{1}) = \frac{2 \cdot \binom{24}{12}}{\binom{26}{13}}=\frac{13}{25}$$
And so, we get:
$$\mathbb{P}(A_{1} \cap A_{2} \cap A_{3} \cap A_{4}) = \frac{9139}{20825} \frac{325}{703} \frac{13}{25} \approx 0.1055$$
As desired. So, you did the basic method largely correct, but you did some sloppy calculations, and more importantly, you forgot to take into account that for the first few piles you have a choice of aces.
ETA: I see that the use of the multiplication rule is expected. I'm still going to leave this here as an approach to use for this kind of problem, in general.
Alternate approach. Place the other $48$ cards into four piles, accounting for order; there are $48!$ ways to do this. Then assign each of the four aces to one of the piles; there are $4!$ ways to do this. Finally, each ace has to be placed into one of $13$ different positions in its respective pile; there are $13^4$ ways to do this.
Since there are $52!$ different arrangements of the deck overall, the desired probability is
$$\frac{13^4 \times 4! \times 48!}{52!} = \frac{2197}{20825} \approx 0.10550$$
• And certainly if the multiplication rule weren't required, this application of binomials to "probability of not...." is much cleaner – Carl Witthoft May 29 '18 at 19:03
• Or another way to write this (perhaps even simpler): the first ace can go anywhere in the deck, the second ace in 39 places of the remaining 51 possible locations etc. yielding a final result of $1 * 39/51 * 26/50 * 13/49$ – soktinpk May 29 '18 at 22:06
Another approach using the multiplication rule
The multiplication rule can be very simply applied imagining $52$ slots divided into $4$ piles of $13$
$\small{\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square}$
The $1^{st}$ ace can go to any slot, and given that, the $2^{nd}$ ace has $39$ out of $51$ free slots to go in,
and so on and so forth, thus simply$\quad\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$
This way, we need not bother at all as to how the other $48$ cards are distributed.
A different approach:
There are
-- $\binom{4}{1}$ ways of selecting an ace and $\binom{48}{12}$ ways of selecting other cards to the $1$st deck, while $\binom{52}{13}$ doing so in total;
-- $\binom{3}{1}$ ways of selecting an ace and $\binom{36}{12}$ ways of selecting other cards to the seond deck, while $\binom{39}{13}$ doing so in total;
-- $\binom{2}{1}$ ways of selecting an ace and $\binom{24}{12}$ ways of selecting other cards to the third deck, while $\binom{36}{13}$ doing so in total;
-- this one is easy.
Thus, we have
$$\frac{\binom{4}{1}\binom{48}{12}}{\binom{52}{13}}\times\frac{\binom{3}{1}\binom{36}{12}}{\binom{39}{13}}\times \frac{\binom{2}{1}\binom{24}{12}}{\binom{26}{13}} \approx 0.1055.$$
You have ignored the fact that there are $4$ different aces, so $P(A_1) = \dfrac{4\binom{48}{12}}{\binom{52}{13}}$, etc.
Choose $12$ non-ace cards for the first pile, $12$ non-ace cards for the second pile, $12$ non-ace cards for the third pile and the remaining $12$ non-ace cards for the fourth pile.
Place the four aces in the four piles, one in each pile.
This can be done in $${48 \choose 12}{36 \choose 12}{24 \choose 12}{12 \choose 12}\cdot 4!$$ ways.
This has to be divided with the number of all possibilities: choose $13$ cards for the first pile, $13$ for the second, $13$ for the third and the remaining $13$ for the fourth. This can be done in
$${52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}$$
ways.
The result is
$$\frac{{48 \choose 12}{36 \choose 12}{24 \choose 12}{12 \choose 12}\cdot 4!}{{52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}} = 0.1055$$ | 2019-10-14T15:25:44 | {
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https://stats.stackexchange.com/questions/348452/calculating-percentage-thats-equal-to-a-value/348474 | # Calculating percentage that's equal to a value
Let's say I have a normal distribution with people lengths in which the mean is 170 cm and the standard deviation is 6 cm.
I know you can calculate the lower tail and higher tail of 180 cm with
pnorm(180, mean=170.6, sd=6.75, lower.tail=TRUE/FALSE)
But how would I get the percentage of the people that are equal to 180 cm?
And if I would like to get the percentage of people between 170 and 180 do I just do that with 175 cm?
Sorry I'm new to this and I just can't figure it out.
Since the normal distribution is a continuous distribution, the probability of any value is zero. Thus, the probability to obtain the value 180 is infinitely small, $p(X = 180) = 0\,$.
To calculate the probability of values between 170 and 180 you have to use the cumulative distribution function of the normal distribution, $\Phi\,$. This can be done with the R function pnorm. To obtain $p(170 < X < 180)$ you have to calculate the $p(X < 180)$ and subtract $p(X < 170)\,$.
# probability for X < 180
p_180 <- pnorm(180, 170.6, 6.75)
# probability für X < 170
p_170 <- pnorm(170, 170.6, 6.75)
# probability für 170 < X < 180
p_180 - p_170
# [1] 0.4535434
Hence, $p(170 < X < 180) \approx 0.45\,$.
### A visualization of the concept
You want to obtain the size of the red area (between 170 and 180). Since the full area under the curve is equal to one, the size of the area corresponds to the probability. First, you calculate the area between $-\infty$ and 180. Second, you subtract the size of the area between $-\infty$ and 170. The overlapping area is purple in the below figure.
The R code:
library(ggplot2)
ggplot(data.frame(x = c(150, 190)), aes(x = x)) +
stat_function(xlim = c(150, 180), fun = dnorm, args = list(mean = 170.6, sd = 6.75),
geom = "area", alpha = 0.5, fill = "red") +
stat_function(xlim = c(150, 170), fun = dnorm, args = list(mean = 170.6, sd = 6.75),
geom = "area", alpha = 0.5, fill = "blue") +
coord_cartesian(xlim = c(150, 190)) +
ylab("p")
• Thanks for the answers guys it makes good sense and I feel stupid for asking the exact 180cm question. – Sinan Samet May 27 '18 at 15:59
It is not possible to be exactly 180 cm in length - you are only as accurate as your measurement device, which always has some error. Mathematically, the probability of a normally distributed variable is the area under the curve, and there's no area under a single value.
There are many ways to find $P(170<X<180)$. One way is to find $P(X<180)$ and subtract $P(X<170)$.
You might find this question and answer valuable.
On the one hand, if you want to obtain the percentage of people with height between 170 and 180 you should use:
pnorm(180, mean=170.6, sd=6.75, lower.tail=TRUE/FALSE) -
pnorm(170, mean=170.6, sd=6.75, lower.tail=TRUE/FALSE)
The area you are considering is:
library(ggplot2)
# Obtained the code from
# http://rstudio-pubs-static.s3.amazonaws.com/58753_13e35d9c089d4f55b176057235778679.html
# Return dnorm(x) for 170 < x < 180, and NA for all other x
dnorm_limit <- function(x) {
y <- dnorm(x,mean=170.6,sd=6.75)
y[x < 170 | x >180] <- NA
return(y)
}
# ggplot() with dummy data
p <- ggplot(data.frame(x=c(120,220)), aes(x=x))
p + stat_function(fun=dnorm_limit, geom="area", fill="blue", alpha=0.2) +
stat_function(fun=dnorm,args=list(mean=170.6,sd=6.75))
On the other hand, the probability of someone being exactly 180 is:
pnorm(180, mean=170.6, sd=6.75, lower.tail=TRUE/FALSE) -
pnorm(180, mean=170.6, sd=6.75, lower.tail=TRUE/FALSE)
The obtained probability is 0. This happens because the normal distribution is a continuous probability distribution.
Don't hesitate to ask if you have further doubts. | 2020-06-04T15:40:17 | {
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https://math.stackexchange.com/questions/608592/an-interesting-integral-i-int-limits-11-arctanexdx | # An interesting integral $I = \int\limits_{-1}^{1} \arctan(e^x)dx$
I solved this interesting integral online: $$I = \int\limits_{-1}^{1} \arctan(e^x)dx$$ Now I tried the substitution $u=e^x$ but it lead me nowhere. I was looking at the following post which was solved in a beautiful way Integrate $\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$. From there I found this very interesting article http://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Nelsen39-41.pdf which has the integral I posted at the end as a question to the reader.
Looking at the graph of $\arctan(e^x)dx$ on the interval $-1 \leq x \leq 1$ I conjectured that $I=\frac{\pi}{2}$. I used the following method to prove it: \begin{eqnarray} -e^{-x}&=&\frac{-1}{e^x}\\&=& \frac{-1}{\tan\{ \arctan(e^x) \} }\\&=&-\cot\{ \arctan(e^x) \}\\ &=& \tan \left\{\arctan(e^x)-\frac{\pi}{2} \right\}\\ \end{eqnarray} For the last equality I used the fact that $\cot(\theta) =-\tan\left(\theta -\frac{\pi}{2}\right)$. Now we take the arctan of both sides to obtain: $$\arctan(-e^{-x}) = \arctan(e^x)-\frac{\pi}{2}$$
Finally I use the fact that $\arctan(-\theta)=-\arctan(\theta)$ and add $\frac{\pi}{4}$ to both sides of the last equation to obtain: $$-\arctan(e^{-x}) +\frac{\pi}{4}= \arctan(e^x)-\frac{\pi}{4}$$ So it is established that the function $f(x) = \arctan(e^x)-\frac{\pi}{4}$ is an odd function. Thus $$I_2 = \int\limits_{-1}^{1} \left[ \arctan(e^x)-\frac{\pi}{4} \right]dx = 0$$ Now $$I = \int\limits_{-1}^{1} \arctan(e^x)dx = I_2 + \int\limits_{-1}^{1} \frac{\pi}{4}dx =\frac{\pi}{2}$$ I thought this integral was really interesting and I was wondering if anyone else has any different ways of solving it, possibly with a clever substitution. I was especially amazed at how easily it could be solved because integrals with arctan usually give me a lot of trouble.
Also, I think we can extend this to a broader result where we replace $x$ by any arbitrary odd function $g(x)$ and show that $$I = \int\limits_{-a}^{a} \arctan(e^{g(x)})dx = \frac{a \pi}{2}$$ for any odd function $f(x): (-a,a) \to\Bbb R$. Essentially the proof for this would follow the exact same reason as above right?
So if anyone has another method of computing the original integral I am definitely interested in reading your solutions! Thanks in advance for any input and ideas! Also thanks to Ron Gordon for his nice answer on the question I linked, the answer given there inspired me to look for different ways of trying to solve this integral that I normally would have given up on.
$g(x)$ need to be an odd function, otherwise $f(x)=\arctan(e^{g(x)})-\pi/4$ is not an odd function.
One simple test is at least $f(x=0)$ should be $0$, which is not satisfied by your choice of $g(x)=1$.
• Yes you are right! Thanks, I see that it is necessary now. I am still interested in anyone having another derivation for the original integral $I$ though! – Slugger Dec 16 '13 at 3:08
• As far as I know, without that "odd function" technique, there is no simple ways to calculate this. The [indefinite integral][1] is really messy. [1]: wolframalpha.com/input/?i=integrate%5Barctan%5Be%5Ex%5D%2Cx%5D – MoonKnight Dec 16 '13 at 3:29
was working on something similar
integration by parts, noticing the initial integral is related to a form of the integral of sech
$I = \int\limits_{-1}^{1} \arctan(e^x)dx$
$= [x \arctan(e^x)] - \int\limits_{-1}^{1} x e^x / (1 + e^{2x})dx$
$= [x \arctan(e^x)] - \int\limits_{-1}^{1} x / (e^{-x} + e^{x})dx$
$= [x \arctan(e^x)] - (1/2)\int\limits_{-1}^{1} x sech(x)dx$
is then odd i think and equals $[x \arctan(e^x)]$
I got Nelsen's answer of pi/2, not sure why the x arctan part works out as that yet, but numerically it did
• actually you proved that last part, it was what I was looking for in the first place thanks! – Cato Mar 20 '17 at 18:16
I have found another way to integrate this integral,first you have to know this identity, very useful for integrals with arctan $$\arctan(y)+arctan(1/y) =\frac{\pi}{2}\qquad (1)$$ with x positive or equal to 0, and $$\arctan(y)+arctan(1/y) =\frac{-\pi}{2}$$ for y negative. The first step is separate the integral
$$\int\limits_{-1}^{1} \arctan(e^x)dx=\int\limits_{0}^{1} \arctan(e^x)dx+\int\limits_{-1}^{0} \arctan(e^x)dx\qquad (2)$$ after that in the seconnd integral use this substitution (x=-y) and we get
$$\int\limits_{-1}^{0} \arctan(e^x)dx=\int\limits_{0}^{1} \arctan(e^-y)$$ then wecan use this and put it in (1) (and replace x=y because is the same what variable we use) and we get $$\int\limits_{0}^{1} \arctan(e^x)dx+\int\limits_{0}^{1} \arctan(e^-x)dx$$ after we use (2) and we get $$\int\limits_{0}^{1} \frac{\pi}{2}dx$$=\frac{\pi}{2} Then i tryed to generalize and I resolved the same first integral with x=x^3, if you want you can try to prove it. Finally we can generalize for g(x) if this function is odd
$$\int\limits_{-1}^{1} \arctan(e^{g(x)})dx=\int\limits_{0}^{1} \arctan(e^{g(x)})dx+\int\limits_{-1}^{0} \arctan(e^{g(x)})dx$$ we subtitue (x=-y) in the second integral and omitting steps we get $$\int\limits_{-1}^{1} \arctan(e^{g(x)})dx=\int\limits_{0}^{1} \arctan(e^{g(x)}-e^{g(-x)})dx=\int\limits_{0}^{1} \frac{\pi}{2}dx=\frac{\pi}{2}$$
for a and -a in the limits we can use the same method, if you can´t do it i will explain it,thank you for reading and goodbye. | 2021-03-04T15:12:45 | {
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https://math.stackexchange.com/questions/2212874/strong-induction-to-find-all-possible-combinations-of-two-numbers | # Strong Induction to find all possible combinations of two numbers
Full disclosure, this is a homework question, so I'm only looking for hints not full solutions please.
There is a store which offers two denominations of gift certificates, \$25 and \$40. Determine the possible total amounts that can be formed using the certificates, using strong induction to prove your answer.
My first approach was to write out something like this, let $S$ be a possible amount where $S = 25a + 40b$ and $a, b \in \mathbb{Z}^+$ ($0$ being included in the positive integers). The problem is that I'm not sure how to use this as a propositional statement for strong induction.
Second, I tried to make a table to find a pattern, like this: $$\begin{array}{c|ccccc} 0 & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 0 & 25 & 50 & 75 & 100 \\ 1 & 40 & 65 & 90 & 115 & 140 \\ 2 & 80 & 105 & 130 & 155 & 180 \\ 3 & 120 & 145 & 170 & 195 & 220 \\ 4 & 160 & 185 & 210 & 235 & 260 \\ \end{array}$$
The issue with this approach was that, once again, I could see no particular path towards using strong induction (or any discernible pattern).
I'm not sure how best to move forward, and I haven't been able to find many helpful resources. I'd be very grateful for some help, thanks!
• You'll prove something like: If we can reach $S>N$, then we can reach $S+n$, where $N$ and $n$ are constants. I could post a more detailed outline of a possible proof if you want, but maybe this is enough. – Mastrem Apr 1 '17 at 11:33
Since you are only looking for a hint:
HINT Expand your table by just two more columns, and then look at your table and start writing the possible total amounts in order. So you get 25, 40, 50, 65, 75, 80, .... After a while you will see a pattern ... Use this to form your hypothesis and prove by strong induction.
First, let us "scale down" the problem by a factor of $5$, i.e., that we are working with $5$ and $8$ gift certificates. We will "scale up" our solution afterwards.
Notice that $5$ and $8$ are coprime and thus their non-negative integer linear combinations form a numerical semigroup with two generators. A nice result of numerical semigroups with two generators $m,n$ is that any number bigger than $mn-m-n$ is in the semigroup.
To show this by induction for this particular problem, notice that $mn-m-n = 8*5-8-5 = 27$ is the largest number that can't be written as a positive integer linear combination of $5$ and $8$. Thus, the next $5$ numbers ($28,29,30,31,32$) can be written as a linear combination of $5$ and $8$, this will form your base case. See if you can explicitly find the linear combinations of $5$ and $8$ that will give you those five numbers. Now, consider $n>32$ and suppose that $k$ can be written as a positive integer linear combination of $5$ and $8$ for all $28 \leq k < n$. We want to show the same holds for $n$. Notice that $28 \leq n-5 < n$ and thus $n-5 = 5a+8b$ for $a,b\in \mathbb{Z}^+$. From this, we see that $n= 5(a+1)+8b$ and thus we can write $n$ as a positive integer linear combination of $5$ and $8$, thus completing the induction.
We still need to check the values less than $28$ which we can do by brute force. We see that we can get the following with positive integer linear combinations of $5$ and $8$ for numbers less than $28$, $$\{ 0,5,8,10,15,16,18,23,24,25,26\}.$$
Scaling our solution up by $5$, we can obtain the following amounts: $$\{0,25,40,50,75,80,90,115,120,125,130\} \cup \{140, 145,150,\dots\}$$
First off, if: $$S=40a+50b$$ then: $$S+10=40(a-1)+50(b+1)$$ but only if $a\ge1$. If $a=0$ and $b\ge 3$, then: $$S+10=40\cdot4+50(b-3)$$ So, if $S\ge150$, we can form $S+10$ if we can form $S$. Now, something similair can be done for $S+5$. After you've got that figured out, use induction to prove you can reach all multiples of $5$ greater than a certain number (what number?) and prove that you can only reach multiples of $5$ (why?)
Then you're left with the small cases, which you'll need to check by hand | 2019-08-24T04:36:24 | {
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https://math.stackexchange.com/questions/2386351/why-does-a-n-frac-a-n-1-frac-2a-n-12-converge-to-an-irratio | # Why does $a_n = \frac {a_{n-1} + \frac {2}{a_{n-1}}}{2}$ converge to an irrational number?
There is a problem in my textbook that goes like this
$$a_n = \frac {a_{n-1} + \frac {2}{a_{n-1}}}{2}$$
and
$$a_0 =1$$
for all $n\ge1$.
It is monotonically decreasing sequence of rational numbers and bounded below. However, it cannot converge to a rational number.
Then the task is to find the limit. The problem itself is easy but I don't understand how the author judged the limit to be irrational even before solving the question? Is there any property or did they just know the answer beforehand?
• I'm sure the author worked out the limit before setting the question.... – Lord Shark the Unknown Aug 8 '17 at 6:26
• Assume $a_n=a_{n-1}=p/q$, and you'll find $p^2/q^2=2$. This tells you that the limit isn't rational even without recognising Newton's algorithm. – Toffomat Aug 8 '17 at 9:29
Yep, the author knew the answer beforehand, I would say much longer before.
Because this is just an instance of the Babylonian method of computing square roots, later discovered to be a particular case of Newton's iterative method for the resolution of nonlinear equations.
So with closed eyes, this sequence converges to $\sqrt2$. You can easily verify it be assuming convergence, so that $a_{n-1}$ and $a_n$ become indiscernible, and
$$a=\frac{a+\dfrac2a}{2}$$ or $$a^2=2.$$ As the initial value is $1$, all terms are positive and convergence is to the positive root (if there is convergence, though).
There is a simple way to explain that method, also known as Heron's formula.
Let $s$ be the number of which you want to extract the root, and let $a$ be an approximation by default. Then $$a<\sqrt s\implies a':=\frac sa>\sqrt s$$ so that $\dfrac sa$ is another approximation, by excess. Now if we take the arithmetic mean, we get a new approximation which is closer than the worse of the two,
$$a''=\frac{a+a'}2=\frac{a+\dfrac sa}{2}.$$
As can be shown, when you are close to the root, the sequence converges extremely rapidly.
For example,
$$a=\color{green}{1.41}\implies a'=\color{green}{1.41}84397163121\cdots\implies a''=\color{green}{1.41421}9858156\cdots$$ while the true value is $$\sqrt2=1.4142135623731\cdots$$
The next iteration gives $11$ exact digits.
A note on convergence, for the skepticals (the method has been in use for at least two millenia).
Let $$x_n:=\frac{a_n}{\sqrt s}.$$
We have
$$\frac{x_{n+1}-1}{x_{n+1}+1}=\frac{x_n+\dfrac1{x_n}-2}{x_n+\dfrac1{x_n}+2}=\frac{(x_n-1)^2}{(x_n+1)^2}$$ and by induction
$$\frac{x_n-1}{x_n+1}=\left(\frac{x_0-1}{x_0+1}\right)^{2^n}.$$
This is an exact formula for the $n^{th}$ iterate, which proves convergence from any $x_0$ such that
$$\left|\frac{x_0-1}{x_0+1}\right|<1.$$
This holds for all positive $x_0$.
In passing, this also proves quadratic convergence, i.e. the relative error is squared on every iteration.
• This is very nice but it doesn't seem obvious that the approximations converge. While the average is better than the worse of the two approximations it might be much worse than the better of the two approximations so you might actually be diverging if you choose the average as your new approximation. Or am I missing something? – DRF Aug 8 '17 at 8:06
• @DRF: the OP claimed convergence, so I didn't address this issue. But as the average is better than the worse estimate, the worse estimate always improves and divergence is not possible. – Yves Daoust Aug 8 '17 at 8:11
• @YvesDaoust You only have one estimate when approaching the problem as the OP does. More over in your case you start with one estimate and get two. How do you continue from there? Choose the one you didn't choose last? Then you might be choosing the worse of the two and end up with three. I suppose this isn't really pertinent to the OP's question but the way you wrote the post seems confusing. – DRF Aug 8 '17 at 8:17
• @YvesDaoust I apologize I didn't intend to provide comments in bad faith. I sincerely didn't see why the algorithm should have to converge always when given as the OP described it, and I have never seen this specific algorithm. (I suppose it might not anyway the initial approximation needs to be good enough). Thank you for the extra explanation which I think works very nicely. (oh and a +1) – DRF Aug 8 '17 at 9:24
• @DRF: I discarded my harsh comment :) – Yves Daoust Aug 8 '17 at 9:55
Suppose $f(x)=x^2-2$ so roots are $\pm \sqrt2$ now use Newton method $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ so you will have $$x_{n+1}=x_n-\frac{x_n^2-2}{2x_n}\\=\frac{\frac{2x_n^2-x_n^2+2}{x_n}}{2}\\=\frac{x_n+\frac{2}{x_n}}{2}$$ now take $x_n \to a_n$ so $$a_n = \frac {a_{n-1} + \frac {2}{a_{n-1}}}{2}$$ and note $a_n$ tends to $\sqrt 2 ,if \space a_1>0$ , tends to $-\sqrt 2 ,if \space a_1<0$ by iteration.
$\bf remark:$ there are some equation like this which work with rational numbers and finally get irrational ... for example :$\sum _{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}\\\text{sum of rationals = irrational}$
Rewrite $$a_n = \frac {a_{n-1} + \frac {2}{a_{n-1}}}{2}$$ as $$a_n=a_{n-1}-\frac 12a_{n-1}+\frac 1{a_{n-1}}=a_{n-1}-\frac{a^2_{n-1}-2 }{2a_{n-1}}$$ and recognize the Newton formula for the solution of $x^2-2=0$.
Because if $a$ is the limit then $a^2=2$, which gives $a=\sqrt2$, which is an irrational number.
$$a_{n}-\sqrt2=\frac{(a_{n-1}-\sqrt2)^2}{2a_{n-1}}>0$$ and $$a_n-a_{n-1}=\frac{1}{a_{n-1}}-\frac{a_{n-1}}{2}=\frac{2-a_{n-1}^2}{2a_{n-1}}<0,$$ which gives that our sequence converges. | 2019-07-16T10:45:30 | {
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http://mathhelpforum.com/statistics/111389-balls-urn.html | # Math Help - Balls in an Urn
1. ## Balls in an Urn
Twenty balls are placed in an urn. Five are red, five green, five yellow and five blue. Three balls are drawn from the urn at random without replacement. Write down expressions for
the probabilities of the following events. (You need not calculate their numerical values.)
a) Exactly one of the balls drawn is red.
b) The three balls drawn have different colours.
c) The number of blue balls drawn is strictly greater than the number of yellow balls
drawn.
2. Hello, Aquafina!
Twenty balls are placed in an urn: 5 red, 5 green, 5 yellow, 5 blue.
Three balls are drawn from the urn at random without replacement.
There are: . ${20\choose3} \,=\,1140$ possible outcomes.
Write expressions for the probabilities of the following events.
(You need not calculate their numerical values.)
a) Exactly one of the balls drawn is red.
There are: 5 Reds and 15 Others.
We want 1 Red and 2 Others. .There are: . ${5\choose1}{15\choose2} \:=\:525$ ways.
. . Therefore: . $P(\text{exactly 1 Red}) \:=\:\frac{525}{1140} \:=\:\frac{35}{76}$
b) The three balls drawn have different colours.
Select 3 of the 4 colors: . ${4\choose3} \:=\:4$ ways.
To pick one of each of the colors, there are: . ${5\choose1}{5\choose1}{5\choose1} \:=\:125$ ways.
Hence, there are: . $4\cdot125 \:=\:500$ ways to get 3 colors.
Therefore: . $P(\text{3 colors}) \:=\:\frac{500}{1140} \:=\:\frac{25}{57}$
c) The number of blue balls drawn is strictly greater than the number of yellow balls drawn.
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
3. Originally Posted by Soroban
Hello, Aquafina!
There are: . ${20\choose3} \,=\,1140$ possible outcomes.
There are: 5 Reds and 15 Others.
We want 1 Red and 2 Others. .There are: . ${5\choose1}{15\choose2} \:=\:525$ ways.
. . Therefore: . $P(\text{exactly 1 Red}) \:=\:\frac{525}{1140} \:=\:\frac{35}{76}$
Select 3 of the 4 colors: . ${4\choose3} \:=\:4$ ways.
To pick one of each of the colors, there are: . ${5\choose1}{5\choose1}{5\choose1} \:=\:125$ ways.
Hence, there are: . $4\cdot125 \:=\:500$ ways to get 3 colors.
Therefore: . $P(\text{3 colors}) \:=\:\frac{500}{1140} \:=\:\frac{25}{57}$
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
Hi I got the same answers for i) and ii) but not iii). First of all, can you explain your method in ii?
I basically did:
P(diff colours) = 5/20 * 5/19 * 5/18 * 3! * 4C3
I get why you multiplied by 4, thats my 4C3, but how do you get 125 ways?
for iii)
I did P(1 blue, no yellow) as well
Also, P(2 blue, no yellow)
4. I get why you multiplied by 4, thats my 4C3, but how do you get 125 ways?
You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:
Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
5. Originally Posted by ANDS!
You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:
Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
Thanks.
for iii)
I did P(1 blue, no yellow) as well
Also, P(2 blue, no yellow)
Why aren't these included in the answer given in the post above?
6. Soroban answered that for you - just labeled it part C. Well I mean he/she is using the labels you are - lol. But it's there. Just add your probabilities and you will get the same answer they did.
7. Soroban only considered these 2 cases:
Originally Posted by Soroban
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
Not the ones I mentioned as well. When I had worked the answer out, I included these 2 cases, along with the cases I mentioned to get the answer.
I want to know why those are not needed.
8. I think I understand your question. You are wondering why he doesn't consider you drawing 1 or 2 blue balls and the rest red or green? You are right. You would need to take those draws into consideration (assuming that haven't stated that its a given red and green balls WERENT drawn).
9. Originally Posted by ANDS!
I think I understand your question. You are wondering why he doesn't consider you drawing 1 or 2 blue balls and the rest red or green? You are right. You would need to take those draws into consideration (assuming that haven't stated that its a given red and green balls WERENT drawn). | 2014-04-19T13:14:48 | {
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http://math.stackexchange.com/questions/422196/left-frac12-right-sqrt-pi/422226 | # $\left(-\frac{1}{2}\right)! = \sqrt{\pi}?$ [duplicate]
I recently learned that $\left(-\frac{1}{2}\right)! = \sqrt{\pi}$ but I don't understand how that makes sense. Can someone please explain how this is possible?
Thanks!
-
Look up the gamma function on Wikipedia. – Ted Shifrin Jun 16 '13 at 20:31
That's because the factorial that you're familiar with is only for positive numbers; it's an abuse of notation. – Chris Gerig Jun 16 '13 at 20:41
## marked as duplicate by O.L., Ayman Hourieh, Amzoti, Joe, sdcvvcJun 16 '13 at 22:52
In order to extend the factorial function to any real number, we introduce the Gamma Function, which is a strange object defined as follows:
$$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t} \, dt$$
The gamma function comes with the special property that $n!=\Gamma(n+1)$ for natural numbers $n$, so to evaluate $(-1/2)!$, (which by itself is not technically defined) we define it to be $\Gamma(1/2)$ and hence we evaluate the integral
$$(-1/2)!:=\Gamma(1/2)=\int_0^\infty\frac{e^{-t}}{\sqrt{t}} \,dt$$ To evaluate this integral, we make the substitution $u=\sqrt{t}$, which results in the well known Gaussian integral:
$$\int_0^\infty \frac{e^{-t}}{\sqrt{t}}dt=2\int_0^\infty \frac{e^{-u^2}}{u}u \, du=\int_{-\infty}^\infty e^{-u^2} \, du=\sqrt{\pi}$$
-
I don't know where you've seen this notation. One thing you surely know is the socalled Gamma Function $\Gamma (z)$. This function is a complex function with a host of properties. One of its properties is that if evaluated on $\mathbb{N}$ it coincides with the factorial function, this is, $\Gamma (n+1) = n!$ if $n\in \mathbb{N}$. Also, one can find that $\Gamma (1/2) = \sqrt{\pi}$. So maybe by using a notation I don't know about someone could write $(-1/2)!$ instead of $\Gamma (1/2)$.
-
I posted this very question and an answer to it: Why is $\Gamma(1/2)=\sqrt{\pi}$?
What follows is the answer I posted there. Several others also posted good answers.
If there's any justice in the universe, someone must have asked here how to show that $$\int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt{2\pi}.$$ Let's suppose that has been answered here. Let (capital) $X$ be a random variable whose probability distribution is $$\frac{e^{-x^2/2}}{\sqrt{2\pi}}\,dx.$$ Consider the problem of finding $\mathbb E(X^2)$. It is $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty x^2 e^{-x^2/2}\,dx = \text{(by symmetry)} \frac{2}{\sqrt{2\pi}} \int_0^\infty x^2 e^{-x^2/2}\,dx$$ $$\sqrt{\frac2\pi}\int_0^\infty xe^{-x^2/2}\Big(x\,dx\Big) = \sqrt{\frac2\pi}\int_0^\infty \sqrt{u}\ e^{-u}\,du = \sqrt{\frac2\pi}\ \Gamma\left(\frac32\right) = \frac12\sqrt{\frac2\pi} \Gamma\left(\frac12\right).$$ So it is enough to show that this expected value is $1$. That is true if the sum of two independent copies of it has expected value $2$. So: $$\Pr\left(X^2+Y^2<w\right) = \frac{1}{2\pi}\iint\limits_\mathrm{disk} e^{-(x^2+y^2)/2}\,dx\,dy$$ where the disk has radius $\sqrt{w}$. This equals $$\frac{1}{2\pi}\int_0^{2\pi}\int_0^{\sqrt{w}} e^{-\rho^2/2} \,\rho\,d\rho\,d\theta = \int_0^{\sqrt{w}} e^{-\rho^2/2} \,\rho\,d\rho.$$ This last equality holds because we are integrating with respect to $\theta$ something not depending on $\theta$. Differentiating this with respect to $w$ gives the probability density function of the random variable $X^2+Y^2$: $$e^{-w/2}\sqrt{w}\frac{1}{2\sqrt{w}} = \frac{e^{-w/2}}{2}\text{ for }w>0.$$ So $$\mathbb E(X^2+Y^2) = \int_0^\infty w \frac{e^{-w/2}}{2}\,dw =2.$$
-
factorial of negative numbers are $defined$ in terms of Gamma function, i.e. $(x!) := \Gamma(x+1)\forall x\in\mathbb R$, except for $x$ to be negative integer. this is because the two of them agree on positive integers, and so this is just a convention.
Actually, $\Gamma(n+1)=n!$ for $n\in\mathbb{Z}$ and neither exist for $n\lt0$. – robjohn Jun 16 '13 at 21:33
Better. However, this is a bit more than convention: $\Gamma$ is the only log-convex function so that $\Gamma(1)=1$ and $\Gamma(x+1)=x\Gamma(x)$. – robjohn Jun 16 '13 at 22:28 | 2014-04-19T21:00:16 | {
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http://mathhelpforum.com/pre-calculus/68206-exponential-functions.html | # Math Help - exponential functions
1. ## exponential functions
1.Tiger drops a golf ball from a height of 60 m. After each bounce, the height of the balls bounce is 4/5 the height of the previous bounce. How high does the ball bounce after the tenth bounce? (Set up an exponential equation for the height.)
2.The deer population on the Bruce trail in 2005, 2006, and 2007 was recorded as 225, 270, and 324 respectively. Using exponential functions, predict the number of deer in 2015
2. Originally Posted by william
1.Tiger drops a golf ball from a height of 60 m. After each bounce, the height of the balls bounce is 4/5 the height of the previous bounce. How high does the ball bounce after the tenth bounce? (Set up an exponential equation for the height.)
What distance is covered in the initial drop?
What is 4/5 of this distance? (This is the height reached after the first bounce.)
What is 4/5 of this first-bounce distance? (This is the height reached after the second bounce.)
Keep going until you see a pattern or until you reach the value for after the tenth bounce.
Originally Posted by william
2.The deer population on the Bruce trail in 2005, 2006, and 2007 was recorded as 225, 270, and 324 respectively. Using exponential functions, predict the number of deer in 2015
Are you supposed to use statistical software to find an exponential regression for this?
3. Originally Posted by stapel
What distance is covered in the initial drop?
What is 4/5 of this distance? (This is the height reached after the first bounce.)
What is 4/5 of this first-bounce distance? (This is the height reached after the second bounce.)
Keep going until you see a pattern or until you reach the value for after the tenth bounce.
Are you supposed to use statistical software to find an exponential regression for this?
Thanks
1. how do i set up the exponential equation for that?
2. no graphing software should be needed
4. Originally Posted by william
1. how do i set up the exponential equation for that?
That's the pattern you're looking for.
Originally Posted by william
2. no graphing software should be needed
Okay. Start by setting up variables so the first data point represents the initial value at "t = 0". Then take whatever equation form they've given you (different books use different forms), and plug the other two points into this. You will then have two equations (with the input t-value and the output value already plugged in) in two unknowns, being the growth constant and the base of the exponential.
Solve the system for the unknowns. Then rewrite the equation form with the initial value, the growth constant, and the base plugged in.
If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!
5. Hello, william;!
1.Tiger drops a golf ball from a height of 60 m.
After each bounce, the height of the balls bounce is 4/5 the height of the previous bounce.
How high does the ball bounce after the tenth bounce?
(Set up an exponential equation for the height.)
The height $h$ at the $n^{th}$ bounce is given by: . $h(n) \:=\:60\left(\frac{4}{5}\right)^n$
2.The deer population on the Bruce trail in 2005, 2006, and 2007 was recorded as
225, 270, and 324 respectively.
Using exponential functions, predict the number of deer in 2015
We have: . $\begin{array}{|c|c|} \hline \text{Year} & \text{Pop'n} \\ \hline \hline 2005 & 225 \\ \hline 2006 & 270 \\ \hline
2007 & 324 \\ \hline \end{array}$
Take consecutive ratios: . $\begin{array}{ccc}\frac{270}{225} &=& \frac{6}{5} \\ \\[-4mm] \frac{324}{270} &=& \frac{6}{5} \end{array}$
Each population is $\tfrac{6}{5}$ times the previous year's popullation.
The population $n$ years after 2005 is given by: . $P(n) \:=\:225\left(\frac{6}{5}\right)^{n-1}$
In 2015 $(n = 10)\!:\;\;P(10) \:=\:225\left(\frac{6}{5}\right)^9 \:\approx\;1161$ deer.
6. Originally Posted by Soroban
Hello, william;!
The height $h$ at the $n^{th}$ bounce is given by: . $h(n) \:=\:60\left(\frac{4}{5}\right)^n$
We have: . $\begin{array}{|c|c|} \hline \text{Year} & \text{Pop'n} \\ \hline \hline 2005 & 225 \\ \hline 2006 & 270 \\ \hline
2007 & 324 \\ \hline \end{array}$
Take consecutive ratios: . $\begin{array}{ccc}\frac{270}{225} &=& \frac{6}{5} \\ \\[-4mm] \frac{324}{270} &=& \frac{6}{5} \end{array}$
Each population is $\tfrac{6}{5}$ times the previous year's popullation.
The population $n$ years after 2005 is given by: . $P(n) \:=\:225\left(\frac{6}{5}\right)^{n-1}$
In 2015 $(n = 10)\!:\;\;P(10) \:=\:225\left(\frac{6}{5}\right)^9 \:\approx\;1161$ deer.
Soroban you are a great mathematician, every response from you is top notch.
Thank you to both. | 2016-07-29T14:38:13 | {
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https://math.stackexchange.com/questions/1766953/does-abba-0-imply-ab-ba-0-when-a-is-real-semidefinite-matrix | # Does $AB+BA=0$ imply $AB=BA=0$ when $A$ is real semidefinite matrix?
This is a general question that came to my mind while listening to a lecture(although its framing may make it look like a textbook question).
Suppose that $A$ and $B$ be real matrices. $A$ is symmetric and positive semi-definite$(x^tAx\ge0\ \ \forall x\in \mathbb{R}^n)$.
Let $AB+BA=0$. Does this imply $AB=BA=0$? If yes can you give me some example?
• Perhaps trace would be helpful. For $C$ symmetrical (or skew), we have $$\text{tr}(C^2)=0\implies C=0.$$ – Vim May 1 '16 at 17:29
• The title and first paragraph are useless. Why not just state the question? – zyx May 1 '16 at 17:59
• @zyx: I didn't want to unnecessarily add the self study tag, which other wise I would have to add to such a bookish question – Qwerty May 1 '16 at 18:51
• Normally I don't click on questions on the front page that say "just asking!" or "neat question" or other vagueness. And when the top paragraph doesn't contain any details of the question, it defeats the short summary on the Questions page. Then we are left only with tags, which is better than nothing. But why not have something with AB+BA=0 in the title? – zyx May 1 '16 at 18:58
• Edited the title. (Notifying in case you prefer to reverse the edit.) – zyx May 1 '16 at 19:31
Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$. Then $$0 = A(Bv) + B(Av) = A(Bv)+\lambda Bv\Longrightarrow A(Bv) = -\lambda Bv$$ what this means is that $Bv$ is also an eigenvector of $A$ with eigenvalue $-\lambda$. But since $A$ is positive semi-definite all eigenvalues of $A$ are $\geq 0$. If $\lambda=0$, so we have two possiblities:
• Either $v$ is an eigenvector with eigenvalue zero, in which case $A(Bv) = 0$, This means $ABv = BAv = 0$ for all eigenvectors corresponding to eigenvalue zero.
• Or $v$ is an eigenvector with eigenvalue $\lambda>0$. Suppose $Bv\neq 0$, then $Bv$, by above observation is an eigenvector of $A$ with negative eigenvalue! This is impossible, so $Bv=0$. Therefore no only $ABv = 0$, we also have $BAv = \lambda Bv = 0$.
Since all vectors can be written as a linear composition of these eigenvectors, we have $AB=BA = 0$ in these circumstances.
Extra: If you are also interested in what this $B$ can actually be if $AB+BA = 0$, note that if $w$ is any vector, and $P_0$ is the porjection linear transformation which sends $w$ to the subspace of vectors with eigenvalue zero. Then we know the following:
Write $w = P_0w + (1-P_0)w$, then by second observation above $B(1-P_0)w = 0$. So in the basis in which $A$ is diagonal (since $A$ is symmetric it is diagonalizable), $B$ is necessarily of the form $$B=\begin{pmatrix} B_0 & 0\\ 0 & 0 \end{pmatrix}$$ In other words $B$ satisfies $AB+BA=0$ if and only if $P_0BP_0=B$.
Proposition. Let $A,B\in M_n(\mathbb{C})$; if $A$ is diagonalizable with non-negative eigenvalues and $AB+BA=0$, then
1. $AB=BA=0$.
2. If moreover $A$ is hermitian, then $A,B$ are unitarily simultaneously similar to $A_1=diag(D_p,0_{n-p}),B_1=\begin{pmatrix}0_p&0\\0&T_{n-p}\end{pmatrix}$, where $D$ is diagonal $> 0$ and $T$ is upper triangular.
Proof. For 1. cf. the Hamed's answer.
For 2. Using the syeh_106's proof, $A,B$ are unitarily simultaneously similar to $A_1=diag(D_p,0_{n-p})),B'=\begin{pmatrix}0_p&0\\0&U_{n-p})\end{pmatrix}$, where $D$ is diagonal $> 0$. It remains to triangularize $U$, using Schur's method.
An alternative proof is as follows. Since $A$ is psd, $A=UDU^*$, where $U$ is unitary and $D$ is diagonal. Suppose $A$ has $k$ distinct eigenvalues, $\lambda_1 < \lambda_2 <...< \lambda_k$. Then $D$ may be chosen to be $D=\lambda_1I_{m_1}\oplus\lambda_2I_{m_2}\oplus...\oplus\lambda_kI_{m_k}$, where $m_i$ denotes the multiplicity of $\lambda_i$, and $I_{m_i}$ is an $m_i$-by-$m_i$ identity matrix. (I.e. $D$ is block-diagonal, consisting of $k$ blocks.)
Now, note that $AB+BA=0 \Rightarrow UDU^*B=-BUDU^* \Rightarrow DU^*BU=-U^*BUD.$ Letting $\hat B\triangleq U^*BU$, we have $D\hat B=-\hat B D$. Given the form of $D$, it's easy to verify that $\hat B$ must be block-diagonal, i.e. $\hat B=\hat B_1\oplus\hat B_2\oplus...\oplus\hat B_k$, where $\hat B_i$ is of the same size as $I_{m_i}$ Moreover, since $\lambda_2, ..., \lambda_k$ must be positive, we conclude that $\hat B_2, ..., \hat B_k$ must all be zero matrices. If $\lambda_1$ is also positive (i.e. $A$ is p.d.), $\hat B_1$ must also be zero. So $\hat B=0$ and $D\hat B=\hat B D=0$. On the other hand, if $\lambda_1=0$ (i.e. $A$ is p.s.d.), a direct multiplication clearly also shows $D\hat B=\hat B D=0$.
Since $AB=UD\hat BU^*$ and $BA=U\hat BDU^*$, it follows that $AB=BA=0$. | 2020-04-06T12:53:52 | {
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http://mathhelpforum.com/discrete-math/27062-party-question.html | # Math Help - Party Question
1. ## Party Question
I did this by hand
for n =1, I get 1 way
n=2, 3 ways
n=3, 15 ways.
Any help or advice on how to look at this problem.
2. The number of ways to arrange 2n people into n named groups of two is $
\left( {\begin{array}{c} {2n} \\ {\underbrace {2,2, \cdots ,2}_n} \\ \end{array}} \right) = \frac{{\left( {2n} \right)!}}{{2^n }}$
But these are not named groups so divide by (n!):
$\frac{{\left( {2n} \right)!}}{{2^n \left( {n!} \right)}}$.
3. Originally Posted by Jrb599
I did this by hand
for n =1, I get 1 way
n=2, 3 ways
n=3, 15 ways.
Any help or advice on how to look at this problem.
I figured it out, but it's not easy to explain how, I just worked on it for a bit :/
Lets assume all the people are males so I can call them "he" without feeling bad.
First I came up with the recursive definition, which is relatively easy, you know that the first person can pair with any of the people except himself (2n-1 options) and that will occupy person1 and person1's partner, leaving 2 fewer people, which leaves n-1 pairs. And then you can do that same step again. Now you know they are multiplied, because for every option that any of them chooses, the others can choose every available option. (if you are hazy on this, maybe take a look at trees, or just think about it for a while, or graphically draw the first few pairs, I drew n=3 on my paper to make sure there were really 15 options and I wasn't getting confused)
So my recursive formula looked like this:
$f(n) = (2n-1) * f(n-1)$
Then I did 5 iterations, and saw the pattern:
f(1) = 1
f(2) = 3*1 = 3
f(3) = 5*3*1 = 15
f(4) = 7*5*3*1 = 105
f(5) = 9*7*5*3*1 = 945
And so the pattern is that it multiplies by the next consecutive odd integer. I played around with this for a while, and saw that I could look at it like this:
$f(5)=\frac{9*8*7*6*5*4*3*2*1}{8*6*4*2}$
The numerator is simply 9! which is (2n-1)! and so I looked at the denominator for a while, and realized that if I divided those numbers by 2, I would have 4! Which means that if I multiplied 4! by $2^4$ I would have the denominator. Basically I made it look like this: $8*6*4*2 =4(2)*3(2)*2(2)*1(2) = (4*3*2*1)(2*2*2*2) = 4!*2^4 = (n-1)!*2^{n-1}$
So now I had a formula for the numerator and the denominator, I put them together to get this:
$f(n) = \frac{(2n-1)!}{(n-1)!*2^{n-1}}$
Then I checked it and it worked. (I didn't prove it, though, so you might want to do that)
------------
EDIT: After reviewing Plato's answer, I realized that you can modify our formula like this:
initial formula
$f(n) = \frac{(2n-1)!}{(n-1)!*2^{n-1}}$
Now if we multiply by 2n on the top and bottom, we are multiplying by 1, and thus not changing the problem. This would cause the numerator to become 2n! instead of (2n-1)! because 2n is the next consecutive integer in the factorial. Then on the bottom, take the 2 and multiply it by the $2^{n-1}$ and you get $2^n$ and multiply the n by the $(n-1)!$ and because it is the next consecutive integer in the factorial, it will become $n*(n-1)! = n!$ so:
modify the formula
$f(n) = \frac{(2n-1)!}{(n-1)!*2^{n-1}}*\frac{2n}{2n} = \frac{(2n)!}{(n)!*2^{n}}$
final formula
$f(n) = \frac{(2n)!}{(n)!*2^{n}}$
4. Angel White... That was amazing help
I actually had the formula
1*3*5*---*n,
couldn't think of how to write it.
Thanks so much!
Btw, your explanation was awesome
If wanted to do this for 2n+1 people, and 1 person was left out, then I would get:
$
f(n) = \frac{(2n+1)!}{(n)!*2^{n}}
$
Technically
$
f(n) = (2n-1) * f(n-1)
$
Is a right answer to 2n people, correct?
Also,
Angel_white, I don't think your logic works for the 2n+1 case.
f(n) = (2n+1 -2)*f(n-1) isn't right, also I did -2 cause you have to include he can't pair with the guy you excluded.
5. Originally Posted by Jrb599
Angel White... That was amazing help
I actually had the formula
1*3*5*---*n,
couldn't think of how to write it.
Thanks so much!
Btw, your explanation was awesome
If wanted to do this for 2n+1 people, and 1 person was left out, then I would get:
$
f(n) = \frac{(2n+1)!}{(n)!*2^{n}}
$
Technically
$
f(n) = (2n-1) * f(n-1)
$
Is a right answer to 2n people, correct?
Also,
Angel_white, I don't think your logic works for the 2n+1 case.
f(n) = (2n+1 -2)*f(n-1) isn't right, also I did -2 cause you have to include he can't pair with the guy you excluded.
I think the answer will be
$g(n) = \frac{(2n)!}{(n)!*2^{n}}(2n+1)$
It took me a while to see that, but here is why:
Lets use n=2 for our example. We will have two pairs plus one person for 5 people total, lets represent them with circles like this:
$\bigcirc_1$
$\bigcirc_2 \bigcirc_3$
$\bigcirc_4 \bigcirc_5$
Lets choose circle1 to be the lonely person, then we have 4 circles left, which we can use our formula $f(n) = \frac{(2n)!}{n!*2^{n}}$ to determine. So if circle1 is lonely, then there will be 3 possibilities because f(2)=3
Now lets choose circle2 to be the lonely circle, there will again be 4 dots with 3 possible matchups.
Now circle3,4, &5 will have the same pattern, so what we realize is that our previously discovered formula will be invoked once for every circle attending our party. Since there are (2n+1) circles, we can simply multiply the previous formula by (2n+1) to arrive at the correct answer. This gives us the new equation:
$g(n) = \frac{(2n+1)*(2n)!}{n!*2^{n}}$
And because (2n)! * (2n+1) = (2n+1)!
$g(n) = \frac{(2n+1)!}{n!*2^{n}}$
I am relatively confident that this is correct, but not 100% sure.
EDIT: And upon looking back at your post, I realize that this is the same answer you gave. Good job, you found it much easier than I did ^_^
6. Angel white, if you see my post early, that's what I got too so I would take it as right also. It took me a while to see why, but your post is much better at understanding the even way. Plato's posted helped me see it the odd way. The only reason why I asked for you to confirm is because I thought that you would offer a really good observation. | 2015-09-05T16:49:53 | {
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https://math.stackexchange.com/questions/2540248/find-an-infinite-set-s-and-a-function-g-s-to-s-that-is-surjective-but-not/2540343 | # Find an infinite set $S$ and a function $g : S \to S$ that is surjective but not injective.
Find an infinite set $S$ and a function $g : S \to S$ that is surjective but not injective.
This is all that is given in the problem. Should I fix $S$ to be a certain set, like the integers, or natural numbers, and work from there? Or should I just create a function, like $f(x) = x^2$? Any guidance would be appreciated.
What about this: Let $S$ be $\mathbb{R}$, and let $g : S \to S$ be defined by $f(x) = x^2$. This is surjective but not injective.
EDIT 11/27, 3:53pm CT: I totally confused myself about what it means to be surjective. I'm not sure what I was thinking. Clearly, $f(x) = x^2$ is not surjective since each y-coordinate is not mapped to.
• Try $f)x)=x^3-x.$ – bof Nov 27 '17 at 22:05
• I see. So then in a formal proof what do you let S be? $\mathbb{R}$? – mathmajor Nov 27 '17 at 22:07
• Yeah, I was thinking of $S=\mathbb R.$ If you want $S=\mathbb C$ then you can just use $f(z)=z^2.$ – bof Nov 27 '17 at 22:17
• I'd suggest, just start with natural numbers. Then make a function where, for every natural number $n$, there are 2 different values $x$, $y$, such that $f(x) = n$ and $f(y) = n$. – DanielV Nov 28 '17 at 14:07
Your proposed function won't work because it is not surjective. For example, $-1$ has no preimage but $-1 \in \mathbb{R}$.
Possible direction:
To make it work, for example, you can let $S$ be the set of nonnegative integer.
Let $g(0)=0=g(1)$, $g(2)=1=g(3)$, and et cetera.
Try to come out with a rule to describe what I did.
• Right. So 0-->0, 1-->0, 2-->1, 3-->1, 4-->2, 5-->2, etc... This would be an infinite set S mapped to itself, being surjective and not injective. Now to figure out what the rule is. – mathmajor Nov 27 '17 at 21:59
• Would this be a piecewise function? – mathmajor Nov 27 '17 at 22:10
• Try dividing by $2$, it might help? Well, technically it can be viewed as a piecewise function, if you separate the rules for odd numbers and even number. but there are function to combine it into a single rule. – Siong Thye Goh Nov 27 '17 at 22:14
What about this: Let S be $\mathbb{R}$, and let g : S → S be defined by f(x) = $x^2$. This is surjective but not injective.
This is not surjective. $-1$ is not mapped to by any element in the domain.
Should I fix $S$ to be a certain set, like the integers, or natural numbers, and work from there?
Yes, from what I can understand you're free to choose $S$, and then come up with a function that solves the problem. Your goal is to try to map multiple elements to the same element, here is an example for $\mathbb{N}_0$:
0 1 2 3 4 5 6 7 8 9 10 11 ...
0 0 1 1 2 2 3 3 4 4 5 5 ...
It is generally easier for this kind of problem to first come up with a solution in general in your head, and then derive a formula for that solution.
\begin{align} f:\mathbb Z &\to \mathbb Z \\ n &\mapsto \left \lfloor \dfrac n2 \right \rfloor \end{align}
For every integer, $n$, $f(2n) = n$ and $f(2n+1)=n$
• So let S = $\mathbb{Z}$, and let f: $\mathbb{Z}$ --> $\mathbb{Z}$ be defined by the rule f(2n) = n, and f(2n+1) = n...? I guess I'm confused with what the n --> (n/2) is referring to. – mathmajor Nov 28 '17 at 2:56
• @wannabemathmajor - it's the floor function, the integer part of x. – steven gregory Nov 28 '17 at 5:12
Choose $$~A = \mathbb{N}~$$ and $$~f(n) = max\{1,n − 1\}~$$ for all $$~n ∈ \mathbb{N}~$$.
Then $$f(n +1) = n$$ for all $$n ∈ \mathbb{N}$$, so $$f$$ is surjective.
But $$f(1) = 1 = f(2)$$ and $$1 ≠ 2$$.
Therefore $$f$$ is not injective.
This website provide really good answer https://mathgraphy.com/functions/ | 2019-11-11T19:35:27 | {
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https://math.stackexchange.com/questions/1856135/prove-if-f0-0-and-f0-0-then-f0-geq-0 | # Prove: if $f(0)=0$ and $f'(0)=0$ then $f''(0)\geq 0$
let $$f$$ be a nonnegative and differentiable twice in the interval $$[-1,1]$$
Prove: if $$f(0)=0$$ and $$f'(0)=0$$ then $$f''(0)\geq 0$$
1. Are all the assumptions on $$f$$ necessary for the result to hold ?
2. what can be said if $$f''(0)= 0$$ ?
Looking at the taylor polynomial and lagrange remainder we get:
$$f(x)=f(0)+f'(0)x+\frac{f''(c)x^2}{2}$$
$$f(x)=\frac{f''(c)x^2}{2}$$
Because the function is nonnegative and $$\frac{x^2}{2}\geq 0$$ so $$f''(c)\geq 0$$
For 1., all the data is needed but I can not find a valid reason.
For 2., can we conclude that the function the null function?
Another approach is via the method of contradiction. Assume that $f''(0) < 0$. Then the function $f'$ is strictly decreasing at $0$ which means that there is a neighborhood $I$ of $0$ such that if $x \in I, x < 0$ then $f'(x) > f'(0) = 0$ and if $x \in I, x > 0$ then $f'(x) < f'(0) = 0$. We can obviously choose $I$ of the form $(-h, h)$ and hence observing sign of $f'$ in $(-h, h)$ we see that $f$ is strictly increasing in $(-h, 0]$ and strictly decreasing in $[0, h]$ and since $f(0) = 0$ it follows that $f(x) < 0$ for all $x \in I$ except $x = 0$. And this is contrary to the hypotheses that $f$ is non-negative. Thus we obtain the desired contradiction.
BTW the above argument can be replaced by the following concise argument. Since $f'(0) = 0, f''(0) < 0$ the point $0$ is a local strict maximum of $f$ and hence $f(x) < f(0) = 0$ for all $x$ in some neighborhood of $I$ of $0$ except $x = 0$. And this contradicts that $f$ is non-negative.
The argument in first paragraph actually shows how the vanishing of derivative and sign of second derivative guarantee local maxima/minima and if the reader is well aware of the proof of second derivative test for maxima/minima, it is preferable to adopt the argument in second paragraph.
As can be seen from the arguments above we don't need continuity of $f''$. And we can avoid difficult theorems like Taylor or L'Hospital. The argument used in first paragraph leads to one of the simpler proofs of Taylor's Theorem with Peano's form of remainder.
Update: The condition $f' (0)=0$ is superfluous here. The function is non-negative and $f(0)=0$ and hence $0$ is a point of local minima so that $f'(0)=0$.
Your proof is enough when $f''$ is continuous.
Here's a way without the continuity assumption.
This Taylor expansion $f(x) = f(0) + xf'(0) + \frac{x^2}{2}f''(0) + o(x^2)$ yields here: $$f(x) = \frac{x^2}{2}f''(0) + o(x^2)$$
Either $f''(0)=0$ and we're good, otherwise the previous equality rewrites as $\displaystyle \lim_{x\to 0}\frac{2f(x)}{f''(0) x^2}= 1$
Consequently, $f(x)$ and $f''(0) x^2$ must share the same sign on a neighborhood of $0$.
Since $f\geq 0$ and $x^2\geq 0$ that implies $f''(0)\geq 0$
When $f''(0)=0$, $f(x)=o(x^2)$. I don't see anything more you could say.
• This works, so +1. You might consider referencing the Peano form of the remainder here. -Mark Jul 11, 2016 at 17:38
• How did you get to $\lim_{x\to 0}\frac{2f(x)}{f''(0) x^2}= 1$?
– gbox
Jul 11, 2016 at 19:10
• @LeGrandDODOM it is the definition of Peano form? if we plug the limit to 2f(x) do not the limit should be $0$?
– gbox
Jul 11, 2016 at 19:25
• @LeGrandDODOM In your last comment, $f=o(g)$ if and only if $\frac{f}{g}\to 0$. Jul 11, 2016 at 22:04
• @Dr.MV Yes, typo. What I mean is $f(x) = o(g(x)) \iff \frac{f(x)}{g(x)} \to 0$ by definition and therefore $f(x) = g(x) + o(g(x)) \iff \frac{f(x)}{g(x)} \to 1$ Jul 11, 2016 at 22:18
For 2) you could have $f(x)=x^4$ which satisfies all properties and conclusions including $f''(0)=0$, yet $f(x)$ is not identically zero.
For 1) relaxing the condition that $f(0)=0$, we could look at $f(x)=\cos(x)+3$ (which has instead $f(0)=4$). It satisfies the property that $f(x)$ is non-negative, is twice differentiable on $[-1,1]$ and that $f'(0)=0$. However, $f''(0)=-\cos(0)=-1$ is not non-negative.
Relaxing the condition that $f(x)$ be non-negative on the interval, letting $f(x)=-x^2$ one has $f(0)=0, f'(0)=0$, yet $f''(0)=-2$ is not non-negative
Relaxing the condition that $f$ be twice differentiable makes no sense since then $f''(0)$ can't be talked about.
Try searching yourself as well for a function $f(x)$ which violates only the hypothesis that $f'(0)=0$ which has $f''(0)<0$.
With these counterexamples, we conclude that in fact every hypothesis was necessary.
• $x^3$ is not nonnegative on the interval. Use $x^4$ instead. Jul 11, 2016 at 17:15
• @Mosquite thanks. I moved one degree up from $x^2$ to fix that $f''(0)=0$ and forgot to account for that. Jul 11, 2016 at 17:16
• But are the hypotheses sufficient to guarantee that $f''(0)\ge 0$? Jul 11, 2016 at 17:28
• @Dr.MV I interpreted his question as asking about the after thought questions (were all in fact necessary or could any hypotheses be dropped, and can you conclude if $f''(x)=0$ that $f(x)=0$). His proof seemed sufficient to me, but LaGrandDODOM points out in his answer that the continuity of $f''(x)$ was tacitly assumed in the OP's proof, and can be dropped as well. I see no reason to replicate LaGrandDODOM's proof at this point or to come up with my own. Jul 11, 2016 at 17:32
• The OP's "proof" only shows that $f''(c)\ge 0$, for some $c\in (0,x)$, ]but does not prove that indeed $f''(0)\ge 0$. And yes, you're correct, using the Peano form for the remainder is one very straightway forward. -Mark Jul 11, 2016 at 17:36
The continuity of $f''$ is not really needed to apply Taylor's formula with the Lagrange remainder. Anyway, we may just apply l'Hopital's rule to get: $$f''(0) =\lim_{x\to 0}\frac{f'(x)-f'(0)}{x} = 2\lim_{x\to 0}\frac{f(x)}{x^2}.$$ Since both $f(x)$ and $x^2$ are non-negative over $[-1,1]\setminus\{0\}$, $\;f''(0)\geq 0$ readily follows.
Obviously we cannot deduce the strict inequality $f''(0)>0$, as shown by $f(x)=x^4$.
• @Dr.MV: I think it is better to apply L'Hospital once to get $$2\lim_{x \to 0}\frac{f(x)}{x^{2}} = \lim_{x \to 0}\frac{f'(x)}{x} = \lim_{x \to 0}\frac{f'(x) - f'(0)}{x} = f''(0) \text{ (by defn of }f''(0))$$ This works without continuity of $f''$. Jul 12, 2016 at 5:34
• @ParamanandSingh: thanks, updated. Jul 12, 2016 at 15:29
• Jul 12, 2016 at 15:46
• Thaks for link to Gowers blog post. It is indeed a very good post. BTW proof of Taylor's theorem with Peano remainder given by Gowers is essentially the same as the one I linked in my answer. Jul 13, 2016 at 6:24 | 2022-06-29T01:52:05 | {
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https://math.stackexchange.com/questions/1448427/conditional-probability-against-combinatorics | # Conditional Probability against Combinatorics
Consider a lot consisting of 3 blue balls and 1 red ball.
Suppose I pick 2 balls one after another without replacement, now the probability of 2 balls being blue:
$$\frac{3\choose 2}{4 \choose 2}$$
Now taking different approach using conditional probability, the solution is also:
$$\frac34\times\frac23$$
Now it turns out that both of them evaluate to $1/2$.
Why this happens and does it happen always?
Is solving the problem using combinations an efficient approach always?
Or am I missing something basic out here!
• Without replacement is the keyword here. The combinations approach results in an arrangement without replacement. Hence the answer matches. Try to solve the same problem with replacement=True. – Sriharsha Madala Sep 23 '15 at 18:28
We are talking of drawing without replacement.
For the particular case you have given, the two approaches match, but beware
If the question were probability of getting 1 red, 1 blue
$(a) \dfrac{{3\choose1}{1\choose1}}{4\choose2} = \dfrac12$ , but
$(b) \dfrac34\cdot\dfrac13 = \dfrac14$ and you need to multiply by $2!$ to take into account different orders in which the events can occur.
On the other hand, if the problem asked the probability of a red followed by a blue,
$(b) \dfrac34\cdot\dfrac13 = \dfrac14$ is correct,
and for $(a)$, you need to divide $\dfrac{\binom{3}{1}\binom{1}{1}}{\binom{4}{2}}= \dfrac12$ by $2!$ to get the specified order
Adopting the KISS approach, I use combinations when order is unspecified, and direct multiplication of probabilities when order is specified.
Both approaches are correct. Both methods can be viewed as combinatorial.
In your first approach, your sample space consists of ${4 \choose 2} = 6$ unordered samples, each with probability $1/6.$ Three of the six outcomes correspond to choosing two blue balls, for a probability of $1/2.$ It is OK to use unordered samples because we are interested in the number of balls chosen of each color, not the order in which choices were made.
Maybe the balls are numbered 1, 2, 3, 4, with the first three being blue. Then outcomes that satisfy your event are $\{12, 13, 23\}$. (I use $numerical$ order in this listing because the $actual$ order of listing is irrelevant.
In your second approach, you can consider that you have a sample space consisting of $12$ ordered outcomes, each with probability $1/12.$ Six of the twelve ordered outcomes have two blue balls, so the probability is $1/2.$
This second approach would be necessary if you wanted to answer a question such as "What is the probability I choose a blue ball first and then the red one?"
Numbering the balls as before, the six outcomes out of twelve that satisfy the event $\{\text{Two Blue}\}$ are $\{12, 13, 23, 21, 31, 32\}.$ Here I am not free to use numerical order in making my list because the actual order of selection does matter.
This is one more method to solve this problem:
Consider a Lot consists of {B1,B2,B3,R} B:Blue ball R:Red ball.
Now list out all the combinations we get when we choose 2 balls from the Lot, one after another(without replacement).Let the event be S.
S={(B1,B2),(B1,B3),(B1,R),(B2,B1),(B2,B3),(B2,R),(B3,B2),(B3,B1),(B3,R),(R,B1),(R,B2),(R,B3)}.
Now our event E is getting 2 Blue balls,
E={(B1,B2),(B1,B3),(B2,B1),(B2,B3),(B3,B2),(B3,B1)}.
Now probability of getting event E is,
P(E)=n(S)/n(E);which is 6/12=1/2. n:cardinality of a set(ie.number of elements).
Inference: There is hell a lot of way to get solution to a probability question,all the answers match provided the approach is right.Everyone chooses their own approach,so there is no "efficient" approach. This is also a reason why I(Most of us) love it. | 2019-06-19T03:29:38 | {
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https://mathhelpboards.com/threads/water-balloon-fight.5371/ | # Water Balloon Fight
#### veronica1999
##### Member
7 people are having a water ballon fight. At the same time, each of the 7 people throws a water balloon at one of the other 6 people, chosen at random. What is the probability that there are 2 people who throw the balloon at each other?
I am having a lot of trouble on this one.
First, I calculated the total number of outcomes and that would be 6x6x6x6x6x6x6
I calculated the number of pairs i could make 7C2 =21
besides the pair there will be 5 people who can throw to any of the 6
so the outcome is 21x6x6x6x6x6
From here, I am sure I have to correct the overcounts and I am still trying...
Can I get some hints please?
the answer is 7847/15552
I tried working backwards from the answer and i still couldn't get it.
Thanks.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hi veronica1999!
This is not an easy one.
The only way that I see is systematically counting all possibilities.
This is quite a bit of work, but still doable by hand.
Just now, I wrote a computer program that simply verified all possibilities and I got indeed your answer.
To do it manually, you could count the number of ways that no one throws a water ball at each other.
To find the actual number of ways that (at least) 2 people throw at each other, you'd have to subtract that number from the total of $6^7$.
Let's call the first person A.
Then A can throw at 6 people.
Let call his target B.
Then B can throw at 5 people.
Let's call his target C.
Then C can either throw at A, completing a triangle, which we'll count separately.
Or C can throw at one of the 4 people left, whom we'll call D.
At this point we have:
A closed triangle (ABC) for 6.5.1 ways.
A partial square ABCD for 6.5.4 ways.
If we continue from the close triangle (ABC), the next person can either throw at one of A, B, C, or he can throw at one of the 3 remaining people, which we'll call E.
We will have to count these 2 possibilities separately.
And so on...
#### Opalg
##### MHB Oldtimer
Staff member
This is not an easy one.
I thought so too, until I saw that veronica1999 was actually well on the way to a solution.
7 people are having a water ballon fight. At the same time, each of the 7 people throws a water balloon at one of the other 6 people, chosen at random. What is the probability that there are 2 people who throw the balloon at each other?
First, I calculated the total number of outcomes and that would be 6x6x6x6x6x6x6
I calculated the number of pairs i could make 7C2 =21
besides the pair there will be 5 people who can throw to any of the 6
so the outcome is 21x6x6x6x6x6
From here, I am sure I have to correct the overcounts and I am still trying...
Can I get some hints please?
the answer is 7847/15552
I tried working backwards from the answer and i still couldn't get it.
You have started in exactly the right way. There are $6^7$ possible outcomes, and there are $21\times6^5$ ways of getting a pair. But, as you realised, that includes some double counting. You can eliminate that by using the inclusion-exclusion method. First, you need to subtract the number of ways of getting two pairs who target each other. There are $35\times3$ ways of choosing two pairs out of seven, leaving three leftover who can throw at any of the $6$. That gives $105\times6^3$ outcomes. Finally, you need to add the number of ways of getting three pairs who target each other. There are $7\times5\times3$ ways of choosing three pairs out of seven, leaving one leftover who can throw at any of the $6$. That gives $105\times6$ outcomes.
Thus the number of outcomes in which at least one pair soak each other is $21\times6^5 - 105\times6^3 + 105\times6 = 163\,296 - 22\,680 + 630 = 141\,246$. Divide that by $6^7 = 279\,936$ and do some cancellation to get the answer.
#### chisigma
##### Well-known member
May be it is comfortable to proceed step by step as follows. If n is the number of players, then the possible plays are $n\ (n-1)$. Let's start...
a) n=2. We call the players A and B and the 2 possible plays are...
AB BA
... and the only 'good pair' is AB + BA, so that P=1...
b) n=3. We call the players A,B and C and the possible 6 plays are...
AB BA CA
AC BC CB
We have 3 'good pairs' AB+BA, AC+CA, BC+CB, so that is $\displaystyle P= \frac{1}{2}$...
c) n=4. We call the players A,B,C and D and the possible 12 plays are...
AB BA CA DA
AC BC CB DB
AD BD CD DC
We have 6 'good pairs' AB+BA, AC+CA, AD+DA,BC+CB,BD+DB,CD+DC, so that is $\displaystyle P=\frac{1}{2}$...
d) n=5. We call the players A,B,C,D and E and the possible 20 plays are...
AB BA CA DA EA
AC BC CB DB EB
AD BD CD DC EC
AE BE CE DE ED
We have10 'good pairs' AB+BA, AC+CA, AD+DA,AE+EA,BC+CB,BD+DB,BE+EB,CD+DC,CE+EC,DE+ED so that is $\displaystyle P=\frac{1}{2}$...
It is easy to see that in general we have $n\ (n-1)$ 'possible plays' and $1 + 2 + 3 + ...+\ n-1 = \frac{n\ (n-1)}{2}$ 'good pairs', so that, with the only exception n=2, in any case we have $P=\frac{1}{2}$...
Kind regards
$\chi$ $\sigma$
Last edited:
#### veronica1999
##### Member
I thought so too, until I saw that veronica1999 was actually well on the way to a solution.
You have started in exactly the right way. There are $6^7$ possible outcomes, and there are $21\times6^5$ ways of getting a pair. But, as you realised, that includes some double counting. You can eliminate that by using the inclusion-exclusion method. First, you need to subtract the number of ways of getting two pairs who target each other. There are $35\times3$ ways of choosing two pairs out of seven, leaving three leftover who can throw at any of the $6$. That gives $105\times6^3$ outcomes. Finally, you need to add the number of ways of getting three pairs who target each other. There are $7\times5\times3$ ways of choosing three pairs out of seven, leaving one leftover who can throw at any of the $6$. That gives $105\times6$ outcomes.
Thus the number of outcomes in which at least one pair soak each other is $21\times6^5 - 105\times6^3 + 105\times6 = 163\,296 - 22\,680 + 630 = 141\,246$. Divide that by $6^7 = 279\,936$ and do some cancellation to get the answer.
Thanks!!!!
Now I think I understand how to correct the overcounting.
I am still amazed how you did it so easily.
First, I made the problem simpler and then applied what you did above.
Let's say there are 4 people A,B,C,D
There are a total of 3x3x3x3 outcomes.
And 4C2 x3x3 to make a pair.
In this case, it is easy to see that I overcounted when there are 2 pairs.
I subtract 6 to correct the double counting.
48/81
I tried again with 6 people
A B C D E F
I would have a total of 5x5x5x5x5x5 outcomes.
To make a pair it would be 6C2x5x5x5x5.
To correct for the double counts of 2 pairs
I would have to subtract (6C2X4C2)/2 x5x5.
Then I realized while i was subtracting all the 2 pairs
I subtracted all the 3 pairs so I have to add back 6C2X4C2/2.
Now finally for 7 people
A B C D E F G
There would be a total of 6x6x6x6x6x6x6 outcomes
And to make a pair, 7C2x6x6x6x6x6
Then subtract 7C2x5C2/2 x6x6x6
and then add (7C2X5C2X3C2)/6 x3
It took me a long time to figure out that I had to divide by 2 and 6 for the 2 pairs and 3pairs.
I hope my understanding is correct.
Once again, thanks!!!!! | 2021-12-01T07:05:04 | {
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https://www.physicsforums.com/threads/integrate-sinx-cox-using-double-angle.704389/ | # Integrate sinx*cox* using double angle
1. Aug 5, 2013
### Jude075
1. The problem statement, all variables and given/known data
∫ Sinx cos x dx
2. Relevant equations
3. The attempt at a solution
If you integrate it using substitution, you get -cos2(x)/2but if you use double angle formula to rewrite the problem, it will be ∫1/2 sin(2x), and integrate it, you get -cos(2x)/4. isn't it weird?
2. Aug 5, 2013
### voko
You are forgetting that each of those two integrals must have $+ C$ appended to it. Now recall that $\cos 2x = 2 \cos^2 x - 1$.
3. Aug 5, 2013
### D H
Staff Emeritus
What voko said is exactly right. To elaborate, these two apparently different integrals differ only by a constant.
4. Aug 5, 2013
### Jude075
So if I was to find the indefinite integral, both answers were right?
5. Aug 5, 2013
### voko
None is correct unless you add the indefinite constant. | 2017-08-22T03:30:36 | {
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https://www.physicsforums.com/threads/determine-the-sum-of-the-given-series.762704/ | # Homework Help: Determine the sum of the given series:
1. Jul 22, 2014
### ybhathena
1. The problem statement, all variables and given/known data
Sum starting from n=1 to infinity for the expression, (3/4^(n-2))
What the solutions manual has done is multiply the numerator and the denominator by 4.
12/(4^(n-1))
I don't know what they have done from here on:
12 / (1 - 1/4)
= 16
Can someone explain the solution? Thank you.
2. Jul 22, 2014
### gopher_p
It's a geometric series.
3. Jul 22, 2014
### ybhathena
I'm having trouble seeing how to use the fact that it is a geometric series in the problem. Could you explain further?
4. Jul 22, 2014
### gopher_p
Do you have a formula for the sum of a geometric series?
5. Jul 22, 2014
### ybhathena
I know for a geometric series the formula is Sn = (a(1-r^n)) / 1-r and for an infinite geometric series S = a/1-r
6. Jul 22, 2014
### gopher_p
OK then ... does $\frac{12}{1-\frac{1}{4}}$ look like one of those forms?
7. Jul 22, 2014
### ybhathena
Well the rate is 4, and the first value, a, is 1/4. Also 12 fits no where into any of these formulas so 12/(1-1/4) doesn't look like either of the two formulas for geometric series.
8. Jul 22, 2014
### ybhathena
Oh wait they are treating the entire expression as a geometric series not just the denominator. Is that it? Give me a second to see then.
9. Jul 22, 2014
### ybhathena
Ok I get it now I am finally getting 16 as an answer. My question is now why don't they divide 16 by 4 to get the original value to which the series converges to. (because in the beginning they multiplied the entire expression by 4?)
10. Jul 22, 2014
### gopher_p
They multiplied both the numerator and denominator of the term inside the sum by 4, which is the same as multiplying the whole term inside the sum by 4/4=1. Multiplying by 1 doesn't change anything, so there is nothing to "undo". It goes $$\sum\limits_{n=1}^\infty\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty1\cdot\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty\frac{4}{4}\cdot\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty\frac{12}{4^{n-1}}=\sum\limits_{n=1}^\infty12\cdot\frac{1}{4^{n-1}}=\sum\limits_{n=1}^\infty12\cdot(\frac{1}{4})^{n-1}$$ with the formula for the geometric series applied to the RHS.
Everything's equal, so ...
11. Jul 22, 2014
### ybhathena
Oh yes that is true the constant that is being multiplied is 1 not 4, so it won't change anything. Thank you for your help throughout this process; it is greatly appreciated!
12. Jul 22, 2014
### HallsofIvy
$$\sum_1^\infty \frac{3}{4^{n-2}}= \sum_1^{\infty} \frac{3}{16}\frac{1}{4^n}= \frac{3}{16}\sum_1^\infty \frac{1}{4^n}$$
BUT the formula for the sum of a geometric series: $$\sum_{n=0}^\infty a^n= \frac{1}{1- a}$$ starts with n= 0, not n= 1. There are two ways to handle that.
1) factor a "1/4" out of the series:
$$\frac{3}{64}\sum_{n=1}^\infty \frac{1}{4^{n-1}}= \frac{3}{64}\sum_{i= 0}^\infty \frac{1}{4^i}$$ where i= n-1.
2) Do the entire geometric series from 0 to infinity, then subtract off the n= 0 term:
$$\frac{3}{4}\sum_{n=0}^\infty \frac{1}{4^n}- \frac{3}{4}$$
since when n= 0, $$\frac{1}{4^n}= 1$$.
Last edited by a moderator: Jul 22, 2014
13. Jul 22, 2014
### gopher_p
Some texts give $\sum\limits_{n=1}^\infty ar^{n-1}=\frac{a}{1-r}$ as (one of) the formula(s) for the geometric series. Given the answer from the solutions manual, I'm guessing the topic creator is using one of those texts. | 2018-07-20T01:44:42 | {
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https://math.stackexchange.com/questions/2106804/factor-a4-3a2-2ab1-b2 | # Factor $a^4-3a^2-2ab+1-b^2$
In order to factor $a^4-3a^2-2ab+1-b^2$, I find that $a=1, b=-1$ makes the value of the expression 0. Thus, I assume $b=-a$.
I rewrite the expression on the assumption as: $$a^4-3a^2-2ab+1-b^2$$ $$=a^4-3a^2+2a^2+1-a^2$$ $$=a^4-2a^2+1$$ $$=(a^2-1)^2$$
Then I insert $a+b$, which is another form of the assumption above, into one of the factors. Since $a+b=0$, this insertion should cause no change to whatever relationships in the original expression. I arbitrarily set its coefficient as $1$. A factor, therefore, would be $(a^2+a+b-1)$.
I then divide the original expression by the factor and find that it indeed can be divided without a remainder and I also find the other factor. Thus, $$a^4-3a^2-2ab+1-b^2$$ $$=(a^2+a+b-1)(a^2-a-b-1)$$ Q.E.D.
The problem is that I don't know what I am doing. I tried to use the factor theorem for a bivariate (if this is the proper word for it) expression but I might be fabricating a fortuitous reasoning from the answer. Yes, I knew the answer before I tried, and no, I don't know how I can solve this in other ways. It would be great if someone can tell me if my reasoning can be made more clear.
• Well, rewriting the polynomial as $a^4-2a^2+1-a^2-2ab-b^2=(a^2-1)^2-(a+b)^2$ is simpler. – egreg Jan 21 '17 at 0:08
Hint: consider it as a quadratic in $b\,$:
$$-b^2-2ab + a^4-3a^2+1$$
Its discriminant is:
$$\frac{1}{4}\Delta = a^2 +(a^4-3a^2+1)=a^4-2a^2+1=(a^2-1)^2$$
Then calculate the roots $b_{1,2}$ as expressions in $a$, and the polynomial factors as $-(b-b_1)(b-b_2)$.
• I like this answer because of its purely mechanical nature. Things won't always work out this cleanly, of course, but this is an excellent tool for this problem. – Steven Stadnicki Jan 21 '17 at 0:12
• Thanks, everything's clear. – edmont123-carla Jan 21 '17 at 1:24
Here is probably the most simple method to factor it.
Note that $$a^2-b^2=(a-b)(a+b)$$So $$a^4-3a^2-2ab+1-b^2=a^4-2a^2+1-a^2-2ab-b^2=(a^2-1)^2-(a+b)^2$$ So we can factor it into $$(a^2+a+b-1)(a^2-a-b-1)$$ | 2021-07-30T08:29:13 | {
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https://stats.stackexchange.com/questions/101063/statistics-of-7-game-playoff-series | # Statistics of 7 game playoff series
Background: a friend of mine makes a hobby (as I imagine many do) of trying to predict hockey playoff outcomes. He tries to guess the winning team in each matchup, and the number of games needed to win (for anyone unfamiliar with NHL hockey a series is decided by a best of 7). His record this year after 3 rounds of play (8+4+2=14 best of 7 matchups) is 7 correct/7 incorrect for winning team and 4 correct/10 incorrect for number of games (he only considers the number of games correct if he also picked the winning team).
We got to joking that he's doing no better than blind guessing on the teams question, but that he's substantially beating the odds if one assumes that the probabilities for a 4, 5, 6 or 7 game series are equal (would expect a 12.5% success rate, he's at 28.5%).
This got us wondering what the odds actually are for each possible number of games. I think I've worked it out, but I want to tie up a few loose ends since part of my approach was brute-force scribbling on a big piece of paper. My basic assumption is that the outcome of every game is random with probability $\frac{1}{2}$ for a each team to win.
My conclusion is that:
$$\rm P(4\;games) = \frac{2}{2^4} = 12.5\%\\ P(5\;games) = \frac{8}{2^5} = 25\%\\ P(6\;games) = \frac{20}{2^6} = 31.25\%\\ P(7\;games) = \frac{40}{2^7} = 31.25\%$$
I guided my analysis based on a notion that a 4 game series should have a probability of $\frac{2}{2^4}$, analogous to the odds of flipping 4 coins and getting either 4 heads or 4 tails. The denominators were easy enough to figure out from there. I got the numerators by counting the number of "legal" combinations (WWLWWLL would be illegal since the series would be decided after 5 games, the last 2 games would not be played) of results for a given number of games:
Possible 4 game series (2):
WWWW LLLL
Possible 5 game series (8):
LWWWW WLLLL
WLWWW LWLLL
WWLWW LLWLL
WWWLW LLLWL
Possible 6 game series (20):
LLWWWW WWLLLL
LWLWWW WLWLLL
LWWLWW WLLWLL
LWWWLW WLLLWL
WLLWWW LWWLLL
WLWLWW LWLWLL
WLWWLW LWLLWL
WWLLWW LLWWLL
WWLWLW LLWLWL
WWWLLW LLLWWL
Possible 7 game series (40):
LLLWWWW WWWLLLL
LLWLWWW WWLWLLL
LLWWLWW WWLLWLL
LLWWWLW WWLLLWL
LWLLWWW WLWWLLL
LWLWLWW WLWLWLL
LWLWWLW WLWLLWL
LWWLLWW WLLWWLL
LWWLWLW WLLWLWL
LWWWLLW WLLLWWL
WLLLWWW LWWWLLL
WLLWLWW LWWLWLL
WLLWWLW LWWLLWL
WLWLLWW LWLWWLL
WLWLWLW LWLWLWL
WLWWLLW LWLLWWL
WWLLLWW LLWWWLL
WWLLWLW LLWWLWL
WWLWLLW LLWLWWL
WWWLLLW LLLWWWL
What's a non-brute-force method for deriving the numerators? I'm thinking there may be a recursive definition, so that $\rm P(5\;games)$ can be defined in terms of $\rm P(4\;games)$ and so on, and/or that it may involve combinations like $\rm(probability\;of\;at\;least\;4/7\;W)\times(probability\;of\;legal\;combination\;of\;7\;outcomes)$, but I'm a bit stuck. Initially I thought of some ideas involving $\left(^n_k\right)$ but it seems that only works if the order of outcomes doesn't matter.
Interestingly, another mutual friend pulled out some statistics on 7 game series played (NHL, NBA, MLB 1905-2013, 1220 series) and came up with:
4 Game Series - 202 times - 16.5%
5 Game Series - 320 times - 26.23%
6 Game Series - 384 times - 31.47%
7 Game Series - 314 times - 25.73%
That's actually a pretty good match (at least from my astronomer's point of view!). I'd guess that the discrepancy comes from the outcome of each game having being biased toward a win for one team or the other (indeed, teams are usually seeded in the first round so that the leading qualifying team plays the team that barely qualified, second place plays second last, and so on... and most of the games are in the first round).
• Am not particularly active on CV.SE, so this may need a bit of re-tagging.
– Kyle
Jun 3 '14 at 23:31
For a team to win [the series] in game N, they must have won exactly 3 of the first N-1 games. For game seven, there are $\binom{6}{3} = 20$ ways to do that. There are 2 possible outcomes for game seven, and 20 possible combinations of wins for each of the teams that can win, so 40 possible outcomes. For an N-game series a best-of-seven series to end in N games, the number of possibilities is $2 \binom{N-1}{3}$.
Indeed the order doesn't matter, if you're already given the number of games played. Only the last game matters, and the winner must have 3 previous wins, in any order.
• For an N game series shouldn't it be $2(^{N-1}_{{\rm floor}(N/2)})$, or something like that? Assuming there is an odd number of games, which is only sensible.
– Kyle
Apr 27 '17 at 15:47
• I was using N as the number of games played in a best-of-seven. Eg. for N=4, $2\binom{3}{3} = 2$ gives you the number of possible ways the series can end in 4 games. ie. for each team, the number of ways to choose 3 wins out of 3 games. Apr 27 '17 at 19:15
• Yes, the possibilities of an M-game series decided in N games, should be $2\binom{N-1}{\mathrm{floor}(M/2)}$. This will still work if there's an even number of games, if tied series are not considered decided. Apr 27 '17 at 19:22
• If you are going to be realistic the probability of win should not be 0.5 for each team for each game. There could be a home ice advantage as one example. Apr 27 '17 at 21:20
• @MichaelChernick true, and I touch on this a bit in the last paragraph of the question, but 0.5 as a starting point that can later be adjusted is reasonable.
– Kyle
Apr 28 '17 at 0:17
If p is the probability of winning a single game, and N is the number of games one needs to win the series, then the probability "P" of winning the series is given by:
$$P = \frac{p^N}{\left(N-1\right)!}\cdot\sum_{i=0}^{N-1}\left[\left(\prod_{j=1}^{N-1}\left(i+j\right)\right)(1-p)^i\right]$$
Think of the way a game plays out with points going to each player as a path. Each path that leads to a win has a probability given by the product of the probabilities for the number of wins and losses. The following image tries to illustrate an example.
The pattern for the coefficients is given by (i+N-1) choose (i) where "i" is the number of points the opponent has scored and N is the number of points for the win.
Remember the choose formula is:
$$nCr = \frac{n!}{r!\left(n-r\right)!}$$
so for us:
$$nCr = \frac{\left(i+N-1\right)!}{i!\left(N-1\right)!}=\frac{\left(i+1\right)...\left(i+N-1\right)}{\left(N-1\right)!}=\frac{\prod_{j=1}^{N-1}\left(i+j\right)}{\left(N-1\right)!}$$
• Welcome to the site! Can you edit to explain how you arrived at this answer? Why is it correct? Note also that you can use math typesetting via MathJax. More information: math.meta.stackexchange.com/questions/5020/…
– Sycorax
Jan 4 at 20:22
• It's not clear what question this formula answers: it can't apply to the original post, because it gives the wrong values for $N\gt 1.$
– whuber
Jan 4 at 20:51
• You're right. In my excitement I wrote the form of the equation for N=4. The generic equation involves a product of factors. It was derived from the equation for the choose function. For each N there is a polynomial that gives the proper coefficients from a diagonal of pascal's triangle. I will post the correction when I figure out how to use MathJax Jan 6 at 0:52
• Unfortunately, your explanation is vague and your formula produces values that exceed $1$ and therefore cannot be correct. Try it for $N=4$ and $p=1/2,$ for instance. It says the probability is $$(1/2)^4\left(3!(1/2)^0 + 4!/1!(1/2)^1 + 5!/2!(1/2)^2 + 6!/3!(1/2)^3\right) = (6 + 12 + 15 + 15)/16 \gt 1.$$
– whuber
Jan 6 at 4:18
• Much better, thank you! +1.
– whuber
Jan 6 at 17:41
An alternate way to look at would be a binomial distribution: You need x=3 (exactly 3 successes) in n = 6 (trails) , so if the probability of winning a game is .5 (both teams equally likely) , binomial would say: P(x=3) = 6C3 * (.5)^3 * (.5)^3 = .3125 This would mean there is 31.25% chance of going to a 7 game series. And the probability you win in the 7th game , would follow negative binomial, how many trails = 7 for 4 success, 7-1 C 4-1 * (.5)^3 * (.5)^4 | 2022-01-19T09:12:05 | {
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https://in.mathworks.com/help/symbolic/pade.html | ## Description
example
pade(f,var) returns the third-order Padé approximant of the expression f at var = 0. For details, see Padé Approximant.
If you do not specify var, then pade uses the default variable determined by symvar(f,1).
example
pade(f,var,a) returns the third-order Padé approximant of expression f at the point var = a.
example
pade(___,Name,Value) uses additional options specified by one or more Name,Value pair arguments. You can specify Name,Value after the input arguments in any of the previous syntaxes.
## Examples
### Find Padé Approximant for Symbolic Expressions
syms x
ans =
-(x*(7*x^2 - 60))/(3*(x^2 + 20))
### Specify Expansion Variable
If you do not specify the expansion variable, symvar selects it. Find the Padé approximant of sin(x) + cos(y). The symvar function chooses x as the expansion variable.
syms x y
ans =
(- 7*x^3 + 3*cos(y)*x^2 + 60*x + 60*cos(y))/(3*(x^2 + 20))
Specify the expansion variable as y. The pade function returns the Padé approximant with respect to y.
ans =
(12*sin(x) + y^2*sin(x) - 5*y^2 + 12)/(y^2 + 12)
### Approximate Value of Function at Particular Point
Find the value of tan(3*pi/4). Use pade to find the Padé approximant for tan(x) and substitute into it using subs to find tan(3*pi/4).
syms x
f = tan(x);
y = subs(P,x,3*pi/4)
y =
(pi*((9*pi^2)/16 - 15))/(4*((9*pi^2)/8 - 5))
Use vpa to convert y into a numeric value.
vpa(y)
ans =
-1.2158518789569086447244881326842
### Increase Accuracy of Padé Approximant
You can increase the accuracy of the Padé approximant by increasing the order. If the expansion point is a pole or a zero, the accuracy can also be increased by setting OrderMode to relative. The OrderMode option has no effect if the expansion point is not a pole or zero.
Find the Padé approximant of tan(x) using pade with an expansion point of 0 and Order of [1 1]. Find the value of tan(1/5) by substituting into the Padé approximant using subs, and use vpa to convert 1/5 into a numeric value.
syms x
p11 = subs(p11,x,vpa(1/5))
p11 =
x
p11 =
0.2
Find the approximation error by subtracting p11 from the actual value of tan(1/5).
y = tan(vpa(1/5));
error = y - p11
error =
0.0027100355086724833213582716475345
Increase the accuracy of the Padé approximant by increasing the order using Order. Set Order to [2 2], and find the error.
p22 = subs(p22,x,vpa(1/5));
error = y - p22
p22 =
-(3*x)/(x^2 - 3)
error =
0.0000073328059697806186555689448317799
The accuracy increases with increasing order.
If the expansion point is a pole or zero, the accuracy of the Padé approximant decreases. Setting the OrderMode option to relative compensates for the decreased accuracy. For details, see Padé Approximant. Because the tan function has a zero at 0, setting OrderMode to relative increases accuracy. This option has no effect if the expansion point is not a pole or zero.
p22Rel = subs(p22Rel,x,vpa(1/5));
error = y - p22Rel
p22Rel =
(x*(x^2 - 15))/(3*(2*x^2 - 5))
error =
0.0000000084084014806113311713765317725998
The accuracy increases if the expansion point is a pole or zero and OrderMode is set to relative.
### Plot Accuracy of Padé Approximant
Plot the difference between exp(x) and its Padé approximants of orders [1 1] through [4 4]. Use axis to focus on the region of interest. The plot shows that accuracy increases with increasing order of the Padé approximant.
syms x
expr = exp(x);
hold on
grid on
for i = 1:4
end
axis([-4 4 -4 4])
legend('Order [1,1]','Order [2,2]','Order [3,3]','Order [4,4]',...
'Location','Best')
title('Difference Between exp(x) and its Pade Approximant')
ylabel('Error')
## Input Arguments
collapse all
Input to approximate, specified as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.
Expansion variable, specified as a symbolic variable. If you do not specify var, then pade uses the default variable determined by symvar(f,1).
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You also can specify the expansion point as a Name,Value pair argument. If you specify the expansion point both ways, then the Name,Value pair argument takes precedence.
### Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: pade(f,'Order',[2 2]) returns the Padé approximant of f of order m = 2 and n = 2.
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You can also specify the expansion point using the input argument a. If you specify the expansion point both ways, then the Name,Value pair argument takes precedence.
Order of the Padé approximant, specified as an integer, a vector of two integers, or a symbolic integer, or vector of two integers. If you specify a single integer, then the integer specifies both the numerator order m and denominator order n producing a Padé approximant with m = n. If you specify a vector of two integers, then the first integer specifies m and the second integer specifies n. By default, pade returns a Padé approximant with m = n = 3.
Flag that selects absolute or relative order for Padé approximant, specified as 'absolute' or 'relative'. The default value of 'absolute' uses the standard definition of the Padé approximant. If you set 'OrderMode' to 'relative', it only has an effect when there is a pole or a zero at the expansion point a. In this case, to increase accuracy, pade multiplies the numerator by (var - a)p where p is the multiplicity of the zero or pole at the expansion point. For details, see Padé Approximant.
collapse all
By default, pade approximates the function f(x) using the standard form of the Padé approximant of order [mn] around x = x0 which is
$\frac{{a}_{0}+{a}_{1}\left(x-{x}_{0}\right)+...+{a}_{m}{\left(x-{x}_{0}\right)}^{m}}{1+{b}_{1}\left(x-{x}_{0}\right)+...+{b}_{n}{\left(x-{x}_{0}\right)}^{n}}.$
When OrderMode is relative, and a pole or zero exists at the expansion point x = x0, the pade function uses this form of the Padé approximant
$\frac{{\left(x-{x}_{0}\right)}^{p}\left({a}_{0}+{a}_{1}\left(x-{x}_{0}\right)+...+{a}_{m}{\left(x-{x}_{0}\right)}^{m}\right)}{1+{b}_{1}\left(x-{x}_{0}\right)+...+{b}_{n}{\left(x-{x}_{0}\right)}^{n}}.$
The parameters p and a0 are given by the leading order term f = a0 (x - x0)p + O((x - x0)p + 1) of the series expansion of f around x = x0. Thus, p is the multiplicity of the pole or zero at x0.
## Tips
• If you use both the third argument a and ExpansionPoint to specify the expansion point, the value specified via ExpansionPoint prevails.
## Algorithms
• The parameters a1,…,bn are chosen such that the series expansion of the Padé approximant coincides with the series expansion of f to the maximal possible order.
• The expansion points ±∞ and ±i∞ are not allowed.
• When pade cannot find the Padé approximant, it returns the function call.
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https://math.stackexchange.com/questions/931119/the-floor-and-ceiling-functions-are-equal-for-integer-numbers | # The floor and ceiling functions are equal for integer numbers
The exercise
Proof the following directly: Let $x \in \mathbb{R}$. If $x \in \mathbb{Z}$, then $\lfloor x \rfloor = \lceil x \rceil$.
My problem
I mostly fail completely on the formal part of any kind of proof (here a direct proof). This means I've mostly the right idea but I'm also mostly not able to bring it to paper in the right form.
What I know currently
The direct proof for a implication like $p \implies q$ could be $p \implies s_1, s_1 \implies s_2, s_2 \implies q$. So you typically start with the (rearranged) hypothesis $q$ assuming it's true and repeatedly apply the modus ponsens on your established implications until you reach $q$.
What I've done
When I take a look at the definition of $\lfloor x \rfloor = \max\{k \in \mathbb{Z}|k\leq x\}$ and $\lceil x \rceil = \min\{k \in \mathbb{Z}|k\geq x\}$ it seems to be "pretty clear". Since $x,k \in \mathbb{Z}$ is $x \in \{k \leq x\}$ and $x = \max\{k \leq x\}$ which in turn implies $\lfloor x \rfloor = x$. The same is true for $\lceil x \rceil$, which means that $x \in \{k \geq x\}$ and $x = \min\{k \geq x\}$ which in turn implies $\lfloor x \rfloor = x$. And since $\lfloor x \rfloor = x$ and $\lfloor x \rfloor = x$ is $\lfloor x \rfloor = \lceil x \rceil$.
My question(s)
How can I get this in a formally correct form? And how can I proof that $x \in \{k \leq x\}$, $x = \max\{k \leq x\}$, $x \in \{k \geq x\}$ and $x = \min\{k \geq x\}$ is true?
With $A=\{k:k\in\mathbb{Z},k\leq x\}$, then $x\in A$ because $x\in\mathbb{Z}$ and $x\leq x$. Moreover $x=\max A$ because (i) $x\in A$, (ii) if $k\in A$, then $k\leq x$. You can do the same for $B=\{k:k\in\mathbb{Z}, k\geq x\}$.
• Thanks! The second part ($x = \max A$) seems to be too easy. Should this be enough? Beside that have you any hints about my first question (how to get the formally correct form or how to improve it)? Sep 14 '14 at 17:08
• Your approach is fine. As for $x=\max A$, what I wrote should be enough. Sep 14 '14 at 17:13
That $x \in \{ k \leq x\}$ follows since $x \leq x$. Similarly for $x \in \{ k \geq x\}$. To show that $x = max\{ k \leq x\}$, suppose not. Then there is some $y \in \{ k \leq x\}$ such that $y > x$. But this is impossible, by the definition of the set $\{ k \leq x\}$. So $x$ must be the supremum. But $x$ is also in the set, so it's the maximum. You should be able to do the minimum case now.
I guess at the beginning, the more (without exaggerating!) details you put the better. For example, your proof is fine, but I'd put sentences of the form "x = max{...} since x is the supremum and it's in the set...", just to show that you understand why what you claim is true.
• Thanks! It looks like you're using a proof by contradiction for $x = \max\{k \leq x\}$. Is this okay as a part of a direct proof? Beside that have you any hints about my first question (how to get the formally correct form or how to improve it)? Sep 14 '14 at 17:11
• The proof by contradiction is a bit of an over-kill to be honest, but it's an alternative to the direct proof. Concerning your first question, what level of details are looking for? Sep 14 '14 at 17:19
• Is it enough what I wrote, or should I bring your parts into it? Is there a more cleaner way to write all that stuff down; less words more math (like in "What I know currently")? ;) Sep 14 '14 at 17:23
• I guess at the beginning, the more (without exaggerating!) details you put the better. For example, your proof is fine, but I'd put sentences of the form "x = max{...} since x is the supremum and it's in the set...", just to show that you understand why what you claim is true. Sep 14 '14 at 17:25
• Put your last comment into the answer and you get your check! Also because you were the one with the first answer (13 seconds faster). Sep 14 '14 at 17:29 | 2021-10-19T06:23:47 | {
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https://mathforums.com/threads/how-to-find-the-reaction-force-from-the-joint-from-a-bar-supported-to-a-wall-where-a-block-is-resting.348163/ | # How to find the reaction force from the joint from a bar supported to a wall where a block is resting?
#### Chemist116
The problem is as follows:
The figure from below shows a system at equilibrium. The bar is homogeneous and uniform and has a weight of $14\,N$ and the block which is labeled $Q$ has a weight equal to $28\,N$. Find the reaction force in Newtons experienced by the joint attached to the wall.
The alternatives are given as follows:
$\begin{array}{ll} 1.&14\sqrt{2}\,N\\ 2.&7\sqrt{2}\,N\\ 3.&14\,N\\ 4.&21\,N\\ 5.&28\,N\\ \end{array}$
I'm not sure exactly how to assign the equilibrium condition in this problem and where the forces are acting in the bar. Can someone help me with the right approach and and free body diagram for this thing?.
What I've attempted to do was to use the equilibrium condition as:
$\sum ^n_{i=1}\tau_{i}=0$
By following this
$T\cos53^{\circ}\cdot 6a + (T-mg)\cdot 2a - (14 + 28) = 0$
Then the reaction in the joint would be given by:
$R=T\sin 53^{\circ}$
or will be given by a vertical reaction?
$R= mg-T-T\cos53^{\circ}$
Which of these reactions are referring in this problem?
Am I understanding this correctly?. How exactly should be understood the reaction. Can someone help me here?. I would like someone can explain me the meaning of the reaction experienced by the joint this part is very important hence a free body diagram will help me a lot here.
#### skeeter
Math Team
about the pivot, $\tau_{ccw} = \tau_{cw}$
$T\cos(53) \cdot 6a = 14(3a) + (28-T) \cdot 2a \implies T = 17.5 \text{ N}$
$\sum F_x = 0 \implies T\sin(53) = R_x \implies R_x = 14 \text{ N}$
$\sum F_y = 0 \implies T\cos(53) + R_y - (28-T) - 14 \implies R_y = 14 \text{ N}$
$R = 14\sqrt{2} \text{ N}$
#### Chemist116
about the pivot, $\tau_{ccw} = \tau_{cw}$
$T\cos(53) \cdot 6a = 14(3a) + (28-T) \cdot 2a \implies T = 17.5 \text{ N}$
$\sum F_x = 0 \implies T\sin(53) = R_x \implies R_x = 14 \text{ N}$
$\sum F_y = 0 \implies T\cos(53) + R_y - (28-T) - 14 \implies R_y = 14 \text{ N}$
$R = 14\sqrt{2} \text{ N}$
Can you please include a FBD for this thing?. I'm getting tangled with the direction of the vectors acting in the object.
My major confusion is what's exactly this problem asking?. Is it a reaction caused by the pushing against the wall and experienced by the joint. Or is it the reaction vertically experienced by the joint?. Btw your answer is correct. I think when the question is put without giving any specification it is meant about the total reaction which is the vector sum of both. Am I right with this?
The confusion is exactly which forces are acting vertically?
In other words why is there is a negative sign in front of $-(28-T)$?
The reaction force over the bar where the block is standing isn't going upwards?. Can you help me with this?
$N+T=28$
hence $N=28-T$
When an object is supported over another generates a reaction but the one which is on top experiences the reaction. But what happens to the one which is on bottom. Is it the force with the same sign?. Can you please explain this part please?.
#### skeeter
Math Team
Can you please include a FBD for this thing?
You try one first ...
#### Chemist116
You try one first ...
Okay here's what I got. It took me some time to figure out where are the forces acting. Please revise my work if I'm getting the idea correctly.
I'm assuming that the reaction created by the block standing over the bar will not produce any torque, but the reaction going downwards it will. Thus is is why it appears of the right side of your equation. Am I right with this?.
#### skeeter
Math Team
The magnitude of the reaction force of the block on the beam, $(28-T)$, exerts a clock-wise torque, $\tau = (28-T) \cdot 2a$, about the wall pivot. Note this is a scalar equation. I'm not going to put a negative sign in front to confuse the issue. | 2020-03-29T14:33:48 | {
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https://math.stackexchange.com/questions/4079714/we-choose-5-numbers-from-1-to-100-we-order-them-by-value-what-is-the-exp | # We choose $5$ numbers from $1$ to $100$. We order them by value. What is the expected difference between the second and the third?
I came across this peculiar problem.
We choose $$5$$ numbers from $$1$$ to $$100$$ (with repetition). We order them in decreasing order by value. What is the expected difference between the second and the third?
For example, we draw $$6,67,89,45,33$$. Difference is $$67-45=22$$.
• Are the five values distinct? Mar 27 '21 at 19:55
• The numbers are drawn with repetition so they can be the same. Also 2nd highest and third highest. For that matter, it doesn't matter. Let's define it order them in decreasing order. Mar 27 '21 at 19:57
• Was this an interview question, by any chance? Mar 27 '21 at 20:57
• The solution should be at least 40 pages long and the last 10 pages are unlocked with a key from a special mission in Zelda. Mar 27 '21 at 21:07
• this kind of problem is given by results of "order statistics"
– qwr
Mar 27 '21 at 22:06
If instead of $$n=100$$ possibilities we have a large $$n$$, the expectation converges to $$n/6$$. In fact, if we select numbers without replacement rather than with (which makes a negligible difference when $$n$$ is large), the expectation is exactly $$\frac16(n+1)$$. Here's a way to see that:
First, change the game to be:
We have $$n+1$$ chairs arranged in a circle. Select one chair at random, remove it from the circle. That leaves a chain of $$n$$ chairs. Select 5 of those chairs at random, and count the gap between the second and third selected chair.
It should be clear that this gives the same answer -- the business with making a circle and cutting it into a line is completely irrelevant. However, now we can modify the rules further to
We have $$n+1$$ chairs arranged in a circle. Select six chairs at random. Now pick (still at random) one of the six chairs, remove it, unfold the rest to a line, and count the gap between the second and third selected chair.
Doing the random choices in a different order shouldn't change anything, so we still get the sought-for answer. One more reformulation:
We have $$n+1$$ chairs arranged in a circle. Select six chairs at random. Start from a random chair among the selected six chairs and walk clockwise. Count how many chairs appear between the second and third time you walk past a selected chair.
And finally:
We have $$n+1$$ chairs arranged in a circle. Select six chairs at random. Select one of the six gaps between selected chairs at random, and count how long it is.
In the original formulation the expected size of the different gaps were not obviously the same -- but now all of the apparent asymmetry in that description has been removed. By symmetry the expected length of the random gap can only be one sixth of the total length of the circle (because that is the sum of the 6 possibilities we choose uniformly between in the second choice).
It gets somewhat trickier with replacement because the chair where we break the circle cannot be selected again in the next choices of the original game, so reformulating the game doesn't give us perfect symmetry. That must be why Patrick Stevens' exact answers don't show a simple 6 in the denominator.
By this reasoning we also get $$n/6$$ exactly if you uniformly choose 5 real numbers in $$[0,n)$$ (in which case replacement doesn't matter at all).
By extremely dumb use of Mathematica I can give you some values for smaller $$n$$ than $$100$$ (though the quintic is getting too big for me to go much further than $$30$$). You want $$n=100$$ in the following.
f[n_] := Mean[Subtract @@ Reverse[Sort[#]][[2 ;; 3]] & /@ Tuples[Range[n], {5}]]
For $$n=1 \dots 15$$, this is:
$$\left\{0,\frac{5}{16},\frac{40}{81},\frac{85}{128},\frac{104}{125},\frac{1295}{1296},\frac{400}{343},\frac{1365}{1024},\frac{3280}{2187},\frac{3333}{2000},\frac{2440}{1331},\frac{20735}{10368},\frac{4760}{2197},\frac{12805}{5488},\frac{25312}{10125}\right\}$$
For $$n=20, 25, 30$$, this is $$\{\frac{53333}{16000},\frac{65104}{15625},\frac{809999}{162000}\}$$.
@Roman notes that this is $$\frac{n^4-1}{6n^3}$$, so by engineer's induction, the answer is $$\frac{33333333}{2000000}$$, or precisely $$16.6666665$$.
A proof of this fact is left as an exercise to the reader.
• I write code in python. And answer seems to approach 16.70... It still running... :) Mar 27 '21 at 21:22
• Your numbers are $\frac{n^4-1}{6n^3}$, giving $\frac{33333333}{2000000}$ for $n=100$. Mar 27 '21 at 21:23
• You're quite right - if only I had bothered to run FindSequenceFunction, or do a moment of thinking :P Mar 27 '21 at 21:23
Your problem pertains to the field of Order Statistic.
You have in fact a random sample of five variables (with repetitions allowed) taken from a uniform discrete distribution over $$[1,100]$$ and want to determine the difference between the second and third highest, which is the case treated in the para
You will see that be able to proficiently apply the concepts and tools of this sector of probability/statistics, you have better to approximately convert your problem to a continuous uniform distribution: with five samples over a width of $$100$$ that is a quite reasonable approximation.
The best approximation is obtained when you center the discrete values into a unitary continuous interval, i.e. you pass from $$[1,100]$$ to $$[1/2,100+1/2]$$.
But since you are considering differences you can shift it to $$[0,100]$$, and scale down to the standard $$[0,1]$$.
That said I am not going to repeat here the exposition you can find on the referred Wikipedia article or on many others.
Value equal
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$
I ran into trouble trying to enumerate where the order that the numbers are chosen doesn't matter, so I've gone in the other direction.
$$D = (100)^5.$$
$$N$$ will be calculated by going through all of the various summations, and attaching a weight to them. In the double summation below, the weight is $$(b-a)$$ and the number of entries represents the number of ways having exactly 2 numbers $$\leq a$$ and 1 number $$\geq b$$.
$$\binom{5}{3} \times \sum_{a=1}^{99} [f(a)] \left[\sum_{b=a+1}^{100}(b-a)[g(b)]\right].$$
Since the denominator is computed by assuming that the order that the numbers are selected is relevant, the numerator must be computed in a consistent manner. The first factor reflects the number of ways that 3 slots, can be chosen to represent the 3 smaller numbers.
Here, I avoid the complication of $$a=b$$ by recognizing that since it will be attached the weight of $$(b-a) = 0$$, ignoring this possibility does not affect the enumeration.
$$f(a) = a^3 - (a-1)^3$$, where the second term represents the number of ways of having all three numbers less than $$a$$.
Similarly, $$g(b) = (101-b)^2 - (100 - b)^2.$$ | 2022-01-26T06:41:20 | {
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https://math.stackexchange.com/questions/3273874/can-a-nowhere-continuous-function-have-a-connected-graph | # Can a nowhere continuous function have a connected graph?
After noticing that function $$f: \mathbb R\rightarrow \mathbb R$$ $$f(x) = \left\{\begin{array}{l} \sin\frac{1}{x} & \text{for }x\neq 0 \\ 0 &\text{for }x=0 \end{array}\right.$$ has a graph that is a connected set, despite the function not being continuous at $$x=0$$, I started wondering, doest there exist a function $$f: X\rightarrow Y$$ that is nowhere continuous, but still has a connected graph?
I would like to consider three cases
• $$X$$ and $$Y$$ being general topological spaces
• $$X$$ and $$Y$$ being Hausdorff spaces
• ADDED: $$X=Y=\mathbb R$$
But if you have answer for other, more specific cases, they may be interesting too.
• As Henning points out via example, this is most interesting when $X = \Bbb{R}$ (and possibly where $Y = \Bbb{R}$ too). – Theo Bendit Jun 25 at 14:56
• I wonder whether the Conway base 13 function has a connected graph. – Nate Eldredge Jun 25 at 18:03
• By transfinite induction one can construct a function $f:\mathbb R\to\mathbb R$ whose graph meets every Borel set in the plane whose projection onto the horizontal axis is uncountable. Can such a graph be disconnected? – bof Jun 25 at 19:21
• @TheoBendit Indeed now I see that case $X=Y=\mathbb R$ is significantly more interesitng. I'll add it as another point. – Adam Latosiński Jun 25 at 22:22
• @NateEldredge: It turns out that the graph of the base-13 function is totally disconnected. – Henning Makholm Jun 27 at 13:09
Check out this paper:
F. B. Jones, Totally discontinuous linear functions whose graphs are connected, November 23, (1940).
Abstract:
Cauchy discovered before 1821 that a function satisfying the equation $$f(x)+f(y)=f(x+y)$$ is either continuous or totally discontinuous. After Hamel showed the existence of a discontinuous function, many mathematicians have concerned themselves with problems arising from the study of such functions. However, the following question seems to have gone unanswered: Since the plane image of such a function (the graph of $$y =f(x)$$) must either be connected or be totally disconnected, must the function be continuous if its image is connected? The answer is no.
In particular, Theorem 5 presents a nowhere continuous function $$f:\Bbb R \rightarrow \Bbb R$$ whose graph is connected.
Whether Conway base 13 function is such an example remains unknown. (at least on MSE; see Is the graph of the Conway base 13 function connected?) It turns out the graph of Conway base 13 function is totally disconnected. See this brilliant answer.
Here is an example for $$\mathbb R^2 \to \mathbb R$$:
$$f(x,y) = \begin{cases} y & \text{when }x=0\text{ or }x=1 \\ x & \text{when }x\in(0,1)\text{ and }y=0 \\ 1-x &\text{when }x\in(0,1)\text{ and } y=x(1-x) \\ x(1-x) & \text{when }x\notin\{0,1\}\text{ and } y/x(1-x) \notin\mathbb Q \\ 0 & \text{otherwise} \end{cases}$$
This is easily seen to be everywhere discontinuous. But its graph is path-connected.
A similar but simpler construction, also $$\mathbb R^2\to\mathbb R$$:
\begin{align} g(1 + r\cos\theta, r\sin\theta) = r & \quad\text{for }r>0,\; \theta\in\mathbb Q\cap[0,\pi] \\ g(r\cos\theta, r\sin\theta) =r & \quad \text{for }r>0,\; \theta\in\mathbb Q\cap[\pi,2\pi] \\ g(x,y) =0 & \quad\text{everywhere else} \end{align}
• Very nice examples. They show how easy is to break some continuity with additional dimensions while retaining enough of it to maintain connectedness of the graph. So I've added another case to the question, $f: \mathbb R \rightarrow \mathbb R$, in which such methods won't work. Do you think a function in this case is possible, like the one that John Hughes is constructing? – Adam Latosiński Jun 25 at 22:38
• @AdamLatosiński: I am unsure, and getting nowhere myself. I've been trying to figure out whether the graph of the Conway base-13 function is connected, but without success either way. – Henning Makholm Jun 25 at 22:54
• Does $y/x(1 - x)$ mean $\frac{y}{x(1 - x)}$ or $\frac{y}{x}(1 - x)$? – Bladewood Jun 26 at 16:30
• @Bladewood: The former. – Henning Makholm Jun 26 at 17:06
• I would rewrite it, the standard interpretation would be the opposite. – Apollys Jun 26 at 20:31
There is a simple general strategy for many questions of this type, which is to just try to build a counterexample by transfinite induction. Let's first think about what it means for the graph $$G$$ of a function $$f:\mathbb{R}\to\mathbb{R}$$ to be disconnected. It means there are open sets $$U,V\subset\mathbb{R}^2$$ such that $$U\cap G$$ and $$V\cap G$$ are both nonempty and together they form a partition of $$G$$ (we will say $$(U,V)$$ separates $$G$$ in that case). So, to make $$G$$ connected, we just have to one-by-one rule out every such pair $$(U,V)$$ from separating it.
So, then, here is the construction. Fix an enumeration $$(U_\alpha,V_\alpha)_{\alpha<\mathfrak{c}}$$ of all pairs of open subsets of $$\mathbb{R}^2$$. By a transfinite recursion of length $$\mathfrak{c}$$ we define values of a function $$f:\mathbb{R}\to\mathbb{R}$$. At the $$\alpha$$th step, we add a new value of $$f$$ to prevent $$(U_\alpha,V_\alpha)$$ from separating the graph of $$f$$, if necessary. How do we do that? Well, if possible, we define a new value of $$f$$ such that the corresponding point in the graph $$G$$ will either be in $$U_\alpha\cap V_\alpha$$ or not be in $$U_\alpha\cup V_\alpha$$, so $$U_\alpha\cap G$$ and $$V_\alpha\cap G$$ will not partition $$G$$.
If this is not possible, then $$U_\alpha$$ and $$V_\alpha$$ must partition $$A\times\mathbb{R}$$ where $$A\subseteq\mathbb{R}$$ is the set of points where we have not yet defined $$f$$. Since $$\mathbb{R}$$ is connected, this means we can partition $$A$$ into sets $$B$$ and $$C$$ (both open in $$A$$) such that $$U_\alpha\cap (A\times\mathbb{R})=B\times\mathbb{R}$$ and $$V_\alpha\cap (A\times\mathbb{R})=C\times\mathbb{R}$$. Now since we have defined fewer than $$\mathfrak{c}$$ values of $$f$$ so far in this construction, $$|\mathbb{R}\setminus A|<\mathfrak{c}$$ and in particular $$A$$ is dense in $$\mathbb{R}$$. If $$B$$ were empty, then $$U_\alpha$$ would have empty interior and thus would be empty, and so $$(U_\alpha,V_\alpha)$$ can never separate the graph of $$f$$. A similar conclusion holds if $$C$$ is empty, so let us assume both $$B$$ and $$C$$ are nonempty. It follows that $$\overline{B}$$ and $$\overline{C}$$ cannot be disjoint (otherwise they would be a nontrivial partition of $$\mathbb{R}$$ into closed subsets), so there is a point $$x\in\mathbb{R}\setminus A$$ that is an accumulation point of both $$B$$ and $$C$$. Since $$x\not\in A$$, we have already defined $$f(x)$$. Note now that $$(x,f(x))\not\in U_\alpha$$, since $$U_\alpha$$ would then contain an open ball around $$(x,f(x))$$ and thus would intersect $$C\times\mathbb{R}$$. Similarly, $$(x,f(x))\not\in V_\alpha$$. Thus $$U_\alpha$$ and $$V_\alpha$$ already do not contain the entire graph of $$f$$, and so we do not need to do anything to prevent them from separating it.
At the end of this construction we will have a partial function $$\mathbb{R}\to\mathbb{R}$$ such that by construction, its graph is not separated by any pair of open subsets of $$\mathbb{R}^2$$, and the same is guaranteed to hold for any extension of our function. Extending to a total function, we get a total function $$f:\mathbb{R}\to\mathbb{R}$$ whose graph is connected. But we can of course arrange in this construction for $$f$$ to be nowhere continuous; for instance, we could start out by defining $$f$$ on all the rationals so that the image of every open interval is dense in $$\mathbb{R}$$. In fact, the construction shows that any partial function $$\mathbb{R}\to\mathbb{R}$$ defined on a set of cardinality less than $$\mathfrak{c}$$ can be extended to a total function whose graph is connected. (Or even stronger, you could start with any partial function whose domain omits $$\mathfrak{c}$$ points from every interval, since that is all you need to guarantee that the set $$A$$ is dense at each step.)
• Without the last sentence, you might have ended up with $f(x)=0$ ;) – Hagen von Eitzen Jun 26 at 6:49
• @HagenvonEitzen: There's not actually a need to do anything additional to make the function nowhere continuous. The construction directly implies that that every Jordan curve in the plane will intersect the graph, which includes even very small circles anywhere in the plane. So the graph naturally becomes dense in $\mathbb R^2$. – Henning Makholm Jun 26 at 7:55
• @EricWofsey: But whenever you have a Jordan curve, then the set of points inside and outside the curve form a possible pair of open $U_\alpha$ and $V_\alpha$ for your construction. There's no point in $U_\alpha\cap V_\alpha$ one can add to the graph, so your construction will add a point outside $U_\alpha\cup V_\alpha$ to the graph instead instead. But $\mathbb R^2\setminus(U_\alpha\cup V_\alpha)$ are exactly the points on the Jordan curve. – Henning Makholm Jun 26 at 16:57
• Even simpler, for every nonempty open $S\subseteq \mathbb R^2$ and $s\in S$, there will be an $\alpha$ such that $(U_\alpha,V_\alpha)=(\mathbb R^2\setminus\{s\},S)$. Then the construction must add a point from $U_\alpha\cap V_\alpha \subseteq S$ to $G$. Since $S$ was arbitrary open, this means that $G$ becomes dense. – Henning Makholm Jun 26 at 20:35
• Stronger yet, the construction ensures that $f$ is "strongly Darboux" -- i.e., $f([a,b])=\mathbb R$ for every $a<b$. For every $y$ we can show that $G$ intersects $[a,b]\times\{y\}$: WLOG $f(a)\ne y$ and $f(b)\ne y$; now extend $[a,b]\times\{y\}$ with a ray that goes straight upwards or downwards from $(a,y)$ such as to avoid $(a,f(x))$, and a ray that goes straight up or down from $(b,y)$ and avoids $(b,f(b))$. The resulting curve divides $\mathbb R^2$ into two nontrivial open regions, so just like the Jordan curve it intersects $G$. But the vertical parts don't, by construction. – Henning Makholm Jun 26 at 21:00
Great question, and I don't have an answer for you, but I've got some small thoughts:
By summing up weighted and displaced copies of $$f$$, you can get discontinuities at many places. For instance, you could write $$F(x) = \sum_{n \in \Bbb Z} \frac{f(x-n)}{1+n^2}$$ That'll have an $$f$$-like discontinuity at every integer.
Digression A comment asks whether the graph is still connected. Let me show that it is at $$x = 1$$ as an example, which should be reasonably compelling for other integer points. (For non-integer points, $$F$$ is continuous, so we're fine).
Write \begin{align} F(x) &= \frac{1}{2} f(x-1) + \sum_{n\ne 1 \in \Bbb Z} \frac{f(x-n)}{1+n^2}\\ &= \frac{1}{2} f(x-1) + G_1(x) \end{align} where $$G_1$$ is a function that's continuous at $$x = 1$$.
Let's look at the graph of $$F$$ near $$1$$, say on the interval $$(3/4, 5/4)$$. It's exactly $$K = \{ (x, \frac{1}{2} f(x-1) + G_1(x)) \mid 3/4 < x < 5/4 \}$$
Contrast this with the graph of $$f$$ near $$0$$, which is $$H = \{ (x, f(x)) \mid -1/4 < x < 1/4 \}$$ and which we know (from standard calculus books like Spivak) to be connected.
Now look at the function $$S : K \to H : (x, y) \mapsto (x-1, y - G_1(x))$$ This is clearly continuous and a bijection (and even extends to a bijection from a (vertical) neighborhood of $$K$$ to a neighborhood of $$H$$), so $$K$$ is also connected.
End of digression
And then for numbers with finite base-2-expansions, you can do the same sort of thing: let $$G(x) = \sum_{k \in \Bbb Z, k > 0} \frac{1}{2^k} F(2^k x)$$ and that'll have $$f$$-like discontinuities at all the points with finite base-2 representations, which is a dense set in $$\Bbb R$$.
But I have a feeling that sliding over to the uncountable-set territory is going to be a lot harder.
• This is a good way to get functions which are discontinuous at many points, but are the graph of $F$ and and the graph of $G$ still connected? – Adam Chalumeau Jun 25 at 14:46
• Well...they could only be disconnected at their points of discontinuity. And (for $F$ at least) at those points the graph is (roughly) the sum of something linear (the derivative approximation) and the graph of $f$; applying a shearing operation gets rid of the linear part, and you've got something a lot like the graph of $f$. I'll add details. – John Hughes Jun 25 at 15:16
• @AdamChalumeau: See "Digression" in which I prove that the graph of $F$ is nice. For $G$, it's presumably tougher.. – John Hughes Jun 25 at 15:24 | 2019-07-21T21:58:27 | {
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https://math.stackexchange.com/questions/1796105/how-does-variable-ordering-in-expressions-work-when-creating-functions-from-an-e | # How does variable ordering in expressions work when creating functions from an equation?
I'm having a really hard time understanding some aspects of functions, i've tried looking around on Khan academy and haven't quite found something to answer my question, i'm sure i'm overlooking something stupid but wanted to know.
So let's say i have the following problem:
For a given input value 'n' the function 'g' outputs a value 'm' to satisfy the following equation:
$3m - 5n = 11$
g(n) expresses m as a function of n:
My step by step is as follows:
$3m-5n=11$ From what i gather, the objective is to isolate the m.
To do so, i subtract $5n$ from the left side, and add $5n$ to the right side
$3m = 11 + 5n$
Then, we want to simplify the equation to use only 1 m, not 3, so we divide everything by 3.
$m= \frac{11}{3} + \frac{5}{3}n$
However, from what khan academy says, this is wrong because the proper way to express this is
$m= \frac{5}{3}n + \frac{11}{3}$
My question is, how do you determine where the variables you balance out from whichever side go on the opposite side? I was told in layman's terms the idea was to have the "dynamic" variables come before the "constant" variables, which i've tried, and sometimes i find it's just the opposite, and the "correct" answer is with the dynamic numbers before the constant ones!
What am i missing here?
Thanks.
TLDR: It is personal preference and largely doesn't matter
Notice that $\frac{11}{3}+\frac{5}{3}n = \frac{5}{3}n+\frac{11}{3}$ due to the commutativity of addition. These two answers are not fundamentally different.
What order you write things in is largely personal preference, and can change depending on context.
One of the most common ordering of terms in a polynomial (referred to as monomial ordering) used is lexicographic ordering which follows the following rules:
• Given that $x_1\succ x_2\succ x_3\succ\dots$, comparing a term $ax_1^{\alpha_1}x_2^{\alpha_2}x_3^{\alpha_3}\cdots$ to $bx_1^{\beta_1}x_2^{\beta_2}x_3^{\beta_3}\cdots$ the term to be written first is determined by comparing powers of $x_1$. Whichever has the higher power of $x_1$ should be written first. If equal then compare powers of $x_2$ and so on until a decision can be made.
This is the same as you should be familiar with as "dictionary order" how you see aardvark in the dictionary before appetizer before apple
For example, something written with lexicographic ordering:
$$x^5+3x^2y^7+3x^2y^5z+x^2y^5+y^{20}+5$$
In such an ordering scheme, the constants would come at the end.
This is not the only choice however. Graded lexicographic order before concerning yourself with the lexicographic order, you first prioritize ordering the terms based on total degree. The above would instead have been written as:
$$y^{20}+3x^2y^7+3x^2y^5z+x^2y^5+x^5+5$$
There are many other orderings that are in use as well, but those are the ones I see most often.
In yet different contexts, it might not be feasible to write larger powers on the left as there may be too many to do so. Take for example:
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\dots+\frac{x^n}{n!}+\dots$$
In this case, you will often see the constant term on the far left as opposed to the far right.
In the end, do whatever feels right to you. If you try to be consistent, great, but it should not truly matter and ordering terms is largely done simply to try to organize the visual information in a useful way. If the equation can be read without any problem, then it is written fine however it is. There isn't much of a mathematical difference between competing ways of writing polynomials, only a visual one.
• If it's just personal preference and my answer is still correct then why would khan academy say it's wrong? May 23, 2016 at 1:29
• @zack6849 Poorly coded homework submission software will search for a specific way or ways of writing an answer. Essentially running a "is inputstring*=*answer" check. Better software will check to see that the answer submitted is mathematically equivalent to what it expects and is more lenient with formatting. If there is a professor moderating the class/homework and you lose points for submissions simply because the system did not like the way that you formatted it, then you should be able to email him/her and request score modification. May 23, 2016 at 1:34
• @zack6849 if the homework were submitted and written and checked by hand, there is no question in my mind that the two answers are the same and would receive the same credit. If you are experiencing difficulty with the specific software checking your answers online, then complain to the company who wrote the software and be ready for several headaches down the road. May 23, 2016 at 1:36
• Alright, this really threw me off because 90% of the time khan academy does check for things that that, thank you for your advice and assistance! May 23, 2016 at 1:36 | 2022-08-18T02:06:35 | {
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https://math.stackexchange.com/questions/2218506/probability-of-drawing-5-cards-from-a-deck-of-52-that-will-have-the-same-suit | # Probability of drawing 5 cards from a deck of 52 that will have the same suit?
A standard deck of cards has 52 members consisting of 4 suits each with 13 members. Five cards are dealt from the randomly mixed deck. What is the probability that all cards are the same suit?
EDIT: How I went about it before posting this question was doing (1/4) as the first card probability because my thought process was that we'll draw 1 suit out of the 4 for the first probability. Then I proceeded to account of the 2nd dealt card with the probability of (12/51) since 1 card has been dealt already out of the 13 cards for that suit, also subtracting 1 from the total amount of cards able to be dealt.
So for the 3rd card: (11/50)
4th card: (10/49)
5th card: (9/48)
Giving us the total overall probability for drawing 5 cards of the same suit: $$(1/4) * (12/51) * (11/50) * (10/49) * (9/48) = 33/66640$$
EDIT2: My practice quiz given by TA's is still saying I have the incorrect answer. Given how the answer should be: $33/16660$ (explained in numerous ways in the thread), I contacted the TA's to see if maybe the have setup the question incorrectly. Will update when I get an answer back.
EDIT3: Got an answer back from my TA's who tested the test. They did have the answer wrong on their end. Everyone who helped me was correct!
• How many ways are there of drawing five cards from the same suit? Note that there are four suits, so the number of ways of drawing five cards from the same suit is four times, say, the number of ways of drawing five clubs. And how many ways are there of drawing five cards in general? – joeb Apr 5 '17 at 1:05
• When you pose a computational question such as this, you should show what you have attempted and indicate where you are stuck so that you receive responses appropriate to your skill level. – N. F. Taussig Apr 5 '17 at 1:12
• @N.F.Taussig, you're right, I apologize. I was just in a bit of a hurry, but I'll edit in on what I did initially. – ProjectDefy Apr 5 '17 at 3:26
The first card has probability $\frac{52}{52}$ of having the same suit as any previously drawn cards (because there are none). This means there is a 100% chance of the first card meeting our criteria.
The second card has probability $\frac{12}{51}$ because there are twelve left out of 51 total that match the suit of the first card.
The third, fourth and fifth cards have probabilities $\frac{11}{50}$, $\frac{10}{49}$, and $\frac{9}{48}$ because there are less and less of the suit of the first card as well as less cards to choose from.
Because I need the first thing to happen AND the second thing to happen AND, ..., I need to multiply the probabilities: $\big(\frac{52}{52}\big)\big(\frac{12}{51}\big)\big(\frac{11}{50}\big)\big(\frac{10}{49}\big)\big(\frac{9}{48}\big)$
You obviously don't need that first fraction, but I think it adds clarity. Five cards, each with their own probability.
• I went ahead and implemented the equation and for some reason my practice quiz is still saying it's wrong. I definitely see your reasoning, and along with others giving the same solution, I have a feeling my TA's setup the practice quiz incorrectly. *ps, I also added how I initially tried to solve it to give some background on my process. – ProjectDefy Apr 5 '17 at 3:34
We know that the first card determines the suit, the second must be one of the $12$ out of $51$ remaining cards with that suit, the third must be one of the $11$ out of $50$ remaining cards with the same suit, etc. The reason the the numerator and denominator keep decrementing is because each time we choose a card, it takes one card away from the pool of the suit and the pool of total cards. So we should get the probability to be $$\prod_{n=1}^4 \frac{13-n}{52-n}$$ which equates to $\frac{33}{16660}$.
• Along with the others who also gave the same answer, for some reason my practice quiz given by my TA's is saying it's wrong. I do see the reasoning behind what everyone is saying, but no luck. I'm wondering at this point if they setup the question wrong. *ps, I also added how I initially tried to solve it to give some background on my process. – ProjectDefy Apr 5 '17 at 3:33
Let's use binomial coefficients to determine the number of ways that this is possible. Then I'll leave it to you to [use binomial coefficients to] divide this by the total number of outcomes to arrive at a probability.
For a 5-card poker hand, this is how we can draw n cards of the same suit.
$${13\choose n} \cdot{39 \choose 5-n}$$
The left factor is the chosen/given suit, and the right factor is all three of the other suits.
If we set $n=5$, that will give us the number of ways to draw all 5 in one of the suits. As there are 4 suits, we multiply the result by 4.
Hint -
Probability of getting 5 cards of same suit -
$$\frac{\binom {13}5}{\binom{52}{5}}$$
Now multiply with 4 as we have 4 suits.
$$4 \times \frac{\binom {13}5}{\binom{52}{5}}$$
• I went ahead and tried your solution (answer: 33/16660), and for some reason it's still saying it's wrong. I do believe this solution (along with others) are correct, but maybe the TA's setup the question wrong online... Will update the main post once I get a word back. *ps, I also added how I initially tried to solve it to give some background on my process. – ProjectDefy Apr 5 '17 at 3:36 | 2019-09-16T14:02:55 | {
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https://math.stackexchange.com/questions/1784516/name-of-the-union-of-a-set-with-its-holes | # Name of the union of a set with its holes
Given an arbitrary connected and compact set $S$ with holes in it, is there a name for the simply connected set formed by the union of $S$ and its holes?
For example, let $S = \{x\in \mathbb{R}^n\ |\ 0 < a \leq||x||^2_2 \leq b\}$, its hole is $H = \{x\in \mathbb{R}^n\ |\ ||x||^2_2 < a\}$. Is there a name for $S \cup H$?
If there is not a specific definition for it, can I just call this set "the union of $S$ and its holes", or is there a more precise way of describing it?
By a set with holes I refer to sets such as this, which is taken from the Wikipedia page on simply connected sets. Here is my attempt to properly define a hole (at least in $\mathbb{R}^n$), inspired by this answer to another question.
Given a connected and compact subset $S \subset \mathbb{R}^n$, its boundary $\partial S$ can be written as the finite union of connected and compact hypersurfaces $F_i \subset \mathbb{R}^{n-1}$, such that $\partial S = \bigcup_{i=1}^k F_i$. Each $F_i$ cuts $\mathbb{R}^n$ in exactly two pieces $A_i \subset \mathbb{R}^n$ and $B_i \subset \mathbb{R}^n$, such that the closure of $A_i$ is compact and the closure of $B_i$ is not compact. Then, we say that $A_i$ is inside $F_i$, and $B_i$ is outside. A hole $H_i$ of $S$ is any $A_i$ such that $A_i \cap S = \emptyset$. Then, the set of all holes is $H = \bigcup H_i$.
If $S \cup H = S$, then is said that $S$ has no holes. This condition is equivalent to $H=\emptyset$.
If $S \cup H \neq S$, then I want to know if $S \cup H$ has a proper name. This condition is equivalent to $H\neq\emptyset$, and $H$ is "filling the holes" of $S$.
• What exactly is meant by a hole ? – Shailesh May 14 '16 at 2:49
• I tried to edit my question to address yours. This is how I would describe the process of making a set with holes: take a simply connected set, remove subsets in its interior so that the resulting set is still connected. This subsets are now the so called holes of the resulting set. – jcmonteiro May 14 '16 at 3:11
• Yes, but if all I've been given is a space, how do I find the "holes" to fill in? – user98602 May 14 '16 at 3:23
• My point is: there is a space, it might have these "holes", but all I want to do is to name the union of the space and its "holes". Much like the union of a set and its boundary is called closure. – jcmonteiro May 14 '16 at 3:28
• The set that I describe has these holes, so it cannot be simply connected by definition. I say that it is connected just because the proof I am building needs it to be. Otherwise, I understand that $S$ could even be not connected and the definitions given so far would still hold. – jcmonteiro May 14 '16 at 20:06
A concise definition of the object you describe is "the complement of the unbounded component of the complement of $S$".
In two dimensions, the term simply connected hull is sometimes used: for example, in a MathOverflow post and in other places you can find with a search.
But in dimensions $\ge 3$ "simply connected" would be misleading: filling in a torus $\mathbb{T}^2$ creates a solid torus, which is still not simply connected.
Another term in use is filled-in set, for example Filled Julia set. This one can be used in any dimension.
• I am not a mathematician, so just to be sure. Would the definition "the complement of the unbounded component of the complement of $S$" still be valid in higher dimensions ($\geq 3$)? It seems to me that it would. Besides, calling these sets filled-in seems appropriate. – jcmonteiro May 14 '16 at 19:21
• Yes, this works for filling in holes in all dimensions. – user147263 May 14 '16 at 19:37
I assume you mean a compact set in $\mathbb{R}^n$ (or some metric space) so that "bounded" makes sense. Also, that in your clarification you probably meant to say something like the following (but see the NOTE at the end)
"take a COMPACT simply connected set $T$ and remove ( open) subsets of the interior to get a compact connected subset $S$."
The goal being to describe $T$ in relation to $S.$
sandwich gives a good answer and from it we see that the key is that "hole" is something like "(a subsequently removed) bounded component of the interior of $T$ " Which becomes "bounded component of the complement of $S.$"
One could avoid double complements by saying "the union of $S$ and the bounded components of the complement of $S.$" That makes sense even in (some) situations which are not compact. For example
"Let $T$ be a (finite union of) CLOSED simply connected set(s) and remove BOUNDED subsets of the interior.."
Of course that is the same as "the complement of (the union of) the unbounded component(s) of the complement of $S.$"
NOTE: Really I (and you) should say something like "remove a finite number of open subsets each of which has compact closure." Although one could also allow the case of an infinite number of holes, perhaps with disjoint closures. It depends somewhat on the exact situation you wish to model.
LATER: In answer to your actual question: I am not aware of a name for what you want, although that does not at all imply there isn't one. You could call it the simply connected closure and define it as the intersection of all simply connected closed sets containing $S.$ That would definitely work provided that such an intersection is closed and simply connected. It does not immediately give one an intuitive picture of what that thing is.
• Yes, the set I described in that comment should be compact. Besides, your definition is very nice. I guess that I could also define a hole $H_i$ of $S$ as a bounded connected subset of the complement of $S$. Then $H(S)$ is the union of all such sets. Would this be equivalent to the definition I originally posted? – jcmonteiro May 14 '16 at 21:19
• I worry that there might be complex topological situations beyond what you (and I) are thinking of. For example let $T$ be the disk of radius $\sqrt{2}$ centered at the origin (to avoid rational points on the boundary.) Let $x_1,\cdots$ be the rational points inside and $H_i$ the open disk of radius $r_i$ around $x_i$ where $r_i$ is the smaller of $4^{-i}$ and half the distance to the boundary. So some of the holes might overlap. Maybe that is not a problem. I'd feel on safer ground if the number of holes was finite or they were convex sets each with positive separation from the others. – Aaron Meyerowitz May 14 '16 at 21:50
• Unfortunately, I cannot force the holes to be convex (even though they will be in most cases). Assuming you are constructing the set $S$ by taking the original disk and removing those $H_i$, I don't get the problem of overlapping holes. What I would do is just call those original "removed chunks" $D_i$ and proceed to define the holes as before. Thus, overlapping "chunks" would all fall into the same hole. I would also use the same argument for intersecting $D_i$ (if there are any). That is why I only require a compact connected set. – jcmonteiro May 14 '16 at 22:15 | 2019-10-21T23:06:21 | {
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https://math.stackexchange.com/questions/679096/a-times-b-is-an-open-set-in-bbb-r2-implies-a-and-b-are-both-open-in | # $A \times B$ is an open set in $\Bbb R^2 \implies A$ and $B$ are both open in $\Bbb R$; $A,B \neq \emptyset$
I am studying Analysis on my own. Reading The Elements of Real Analysis by Bartle. Came across the above problem and I came up with the following solution but am very unsure about it. Would be very grateful if someone could look through it and verify it for me. Thanks in advance.
My Attempt:
Let $A$, $B \subseteq \Bbb R$ and let $A \times B$ be open in $\Bbb R^2$.
Let $x \in A$ be arbitrary. $\exists b \in B$ since $B$ is non-empty. Since $A \times B$ is open in $\Bbb R^2$, $\;\exists r_{x, \ b} \gt 0$ such that the set $B_{x, \ b} = \{ (z_1, z_2) \ | \ \left|{\left| { (z_1, z_2) - (x, b) } \right |}\right| \lt r_{x, \ b} \}$ is contained in $A \times B$.
Consider the set $A_x = \{ z \ | \ \left|{z - x}\right| \lt r_{x, \ b}\}$. It can be seen that $(z, b) \in B_{x, \ b} \subseteq A \times B \;\; \forall z \in A_x$. This will only be true if $A_x \subseteq A$. As per its definition, $A_x$ is an open ball.
Since $x \in A$ was arbitrary, we have shown that for each point in $A$ there is an open ball entirely contained in $A \implies A$ is an open set.
It can be similarly shown that $B$ is open by taking an arbitrary point in $B$ and a fixed one in $A$.
Q.E.D.
I have the following doubts..
$Q_1$ : Are the above arguments rigorous enough?? Because I solved this purely on geometric imagination thinking of a circle on the plane.
$Q_2$ : According to my proof above it can be shown in a similar way that $A_1 \times A_2 \times ...\times A_p$ is open in $\Bbb R^p \implies$ Each of $A_1, A_2.. A_p$ is open in $\Bbb R$(further worsening my doubts). Is that true in general??
• I think a lot of your doubt comes from the fact your grasp of both the cartesian product and the product topology is still shaky.A big flaw in your arguement is you simply assumed A and B are not the empty set-there's nothing in your arguement that shows that. By definition, if either A or B is the empty set, then A X B is empty! Overall,the basic logic looks ok,but A is an open set only if you can ensure the ball lies entirely on the real line in A for any z in A.Hint:Choose r*= min{Rz,Rb} where Rb is the radius of the open ball in AX B and Rz is the radius of the open ball in A. – Mathemagician1234 Feb 17 '14 at 5:54
• @Mathemagician1234 Well you're right. The result is still true if either $A$ or $B$ is empty. Well I thought I did use the non-void condition in assuming there is an element $b$ in $B$. About the hint: We do not know there is an open ball in $A$ yeah to come up with $R_z$? – Ishfaaq Feb 17 '14 at 6:02
• @Mathemagician1234: If you meant that I had not considered the case where $A$ and $B$ are empty - the question requires me to solve assuming they are both non-void.. – Ishfaaq Feb 17 '14 at 6:08
• The way you worded the question doesn't make that clear. But I think it's worth covering all cases if you're going to write up a full response. Notice since the empty set is open, the implication is true in that case-in effect,this is the trivial case of the answer to your question. Little things like this matter in mathematics. – Mathemagician1234 Feb 17 '14 at 6:36
• As for the hint, for every z,x in A, your definition is clearly an open ball in A. But you made z arbitrary in A. You defined Bx, b={(z1,z2) | ∣∣∣∣(z1,z2)−(x,b)∣∣∣∣<rx, b} is contained in A×B. We know for each ordered pair in A X B, the first coordinate lies in A by definition of the Cartesian product. So now let z=z1 in the ordered pair (z1,z2) in Bx. Hint: Write the norm that defines Bx explicitly in terms of the distance function in R2 by the length of the vectors.Then you'll see the resulting set must lie entirely in A and is indeed an open ball. – Mathemagician1234 Feb 17 '14 at 7:04
I think the proof is, logically, just fine.
Stylistically, I would make a few changes, none of them major or all that significant:
First, it seems weird to use $x$ and $b$ as your chosen elements of $A$ and $B$. I'd use $a$ and $b$.
Second, the notation is subscript heavy. For example, I'd probably just write $r$ instead of $r_{x,b}$ (or $r_{a,b}$). A lot of this depends on the mathematical maturity of you/your intended audience. The notation $r_{x,b}$ is wonderful at reminding you that $r$ depends on the point $(x,b)$, but if you and your audience don't need the reminder, it's kind of cumbersome.
Third, I'd write "consider the open ball $A_x = ...$," as opposed to "consider the set $A_x = ...$. This allows you to completely avoid the last line in that paragraph.
Lastly, rather than write "It can be seen that $(z,b)\in B_{x,b}...$" I would probably write the computation which proves it.
Q2: You're right that a small modification of your proof shows that if $A = A_1 \times \ldots \times A_p\subseteq \mathbb{R}^p$ is open with all of the $A_i$ nonepmty, then every $A_i$ must be open in $\mathbb{R}$.
As an aside, if you consider product topologies in general, then it's easy to see that the projection maps are open on the standard base, and hence open maps. This implies that for any open subset $O \subset \mathbb{R}^2$, $\pi_1[O]$ and $\pi_2[O]$ are open subsets of $\mathbb{R}$. This holds even in any finite or infinite product. So it's not a metric fact, in a way, but more topological. Of course, the relevant fact is that the standard metric on $\mathbb{R}^n$ actually indices the product topology, which is not a priori clear. | 2019-12-10T00:01:26 | {
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https://math.stackexchange.com/questions/1125834/do-there-exist-pro-p-groups-with-finite-quotients-of-non-p-power-order | # Do there exist pro-$p$ groups with finite quotients of non $p$ power order?
We define a pro-$p$ group to be a projective (i.e. inverse) limit of $p$-groups.
My question is exactly as stated in the title:
If a subgroup $H$ of a pro-$p$ group $G$ has finite quotient, must $|G/H|$ be some power of $p$?
If we restrict ourselves to open subgroups, then I believe this is the case because the open subgroups in the projective limit form an open basis of the identity. However for general finite index subgroups I am not sure.
No such subgroup exists. Pro-$p$ groups with the second definition (i.e. inverse limit of discrete, finite $p$-groups) can be easily shown to be equivalent to the first definition:
$$G/N\cong P$$
where $N$ is open, normal and $P$ is a finite $p$-group.
How so? Since $N$ is normal it is the union of some open subgroups from a basis of open normal subgroups around the identity, however we know that if
$$G=\varprojlim_{i\in I} G_i$$
with each $G_i$ a discrete, finite $p$-group, then such a basis is given by
$$U_i=\pi_i^{-1}(G_i)$$
If
$$N=\bigcup_{i\in J\subseteq I} U_i$$
select any $i_0\in J$ then we have a surjective homomorphism:
$$G/U_{i_0}\to G/N$$
with kernel $N/ U_{i_0}$, hence $G/N$ is the homomorphic image of a finite $p$-group, and is hence a $p$-group.
(Edit) For I think, following the comments, I should include the nitty gritty of the reduction to the closed case since there's enough confusion to merit it. Throughout we use $|\cdot|$ for the order of an element and of a subgroup, understood in the generalized sense of profinite groups (i.e. supernatural numbers)
The basic idea: just use the Lagrange theorem for profinite groups.
The problem: indices are only defined for closed subgroups.
If $|A|=p^nm$ with $(p,m)=1$ and select $a\in A$ such that $|a|\big| m$, which is possible by Cauchy's theorem. Name the projection map $\pi:G\to G/N\cong A$ (first isomorphism theorem) is surjective, we may select a lift $\stackrel{\sim}{a}\in G$, and by definition $\left|\pi\left(\stackrel{\sim}{a}\right)\right|=|a|$, we have just changed our context from $A$ to a subgroup of order coprime to $A$--namely $\langle a\rangle$--so that we may assume that $p\not\big| |A|$ rather than the weaker condition $|A|\ne p^n$. Denote by $H$ the closure of $\langle\stackrel{\sim}{a}\rangle$ in $G$. Then $|\stackrel{\sim}{a}|=m$ in $H/H\cap N$, which is finite because $N\cap H$ is clearly relatively open in $H$.
Since $H$ is closed, $m\big||H|$, and by Lagrange $|H|\; \bigg|\; |G|$. However $(m,p)=1$ and the only prime dividing $|G|$ is $p$, hence $m=1$, so that $\langle a\rangle\le A$ is the trivial subgroup and $|A|=p^n$.
• Doesn't the universal property go the other way? – Alexander Jan 30 '15 at 4:31
• But doesn't this only work for normal subgroups? – Alexander Jan 30 '15 at 5:14
• @Alexander You're assuming that $A$ is the image of $G$ under a homomorphism, and kernels of homomorphisms are always normal... And even so, the open normal subgroups still form a fundamental set of neighborhoods for the topology around the identity, that's how the inverse limit topology is defined. – Adam Hughes Jan 30 '15 at 5:16
• Yeah, sorry. I guess I meant open subgroups. Like what if the subgroup is not open but still finite index? – Alexander Jan 30 '15 at 6:14
• @Alexander that's impossible, profinite groups are compact, and if $U\subseteq G$ is open then we note that $$\bigcup_{g\in G} gU$$ is an open cover of $G$, so it has a finite subcover. i.e. $G=U\cup g_1U\cup\ldots\cup g_nU$ i.e. $U$ has finite index, since these are all the left cosets of $U$, and they are finite in number. If the group is finite index, then note that it is the complement of the union of the other left cosets, which are all closed, hence it is open. In fact $U$ is finite index, closed iff $U$ is open. – Adam Hughes Jan 30 '15 at 6:19
Look at Proposition 4.2.3 in the book by Ribes-Zalesskii, the whole section 4.2. will be interesting for you. Edit: the question is answered affirmatively there.
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Edward Jiang Jan 31 '15 at 21:12
• @Edward I agree, the post does not provide an answer, but the cited Proposition does. I did not require any clarification from the author nor did I intend to critisize the original post. To the contrary, I consider this an interesting question. – user 59363 Feb 1 '15 at 10:07
• @user59363 can you edit to provide more explicit support for the affirmative? Your answer may be accurate but is not approachably informative. – Joffan Feb 1 '15 at 10:20 | 2019-05-21T09:22:31 | {
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https://www.physicsforums.com/threads/analyzing-the-graphs-of-greatest-integer-functions.949820/ | # Analyzing the graphs of Greatest Integer Functions
Gold Member
## Homework Statement
Consider $u\left(x\right)=2\left[\frac{-x}{4}\right]$
(a) Find the length of the individual line segments of the function,
(b) Find the positive vertical separation between line segments.
## Homework Equations
The output of Greatest Integer Functions are always integers.
## The Attempt at a Solution
Length:
The text states that the coefficient of x within the greatest integer symbols is the length of the individual line segments of the graph.
In $u\left(x\right)=2\left[\frac{-x}{4}\right]$, the coefficient of x is $\frac{-1}{4}$.
However, the solution for the length of the graph states that length=4.
It explains this by stating that there's a decrease of 1 for every increase of 4 in the variable x.
This would make sense if we were talking about the slope of a line, but it doesn't make any sense at all in this context.
And since we're talking about the length of a line segment, does the negation matter?
Vertical Separation:
The text states that the coefficient of the greatest integer function is the positive vertical separation between line segments.
This is a straight forward statement, and the vertical separation=2, but I don't see why this leading coefficient determines this.
Can anyone help me get a better idea of what is going on with the graphs of these functions?
Related Precalculus Mathematics Homework Help News on Phys.org
andrewkirk
Homework Helper
Gold Member
I haven't seen the notation you are using but I presume that the square brackets [...] denote the function that gives the greatest integer that is less than or equal to the value of the contents of the brackets. In my experience that is usually indicated by $\lfloor...\rfloor$ and is called the 'floor' function.
If that is the case then the function value is piecewise constant and steps down at each multiple of 4, ie at ....,-8, -4, 0, 4, 8, ....
So each line segment is horizontal and extends for the period that x takes to increase by 4. So its length is 4.
How much does it step down by each time? Well $\lfloor\frac{-x}4\rfloor$ always has an integer value and decreases by 1 each time $x$ reaches a new, higher multiple of 4. So that's a step size of 1, but then it is multiplied by 2 - the leading coefficient - so the step size (the vertical separation) is 2.
The impact of the negation is to make the steps go down as $x$ increases, rather than up. But it doesn't affect the step length or height.
Gold Member
Yes we are talking about the same function. My text has the it denoted as something similar to but not exactly like [[x]]. But I looked up on Wikipedia the "floor function" that you stated and we're talking about the same thing.
That was a great explanation, thank you.
Let me see if I understand:
The length is 4 because if we were to create a table of values (leaving the leading coefficient of 2 out for simplicity's sake), it would take 4 integer x inputs for the line segment to be complete and move up or down to the next segment? For example, with an input of x = -8 gives an output of 2, and inputs x = -7, -6, -5, -4 all give an output of 1. Then when x = -3, the output is zero so there's a new line segment vertically shifted.
For the vertical displacement: the greatest integer function output is always an integer. So when we take the output of the function's argument, and multiply it by 2, we are vertically shifting by that value 2.
andrewkirk | 2020-04-04T16:34:52 | {
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https://math.stackexchange.com/questions/469800/is-there-an-and-symbol | # Is there an “and” symbol?
I was wondering if there was a symbol for "and"? For example, I want to say something like
$\therefore a = b \, \text{and} \, c = d$
Then can I replace "and" with a symbol?
Thanks!
• Though your mathematical argument may be much more legible if you write "Therefore, $a=b$ and $c=d$." (Unless you specfically want to prove a statement formally expressed using first order logic - but then you'd already know the $\land$ symbol ...) – Hagen von Eitzen Aug 17 '13 at 15:03
• Oh I see. Thanks, I didn't know that. Is it general practice to use symbols only when you're proving something? or do you use them at convenience? – Jeel Shah Aug 17 '13 at 15:08
• Just use a cost/benefit analysis. Costs: 1. People unfamiliar with this notation (like yourself a matter of minutes ago) will have to deduce (or guess) what it means. 2. \wedge takes more effort to type in LaTeX than and. Benefits: 1. Could potentially result in a better line break. 2. Looks more mathsy (useful for grant writing). Anything I've missed? – Douglas S. Stones Aug 17 '13 at 15:21
• I agree that its better not to use too many logic symbols. Then people end up having to try and decipher all the symbols if there are too many. But sometimes they are helpful if a statement has ambiguities in interpretation (then the logic symbols will help to write exactly what you mean). – Pratyush Sarkar Aug 17 '13 at 15:57
• Line breaking in the middle of an equation is close to the worst line breaking there is! – Mariano Suárez-Álvarez Aug 17 '13 at 23:56
## 2 Answers
"And": The logical symbol for and is given by $\land$ (known as the "wedge" symbol). And for future reference, the logical symbol for or is given by $\lor$, sometimes called "vee".
In your case, you can express $a = b\;\; \text{and}\;\; c = d$ as: $$a = b\, \land\, c = d$$
• For formatting in latex/mathjax: $P \land Q$ is given by P \land Q or by P \wedge Q. – amWhy Aug 17 '13 at 15:05
• +1 How does this symbol wedge spell in speaking, Amy? Can I spell it as "$P$ and $Q$"? – mrs Aug 18 '13 at 5:25
• @amWhy: Can use another TU! +1 – Amzoti Aug 19 '13 at 12:09
• @BabakS. Yes, it's simply read "P and Q". The name wedge is just a "name" of the symbol. – amWhy Aug 19 '13 at 12:15
Yeah. The symbol $\land$ is the most common. You can look at all types of logic symbols here: http://en.wikipedia.org/wiki/List_of_logic_symbols | 2020-07-04T21:59:27 | {
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http://math.stackexchange.com/questions/377868/taylor-series-for-ex-where-x-1-estimating-the-error | # Taylor Series for $e^x$ where $x = 1$, estimating the Error
I'm trying to calculate $e$ to a certain number of digits. The Maclaurin Series expansion of $\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$. When $x = 1$ we can approximate the value of $e$ by evaluating $\displaystyle\sum_{n=0}^m\frac1{n!}$. How can I find the error in the approximation when I compute the sum to $m$? If I take the output from the series, how many digits (truncated) of my approximation to $e$ are correct if I evaluate it to $m$ terms?
In general, how can we find the error when evaluating such a series?
Another good series, Brother's Formula (converges to $e$ faster than the one noted earlier):
$$\sum_{n=0}^\infty \frac{2n+2}{(2n+1)!}$$
Edit: To find the digits that are correct, add the upper bound for the error to the estimate and the digits before the first digit that changed are correct. For example, let's say that we have found $3.1234122$ as an estimate to some series. Let's say that we compute the error to be less than $0.0001879$. To find the digits that are correct in the estimate:
$$3.1234122 + 0.0001879 = 3.1236001$$ Thus we can see that the digits that we can be sure to be correct are $3.123$ so the first 4 digits of the number are $3.123$.
-
I've written another answer on this subject for Taylor series, and use $e^x$ as an example.
For Brother's formula, I've never seen it before but we can use some standard tricks to represent it.
I can first define $f(x) = (e^x - e^{-x})/2$ to cancel out all of the even terms out of the Taylor series for $e^x$:
$$f(x) = \sum_{n=0}^{+\infty} \frac{x^{2n+1}}{(2n+1)!}$$
If you're savvy, you might recognize this (either the series or the definition of $f(x)$) as the hyperbolic sine function: $f(x) = \sinh x$.
A common trick is to differentiate the series to bring the exponent on $x$ out as a coefficient. But this series is not quite right. Define
$$g(x) = x f(x) = \sum_{n=0}^{+\infty} \frac{x^{2n+2}}{(2n+1)!}$$
Now, define
$$h(x) = g'(x) = \sum_{n=0}^{+\infty} \frac{(2n+2) x^{2n+1}}{(2n+1)!}$$
So, we can apply the ideas of estimating the error of a Taylor series towards the calculation of $h(1)$, where
$$h(x) = \sinh x + x \cosh x = \frac{1}{2} \left( e^x - e^{-x} + x e^x + x e^{-x} \right)$$
Or, you might analyze the error on the Taylor series for $\sinh x$ and $\cosh x$ individually, and then combine them into an analysis of $h(x)$.
-
Strangely, that answer allows evaluation for the Mclaurin Series when $0<x<1$ and $x>1$, but not $x=1$, or did you just leave out the equals part on one of the ranges? – Justin May 1 '13 at 4:57
@gangqinlaohu: I fully expect that you could take any section of that analysis, and replace all of the $<$ with $\leq$ on one side of the inequalities. I'm just used to exluding the endpoints in these sorts of calculations because weird things often happen at endpoints. – Hurkyl May 1 '13 at 5:05
I don't quite understand all of the content from either answer. In your other answer, what is $c$? Also, the math is a bit confusing. In this answer, why did you define $f(x)$ to be $sinh(x)$? If I understand correctly, you are trying to remove all even terms from the series expansion of $e^x$ (which seems right for Brother's Formula). You then multiply by $x$ to make it fit the form you want. You then differentiate to find $h(x)$ which gives $e$ when $x=1$ but where did the formula for the error come from? What are these "ideas" of estimating the error of a Taylor Series? – Justin May 1 '13 at 5:15
There is a simple way to bound the error. Suppose you did your aproximation until some $m$, I mean, you computed $\displaystyle\sum_{n=0}^m\frac{1}{n!}$. Your exact error is $\displaystyle\sum_{n=0}^\infty\frac{1}{n!}-\sum_{n=0}^m\frac{1}{n!}=\sum_{n=m+1}^\infty\frac{1}{n!}$.
But note that $$\sum_{n=m+1}^\infty\frac{1}{n!}<\sum_{n=1}^\infty\frac{1}{(m+1)^n}=\frac{1}{m}.$$ What is an acceptable(if the idea is just to get some bound, not necessarily a good bound) error bound.
-
Improving @Integral's bound:
\begin{align} \sum_{n=m+1}^\infty \frac{1}{n!} &= \frac{1}{(m+1)!} \left( \frac11 + \frac1{m+2} + \frac1{(m+2)(m+3)} + \cdots \right) \\ &\le \frac{1}{(m+1)!} \left( \frac11 + \frac1{m+2} + \frac1{(m+2)^2} + \cdots\right) \\ &= \frac{1}{(m+1)!} \frac{1}{1-\frac1{m+1}} = \frac{1}{(m+1)!} \frac{m+1}{m} = \frac1{m\cdot m!} \end{align}
(comparing with a geometric series)
-
In fact, you can show that $\lim_{n\to\infty}\left(e-\sum_{k=0}^n\frac{1}{k!}\right)n\,n!=1$. – TZakrevskiy Nov 24 '15 at 11:42 | 2016-04-29T20:01:00 | {
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https://math.stackexchange.com/questions/1350953/symmetric-matrices-and-eigenvalues | # Symmetric matrices and eigenvalues
If the eigenvalues of a symmetric matrix $A$ are greater than 0, show that $v^{\top}Av > 0$ for every $v \ne 0$
I am trying to prove this as follows:
If $v$ is an eigenvector of $A$, then $Av = \lambda v$ and we can rewrite the expression as follows: \begin{align*} v^\top A v = v^\top \lambda v= \lambda \left(v^\top v\right) \end{align*} As $\lambda > 0$ and $\left(v^\top v\right) = \vert\vert v\vert \vert_2 > 0$ for all $v \ne 0$. Therefore, $v A v > 0$ has to be true.
What if $v$ is not an eigenvector of $A$? is it possible? I know this should be a proof for symmetric matrices to be positive-definite, but how can I strengthen this proof?
• No the condition $Av = \lambda v$ is for eigenvectors. Your proof looks fine to me. – IAmNoOne Jul 6 '15 at 4:45
• Hint - A symmetric real (thanks @hardmath) matrix $A$ is diagonalizable over the reals. Furthermore, all of the eigenspaces (corresponding to distinct eigen values) are mutually orthogonal. – peter a g Jul 6 '15 at 4:48
• It might be useful to clarify that $A$ is real symmetric. – hardmath Jul 6 '15 at 4:50
• $vAv$ ? I think it should be $v^TAv$ – Chiranjeev_Kumar Jul 6 '15 at 4:50
• @Chiranjeev: Yes, thanks for pointing this out. – amaatouq Jul 6 '15 at 4:52
By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e $$P^{T}AP=\pmatrix{\mu_1 \\ & \ddots \\ && \mu_n}$$ where $\mu_i>0$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$.
For any $v$, let $v=Pu$. Then $$v^TAv=u^TP^TAPu=\sum_{k=1}^n\mu_ku_k^2>0$$
• Yup! That's how to show this one. +1 – Mark Viola Jul 6 '15 at 5:38
If $A$ is a real symmetric matrix, then you can form an orthonormal basis for $\mathbb{R}^n$ from its eigenvectors. What happens when you expand $v$ in this basis?
@hermes already provided the way forward. I thought that it might be instructive to see the use of tensor notation herein. So, here we go ...
Let $U$ be the orthogonal matrix that diagonalizes $A$, such that $UU^T=U^TU=I$ and $(U^TAU)_{ij}=\lambda_i\delta_{ij}$, where $\lambda_i$ is the $i$'th eigenvalue of $A$ and $\delta_{ij}$ is the Kronecker Delta.
Using tensor notation, with summation implicit over repeated indices, we have
\begin{align} (v^TAv)_{in}&=v_iU_{ij}U^T_{jk}A_{k\ell}U_{\ell m}U^T_{mn}v_n\\\\ &=v_iU_{ij}\lambda_j\delta_{jm}U^T_{mn}v_n\\\\ &=v_iU_{im}\lambda_mU^T_{mn}v_n\\\\ &=\lambda_m\left(v_iU_{im}\right)\left(v_nU_{nm}\right)\\\\ &=\sum_{m}\lambda_m\left(v_iU_{im}\right)^2\\\\ &>0 \end{align}
as expected! | 2019-06-27T08:08:44 | {
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https://mathoverflow.net/questions/287182/picard-group-under-base-change-for-algebraically-closed-fields | # Picard group under base change for algebraically closed fields
Let $k \subset K$ be an extension of algebraically closed fields of characteristic $0$ (e.g. $\overline{\mathbb{Q}} \subset \mathbb{C}$).
Let X be a smooth variety over $k$. Then is the natural map $$\mathrm{Pic}(X)\,/\,n \to \mathrm{Pic}(X_K)\,/\,n$$ and isomorphism for all $n \in \mathbb{Z}$?
Here for an abelian group $A$ I denote by $A \, / \, n := A/nA$. Also $X_K$ denotes the base change of $X$ to $K$.
I can prove this if $X$ is proper using the theory of the Picard scheme. Here we have $\mathrm{Pic}^0(X)\,/\,n = 0$ (the set of points on an abelian variety over an algebraically closed field is divisible). The quotient $\mathrm{Pic}(X)/\mathrm{Pic}^0(X)$ is the Neron-Severi group, which is a finitely generated abelian group scheme hence its points don't change upon base change to $K$.
This proof breaks down for non-proper $X$ as the Picard functor is not representable in general.
• Yes. By resolution of singularities we can realize $X$ as dense open in a smooth proper $k$-scheme $X'$, so by smoothness ${\rm{Pic}}(X')\to {\rm{Pic}}(X)$ is surjective. Hence, ${\rm{Pic}}(X')/(n)\to {\rm{Pic}}(X)/(n)$ is surjective and likewise over $K$, so the isomorphism property for $X'$ implies surjectivity of the map for $X$. Injectivity follows by the usual spreading-out arguments (descending a relation "$L_K \simeq N^{\otimes n}$" over $X_K$ to one over $X_A$ for a finitely generated $k$-subalgebra $A \subset K$, etc.). – nfdc23 Nov 28 '17 at 15:56
• @Daniel: Do you mean "an extension of algebraically closed fields"? Do you mean $\overline{\Bbb Q}\subset \Bbb C$? – Mikhail Borovoi Nov 28 '17 at 16:05
• @Borovoi: Yes of course; I've corrected the typo. – Daniel Loughran Nov 28 '17 at 16:08
$\require{AMScd}$ The following result holds in much more generality than asked by the OP. There is no characteristic zero/smoothness assumption.
Proposition: Let $X/k$ be a qcqs scheme with $k$ separably closed. Let $K/k$ be a separable extension with $K$ separably closed. Let $n \in \mathbf{N}$ be a positive integer that is prime to the characteristic of $k$. Then the natural map $$\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} \to \frac{\operatorname{Pic}(X_K)}{n\operatorname{Pic}(X_K)}$$ is an isomorphism.
The proof goes as follows. The Kummer sequence gives the following commutative diagram with exact rows: $\require{AMScd}$ \begin{CD} 0 @>{}>>\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} @>{}>> H^2(X, \mu_n)@>{}>> H^2(X,\mathbf{G}_m) \\ {} @VVV @VVV @VVV {} \\ 0 @>{}>> \frac{\operatorname{Pic}(X_K)}{n\operatorname{Pic}(X_K)} @>{}>> H^2(X_K, \mu_n) @>{}>> H^2(X_K, \mathbf{G}_m) \end{CD}
The middle vertical arrow is an isomorphism by the smooth base change theorem in \'{e}tale cohomology (Corollary 1.3.1 of this):
Smooth Base Change: Suppose $K/k$ is an extension of separably closed fields, $X$ a qcqs $k$-scheme and $\mathscr{F}$ a torsion sheaf with torsion orders not divisible by the characteristic of $k$.Then, the base change map $$H^i(X,\mathscr{F}) \to H^i(X_K, \mathscr{F}_K)$$ is an isomorphism for all $i \geq 0$.
Therefore to prove the proposition, we reduce to proving the following injectivity statement concerning Brauer groups:
Proposition: Let $k \subseteq K$ be a separable extension of fields with $k$ separably closed. Let $X/k$ be a qcqs scheme. Then $H^2(X,\mathbf{G}_m)$ injects into $H^2(X_K, \mathbf{G}_m)$.
To prove this, write $K$ as a direct limit of smooth $k$-algebras. Since cohomology commutes with direct limits for qcqs schemes, if a (cohomological) Brauer class $\alpha$ in $H^2(X,\mathbf{G}_m)$ dies in $H^2(X_K,\mathbf{G}_m)$ it must die in $H^2(X_R, \mathbf{G}_m)$ for some smooth $k$-algebra $R$. By smoothness and using that $k$ is separably closed, $\operatorname{Spec} R$ has a $k$-rational point. Therefore the canonical map $f: X_{R} \to X$ has a section $g$ and so on cohomology the composition $$H^2(X,\mathbf{G}_m) \stackrel{f^\ast}{\to} H^2(X_R,\mathbf{G}_m) \stackrel{g^\ast}{\to} H^2(X,\mathbf{G}_m)$$ is the identity. We conclude that $\alpha = 0$ as desired.
Finally we remark that in the injectivity statement on Brauer groups, only the ground field is required to be separably closed. There is no requirement on $K$, only that $K/k$ be separable.
Edit: The proof also shows that for any \'{e}tale sheaf $\mathscr{F}$ on $X$ such that multiplication by $n$ is surjectve, then $$\frac{H^i(X,\mathscr{F})}{nH^i(X,\mathscr{F})} \to \frac{H^i(X_K,\mathscr{F})}{nH^i(X_K,\mathscr{F})}$$ is an isomorphism for all $i \geq 1$.
• Nice answer. I have a silly question though. Why do you say "By smoothness and [...]"? Does it not suffice that $R$ is a reduced $k$-finitely generated algebra for Spec $R$ to have a $k$-point? – Ariyan Javanpeykar Nov 29 '17 at 23:23
• @AriyanJavanpeykar: What if $R$ is a purely inseparable finite extension of $k$? – Daniel Litt Nov 29 '17 at 23:37
• @DanielLitt Thanks. That was bad of me. One needs that $R$ is geometrically reduced. – Ariyan Javanpeykar Nov 30 '17 at 0:18
• @AriyanJavanpeykar Yes. See Lemma 1.7.4. here. – Ben Lim Nov 30 '17 at 0:42
You may consult Florence Lecomte's paper: Rigidité des groupes de Chow, Duke Math. J. 53 (1986), no. 2, 405-426.
There is also Uwe Jannsen's preprint http://www.mathematik.uni-regensburg.de/Jannsen/home/Preprints/RIGBAT.pdf (rigidity also for other functors) | 2020-01-19T07:08:18 | {
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https://math.stackexchange.com/questions/927902/does-a-positive-semidefinite-matrix-always-have-a-non-negative-trace | # Does a positive semidefinite matrix always have a non-negative trace?
A simple question:
If $A$ is a positive semidefinite matrix ($A\succeq 0)$, does it imply that $\mbox{Tr}(A)\geq 0$, where the $\mbox{Tr}(\cdot)$ denotes the trace.
If not, any counter-example? Thanks.
• Suppose a matrix has a negative diagonal element. Can you see that the matrix cannot be positive semidefinite? – Daniel Fischer Sep 11 '14 at 19:03
• I know most likely this statement is wrong, but I just could not figure out an counter-example. – MIMIGA Sep 11 '14 at 19:06
• @MIMIGA the statement is correct: if $A \succeq 0$, then $\mathrm{Tr}(A) \geq 0$. – Omnomnomnom Sep 11 '14 at 19:12
• @Omnomnomnom Is the converse true? This is, if $\mbox{Tr}(A)\geq 0$, then $A\succeq 0$. – Diego Fonseca Aug 31 '16 at 15:19
• @DiegoFonseca No. Consider $$A = \pmatrix{2&0\\0&-1}$$ – Omnomnomnom Aug 31 '16 at 16:11
Suppose that $A = [a_{ij}]_{i,j=1}^n$ is such that $a_{ii} < 0$ for some $i$. Let $e_i$ be the $i$th standard basis vector; that is, $$e_i = (\overbrace{0,\cdots,0}^{i-1},1,0,\dots,0)$$ then $e_i^T Ae_i = a_{ii} < 0$, which means that $A$ is not positive semidefinite.
So, if $A$ is positive semidefinite, then all diagonal elements are non-negative, which means that the trace is non-negative. | 2019-05-23T14:58:19 | {
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http://math.stackexchange.com/questions/538402/derivatives-of-frac-csc-xe-x-and-ln-left-frac3x2-sqrt3x2-r/538550 | # Derivatives of $\frac{\csc x}{e^{-x}}$ and $\ln\left(\frac{3x^2}{\sqrt{3+x^2}}\right)$
I have tried to mainly ask thoughtful conceptual questions here, but now I am reduced to asking for help on a specific problem that I have been wrestling with for over an hour.
Disclaimer: I am not a lazy student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook. I am deeply grateful to the members of this community for their time.
$$f(x)= \frac{\csc(x)}{e^{-x}}$$ $$f'(x)= ?$$
Answer key says choice (1): $$e^x\csc2x(1-2\cot2x)$$
2) My answer does not match the answer key. Is there a typo?
Any guidance is enjoyed...it's driving me nuts!
$$\frac{d}{dx}\ln\left(\frac{3x^2}{\sqrt{3+x^2}}\right)$$
Answer key says choice (2): $$\frac{x^2+6}{x^3+3x}$$
-
Please, try to make the title of your questions more informative. Note that our maths renderer MathJax also works in titles. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. – Lord_Farin Oct 24 '13 at 17:39
The first is probably a typo, they thought they had $\csc(2x)$ on top. By the way, your work (which is correct for $\frac{\csc x}{e^{-x}}$) would have been somewhat easier if you had made the preliminary simplification to $e^x\csc x$. For the second, I would use "laws of logs" to simplify the expression before differentiating. Lots of work saved. – André Nicolas Oct 24 '13 at 17:44
The textbook answer to the second question is right. You missed a 2 in the first line of your work for the $-u \cdot \frac{dv}{dx} \space$part. It should have been $(-3x^2)\cdot \frac{2x}{2\sqrt{3+x^2}}$ – K. Rmth Oct 24 '13 at 17:51
Andre! Yes, the first question had a typo. Good call, it's very obvious now! – JackOfAll Oct 24 '13 at 18:05
K.Rmth! Yes, I forgot the chain rule at that step. Let me add back the (2x) and see if I can finish it the hard way. (Using Log rules up front sure make it easier) – JackOfAll Oct 24 '13 at 19:06
For 1, I would recommend re-writing it as: $f(x)=\frac{e^x}{\sin{x}}$.
Then using the quotient rule with $u=e^x, v=\sin{x},$ we obtain:
$f'(x)=\frac{e^x\sin{x}-e^x\cos{x}}{\sin^2{x}}=\frac{e^x(\sin{x}-\cos{x})}{\sin^2{x}}=\frac{e^x}{\sin{x}}\frac{\sin{x}-\cos{x}}{\sin{x}}=e^x\csc{x}(1-\cot{x}).$
I do not think this is equivalent to what was given in the textbook, although Wolfram Alpha agrees with my answer so I am inclined to think it is the correct one.
As for 2, it makes it considerably simpler to use log laws first, to show that: $\ln{\frac{3x^2}{\sqrt{3+x^2}}}=\ln{3}+2\ln{x}-\frac{1}{2}\ln{(x^2+3)}=g(x)$, say.
Then using the standard derivative rules for logs we get: $g'(x)=\frac{2}{x}-\frac{x}{x^2+3}=\frac{x^2+6}{x^3+3x}$, as required.
-
Thank you for the replies.
For #1, it looks like the question meant to say $csc(2x)$ Looks like I solved the original question correctly, and I redid the revised problem to match the given solution. Check!
For #2, yes, the log rules make it a lot easier. I was able to do it correctly that way. I also fixed my chain rule error and solved it correctly the "hard way". Damn, that took me a few tries and quite some time. But, I am stronger for it.
Thanks!!
- | 2015-04-18T13:15:03 | {
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http://www.billthelizard.com/2010/01/sicp-exercise-116-fast-exponentiation.html | ## Monday, January 11, 2010
### SICP Exercise 1.16: Fast Exponentiation
From SICP section 1.2.4 Exponentiation
Exercise 1.16 asks us to design an exponentiation procedure that evolves an iterative process using successive squaring and uses a logarithmic number of steps. A hint tells us to use the observation that (bn/2)2 = (b2)n/2 and to keep, along with the exponent n and base b, an additional state variable a, and to define the state transformation in such a way that the product a * bn is unchanged from state to state. The value of a should start at 1, and should contain the final answer by the end of the process.
Earlier in section 1.2.4, we were given the procedure:
(define (even? n) (= (remainder n 2) 0))
which tests whether an integer is even. This will come in handy again for this exercise, since we'll need to adjust a by a different amount when n is even than when n is odd.
We'll also use the following simple procedure for squaring a number:
(define (square x) (* x x))
With those building blocks in place, we can define our iterative procedure.
(define (expt-iter b n a) (cond ((= n 0) a) ((even? n) (expt-iter (square b) (/ n 2) a)) (else (expt-iter b (- n 1) (* a b)))))
We call the procedure expt-iter with the arguments base b, exponent n, and the state variable a. The first thing we do is check to see if the exponent is 0. If so, we've reached the end of the procedure and just return the value of a. Next we check to see if n is even. If it is, we use the observation that (bn/2)2 = (b2)n/2 to transform state information by squaring b and dividing n by 2. Finally, if n is odd we transform state information by reducing n by 1 and multiplying a by the base b.
The only thing that's left to do is define a procedure that calls expt-iter with the correct initial values.
(define (fast-expt b n) (expt-iter b n 1))
Here are a few sample runs of the procedure:
> (fast-expt 2 2)4> (fast-expt 2 3)8> (fast-expt 2 10)1024> (fast-expt 3 3)27> (fast-expt 5 5)3125
You can try running the procedure with large values of n to verify that it really is fast.
Related:
For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.
pbewig said...
The ipow function in my Standard Prelude does this using exactly the same algorithm:
(define (ipow b e)
(cond ((zero? e) 1)
((even? e) (ipow (* b b) (/ e 2)))
(else (* b (ipow (* b b) (/ (- e 1) 2))))))
Bill the Lizard said...
pbewig,
There's a lot of really useful stuff in your Standard Prelude. That will probably be the first place I look for hints if I get stuck on any SICP exercises. I can already tell it will come in handy when I get to the next chapter, which deals with data abstraction.
geeknanny said...
pbewig's ipow function isn't tail-recursive, is it? The else condition calls ipow and then multiplies the result by b. Rewriting the function to be tail-recursive is the whole point of the exercise, yes?
Bill the Lizard said...
geeknanny,
You're absolutely right. The tail-call version will be more efficient in both time and (stack) space for most exponents. (They'll run virtually the same for exponents that are powers of 2.)
I hadn't noticed this before, thanks for pointing it out.
Anonymous said...
If you subtract 1 from n, the product b*n is not invariant. Is this correct?
Anonymous said...
I apologize, there is a problem with my browser, the text shows up as:
"the product a bn is unchanged"
In your example: a * (b ^ n) is invariant, which is correct. | 2016-02-08T23:02:37 | {
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https://math.stackexchange.com/questions/2784334/construct-a-function-on-a-bounded-interval-on-bbbr-which-is-continuous-ever | # Construct a function on a bounded interval on $\Bbb{R}$ which is continuous everywhere but differentiable only at irrationals.
Initially, I constructed a function with the help of famous Thomae's function, but later I found it to be wrong. Firstly, I did-
Let $f:[0,1]\to\Bbb{R}$ be the Thomae's function defied by $$f(x)=\begin{cases}0& \text{if x is irrational} \\{1 \over q} &\text{if x is rational, x={p \over q} with p\in\Bbb{Z}, q\in\Bbb{N} are coprime}\end{cases}$$ This function is Riemann Integrable on $[0,1]$. Now, define $F:[0,1]\to\Bbb{R}$ by $F(x)=\int_{[0,x]} f\:\forall x\in[0,1]$
But here $\int_{[0,x]} f=0\:\forall x\in[0,1]\implies F(x)=0\:\forall x\in[0,1]\implies F$ is differentiable everywhere on $[0,1]$.
But, as per the question $F$ is not the required function.
Can anybody give me an example of such a function with some proper arguments?
• I feel there is no such curve. But since Weierstrass discovered a function having the property of being continuous everywhere but differentiable nowhere the geometric intuition lost a good role in the analysis. – Piquito May 16 '18 at 22:42
• To be clear: you want it to be differentiable at every irrational, and nowhere else? – Robert Israel May 16 '18 at 23:45
• @RobertIsrael, this is a problem from the book of Terence Tao Analysis 1 – Biswarup Saha May 17 '18 at 4:43
I provide here an explicit construction (in a more general case).
Consider $D = (a_n)_{n \ge 1}$ a countable subset of $[0,1]$. Then the following function is differentiable exactly on $[0,1] \backslash D$ : $$f : x \in [0,1] \mapsto \sum\limits_{n=1}^{+\infty} \frac{|x-a_n|}{2^n}.$$
In the rest of the proof, $[a,]$ will denote the points between $a$ and $b$ regardless of which is greater (even if $a > b$, $[a,]$ is not empty ; $[a,b]=[b,a]$). Take any $x \in [0,1]$. Note that if $a > \max(x,y)$ then $|y-a|-|x-a|=x-y$, if $a < \min(x,y)$ then $|y-a|-|x-a|=y-x$, and if $a$ is between $x$ and $y$, $||y-a|-|x-a||\le 2|y-x|$.
Case 1: $x \notin D$. Let $n_0 \in \mathbb{N}$. For $y$ in some neighborhood of $x$, $[x,y]$ does not contain $a_1,...,a_{n_0}$. Then $\frac{f(y)-f(x)}{y-x} = \sum\limits_{a_n > \max(x,y)} \frac{1}{2^n} + \sum\limits_{a_n < \min(x,y)} \frac{-1}{2^n} + \sum\limits_{a_n \in [x,y]} \frac{|y-a_n|-|x-a_n|}{2^n}$. And $\big| \sum\limits_{a_n \in [x,y]} \frac{|y-a_n|-|x-a_n|}{2^n|y-x|}\big| \le \sum\limits_{n>n_0} \frac{2}{2^n}=\frac{1}{2^{n_0-1}}$. Hence $f$ is differentiable at $x$ and $f'(x) = \sum\limits_{a_n > x} \frac{1}{2^n}-\sum\limits_{a_n < x} \frac{1}{2^n}$.
Case 2: $x = a_m \in D$. Let $n_0 \in \mathbb{N}$. For $y$ in some neighborhood of $x$, $]x,y]$ does not contain$a_1,...,a_{n_0}$. Then as before, we have $\frac{f(y)-f(x)}{y-x}=\sum\limits_{a_n > \max(x,y)} \frac{1}{2^n} - \sum\limits_{a_n < \min(x,y)} \frac{1}{2^n} + \frac{|y-a_m|}{2^m (y-a_m)} + C(y)$ where $|C(y)| \le \frac{1}{2^{n_0-1}}$. Denoting $K = \sum\limits_{a_n>x}\frac{1}{2^n}-\sum\limits_{a_n<x}\frac{1}{2^n}$, we find that the left limit is $\frac{f(y)-f(x)}{y-x} \underset{y\to x^-}{\longrightarrow} K - \frac{1}{2^m}$, while the right limit is $\frac{f(y)-f(x)}{y-x} \underset{y \to x^+}{\longrightarrow} K - \frac{1}{2^m}$. Hence $f$ is not differentiable at $x$.
Conclusion given $D = \{a_n \ |\ n \ge 1\}$ a countable subset of $[0,1]$, $f : x \mapsto \sum\limits_{n=1}^{+\infty}\frac{|x-a_n|}{2^n}$ is differentiable exactly at $[0,1]\backslash D$.
• +1 I haven't checked the details, but I'm pretty sure this works, and I'm surprised I'd forgotten its history. Schwarz gave an example of such a function in 1873, using roughly the same method (which is basically an application of Hankel's 1870 method of "condensation of singularities"). My next 3 comments quote from my answer to Is Kline right that Cauchy believed that continuous functions must be differentiable? (middle of Note for [3]). – Dave L. Renfro May 17 '18 at 14:20
• Gilbert's approach was based on Lamarle's methods and was intended to conclusively prove that a continuous function is differentiable except for an isolated set of points (see Gilbert's final theorem at the bottom of p. 31). However, after Gilbert learned of the results in Hermann Amandus Schwarz's 1873 paper Neues beispiel einer stetigen nicht differentiirbaren function [New example of a continuous non-differentiable function] French version, (continued) – Dave L. Renfro May 17 '18 at 14:21
• which gives a particularly simple example of a strictly increasing continuous function that is not differentiable at each nonzero dyadic rational number (in fact, at each such number the right derivative is $+\infty$ and the left derivative is positive and finite), Gilbert came to realize that Hankel's examples were possible. (Schwarz presented his example on 19 August 1873, but since Gilbert's Rectification was presented in a session dated 7 June 1873, Gilbert obviously knew about it before Schwarz's presentation. (continued) – Dave L. Renfro May 17 '18 at 14:21
• Incidentally, although Weierstrass' presentation of an everywhere non-differentiable continuous function was on 18 July 1872, I can find no mention of Weierstrass in Gilbert's Rectification.) – Dave L. Renfro May 17 '18 at 14:22
Yes, such a function exists. In fact, Zahorski [1] (see also p. 103 here) proved that if $E$ can be written as $E = G \cup Z,$ where $G$ is a $\mathcal{G}_{\delta}$ set and $Z$ is a $\mathcal{G}_{\delta \sigma}$ Lebesgue measure zero set, then there exists a continuous function $f: {\mathbb R} \rightarrow {\mathbb R}$ such that the set of points at which $f$ does not have a finite two-sided derivative is equal to $E.$ (Zahorski’s proof was simplified in [2].) Choosing $G$ to be the set of irrational numbers and $Z$ to be the empty set gives the result you want.
Off-hand, I don’t know of a simple explicit description of such a function for the set of irrational numbers (I've probably never previously looked for one or thought much about it), but I believe Lemma 3 on p. 151 of Zahorski’s paper gives the closest special case in his paper for what you want --- the construction of a continuous function whose non-differentiability set is any specified $\mathcal{G}_{\delta}$ set that contains no intervals. If I happen to come across such a function (continuous, with irrationals as its non-differentiability set) at some later time, I'll try to remember to come back here and mention it.
I’ve posted several answers/essays that give more details, references, and related results pertaining to this topic --- some of those are listed below (in chronological order).
[1] Zygmunt Zahorski, Sur l'ensemble des points de non-dérivabilité d'une fonction continue [On the set of points of non-differentiability of a continuous function], Bulletin de la Société Mathématique de France 74 (1946), 147-178. paper at EUDML and paper at NUMDAM
[2] George Piranian, The set of nondifferentiability of a continuous function, American Mathematical Monthly 73 #4 (April 1966) [Part II: Papers in Analysis, Herbert Ellsworth Slaught Memorial Papers #11], 57-61.
Differentiability of the Ruler Function (13 December 2006 sci.math post)
Continuous functions are differentiable on a measurable set?
Set of zeroes of the derivative of a pathological function
Characterization of sets of differentiability
Points of differentiability of $f(x) = \sum\limits_{n : q_n < x} c_n$
Monotone Function, Derivative Limit Bounded, Differentiable – 2 | 2019-05-26T02:02:44 | {
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https://math.stackexchange.com/questions/3790744/if-0-9999-ldots-1-then-why-is-lim-n-to-infty-frac-tan89-n-text9 | # If $0.9999\ldots=1$, then why is $\lim_{n\to\infty}\frac{\tan(89.[n\,\text{"$9$"s}]^\circ)}{\tan(89.[(n-1)\;\text{"$9$"s}]^\circ)}$ not equal to $10$?
If $$0.9999\ldots=1$$, then why is this limit not equal to $$10$$? $$L = \lim_{n \to \infty} \frac{\tan(89.\overbrace{9999...}^{\text{n times}} \space ^\circ)}{\tan(89.\underbrace{999...}_{\text{n-1 times}} \space ^\circ)}$$
We can rewrite this limit as $$\begin{gather} L = \lim_{n \to \infty} \frac{\tan\left( \frac \pi 2 - \frac{\pi}{180 \times 10^n}\right)}{\tan\left( \frac \pi 2 - \frac{\pi}{180 \times 10^{n-1}} \right)} \\ L = \lim_{n \to \infty} \frac{\tan\left( \frac{\pi}{180 \times 10^{n-1}}\right)}{\tan\left( \frac{\pi}{180 \times 10^{n}}\right)} \end{gather}$$ let $$t = \frac{\pi}{180 \times 10^n}, t \to 0$$ $$L = \lim_{t \to 0}\frac{\tan10t}{\tan t} \\ \boxed{L = 10}$$
However, according to the well-known proof 0.9999 = 1, shouldn't the limit be $$\frac{\tan(90^\circ)}{\tan(90^\circ)}$$, which is undefined? Where am I going wrong here?
• There is a condition on the "rule" $$\lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}\,.$$ Is that condition satisfied here? Aug 14, 2020 at 14:29
• Think on a simpler example: $\lim_{x\to \infty}\frac{10x}{x} = \lim_{x\to \infty} 10 = 10$. However $\frac{\lim_{x\to\infty} 10x}{\lim_{x\to\infty} x} = \frac{\infty}{\infty}$ is undefined. Aug 14, 2020 at 14:33
• Well, it doesn't tell us that the limit is $10$, nor that it exists at all, for that we must take a closer look at the quotients. But it tells us that separating numerator and denominator doesn't help us at all here. Aug 14, 2020 at 14:36
• @Novice $\tan(\frac \pi 2 - \theta) = \cot \theta = \frac 1 {\tan\theta}$ Aug 14, 2020 at 14:38
• "why is this limit not equal to 10?": hem, it is equal to $10$.
– user65203
Aug 14, 2020 at 15:15
$$\lim_{x\to a}f(x)$$ you don't care about $$f(a)$$ (which could be defined or undefined), but only about $$f(x)$$ for $$x\ne a$$.
We cannot simply assume that both of the limits $$\lim_{n \rightarrow \infty}f(x)$$ and $$\lim_{n \rightarrow \infty}g(x)$$ exist, so the formula $$\lim_{n \rightarrow \infty} \dfrac{f(x)}{g(x)}= \dfrac{\lim_{n \rightarrow \infty}f(x)}{\lim_{n \rightarrow \infty}g(x)}$$ doesn't have to work.
As a result, you can't think that if $$89,999\ldots\approx90$$ then in the formula which you analyse is the same in numerator and denominator and the result is simply $$1$$. You ought to go deeper and that is what you have done after.
As pointed out in the comments, $$\lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$$ only when $$\lim f(x)$$ and $$\lim g(x)$$ exist. In this case, both of these limits do not exist, which is why $$\lim_{n \to \infty} \frac{\tan(89.\overbrace{9999...}^{\text{n times}} \space ^\circ)}{\tan(89.\underbrace{999...}_{\text{n-1 times}} \space ^\circ)} \ne \frac{\tan 90^\circ}{\tan 90^\circ}$$ The limit must be evaluated as shown in the question to obtain the answer, which is 10. | 2022-05-23T15:53:06 | {
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http://math.stackexchange.com/questions/47734/prove-that-62n33n2n/47736 | # Prove that $6|2n^3+3n^2+n$
My attempt at it: $\displaystyle 2n^3+3n^2+n= n(n+1)(2n+1) = 6\sum_nn^2$ This however reduces to proving the summation result by induction, which I am trying to avoid as it provides little insight.
-
if you want to use induction then use the fact that for your polynomial $p(n)=n(n+1)(2n+1)$ the following property holds: $$p(n+1)=6(n+1)^2+p(n)$$ – miracle173 Jun 26 '11 at 9:56
If you write this as $2n^3+3n^2+n\equiv 0 \mod 6$ then you only need to check $n=0,1,2,3,4,5$.
Alternatively, write as $\dfrac{2n(2n+1)(2n+2)}{4}$ where the numerator obviously has a multiple of 3, a multiple of 4 and another multiple of 2, so is divisible by 24, meaning the expression is divisible by 6.
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Dear @henry, could you incorporate the comment into your answer? All other solutions are correct, but that seemed the most elegant and closest to what i was looking for. (re: 3 consecutive) – kuch nahi Jun 26 '11 at 7:59
You have $2n^3+3n^2+n=n(n+1)(2n+1)$, and $2\mid n(n+1)$. If $3\mid n(n+1)$, then you're done. Otherwise, $n\not\equiv 0\pmod{3}$ and $n\not\equiv -1\pmod{3}$, so $n\equiv 1\pmod{3}$. Then $2n+1\equiv 3\equiv 0\pmod{3}$, so $3\mid 2n+1$ and you get the result.
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HINT $\rm\ f(n) =\: 3\ (n^2+n) + 2\ (n^3-n)\ =\ 3\ n\ (n+1)\ +\ 2\ (n-1)\ n\ (n+1)\:.\:$ But $2$ divides one of $\rm\:n,\:n+1\:$ and $3$ divides one of $\rm\:n-1,\:n,\:n+1\:.\:$ Or, said in terms of binomial coefficients,
$$\rm f(n)\ =\ 6\ {n+1\choose 2}\ +\ 12\ {n+1\choose 3}\quad\text{is a multiple of}\ \ 6$$
In fact this generalizes widely: it is a classical result of Polya and Ostrowski (1920) that every integer valued polynomial, i.e. every $\rm\:f(x)\in \mathbb Q[x]\:$ with $\rm\:f(\mathbb Z)\subset \mathbb Z\:,\:$ is an integral linear combination of binomial coefficients. See this answer for references (and a similar problem).
-
Yet another way to look at it is as follows:
First, as several others have already noted, $n(n+1)$ is divisible by $2$, so we just need to check for divisibility by $3$. Now, $n \equiv 0,1, \text{ or } 2 (\text{mod } 3)$. In the case of $n \equiv 0 (\text{mod } 3)$, the problem is trivial. In the case of $n \equiv 2 (\text{mod } 3)$, $n+1 \equiv 0 (\text{mod } 3)$. There is but one last case, but that too is covered: $2n+1 \equiv 0 (\text{mod } 3)$ if $n \equiv 1 (\text{mod } 3)$. To summarize...
$n \equiv 0 (\text{mod } 3) \implies n \equiv 0 (\text{mod } 3)$
$n \equiv 1 (\text{mod } 3) \implies 2n+1 \equiv 0 (\text{mod } 3)$
$n \equiv 2 (\text{mod } 3) \implies n+1 \equiv 0 (\text{mod } 3)$
How's that? :)
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Yet another way to see it: consider n(n+1)(2n+1). Write the third factor as (2(n+2)-3), so we have n (n+1) (2(n+2)-3). Since one of n, n+1 must be even, the product is divisible by 2.
One of n, n+1, n+2 must be divisible by 3. Note that n+2 is divisible by 3 if and only if 2(n+2)-3 is divisible by three, so this means that one of n, n+1, 2(n+2)-3 is divisible by three, and hence so is their product.
Since 2 and 3 are relatively prime, we have that n(n+1)(2n+1) is divisible by their product, 6.
-
Note that either $n$ or $n+1$ is divisible by 2. Now if $n=3k+c$, then $n+1 = 3k+c+1$ and $(2n+1) = 6k + 2c+1$. If $c=0$ then $n$ is divisible by 3, if $c=1$ then $2n+1$ is divisible by 3, and if $c=2$ then ...
-
HINT $\rm\quad 6\ |\ f(0) = 0\:$ and $\rm\ 6\ |\ f(k+1)-f(k)\:=\ 6\ (k+1)^2\:$ so $\rm\:6\ |\ f(n)\:$ by telescopic induction
$$\rm f(n)\ =\ (f(n)-f(n-1))\ +\ (f(n-1)-f(n-2))\ +\ \cdots\ +\ (f(1)-f(0))\ +\ f(0)$$
I.e. $\rm\ f\:$ is constant mod $6$, $\rm\:\ f(k+1)\equiv f(k)\$ hence $\rm\ f(n)\equiv f(0) = 0\:.$
You can find many examples of telescopy in my prior posts here.
-
Since $\displaystyle\frac{2n^3+3n^2+n}{6}=2\binom{n}{3}+3\binom{n}{2}+\binom{n}{1}$, it is easy to see that $6|(2n^3+3n^2+n)$.
Here is a general collection of results that can be applied in cases like this.
Define the combinatorial polynomial of degree $k$: $C_k(n)=\binom{n}{k}$. Let $L_k$ be the set of integral linear combinations of combinatorial polynomials of degree at most $k$. That is, $$f\in L_k \Leftrightarrow f=\sum\limits_{j=0}^ka_kC_k\text{ for some }a_k\in\mathbb{Z}$$ Claim: Let $\{ a_j : j = 0\dots k \}$ be a set of $k+1$ integers, then there exists a $P\in L_k$ such that $P(j) = a_j$ for $j = 0\dots k$.
Proof: Since $a_0C_0(0)=a_0$, the claim is true for $k=0$.
Suppose the claim is true for some $k$. Let $\{ a_j : j = 0\dots k+1 \}$ be a set of $k+2$ integers and let $Q$ be an element of $L_k$ so that $Q(j) = a_j$ for $j = 0\dots k$. Since $b = a_{k+1} - Q(k+1)$ is an integer and $$C_{k+1}(j) = \left\{\begin{array}{ll}0&\text{for }j=0\dots k\\1&\text{for }j=k+1 \end{array}\right.$$ $P(j) = Q(j) + b C_{k+1}(j)$ is an element of $L_{k+1}$, $P(j) = Q(j) = a_j$ for $j = 0\dots k$, and \begin{align} P(k+1) &= Q(k+1) + b C_{k+1}(k+1)\\ &= Q(k+1) + (a_{k+1} - Q(k+1))\\ &= a_{k+1} \end{align} Thus, the claim is true for $k+1$.$\hspace{.25in}\square$
Theorem: a polynomial, $P:\mathbb{Z}\mapsto\mathbb{Z}$ if and only if $P\in L_k$ for some $k$.
Proof: Because $C_k:\mathbb{Z}\mapsto\mathbb{Z}$, it is easy to see that any $f\in L_k$ sends $\mathbb{Z}\mapsto\mathbb{Z}$.
Let $Q$ be a polynomial of degree $k$ that maps $\mathbb{Z}\mapsto\mathbb{Z}$. Let $P$ be a polynomial in $L_k$ such that $P(j) = Q(j)$ for $j = 0\dots k$. Since a polynomial of degree $k$ is determined by its values at $k+1$ points, we must have that $P = Q$; that is, $Q\in L_k$.$\hspace{.25in}\square$
So we have a characterization of all polynomials that map $\mathbb{Z}\mapsto\mathbb{Z}$. We also have
Corollary: If a polynomial of degree $k$ maps $k+1$ consecutive integers to integers, it maps all integers to integers.
Proof: Suppose $P$ is a polynomial of degree $k$ and $P:\{m,m+1,m+2,\dots,m+k\}\mapsto\mathbb{Z}$. The Claim above assures that there is a $Q\in L_k$ so that $Q(j)=P(m+j)$ for $j=0,1,2,\dots,k$. Since a polynomial of degree $k$ is determined by its values at $k+1$ points, we must have that $P(j)=Q(j-m):\mathbb{Z}\mapsto\mathbb{Z}$.$\hspace{.25in}\square$
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Well you can divide $n$ by $3$ using the usual division with remainder to get $n = 3k + r$ where $r = 0, 1$ or $2$. Then just note that if $r = 0$ then $3$ divides $n$ so $3$ divides the product $n(n+1)(2n+1)$.
If $r = 1$ then $2n + 1 = 2(3k+1) + 1 = 6k+3 = 3(2k+1)$ so again $3$ divides $2n+1$ so it divides the product $n(n+1)(2n+1)$. And similarly you can check that if $r = 2$ then something like this happens so in all possible cases $3$ divides $n(n+1)(2n+1)$
And then you can do the same process but with 2 instead, that is, writing $n = 2k + r$ where now $r = 0$ or $1$.
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To prove that an expression in terms of $n$ is divisible by 6, it may be helpful to look at the cases where $n=6k$, $n=6k+1$, $n=6k+2$, $n=6k+3$, $n=6k+4$, and $n=6k+5$, where $k\in\mathbb{Z}$.
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This sounds cumbersome. We know that one of the three factors in $n(n+1)(2n+1)$ must be even. Is there a way to show that one of them must be divisible by three as well? (If they were consecutive, this would have been obvious) – kuch nahi Jun 26 '11 at 7:55
@kuch nahi: Ahh, if you already know that you've got a factor of 2 and just need to check for the factor of 3, then you can look at only 3 cases: $n=3k$, $n=3k+1$, and $n=3k+2$. – Isaac Jun 26 '11 at 7:57
@kuch nahi: $2n(2n+2)(2n+1)/4$ has three consecutive – Henry Jun 26 '11 at 7:58
$n(n+1)$ is divisible by $2$.
And $2(2n^3+3n^2+n) = n(n+1)(4n+2) \equiv n(n+1)(n+2) \mod 3$ is divisible by 3.
-
to check for divisibility by 6 a number must be divisible by both 2 and 3 so we will prove that $2n^3 + 3n^2 + n = n (2n^2 + 3n +1) = n (n+1) (2n+1)$
If $n$ is even then 2 divides $n$ and $n+1$ will be odd so $n+1$ can be $3k+2$ or $3k$ where $k$ is some integer.
If $3k+1 = n+1$ as it would make $n$ itself a multiple of 6 as our assumption that $n$ is even
So if $n+1$ is $3k$ it can be divided by 3 so no problem. But if $n+1 = 3k+2$ then $2n+1$ will be $2(3k+1) +1 = 6k+3$ which is divisible by 3 and hence by 6.
Case 2: If $n$ is odd then $n+1$ is even and thus divisible by 2.
• if $n$ is $3k$ then it is divisible by 3 so no issues
• if $n$ is $3k+2$ as it makes $n+1=3k+3$ which is itself a multiple of 6 as our earlier assumption is that $n+1$ is even
• if $n$ is $3k+1$ then $2n+1$ becomes $2(3k+1)+1=6k+3$ which is divisible by 3 and hence proved
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I've tried to made your post more readable by adding formatting, LaTeX markup, paragraphs.... I hope I did not unintentionally changed meaning somewhere - of course, you should edit your post again, if you think it's needed. – Martin Sleziak Jan 7 '12 at 17:57 | 2014-10-22T00:28:40 | {
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https://math.stackexchange.com/questions/3120256/sum-to-infinity-of-a-series | # Sum to infinity of a Series
Problem: Find the sum of the series $$x+2x^2+3x^3+4x^4+... = \sum\limits_{n=1}^\infty nx^n$$
My solution is as follows: Let P be the above series. $$P=x(1+2x+3x^2+4x^3+...) =x(1+x+x^2+x^3+x^4+...x+2x^2+3x^3+4x^4) =x\left(\frac{1}{1-x}+P\right)$$
Then by some trivial algebraic manipulation we obtain $$P=\frac{x}{(1-x)^2}$$
However, I am mainly concerned about a technical issue in my solution. From the first to the second step, I sort of "rearranged" /“split” the terms in my series so as to massage it into a desired form. Yet I am unable to prove that the series in the bracket is convergent or even absolutely convergent, so I am scared that a permutation of the terms might affect the limit that the series actually approaches. (I have heard of the Riemman Series Theorem before). Is there thus any way I can prove the convergence of the series in the brackets? I may even be going in circles and have confused myself in the process.
Any help will be much appreciated. Thanks in advance.
You can find where the series converges using D'Alembert's convergence test. Namely,
$$\lim\limits_{n\to\infty}\frac{|(n+1)x_{n+1}|}{|nx_n|} = \lim\limits_{n\to\infty}\left| \frac{n+1}{n}\right||x| = |x|$$ If this limit is less than 1 (i.e. $$|x|< 1$$), then the series converges. When $$|x|>1$$, then the series diverges. For $$|x|=1$$, the test is inconclusive, but we can verify by plugging in $$x=1$$ and $$x=-1$$ respectively, that those series diverge. Another special case to consider is $$x=0$$, for which the series obviously converges (and its sum is 0).
Thus, we conclude that this series converges iff $$|x|<1$$.
Let $$P_N=\sum\limits_{n=1}^Nnx^n$$ Note that $$P=\lim\limits_{N\to\infty}P_N$$
We have $$P_N=x\cdot \sum\limits_{n=1}^N nx^{n-1} = x \cdot \left(\underbrace{1+x+...+x^{N-1}}+\underbrace{x+2x^2+...+(N-1)x^{N-1}}\right) =$$ $$= x \cdot \left( \sum\limits_{n=0}^{N-1} x^n + \sum\limits_{n=1}^{N-1} nx^n\right) = x \cdot \left( \frac{1-x^N}{1-x} + P_{N-1} \right)$$
Letting $$N\to\infty$$, we arrive at the same result as you did. You can also notice that I did the same manipulations as you did, the only difference is that I was considering finite sums.
There is a theorem that states that if a series converges absolutely (the one in your example does), then any series obtained by any permutation of the summands converges and it has the same sum as the starting series.
If you find yourself unsure whether you can permute the summands, try to consider finite sums, because then you can use any manipulations that apply to finite sums, including permutation of elements. After that, just take the limit as $$N \to \infty$$.
For $$|x|<1$$ we have $$P=x(1+2x+3x^2+...+)=x(x+x^2+x^3+...)'=x\cdot\left(\frac{x}{1-x}\right)'=...$$ Can you end it now?
Hint: For the finite $$\sum_{i=1}^{n}ix^i$$ we get $$\sum_{i=1}^{n}i x^i=\frac{(n x-n-1) x^{n+1}+x}{(1-x)^2}$$ | 2021-04-12T00:52:36 | {
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https://www.physicsforums.com/threads/two-forces-two-angles-net-force.289544/ | # Two forces, two angles; Net force = ?
1. Feb 3, 2009
### MaximaMan
Greetings everyone! Thanks in advance for the help!!!
1. The problem statement, all variables and given/known data
Two parts to the problem. For both parts it asks for the magnitude of the net force and the direction of angle from the X axis.
Part A:
Two forces, F1 and F2, act on an object at rest at a 90* angle. F1 = 30N, F2 = 11N, M = 9kg
|F1
|
|
|____F2___
Part B:
F1/
./
/____ F2
Theta is 60*
3. The attempt at a solution
Part A:
For magnitude, I got 31.8N.
F-net = sqrt [ f1^2 + f2^2 ]
Angle: 24*
Part B:
Magnitude:
F-net = F1 + F2
F1 = (30N) cos 0 + (30N) sin 0 => (30N) 1 + 0 => 30N
F2 = (11N) cos 60 + (11N) sin 60 => (11N) 1/2 + (11N) 0.8660 => 15N
Therefore, Fnet = 45N
Angle: 16*
I really dont think I came close to the answers...but someone let me know, please!
2. Feb 3, 2009
### Hannisch
Part A:
The way you've calculated your magnitude is correct, so I suppose your answer is as well. (although I can't for my life figure out why on earth you're given mass = 9kg.)
How did you calculate the angle? Because I don't get the angle to be 24 degrees.. (it's a bit because the y-component is larger than the x-component - how can the net force be closer to the x than the y, if you understand what I mean?)
Part B:
Here you need to find the x and y components of the force F1, first. This isn't what you've done. The x-component of it will work together with F2 in the horizontal, and the y-component will be alone in the vertical. The total x and y components will be at 90 degrees to one another and you find the net force from there.
3. Feb 3, 2009
### v0id19
I agree with your magnitude in Part A, but like Hannisch said, I don't think your angle is right. Remember that $$arctan(\frac{opposite}{adjacent})=\theta$$. Also, it is imperative that you define the reference point for your angle! Is it 24o above the horizontal, below the horizontal, or is it measured from the vertical? (24 doesn't work for any of these references, but I hope you see my point).
4. Feb 3, 2009
### MaximaMan
Wow Im sorry. Somehow I mistyped. F1 is on the X-axis(0*) and F2 is the other Tangent at the angle, not the other way around. That is how I got 24* (though I still think it is probably wrong).
Thanks for the help guys, im trying to get this as best I can.
side note, void, the question itself from the HW states the reference point for the angle. When it asks to solve, it solely asks for the numerical value. But I agree, reference point is always a necessity!
5. Feb 4, 2009
### MaximaMan
Okay, so I redid everything. This is what I got.
PART A:
Magnitude: 31.95
Angle: sin x = 11 / 31.95 => 20.1*
Part B:
Magnitude: 36.7
Angle: sin x = 9.5/36.7 => 15*
6. Feb 4, 2009
### Hannisch
I say I agree with you on those values. Intuatively they make sense (net force should get greater in B, angle should get smaller since they're closer together, etc) and with my quick calculations it seems correct.
7. Feb 4, 2009
### MaximaMan
Okay, I am done with workin on that section. I am onto a new question and I got all parts of the answer right so far, except the last part. The question is:
Consider two vectors by A = 5 i - 3 j and B = - i - 6 j.
(a) Calculate A + B
4i + -9j
(b) Calculate A - B
6i + 3j
(c) Calculate |A + B|
9.84
(d) Calculate |A - B|
6.7
(e) What is the direction of A + B? (from the +x axis)
I did: sin x = 4/9.84 => x = 24 => 180 - 24 => 156*
(f) What is the direction of A - B? (from the +x axis) | 2017-12-11T04:29:45 | {
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https://math.stackexchange.com/questions/3437363/prove-bigcap-a-n-cup-b-n-supset-bigcap-a-n-cup-bigcap-b-n | # Prove $\bigcap (A_n \cup B_n) \supset (\bigcap A_n)\cup(\bigcap B_n)$
Prove that:
$$\bigcap (A_n \cup B_n) \supset \left(\bigcap A_n\right) \cup \left(\bigcap B_n\right)$$
Also find an example when there is no equality. Ok, so
$$\bigcap A_n := \{a \mid \forall Y \in A: a \in Y\}$$
$$\bigcap B_n := \{b \mid \forall X \in B: b \in X\}$$
$$\bigcup A_n := \{c \mid \exists Z \in A: c \in Z\}$$
$$\bigcup B_n := \{d \mid \exists T \in B: d \in T\}$$
Then how can I proceed?
• Normally one uses $\displaystyle \text{“}\cup\text{''}$ in things like $$A\cup B\quad\text{and}\quad A_1 \cup \cdots \cup A_n$$ and $\displaystyle \text{“}\bigcup\text{''}$ in things like $$\bigcup_{k=1}^n A_k.$$ I edited the question accordingly. $\qquad$ – Michael Hardy Nov 16 '19 at 0:50
Let $$S = \bigcap(A_n\cup B_{n})$$ and let $$T=(\bigcap A_n)\cup (\bigcap B_n)$$ denote the left and right hand sides. We want to prove that $$T\subset S$$. So let $$t\in T$$; we must show that $$t\in S$$.
Because $$t\in T$$, $$t\in \bigcap A_n$$ or $$t\in \bigcap B_n$$. That is, either $$t\in A_n$$ for all $$n$$, or $$t\in B_n$$ for all $$n$$. In either case, we have $$t\in A_n\cup B_n$$ for all $$n$$, which means $$t\in S$$.
• Normally one uses $\displaystyle \text{“}\cup\text{''}$ in things like $$A\cup B\quad\text{and}\quad A_1 \cup \cdots \cup A_n$$ and $\displaystyle \text{“}\bigcup\text{''}$ in things like $$\bigcup_{k=1}^n A_k.$$ I edited this answer accordingly. $\qquad$ – Michael Hardy Nov 16 '19 at 0:52
• I see. Could you please give me a hand on how to prove the equality instead of inclusion when $\forall n \in N: A_{n+1} \subset A_n$ and same for $B_n$? And about the inclusion, an example when it becomes an equality? – alladinsane Nov 16 '19 at 8:59
For each $$n$$, $$\; A_n\subset A_n\cup B_n$$, so $$\bigcap_n A_n \subset \bigcap_n( A_n \cup B_n),$$ and similarly for $$B_n$$. | 2020-02-29T10:53:40 | {
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http://math.stackexchange.com/questions/368603/how-do-i-write-a-trig-function-that-includes-inverses-in-terms-of-another-variab | # How do I write a trig function that includes inverses in terms of another variable?
It's been awhile since I've used trig and I feel stupid asking this question lol but here goes:
Given: $z = \tan(\arcsin(x))$
Question: How do I write something like that in terms of $x$?
Thanks! And sorry for my dumb question.
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– Martin Sleziak Oct 19 at 13:18
## 2 Answers
$$z = \tan[\arcsin(x)]$$
$$\arctan(z) = \arctan[\tan(\arcsin(x)] = \arcsin(x)$$
$$\sin[\arctan(z)] = \sin[\arcsin(x)] = x$$
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So you basically just do the inverse of each side until x is by itself? – ModdedLife Apr 21 '13 at 21:29
I'm assuming you want x as a function of $z$? To isolate $x$? – amWhy Apr 21 '13 at 21:34
Yes, that is correct. – ModdedLife Apr 21 '13 at 21:35
Thenwh at I've written above will indeed give you x as a function of $z$: $$x = \sin[\arctan(z)]$$ – amWhy Apr 21 '13 at 21:39
Ok thanks. So if the question was worded something like: "Find the value of z = tan(arcsin(x)) in terms of x." does this mean the same thing that you have shown here? Or would that be more like what Andre showed as the answer? – ModdedLife Apr 21 '13 at 21:42
Draw a triangle $ABC$, right-angled at $C$. Let us suppose that $\arcsin x=\angle A$.
Decide that the hypotenuse $AB$ is of length $1$ (it doesn't matter). Write $1$ next to the hypotenuse.
Then the side $BC$ opposite $\angle A$ has length $x$. The remaining side $AC$, by the Pythagorean Theorem, is $\sqrt{1-x^2}$. Write the appropriate lengths on the diagram.
Finally, read off $\tan A$ from the picture. It is $\frac{x}{\sqrt{1-x^2}}$.
This is not quite all there is to it. We need to check whether the result makes sense for all $x$. The only possible issues that can arise for angles not between $0$ and $\frac{\pi}{2}$ are issues of sign. Since $\arctan x$ is conventionally always between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and on this interval $\sin$ has the same sign as $\tan$, there is no problem.
If you want something like $\csc(\arctan x)$, do something similar. We could let the hypotenuse be $1$. But it may be easier to let the leg adjacent to $\angle A$ be $1$, and the leg opposite $\angle A$ be $x$. Then the hypotenuse is $\sqrt{1+x^2}$, and now you can read off the cosecant from the picture.
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Ok thank you. So would that be "the value of z in terms of x?" – ModdedLife Apr 21 '13 at 21:33
OK this made perfect sense now. Thanks for your help! – ModdedLife Apr 21 '13 at 21:53
The expression you write already gives the value of $z$ in terms of $x$, but in a complicated way. There are quite a few situations in which we want an algebraic simplification of your expression. This happens most frequently when we have made a trigonometric substitution to evaluate an integral. But it is very possible that you might be asked for this kind of simplification in a "precalculus" course. – André Nicolas Apr 21 '13 at 22:00 | 2015-11-26T14:57:13 | {
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https://www.mathworks.com/help/symbolic/symbolic-summation.html | ## Symbolic Summation
Symbolic Math Toolbox™ provides two functions for calculating sums:
• sum finds the sum of elements of symbolic vectors and matrices. Unlike the MATLAB® sum, the symbolic sum function does not work on multidimensional arrays. For details, follow the MATLAB sum page.
• symsum finds the sum of a symbolic series.
### Comparing symsum and sum
You can find definite sums by using both sum and symsum. The sum function sums the input over a dimension, while the symsum function sums the input over an index.
Consider the definite sum $S=\sum _{k=1}^{10}\frac{1}{{k}^{2}}.$ First, find the terms of the definite sum by substituting the index values for k in the expression. Then, sum the resulting vector using sum.
syms k
f = 1/k^2;
V = subs(f, k, 1:10)
S_sum = sum(V)
V =
[ 1, 1/4, 1/9, 1/16, 1/25, 1/36, 1/49, 1/64, 1/81, 1/100]
S_sum =
1968329/1270080
Find the same sum by using symsum by specifying the index and the summation limits. sum and symsum return identical results.
S_symsum = symsum(f, k, 1, 10)
S_symsum =
1968329/1270080
### Computational Speed of symsum versus sum
For summing definite series, symsum can be faster than sum. For summing an indefinite series, you can only use symsum.
You can demonstrate that symsum can be faster than sum by summing a large definite series such as $S=\sum _{k=1}^{100000}{k}^{2}.$
To compare runtimes on your computer, use the following commands.
syms k
tic
sum(sym(1:100000).^2);
toc
tic
symsum(k^2, k, 1, 100000);
toc
### Output Format Differences Between symsum and sum
symsum can provide a more elegant representation of sums than sum provides. Demonstrate this difference by comparing the function outputs for the definite series $S=\sum _{k=1}^{10}{x}^{k}.$ To simplify the solution, assume x > 1.
syms x
assume(x > 1)
S_sum = sum(x.^(1:10))
S_symsum = symsum(x^k, k, 1, 10)
S_sum =
x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x
S_symsum =
x^11/(x - 1) - x/(x - 1)
Show that the outputs are equal by using isAlways. The isAlways function returns logical 1 (true), meaning that the outputs are equal.
isAlways(S_sum == S_symsum)
ans =
logical
1
For further computations, clear the assumptions.
assume(x, 'clear') | 2022-11-27T18:03:33 | {
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http://mathhelpforum.com/statistics/9000-simple-chance-problem-print.html | # simple chance problem
• December 17th 2006, 12:33 PM
emptyglasses
simple chance problem
I need help with figuring out what method to use to figure out this problem..
Here is the problem.
A fair coin was flipped 10 times; compute the chance that the number of heads is less than 4.
• December 17th 2006, 01:02 PM
Soroban
Hello, emptyglasses!
You need the Binomial Probability formula . . . several times.
Quote:
A fair coin was flipped 10 times.
Compute the chance that the number of heads is less than 4.
We must find these separate probabilities . . . and add them.
$P(3H)\:=\:\binom{10}{3}\left(\frac{1}{2}\right)^3\ left(\frac{1}{2}\right)^7 \:=\:\frac{120}{1024}$
$P(2H) \:=\:\binom{10}{2}\left(\frac{1}{2}\right)^2\left( \frac{1}{2}\right)^8 \:=\;\frac{45}{1024}$
$P(1H) \:=\:\binom{10}{1}\left(\frac{1}{2}\right)^1\left( \frac{1}{2}\right)^9 \:=\:\frac{10}{1024}$
$P(0H)\:=\:\binom{10}{0}\left(\frac{1}{2}\right)^0\ left(\frac{1}{2}\right)^{10}\:=\:\frac{1}{1024}$
Therefore: . $P(H < 4) \;=\;\frac{120}{1024} + \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024} \;=\;\frac{176}{1024} \;=\;\boxed{\frac{11}{64}}$
• December 17th 2006, 01:30 PM
emptyglasses
Is there a way of figuring this out without doing all of that computing?
Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!
• December 17th 2006, 01:36 PM
ThePerfectHacker
Quote:
Originally Posted by emptyglasses
Is there a way of figuring this out without doing all of that computing?
Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!
I devised my own approach to this problem.
Such that can be programmed into a computer.
I presume there is a way to do it via normal distribution but I never learned probability.
My approah is to find,
$\int_0^{40} \frac{100!}{40!\cdot \Gamma (41-x)}.5^{100} dx$
Where, $s>0$
$\Gamma (s)=\int_0^{\infty} e^{-t}t^{s-1}dt$ (infamous Gamma function).
I know it looks ugly but programs can do this easily. Taking integral by Simpon's rule. Thus, all you need to do is tell your computer to find this integral.
• December 17th 2006, 06:53 PM
eureka
Quote:
Originally Posted by emptyglasses
Is there a way of figuring this out without doing all of that computing?
Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!
Hi emptyglasses;
Could i suggest to use Normal distribution method. Normal distribution can be used as an approximation to the binomial distribution (as suggested by soroban) when:
- Number of trail is large (greater than 30)
- probability of success at any trail is not too small or large (the closer to 0.5 the better)
Here you go:
First you will need to find the expected mean: number of trail x probability of success at any trail = 100 x 0.5 = 50
Next, find the standard deviation = squaroot 100 x 0.5 x 0.5 = 5
Then you can find the probability by; P(z < (39.5 - 50)/5) = P(z < -2.1) =0.0179 (find from Normal disribution tables)
• December 19th 2006, 04:21 AM
CaptainBlack
Quote:
Originally Posted by ThePerfectHacker
I devised my own approach to this problem.
Such that can be programmed into a computer.
I presume there is a way to do it via normal distribution but I never learned probability.
My approah is to find,
$\int_0^{40} \frac{100!}{40!\cdot \Gamma (41-x)}.5^{100} dx$.
Are you sure that is what you use? What about:
$\int_0^{40} \frac{100!}{\Gamma(x+1)\cdot \Gamma (101-x)}(0.5)^{100} dx$?
RonL
• December 19th 2006, 04:32 AM
CaptainBlack
Quote:
Originally Posted by CaptainBlack
Are you sure that is what you use? What about:
$\int_0^{40} \frac{100!}{\Gamma(x+1)\cdot \Gamma (101-x)}(0.5)^{100} dx$?
RonL
It is a nice excecise to compare the normal approximation, the above
approximation and the exact value. These are 0.0179, 0.0227 and 0.0176
respectivly.
Though if we apply a continuity correction to the integral above so we
integrate from -0.5 to 39.5 the integral changes from 0.0227 to 0.0178
which is indistinguishable from the normal approximation in this case.
RonL
• December 19th 2006, 05:54 AM
ThePerfectHacker
Quote:
Originally Posted by CaptainBlank
It is a nice excecise to compare the normal approximation, the above
approximation and the exact value. These are 0.0179, 0.0227 and 0.0176
respectivly.
Though if we apply a continuity correction to the integral above so we
integrate from -0.5 to 39.5 the integral changes from 0.0227 to 0.0178
which is indistinguishable from the normal approximation in this case.
RonL
Thank you for the correction. But my purpose in the post is that it amazingly turns out that we use integration. The reason why the Gamma function appears, as you guessed, is to generalize the factorial. I have a question, is this approach, known? In probability theory.
• December 19th 2006, 06:10 AM
CaptainBlack
Quote:
Originally Posted by ThePerfectHacker
Thank you for the correction. But my purpose in the post is that it amazingly turns out that we use integration. The reason why the Gamma function appears, as you guessed, is to generalize the factorial. I have a question, is this approach, known? In probability theory.
I have a sneaking suspicion that it is virtualy identical to the normal approximation (certainly in this case it appears to be)
I think the problem is that if you have the computational resources to
to the integral, you also have the resources to do the exact calculation.
RonL | 2014-07-14T09:09:05 | {
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https://math.stackexchange.com/questions/2489395/irrationality-of-sqrt15 | # Irrationality of $\sqrt{15}$
Could someone verify the correctness of this proof for the irrationality of $\sqrt{15}$?
Assume $\sqrt{15}\in\mathbf{Q}$, then $\sqrt{15}=\frac{p}{q}$ with $p,q\in\mathbf{Z}$ ($q\ne0$ and $\gcd(p,q)=1$). $\implies 15q^2=p^2 \implies 15\mid p^2 \implies 3\mid p^2 \implies 3\mid p$ (Euclid's Lemma)
Now we write $p=3k$ for $k\in\mathbf{Z}$, then we have $15q^2=9k^2 \implies 5q^2=3k^2 \implies 3\mid 5q^2$. Since $\gcd(3,5)=1$ (this lemma: if $a\mid bc$ and $\gcd(a,b)=1$ then $a\mid c$) this gives $3\mid q^2 \implies 3\mid q$ (Euclid's Lemma).
Therefore $3\mid p$ and $3\mid q$. Contradiction.
• That's a great proof. – Yanko Oct 25 '17 at 16:18
• looks correct to me well done – Isham Oct 25 '17 at 16:34
• Agreed and well explained – imranfat Oct 25 '17 at 16:53
• Looks good. One comment though, when writing "Contradiction." it's good to mention what the contradiction is, in this case "Contradiction with $\gcd(p,q)=1$" – rtybase Oct 25 '17 at 18:12
I like your proof. As a further observation, you could show that your proof generalizes: you could have replaced $$15$$ with any product $$p_0 r$$ where $$p_0$$ is prime and $$r$$ is relatively prime to $$p_0$$. So you have really proven the following:
Proposition. Let $$n$$ be an integer and let $$p_0$$ be a prime number such that $$p_0$$ divides into $$n$$ exactly once (that is, $$p_0$$ divides $$n$$ but $$p_0^2$$ does not divide $$n$$). Then $$\sqrt{n}$$ is irrational. | 2019-06-20T17:43:43 | {
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http://mathoverflow.net/questions/134405/a-question-in-fourier-analysis | # A question in Fourier analysis
I recently came across this problem:
Let $T=\mathbb R /2 \pi \mathbb Z$ the circle, with its proabability Haar measure $\mu$. Any integrable function $f : T \rightarrow \mathbb C$ has Fourier coefficients $c_n(f)= \int_T f(t) e^{-int} \frac{dt}{2\pi}$, and we define $$||f||_{A(T)} = \sum_{n \in \mathbb Z} |c_n(f)| \leq + \infty$$ its algebra norm. (This definition is standard)
Now let $E$ be a measurable subset of $T$. I want to compute, or estimate, $$\alpha(E) = \inf \{ ||f||_{A(T)}\ | \ f : T \rightarrow \mathbb R, \ f_{|E} \leq 0, \int_T f = 1 \}$$
An easy lower bound for $\alpha(E)$ is $\alpha(E) \geq \frac{1}{1-\mu(E)}$, but Id like to know as many thing as possible about $\alpha(E)$. For example, here is a basic question:
When $E$ is an interval such that $\mu(E) \rightarrow 1$, is that lower bound sharp?
Also, has this problem, or something related, been studied ? Does $\alpha(E)$ have a name ?
-
Out of curiosity: is it important for your purposes that $f\leq 0$ on $E$, rather than $f=0$ on $E$? – Yemon Choi Jun 22 '13 at 3:44
@Yemon: Yes, kind of. I mean, I am mostly interested in upper bound on $\alpha(E)$, so any such bound for the variant $\alpha'(E)$ you suggest, where the inf is taken over function which are zero outside $E$, obviously give a bound that interests me for $\alpha(E)$. But in my problem, I have no need to impose the condition $f_{|E}=0$, so if I can get better bound without that, the better. – Joël Jun 24 '13 at 14:34
Yemom and Noam, thanks for both your great answers. I accept Noam's as it gives a better upper bound, discusses also lower bounds, and gives plenty of ideas but already Yemon's one was very helpful. – Joël Jun 24 '13 at 14:38
By the way, from the fact that neither Yemon nor Noam nor anyone else here mentioned anything related to this, it seems more likely than before than this notion of $\alpha(E)$ (or its variant $\alpha'(E)$ defined with the condition $f_{|E}=0$) has no official name and was not extensively studied. Or am I wrong? – Joël Jun 24 '13 at 14:44
Joel: I have not seen either $\alpha$ or $\alpha'$ before, but somehow $\alpha'$ "looks closer to familiar settings" than $\alpha$ - something to do with finding elements of an ideal that satisfy a linear constraint. If I see anything I will leave a comment or an update – Yemon Choi Jun 25 '13 at 5:43
Yemon Choi (henceforth "YC") answered the question to within a constant factor with a function $f_\delta$ showing that $\alpha(E_\delta) \leq 2/\delta$ where $E_\delta$ is an interval of measure $1-\delta$ (YC's notation). That's optimal asymptotically $-$ and even exactly if $2/\delta \in {\bf Z}$ $-$ for functions $f$ with nonnegative Fourier coefficients $c_n(\phantom.f)$. But without this additional hypothesis on the $c_n(\phantom.f)$ the constant can be improved somewhat: we give another piecewise linear function (still vanishing on $E_\delta$) that reduces the constant from $2$ to below $1.9$, and that's surely not optimal either. But it certainly can't go all the way down to Joël's lower bound of $1/\delta$: the constant must be at least $3/2$, and I think it must be at least $\pi/2$ if $f=0$ on $E_\delta$ as YC suggested.
If $\| \phantom. f \|_{A(T)} \lt \infty$ then $\sum_n c_n(\phantom.f) e^{int}$ converges absolutely to a continuous function with Fourier coefficients $c_n$, so we may assume $f$ is continuous. If moreover each $c_n(\phantom.f) \geq 0$ then $\| \phantom. f \|_{A(T)} = \sum_{n\in{\bf Z}} c_n = f(0)$. Now suppose $f(t) \leq 0$ for $t \in E_\delta$, and let $N$ be the integer $\lfloor 2/\delta \rfloor$. We have $f(m/N) \leq 0$ for all integers $m \not\equiv 0 \bmod N$. Therefore $$f(0) \geq \sum_{m \phantom. \bmod \phantom. N} f(m/N) = N \sum_{N\phantom.|\phantom.n} c_n(\phantom.f) \geq N c_0(\phantom.f) = N$$ because $c_0 = \int_T f \phantom. d\mu = 1$; this proves our claim.
To give an example with smaller $\| \phantom. f \|_{A(T)}$, consider the following generalization: YC's function $f_\delta$ is the self-convolution of uniform meausre on an interval of length $\delta/2$; for $0 \lt b \lt 1$ define $f_{\delta,b}$ to be the convolution of uniform measures on intervals of length $b\delta$ and $(1-b)\delta$. Then $$c_n(\phantom.f_{\delta,b}) = \frac{\sin(\pi b \delta n)}{\pi b \delta n} \frac{\sin(\pi (1-b) \delta n)}{\pi (1-b) \delta n}.$$ Hence $\delta \sum_n \left| c_n(\phantom.f_{\delta,b}) \right|$ is a Riemann sum for $$\int_{-\infty}^\infty \left| \frac{\sin(\pi b s)}{\pi b s} \frac{\sin(\pi (1-b) s)}{\pi (1-b) s} \right| \phantom. ds.$$ Numerical calculation shows this integral decreasing from $2$ at $b=1/2$ (YC's function) down to about 1.9 at $b \sim 0.38$, then up slightly and back down to just below $1.9$ at $b \sim 0.29$ before increasing and crossing $2$ at $b\sim 0.14$ (it must blow up at $b \rightarrow 0$). We do a bit better by averaging $f_{\delta,0.29}$ with $f_{\delta,0.38}$, getting just below $1.875 = \frac{15}{8}$.
This kind of analysis suggests that the asymptotic value of $\delta \cdot \alpha(E_\delta)$ should be the infimum, call it $\alpha_0$, of $\int_{-\infty}^\infty \bigl|\phantom.\hat f(s)\bigr| ds$ over functions $f: {\bf R} \rightarrow {\bf R}$ such that both $f$ and its Fourier transform $$\hat f(s) = \int_{-\infty}^\infty e^{-2\pi i s t} f(t) dt$$ are absolutely integrable and satisfy $\hat f(0) = 1$ and $f(t) \leq 0$ for all $t$ with $|t| \geq 1/2$. I don't quite have a proof of this, but all examples and bounds so far translate immediately to $\alpha_0$. For example, if we also assume that $\hat f(s) \geq 0$ for all $s$ then $f(0) \geq 2$ follows from Poisson summation: $$f(0) \geq \sum_{m \in {\bf Z}} f(m/2) = 2 \sum_{s \in {\bf Z}} f(2n) \geq 2\hat f(0) = 2.$$ This is the 1-dimensional case of the inequality in Henry Cohn's thesis on upper bounds of sphere-packing densities.
The dictionary between bounds on $\delta \alpha(E_\delta)$ as $\delta \rightarrow 0$ and bounds on $\alpha_0$ also works for the following improved lower bound, which I'll give only in the $\alpha_0$ setting. Here Joël's "easy lower bound" is $\alpha_0 \geq 1$, proved by noting that $$1 = \int_{-\infty}^\infty f(t) dt \leq \int_{-1/2}^{1/2} f(t) dt$$ implies $f(t) \geq 1$ for some $t \in [-1/2, +1/2]$, and thus $\bigl\| \phantom. \hat f \bigr\|_1 \geq 1$. But clearly equality cannot hold here. To exploit this, we write
$$\int_{-1/2}^{1/2} f(t) dt = \langle \phantom. f, \chi \rangle = \langle \phantom. \hat f, \hat\chi \rangle \leq \bigl\| \phantom. \hat f \bigr\|_1 \bigl\| \hat\chi \bigr\|_\infty$$ where $\chi$ is the characteristic function of $[-1/2,1/2]$. Now $\bigl\| \hat\chi \bigr\|_\infty = 1$, but only because $\hat\chi(0) = 1$. So we can improve the estimate by writing $\langle \phantom. f, \chi \rangle \leq \langle \phantom. f, \psi \rangle$ for any function (or even any measure) $\psi$ such that $\hat\chi - \hat\psi$ is nonnegative and supported outside $(-1/2, 1/2)$; if $\bigl\| \hat\psi \bigr\|_\infty \lt 1$ then $1 \bigl/ \bigl\| \hat\psi \bigr\|_\infty \bigr.$ is an improved lower bound on $\alpha_0$. A choice that gives $\alpha_0 \geq 3/2$ is $$\hat\psi(s) = \hat\chi(s) - \frac13 \cos(\pi s) = \frac{\sin \pi s}{\pi s} - \frac13 \cos(\pi s) = \frac23 - \frac1{180} (\pi s)^4 + - \cdots,$$ so $\langle \phantom. f, \psi \rangle = \langle \phantom. f, \chi \rangle - \frac16\bigl( \phantom.f(1/2) + f(-1/2)\bigr) \geq \langle \phantom. f, \chi \rangle \geq 1$; then $\bigl\| \hat\psi \bigr\|_\infty \lt 2/3$, so $\alpha_0 \geq 3/2$ as claimed.
-
Just to get the ball rolling: for any $\delta \in(0,1)$, the function $f_\delta(t)$ which is supported on the interval $[-\delta\pi,\delta\pi]$ and looks like $1- (\delta\pi)^{-1}|t|$ inside it, has norm $1$ in $A(T)$ -- one either sees this using positive-definiteness or a trick with convolving characteristic functions in $L^2(T)$, or by a direct calculation one can show the Fourier coefficients are all positive.
[The following calculation was originally out by a factor of $2$ -- thanks to Noam Elkies for pointing this out.]
Clearly $\int_T f_\delta = \delta/2$. Therefore, by considering $2\delta^{-1}f_\delta$, and $E_\delta= T \setminus [-\delta\pi,\delta\pi]$, $$\alpha( E_\delta ) \leq 2\delta^{-1} = \frac{2}{1- \mu(E_\delta)}$$ which is a factor of $2$ out from the lower bound you state in your specific question. (My initial suspicion is that your lower bound is not exactly sharp, but I am hesitant to claim right now that the inequality above is equality.)
-
That works within a factor of 2: $\int_T f_\delta$ is actually $\delta/2$, because $\mu([-\delta\pi,\delta\pi]) = \delta$ and the average of $f_\delta$ over that interval is $1/2$. So we must use $2\delta^{-1} f_\delta$, and deduce the bound $\alpha(E_\delta) \leq 2\delta^{-1}$. – Noam D. Elkies Jun 21 '13 at 22:16
Thanks, Noam. I've edited with the correction. – Yemon Choi Jun 21 '13 at 22:28 | 2015-02-28T01:56:18 | {
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http://math.stackexchange.com/questions/113687/different-interpretations-the-same-result | # Different interpretations, the same result!
In a box there are 10 white ball and 10 black balls.Two balls are successively drawn.What is the probability to get two balls of different colors?
My interpretation is that the order doesn't matter, because the important is that the balls have different colors.So to find the possible cases I computed a combination, $^{20}C_{2}$. To find the favorable cases I multiplied $10$ white balls by $10$ black balls.
The final mathematical expression stays: $\frac{10^2 \cdot 2}{20 \cdot 19}$
The book's interpretation is the following:
For the possible cases, there are $20$ for the first ball and $19$ for the second.So there are $20 \cdot 19$ possible cases.Regarding the favorable cases the book says that there are $10 \cdot 10 \cdot 2$.
It's clear that by the book's interpretation the order matters.The two final mathematical expressions are the same.How can this happen if the interpretations were different?
-
Because the final answer (i.e. the probability), should not depend on how you solve the problem. You just listed two different ways to solve the same problem, so the answers coincide. – Tom Artiom Fiodorov Feb 26 '12 at 19:33
We can start by looking at all patterns from arranging all twenty balls. There are $20!$ ways of ordering all twenty balls. Of these $10 \times 10 \times 18!$ have a black followed by a white at the beginning and $10 \times 10 \times 18!$ have a white followed by a black at the beginning, so the proportion of patterns meeting the condition is $\dfrac{10 \times 10 \times 18!+10 \times 10 \times 18!}{20!}$.
The book then decides that all the arrangements of the last eighteen can be treated as equivalent and grouped together as equivalence classes, with each equivalence class having $18!$ elements, to give a proportion of equivalence classes of $\dfrac{10 \times 10 +10 \times 10 }{20 \times 19}$.
You instead decide that all arrangements of the last eighteen and all arrangements of the first two can be treated as equivalent and grouped together as equivalence classes, with each equivalence class having $18!\times 2!$ elements, to give a proportion of equivalence classes of $\dfrac{10 \times 10 }{20 \times 19/2!}$. | 2015-10-08T22:53:23 | {
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https://math.stackexchange.com/questions/926795/find-the-last-non-zero-digit-of-28 | # Find the last non zero digit of 28!.
Find the last non zero digit of 28!.
It is very hard to multiply and find the last nonzero digit. I just wanna know that, is there any easy technique to solve this type of problem?
• Consider factors of $10$ separately (you can ignore them). Build factors of $10$ by pairing a factor of $5$ with a factor of $2$. If you pair say the factor $5$ from $15$ and a factor $2$ from $12$, that leaves you with the factors $3$ and $6$ that you cannot discard. Multiply the remaining factors modulo $10$ (that means you need only remember the last digit of each partial product). – Daniel Fischer Sep 10 '14 at 21:14
• @DanielFischer This approach can be improved by noticing that there will always be more factors of 2 than factors of 5. So one can increase a counter when encountering a 2 and decreasing it when encountering a 5. And since the last digit of $2^1$ and $2^5$ is the same, the counter can be reduced mod 4. – kasperd Sep 11 '14 at 10:33
Note that $$5\times 10\times 15\times 20\times 25=5\times(2\cdot 5)\times (3\cdot 5)\times (2^2\cdot 5)\times 5^2=2^3\cdot 3\cdot 5^6$$ and that $$10^6=2^6\cdot 5^6=\frac{8\cdot 5\cdot 10\cdot 15\cdot 20\cdot 25}{3}=8\cdot 5\cdot 10\cdot 5\cdot 20\cdot 25.$$
So, in mod $10$, we have \begin{align}\frac{28!}{10^6}&\equiv 1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 9\cdot 1\cdot 2\cdot 3\cdot 4\cdot 3\cdot 6\cdot 7\cdot 8\cdot 9\cdot 1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 8\\&\equiv (1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7)^3\cdot 3\cdot 8^2\cdot 9^2\\&\equiv \{1\cdot 2\cdot 3\cdot 4\cdot (-4)\cdot (-3)\}^3 \cdot 3\cdot 4\cdot 1\\&\equiv 2\cdot 2\\&\equiv 4.\end{align}
• Nice solution. You dropped the ! from $28!/10^6$, but it won't let me make a one-character edit. – Bungo Sep 10 '14 at 21:56
• @Bungo: Thanks:) – mathlove Sep 10 '14 at 21:57
• @Bungo in cases of need you can add an HTML <!-- comment --> explaining why your too-short edit really is good, and then it won't be a too-short edit anymore :) – hobbs Sep 11 '14 at 0:58
• @hobbs: Thanks for the tip! – Bungo Sep 11 '14 at 1:08
Here is a technique:
1. Factor all numbers $1,\dots,28$ as a product of twos, fives and other numbers. (Other numbers meaning numbers not divisable by 2 or 5.)
2. Calculate the number of fives and remove them, and also remove an equal number of twos. (Each zero at the end is due to a factor of $10=5\times2$, and there are always more twos than fives.)
3. For the remaining factors (including the remaining twos), calculate them modulo 10 (simply remove tens) and multiply them. You can also make any other changes modulo 10, like turning a $9$ into $-1$.
4. The last digit of this product will be the last nonzero digit of $28!$
Example in the case of $11!$:
1. $1=1$, $2=2$, $3=3$, $4=2^2$, $5=5$, $6=2\times3$, $7=7$, $8=2^3$, $9=9$, $10=2\times5$, $11=11$.
2. There are two fives.
3. The remaining factors are $1$, $2$ (6 times), $3$ (twice), $7$, $9$ and $11$.
4. The product is $1\times2^6\times3^2\times(-3)\times(-1)\times1=2^63^3=1728$. The last nonzero digit of $11!$ is thus 8. (Check: $11!=39916800$.)
This technique is not optimal, but it works. Once you figure out why and how it works, you can probably find a way to make it more efficient if you want to do such calculations again. Note that in the last step it suffices to calculate the product modulo 10, so you can throw away unnecessary tens and hundreds.
Perhaps a small reduction in the calculations in other answers. First, we use the standard technique for finding the power of a prime which divides a factorial: $$\Bigl\lfloor\frac{28}{5}\Bigr\rfloor+\Bigl\lfloor\frac{28}{25}\Bigr\rfloor =5+1=6$$ so $5^6$ is a factor of $28!$ (and $5^7$ is not), and likewise $$\Bigl\lfloor\frac{28}{2}\Bigr\rfloor+\Bigl\lfloor\frac{28}{4}\Bigr\rfloor +\Bigl\lfloor\frac{28}{8}\Bigr\rfloor +\Bigl\lfloor\frac{28}{16}\Bigr\rfloor=14+7+3+1=25\ ,$$ so $2^{25}$ is a fcator of $28!$. Therefore $28!$ ends with $6$ zeros and we need the last digit of $28!/10^6$. Simplifying modulo $10$, we have $$2^\equiv2^5\equiv2^9\equiv2^{13}\equiv2^{17}\ ,\quad 3^4\equiv1\ ,\quad 7^4\equiv1\ ,\quad 9^2\equiv1\ ;$$ cancelling $2$s and $5$s and reducing the resuts modulo $10$ gives \eqalign{28!/10^6 &\equiv2^{19}1.1.3.1.1.3.7.1.9.1.1.3.3.7.3.1.7.9.9.1.1.1.3.3.1.3.7.7\cr &\equiv2^3.3^8.7^5.9^3\cr &\equiv8.1.7.9\cr &\equiv4\ .\cr}
Here is a general technique that calculates the digit in question in polylog time. The technique is applied to $28!$ at the end since that is the specific question here. I apply it to $1000!$ to emphasize that it can be used for much larger numbers.
The idea is to extract all powers of $5$ and $2$ from $n!$, and write it as $10^A2^BC$ with $C$ odd and not a multiple of $5$. Then $A$ is the number of trailing zeros. And whatever $2^BC$ is modulo 10 is the preceding digit. To that end, these two reduction formulas are useful. They use $$n\underset{2}{!}=1\cdot3\cdot5\cdots(\text{last odd number}\leq n)$$ and $$n\underset{5}{!}=1\cdot3\cdot7\cdot9\cdot11\cdot13\cdot17\cdots(\text{last such number}\leq n)$$ and the reduction rules (which are easy to verify) are \begin{align} n! &= n\underset{2}{!} \times \left\lfloor\frac{n}{2}\right\rfloor! \times 2^{ \left\lfloor\frac{n}{2}\right\rfloor}\\ n\underset{2}{!} &= n\underset{5}{!} \times \left(2\left\lfloor\frac{n+5}{10}\right\rfloor-1\right)\underset{2}{!} \times 5^{\left\lfloor\frac{n+5}{10}\right\rfloor}\hspace{2pc}n\geq5 \end{align} The first of these is something you may have seen before. The second is the result of trying to find a similar formula that excludes multiples of $5$. A number of the form $n\underset{5}{!}$ is only important to us mod $10$, and $n\underset{5}{!}\equiv (1\times3\times7\times9)^{\left\lfloor\frac{n}{10}\right\rfloor}\times\epsilon\equiv 9^{\left\lfloor\frac{n}{10}\right\rfloor}\times\epsilon$, where $\epsilon$ is either $1$, $3$ or $9$ depending on $n$ the last digit of $n$. These two reduction rules together with the modular reduction of $n\underset{5}{!}$ cut the size of $n$ in half, and repeated application will bring the size of $n$ down fairly fast. We have:
\begin{align} 1000!&\equiv1000\underset{2}{!}\times500!\times2^{500}\\ &\equiv1000\underset{5}{!}\times199\underset{2}{!}\times5^{100}\times500!\times2^{500}\\ &\equiv9^{100}\times500!\times199\underset{2}{!}\times2^{500}\times5^{100} \\ &\equiv500!\times199\underset{2}{!}\times2^{500}\times5^{100}\\ & \text{...note we've cut the largest number in play from 1000 to 500...}\\ & \text{...repeated factorial and modular reductions...}\\ &\equiv 2^{994}\times5^{249}\\ &\equiv10^{249}\times2^{741}\\ &\equiv10^{249}\times2^1 \end{align}
So there are $249$ zeros, and the preceding digit is a $2$ (in $1000!$).
Applied to $28!$:
\begin{align} 28! &\equiv28\underset{2}{!}\times14!\times2^{14}\\ &\equiv\left(28\underset{5}{!}\times5\underset{2}{!}\times5^3\right)\times14!\times2^{14}\\ &\equiv9^2\times5\underset{2}{!}\times5^3\times14!\times2^{14}\\ &\equiv10^3\times2^{11}\times5\underset{2}{!}\times14!\\ &\equiv10^3\times2^{11}\times15\times14!\\ &\equiv10^4\times2^{10}\times3\times14!\\ &\equiv10^4\times2^{10}\times3\times\left(14\underset{2}{!}\times7!\times2^7\right)\\ &\equiv10^4\times2^{17}\times3\times14\underset{2}{!}\times7!\\ &\equiv10^4\times2^{17}\times3\times\left(14\underset{5}{!}\times1\underset{2}{!}\times5^1\right)\times7!\\ &\equiv10^5\times2^{16}\times3\times14\underset{5}{!}\times7!\\ &\equiv10^5\times2^{16}\times3\times\left(9\times3\right)\times7!\\ &\equiv10^5\times2^{16}\times7!\\ &\equiv10^5\times2^{16}\times5040\\ &\equiv10^6\times2^{16}\times504\\ &\equiv10^6\times2^{16}\times4\\ &\equiv10^6\times2^{18}&2^{a+4n}\equiv2^a\text{ for }a\geq1\\ &\equiv10^6\times2^{2}\\ &\equiv10^6\times4\\ \end{align}
So $28!$ ends with six $0$s and it's first nonzero digit (from the right) is a $4$. | 2020-02-19T20:04:34 | {
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http://www.maa.org/publications/maa-reviews/pi-unleashed | # Pi Unleashed
###### Jörg Arndt and Christoph Haenel
Publisher:
Springer
Publication Date:
2001
Number of Pages:
276
Format:
Paperback with CDROM
Price:
52.95
ISBN:
978-3540665724
Category:
General
[Reviewed by
Carl D. Mueller
, on
10/3/2002
]
This number will probably look familiar to the readers of this review, and most readers can probably point to at least one digit that is not what was expected. In fact, the number above is the beginning of the decimal representation of the sum of the first 50,000 terms of the series 4 (1 - 1/3 + 1/5 - 1/7 + ... ). The full infinite series adds to π but does so very very slowly. Since this is an alternating series, it is easy to determine that the partial sum of these 50,000 terms should be within about 0.00004 of π. So, the fact that it is close to but not exactly π is not surprising. What is surprising is that so many of the decimal places are correct! Only three of the digits listed above are incorrect. The reasons for this were explained in an article in the American Mathematical Monthly [Vol. 96 (1989), 681-687] and are contained in the book being reviewed here.
Let me begin this review by stating simply that I know of no better book than this one for those interested in everything π. While the book itself has fewer than 300 pages, it is packed with information, formulas, historical notes, etc. Just about anything you could want is here. The following list of chapter titles should give an idea of what is here: The State of π Art, How Random is π?, Shortcuts to π, Approximations for π and Continued Fractions, Arcus Tangens, Spigot Algorithms, Gauss and π, Ramanujan and π, The Borweins and π, The BBP Algorithm, Arithmetic, Miscellaneous, The History of π, Historical Notes, The Future: π Calculations on the Internet, π Formula Collection, Tables. In addition, there is a brief appendix: Documentation for the hfloat Library. There is a very good index, and, for those interested in more details than this book can provide, there is a lengthy bibliography with 123 references.
The CD-ROM that comes with the book is also quite full. Much of the space is taken up with raw data. The first one million decimal places of π, the first one million hexadecimal places of π, the first 20,000 decimal places of e, and the first 10,000 decimal places of log(2) are included in dvi and ps formats. And, the first 400 million digits of π are included in a set of 40 text files. But, the real prizes on this CD-ROM are the libraries and algorithms. The hfloat (huge float) package is included as are apfloat (a C++ arbitrary precision library by Mikko Tommila) and cln (a class library for numbers by Bruno Haible). Also, Carey Bloodworth's piagm package for high precision (π) calculations is included. There is source code for the BBP algorithm, arctan formulas for calculating digits of π, the spigot algorithm, monte carlo algorithms, and more. Executables (mostly for DOS or Windows) are included with most of the source code.
π Unleashed can be used as a reference book or it can be read straight through as one would read a textbook or a novel. It begins with an overview of π. The latest record is given, several formulas and algorithms are presented and discussed briefly, and the question of whether or not π is normal is asked. Later chapters expand on these and other issues. A brief chapter discusses what is known thus far about the distribution of the digits in the decimal representation of π and presents some oddities and statistical results. Separate chapters are devoted to the roles of Gauss, Ramanujan, and the Borweins in the history of π calculation.
Several algorithms are discussed in some detail. The spigot algorithm for calculating π is presented as is a similar algorithm for calculating e. This is an interesting algorithm in that the digits are not computed all at once but sequentially in chunks of several digits. This means that large storage buffers are not necessary and makes it an ideal algorithm for desktop computers. What makes this book especially nice is that when algorithms are presented, they are explained thoroughly. Not only is the mathematics explained (so that we understand that the result will indeed be π) but so too is the programming logic (so that we see what makes the algorithm efficient). The amazing BBP algorithm (which gives isolated hexadecimal digits of π) is also described. Code for both of these algorithms is included on the CD-ROM provided with the book.
There is a chapter on arithmetic. This may seem strange in a book on π, but when a computer is used to calculate billions of digits of π, it is not enough to have an efficient π algorithm. It is also necessary to perform arithmetic operations as efficiently as possible. In particular, an efficient method for multiplication of long numbers can have an enormous effect on the time required for such calculations. Ordinary multiplication of N digit numbers requires a time proportional to N squared. A technique called Karatsuba multiplication (described in the book) requires a time proportional to N raised to the 1.583 power. While this is an enormous savings when N is large, a third method called FFT multiplication requires a time proportional to only N log(N). This gives an astonishing savings over either of the other methods. FFT multiplication is discussed in the book and there is a tutorial on the method on the CD-ROM. Techniques for rapid division and root extraction are also discussed in the book.
Two chapters are especially nice for those who enjoy tables of various sorts. An entire chapter is devoted to π formulas. There are 154 different formulas for computing π and related constants. The final chapter of the book contains tables of digits: 5000 decimal digits of π, 5000 hexadecimal digits of π, 2000 continued fraction elements (decimal) of π, and for good measure, 100 digits of many constants relating to π and e.
There is more in the book: fascinating π trivia, intriguing formulas, and predictions for the future. There is far too much to describe in a brief review. Of course, no book is perfect and this one has a few minor errors (perhaps occurring in the translation process) such as the description of the constant e as a limiting balance in an interest bearing account earning "100% interest per interest period" as "this interest period became ever shorter." Clearly, this should read 100% interest per year. But these mistakes are minor and do not detract from the overall quality of the book. This book is a must have for π aficionados and would fit comfortably in the mathematics collection of high school and university libraries. I recommend it highly.
Carl D. Mueller ([email protected]) is Associate Professor of Mathematics at Georgia Southwestern State University in Americus, GA.
1. The State Of Pi Art 1
2. How Random Is pi? 21
2.1 Probabilities 21
2.2 Is pi normal? 21
2.3 So is pi not normal? 24
2.4 The 163 phenomenon 25
2.5 Other statistical results 28
2.6 The Intuitionists and pi 30
2.7 Representation of continued fractions 32
3. Shortcuts To pi 35
3.1 Obscurer approaches to pi 35
3.2 Small is beautiful 37
3.3 Squeezing pi through a sieve 38
3.4 pi and chance (Monte Carlo methods) 39
3.5 Memorabilia 44
3.6 Bit for bit 47
3.7 Refinements 49
3.8 The pi room in Paris 50
4. Approximations For pi And Continued Fractions 51
4.1 Rational approximations 51
4.2 Other approximations 55
4.3 Youthful approximations 63
4.4 On continued fractions 64
5. Arcus Tangens 69
5.1 John Machin's arctan formula 69
5.2 Other arctan formulae 72
6. Spigot Algorithms 77
6.1 The spigot algorithm in detail 78
6.2 Sequence of operations 80
6.3 A faster variant 82
6.4 Spigot algorithm for e 84
7. Gauss And pi 87
7.1 The pi AGM formula 87
7.2 The Gauss AGM algorithm 90
7.3 Schönhage variant 92
7.4 History of a formula 94
8. Ramanujan And 103
8.1 Ramanujan's series 103
8.2 Ramanujan's unusual biography 105
8.3 Impulses 110
9. The Borweins And pi 113
10. The BBP Algorithm 117
10.1 Binary modulo exponentiation 120
10.2 A C program on the BBP series 123
10.3 Refinements 126
11. Arithmetic 131
11.1 Multiplication 131
11.2 Karatsuba multiplication 132
11.3 FFT multiplication 135
11.4 Division 145
11.5 Square root 146
11.6 nth root 149
11.7 Series calculation 150
12. Miscellaneous 153
12.1 A pi quiz 153
12.2 Let numbers speak 154
12.3 A proof that pi = 2 155
12.4 The big change 155
12.5 Almost but not quite 156
12.6 Why always more? 158
12.7 pi and hyperspheres 158
12.8 Viète × Wallis = Osler 160
12.9 Squaring the circle with holes 162
12.10 An (in)finite funnel 164
13. The History Of pi 165
13.1 Antiquity 167
13.2 Polygons 17 0
13.3 Infinite expressions 185
13.4 High-performance algorithms 198
13.5 The hunt for single pi digits 203
Table: History of pi in the pre-computer era 205
Table: History of pi in the computer era 206
Table: History of digit extraction records 207
14. Historical Notes. 209
14.1 The earliest squaring the circle in history? 209
14.2 A pi law 211
14.3 The Bieberbach story 213
15. The Future: pi Calculations On The Internet 215
15.1 The binsplit algorithm 215
15.2 The pi project on the Internet 219
16. pi Formula Collection223
17. Tables 239
17.1 Selected constants to 100 places (base 10) 239
17.2 Digits 0 to 2,500 of pi (base 10) 240
17.3 Digits 2,501 to 5,000 of pi (base 10) 241
17.4 Digits 0 to 2,500 of pi (base 16) 242
17.5 Digits 2,501 to 5,000 of pi (base 16) 243
17.6 Continued fraction elements 0 to 1,000 of pi 244
17.7 Continued fraction elements 1,001 to 2,000 of pi 245
A. Documentation For The hfloat Library 247
A.1 What hfloat is (good for) 247
A.2 Compiling the library 248
A.3 Functions of the hfloat library 248
A.4 Using hfloats in your own code 250
A.5 Computations with extreme precision 250 | 2014-03-09T19:28:22 | {
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https://math.stackexchange.com/questions/3175288/on-determinants-of-a-certain-class-of-matrices | # On determinants of a certain class of matrices
$$\newcommand{\M}{\mathcal{M}}$$Recently I encountered a certain class of matrices whose determinants behave in an interesting manner. Define $$\M(n,k)$$ for positive integers $$n,k$$ with $$k\leq n$$ to be the real $$n\times n$$ matrix with all $$1$$s on the diagonal, all $$1$$s for $$k-1$$ entries to the right of the diagonal on each row, and $$0$$s everywhere else. Note that if there are less than $$k-1$$ entries to the right of the diagonal, then the $$1$$s carry over to the leftmost columns. For example: $$\M(4,2)=\begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\1&0&0&1\end{bmatrix}\quad\text{and}\quad\M(3,1)=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.$$
At first, I believed that $$\M(n,k)$$ would always be nonsingular when $$k, but this turned out to be false. The smallest example where this conjecuture fails is $$\M(4,2)$$, whose determinant is $$0$$. After even more numerical testing I have been led to believe the following new conjecture:
The determinant $$\det\M(n,k)$$ is $$0$$ if and only if $$\gcd(n,k)>1$$. If $$n,k$$ are coprime, then $$\det\M(n,k)=k$$.
I have tested for all $$k\leq n$$ up to $$n=9$$ with the help of a computer without any counterexamples. Does anyone have an idea how this conjecture might be proven?
Here are some partial results. It is trivial that $$\det\M(n,k)$$ is zero when $$k=0$$ or $$n$$, and that it is $$1$$ when $$k=1$$. I can prove the result when $$k=2$$ as well: Let $$U(n)$$ be the $$n\times n$$ matrix with all entries $$0$$ except a $$1$$ at the bottom leftmost corner. Then, we have the recurrence relation $$\det\M(n,2)=\det\big(\M(n-1,2)-U(n-1)\big)+(-1)^{n+1}\det\big(\M(n-1,2)^t-U(n-1)^t\big)$$ which of course simplifies to $$\det\M(n,2)=(1+(-1)^{n+1})\det\big(\M(n-1,2)-U(n-1)\big).$$ Upon verifying $$\det\M(2,2)=0$$, it's clear that this proves the conclusion for $$k=2$$ if we assume that $$\det\big(\M(n-1,2)-U(n-1)\big)=1$$. Call the LHS $$x(n-1)$$. It is easy to compute that $$x(2)=1$$, from which it follows by induction $$x(n)=1$$ for all $$n$$, since $$x(n+1)=x(n)$$.
Unfortunately, I am at a complete loss in any of the cases other than $$k=0,1,2,n$$. It seems like (for sufficiently large $$n$$) as $$k$$ increases from $$3$$ until $$\lfloor n/2\rfloor$$, the proof going by the same thinking as my proof for $$k=2$$ would get increasingly complicated until it becomes hopeless to even attempt. (I could be wrong, of course.) Not to mention the general case for any positive integer $$n$$.
Does anyone have an idea how to attack the general problem? Thanks in advance!
• Your matrix is a circulant matrix, its determinant equals to $\prod_{\ell=0}^{n-1} P(\omega^\ell)$ where $\omega$ is the $n^{th}$ of unity and $P(x) = 1 + x + \cdots + x^{k-1}$. – achille hui Apr 5 at 0:37
• @achillehui Thank you for the great hint! – YiFan Apr 5 at 1:53
$$\newcommand{\M}{\mathcal{M}}$$Thanks to @achillehui's suggestion, I am able to prove the conjecture, and I present the proofs below. For any $$n\times n$$ circulant matrix $$C$$ with column vectors being the permutations of a certain vector $$c=(c_0,c_1,\dots,c_{n-1})^T$$, and with the associated polynomial $$p(x)=c_0+c_1x+\dots+c_{n-1}x^{n-1}$$, the following is known.
The determinant of $$C$$ is given by $$\displaystyle\prod_{j=0}^{n-1}p(\zeta_n^j)$$, where $$\zeta_n:=\exp(2\pi i/n)$$ is the primitive $$n$$th root of unity.
The matrices $$\M(n,k)$$ are circulant, with associated polynomial $$p(x)=1+x+\dots+x^{k-1}$$. Note that whenever $$x\neq 1$$, we have $$p(x)=(x^k-1)/(x-1)$$, so we may write $$\det\M(n,k)=\prod_{j=0}^{n-1}p(\zeta_n^j)=p(1)\prod_{j=1}^{n-1}\frac{\zeta_n^{jk}-1}{\zeta_n^j-1}.$$ In the case where $$\gcd(n,k)=:d>1$$, we know that $$\zeta_n^k$$ is no longer a primitive $$n$$th root of unity, but it is a primitve $$(n/d)$$th root of unity. Since $$d>1$$, the integer $$n/d$$ lies in the set $$\{1,2,\dots,n-1\}$$ from which $$j$$ takes values in the product, so there is a term in the product where $$j=n/d$$. For this term, the numerator is $$\zeta_n^{jk}-1=\zeta_{n/d}^{n/d}-1=0$$, so that the whole product is $$0$$ and the determiant is $$0$$. Conversely, when $$d=1$$, $$\zeta_n^k$$ will always be a primitive $$n$$th root of unity, and their $$j$$th powers will never be $$1$$ (for $$1\leq j\leq n-1$$). Hence the product is nonzero.
The stronger conjecture that $$\det\M(n,k)=k$$ when $$\gcd(n,k)=1$$ can be proven in this way as well. It is trivial that $$p(1)=k$$, so it suffices to show the product $$\prod(\zeta_n^{jk}-1)/(\zeta_n^j-1)$$ is $$1$$. The reason this is true is because of cancellation. The element $$k$$ in the additive group $$\mathbb Z/n\mathbb Z$$ (which is isomorphic to the multiplicative group of roots of unity) always generates the group whenever $$n,k$$ are coprime. Furthermore, each element is generated in a unique way. Therefore, as $$j$$ takes all nonzero values in $$\mathbb Z/n\mathbb Z$$, every element is represented exactly once, so the cancellation is perfect, proving the result.
If you use row operations, you can see where the periodicity comes in. As an example take $$M(5,3)$$; subtract the first row from the two bottom ones, then add the second row, etc.:
$$\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\{\color{red}1}&0&0&1&1\\{\color{red}1}&{\color{red}1}&0&0&1\end{pmatrix}\mapsto\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\0&{\color{red}-\color{red}1}&{\color{red}-\color{red}1}&1&1\\0&0&\color{red}-\color{red}1&0&1\end{pmatrix}\mapsto\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\0&0&0&2&1\\0&0&-1&0&1\end{pmatrix}$$
Let us just track the $$1$$s in the lower left corner as you perform row subtraction or addition. You'll see that the triangle of $$1$$s moves right with a period of $$k$$ as seen below (omitting the $$0$$s):
$$k=2,\qquad 1\mapsto\quad-1\quad\mapsto 1$$ $$k=3,\qquad \begin{matrix}1\\1&1\end{matrix}\quad\mapsto\quad \begin{matrix}-1&-1\\&-1\end{matrix}\quad\mapsto\quad\begin{matrix}&1\\-1\end{matrix}\quad\mapsto\quad\begin{matrix}1\\1&1\end{matrix}$$
$$k=4,\qquad\begin{matrix}1&&\\1&1&\\1&1&1\end{matrix}\quad\mapsto\quad\begin{matrix}-1&-1&-1\\&-1&-1\\&&-1\end{matrix}\quad\mapsto\quad\begin{matrix}&&1\\-1&-1\\&-1\end{matrix}\quad\mapsto\quad\begin{matrix}&&1\\&1&1\\-1\end{matrix}\quad\mapsto\quad\begin{matrix}1\\1&1\\1&1&1\end{matrix}$$
When the $$1$$s have moved by $$k$$ steps, they leave a matrix of $$0$$s behind. Since $$\det\begin{pmatrix}A&B\\O&C\end{pmatrix}=\det A\det C$$, and since the upper left matrix $$A$$ has determinant 1, it follows that $$\det M(n,k)=\det M(n-k,k)$$. So what's left is to check what happens when this triangle of $$1$$s meets the rest of the matrix as it moves right. The end result is one of $$M(k,k)$$, $$M(k+1,k)$$,..., $$M(2k-1,k)$$. Clearly if $$n$$ is a multiple of $$k$$ we reach the matrix $$M(k,k)$$ of all $$1$$s which has determinant $$0$$. I believe you can take it from here.
• Sorry, I'm afraid I don't understand what you're trying to say. What do you mean by "as you perform row subtraction or addition"? Which rows are added/subtracted from which others? I don't understand what your diagram is trying to say, nor to I get anything at all in the last paragraph. Maybe you want to rephrase it? – YiFan Apr 6 at 1:10 | 2019-08-22T10:05:13 | {
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http://mathematica.stackexchange.com/questions/37266/even-fibonacci-numbers/37268 | # Even Fibonacci numbers
Today, I found the Euler Project. Problem #2 is
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
I have two solutions:
For[i = 1, i < 100, i++, If[Fibonacci[i] > 4000000, Break[]]];
Plus @@ Select[Fibonacci /@ Range[i - 1], Mod[#, 2] == 0 &]
4613732
Plus @@ Select[
Fibonacci /@ Select[Range@100, Fibonacci@# < 4000000 &],
Mod[#1, 2] == 0 &]
4613732
However, I feel my solutions are not good and efficient. So my question is: can you show me more concise and efficient methods?
Also, how cam I test whether a number is a Fibonacci number？
-
Since you already have the answer, you may access the problem 2 forum on the Euler project. Here you will find the most efficient algorithms from other users. The third post there demonstrates an algorithm that can be easily calculated without a program. – trav1s Nov 19 '13 at 5:38
The other methods described work well, but I am partial to exact answers. So, here it is. First, note that the Fibbonacci sequence has a closed form
$$F_n = \frac{\Phi^n - (-\Phi)^{-n}}{\sqrt{5}}$$
where $\Phi$ is the GoldenRatio. Also, as observed elsewhere, we are looking only at every third term, which gives the sum
$$\sum^k_{n = 1} \frac{\Phi^{3n} - (-\Phi)^{-3n}}{\sqrt{5}}$$
which is just the sum of two geometric series. So, after a few steps, this simplifies to
$$\frac{1}{\sqrt{5}} \left[ \Phi^3\frac{1 - (\Phi^3)^{k}}{1-\Phi^3} + \Phi^{-3}\frac{1 - (-\Phi^{-3})^{k}}{1 + \Phi^{-3}} \right]$$
where $k$ is the index of the highest even Fibbonacci number. To find $k$, we can reverse the sequence,
n[F_] := Floor[Log[F Sqrt[5]]/Log[GoldenRatio] + 1/2]
n[4000000]
(* 33 *)
So, $k = 11$. In code,
Clear[evenSum];
evenSum[(k_Integer)?Positive] :=
Round @ N[
With[
{phi = GoldenRatio^3, psi = (-GoldenRatio)^(-3)},
(1/Sqrt[5])*(phi*((1 - phi^k)/(1 - phi)) - psi*((1 - psi^k)/(1 - psi)))
],
Max[\$MachinePrecision, 3 k] (* needed for accuracy *)
]
evenSum[11]
(* 4613732 *)
which is confirmed by
Total @ Fibonacci @ Range[3, 33, 3]
(* 4613732 *)
Edit: The above implementation of n suffers round off problems for large numbers, for example n[9000] gives 21, but Fibonacci[21] is 10946. But, wikipedia gives a better choice
Clear[n2];
n2[fn_Integer?Positive] :=
Floor@N@Log[GoldenRatio, (fn Sqrt[5] + Sqrt[5 fn^2 - 4])/2]
n2[10945]
(* 20 *)
-
Had to fix the formulas, I dropped quite a few minus signs in the original version. – rcollyer Nov 18 '13 at 18:04
Related to your error, of course facebook.com/physic.department/posts/360869444018997 – belisarius Nov 18 '13 at 18:25
@belisarius :)) ingenious. – Stefan Nov 18 '13 at 18:27
@belisarius according to my adviser, his adviser was infamous for making multiple canceling errors. His results were correct, the derivation not usually. :) – rcollyer Nov 18 '13 at 18:31
Could you explain how these functions n and evenSum should be used for big numbers? I tried to test your solution unsucessfully. While with mine approach I could simply find sums up to 10^10000. – Artes Nov 18 '13 at 18:36
Straight iteration over the even valued Fibonacci numbers is fast.
fibSum[max_] := Module[
{tot, n, j},
tot = 0;
n = 0;
j = 3;
While[n = Fibonacci[j]; n <= max, j += 3; tot += n];
tot]
Or one can use matrix products. This seems to be about the same speed. It has the advantage of not requiring a built in Fibonacci function.
fibSum2[max_] := Module[
{tot, n, fp, mat, mat3},
mat = {{0, 1}, {1, 1}};
tot = 0;
n = 0;
fp = mat.mat;
mat3 = mat.mat.mat;
While[n = fp[[2, 2]]; n <= max, fp = fp.mat3; tot += n];
tot]
These handle 10^10000 in slightly under a second on my desktop running Mathematica 9.0.1.
Here is a test for whether an integer is a Fibonacci number (a Fib detector?)
fibQ[n_] :=
With[{k =
Round[Log[GoldenRatio, N[n, 20 + Length[IntegerDigits[n]]]] +
Log[GoldenRatio, Sqrt[5]]]},
n == Fibonacci[k]]
-
You need to correct 10^10000. – Ramchandra Apte Nov 19 '13 at 10:35
[Correct it how? Will a good spanking suffice?] – Daniel Lichtblau Nov 19 '13 at 15:22
My mistake; I read it as 10^10000 numbers in slightly under a second. – Ramchandra Apte Nov 19 '13 at 16:06
Sum[Fibonacci[n],
{n, 3, InverseFunction[Fibonacci][4000000], 3}]
(* 4613732 *)
-
That's tyte, + goes w/o saying. – alancalvitti Oct 30 '14 at 18:18
Your solution looks fine. You can take advantage of the fact that every third Fibonacci number is even, which makes it a little faster. Fibonaccis are cheap to compute and they quickly exceed 4 million. Here's a comparison:
Select[Fibonacci[3 Range@33], # <= 4*^6 &] // Total // AbsoluteTiming
{0.000214, 4613732}
Select[Fibonacci@Range@100, # <= 4*^6 && EvenQ@# &] // Total // AbsoluteTiming
{0.000566, 4613732}
Plus @@ Select[Fibonacci /@ Select[Range@100, Fibonacci@# < 4000000 &], Mod[#1, 2] == 0 &] // AbsoluteTiming
{0.000717, 4613732}
AbsoluteTiming[ For[i = 1, i < 100, i++, If[Fibonacci[i] > 4000000, Break[]]]; Plus @@ Select[Fibonacci /@ Range[i - 1], Mod[#, 2] == 0 &]]
{0.000526, 4613732}
-
Here I'd like to show my solution.
Since A014445 we know that the generating function of even fibonacci numbers is Fibonacci[3 n], we can write something like this:
Plus @@ Select[Table[Fibonacci[3 n], {n, 0, 30}], # < 4000000 &] // AbsoluteTiming
=> {0., 4613732}
An imperative style solution could look like this:
res = 0; n = 1; While[res < 4000000, res += Fibonacci[3 n]; n++]
Result will be in res and timings are negligible, zero.
Edit: To bring that into a cogent form. Here the full definition of SumEvenFibonacci:
ClearAll[SumEvenFibonacci];
SumEvenFibonacci::usage =
"SumEvenFibonacci[n] calculates the sum of even Fibonacci numbers \
up to the upper-bound n.";
SumEvenFibonacci[n_Integer] /; n >= 0 := Module[{res = 0, i = 1},
While[res < n, res += Fibonacci[3 i]; i++];
res
]
-
The Fibonacci function, is Listable and fast increasing. We can observe a simple pattern, every third element is even (see e.g. Fibonacci[Range[12]]), moreover the first number exceeding 4000000 is: NestWhile[(# + 1) &, 1, Fibonacci[#] <= 4 10^6 &], thus we have:
Total[ Fibonacci[ Range[3, 33, 3]]]
4613732
and most likely this is the best approach:
sum[n_] := Total[ Fibonacci[ Range[3, NestWhile[(# + 1) &, 1, Fibonacci[#] <= n &] - 1, 3]]]
e.g.
sum[10^600]; // AbsoluteTiming
{0.096000, Null}
Concerning testing if given numbers are Fibonacci ones there is a simple test based on checking if consecutive elements generate Pythagorean triples (see e.g. this answer)
And @@ (#1^2 + #2^2 == #3^2 & @@@
Array[{ Fibonacci[#] Fibonacci[# + 3],
2 Fibonacci[# + 1] Fibonacci[# + 2],
Fibonacci[# + 1]^2 + Fibonacci[# + 2]^2} &, 1000])
True
-
You already got my +1, but, yeah, it looks a little cretinous. – rcollyer Nov 18 '13 at 21:14
@rcollyer I like your approach, but I still cannot use it successfully. I guess Daniel Lichtblau's solution is the most efficient for big numbers, however I can see that another reliable (mine) solution is the least appreciated. – Artes Nov 18 '13 at 21:20
A little poking around indicates that n gives the nearest Fibbonacci number, not the one immediately below it, e.g. n[9000] returns 21, not 20. Likely a similar rounding issue is occurring in evenSum, but it tracks Daniel's exactly for a long time. So, I'm not sure of the specific issue. – rcollyer Nov 18 '13 at 21:30
## protected by Verbeia♦Nov 17 '14 at 1:30
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site. | 2015-10-04T15:38:55 | {
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https://math.stackexchange.com/questions/1697991/do-diagonal-matrices-always-commute/1698015 | # Do Diagonal Matrices Always Commute?
Let $A$ be an $n \times n$ matrix and let $\Lambda$ be an $n \times n$ diagonal matrix. Is it always the case that $A\Lambda = \Lambda A$? If not, when is it the case that $A \Lambda = \Lambda A$?
If we restrict the diagonal entries of $\Lambda$ to being the equal (i.e. $\Lambda = \text{drag}(a, a, \dots, a)$), then it is clear that $A\Lambda = AaI = aIA = \Lambda A$. However, I can't seem to come up with an argument for the general case.
• Have you tried simple examples? – Friedrich Philipp Mar 15 '16 at 1:18
• When the diagonal matrix is on the right, it scales the columns of the matrix it is multiplying. when the diagonal matrix is on the left, it scales the rows. Since column-scaling and row scaling are different operations, there are only very limited circumstances that the matrices will commute. – Nick Alger Mar 15 '16 at 1:30
If all the diagonal entries of$\Lambda$ are distinct, it commutes only with diagonal matrices.
In contrast, for each $k$ consecutive equal diagonal entries in $\Lambda,$ we may allow $A$ to have anything at all in the corresponding $k$ by $k$ square block with both corners on the main diagonal.
This means that the set of matrices that commute with $\Lambda$ has a minimum dimension $n$ and a maximum dimension $n^2.$ Suppose we have $r$ different diagonal entries, and there are $k_i$ copies of diagonal entry $\lambda_i.$ Each $k_i \geq 1,$ and we have $$k_1 + k_2 + \cdots + k_r = n.$$ Then by the block construction I mentioned above, the dimension of the space of matrices that commute with $\Lambda$ is $$k_1^2 + k_2^2 + \cdots + k_r^2.$$ The minimum is when $r=n,$ so all $k_i = 1,$ and the dimension is $n$
The maximum is when $r=1,$ and $k_1=n,$ the matrix is a scalar multiple of the identity matrix, and the dimension is $n^2.$
• Do the copies have to be right next to each other? e.g. diag(1,1,1,2,2,7,8,8)? – Hrit Roy Feb 8 '18 at 5:25
• @HritRoy, no. You can change your basis to express the copies next to each other if you want to. But it is not necessary – Vladimir Vargas Apr 27 '18 at 22:22
It is possible that a diagonal matrix $\Lambda$ commutes with a matrix $A$ when $A$ is symmetric and $A \Lambda$ is also symmetric. We have
$$\Lambda A = (A^{\top}\Lambda^\top)^{\top} = (A\Lambda)^\top = A\Lambda$$
The above trivially holds when $A$ and $\Lambda$ are both diagonal.
A diagonal matrix will not commute with every matrix.
$$\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}*\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$
But:
$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} * \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}.$$
The claim is false in general. Take $A = \begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}$ $\Lambda = \begin{bmatrix}2 & 0\\0 & 3\end{bmatrix}$. Then
$A \Lambda = \begin{bmatrix}2 & 6\\6 & 12\end{bmatrix}$
$\Lambda A = \begin{bmatrix}2 & 4\\9 & 12\end{bmatrix}$
On a more useful note, you can look up commuting matrices on Wikipedia.
We want to find $\Lambda$ such that $A\Lambda=\Lambda A$. Then we have $$\left( \begin{array}{cc} \sum a_{1n}\lambda_{n1} & \sum a_{1n}\lambda_{n2} & \sum a_{1n}\lambda_{n3} & ... \\ \sum a_{2n}\lambda_{n1} & \sum a_{2n}\lambda_{n2} & \sum a_{2n}\lambda_{n3} & ...\\ \sum a_{3n}\lambda_{n1} & \sum a_{3n}\lambda_{n2} & \sum a_{3n}\lambda_{n3} & ...\\ ... & ... & ... & ...\end{array} \right)=\left( \begin{array}{cc} \sum a_{n1}\lambda_{1n} & \sum a_{n1}\lambda_{2n} & \sum a_{n1}\lambda_{3n} & ... \\ \sum a_{n2}\lambda_{1n} & \sum a_{n2}\lambda_{2n} & \sum a_{n2}\lambda_{3n} & ...\\ \sum a_{n3}\lambda_{1n} & \sum a_{n3}\lambda_{2n} & \sum a_{n3}\lambda_{3n} & ...\\ ... & ... & ... & ...\end{array} \right)$$ Hence for the equality to hold, both $A$ and $\Lambda$ must be symmetric since $a_{in}=a_{ni}$ and $\lambda_{in}=\lambda_{ni}$ for $i=1,2,3,...$ A diagonal matrix is a special case of this.
• Hello, what is the name of this theorem? Or where is it in a book? I want to reference it. Thanks a lot. – Vladimir Vargas Apr 27 '18 at 22:09
• I am not aware of a theorem whose result is this since it is just derived from the definition. – TheSimpliFire Apr 28 '18 at 8:20
• "both $A$ and $\Lambda$ must be symmetric" <-- not true as stated (it's not a necessary condition); also, $\Lambda$ is symmetric already (being diagonal). – darij grinberg Jan 12 at 22:45
The answer from @AOK is not generally true. Obviously for diagonal identical matrix $$\Lambda$$ is valid only when $$A$$ is symmetric matrix. | 2019-06-19T21:08:22 | {
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https://math.stackexchange.com/questions/1103529/applying-linear-algebra-to-solve-a-problem-in-mechanical-equilibrium | # Applying linear algebra to solve a problem in mechanical equilibrium
I came across the following problem in "Introduction to Applied Mechanics" by Gilbert Strang, and am a little confused about the solution to this problem. The following figure shows the problem.
The following vectors are defined for the problem,
1. $\mathbf{x} = [x_1, x_2, x_3]^T$ represents the displacements of the masses.
2. $\mathbf{e} = [e_1, e_2, e_3, e_4$ represents the elongations of the springs.
3. $\mathbf{y} = [y_1, y_2, y_3, y_4]$ represents the force in the springs.
4. $\mathbf{f} = [f_1, f_2, f_3]$ represents the force on the masses.
Let us assume that upward direction is positive, i.e. a displacement in the upward displacement will result in a positive $x_i$.
Thus the elongations of the $i^{th}$ spring is given by, $e_i = x_{i-1} - x_{i}$. This means that an upward movement of the mass above the spring, and the downward movement of the mass below the spring will result in elongating the spring, and thus positive elongation. It must be noted that $x_0, x_4 = 0$ as these correspond to the boundary conditions represented by the walls on which the ends of springs 1 and 3 and clamped.
$$\implies \mathbf{e} = \left( \begin{array}{ccc} -1 & 0 & 0\\ 1 & -1 & 0\\ 0 & 1 & -1\\ 0 & 0 & 1\\ \end{array} \right) \mathbf{x} = \mathbf{A}\mathbf{x}$$
The above equation makes sense, as a negative $x_1$ (movement downwards), will correspond to an elongation of spring 1. Similarly, a positive $x_3$ (movement upwards) will correspond to elongation of spring 4.
The forces in the springs (positive for tension, and negative for compression) is given by the following,
$$\mathbf{y} = \left( \begin{array}{cccc} k_1 & 0 & 0 & 0\\ 0 & k_2 & 0 & 0\\ 0 & 0 & k_3 & 0\\ 0 & 0 & 0 & k_4\\ \end{array} \right) \mathbf{e} = \mathbf{C}\mathbf{e}$$
Now the force on the $i^{th}$ mass is $f_i = y_i - y_{i+1}$. If the spring above the $i^{th}$ mass is under tension then $y_i$ will be positive, and will tend to pull the mass upwards, and thus it has a positive contribution $y_i$ to the force on the mass. The opposite is true for the spring below the mass, thus has a negative contribution $y_{i+1}$.
$$\mathbf{f} = \left( \begin{array}{cccc} 1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ \end{array} \right) \mathbf{y} = \mathbf{B}\mathbf{e}$$
The problem here is that $\mathbf{B}$ must equal $\mathbf{A}^T$, but according the conventions I have followed it does not. In fact $\mathbf{B} = -\mathbf{A}^T$.
Can someone tell me where I am going wrong?
• To start "negative x1 (movement downwards)" seems to be in contradiction to the diagram or at least with the way I'm used to it drawn with the displacement arrow in the direction of the positive x value. Also it's not clear where your masses come into these equations. Have you omitted those on purpose here or you just haven't gotten to that part? – Fizz Jan 14 '15 at 5:13
• @RespawnedFluff I have not yet gotten to the point with the masses. At this point I am only looking at the basic set of equations relating the displacement of the masses, the spring elongations and the resulting forces on the mass. I am supposed to get $\mathbf{A}^T\mathbf{e}$ for $\mathbf{f}$, instead I get $-\mathbf{A}^T\mathbf{e}$. And regarding your point about the direction of movement of the masses, I get the same results even if I choose the downward movement as positive as shown in the figure. – siva82kb Jan 16 '15 at 5:05
Like I said in my comment the problem is with your $\mathbf{A}$. You've actually made me check out Strang's book. His $\mathbf{A}$ (given on p. 41) is minus your $\mathbf{A}$ (because his $x$ axis is actually in the direction depicted in his diagram), i.e.:
$$\mathbf{e} = \left( \begin{array}{ccc} 1 & 0 & 0\\ -1 & 1 & 0\\ 0 & -1 & 1\\ 0 & 0 & -1\\ \end{array} \right) \mathbf{x} = \mathbf{A}\mathbf{x}$$
And he has $e_i = x_{i} - x_{i-1}$ (the opposite of your equation), which makes sense because $x_0=0$ and the obvious physical intuition is that $e_1>0$, i.e. the first/top spring is elongated. Besides the obvious "why would want to do that?", you cannot just pick arbitrary signs for the elongation and force in Hooke's law; they are correlated (in sign) or else the force changes its meaning.
The $\mathbf{A} = \mathbf{B}^\mathrm{T}$ equality holds only if the axes for measuring $\mathbf{x}$ and the forces $\mathbf{f}$ is the same, including orientation/sign. You have flipped the sign/direction of one axis (i.e. of $\mathbf{x}$) but not the other one ($\mathbf{f}$), i.e. your forces are still positive downwards as in Strang's book.
To see this more easily, consider the reduced/simplified problem where you only have one object between two springs one from top, the other from the bottom:.
Then since $\mathbf{x} = (x_1)$,
$$\mathbf{e} = \left( \begin{array}{ccc} 1 \\ -1 \\ \end{array} \right)\mathbf{x} = \left( \begin{array}{ccc} x_1 \\ -x_1 \\ \end{array} \right)$$
So the elongations of the two springs necessarily have equal magnitude, but have opposite signs: $e_1=x_1$, $e_2=-x_1$ in the same reference system. Obviously
$$\mathbf{y} = \left( \begin{array}{ccc} k_1x_1 \\ -k_2x_1 \\ \end{array} \right)$$
But the external force $$\mathbf{f} = \left( \begin{array}{ccc} 1 & -1\\ \end{array} \right)\mathbf{y} = k_1x_1+k_2x_1$$
This is physically ok because from an external point of view the two springs exert forces in the same direction since compressing one spring means elongating the other by the same amount. So any other force that has to work against these two springs, like the gravitational force due to the mass $m_1$, has to equal (be in equilibrium with) $\mathbf{f}$, i.e.
$$m_1g = \mathbf{f} = k_1x_1 + k_2x_1$$
This is as if the springs were actually in parallel if both were attached to the same surface; you can visualize this by rotating the bottom surface (while keeping the spring attached) until you glue it to the top one.
In matrix form, the dependence of $\mathbf{f}$ on $\mathbf{x}$ is via the $1\times 1$ stiffness matrix $(k_1+k_2)$:
$$\mathbf{f} = \left( \begin{array}{ccc} 1 & -1\\ \end{array} \right) \left( \begin{array}{ccc} k_1 & 0 \\ 0 & k_2 \\ \end{array} \right) \left( \begin{array}{ccc} 1 \\ -1 \\ \end{array} \right)\mathbf{x} = \left( \begin{array}{ccc} 1 & -1\\ \end{array} \right) \left( \begin{array}{ccc} k_1 \\ -k_2 \\ \end{array} \right) \mathbf{x} = (k_1+k_2)\mathbf{x}$$
So let's see now what happens in this simpler system if we consider $x_1$ positive going up instead, i.e.
$$\mathbf{e} = \left( \begin{array}{ccc} -1 \\ 1 \\ \end{array} \right) \mathbf{x}$$
You'll obviously get the stiffness matrix to change sign, i.e. $\mathbf{f} = -k_1x_1 - k_2x_1$, which corresponds to expressing Hooke's law (for both springs) in terms of the reaction force. Exactly the same thing happens in the general case: if you flip the axis of reference for $\mathbf{x}$ and keep the one for $\mathbf{f}$ unchanged you get a change of sign in the stiffness matrix. This is actually explained a bit later in the book as "Positivity means that when all the forces $f$ go in one direction, so do all the displacements $x$." | 2021-06-23T21:59:44 | {
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https://math.stackexchange.com/questions/178222/latin-squares-of-even-order-sub-squares | Latin squares of even order-sub squares
Consider Latin squares of even order that is not of form $2^x$, where every cell is involved in a $2\times 2$ sub square. Here is one such square for order 6:
0 1 2 3 4 5
1 0 3 4 5 2
2 4 0 5 1 3
3 5 1 2 0 4
4 2 5 1 3 0
5 3 4 0 2 1
Notice that rows correspond to each other in pairings. Cells in the first row make sub squares with cells in the third row, cells in the second row make sub squares with cells in the fourth, and the fifth and sixth rows make sub squares with each other. These are not the only existent subsquares, but there is a definitive correspondence between pairs of rows for subsquares.
The question- what does a latin square of order 10 with all cells involved in a $2\times 2$ subsquare, and with rows presumably paired like in the order 6 square satisfying the same conditions, look like? And once we can see that square, is there any visible relationship between such squares of even but not power-of-two order and the group tables for $\mathbb{Z}/2^n$, which also satisfy the all-cells-involved-in-two-by-two-subsquares condition? Is there any generalizable pattern for this sort of square at all?
But I suppose my main question is with regards to finding $10\times 10, 12\times 12$ etc examples- especially $10\times 10$. I have not been able to find an example using the pairing rows method- maybe I am just being disorganized.
In the Cayley table of $\mathbb{Z}_n \times \mathbb{Z}_2$, every entry belongs to a $2 \times 2$ subsquare for all even orders -- this resolves the existence question. In fact, a much more general statement is true:
Lemma: If $G$ is a group and $H$ is a subgroup of $G$, then the Cayley table of $G$ partitions into $|H| \times |H|$ subsquares induced by the double-cosets of $H$.
So, for example, $S_4$ is a group of order $24$, and every entry in its Cayley table (a Latin square of order $24$) belongs to some $2 \times 2$ subsquare, some $3 \times 3$ subsquare, some $4 \times 4$ subsquare, and so on for the other subgroup sizes.
Non-group examples can be formed by a direct product construction. Take a Latin square of order $n$, and replace each entry by a $2 \times 2$ subsquare as indicated below:
1 2 3 1 2 3 4 5 6
2 3 1 -> 2 1 4 3 6 5
3 1 2 3 4 5 6 1 2
4 3 6 5 2 1
5 6 1 2 3 4
6 5 2 1 4 3
and then we can "turn" the subsquares using the switching operation:
1 2 -> 2 1
2 1 1 2
As usual, we can permute the rows/columns/symbols arbitrarily and preserve this property (along with taking "matrix transpose" and "replacing each row by its inverse", and combinations of these).
For odd $n$, the question was resolved by:
• P. Danziger and E. Mendelsohn, Intercalates everywhere, London Math. Soc. Lecture Note Ser. 245 (1997). [Thanks to Ian Wanless for pointing me to this paper.]
Hence, we have the following:
Theorem: For all $n \geq 1$ except $n \in \{1,3,5,7\}$ there is a Latin square of order $n$ in which each entry belongs to some $2 \times 2$ subsquare.
Going the other direction, i.e. finding a general description of Latin squares whose entries belong to $2 \times 2$ subsquares, I suspect, is more-or-less impossible. For small orders (orders $\leq 9$ or maybe $10$), it would be possible to exhaustively find a main class representative in each case -- I would expect this to be a particularly long calculation, and likely, not particularly satisfying.
It would also be possible to find related results using in the following (and their references):
• B. D. McKay and I. M. Wanless, Most Latin Squares Have Many Subsquares, Journal of Combinatorial Theory, Series A 86, 323 347 (1999).
• J. Browning, D. S. Stones and I. M. Wanless, Bounds on the number of autotopisms and subsquares of a Latin square, Combinatorica, to appear.
• I happen to be working on a proof that it is not possible for odd order. – Xuan Huang Aug 3 '12 at 1:49
$$\matrix{0&1&2&3&4&5&6&7&8&9\cr1&0&3&2&5&4&7&6&9&8\cr2&3&4&5&6&7&8&9&0&1\cr3&2&5&4&7&6&9&8&1&0\cr4&5&6&7&8&9&0&1&2&3\cr5&4&7&6&9&8&1&0&3&2\cr6&7&8&9&0&1&2&3&4&5\cr7&6&9&8&1&0&3&2&5&4\cr8&9&0&1&2&3&4&5&6&7\cr9&8&1&0&3&2&5&4&7&6\cr}$$ 1st row paired with 2nd, 3rd with 4th, 5th with 6th, etc.
• I feel like an idiot. Are any different Latin squares possible? – Xuan Huang Aug 3 '12 at 1:03
• Perhaps more like the six by six I showed? – Xuan Huang Aug 3 '12 at 1:05
• Or is my square just a row column permutation of your sort of circulant? – Xuan Huang Aug 3 '12 at 1:11 | 2019-05-21T16:57:35 | {
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https://www.spoj.com/problems/BOTTOM/ | ## BOTTOM - The Bottom of a Graph
no tags
We will use the following (standard) definitions from graph theory. Let $V$ be a nonempty and finite set, its elements being called vertices (or nodes). Let $E$ be a subset of the Cartesian product $V \times V$, its elements being called edges. Then $G = (V, E)$ is called a directed graph.
Let $n$ be a positive integer, and let $p = (e_1, \ldots, e_n)$ be a sequence of length $n$ of edges $e_i \in E$ such that $e_i = (v_i, v_{i+1})$ for a sequence of vertices ($v_1, \ldots, v_{n+1}$). Then $p$ is called a path from vertex $v_1$ to vertex $v_{n+1}$ in $G$ and we say that $v_{n+1}$ is reachable from $v_1$, writing $(v_1 \to v_{n+1})$.
Here are some new definitions. A node $v$ in a graph $G = (V, E)$ is called a sink, if for every node $w$ in $G$ that is reachable from $v$, $v$ is also reachable from $w$. The bottom of a graph is the subset of all nodes that are sinks, i.e., $\mathrm{bottom}(G) = \{v \in V \mid \forall w \in V : (v \to w) \Rightarrow (w \to v) \}$. You have to calculate the bottom of certain graphs.
### Input Specification
The input contains several test cases, each of which corresponds to a directed graph $G$. Each test case starts with an integer number $v$, denoting the number of vertices of $G = (V, E)$, where the vertices will be identified by the integer numbers in the set $V = \{1, \ldots, v\}$. You may assume that $1 \le v \le 5000$. That is followed by a non-negative integer $e$ and, thereafter, $e$ pairs of vertex identifiers $v_1, w_1, \ldots, v_e, w_e$ with the meaning that $(v_i, w_i) \in E$. There are no edges other than specified by these pairs. The last test case is followed by a zero.
### Output Specification
For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.
### Sample Input
3 3
1 3 2 3 3 1
2 1
1 2
0
### Sample Output
1 3
2
hide comments
< Previous 1 2 3 4 5 6 7 Next > aman_sachin200: 2018-06-17 22:00:51 Nice one!!!Try CAPCITY and TOUR after this! sherlock11: 2018-06-08 10:29:56 if u want a clear understanding of SCC then this problem and CAPCITY are the problems that u are looking for.............if u are new with SCC then first read the concepts (kosaraju's algo) and then ......well u know what to do after that.............AC:) karthik1997: 2017-12-18 09:18:35 Applied Kosaraju's algorithm . Really good problem . :) Hey @asib_133012 The scc's are [3] , [4] and [1,2] . Since 3-4 edge exists , you cannot take 3 as sink . All other scc's have nodes that are actually sinks . So the bottoms are 1,2,4 -> output :) Last edit: 2017-12-18 09:19:28 vib_s02: 2017-10-29 09:46:51 @justforpractic I think that answer should be 1 2 minaamir26: 2017-08-17 06:52:31 too strict time for java users asib_133012: 2017-05-25 12:43:02 4 4 1 2 2 1 3 2 3 4 0 In this case what should be the ans? and why? please explain someone.thanks in advance ayush_1997: 2017-03-10 21:17:24 learned the concept of sink vertex :) flyingduchman_: 2017-03-03 16:55:41 A correct algorithm : (Not easily available on the internet) Theorem : A strongly connected component (all vertices) is sink(i.e. all vertices are sink) if every vertex in a scc is connected to no other scc than itself's. Find all strongly connected component(scc)s by Kosaraju or Tarjan's algorithm and store them. Now, for every vertex in a scc, if you find an adjacent vertex (in the adjacency list) let's call it "v" such that v is not in that scc, then discard the whole scc. That is, when a vertex of a scc(No.1) is adjacent to a different scc(No.2) then discard scc(No.1), otherwise take all vertices of scc(No.1) to the "result" array or container. Do this for all scc s. After that, sort the "result" array and print it. Last edit: 2017-03-03 17:01:37 tni_mdixit: 2017-02-24 20:23:14 just try to read the problem carefully...that's it! rraj001: 2017-01-18 15:57:43 Good one for SCC
Added by: Wanderley Guimarăes Date: 2007-09-21 Time limit: 0.254s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: All except: ERL JS-RHINO Resource: University of Ulm Local Contest 2003 | 2018-06-23T06:19:24 | {
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https://math.stackexchange.com/questions/2836626/find-the-minimum-number-of-terms-needed-to-approximate-the-series-correct-to | # Find the minimum number of terms needed to approximate the series … correct to 1DP
The series is $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}k}{k^2+1}$
The minimum number of terms needed should be given by $\lvert a_{n+1} \rvert$
Thus $\lvert \frac{(-1)^{k+2}(k+1)}{(k+1)^2+1} \rvert<0.01$
However, this is only satisfied after the $99th$ term, whereas the answer says the $19th$ term.
• You could sum 100 terms that are around $0.009$ which matches your criterion, but that will still affect the decimal place. This will have something more to do with the rate that the series terms get smaller. – Benedict W. J. Irwin Jun 30 '18 at 10:34
• Yes, now that I think about it, the answer is still wrong in a sense that it will still be affecting that decimal point after 19 terms, however, at 19 terms exactly it is correct to 1DP. I believe, there is another method to solve this rather than using the truncation error, but I am not quite sure. – Anthony P Jun 30 '18 at 10:38
• Since the series is alternating, solve for $k$, $\frac{k+1}{(k+1)^2+1}=0.05$ (don't miss marty cohen's answer). – Claude Leibovici Jul 1 '18 at 14:07
## 1 Answer
For 1dp the bound should be.05, not .01. | 2020-01-19T10:40:07 | {
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https://www.physicsforums.com/threads/lenth-of-line-around-an-object.28381/ | # Lenth of line around an object?
1. May 30, 2004
### Physics is Phun
Lets say i had a cylinder x long with y radius with a string attached to the edge and I wrapped a string around it z number of time until the string reached the other end. How would I go about finding the length of this string. What if the object was something else like a cone or shere or a pyramid. thanks.
2. May 30, 2004
### Grizzlycomet
Well, finding the length of the string for one twist around the cylinder is easy using the formula for the circumference of a circle, pi times diametre. Then just multiply that number with the number of twists needed.
3. May 30, 2004
### Chen
The area that a line of length L and width D is S = LD. The problem is you don't have the width of the line (its thickness), D. But you have Z, the number of times you wrap the line, so you can use it to find D = X/Z. Going back to the area S, that's also the area of the zone the line covers which is S = 2πYX. So finally you have:
$$S = LD = L\frac{X}{Z} = 2\pi YX$$
$$L = 2\pi YZ$$
Grizzlycomet's method is of course shorter and easier, but this is a more general description of how to go about solving such problems when dealing with objects more complicated than cylinders.
4. Jun 1, 2004
### Physics is Phun
I am afraid I am confused and I may have confused you with my question. So I have a cylinder and a piece of string. If I just wrap the string around the can the length of the string will be the circumference. I want to find the length of the string when it is spiraled around the cylinder from one end to the other. Picture a barber shop pole or the stripes on a candy cane.
I have tried doing this with a tin can and fishing line. The fishing line has basically no thickness so I am not worried with that. The circumference of the can I measured 32.5cm, the length 16.5cm. When I wrap the fishing line around the can once (like a barber shop pole) I measure 36cm, when I wrap it twice I get 66cm. Now this does match with the equation L = 2pi *Y*X where y is the radius and x is the length. The length of the string is 3cm longer when wraped around the cylinder end to end than just the circumference I can't see is any relation to the length.
5. Jun 1, 2004
### jcsd
Well the surface of a Cylinder is Euclidan, so imagine drawing the line your string makes and unwrapping the surface of the cyclinder, what would it look like?
If you wnat the string to go from top to bottom (or bottom to top) and once around the cylinder it's lenght will be given by Pytahgoras's theorum where r is the radius and h the height:
$$L^2 = 4\pi^2r^2 + h^2$$
For n twists it will be:
$$L^2 = n^24\pi^2r^2 + {h^2}$$
Last edited: Jun 1, 2004
6. Jun 1, 2004
### robphy
Imagine a label on the can. If you unroll it, it's a rectangle with the same height of the can and the width of one circumference. The string wrapped once like a barber-pole on the can would be the diagonal of that rectangle.
sqrt(32.5^2+16.5^2)=36.44
If you wrap the string twice, that's the diagonal of a rectangle that is twice the width of label.
sqrt( (2*32.5)^2+16.5^2)=67.06
for 3 turns, sqrt( (3*32.5)^2+16.5^2)=98.88
7. Jun 1, 2004
### Gokul43201
Staff Emeritus
Clearly this problem is easier to solve (jcsd's post) than if the object were a cone or sphere or pyramid. Also, you would have to add more information to completely define the path of the string.
For a sphere you could require theta(phi,n+1) - theta(phi, n) = constant, where theta = polar angle (latitude), phi = azimuthal angle (longitude) and n = turn number
8. Jun 1, 2004
### jcsd
The only finite value allowed for n on a pyramid is 0, if you use the conditon that the line must be straight and have the same gradient on all faces, so other information must be specified.
I haven't worked out the equation yet but I'm pretty sure it's doable on a cone with the information specified (it is after all a conical helix), but it involves a little differentiation.
Last edited: Jun 1, 2004
9. Jun 1, 2004
### Gokul43201
Staff Emeritus
For the pyramid :
the length of the line will converge even if n diverges...so I don't think is so bad to assume a constant gradient.
This requirement of a constant gradient was not stated in the original post. It seemed reasonable to assume it in the case of the cylinder, with the slope being dy/dx. In other cases you might want to use dy/d(phi) or d(theta)/d(phi) to quantify the slope. So you want specify how you calculate slope and whether or not it needs to be constant or some other function. | 2018-04-21T10:07:14 | {
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https://math.stackexchange.com/questions/1099238/how-many-ways-can-you-put-8-red-6-green-and-7-blue-balls-in-4-indistinguishable | # How many ways can you put 8 red, 6 green and 7 blue balls in 4 indistinguishable bins?
1. Assume all balls with the same color are indistinguishable.
2. The order in which balls are put in a bin does not matter.
3. No bins are allowed to have the same distribution of balls!
For example, this configuration
{RRGGGB} {RRRRGBBBB} {RGB} {RGB}
is not ok, because the two last bins contains the same distribution of balls: one Red, one Green, one Blue.
Actualy I'm most intressted in the procedure for solving this problem where the number of balls, colors and bins are much larger numbers.
• Is this from some programming contest? – Aryabhata Jan 10 '15 at 18:43
• No, not what I know of. But I think the solution to this problem can be used to find all ways a number can be written as a product of k distinct integers, where k is the number of bins and balls corresponds to prime factors. – Penlect Jan 10 '15 at 19:04
• What exactly do you mean by indistinguishable bins? – Nicholas Pipitone Jan 10 '15 at 19:24
• I just mean that the bins are unlabeled. If two bins switched position, that should not be counted as a "new way". – Penlect Jan 10 '15 at 20:18
This problem is a straightforward application of the Polya Enumeration Theorem. Suppose we treat the case of $r$ red balls, $g$ green balls and $b$ blue balls and $n$ indistinguishable bins where no bins are left empty.
Recall the recurrence by Lovasz for the cycle index $Z(P_n)$ of the set operator $\mathfrak{P}_{=n}$ on $n$ slots, which is $$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l}) \quad\text{where}\quad Z(P_0) = 1.$$
We need to employ the set operator because the elements of a distribution are supposed to be unique.
We have for example, $$Z(P_3) = 1/6\,{a_{{1}}}^{3}-1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}$$ and $$Z(P_4) = 1/24\,{a_{{1}}}^{4}-1/4\,a_{{2}}{a_{{1}}}^{2} +1/3\,a_{{3}}a_{{1}}+1/8\,{a_{{2}}}^{2}-1/4\,a_{{4}}.$$
Applying PET and PIE it now follows almost by inspection that the desired count is given by the term $$\sum_{q=1}^n (-1)^{n-q} [R^r G^g B^b] Z(P_q)\left(\frac{1}{1-R}\frac{1}{1-G}\frac{1}{1-B}\right).$$
We need to employ PIE here because the substituted cycle index will include empty slots, which we are not counting in this problem.
As an example of what these substituted cycle indices look like consider $$Z(P_3)\left(\frac{1}{1-R}\frac{1}{1-G}\frac{1}{1-B}\right) \\ = 1/6\,{\frac {1}{ \left( 1-R \right) ^{3} \left( 1-G \right) ^{3} \left( 1-B \right) ^{3}}} \\-1/2\,{\frac {1}{ \left( -{R}^{2}+1 \right) \left( -{G}^{2}+1 \right) \left( -{B}^{2}+1 \right) \left( 1-R \right) \left( 1-G \right) \left( 1-B \right) }} \\+1/3\,{\frac {1}{ \left( -{R}^{3}+1 \right) \left( -{G }^{3}+1 \right) \left( -{B}^{3}+1 \right) }} .$$
It should be clear that coefficient extraction here is fast and efficient using the Newton binomial series which says that $$[Q^{\mu k}] \left(\frac{1}{1-Q^\mu}\right)^\nu = {k+\nu-1\choose \nu-1}.$$
The answer for eight red, six green and seven blue balls in four indistinguishable bins turns out to be $$60040.$$
The following Maple code implements two routines, q1 and q2. The first of these computes the value for $(r,g,b)$ and $n$ by brute force (enumerate all configurations) and can be used to verify the correctness of the PET formula for small values of the parameters. The second one uses the PET for instant computation of the desired coefficient.
Observe that we can compute values that are utterly out of reach of brute force enumeration, e.g. with ten balls of each color and five bins we obtain $$7098688.$$
With twenty balls of each color and six bins we get $$194589338219.$$
with(combinat);
pet_cycleind_set :=
proc(n)
option remember;
if n=0 then return 1; fi;
expand(1/n*
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
allparts :=
proc(val, size)
option remember;
local res, els, p, pp, q;
res := [];
for p in partition(val) do
if nops(p) <= size then
pp := [seq(q, q in p),
seq(0, q=1..(size-nops(p)))];
res := [op(res), pp];
fi;
od;
res;
end;
q1 :=
proc(RC, GC, BC, n)
option remember;
local p, q, pr, pg, pb, res,
sr, sg, sb, dist;
pr := [];
for p in allparts(RC, n) do
pr := [op(pr), seq(q, q in permute(p))];
od;
pg := [];
for p in allparts(GC, n) do
pg := [op(pg), seq(q, q in permute(p))];
od;
pb := [];
for p in allparts(BC, n) do
pb := [op(pb), seq(q, q in permute(p))];
od;
res := {};
for sr in pr do
for sg in pg do
for sb in pb do
dist :=
{seq(R^sr[pos]*G^sg[pos]*B^sb[pos],
pos=1..n)};
if nops(dist) = n and
not member(1, dist) then
res := res union {dist};
fi;
od;
od;
od;
nops(res);
end;
q2 :=
proc(RC, GC, BC, n)
option remember;
local els, vals, sind;
vals := [];
for els to n do
sind :=
pet_varinto_cind(1/(1-R)/(1-G)/(1-B),
pet_cycleind_set(els));
vals :=
[op(vals),
coeftayl(
coeftayl(
coeftayl(sind, R=0, RC),
G=0, GC),
B=0, BC)];
od; | 2019-04-20T06:30:26 | {
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Bhatnagar, G. Inverse Relations, Generalized Bibasic Series, and their U(n) Extensions. 4. In particular, we can determine the sum of binomial coefficients of a vertical column on Pascal's triangle to be the binomial coefficient that is one down and one to the right as illustrated in the following diagram: 1 à 8 (en) John Riordan (en), Combinatorial Identities, R. E. Krieger, 1979 (1 re éd. Other shorthands For the here most common binomial-coefficient binomial(r,c) I use for brevity bi(r,c) := binomial(r,c) ch(r,c) := binomial(r,c) // I'll delete this abbreviation while rewriting the articles Book Description. = \frac{n!}{k!(n-k)!} The factorial formula facilitates relating nearby binomial coefficients. Recall from the Binomial Coefficients page that the binomial coefficient $\binom{n}{k}$ for nonnegative integers $n$ and $k$ that satisfy $0 \leq k \leq n$ is defined to be: We will now look at some rather useful identities regarding the binomial coefficients. Binomial Coefficient Identity, Double Series, Floor Function. Les coefficients binomiaux sont importants en combinatoire, ... Combinatorial Identities, A Standardized Set of Tables Listing 500 Binomial Coefficient Summations, 1972 (lire en ligne) (en) Henry W. Gould, Tables of Combinatorial Identities, edited by J. Quaintance, 2010, vol. Identities. So I want to show you some surprising identities involving the binomial coefficient. Its simplest version reads (x+y)n = Xn k=0 n k xkyn−k whenever n is any non-negative integer, the numbers n k = n! Seeking a combinatorial proof for a binomial identity. Identities. Galaxy Clustering." Identities. and, with a little more work, Moreover, the following may be useful: Series involving binomial coefficients. Identities involving binomial coefficients. The Art of Proving Binomial Identities accomplishes two goals: (1) It provides a unified treatment of the binomial coefficients, and (2) Brings together much of the undergraduate mathematics curriculum via one theme (the binomial coefficients). Today we continue our battle against the binomial coefficient or to put it in less belligerent terms, we try to understand as much as possible about it. A combinatorial interpretation of this formula is as follows: when forming a subset of $k$ elements (from a set of size $n$), it is equivalent to consider the number of ways you can pick $k$ elements and the number of ways you can exclude $n-k$elements. Listing them all here would be superfluous, but we’ll prove two popular ones: = \frac{n^{\underline{k}}}{k!} Yes, we can, but that's not the point. }{(k - 1)! From MathWorld--A Wolfram Web Resource. Products and sum of cubes in Fibonacci. See pages that link to and include this page. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. The name Gaussian binomial coefficient stems from the fact [citation needed] that their evaluation at q = 1 is → = for all m and r. The analogs of Pascal identities for the Gaussian binomial coefficients are = (−) + (− −) and = (−) + − (− −). Binomial is a polynomial having only two terms in it. 2 Chapter 4 Binomial Coef Þcients 4.1 BINOMIAL COEFF IDENTITIES T a b le 4.1.1. In particular, we can determine the sum of binomial coefficients of a vertical column on Pascal's triangle to be the binomial coefficient that is one down and one to the right as illustrated in the following diagram: 1, 181-186, 1971. ((n - (n-k))!} Wikidot.com Terms of Service - what you can, what you should not etc. Let's arrange the binomial coefficients $${n \choose k}$$ into a triangle like follows: There are lots of patterns hidden away in the triangle, enough to fill a reasonably sized book. Such rela-tions are examples of binomial identities, and can often be used to simplify expressions involving several binomial coe cients. = \frac{n}{k} \cdot \frac{(n - 1)! Here are just a few of the most obvious ones: The entries on the border of the triangle are all 1. http://www.combinatorics.org/Volume_3/Abstracts/v3i2r16.html, https://mathworld.wolfram.com/BinomialIdentity.html. Electronic J. Combinatorics 3, No. Let m = 0. The binomial coefficient (n; k) is the number of ways of picking k unordered outcomes from n possibilities, also known as a combination or combinatorial number. 37-49, 1993. Mathematica says it is true, but how to show it? New York: Wiley, p. 18, 1979. Identities involving binomial coefficients. The right side counts the same parameter, because there are ways of choosing … Recursion for binomial coefficients A recursion involves solving a problem in terms of smaller instances of the same type of problem. True . For instance, if k is a positive integer and n is arbitrary, then (5) and, with a little more work, Moreover, the following may be useful: For constant n, we have the following recurrence: Series involving binomial coefficients. Binomial Coe cients and Generating Functions ITT9131 Konkreetne Matemaatika Chapter Five Basic Identities Basic Practice ricksT of the radeT Generating Functions Hypergeometric Functions Hypergeometric ransfoTrmations Partial Hypergeometric Sums. Proposition 4.1 (Complementation Rule). The formula is obtained from using x = 1. Ohio State University, p. 61, 1995. Here we will learn its definition, examples, formulas, So I want to show you some surprising identities involving the binomial coefficient. Contents 1 Binomial coe cients 2 Generating Functions Intermezzo: Analytic functions Operations on Generating Functions Building … ∼: asymptotic equality, (m n): binomial coefficient, π: the ratio of the circumference of a circle to its diameter and n: nonnegative integer Referenced by: §26.5(iv) (1 + x−1)n.It is reflected in the symmetry of Pascal's triangle. This interpretation of binomial coefficients is related to the binomial distribution of probability theory, implemented via BinomialDistribution. 8. Append content without editing the whole page source. Riordan, J. Combinatorial Listing them all here would be superfluous, but we’ll prove two popular ones: Find out what you can do. 1968, John Wiley & Sons) 1996. http://www.combinatorics.org/Volume_3/Abstracts/v3i2r16.html. 1972, Item 42). Comtet, L. Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. \cdot (n - k) \cdot (n - k - 1) \cdot ... \cdot 2 \cdot 1} \\ = \frac{n}{k} \cdot \frac{(n-1) \cdot (n - 2 \cdot) ... \cdot (n - k + 1)}{(k-1)!} Every regular multiplicative identity corresponds to an RMI-diagram. En mathématiques, les coefficients binomiaux, ... Combinatorial Identities, A Standardized Set of Tables Listing 500 Binomial Coefficient Summations, 1972 (lire en ligne) (en) Henry W. Gould, Tables of Combinatorial Identities, edited by J. Quaintance, 2010, vol. It's hard to pick one of its 250 pages at random and not find at least one binomial coefficient identity there. We provide some examples below. Corollary 4. 30 and 73), and. Recursion for binomial coefficients A recursion involves solving a problem in terms of smaller instances of the same type of problem. Section 4.1 Binomial Coeff Identities 3. Explore anything with the first computational knowledge engine. We will prove Theorem 2 in two different ways. Contents 1 Binomial coe cients 2 Generating Functions Intermezzo: Analytic functions Operations on Generating Functions Building … MULTIPLICATIVE IDENTITIES FOR BINOMIAL COEFFICIENTS As we have seen, the proof of (10) is straightforward. k!(n−k)! Since the binomial coecients are dened in terms of counting, identities involv- ing these coecients often lend themselves to combinatorial proofs. Binomial Coefficients (3/3): Binomial Identities and Combinatorial Proof - Duration: 8:30. Umbral Calculus. 1994, p. 203). The following relations all hold. Discrete Math. Binomial Coefficients and Identities (1) True/false practice: (a) If we are given a complicated expression involving binomial coe cients, factorials, powers, and fractions that we can interpret as the solution to a counting problem, then we know that that expression is an integer. Identities involving binomial coefficients. The extended binomial coefficient identities in Table 2 hold true. ∼: asymptotic equality, (m n): binomial coefficient, π: the ratio of the circumference of a circle to its diameter and n: nonnegative integer Referenced by: §26.5(iv) This notion of symmetry between q-binomial numbers illustrates identities similar to those found when working with binomial coe cients. The expression formed with monomials, binomials, or polynomials is called an algebraic expression. \displaystyle{\binom{n}{k} = \frac{n^{\underline{k}}}{k! Corollary 1.4. We present some identities that have combinatorial proofs. share | cite | improve this question | follow | edited May 19 at 15:42. Binomial Expansion. The binomial coefficients satisfy the identities: (5) (6) (7) Sums of powers include (8) (9) (10) (the Binomial Theorem), and (11) where is a Hypergeometric Function (Abramowitz and Stegun 1972, p. 555; Graham et al. In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written $\tbinom{n}{k}. Identities involving binomial coefficients. The binomial coefficient is the multinomial coefficient (n; k, n-k).$ It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and it is given by the formula Saslaw, W. C. "Some Properties of a Statistical Distribution Function for Weisstein, Eric W. "Binomial Identity." Astrophys. Every regular multiplicative identity corresponds to an RMI-diagram. enl. Unlimited random practice problems and answers with built-in Step-by-step solutions. In general, a binomial identity is a formula expressing products of factors as a sum over terms, each including a binomial coefficient (n; k). Each of these is an example of a binomial identity: an identity (i.e., equation) involving binomial coefficients. We have, for example, for The combinatorial proof goes as follows: the left side counts the number of ways of selecting a subset of of at least q elements, and marking q elements among those selected. For Nonnegative Integers and with , (12) Taking gives (13) Another identity is (14) (Beeler et al. The first proof will be a purely algebraic one while the second proof will use combinatorial reasoning. The binomial coefficients satisfy the identities: (5) (6) (7) Sums of powers include (8) (9) (10) (the Binomial Theorem), and (11) where is a Hypergeometric Function (Abramowitz and Stegun 1972, p. 555; Graham et al. C. F. Gauss (1812) also widely used binomials in his mathematical research, but the modern binomial symbol was introduced by A. von Ettinghausen (1826); later Förstemann (1835) gave the combinatorial interpretation of the binomial coefficients. Something does not work as expected? For all real numbers a and b, I;]= l.“bl* Proposition 4.2 (Iterative Rule). Proving Binomial Identities (2 of 6: Proving harder identities by substitution and using Theorem) For all n 0 we have h n 0 i = hn n i (4) Our rst proof of Corollary 1.4. Roman coefficients always equal integers or the reciprocals of integers. Math. Binomial identities, binomial coefficients, and binomial theorem (from Wikipedia, the free encyclopedia) In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. identities (Riordan 1979, Roman 1984), some of which include, (Abel 1826, Riordan 1979, p. 18; Roman 1984, pp. For instance, if k is a positive integer and n is arbitrary, then. En mathématiques, et plus précisément en algèbre, le théorème binomial d'Abel, ... Combinatorial Identities, A Standardized Set of Tables Listing 500 Binomial Coefficient Summations, 1972 (lire en ligne), p. 15, (1.117), (1.118) et (1.119) (en) Henry W. Gould et J. Quaintance (ed. Each of these is an example of a binomial identity: an identity (i.e., equation) involving binomial coefficients. Can we find a nice expression for the sum? ed. . Prof. Tesler Binomial Coefficient Identities Math 184A / Winter 2017 9 / 36. (13). 1, 159-160, 1826. The factorial formula facilitates relating nearby binomial coefficients. General Wikidot.com documentation and help section. The factorial formula facilitates relating nearby binomial coefficients. Still it's a … \binom {n-1}{k} - \binom{n-1}{k-1} = \frac{n-2k}{n} \binom{n}{k}. The #1 tool for creating Demonstrations and anything technical. Here we use the multiplication principle, namely that if choosing an object is equivalent to making a series of choices and the number of options at each step does not depend on the previous choices, then the number of objects is simply the product of the number of options at each step.. 2.2 Binomial coefficients. 4.1 Binomial Coef Þ cient Identities 4.2 Binomial In ver sion Operation 4.3 Applications to Statistics 4.4 The Catalan Recurrence 1. = \binom{n}{k} \quad \blacksquare \end{align}, \begin{align} \quad \binom{n}{k} = \frac{n!}{k! When studying the binomial coe cients, we proved a powerful theorem called the Binomial The-orem. The converse is slightly more difficult. Retrouvez The Art of Proving Binomial Identities et des millions de livres en stock sur Amazon.fr. this identity for all in a field of field characteristic Proposition 4.1 (Complementation Rule). = \frac{n}{k} \cdot \frac{(n - 1) \cdot (n - 2) \cdot ... \cdot 2 \cdot 1}{(k - 1)! Multinomial returns the multinomial coefficient (n; n 1, …, n k) of given numbers n 1, …, n k summing to , where . Combinatorial identities involving binomial coefficients. Properties of Roman coefficients Several binomial coefficients identities extend to Roman coefficients. New York: Academic Press, pp. The right side counts the same parameter, because there are ways of choosing … in Œuvres Complètes, 2nd ed., Vol. Achetez neuf ou d'occasion. A. L. Crelle (1831) used a symbol that notates the generalized factorial . Binomial Coefficient Identities. Recollect that and rewrite the required identity as In this form it admits a simple interpretation. Recall thatn^{\underline{k}}$represents a falling factorial. Properties of Roman coefficients Several binomial coefficients identities extend to Roman coefficients. There exist very many summation identities involving binomial coefficients (a whole chapter of Concrete Mathematics is devoted to just the basic techniques). ed. theorem, for . J. En mathématiques, les coefficients binomiaux, ... Combinatorial Identities, A Standardized Set of Tables Listing 500 Binomial Coefficient Summations, 1972 (lire en ligne) (en) Henry W. Gould, Tables of Combinatorial Identities, edited by J. Quaintance, 2010, vol. Book Description. In Algebra, binomial theorem defines the algebraic expansion of the term (x + y) n. It defines power in the form of ax b y c. The exponents b and c are non-negative distinct integers and b+c = n and the coefficient ‘a’ of each term is a positive integer and the value depends on ‘n’ and ‘b’. We have, for example, for The combinatorial proof goes as follows: the left side counts the number of ways of selecting a subset of of at least q elements, and marking q elements among those selected. For example, The 2-subsets of {1,2,3,4} … Can we find a nice expression for the sum? The difficulty here is that we cannot simply copy down the lower indices in the given identity and interpret them as coordinates of points in an RMI-diagram. The binomial coefficient has associated with it a mountain of identities, theorems, and equalities. J. reine angew. Bibliographie (en) Henry W. Gould , Combinatorial Identities, A Standardized Set of Tables Listing 500 Binomial Coefficient Summations, 1972 (lire en ligne) (en) Henry W. Gould, Tables of Combinatorial Identities, edited by J. Quaintance, 2010, vol. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. Yes, we can, but that's not the point. combinatorics summation binomial-coefficients. Choisir vos préférences en matière de cookies. 1. pp. 102-103, 1994, p. 203). View wiki source for this page without editing. For all real numbers a and b, I;]= l.“bl* Proposition 4.2 (Iterative Rule). Change the name (also URL address, possibly the category) of the page. For instance, if k is a positive integer and n is arbitrary, then Some identities satisfied by the binomial coefficients, and the idea behind combinatorial proofs of them. = \binom{n - 1}{k - 1}$, Creative Commons Attribution-ShareAlike 3.0 License. Another important application is in the combinatorial identity known as Pascal's rule, which relates the binomial coefficient with shifted arguments according to . The Art of Proving Binomial Identities accomplishes two goals: (1) It provides a unified treatment of the binomial coefficients, and (2) Brings together much of the undergraduate mathematics curriculum via one theme (the binomial coefficients). \end{align}, \begin{align} \quad \binom{n}{k} = \frac{n!}{k!(n-k)!} Click here to toggle editing of individual sections of the page (if possible). We wish to prove that they hold for all values of $$n$$ and $$k\text{. 29-30 and 72-75, 1984. Theorem 2 establishes an important relationship for numbers on Pascal's triangle. Some of the most basic ones are the following. Xander Henderson ♦ 20.8k 11 11 gold badges 47 47 silver badges 71 71 bronze badges. View and manage file attachments for this page. Roman (1984, p. 26) defines "the" binomial identity as the equation. Reprinted Click here to edit contents of this page. View/set parent page (used for creating breadcrumbs and structured layout). enl. \cdot (n - k)!} The factorial formula facilitates relating nearby binomial coefficients. 1 à 8 (en) John Riordan (en), Combinatorial Identities, R. E. Krieger, 1979 (1 re éd. Naturally, we might be interested only in subsets of a certain size or cardinality. Strehl, V. "Binomial Identities--Combinatorial and Algorithmic Aspects." }$$ These proofs can be done in many ways. Abel (1826) gave a host of such MULTIPLICATIVE IDENTITIES FOR BINOMIAL COEFFICIENTS As we have seen, the proof of (10) is straightforward. In general, a binomial identity is a formula expressing products of factors as a sum over terms, each including a binomial coefficient . Discr. Strehl, V. "Binomial Sums and Identities." W. Volante W. Volante. Michael Barrus 17,518 views. Subsection 5.3.2 Combinatorial Proofs. For instance, if k is a positive integer and n is arbitrary, then. The difficulty here is that we cannot simply copy down the lower indices in the given identity and interpret them as coordinates of points in an RMI-diagram. We wish to prove that they hold for all values of $$n$$ and $$k\text{. Theorem 2.1. Our goal is to establish these identities. \binom{n}{k} = \frac{n+1-k}{k} \binom{n}{k-1}. I feel I exhausted all identities/properties of binomials without success. = \frac{n}{k} \cdot \frac{(n - 1)^{\underline{k-1}}}{(k - 1)!} Definition. On the other hand, if the number of men in a group of grownups is then the number of women is , and all possible variants are expressed by the left hand side of the identity. Unfortunately, the identities are not always organized in a way that makes it easy to find what you are looking for. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. To prove (i) and (v), apply the ratio test and use formula (2) above to show that whenever is not a nonnegative integer, the radius of convergence is exactly 1. Prof. Tesler Binomial Coefficient Identities Math 184A / Winter 2017 9 / 36. 1881. So for example, what do you think? \binom{n}{h}\binom{n-h}{k}=\binom{n}{k}\binom{n-k}{h}. These proofs are usually preferable to analytic or algebraic approaches, because instead of just verifying that some equality is true, they provide some insight into why it is true. Binomial identities, binomial coefficients, and binomial theorem (from Wikipedia, the free encyclopedia) In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. Iff the sequence satisfies §4.1.5 in The (The q-Binomial Theorem) For all n 1 we have Yn j=1 Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. 4 Chapter 4 Binomial Coef Þcients Combinatorial vs. Alg ebraic Pr oofs Symmetr y. Once again we will prove Theorem 3 in two different ways like before. Today we continue our battle against the binomial coefficient or to put it in less belligerent terms, we try to understand as much as possible about it. If you want to discuss contents of this page - this is the easiest way to do it. \(\binom{n}{k}$$ is the coefficient of $$x^{n-k}y^k$$ in the expansion of $$(x+y)^n$$ $$\binom{n}{k}$$ is the number subsets of size $$k$$ from a set of size $$n$$ $$\dots$$ there are many more ways of viewing binomial coefficients. Proof. The number of possibilities is , the right hand side of the identity. The above formula for the generalized binomial coefficient can be rewritten as ) = ∏ = (+ −). Foata, D. "Enumerating -Trees." In Maths, you will come across many topics related to this concept. So for example, what do you think? The binomial coefficient has associated with it a mountain of identities, theorems, and equalities. The binomial coefficients arise in a variety of areas of mathematics: combinatorics, of course, but also basic algebra (binomial … Check out how this page has evolved in the past. (13). sequence known as a binomial-type sequence. The converse is slightly more difficult. 1972, Item 42). 8:30. Binomial coefficients are generalized by multinomial coefficients. 1 à 8 (en) John Riordan (en), Combinatorial Identities, R. E. Krieger, 1979 (1 re éd. x. x x in the expansion of. and, with a little more work, Moreover, the following may be useful: Series involving binomial coefficients. For Nonnegative Integers and with , (12) Taking gives (13) Another identity is (14) (Beeler et al. Combinatorial identities involving binomial coefficients. \cdot (n - k) \cdot (n - k - 1) \cdot ... \cdot 2 \cdot 1} = \frac{n \cdot (n - 1) \cdot ... \cdot (n - k + 1)}{k!} Ekhad, S. B. and Majewicz, J. E. "A Short WZ-Style Proof of Abel's Identity." The factorial formula facilitates relating nearby binomial coefficients. Dordrecht, Binomial Coefficients and Identities Terminology: The number r n is also called a binomial coefficient because they occur as coefficients in the expansion of powers of binomial expressions such as (a b)n. Example: Expand (x+y)3 Theorem (The Binomial Theorem) Let x … Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. 0, then is an associated Our goal is to establish these identities. 2, R16, 1, Binomial Coe cients and Generating Functions ITT9131 Konkreetne Matemaatika Chapter Five Basic Identities Basic Practice ricksT of the radeT Generating Functions Hypergeometric Functions Hypergeometric ransfoTrmations Partial Hypergeometric Sums. \end{align}, \begin{align} \quad \binom{n}{k} = \frac{n}{k} \cdot \binom{n-1}{k-1} \quad \blacksquare \end{align}, \begin{align} \quad \binom{n}{k} \cdot k = n \cdot \binom{n-1}{k-1} \\ \quad \binom{n}{k} = \frac{n}{k} \cdot \binom{n-1}{k-1} \quad \blacksquare \end{align}, Unless otherwise stated, the content of this page is licensed under. For constant n, we have the following recurrence: 1. Added: Another useful reference is John Riordan's Combinatorial Identities. Below is a construction of the first 11 rows of Pascal's triangle. Binomial Identities While the Binomial Theorem is an algebraic statement, by substituting appropriate values for x and y, we obtain relations involving the binomial coe cients. Knowledge-based programming for everyone. Theorem 2 establishes an important relationship for numbers on Pascal's triangle. 341, 588-598, 1989. Recall from the Binomial Coefficients page that the binomial coefficient for nonnegative integers and that satisfy is defined to be: (1) We will now look at some rather useful identities regarding the binomial coefficients… Watch headings for an "edit" link when available. Ph.D. thesis. MathOverflow . (n - k)!} ( 1 + x) n: (1+x)^n: (1+x)n: ( 1 + x) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n, (1+x)^n = n_ {c_ {0}} + n_ {c_ {1}} x + n_ {c_ {2}} x^2 + \cdots + n_ {c_ {n}} x^n, (1+x)n = nc0. 4. We present some identities that have combinatorial proofs. Practice online or make a printable study sheet. The factorial definition lets one relate nearby binomia… }\) These proofs can be done in many ways. The formula is obtained from using x = 1. For other uses, see NCK (disambiguation). Moreover, the following may be useful: 1. }}$,$\displaystyle{\binom{n}{k} = \binom{n}{n-k}}$,$\displaystyle{\binom{n}{k} = \frac{n}{k} \cdot \binom{n-1}{k-1}}$,$\frac{(n - 1)^{\underline{k-1}}}{(k - 1)!} Math. More resources available at www.misterwootube.com. \begin{align} \quad \binom{n}{k} = \frac{n!}{k!(n-k)!} Walk through homework problems step-by-step from beginning to end. Netherlands: Reidel, p. 128, 1974. Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. For instance, if k is a positive integer and nis arbitrary, then and, with a little more work, 1. Abel, N. H. "Beweis eines Ausdrucks, von welchem die Binomial-Formel ein einzelner Fall ist." Roman, S. "The Abel Polynomials." Maple Technical Newsletter 10, It is powerful because it allows us to easily nd many more binomial coe cient identities. Examples open all close all. The symbols _nC_k and (n; k) are used to denote a binomial coefficient, and are sometimes read as "n choose k." (n; k) therefore gives the number of k-subsets possible out of a set of n distinct items. are the binomial coefficients, and n! 6. Its simplest version reads (x+y)n= Xn k=0 n k xkyn−k Roman coefficients always equal integers or the reciprocals of integers. Here we are going to nd the q-analog of the Binomial Theorem, aptly named the q-Binomial Theorem. asked Apr 29 at 16:27. ), Tables of Combinatorial Identities, vol. 136, 309-346, 1994. Binomial coefficients are the ones that appear as the coefficient of powers of. 1 à 8 (en) John Riordan , Combinatorial Identities, R. E. Krieger, 1979 (1 re éd. Hints help you try the next step on your own. Join the initiative for modernizing math education. \quad \blacksquare \end{align}, \begin{align} \quad \binom{n}{n-k} = \frac{n!}{(n-k)! The prototypical example is the binomial The binomial coefficients arise in a variety of areas of mathematics: combinatorics, of course, but also basic algebra (binomial … It is required to select an -members committee out of a group of men and women. The factorial formula facilitates relating nearby binomial coefficients. Notify administrators if there is objectionable content in this page. For instance, we know that n 0 = n n. In fact, this identity transfers to the q-analog of the binomial coe cients, which leads us to our next corollary. Identities with binomials,Bernoulli- and other numbertheoretical numbers Mathematical Miniatures 1.1.3. Proof. Binomial Coefficient – Harmonic Sum Identities Associated to Supercongruences; Euler's Pentagonal Number Theorem Implies the Jacobi Triple Product Identity; On Directions Determined by Subsets of Vector Spaces over Finite Fields; A Remark on a Paper of Luca and Walsh ; On the Tennis Ball Problem; On the Conditioned Binomial Coefficients; Convolution and Reciprocity Formulas for … = \frac{n \cdot (n - 1) \cdot ... \cdot 2 \cdot 1}{k! For instance, if k is a positive integer and n is arbitrary, then (5) and, with a little more work, Moreover, the following may be useful: For constant n, we have the following recurrence: Series involving binomial coefficients. "nCk" redirects here. https://mathworld.wolfram.com/BinomialIdentity.html. | 2022-08-08T01:49:04 | {
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https://pabloaguiar.me/post/multiples-of-3-and-5/ | # Multiples of 3 and 5
This is the first post of a series entitled Project Euler In Under 0.1ms
### Project Euler #1
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
### Common (and slow) solution
The most common, naive solution involves a simple loop from 0 to x, checking whether n is multiple of 3 or 5 and then adding n to the sum. Pretty straightforward:
def multiples_three_five(x):
s = 0
for n in range(x):
if n % 3 == 0 or n % 5 == 0:
s += n
return s
# Time spent: 497.477ms
### Stay put as a Pythonista
Things get worse if we’re not true Pythonistas and dare to use filter. It takes almost twice the time to complete compared to the previous one:
def multiples_three_five(x):
return sum(filter(lambda n: n % 3 == 0 or n % 5 == 0, range(x)))
# Time spent: 792.545ms
### A slightly better approach
Instead of looping up to x, we could loop until the maximum multiple of 3 or 5 that is less than x. As we want the maximum one, it should be multiple of 3, since 3 < 5. Just remember not to re-add “duplicate numbers”, those that are multiples of both 3 and 5:
def multiples_three_five(x):
s = 0
for n in range(1, int(x/3) + 1):
p = n * 3
if p < x:
s += p
p = n * 5
if p % 3 and p < x:
s += p
return s
# Time spent: 289.934ms
### A better approach
But, wait! We could just loop from 0 to x, 3 by 3, summing all values, then do the same 5 by 5 – but not re-adding dupes:
def multiples_three_five(x):
s = 0
for n in range(3, x, 3):
s += n
for n in range(5, x, 5):
if n % 3:
s += n
return s
# Time spent: 152.503ms
### Same approach improved a bit
We’re good re-adding duplicate numbers as long as we subtract them later:
def multiples_of_three_and_five(x):
s = 0
for n in range(3, x, 3):
s += n
for n in range(5, x, 5):
s += n
for n in range(15, x, 15):
s -= n
return s
# Time spent: 129.046ms
### Same approach improved some more
We can gain some precious milliseconds benefitting from Python’s fast addition. The following is around three times faster than the previous one:
def multiples_of_three_and_five(x):
return sum(range(3, x, 3)) + sum(range(5, x, 5)) - sum(range(15, x, 15))
# Time spent: 38.212ms
WOW! From 497.477ms down to 38.212ms! But even so, this is far from being fast. We’ve just saved hundreds of milliseconds. We need a completely different approach.
### Going under 0.1ms
Look at what we’ve got so far. Each of those sums is the sum of a series! And, as young Gauss brilliantly found out, we can write them as mathematical equations. Kudos, Johann Carl Friedrich Gauss.
Take S, for instance, the sum of all natural numbers from 1 to n. We can define it as:
$$^{(eq.\ 1)}\ \ \ S=\sum_{n=1}^{x}{n}=1+2+3+4+…+x = {x(x+1)\over2}$$
Analogously, we can state the following:
$$^{(eq.\ 2)}\ \ \ 5S=5\sum_{n=1}^{x}{n}=5+10+15+20+…+5x = 5{x(x+1)\over2}$$
Considering the above, to sum all natural numbers below 25 that are multiples of 5, we could just sum all of them up to 20, the 4th element (i.e. floor((25 - 1) / 5) = 4). Thus, defining x' = x - 1, we can use the following to obtain the sum of all multiples of 5 below x:
$$^{(eq.\ 3)}\ \ \ 5S'=5\sum_{n=1}^{\left\lfloor\frac{x'}{5}\right\rfloor}n=5\ \frac{{\left\lfloor\frac{x'}{5}\right\rfloor}({\left\lfloor\ \frac{x'}{5}\right\rfloor}+1)}{2}$$
Then we can take:
$$^{(eq.\ 4)}\ \ \left\lfloor\frac{x'}{5}\right\rfloor=\frac{x'-(x'\mod\ 5)}{5}$$
Replace it into equation 3 to end up with:
$$^{(eq.\ 5)}\ \ \ 5S'=\frac{\{x'-[(x')\mod\ 5]\}\{x'-[(x')\mod\ 5]+5\}}{10}$$
And solve the problem really fast:
def multiples_of_three_and_five(x):
x -= 1
s = ((x - x % 3) * (x - x % 3 + 3)) // 6
s += ((x - x % 5) * (x - x % 5 + 5)) // 10
s -= ((x - x % 15) * (x - x % 15 + 15)) // 30
return s
# Time spent: 0.019ms | 2019-08-23T23:59:40 | {
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https://math.stackexchange.com/questions/1179019/prove-with-induction-that-11-divides-102n-1-for-all-natural-numbers/1179145 | # Prove with induction that $11$ divides $10^{2n}-1$ for all natural numbers.
$$10^{2(k+1)}-1 = 10^{2k+2}-1=10^{2k}\cdot10^{2}-1$$
I feel like there's something in that last part that should make it work, but I can't grasp it. Am I missing something obvious? Am I going in the completely wrong direction? Any help would be appreciated.
• $10^{2k+2}-1=10^2(10^{2k}-1)+10^2-1$. Mar 7 '15 at 1:46
• @AndréNicolas: I have no idea how I would've thought of that. Thank you very much.
– Bob
Mar 7 '15 at 2:10
• You are welcome. If one works with induction on exponentials, tricks of this general kind become familiar. Mar 7 '15 at 2:16
• In my answer I show how you can discover the proof of @André and oujdid, by a mechanical method. Mar 7 '15 at 5:05
In a comment you remark that you have no idea how Andre thought of the inductive proof. Below we will show that it is not a magic trick but, rather, a special case of a simple general method that works for inductions of this type. Below we show extremely explicitly how the inductive steps in the proofs by Andre and oujdid are both precisely special cases of proofs of the Congruence Product Rule. You need not be familiar with congruences to understand this since we also write it in equivalent $$\rm\color{#0a0}{divisibility\ form}$$. Recall that $$\rm\ n\mid k\,$$ means $$\rm\,n\,$$ divides $$\rm\,k.\,$$
\begin{align} {\bf Claim}\rm\qquad\ \ 10^2\!&\rm\equiv 1, \, 10^{2k}\!\equiv 1\ \, \Rightarrow\,\ 10^{2(k+1)}\!\equiv 1\, \pmod{\!11}\\[.3em] {\bf Lemma}\rm\qquad\ A&\rm\equiv a,\ \, B\equiv b\quad \Rightarrow\quad\,\ AB\equiv ab\, \pmod{\!n}\ \ \ \ [\rm\color{#c00}Congruence \ \rm\color{#c00}Product\ \rm\color{#c00}Rule]\\[.3em] \rm\quad\ \ i.e.\quad\! n\mid A&\rm -a,\ \ B\,-\,b\ \Rightarrow\,\ n\mid \,AB\,-\,ab\qquad\qquad\ \ \ \ \ \ \:\! [\rm\color{#c00}{CPR}\ in\ \color{#0a0}{divisibility\ form}]\\[.5em] {\bf Proof}\quad\ \ \ \rm n\mid A&\rm -a,\ \ B\,-\,b\,\Rightarrow\,\ n\mid \ A\ (\ B\,-\,b)+\,(\ A-a)b\, =\ \,A\, B\,-\,a\,b\\[.3em] \rm\quad\ e.g.\,\ 11\mid 10^2&\rm -\!1,\ 10^{2k}\!-\!1\,\Rightarrow 11\mid\! 10^2(10^{2k}\!-\!1)+(10^2\!-\!1)1\,= 10^{2(k+1)}\!-1\ \ \ [Andre]\\[.5em] {\bf Proof}\quad\ \ \ \rm n\mid A&\rm -a,\ \ B\,-\,b\,\Rightarrow\,\ n\mid (\, A-a)\,\ B\ \ +a\ (\ B-b)\, =\ \,A\, B\,-\,a\,b\\[.3em] \rm\quad\ e.g.\,\ 11\mid 10^2&\rm -\!1,\ 10^{2k}\!-\!1\,\Rightarrow 11\mid(10^2\!-\!1)10^{2k}\!+\! 1(10^{2k}\!\!-\!1) = 10^{2(k+1)}\!-1\ \ \ [oujdid] \end{align}
So the inductive steps of Andre and oujdid - which appear to have been pulled out of hat like magic - are actually nothing but special cases of the proofs of $$\,\small \rm\color{#c00}{CPR} =$$ Congruence Product Rule. Once we know this rule, there is no need to repeat the entire proof inline every time that we employ it. Rather, we can simply invoke the rule as a Lemma (in $$\rm\color{#0a0}{divisibility\ form}$$ if congruences are not yet known). That done, the inductive step has vivid arithmetical structure, being simply the computation of a product $$\, 10^2\!\cdot 10^{2k}\equiv 10^{2(k+1)}.\,$$ No longer is the innate arithmetical structure of the induction obfuscated by the details of the proof of the Product Rule - since the proof has been encapsulated into a Lemma for convenient reuse.
In much the same way, congruences often allow us to impart intuitive arithmetical structure onto inductive proofs - allowing us to reuse our well-honed grade-school skills manipulating arithmetical equations (vs. more complex divisibility relations). Often introduction of congruence language will serve to drastically simplify the induction, e.g. reducing it to a trivial induction such as $$\, 1^n\equiv 1,\,$$ or $$\,(-1)^{2n}\equiv 1,\,$$ which is the essence above, i.e. $$\,{\rm mod}\ 11\!:\,\ 10\equiv -1\,\Rightarrow\,10^{2n}\equiv (-1)^{2n}\equiv 1\,$$ by the Congruence Power Rule, whose proof is an immediate inductive extension of the Product Rule exactly as in the proof of above OP special case. In number theory we typically make such inferences by applying the Power Rule (or the more general Polynomial Congruence Rule, i.e. $$\,a\equiv b\,\Rightarrow\, P(a)\equiv P(b)\,$$ for any polynomial $$P(x)$$ with integer coefficients).
• And, as an instance, note that the usual proof of the calculus Product Rule uses the same device. Mar 7 '15 at 5:18
• @André Indeed, I write about that here (too briefly) Mar 7 '15 at 5:24
Hint did you observe that $10^2=99+1$ so you can write: $$10^{2k+2}-1=99.10^{2k}+10^{2k}-1$$
and apply the iduction hypothesis
• If you observe that, it's finished with congruences, without any induction. Mar 7 '15 at 1:49
• Did you mean $10^{2k+2}-1$ as the left side? Mar 7 '15 at 1:51
• it's really easy so using induction or congruence yields the result, but he pointed that he is trying to prove it by induction so i'm supposed to follow his work not to ignore what he did
– user217773
Mar 7 '15 at 2:07
• @Bernard In my answer I show that this proof is a special case of the proof of the Congruence Product Rule. Mar 7 '15 at 5:38
One really doesn’t need induction for this:
\begin{align*} 10^{2n}-1&=\underbrace{999999\ldots999999}_{2n\text{ nines}}\\ &=\underbrace{99\,99\,99\,\ldots\,99\,99\,99}_{n\text{ copies of }99}\\ &=11\cdot\underbrace{09\,09\,09\,\ldots\,09\,09\,09}_{n\text{ copies of }09}\\ &=11\cdot 9\underbrace{090909\ldots090909}_{n-1\text{ copies of }09}\;. \end{align*}\tag{1}
However, if you’re required to use induction you can use the calculation above to work out what should happen in the induction step. It’s pretty clear from $(1)$ that $10^{2n+2}-1$ is going to be $11$ times the number formed by a $9$ followed by $n$ copies of $09$:
\begin{align*} 10^{2n+2}-1&=11\cdot 9\underbrace{090909\ldots090909}_{n\text{ copies of }09}\\ &=11\cdot 9\underbrace{090909\ldots090909}_{n-1\text{ copies of }09}09\\ &=11\left(100\cdot9\underbrace{090909\ldots090909}_{n-1\text{ copies of }09}+9\right)\\ &=100\cdot11\cdot9\underbrace{090909\ldots090909}_{n-1\text{ copies of }09}+11\cdot9\\ &=100\left(10^{2n}-1\right)+11\cdot 9\;, \end{align*}
so if $10^{2n}-1$ is a multiple of $11$, so is $10^{2n+2}$.
Now you can clean up the induction step by eliminating all of the explicit decimal representations and going directly (or nearly so) from $10^{2n+2}$ to $100\left(10^{2n}-1\right)+11\cdot9$; the digit-tracking can then be regarded as simply a way to figure out what’s going on.
Just in case, a solution without induction (or at least any induction is buried somewhere):
We have $[10]_{11} = [-1]_{11}$, and so $[10^{2n}]_{11} = [(-1)^{2n}]_{11} = [1]_{11}$
Then $[10^{2n}-1]_{11} = 0$ and so $11 \mid 10^{2n}-1$.
Notation: $[a]_b = a+b \mathbb{Z}$.
• Simpler solution inspired by Bill's answer. Mar 7 '15 at 5:36
First, show that this is true for $n=1$:
$10^{2}-1=11\cdot9$
Second, assume that this is true for $n$:
$10^{2n}-1=11k$
Third, prove that this is true for $n+1$:
$10^{2(n+1)}-1=$
$10^{2n+2}-1=$
$100(10^{2n})-1=$
$100(\color{red}{10^{2n}-1})+99=$
$100(\color{red}{11k})+99=$
$11(100k+9)=$
Please note that the assumption is used only in the part marked red.
For $n = 1$, $10^{2n}-1 = 99$, which is divisible by $11$. Let us assume that the given statement is true for $n = m$, i.e. $10^{2m}-1$ is divisible by $11$. Now, $10^{2m+2}-1 = 10^{2m}\times 100-1 = 10^{2m}\times (11\times 9 + 1) - 1 = (10^{2m}\times 11\times 9) + 10^{2m}-1$. But $10^{2m}-1$ is divisible by $11$. So the result follows.
A variant for the induction: suppose the assertion true for all $k\le n$ and let's prove it for $n+1$. We have: $$10^{2(n+1)}-1=(10^{n+1}-1)(10^{n+1}+1)$$ If $n$ is even, $n+1$ is odd, and $10^{n+1}+1\,$ is divisible by $10+1$.
If $n$ is odd, write $n=2l-1$; Then $$10^{2(n+1)}-1=10^{4l}-1=(10^{2l}-1)(10^{2l}+1)$$ and $10^{2l}-1$ is divisible by $11$ by the inductive hypothesis.
Ok this is not a strict inductive proof. More like an algorithmic proof or something.
Conjugate rule $a^2 - b^2 = (a+b)(a-b)$, $a = 10^n$, $b = 1$,
1) if $n$ even continue with conjugate rule on the factor $a-b$ which is on the same form,
2) once $n$ is odd , write term $a+b = 10^{2k+1}+1$ and take mod 11, $(-1)^{2k+1} + 1 = 0 (mod 11)$ for all integers $k$. | 2021-09-17T08:08:04 | {
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https://brilliant.org/discussions/thread/simple-numbr-puzzle/ | # Simple number puzzle
You have to solve the following:
$$0$$ _ $$0$$ _ $$0 = 6$$
$$1$$ _ $$1$$ _ $$1 = 6$$
$$2$$ _ $$2$$ _ $$2 = 6$$
$$3$$ _ $$3$$ _ $$3 = 6$$
$$4$$ _ $$4$$ _ $$4 = 6$$
$$5$$ _ $$5$$ _ $$5 = 6$$
$$6$$ _ $$6$$ _ $$6 = 6$$
$$7$$ _ $$7$$ _ $$7 = 6$$
$$8$$ _ $$8$$ _ $$8 = 6$$
$$9$$ _ $$9$$ _ $$9 = 6$$
You can put as many operations in the spaces but they must not contain a digit/letter. e.g.
You can have $$2 \times \sqrt{2} + 2!$$ but not $$2^2 + \sqrt [3]{2} + 2e$$.
You can only use factorials and double factorials, BODMAS/PEMDAS operations and square roots.
Good luck.
Extension: Find the smallest positive integral value of $$n$$ such that $$n$$ _ $$n$$ _ $$n = 6$$ has no solutions.
Note by Sharky Kesa
3 years, 8 months ago
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# up as a code block.
print "hello world"
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2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
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Sort by:
Answer below. If you want to do it yourself, don't scroll down: $.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$$.$ $$(0!+0!+0!)!$$
$$(1+1+1)!$$
$$2+2+2$$
$$3!+3-3$$
$$\sqrt { 4 } +\sqrt { 4 } +\sqrt { 4 }$$
$$(5\div5)+5$$
$$6+6-6$$
$$7-(7\div7)$$
$$\left( \sqrt { 8+8 } !\div 8 \right) !$$
$$\sqrt { 9 } +\sqrt { \sqrt { 9 } \times \sqrt { 9 } }$$
Legid?
- 3 years, 8 months ago
Try this
$$9$$_ $$9$$ _ $$9 = 10$$
- 3 years, 1 month ago
$$-(-9\div9-9)$$
- 3 years, 1 month ago
Sir, you are genius!
- 3 years, 1 month ago
But $$--$$ is $$+$$...
- 3 years, 1 month ago
Try the extension.
- 3 years, 1 month ago
After the crazy $$logic$$ question by $$\pi$$... I'd rather not
- 3 years, 1 month ago
Why? Those questions are awesome. Plus, this question shouldn't be too difficult.
- 3 years, 1 month ago
Ahh... but that was only the first part o the problem. You have to find as many different solutions as you can.
- 3 years, 8 months ago
Nice!...
- 3 years, 8 months ago
You should try it yourself! Its fun! :D
- 3 years, 8 months ago | 2018-07-20T14:42:58 | {
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http://math.stackexchange.com/questions/86644/determinant-of-a-specially-structured-matrix/86647 | # Determinant of a specially structured matrix
I have the following $n\times n$ matrix:
$$A=\begin{bmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots& &\ddots&\vdots\\b&\cdots&b&a\end{bmatrix}$$
where $0 < b < a$.
I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$. This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out. Any help would be appreciated.
-
Please do not use math displays in titles. – Mariano Suárez-Alvarez Nov 29 '11 at 6:23
This isn't a duplicate of math.stackexchange.com/questions/84206/… but the answer to this question appears in the answers to that question. – Gerry Myerson Nov 29 '11 at 6:23
There's a geometric, rather than algebraic, way of viewing it that makes it easy to understand. I've posted it below. – Michael Hardy Nov 29 '11 at 15:56
I've been away for a while -- I apologize for the late acceptance of an answer (I've looked at this before I left, but haven't had time to pick the best answer.) Also, @MarianoSuárez-Alvarez, I apologize for putting the matrix into the title... Thanks for correcting me on that. – M.B.M. Dec 13 '11 at 6:40
Add row 2 to row 1, add row 3 to row 1,..., add row $n$ to row 1, we get $$\det(A)=\begin{vmatrix} a+(n-1)b & a+(n-1)b & a+(n-1)b & \cdots & a+(n-1)b \\ b & a & b &\cdots & b \\ b & b & a &\cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \ldots & a \\ \end{vmatrix}$$ $$=(a+(n-1)b)\begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ b & a & b &\cdots & b \\ b & b & a &\cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \ldots & a \\ \end{vmatrix}.$$ Now add $(-b)$ of row 1 to row 2, add $(-b)$ of row 1 to row 3,..., add $(-b)$ of row 1 to row $n$, we get $$\det(A)=(a+(n-1)b)\begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & a-b & 0 &\cdots & 0 \\ 0 & 0 & a-b &\cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & a-b \\ \end{vmatrix}=(a+(n-1)b)(a-b)^{n-1}.$$
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I like this solution due to its simplicity and elegance. Other solutions are great too. – M.B.M. Dec 13 '11 at 6:42
nice solution. easy to understand. – prasad Apr 22 '13 at 6:33
SFAICT this route hasn't been mentioned yet, so:
Consider the decomposition
$$\small\begin{pmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots&&\ddots&\vdots\\b&\cdots&b&a\end{pmatrix}=\begin{pmatrix}a-b&&&\\&a-b&&\\&&\ddots&\\&&&a-b\end{pmatrix}+\begin{pmatrix}\sqrt b\\\sqrt b\\\vdots\\\sqrt b\end{pmatrix}\cdot\begin{pmatrix}\sqrt b&\sqrt b&\cdots&\sqrt b\end{pmatrix}$$
Having this decomposition allows us to use the Sherman-Morrison-Woodbury formula for determinants:
$$\det(\mathbf A+\mathbf u\mathbf v^\top)=(1+\mathbf v^\top\mathbf A^{-1}\mathbf u)\det\mathbf A$$
where $\mathbf u$ and $\mathbf v$ are column vectors. The corresponding components are simple, and thus the formula is easily applied (letting $\mathbf e$ denote the column vector whose components are all $1$'s):
\begin{align*} \begin{vmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots&&\ddots&\vdots\\b&\cdots&b&a\end{vmatrix}&=\left(1+(\sqrt{b}\mathbf e)^\top\left(\frac{\sqrt{b}}{a-b}\mathbf e\right)\right)(a-b)^n\\ &=\left(1+\frac{nb}{a-b}\right)(a-b)^n=(a+(n-1)b)(a-b)^{n-1} \end{align*}
where we used the fact that $\mathbf e^\top\mathbf e=n$.
-
This is very pretty! – M.B.M. Jul 28 '12 at 5:35
Can't you put $\mathbf u:=\mathbf e$ and $\mathbf v:=b\mathbf e$ (or vice versa) to avoid the radicals? – yo' Oct 2 '15 at 20:09
@yo, yes, that can be done. – J. M. Oct 6 '15 at 12:49
Subtract the bottom row from each of the other rows, then expand along some convenient row or column.
-
This is indeed an easy problem. Let $J$ be the square matrix with every entry equal to $1$. Your problem is equivalent to finding the determinant of $\lambda I + \mu J$ for arbitrary $\lambda, \mu$. Let $v=\left(\frac1{\sqrt{n}}, \frac1{\sqrt{n}},\ldots,\frac1{\sqrt{n}}\right)^\top$ and $e=(1,0,\ldots,0)^\top$. Then $J=nvv^\top$. Take any orthogonal matrix with its first column equal to $v$. Then $V^\top(\lambda I + \mu J)V = \lambda I+\mu nee^\top = \textrm{diag}\left(\lambda+\mu n,\lambda,\ldots,\lambda\right)$. Hence $\det(\lambda I + \mu J) = (\lambda+\mu n)\lambda^{n-1}$. Put $\lambda=a-b$ and $\mu=b$, we get the answer to your question as $[a+(n-1)b](a-b)^{n-1}$.
-
Hint: We can assume that the ground ring is a field, and that $b\neq0$. Consider the subspace formed by the vectors with equal coordinates, and the subspace formed by the vectors whose coordinates add up to $0$; note that these two subspaces are eigenspaces; compute the corresponding eigenvalues; and conclude.
EDIT. To find the eigenspaces of $A$ you can add $b-a$ to it, to make all the entries equal to $b$. (Adding a scalar to $A$ doesn't affect the eigenspaces.)
-
The matrix can be diagonalized. All it takes is a bit of geometry. We have
$$A=\begin{bmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots& &\ddots&\vdots\\b&\cdots&b&a\end{bmatrix}.$$
This is a linear combination of the matrices $P$ and $Q=I-P$ where $P$ is the matrix of the orthogonal projection onto the $1$-dimensional space of column vectors in which all scalar components are equal, i.e. the space $$\left\{\begin{bmatrix} x \\ x \\ x \\ \vdots \\ x \end{bmatrix} : x \text{ is a scalar} \right\}.$$ We have $$P = \begin{bmatrix} 1/n & 1/n & \ldots & 1/n \\ 1/n & 1/n & \ldots & 1/n \\ \vdots & \vdots & & \vdots \\ 1/n & 1/n & \ldots & 1/n \end{bmatrix}$$ (all entries are $1/n$), so that $$P \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} \bar{x} \\ \bar{x} \\ \bar{x} \\ \vdots \\ \bar{x} \end{bmatrix}$$ where $\bar{x} = (x_1+\cdots+x_n)/n$ is the average of the components, and $$Q \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} x_1-\bar{x} \\ x_2-\bar{x} \\ x_3-\bar{x} \\ \vdots \\ x_n-\bar{x} \end{bmatrix}.$$ We want $$A = \alpha P + \beta Q.$$ Looking at the diagonal elements we have $$\alpha\cdot\frac 1n + \beta \left(1-\frac 1n\right) = a,$$ and from the off-diagonal elements we get $$\alpha\cdot\frac 1n - \beta \cdot\frac 1n =b.$$ Hence $$\alpha = a + (n-1)b \qquad\text{and}\qquad\beta= a-b.$$
Since $P$ projects orthogonally onto a $1$-dimensional subspace and $Q$ is the complementary orthogonal projection onto an $(n-1)$-dimensional subspace, the matrix $\alpha P+\beta Q$ can be diagonalized as $$\begin{pmatrix} \alpha \\ & \beta \\ & & \beta \\ & & & \beta \\ & & & & \ddots \\ & & & & & \beta \end{pmatrix}.$$ The determinant is therefore $$\alpha\beta^{n-1}.$$
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Consider the $n\times n$ matrix $B$ with entries $-b$ everywhere, except on the main diagonal where it has entries $0$. Now $\det A=\det(aI-B)$ is just the value of the characteristic polynomial $\chi_B\in K[X]$ at $X=a$. For $X=b$ the matrix $bI-B$ clearly has rank at most$~1$ (all columns are equal), so by rank-nullity the eigenspace of $B$ for the eigenvalue $b$ has dimension at least $n-1$, and $\chi_B$ is divisible by $(X-b)^{n-1}$. The final eigenvalue must make their sum $\operatorname{tr}(B)=0$ so it is $-(n-1)b$, and the final factor of $\chi_B$ is $X+(n-1)b$. So$$\det A=\chi_B[X:=a] = (a-b)^{n-1}(a+(n-1)b).$$
- | 2016-02-10T20:40:10 | {
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https://discuss.codechef.com/t/chn08-editorial/12127 | # CHN08 - Editorial
Contest
Author: ???
Tester: Kevin Atienza, Jingbo Shang
Editorialist: Kevin Atienza
### PREREQUISITES:
Modulo, recurrences, pattern matching
### PROBLEM:
A function f on integers is defined:
• f(1) = A, f(2) = B.
• For all integers x \ge 2, f(x) = f(x-1) + f(x+1).
Find f(N) \bmod (10^9 + 7).
### QUICK EXPLANATION:
f is periodic with period 6, so f(N) is equal to f(N \bmod 6). (Just define f(0) = A-B.)
### EXPLANATION:
At first glance, this problem seems to require standard techniques in solving general linear recurrences. While those techniques will work, they are quite overkill for this problem. Simple pattern spotting will do (This solution is much faster to code and very easy to find, so you’ll probably be able to get AC earlier with it. The few minutes you save is sometimes significant.)
The key is to simply compute the first few terms. As an example, let’s say A = 11 and B = 9. Since f(x+1) = f(x) - f(x-1) (a rearrangement of the above), we have:
• f(\,\,3) = f(\,\,2) - f(\,\,1) = -2
• f(\,\,4) = f(\,\,3) - f(\,\,2) = -11
• f(\,\,5) = f(\,\,4) - f(\,\,3) = -9
• f(\,\,6) = f(\,\,5) - f(\,\,4) = 2
• f(\,\,7) = f(\,\,6) - f(\,\,5) = 11
• f(\,\,8) = f(\,\,7) - f(\,\,6) = 9
• f(\,\,9) = f(\,\,8) - f(\,\,7) = -2
• f(10) = f(\,\,9) - f(\,\,8) = -11
• f(11) = f(10) - f(\,\,9) = -9
• f(12) = f(11) - f(10) = 2
At this point, you might notice that the sequence repeats every 6 terms. Once you notice this, you might think to try to prove it rigorously before coding it. Though proving it is simple, you don’t have to do it; you just pray to the god of pattern matching and hope that this pattern will go on forever (or you trust your instinct). Though this doesn’t always work, with experience and intuition, you’ll eventually learn which of the unproven statements / observations you make may actually be true.
Note: Don’t forget to ensure that f(N) \bmod (10^9 + 7) is positive! Also, f(N) can be < -(10^9 + 7), so adding (10^9 + 7) just once isn’t enough to make f(N) positive.
# Proving it
But as just mentioned, proving it is simple. Let’s first generalize the observation above, by actually using variables A and B:
• f(3) = f(2) - f(1) = B - A
• f(4) = f(3) - f(2) = -A
• f(5) = f(4) - f(3) = -B
• f(6) = f(5) - f(4) = A - B
• f(7) = f(6) - f(5) = A
• f(8) = f(7) - f(6) = B
• f(9) = f(8) - f(7) = B - A
• f(10) = f(9) - f(8) = -A
• f(11) = f(10) - f(9) = -B
• f(12) = f(11) - f(10) = A - B
Let’s now prove that this pattern will go on forever:
Claim: For n \ge 1, f(n) = f(n+6).
Proof:
Clearly, this is true for n = 1 and n = 2 (because f(1) = f(7) = A and f(2) = f(8) = B). Now, assume n > 2. Assume by induction that it is true for all 1 \le n' < n. So it must be true for n-2 and n-1. Then:
\begin{aligned} f(n) &= f(n-1) - f(n-2) \\\ &= f((n-1)+6) - f((n-2)+6) \\\ &= f(n+5) - f(n+4) \\\ &= f(n+6) \end{aligned}
This proves the claim.
End of proof
The general idea is that because each term f(n) depends only on the previous two terms (f(n-2), f(n-1)), once two consecutive terms repeat, the whole function will repeat forever. In our case, since (A,B) repeated, we’re sure that the following sequence will just repeat previous terms after that.
This also works even if f(n) is not a linear recurrence! If f(n) = g(f(n-2), f(n-1)) for some function g, then regardless of the nature of g, as long as you find a repeat of the pair (f(n-2),f(n-1)), the function f will repeat after that point. This also works if f was a recurrence that depended on more than two previous terms, say k \ge 2, though this time you’ll need to keep track of a k-tuple of values instead of a pair.
# More properties
Let’s analyze f more deeply. Using generating functions, we can actually find a neat formula for f. (though it doesn’t necessarily yield any algorithmic improvements, it is interesting nonetheless.) It will also explain why every three terms almost repeats, except that you have to flip the sign first. To make things cleaner, in the following we’ll write f_n as a synonym of f(n). We’ll also define f(0) = A - B, so the recurrence still holds.
Let F(x) = f_0 + f_1x + f_2x^2 + f_3x^3 + \ldots. Then:
\begin{array}{rrrrrrr} F(x) = & f_0 &+ f_1x &+ f_2x^2 &+ f_3x^3 &+ f_4x^4 &+ \ldots \\\ xF(x) = & & f_0x &+ f_1x^2 &+ f_2x^3 &+ f_3x^4 &+ \ldots \\\ x^2F(x) = & & & f_0x^2 &+ f_1x^3 &+ f_2x^4 &+ \ldots \\\ \end{array}
Now, consider F(x) - xF(x) + x^2F(x), by adding like terms together. Since f_n = f_{n-1} - f_{n-2}, almost all terms will cancel out. The only nonzero terms are the constant and linear term. Thus, we have:
\begin{aligned} F(x) - xF(x) + x^2F(x) &= f_0 + f_1x \\\ (1 - x + x^2)F(x) &= f_0 + f_1x \\\ F(x) &= \frac{f_0 + f_1x}{1 - x + x^2} \\\ F(x) &= \frac{A-B + Ax}{1 - x + x^2} \end{aligned}
Now, let’s try to factorize 1 - x + x^2. It has two roots: \dfrac{1 \pm \sqrt{3}i}{2}. Let us use the symbol \psi and \bar{\psi} for \dfrac{1 + \sqrt{3}i}{2} and \dfrac{1 - \sqrt{3}i}{2}, respectively (\bar{x} denotes complex conjugation). Thus, 1 - x + x^2 must be equal to (x - \psi)(x - \bar{\psi}). But since \psi\bar{\psi} = 1, we get:
\begin{aligned} 1 - x + x^2 &= (x - \psi)(x - \bar{\psi}) \\\ &= \psi\bar{\psi}(x - \psi)(x - \bar{\psi}) \\\ &= (\bar{\psi}x - \bar{\psi}\psi)(\psi x - \psi\bar{\psi}) \\\ &= (\bar{\psi}x - 1)(\psi x - 1) \\\ &= (1 - \psi x)(1 - \bar{\psi}x) \end{aligned}
Note that \psi and \bar{\psi} are both primitive 6 th roots of unity, so \psi^6 = \bar{\psi}^6 = 1. (verify)
Back to F(x), we get:
\begin{aligned} F(x) &= \frac{A-B + Ax}{1 - x + x^2} \\\ F(x) &= \frac{A-B + Ax}{(1 - \psi x)(1 - \bar{\psi}x)} \end{aligned}
At this point, let’s use the method of partial fractions, to obtain
F(x) = \frac{C}{1 - \psi x} + \frac{D}{1 - \bar{\psi}x}
where C and D are constants that we don’t know yet. This form is very useful to us because of the equality:
\frac{1}{1 - rx} = 1 + rx + r^2x^2 + r^3x^3 + \ldots
which means that the coefficient of x^n in F(x) is just C\psi^n + D\bar{\psi}^n, giving us a formula for f(n). Right away, this tells us that f(n) is periodic with period 6, because \psi and \bar{\psi} are both primitive 6 th roots of unity. It also explains why f(n) = -f(n+3): It’s because \psi^3 = \bar{\psi}^3 = -1.
To compute C and D, simply add the two fractions together and equate coefficients:
\begin{aligned} F(x) &= \frac{C}{1 - \psi x} + \frac{D}{1 - \bar{\psi}x} \\\ F(x) &= \frac{(C + D) - (C\bar{\psi} + D\psi)x}{(1 - \psi x)(1 - \bar{\psi}x)} \end{aligned}
Thus, we get C + D = A - B and C\bar{\psi} + D\psi = A. By solving this system, we get
D = -\frac{A(\psi-2)+B(\psi+1)}{3}
C = -\frac{A(\bar{\psi}-2)+B(\bar{\psi}+1)}{3}
This gives us the following formula for f(n):
f(n) = -\frac{1}{3}\left[\left(A(\psi-2)+B(\psi+1)\right)\psi^n + \left(A(\bar{\psi}-2)+B(\bar{\psi}+1)\right)\bar{\psi}^n\right]
In general, for any periodic sequence g(n) with period k, g always has some formula that looks like:
g(n) = a_0 + a_1\omega_k^n + a_2\omega_k^{2n} + a_3\omega_k^{3n} + \ldots + a_{k-1}\omega_k^{(k-1)n}
where \omega_k is a primitive k th root of unity (so \{1, \omega_k, \omega_k^2, \ldots, \omega_k^{k-1}\} is the complete set of k th roots of unity).
O(1)
### AUTHOR’S AND TESTER’S SOLUTIONS:
[setter][333]
[tester][444]
[editorialist][555]
[333]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server.
[444]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server.
[555]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server.
Why is my answer not working? I have implemented what you just said and the test cases work fine:
#include <iostream>
using namespace std;
int main(){
int i;
cin >> i;
int a,b,N;
for (int j = 0; j< i; j++){
cin >> a;
cin >> b;
cin >> N;
N=(N+6%6+6)%6;
if (N==0){
cout << (a-b+1000000007)%1000000007 << endl;
}
if (N==1){
cout << (a+1000000007)%1000000007 << endl;
}
if (N==2){
cout << (b+1000000007)%1000000007 << endl;
}
if (N==3){
cout << (b-a+1000000007)%1000000007 << endl;
}
if (N==4){
cout << (1000000007-a)%1000000007 << endl;
}
if (N==5){
cout << (1000000007-b)%1000000007 << endl;
}
}
return 0;
}
Whats wrong with this code?
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define vi vector<int>
#define all(a) a.begin(),a.end()
#define t int t; cin>>t; while(t--)
#define show(a) for(i=0;i<a.size();i++) cout<<a[i]<<" ";
#define s(n) scanf("%d",&n);
#define ss(n,m) int n,m;scanf("%d%d",&n,&m);
#define p printf
#define u(i,a,b) for(i=a;i<b;i++)
#define d(i,a,b) for(i=a;i>b;i--)
#define mod 1000000007
#define in vi a(n);u(i,0,n) s(a[i])
#define pb push_back
int main()
{
int n;
t
{
ss(a,b)
s(n) n=n%6;
vi A{a-b,a,b,b-a,-a,-b};
p("%d\n",(mod+A[n])%mod);
}
}
@savage_coder
Overflow is being savage to you. Try changing data type to long long int and see.
Also, (mod + A[n]) need not necessarily positive always. make it (mod + A[n]%mod)%mod and see.
Help me,please;what’s wrong with the code?
#include<iostream>
using namespace std;
long long int f(long long int index[],int n) {
const long long int m = 1'000'000'007;
auto x = index[n % 6];
return (x%m + m) % m;
}
int main() {
int n,a, b, c;
cin >> n;
while (n--) {
cin >> a >> b >> c;
long long int arr[] = { a,b,b - a,-a,-b,a - b };
cout << f(arr, c-1);
}
}
I would appreciate if anyone could help me to show me the problem with my code. I got the wrong answer but I think my logic is correct. | 2020-09-25T15:44:26 | {
"domain": "codechef.com",
"url": "https://discuss.codechef.com/t/chn08-editorial/12127",
"openwebmath_score": 0.9550616145133972,
"openwebmath_perplexity": 3470.341465561802,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9840936087546923,
"lm_q2_score": 0.8577681049901037,
"lm_q1q2_score": 0.8441241099143849
} |
https://bathmash.github.io/HELM/30_2_gauss_elim-web/30_2_gauss_elim-webse2.html | ### 2 Partial pivoting
Partial pivoting is a refinement of the Gaussian elimination procedure which helps to prevent the growth of rounding error.
#### 2.1 An example to motivate the idea
Consider the example
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 1{0}^{-4}\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right].$
First of all let us work out the exact answer to this problem
$\begin{array}{rcll}\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]& =& {\left[\begin{array}{cc}\hfill 1{0}^{-4}\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]}^{-1}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]& \text{}\\ & =& \frac{1}{2×1{0}^{-4}+1}\left[\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1{0}^{-4}\hfill \end{array}\right]\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]& \text{}\\ & =& \frac{1}{2×1{0}^{-4}+1}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1+1{0}^{-4}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0.999800...\hfill \\ \hfill 0.999900...\hfill \end{array}\right].\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
Now we compare this exact result with the output from Gaussian elimination. Let us suppose, for sake of argument, that all numbers are rounded to 3 significant figures. Eliminating the one non-zero element below the diagonal, and remembering that we are only dealing with 3 significant figures, we obtain
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 1{0}^{-4}\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1{0}^{4}\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1{0}^{4}\hfill \end{array}\right].$
The bottom equation gives ${x}_{2}=1$ , and the top equation therefore gives ${x}_{1}=0$ . Something has gone seriously wrong, for this value for ${x}_{1}$ is nowhere near the true value 0.9998…found without rounding.The problem has been caused by using a small number ( $1{0}^{-4}$ ) to eliminate a number much larger in magnitude ( $-1$ ) below it.
The general idea with partial pivoting is to try to avoid using a small number to eliminate much larger numbers.
Suppose we swap the rows
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1{0}^{-4}\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$
and proceed as normal, still using just 3 significant figures. This time eliminating the non-zero below the diagonal gives
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$
which leads to ${x}_{2}=1$ and ${x}_{1}=1$ , which is an excellent approximation to the exact values, given that we are only using 3 significant figures.
#### 2.2 Partial pivoting in general
At each step the aim in Gaussian elimination is to use an element on the diagonal to eliminate all the non-zeros below. In partial pivoting we look at all of these elements (the diagonal and the ones below) and swap the rows (if necessary) so that the element on the diagonal is not very much smaller than the other elements.
##### Key Point 3
Partial Pivoting
This involves scanning a column from the diagonal down. If the diagonal entry is very much smaller than any of the others we swap rows. Then we proceed with Gaussian elimination in the usual way.
In practice on a computer we swap rows to ensure that the diagonal entry is always the largest possible (in magnitude). For calculations we can carry out by hand it is usually only necessary to worry about partial pivoting if a zero crops up in a place which stops Gaussian elimination working. Consider this example
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill -6\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \\ \hfill {x}_{3}\hfill \\ \hfill {x}_{4}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill -4\hfill \\ \hfill 1\hfill \\ \hfill 12\hfill \\ \hfill 0\hfill \end{array}\right].$
The first step is to use the 1 in the top left corner to eliminate all the non-zeros below it in the augmented matrix
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 2\hfill & \hfill -6\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 12\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\begin{array}{c}\hfill \\ R2-2×R1\hfill \\ R3+R1\hfill \\ \hfill \\ \hfill \end{array}⇒\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 5\hfill & \hfill 5\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right].$
What we would like to do now is to use the boxed element to eliminate all the non-zeros below it. But clearly this is impossible. We need to apply partial pivoting. We look down the column starting at the diagonal entry and see that the two possible candidates for the swap are both equal to $-1$ . Either will do so let us swap the second and fourth rows to give
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 5\hfill & \hfill 5\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right].$
That was the partial pivoting step. Now we proceed with Gaussian elimination
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 5\hfill & \hfill 5\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right]\begin{array}{c}\hfill \\ \hfill \\ R3-R2\hfill \\ \hfill \end{array}⇒\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 4\hfill & \hfill 4\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right].$
The arithmetic is simpler if we cancel a factor of 4 out of the third row to give
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right].$
And the elimination phase is completed by removing the $-3$ from the final row as follows
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 9\hfill \end{array}\right]\begin{array}{c}\hfill \\ \hfill \\ \hfill \\ R4+3×R3\hfill \end{array}⇒\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill & \hfill 15\hfill \end{array}\right].$
This system is upper triangular so back substitution can be used now to work out that ${x}_{4}=3$ , ${x}_{3}=-1$ , ${x}_{2}=2$ and ${x}_{1}=1$ .
The Task below is a case in which partial pivoting is required.
[For a large system which can be solved by Gauss elimination see Engineering Example 1 on page 62].
Transform the matrix
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill -3\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right]$
into upper triangular form using Gaussian elimination (with partial pivoting when necessary).
The row operations required to eliminate the non-zeros below the diagonal in the first column are
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill -3\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right]\begin{array}{c}\hfill \\ R2+3×R1\hfill \\ R3-4×R1\hfill \end{array}⇒\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 11\hfill & \hfill -11\hfill \end{array}\right]$
which puts a zero on the diagonal. We are forced to use partial pivoting and swapping the second and third rows gives
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 11\hfill & \hfill -11\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$
which is in the required upper triangular form.
##### Key Point 4
When To Use Partial Pivoting
1. When carrying out Gaussian elimination on a computer, we would usually always swap rows so that the element on the diagonal is as large (in magnitude) as possible. This helps stop the growth of rounding error.
2. When doing hand calculations (not involving rounding) there are two reasons we might pivot
1. If the element on the diagonal is zero, we have to swap rows so as to put a non-zero on the diagonal.
2. Sometimes we might swap rows so that there is a “nicer" non-zero number on the diagonal than there would be without pivoting. For example, if the number on the diagonal can be arranged to be a 1 then no awkward fractions will be introduced when we carry out row operations related to Gaussian elimination.
##### Exercises
1. Solve the following system by back substitution $\begin{array}{rcll}{x}_{1}+2{x}_{2}-{x}_{3}& =& 3& \text{}\\ 5{x}_{2}+6{x}_{3}& =& -2& \text{}\\ 7{x}_{3}& =& -14\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
1. Show that the exact solution of the system of equations
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 1{0}^{-5}\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 4\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 10\hfill \end{array}\right]$ is $\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill -0.99998\hfill \\ \hfill 2.00001\hfill \end{array}\right]$ .
2. Working to 3 significant figures, and using Gaussian elimination without pivoting, find an approximation to $\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]$ . Show that the rounding error causes the approximation to ${x}_{1}$ to be a very poor one.
3. Working to 3 significant figures, and using Gaussian elimination with pivoting, find an approximation to $\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]$ . Show that the approximation this time is a good one.
2. Carry out row operations (with partial pivoting if necessary) to reduce these matrices to upper triangular form.
1. $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill -4\hfill & \hfill -3\hfill & \hfill -3\hfill \\ \hfill -1\hfill & \hfill 13\hfill & \hfill 1\hfill \end{array}\right]$ ,
2. $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -4\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 5\hfill & \hfill -4\hfill \end{array}\right]$ ,
3. $\left[\begin{array}{ccc}\hfill -3\hfill & \hfill 10\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 10\hfill & \hfill -4\hfill \end{array}\right]$ .
(Hint: before tackling (c) you might like to consider point 2(b) in Key Point 4.)
1. From the last equation we see that ${x}_{3}=-2$ . Using this information in the second equation gives us ${x}_{2}=2$ . Finally, the first equation implies that ${x}_{1}=-3$ .
1. The formula ${\left[\begin{array}{cc}\hfill a& \hfill b\\ \hfill c& \hfill d\end{array}\right]}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}\hfill d& \hfill -b\\ \hfill -c& \hfill a\end{array}\right]$ can be used to show that
$\phantom{\rule{2em}{0ex}}{x}_{1}=-\frac{50000}{50001}=-0.99998\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{x}_{2}=\frac{200005}{100002}=2.00001$ as required.
2. Carrying out the elimination without pivoting, and rounding to 3 significant figures we find that ${x}_{2}=2.00$ and that, therefore, ${x}_{1}=0$ . This is a very poor approximation to ${x}_{1}$ .
3. To apply partial pivoting we swap the two rows and then eliminate the bottom left element. Consequently we find that, after rounding the system of equations to 3 significant figures, ${x}_{2}=2.00$ and ${x}_{1}=-1.00$ . These give excellent agreement with the exact answers.
1. The row operations required to eliminate the non-zeros below the diagonal in the first column are as follows
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill -2& \hfill 4\\ \hfill -4& \hfill -3& \hfill -3\\ \hfill -1& \hfill 13& \hfill 1\\ \hfill \end{array}\right]\begin{array}{c}\hfill \\ R2+4×R1\hfill \\ R3+1×R1\hfill \end{array}⇒\left[\begin{array}{ccc}\hfill 1& \hfill -2& \hfill 4\\ \hfill 0& \hfill -11& \hfill 13\\ \hfill 0& \hfill 11& \hfill 5\\ \hfill \end{array}\right]$
Next we use the element in the middle of the matrix to eliminate the value underneath it. This gives
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill -2& \hfill 4\\ \hfill 0& \hfill -11& \hfill 13\\ \hfill 0& \hfill 0& \hfill 18\\ \hfill \end{array}\right]\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}$ which is of the required upper triangular form.
2. We must swap the rows to put a non-zero in the top left position (this is the partial pivoting step). Swapping the first and second rows gives the matrix
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill -4& \hfill 2\\ \hfill 0& \hfill -1& \hfill 2\\ \hfill -2& \hfill 5& \hfill -4\end{array}\right].$
We carry out one row operation to eliminate the non-zero in the bottom left entry as follows
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill -4& \hfill 2\\ \hfill 0& \hfill -1& \hfill 2\\ \hfill -2& \hfill 5& \hfill -4\\ \hfill \end{array}\right]\begin{array}{c}\hfill \\ \hfill \\ R3+2×R1\hfill \end{array}⇒\left[\begin{array}{ccc}\hfill 1& \hfill -4& \hfill 2\\ \hfill 0& \hfill -1& \hfill 2\\ \hfill 0& \hfill -3& \hfill 0\\ \hfill \end{array}\right]$
Next we use the middle element to eliminate the non-zero value underneath it. This gives
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill -4& \hfill 2\\ \hfill 0& \hfill -1& \hfill 2\\ \hfill 0& \hfill 0& \hfill -6\\ \hfill \end{array}\right]\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}$ which is of the required upper triangular form.
3. If we swap the first and second rows of the matrix then we do not have to deal with fractions. Having done this the row operations required to eliminate the non-zeros below the diagonal in the first column are as follows
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill -3& \hfill 2\\ \hfill -3& \hfill 10& \hfill 1\\ \hfill -2& \hfill 10& \hfill -4\\ \hfill \end{array}\right]\begin{array}{c}\hfill \\ R2+3×R1\hfill \\ R3+2×R1\hfill \end{array}⇒\left[\begin{array}{ccc}\hfill 1& \hfill -3& \hfill 2\\ \hfill 0& \hfill 1& \hfill 7\\ \hfill 0& \hfill 4& \hfill 0\\ \hfill \end{array}\right]$
Next we use the element in the middle of the matrix to eliminate the non-zero value underneath it. This gives
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill -3& \hfill 2\\ \hfill 0& \hfill 1& \hfill 7\\ \hfill 0& \hfill 0& \hfill -28\\ \hfill \end{array}\right]\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}$ which is of the required upper triangular form. | 2022-11-29T00:32:16 | {
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https://math.stackexchange.com/questions/3047809/probability-that-the-10th-drawn-ball-is-black | # Probability that the 10th drawn ball is black.
An urn contains $$7$$ white and $$13$$ black balls. We draw a ball from an urn, put it back and add $$2$$ additional balls of the same color (as the one we have just drawn and put back). We repeat this $$10$$ times (so the numbers of balls change throughout the whole experiment). What is the probability that the $$10$$th drawn ball is black?
I think it should be $$13/20$$. Basically, I think that no matter how many times the experiment is repeated, the probability of picking up the black ball at the end is $$13/20$$. I was trying induction but got stuck.
• I believe you can prove this by simply looking at the expected number of white and black balls. With the linearity of the expected value, it should be easy to prove by induction. Dec 20, 2018 at 18:16
• The probability of getting black ball the second time is $13/20$. $P(B_2) = P(B_2 | B_1) P(B_1) + P(B_2 | W_1) P(W_1) = \frac{15}{22} \cdot \frac{13}{20} + \frac{13}{22} \cdot \frac{7}{20} = \frac{286}{440} = \frac{13}{20}$. Dec 20, 2018 at 18:17
• @DonThousand What makes you say that? The probability that the second ball is black is 13/20 as well. Dec 20, 2018 at 18:20
• Dec 20, 2018 at 18:20
• Thank you, @saulspatz Dec 20, 2018 at 18:23
Direct computation shows that this is true for one round of the game, regardless of the starting configuration. (and for any "shift" where, in your problem, the "shift" is $$3$$).
Inductively, suppose we have shown that the probability is unchanged for $$n-1$$ rounds of the game (regardless of the starting configuration).
Now suppose we start with $$w$$ White and $$b$$ Black (denoting that by $$(w,b)$$). On round two we either have $$(w+3,b)$$ with probability $$\frac w{b+w}$$ or $$(w,b+3)$$ with probability $$\frac b{b+w}$$. By the inductive hypothesis the probability is then unchanged for at least the next $$n-1$$ rounds.
Of course, if you are in state $$(w+3,b)$$ the (new) probability of drawing a black ball is $$\frac b{b+w+3}$$ and if you are in state $$(w,b+3)$$ it is $$\frac {b+3}{b+w+3}$$.
Thus the probability on the $$n^{th}$$ round of the original game is $$\frac w{b+w}\times \frac b{b+w+3}+\frac b{b+w}\times \frac {b+3}{w+b+3}=\frac {b(b+w+3)}{(b+w)(b+w+3)}=\frac b{b+w}$$
as desired.
(To be sure, the $$3$$ is incidental as well. As the computation makes clear, we could replace it with any natural number).
The comments suggest that the logic of this argument is unclear, so let me elaborate. I am inductively supposing that we know the desired claim for any starting configuration of White and Black balls for $$n-1$$ rounds. I then take the desired configuration and look at the second round. To be sure, the probability will have changed depending on the first draw. Inductively, though, I know that the new probability will be constant for the next $$n-1$$ rounds. So it suffices to consider the two possible branch paths, weighted by the probability of being on one or the other. Of course, the probability of being on the "White" branch is $$\frac w{b+w}$$ and the new probability of drawing black along that branch is $$\frac b{b+w+3}$$. That explains the first term of my sum. The second term, corresponding to the "Black" branch, is entirely similar.
• you want a $b+w$ in the final denominator, I think
– Ned
Dec 20, 2018 at 18:34
• You meant $b/(b+w)$ in the last line of the calculation. Dec 20, 2018 at 18:36
• I have some problems with this proof. It seems like you assume sth like: ok, at the $n-1$ round the probability of drawing a black ball is $\frac{b}{b+w}$. But then for calculating the probability for the $n$th round, you assume that you actually have $b+w$ balls in the urn which seems a bit weird for me. I am not telling your proof is incorrect. I am just saying I am not fully convinced. Dec 20, 2018 at 18:47
• I have doubts about this. There aren't $b$ and $w$ balls in the urn after $n$ draws. Even if we assume there are $kb$ black and $kw$ white balls, at the next draw there will be $kw+kb+3,$ and if I'm not mistaken, the algebra doesn't work out to $b/(b+w).$ Dec 20, 2018 at 18:48
• @Ned Yes, thank you.
– lulu
Dec 20, 2018 at 19:00
Here's a conceptual solution based on symmetry which also works regardless of how many extras you put back at each stage. First note that you can play the games with more than 2 colors and ask the same question.
Second note that if two colors have the same number of initial balls, then the probabilities of obtaining those colors on the $$j$$-th draw are the same (by symmetry, nothing favors one of the two colors over the other).
Now given the initial configuration of $$w$$ white and $$b$$ black, imagine that the balls are all different colors (maybe, $$w$$ shades of white and $$b$$ shades of black). On draw $$j$$ each of the $$w+b$$ shades is equally likely to be drawn, since we started with the same number (namely $$1$$) of each. But that says the probability of drawing a shade of Black is $$b/(w+b)$$.
Here is I what I have found based on a problem on Polya's urn scheme but I elaborate this formally below
Proof for the specific case
We start with $$7$$ white balls and $$13$$ black balls. On each round we add additional $$2$$ balls according to the rules described in the question.
Think that for each round you either draw a ball that was there in the previous round or you draw a ball that was added in the previous round.
Assume inductively that on round $$n$$ the probability of drawing a black ball is $$p$$ ($$:= \frac{b}{b+w}$$). Let's see what happens on round $$n+1$$. Let $$k$$ denote the number of balls in the $$n+1$$ round. So with probability $$\frac{k-2}{k}$$ we draw an old ball (that was there in the $$n$$th round) and with probability $$\frac{2}{k}$$ we draw a new ball (one of the two that were added on the $$n$$th round). By the inductive hypothesis, the probability of drawing a black ball from the old ones is $$p$$. The probability of drawing a black ball from the new ones is simply $$p$$ (because you either have drawn a black ball with probability $$p$$ on the $$n$$th round and hence necessarily a black one on the $$n+1$$th round, or you have drawn a white one on the $$n$$th round and hence a white one on the $$n+1$$th round).
All in all, the probability of drawing a black ball on the $$n+1$$th round is:
$$p \cdot \frac{k-2}{k} + p \cdot \frac{2}{k} = p \cdot ( \frac{k-2}{k} + \frac{2}{k}) = p$$
Proof for the general case
You can use this idea to give a general inductive proof for the number $$k$$ of balls that you add each time to the urn and arbitrary initial configuration $$b,w$$ of black and white balls, respectively.
Let $$b,w$$ be our starting configuration. On the first round the probability of drawing a black ball is $$\frac{b}{b+w}$$.
Now, assume inductively that on the $$n$$th round the probability of drawing a black ball is $$\frac{b}{b+w}$$. The probability of drawing a black ball on the $$(n+1)$$th round is:
$$P(B_{n+1}) = P(B_n) P(B_{n+1} | B_n) + P(W_n) P(B_{n+1} | W_n) ,$$
where $$B_i$$, $$W_i$$ are the events of drawing a black ball and the white ball on the $$i$$th round, respectively.
We have:
$$P(B_{n+1}) = \frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{w}{b+w}\frac{(b+w+(n-1)k)\frac{b}{b+w}}{b+w+nk}=$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{w}{b+w}\frac{b}{w}\frac{(b+w+(n-1)k)\frac{b}{b+w}\frac{w}{b}}{b+w+nk}=$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{b}{b+w}\frac{(b+w+(n-1)k)\frac{w}{b+w}}{b+w+nk}=$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k + (b+w+(n-1)k)\frac{w}{b+w}}{b+w+nk}$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k)(\frac{b}{b+w} + \frac{w}{b+w}) + k}{b+w+nk}$$ $$\frac{b}{b+w} \frac{(b+w + (n-1)k) + k}{b+w+nk}$$ $$\frac{b}{b+w} \frac{b+w + nk}{b+w+nk} = \frac{b}{b+w}$$
This ends the proof. | 2023-02-08T12:45:10 | {
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https://www.physicsforums.com/threads/completing-the-square.281868/ | # Completing the square
1. Dec 28, 2008
### schlynn
1. The problem statement, all variables and given/known data
I need to know how to complete the square, I need to know how to do this to get
(this is to find the vertex of the quadratic function and the x intercept(s)
f(x)=aX2+bX+c to f(x)=a(x-h)2+k
It would be helpfull to see walkthroughs for when a=1 and when "a" does not equal 1.
questions:
-3x2+9x+(1/4)
and
x2-6x-2
2. Relevant equations
I dont think theres any, just that to get y set x=0 and to get x set y=0, for the intersepts that is.
h=horizontal shift in the graph
k=verticle shift in the graph
there is this shortcut too....
f(x)=a(x-h)2+k
and the vertex will be
((-b/2a),c-(b2/4a))
3. The attempt at a solution
Well, I'm learning this all out of school, and the book doesn't really explain how to do this very well, so I have no idea what do to. Could someone make walkthroughs on how to solve both of the question? This will help me tremendusly as I have been stuck on what to do now.
2. Dec 28, 2008
$$g(x) = x^2 + 6x -120$$
• Here $$b = 6$$. Start by calculating half it
$$\frac b 2 = 3$$
• Add and immediately subtract the square of 3, as shown here. (I've grouped the terms
together for easier reading.) This changes the look of the right-hand-side, but since 9-9=0,
no numerical change has occurred
$$g(x) = \left( x^2 + 6x + 3^2 \right) - 3^2 - 120$$
• The portion in parentheses is $$(x-3)^2$$, so we have
$$g(x) = (x-3)^2 - 129$$
• The vertex is the point $$(3,-129)$$. The final step is to find the intercepts.
\begin{align*} (x-3)^2 - 129 & = 0\\ (x-3)^2 & = 129 \\ x-3 & = \pm \sqrt{\, 129} \\ x & = 3 \pm \sqrt{\, 120} \end{align*}
The intercepts are $$(3 \pm \sqrt{\, 129}, 0 )$$.
For the second question
$$h(x) = 3x^2 + 12x - 111$$
Here is the work to complete the square.
\begin{align*} h(x) & = 3x^2 + 12x - 111 \\ & = 3 \left(x^2 + 4x \right) - 111\\ & = 3 \left( \left(x^2 + 4x + 2^2\right) - 2^2 \right) - 111 \\ & = 3 ((x+2)^2 - 4) - 111 \\ & = 3(x+2)^2 - 12 - 111 = 3(x+2)^2 - 123 \end{align*}
The vertex is the point $$(-2, -123)$$. To find the intercepts
\begin{align*} 3(x+2)^2 - 123 & = 0\\ 3(x+2)^2 & = 123\\ (x+2)^2 & = \frac{123} 3 = 41\\ x+2 & = \pm \sqrt{\, 41}\\ x & = -2 \pm \sqrt{\, 41} \end{align*}
The two intercepts are $$(-2 \pm \sqrt{\, 41}, 0)$$.
3. Dec 28, 2008
### Unco
Just a minor sign error there, Statdad.
4. Jan 3, 2009
### Gregg
$$ax^2 + bx + c = a(x^2 + \frac{b}{a} + \frac{c}{a})$$
$$ax^2 + bx + c = a((x + \frac{b}{2a})^2 +\frac{c}{a})- \left[\frac{b}{2a}\right]^2$$
$$ax^2 + bx + c =a(x + \frac{b}{2a})^2 - \left[\frac{b}{2a}\right]^2 + c$$
Vertex for any function like
$$(x+a)^2 + b= \left[\stackrel{-a}{ b}\right]$$
The reason for this is because for the function to be at its lowest value, given that the first part is squared it has to be equal to 0. The x value for this co-ordinate therefore has to be the negative of a. b is more obvious.
Intercepts
You can do this two ways (assuming that because you have completed the square therefore you cannot factorise) using the quadratic formula or an easier way, manipulating the completed square you have.
$$(x+a)^2 - b = 0$$
$$(x+a)^2 = b$$
$$(x+a) = \sqrt{b}$$
$$x = -a \pm \sqrt{b}$$
What if b is positive
$$(x+a)^2 + b = 0$$
If b is positive then there is no solution for the function = 0. This is because $$(x+a)^2$$ does not intersect the x axis but touches it at (-a,0). Any increment to such a function will not have real roots.
You can prove this using the discriminant of the quadratic $$b^2 - 4ac$$.
Edit:
$$-3x^2+9x+\frac{1}{4}$$
1. Take out the -3 for all the expressions
$$ax^2 + bx + c = a(x^2 + \frac{b}{a} + \frac{c}{a})$$
2. Now complete the square within the brackets remembering that the co-efficient of x, b is now actually $$\frac{b}{a}$$ and becomes $$\frac{b}{2a}$$.
$$ax^2 + bx + c = a((x + \frac{b}{2a})^2 +\frac{c}{a}) - \left[\frac{b}{2a}\right]^2$$
3. You can now muliply the constant out to get.
$$ax^2 + bx + c =a(x + \frac{b}{2a})^2 - \left[\frac{b}{2a}\right]^2 + c$$
$$x^2-6x-2$$ You will be able to do now.
Last edited: Jan 3, 2009 | 2017-04-28T04:37:50 | {
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http://math.stackexchange.com/questions/261825/combination-with-repetition | Combination with repetition
Suppose a donuts shop has $20$ varieties of donuts. how many ways are there to choose at least two kinds of donuts in dozen donuts ?
this is combination with repetition question and we have total of $C\left(12+20-1,12\right)$ cases and we want to subtract it from the cases if we have one kind of donuts so this is $20$ so answer will be $C\left(31,12\right)-20$.
Yes, it’s fine. (But you’re not subtracting $\binom{31}{12}$ from $20$; you’re subtracting $20$ from $\binom{31}{12}$. Subtract $a$ from $b$ means $b-a$.) – Brian M. Scott Dec 19 '12 at 1:58 | 2013-05-20T13:26:40 | {
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https://dsp.stackexchange.com/questions/74060/different-values-when-calculating-signals-energy-by-hand-vs-matlab | # Different values when calculating signal's energy by hand vs MATLAB
Given the following signal:
$$x(n) = \left\{ \begin{array}{ll} u(0.01n-0.025)2.3704e^{(-0.287682n)}, & n\ge 0 \\ u(-(0.01n+0.025))2.3704e^{(0.287682n)}, & n< 0 \\ \end{array} \right.$$
I'm finding the energy using the summation formula:
$$E_{x(n)}=\sum_{n=-\infty}^\infty |x(n)|^2=\sum_{n=0}^\infty(2.3704e^{(-0.287682n)}u(0.01n-0.025))^2+\sum_{n=-\infty}^0 (2.3704e^{(0.287682n)}u(-(0.01n+0.025)))^2 =\sum_{n=0}^\infty(2.3704e^{(-0.287682n)}u(0.01n-0.025))^2+\sum_{n=0}^\infty(2.3704e^{(-0.287682n)}u(-(0.01n+0.025)))^2 = 2\times2.3704^2\times \sum_{n=0}^\infty (e^{(-0.287682\times 2)}u(0.01n-0.025))^n$$
So due to the sampling frequency and other information given about the problem, the sampling doesn't start until $$n=3$$ so I adjusted the limits of summation and removed the u(n) function:
$$=2\times2.3704^2\times \sum_{n=3}^\infty (e^{(-0.287682\times 2)})^n$$
And then adjusted the series to begin at $$n=1$$ so I could use the summation formula:
$$=2\times2.3704^2\times(\sum_{n=1}^\infty (e^{(-0.287682\times 2)})^n - (e^{(-0.287682\times 2)})^1 - (e^{(-0.287682\times 2)})^2)$$
So the series converges since the common ratio $$<1$$ and the sum of the series is:
$$S_{\infty}=\frac{a_1}{1-r}=\frac{(e^{(-0.287682\times 2)})^1}{1-e^{(-0.287682\times 2)}}=1.2857$$
Replacing the number above in the energy summation equation, I get:
$$E_{x(n)}=2\times 2.3704^2\times(1.2857-0.5625-0.3164)=4.5715$$
However, when I use MATLAB to find the energy of the signal using the following code:
E = 0
for i=1:fs+1
E = E + abs(x(i))^2
end
where fs is the sampling frequency, the code yields the following result:
$$E=2.2858$$
which doesn't match my mathematical method. However, I noticed that the answer that MATLAB gives me is half the answer of what I got by hand. I also know with certainty that the MATLAB answer is correct.
So how do I go from $$4.5715$$ to $$2.2858$$? What's the step/reasoning that I'm missing?
Thank you!
• Why do you calculate signal energy of fs+1 points? Shouldn't it be as long as possible? – ZR Han Mar 26 at 6:04
• @ZRHan because x() is a vector and vectors start their indices at 1 rather than 0 in MATLAB; x() is also the time-index vector in terms of n where n includes 0 so the +1 in the fs+1 line is to compensate for that – Kevin KZ Mar 26 at 15:26
• @ZRHan nvm, I just realized my error and what you mean. Thank you! – Kevin KZ Mar 26 at 15:34
Since I don’t know rest of the given code, I’ve decided to calculate the energy other way round.
The MATLAB code for calculating the given function’s energy is given below:
syms n; % Symbol assignment for summation
Energy_1 = symsum(power((2.3704 * exp(0.287682 * n)), 2), n, -Inf, -3); % Energy for the left-hand side of the sample domain
Energy_2 = symsum(power((2.3704 * exp(-0.287682 * n)), 2), n, 3, Inf ); % Energy for the right-hand side of the sample domain
Energy = Energy_1 + Energy_2 % Overall energy
According to this code above, the resultant overall energy becomes $$E = 4.5715$$.
As the number of samples increases, the result gets accurate. Also, the coefficient in the exponential’s power affects the accuracy of the result too, but this is not the main case.
• Ok I'm dumb I just figured out why; an issue with my code. The x(i) vector in my code contains the samples from -inf to +inf but i goes from 1 to half the samples, so it's only summing up half the samples, which obviously gives half the energy. I feel so stupid rn haha thank you for your help – Kevin KZ Mar 26 at 15:33
• @Kevin KZ That is okay, no problem. – Karakoncolos Mar 26 at 16:18 | 2021-07-27T10:58:37 | {
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https://math.stackexchange.com/questions/1079464/difference-of-elements-from-measurable-set-contains-open-interval | # Difference of elements from measurable set contains open interval
Let $A\subset\mathbb{R}$ be a measurable set s.t $,m(A)>0$. Prove that the set $$B=\{x-y\mid x,y\in A\}$$contains nonempty open interval around 0.
I thought to take an interval in $A$, $I=(x-\frac{\epsilon}{2},x+\frac{\epsilon}{2})\subset A$ and hence taking $y$ values from I we get an epsilon - neighborhood of $0$ but I'm quite not sure that I can assume the existence of such I.
How can I prove the existence of I?
• No, you cannot. There are nowhere dense measurable sets of positive measure. – Andrés E. Caicedo Dec 24 '14 at 4:09
• FIne. so can you please give a hint for alternative approach? – user65985 Dec 24 '14 at 4:10
Here is an interesting proof I came across sometime ago. It begins with a Lemma:
Let $$F$$ be a compact set and $$U$$ open such that $$F \subset U$$. Then $$\exists$$ an open neighbourhood $$V$$ of $$0$$ such that $$V+F \subset U$$
Proof: For each $$x\in F, \exists \epsilon > 0$$ such that $$(x-\epsilon,x+\epsilon) \subset U$$. Let $$V_x = (-\epsilon/2,\epsilon/2)$$, then $$F \subset \bigcup_{x\in F} x+V_x$$ Since $$F$$ is compact, $$\exists x_1,x_2,\ldots, x_n\in F$$ such that $$F \subset \bigcup_{i=1}^n x_i + V_{x_i}$$ Take $$V := \bigcap_{i=1}^n V_{x_i}$$ Then $$V+F \subset \cup_{i=1}^n V+(x_i + V_{x_i}) \subset \cup_{i=1}^n x_i+(V_{x_i}+V_{x_i}) \subset U$$
Now we may prove the result:
1. Since $$A$$ is measurable, and $$m(A) > 0$$, there is a compact set $$F \subset A$$ such that $$m(F) > 0$$. So, we may assume that $$A$$ itself is compact.
2. Since $$A$$ is measurable and compact, $$m(A) < \infty$$ and $$\exists U$$ open such that $$A\subset U$$ and $$m(U) < 2m(A) \qquad\qquad \text{(1)}$$
3. By the Lemma, $$\exists V$$ an open neighbourhood of $$0$$ such that $$V+A \subset U$$. Now, we claim that $$V \subset A-A$$. If $$v\in V$$, then it suffices to prove that $$(v+A)\cap A \neq \emptyset$$ (why?), so suppose $$v+A\cap A = \emptyset$$, then since $$v+A \subset U$$, we have $$m(U) \geq m(v+A) + m(A) \geq 2m(A)$$ This contradicts (1) and we are done.
HINT: Use the fact that for any $\epsilon>0$ there are a compact set $K\subseteq A$ and an open set $U\supseteq A$ such that $m(U)-m(K)<\epsilon$. Let
$$d=\inf\{|x-y|:x\in K\text{ and }y\in\Bbb R\setminus U\}\;.$$
• Why is $d>0$?
Fix $\delta\in(0,d)$.
• Why is $K+(-\delta,\delta)\subseteq U$?
Suppose that $|r|<\delta$ and $K\cap(K+r)=\varnothing$.
• Calculate $m\big(K\cup(K+r)\big)$ in terms of $m(K)$.
• Use the fact that $K\cup(K+r)\subseteq U$ to get an upper bound on $m\big(K\cup(K+r)\big)$ in terms of $m(K)$ and $\epsilon$.
• Derive a contradiction by choosing $\epsilon$ sufficiently small.
Another approach: WLOG, take $$A$$ to have finite measure. Let $$\mathcal C$$ be a covering of $$A$$ in disjoint open intervals such that $$\left| \bigcup_{I \in \mathcal C} I\right| < \frac 43 |A|$$ deduce that there exists an $$I \in \mathcal C$$ such that $$|A \cap I| \geq \frac 34 |I|$$. Denote $$A' = A \cap I$$. Note that $$A'$$ is bounded. Let $$I = (a,b)$$.
Now, suppose that $$d \in \Bbb R$$ is such that $$(d+A') \cap A' = \emptyset$$. Then $$2|A'| = |d+A'| + |A'| = |(d+A')\cup A'|$$ If $$d > 0$$, then $$(d+I)\cup I \subset (a,d+b)$$.
If $$d < 0$$, then $$(d+I) \cup I \subset (d+a,b)$$.
In either case, we have $$|(d + I) \cup I| \leq |d| + |I|$$. That is, for any such $$d \in \Bbb R$$, we have $$2|A'| \leq |(d+I) \cup I| \leq |d| + |I|$$ And if $$(d+A') \cap A' = \emptyset$$, then $$|(d+A') \cup A'| = 2|A'|$$.
So $$(d+A') \cap A' = \emptyset$$ implies that $$\frac 32|I| \leq 2|A'| \leq |d| + |I| \implies |d| \geq \frac 12|I|.$$ Contrapositively: if $$|d| < \frac 12 |I|$$, then $$(d + A') \cap A' \neq \emptyset$$.
That is, every $$d \in (-|I|/2,|I|/2)$$ can be written as $$x-y$$ for $$x,y \in A' \subset A$$.
• Hi why is $2|A|\leq |d|+|I|$ – Jhon Doe Sep 12 '19 at 6:10
• @JhonDoe I fixed a few typos and added some explanation; see my latest edit. – Ben Grossmann Sep 12 '19 at 8:05
• Also why is $|(d+I) \cup I| \leq 2|A'| \leq |d| + |I|$. Isn't I a super set of A'. Thanks – Jhon Doe Sep 12 '19 at 8:29
• @JhonDoe well spotted; that was another silly mistake – Ben Grossmann Sep 12 '19 at 10:10 | 2021-06-22T18:12:54 | {
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https://math.stackexchange.com/questions/2113686/probability-of-at-least-one-boolean-random-variable-being-true-in-a-network-of-p | Probability of at least one boolean random variable being true in a network of positively-correlated boolean variables
My question
I have a set of $N$ random boolean variables $X_1, \ldots, X_N$ (each can be $1$ or $0$). For every $i \in [1, N]$, I know that
$$P(X_i = 1) = p^*$$
Now, I know that the variables are positively correlated, i.e., for every $i, j \in [1, N]$ I have:
$$P(X_i = 1 | X_j = 1) \geq p^*$$ $$P(X_i = 1 | X_j = 0) \leq p^*$$
(but I can't find an expression for those conditional probabilities)
Is the probability of at least one variable being $1$ larger or smaller than if they were independent?
The extreme case
Consider the case where all the variables are completely correlated, so that for every $i, j \in [1, N]$
$$P(X_i = 1 | X_j = 1) = 1$$ $$P(X_i = 1 | X_j = 0) = 0$$
Then the probability $p$ of at least one random variable being $1$ is $p = p^*$ (either all of them are $1$ or none is).
While in the case where they are completely independent, i.e.,
$$P(X_i = 1 | X_j = 1) = p^*$$ $$P(X_i = 1 | X_j = 0) = p^*$$
I have
$$p = 1 - (1 - p^*)^N$$
and since $(1 - p^*) < 1 \implies (1 - p^*)^N < (1 - p^*)$ I have
$$p > p^*$$
So in the limiting case I know that being positively correlated makes the probability of at least one variable being $1$ smaller. Is this true for any positive correlation?
Is the probability of at least one variable being $1$ larger or smaller than if they were independent?
NB: For convenience I write $p$ instead of your $p^*$. Now the probability in question is $1-P(X_1=0,...,X_N=0)$, which equals $1-(1-p)^N$ if all the $X_i$ are mutually independent.
Theorem: Suppose $(X_1,...,X_N)$ is distributed on $\{0,1\}^N$ such that $P(X_i=1)=p\in(0,1)$ for all $1\le i\le N$, and $P(X_i=1\mid X_j=1)\ge p$ for all $1\le i\lt j\le N$. If $N=2$, then
$$1-P(X_1=0,...,X_N=0) \le 1-(1-p)^N$$ but if $N\ge 3$, then the preceding inequality holds for some distributions and fails for others.
Proof:
Note that because $P(X_i=1\mid X_j=1)\ge p$, the correlation coefficient between any two distinct $X_i$ and $X_j$ is nonnegative: \begin{align}\rho_{ij} &=\frac{E(X_iX_j)-(EX_i)(EX_j)}{E(X_iX_i)-(EX_i)(EX_i)}=\rho_{ji}\\ \\ &= \frac{E(X_iX_j)-p^2}{p-p^2}\\ &\ge 0 \end{align} because $$EX_iX_j = P(X_i=1,X_j=1) = P(X_i=1\mid X_j=1)P(X_j=1)=P(X_i=1\mid X_j=1)\cdot p\ge p^2.$$
Notation: Let $P(*)$ denote the sum of all $2^N$ joint probabilities. Let $P(*1_i)$ denote the sum of just those joint probabilities that have a $1$ in the $i$th position. Similarly, let $P(*1_i0_j)$ denote the sum of just those joint probabilities that have a $1$ in the $i$th position and a $0$ in the $j$th position, etc. We have the following general formulas: \begin{align}1 &= P(*)\tag{1}\\ p &= P(*1_i)=P(*1_i0_j)+P(*1_i1_j)\tag{2}\\ P(*1_i1_j)&=E(X_iX_j)= p^2 + (p-p^2)\rho_{ij}\tag{3}\\ P(*1_i0_j)&= 1-P(*1_i1_j)=(p-p^2)(1-\rho_{ij})\tag{4}. \end{align}
For convenience we also write $P(x_1...x_N)$ to denote $P(X_1=x_1,...X_N=x_N)$.
Case $N=2$
We show that it always holds that $1-P(00)\le [1-P(00)]_\text{independent}$: \begin{align}1 &= P(*) = P(00)+P(11)+P(01)+P(10)\\ 1-P(00)&=P(11)+P(01)+P(10)\\ &= [p^2 + (p-p^2)\rho_{12}]+[(p-p^2)(1-\rho_{21})]+[(p-p^2)(1-\rho_{12})]\\ &= 2p-p^2-(p-p^2)\rho_{12}\\ &\color{blue}{\le} 2p-p^2=1-(1-p)^2=[1-P(00)]_\text{independent} \end{align} because $p-p^2\ge 0$ and $\rho_{12}\ge 0$.
Case $N\ge 3$
We show the proof for $N=3$ (the higher-dimensional cases being similar), that it is not generally true that $1-P(000)\le [1-P(000)]_\text{independent}$; that is, $(X_1,X_2,X_3)$ may be jointly distributed such that $1-P(000)> 1-(1-p)^3$:
\begin{align}1 = P(*)&= P(000) + P(111) + P(*1_10_2)+P(*1_20_3)+P(*1_30_1)\\ 1-P(000)&=P(111) + P(*1_10_2)+P(*1_20_3)+P(*1_30_1)\\ &=P(111)+[(p-p^2)(1-\rho_{12})]+[(p-p^2)(1-\rho_{23})]+[(p-p^2)(1-\rho_{31})]\\ &=P(111)+(p-p^2)(3-\rho_{12}-\rho_{23}-\rho_{31}) \end{align} It suffices to take $p=\frac{1}{2}$ and $\rho_{12}=\rho_{23}=\rho_{31}(=\rho\text{, say)},$ in which case \begin{align}1-P(000)&=P(111)+\frac{3}{4}(1-\rho). \end{align} Thus, the proof will be accomplished if we can find a joint distribution such that $$P(111)+\frac{3}{4}(1-\rho)>1-(1-p)^3=\frac{7}{8}$$ i.e., such that $$P(111)>\frac{1}{8}+ \frac{3}{4}\rho.\tag{5}$$ Now from (3) and (4) we can deduce the following: \begin{align}P(110)=P(011)=P(101)&=\frac{1}{4}(1+\rho)-P(111)\tag{6a}\\ P(001)=P(101)=P(100)&=P(111)-\frac{1}{2}\rho.\tag{6b} \end{align} Because each $P(x_1x_2x_3)$ must lie in the unit interval, we therefore need to find a value for $P(111)$ that satisfies, in addition to (5), also the following: \begin{align}\frac{1}{2}\rho\le P(111)\le \frac{1}{4}(1+\rho). \tag{7}\end{align} There will be solutions to both (5) and (7) iff \begin{align}\frac{1}{8}+ \frac{3}{4}\rho < \frac{1}{4}(1+\rho)\end{align} i.e., iff $\rho <\frac{1}{4}.$
Hence, for $N=3$, any joint distribution with $p=\frac{1}{2}$, $\ \ 0<\rho_{12}=\rho_{23}=\rho_{31}=\rho<\frac{1}{4}$, and $\frac{1}{2}\rho\le P(111)\le \frac{1}{4}(1+\rho)$ will be an example such that $1-P(000)> [1-P(000)]_\text{independent}$.
As an explicit example, we may take $p=\frac{1}{2}$, $\rho=\frac{1}{8}$, and $P(111)=\frac{1}{4}$. Then $$P(110)=P(011)=P(101)=\frac{1}{32},$$ $$P(001)=P(101)=P(100)=\frac{3}{16},$$ and $$1-P(000)=\frac{29}{32}>1-(1-\frac{1}{2})^3=\frac{7}{8}.$$ | 2019-12-14T13:25:51 | {
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https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_11 | # 2007 AIME I Problems/Problem 11
## Problem
For each positive integer $p$, let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$. For example, $b(6) = 2$ and $b(23) = 5$. If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000.
## Solution 1
$\left(k- \frac 12\right)^2=k^2-k+\frac 14$ and $\left(k+ \frac 12\right)^2=k^2+k+ \frac 14$. Therefore $b(p)=k$ if and only if $p$ is in this range, or $k^2-k. There are $2k$ numbers in this range, so the sum of $b(p)$ over this range is $(2k)k=2k^2$. $44<\sqrt{2007}<45$, so all numbers $1$ to $44$ have their full range. Summing this up with the formula for the sum of the first $n$ squares ($\frac{n(n+1)(2n+1)}{6}$), we get $\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44+1)}{6}=58740$. We need only consider the $740$ because we are working with modulo $1000$.
Now consider the range of numbers such that $b(p)=45$. These numbers are $\left\lceil\frac{44^2 + 45^2}{2}\right\rceil = 1981$ to $2007$. There are $2007 - 1981 + 1 = 27$ (1 to be inclusive) of them. $27*45=1215$, and $215+740= \boxed{955}$, the answer.
## Solution 2
Let $p$ be in the range of $a^2 \le p < (a+1)^2$. Then, we need to find the point where the value of $b(p)$ flips from $k$ to $k+1$. This will happen when $p$ exceeds $(a+\frac{1}{2})^2$ or $a(a+1)+\frac{1}{4}$. Thus, if $a^2 \le p \le a(a+1)$ then $b(p)=a$. For $a(a+1) < p < (a+1)^2$, then $b(p)=a+1$. There are $a+1$ terms in the first set of $p$, and $a$ terms in the second set. Thus, the sum of $b(p)$ from $a^2 \le p <(a+1)^2$ is $2a(a+1)$ or $4\cdot\binom{a+1}{2}$. For the time being, consider that $S = \sum_{p=1}^{44^2-1} b(p)$. Then, the sum of the values of $b(p)$ is $4\binom{2}{2}+4\binom{3}{2}+\cdots +4\binom{44}{2}=4\left(\binom{2}{2}+\binom{3}{2}+\cdots +\binom{44}{2}\right)$. We can collapse this to $4\binom{45}{3}=56760$. Now, we have to consider $p$ from $44^2 \le p < 2007$. Considering $p$ from just $44^2 \le p \le 1980$, we see that all of these values have $b(p)=44$. Because there are $45$ values of $p$ in that range, the sum of $b(p)$ in that range is $45\cdot44=1980$. Adding this to $56760$ we get $58740$ or $740$ mod $1000$. Now, take the range $1980 < p \le 2007$. There are $27$ values of $p$ in this range, and each has $b(p)=45$. Thus, that contributes $27*45=1215$ or $215$ to the sum. Finally, adding $740$ and $215$ we get $740+215=\boxed{955}$.
~firebolt360
## See also
2007 AIME I (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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https://math.stackexchange.com/questions/2970045/problem-on-numbered-books-order-in-a-shelf | Problem on numbered books order in a shelf
Problem: We have 40 books on a shelf randomly arranged. Three books are a series and have numbers - 1,2,3. Need to find the probability that they'll be arranged at ascending order, like 1 comes earlier than 2, and 2 earlier than 3, but they need not to be right after each other. Like this case is possible (1.....2......3)
Now I managed to find the probability when 123 come right after each other. It is about $$0.00064$$. But I can't find a strategy to solve the above mentioned one. Help would be very appreciated.
• Please note that --ignoring the spaces in between them-- there are only six ways to order these three books: (123), (132), (213), (231), (312) and (321). Since the books are randomly arranged, each of this six orderings must have the same probability. Which is therefore $1/6$. So that is your answer. – M. Wind Oct 25 '18 at 4:37
• @M.Wind You should post your comment as an answer. – N. F. Taussig Oct 25 '18 at 8:49
Note that the symmetry argument of M. Wind produces an elegant solution.
Here, I present a solution "on combinatorical foot":
• First, arrange the $$37$$ other books: $$\color{blue}{37!}$$
• These $$37$$ books offer $$38$$ "slots" where to put the $$3$$ books in the given specific order: $$\color{green}{38}$$
• In each "slot" $$s_i$$ $$(i=1,\ldots, \color{green}{38})$$ you can put $$s_i =0,1,2,3$$ books, such that $$s_1 + \cdots +s_{\color{green}{38}} = \color{orange}{3}$$. As the order of the books is fixed, you get as the number of ways (using the formula for combinations with repetitions): $$\binom{\color{green}{n}+\color{orange}{k}-1}{\color{orange}{k}}=\color{blue}{\binom{40}{3}}$$
It follows: $$P(\text{books of series appear in ascending order}) = \frac{\color{blue}{37!\cdot \binom{40}{3}}}{40!} = \frac{1}{6}$$
• You made a computational mistake. Note that your $n = 38$ and your $k = 3$, so $\binom{n + k - 1}{k} = \binom{38 + 3 - 1}{3} = \binom{40}{3}$. The answer reduces to $1/6$, which can be more easily obtained by using a symmetry argument. – N. F. Taussig Oct 25 '18 at 8:48
• @N.F.Taussig: OOOps. Thanks. Sometimes adding and subtracting isn't that easy ;-) – trancelocation Oct 25 '18 at 8:56
Please note that --ignoring the spaces in between them-- there are only six ways to order these three books: (123), (132), (213), (231), (312) and (321). Since the books are randomly arranged, each of this six orderings must have the same probability. Which is therefore 1/6. So that is your answer. | 2020-04-01T11:42:40 | {
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http://mathhelpforum.com/pre-calculus/182281-inequalities.html | # Math Help - Inequalities
1. ## Inequalities
First part of the question says: Solve the inequality $\frac{(x+1)(4-x)}{(3x+1)^2}\geqslant 0$
And after solving, the answer is $-1\leqslant x\leqslant 4, x\neq \frac{1}{3}$
Then the next part of the question says: Hence deduce the solution to the inequality $\frac{(\sqrt{x}+1)(4-\sqrt{x})}{(3\sqrt{x}+1)^2}\geqslant 0$
On comparing, i replaced x in the answer to the first part with $\sqrt{x}$
so, $-1\leqslant \sqrt{x}\leqslant 4, \sqrt{x}\neq -\frac{1}{3}$
$-1\leqslant \sqrt{x}$ and $\sqrt{x}\leqslant 4$
$x\geqslant 0$ and $x\leqslant 16$
therefore, $0\leqslant x\leqslant 16$
Yes, that's right. The answer is correct but shouldn't the answer $x\neq -\frac{1}{3}$ also be dealt with?
So shouldn't the right answer be... $0\leqslant x\leqslant 16, x\neq \frac{1}{9}$ ? why is this not the answer? Please advise, thanks
2. The difficulty is that your solution to the original inequality is wrong. There is no reason to say $x\ne \frac{1}{3}$. It is $x= -\frac{1}{3}$ that would cause the left side to not exist.
3. sorry that was a typo, so why shouldn't the solution include $\sqrt{x}\neq-\frac{1}{3}$
$x\neq \frac{1}{9}$
4. $\sqrt{x}$ is never negative so there no need to exclude -1/3. And, if x= 1/9, then [tex]\sqrt{x}= 1/3[/itex], not -1/3.
5. Problem 1
$\frac{(x+1)(4-x)}{(3x+1)^2} \geq 0$
Let's change the equation up a bit to really look at it. All we'll do is pull a -1 out of the $(4-x)$ to give $-(x-4)$, then rearrange it:
$\frac{-(x+1)(x-4)}{(3x+1)^2} \geq 0$
So this is what we already know by looking at it:
1. We have a negative degree two polynomial expression on top
2. This expression has two real 0s so at some point x will be > 0
2. We can easily figure out our zeros now for the equation (-1 and 4)
3. Also it looks like if $x = -\frac{1}{3}$ because we cannot divide by 0, there would be no solution.
$-1 \leq x \leq 4$ and $x \neq -\frac{1}{3}$
Note another way to write this solution:
$-1 \leq x < -\frac{1}{3}$ and $-\frac{1}{3} < x \leq 4$
Problem 2
$\frac{(\sqrt{x}+1)(4-\sqrt{x})}{(3\sqrt{x}+1)^2} \geq 0$
$\frac{-(\sqrt{x}+1)(\sqrt{x}-4)}{(3\sqrt{x}+1)^2} \geq 0$
It's going to be mostly the same except for two differences:
1. Solving for our roots won't be as easy, for example:
$(\sqrt{x} - 4) = 0$
$\sqrt{x} = 4$
$x = 16$
$(\sqrt{x}+1) = 0$
$\sqrt{x} = -1$
There is no solution here, the sqrt(x) will never be negative.
No solution, so only 16 for a root.
So that was a little different but not too bad.
2. There will never be a real value for x that would create a 0 in the bottom of your expression on the right
For real numbers in x: $(3\sqrt{x}+1)^2 \neq 0$
So we don't have to worry about a "no solution" problem on the bottom.
And we understand that sqrt{x} will always remain non-negative. To explain this better, if we put a negative number in for x, we get a complex number in return. If we put a 0 in for x, we get a 0. And if we put a positive number in for x, we get a positive in return. Never a negative result.
We can form our answer now to be $\sqrt{x} \leq 4$
Or in other words, because $\sqrt{x}$ is non-negative: $0 \leq \sqrt{x} \leq 4$
And thusly we have our solution: $0 \leq x \leq 16$
Good luck you were on the right track it just looks like some typos threw you off.
6. $\displaystyle \frac{(x + 1)(4 - x)}{(3x + 1)^2} \geq 0$
You first need to note that $\displaystyle x \neq - \frac{1}{3}$, because that would give a zero denominator.
Since the denominator is always nonnegative, you can multiply both sides by the denominator without changing the inequality sign
$\displaystyle (x + 1)(4 - x) \geq 0$
Quadratic inequalities are nearly always easiest solved by completing the square.
\displaystyle \begin{align*} (x + 1)(4 - x) &\geq 0 \\ -x^2 + 3x + 4 &\geq 0 \\ -(x^2 - 3x - 4) &\geq 0 \\ -\left[x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - 4\right] &\geq 0 \\ -\left[\left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{16}{4}\right] &\geq 0 \\ -\left[\left(x - \frac{3}{2}\right)^2 - \frac{25}{4}\right] &\geq 0 \\ -\left(x - \frac{3}{2}\right)^2 + \frac{25}{4} &\geq 0 \\ -\left(x - \frac{3}{2}\right)^2 &\geq -\frac{25}{4} \\ \left(x - \frac{3}{2}\right)^2 &\leq \frac{25}{4} \\ \left|x - \frac{3}{2}\right|^2 &\leq \frac{5}{2} \\ -\frac{5}{2} \leq x - \frac{3}{2} &\leq \frac{5}{2} \\ -1 \leq x &\leq 4\end{align*}
Now, remembering that $\displaystyle x \neq -\frac{1}{3}$, that means our solution is
$\displaystyle x \in \left[-1, -\frac{1}{3}\right) \cup \left(-\frac{1}{3}, 4\right]$
7. Alternatively,
the fraction is undefined for $x=-\frac{1}{3}$
For all other x, the denominator is positive.
Then for $x\ne\ -\frac{1}{3}$
the solution reduces to $(x+1)(4-x)\ge \0$
which will be true if both factors are negative or both are positive, or either is zero.
Both are negative if $x< -1$ and $x>4$ at the same time, which is not possible.
Both are positive if $-1< x< 4$
Therefore, both factors cannot be negative and so for $x\ne\ -\frac{1}{3}$
$-1\le\ x\le\ 4$ | 2015-03-02T01:38:01 | {
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Oct 22, 2017 · I am using following MATLAB code for implementing 1D diffusion equation along a rod with implicit finite difference method. In this method the nonlinear iterations is only to execute on the coarse grid and the fine-grid solution can be obtained in a single linear step. Code Group 2: Transient diffusion - Stability and Accuracy. m containing a Matlab program to solve the advection diffusion equation in a 2D channel flow with a parabolic velocity distribution (laminar flow). The diffusion equation is simulated using finite differencing methods (both implicit and explicit) in both 1D and 2D domains. MATLAB Central contributions by Mani Mani. In order to help better understand this phenomenon, model systems are needed where the anomalous subdiffusion of the lipid bilayer can be tuned and quantified. In statistics and probability theory, the Gaussian distribution is a continuous distribution that gives a good description of data that cluster around a mean. Spectral methods in Matlab, L. At each time. The Gaussian functions are thus those functions whose logarithm is a concave quadratic function. Johnson 1 NeuroInformatics Center, University of Oregon 2 SCI Institute, University of Utah ABSTRACT We propose a novel difference metric, called the graph diffusion dis-. They would run more quickly if they were coded up in C or fortran. This plugin implement the anisotropic diffusion filter in 2D. The reconstructed 2D image from the analytical 2D Shepp-Logan phantom in the Fourier domain. 2d Heat Equation Using Finite Difference Method With Steady. To add a directory to the MATLAB path, either use the pathtool on the command line, or select "Set Path" from the menus in the main MATLAB window. Heat Conduction in Multidomain Geometry with Nonuniform Heat Flux. The diffusion equation is simulated using finite differencing methods (both implicit and explicit) in both 1D and 2D domains. In particular, when solving PDEs with 3 or more spatial dimensions (i. MATLAB Matlab code for 2D inverse Fourier transforms. Louise Olsen-Kettle The University of Queensland 9. clc clear K=1. 2D heat diffusion with convection boundary conditions. x = 0 c Fe t 2 t 1 t 1 t 2 x C t 3 t 3 > t 2 > t 1 t 2 > t 1 α = D. Earlier this year I wrote up a simulation for on-lattice diffusion in MATLAB. 2 Predator-prey A predator population y eats from a prey population x, the most famous predator prey model (Lotka. Dispersivity. This method can also be applied to a 2D situation. Perona and J. Hancock Fall 2006 1 The 1-D Heat Equation 1. In each case, you should select the root k-Wave Toolbox folder to add to the path. It's free to sign up and bid on jobs. , concentration and temperature) vary as two or more independent variables (e. Compare the numerical results with the exact solution. Search for jobs related to Diffusion equation matlab or hire on the world's largest freelancing marketplace with 17m+ jobs. The toolbox renders variety of finite impulse response (FIR) filter kernels in closed form that can be used to approximate numerical derivatives of a given discrete signals and images. The drift-diffusion model of a semiconductor is frequently used to describe semiconductor devices. edu March 31, 2008 1 Introduction On the following pages you find a documentation for the Matlab. I learned much of this from a very accessible paper by Palmer, Huk and Shadlen:. The matrix of higher order can be solved in MATLAB. matlab simulation diffusion. See Finite volume method for two dimensional diffusion problem. 2 2 CC Du txx C (1) into a diffusion equation by eliminating the advection term. We let C(x,y,z,t) be the density (mass per unit volume) of a diffusing substance X, and let E be any small subregion of the region where diffusion is occurring. Solving 2D Convection Diffusion Equation. m ) Try the Tutorials , or execute one of the sample programs in the /path/to/eidors/examples directory (such as compare_2d_algs(1) ). FEM 1D_Transient_Diffusion. It is very hard to pick up someone else's code, so in this post I will walk through the details of the code line-by-line with some examples of how each sect. Face Liveness Detection From a Single Image via Diffusion Speed Model. A threshold function is used to prevent diffusion to happen across edges, and therefore it preserves edges in the image. FEM 2D_Transient_Diffusion. This paper describes a new MATLAB software package of iterative regularization methods and test problems for large-scale linear inverse problems. The convection-diffusion equation describes the flow of heat, particles, or other physical quantities in situations where there is both diffusion and convection or advection. The introduction of a T-dependent diffusion coefficient requires special treatment, best probably in the form of linearization, as explained briefly here. 0 ⋮ Discover what MATLAB. For upwinding, no oscillations appear. EML4143 Heat Transfer 2 For education purposes. kWaveDiffusion is a class definition for the time-domain solution of the diffusion equation or Pennes' bioheat equation in 1D, 2D, and 3D. Rayleigh Benard Convection File. FEM Introduction. Particle Tracking Model for 2D Taylor Dispersion : Here is a script file taylor. Initially, particles are randomly distributed near the center. docx), PDF File (. Anisotropic Diffusion Filtering Matlab Codes Codes and Scripts Downloads Free. Exploring the diffusion equation with Python Twitter @ clasticdetritus @ moscardellil @ CriticalStress_ @ seis_matters @ MirelandMark @ minigeology @ AAPG That paragraph is a…. MATLAB includes a wide variety of functions that can be used to simulate a random walk. (for more information please read the MATLAB documentation) The first will work in the same workspace as can be seen from the workspace, whereas the second will usually only produce an output to the workspace. Create a 2D Cartesian circle or arc: makeCartSphere: Create a 3D Cartesian sphere: makeCircle: Create a binary map of a circle within a 2D grid: makeDisc: Create a binary map of a filled disc within a 2D grid: makeLine: Create a binary map of a straight line within a 2D grid: makeMultiArc: Create a binary map of multiple arcs within a 2D grid. Random Walk: 2D Let the position of a particle at time be denoted by We assume that the particle changes its position according to Sample paths: %random walk here we have to modify our MATLAB code randWalk1D. Patankar, Suhas V. So I figured I'd implement a 2D random walker in MATLAB. It is an easy method to use when you have a hard time solving a differential equation and are interested in approximating the behavior of the equation. edu March 31, 2008 1 Introduction On the following pages you find a documentation for the Matlab. The 1-D Heat Equation 18. Ask Question Asked 2 years, 11 months ago. In the Matlab command window type: >>run /path/to/eidors/startup. It is available as part of Stanford VISTA Lab's open-source and freely distributed mrVista package. , spatial position and time) change. Chapter 2 Unsteady State Molecular Diffusion 2. The development of this matlab toolbox is in its infancy. system of reaction-diffusion equation that arise from the viscous Burgers equation which is 1D NSE without pressure gradient. Following section shows few examples to illustrate the concept. Create a 2D Cartesian circle or arc: makeCartSphere: Create a 3D Cartesian sphere: makeCircle: Create a binary map of a circle within a 2D grid: makeDisc: Create a binary map of a filled disc within a 2D grid: makeLine: Create a binary map of a straight line within a 2D grid: makeMultiArc: Create a binary map of multiple arcs within a 2D grid. They include EULER. m containing a Matlab program to solve the advection diffusion equation in a 2D channel flow with a parabolic velocity distribution (laminar flow). The specific system you will simulate is known as the Gray-Scott reaction-diffusion system. m: Simulating a reaction-diffusion system (Reaction-diffusion Brusselator - 2D) 29/5/2008: bruxRD2D. This lesson covers the most basic model for predicting the frequency distributions and accuracy in a reaction time (RT) experiment, the random walk or diffusion model. The "UNSTEADY_NAVIER_STOKES" script solves the 2D steady Navier-Stokes equations. GRAPH DIFFUSION DISTANCE : A DIFFERENCE MEASURE FOR WEIGHTED GRAPHS BASED ON THE GRAPH LAPLACIAN EXPONENTIAL KERNEL David K. Diffusion in biological membranes is seldom simply Brownian motion; instead, the rate of diffusion is dependent on the timescale of observation and so is often described as anomalous. HEATED_PLATE, a MATLAB program which solves the steady state heat equation in a 2D rectangular region, and is intended as a starting point for a parallel version. Inspired: 2d diffusion simulator with particle track option Discover Live Editor Create scripts with code, output, and formatted text in a single executable document. Bazant) Department of Mathematics, MIT February 1, 2005 History The term "random walk" was originally proposed by Karl Pearson in 19051. Heat Conduction in Multidomain Geometry with Nonuniform Heat Flux. You can cheat and go directly to lecture 19, 20, or 21. Writing for 1D is easier, but in 2D I am finding it difficult to. Here we look at using matlab to obtain such solutions and get results of design interest. It applies the law of diffusion on pixel intensities to smooth textures in an image. Commented: Jonathan Ayala on 14 Nov 2019 1. Youtube introduction; Short summary; Long introduction; Longer introduction; 1. In the dimension I want to scroll through the data is all fixed at the integers and not in between, so the 2D-slices should only be made. Burgers equation in 1d and 2d in matlab Potential flow in matlab Diffusion in 1d and 2d in matlab Example of weno3 lf and weno5 lf scheme for 1d buckey leverett problem in matlab Steady 1 dimensional convection and diffusion in matlab Load ensight gold file format in matlab. simulate diffusion in 3D. Please notice, however, that the method has meanwhile evolved and is now also available in 3D, and in formulations that do not require the initial segmentation to be topologically correct. EML4143 Heat Transfer 2 For education purposes. So Fick’s First Law fundamentally states that random particle movement from Brownian motion leads to the drift or dispersal of particles from regions of high concentration to low concentrations – and that drift rate, or diffusion rate, is proportional to the gradient of density, but. The software package, called IR TOOLS, serves two related purposes: we provide implementations of a range of iterative solvers, including several recently proposed methods that are not available elsewhere, and we provide a set of large-scale test. Stencil figure for the alternating direction implicit method in finite difference equations. Chapter 2 Unsteady State Molecular Diffusion 2. ML_power_law. 1000 seconds, or 17 min. Lab10_3: Diffusion Eq 2D with Source Haroon Stephen. An assortment of diffusion based image filtering methods: 1. Writing for 1D is easier, but in 2D I am finding it difficult to. Results are verified with Abaqus results; arbitrary input geometry, nodal loads, and. using the master equation here we will use sparse matrices in MATLAB for. Diffusion In 1d And 2d File Exchange Matlab Central. how to model a 2D diffusion equation?. Unsteady Convection Diffusion Reaction Problem File. Mathematically, the problem is stated as. Browse other questions tagged matlab simulation diffusion or ask your own question. , 4 or more when including time) Matlab is not something you. rnChemical Equation Expert calculates the mass mole of the compounds of a selected equation. For the latter, probabilistic tractography maps were generated using the FSL/FMRIB’s Diffusion Toolbox (FDT v. Rycroft (and Martin Z. Here, f U,V (U,V) is typically a nonlinear function of concentrations U and V. MATLAB is an interactive system whose basic data type is the array or matrix. 2D Elliptic PDEs The general elliptic problem that is faced in 2D is to solve where Equation (14. 2) is gradient of uin xdirection is gradient of uin ydirection. Each (x(i), y(i)) coordinate corresponds to one z(i). I am trying to run an ANOVA with 5 variables (1 independent and 4 dependent). MATLAB® FEM solver for diffusion and advection-diffusion equations for modeling of heat transport, diffusion of drugs, chemical reactions, mixing etc. 0) following their pipeline with eddy current correction, diffusion parameters, segmentation and fitting the diffusion tensors 2 (Behrens et al. I am currently coding the 2D heat/diffusion equation in matlab but i'm having trouble adding in the source term. Provide your first answer ever to someone else's question. Neglecting unsteady term in the equation. m files to solve PDEs using spectral methods. Anisotropic filters are a class of filter that reduces noise in an image while trying to preserve sharp edges. txt) or read online for free. anisotropic diffusion in matlab (for 2D and 3D image processing) %ANISODIFF2D Conventional anisotropic diffusion% DIFF_IM = ANISODIFF2D(IM, NUM_ITER, DELTA_T, KAPPA, OPTION) perfoms % conventional anisotropic diffusion (Perona & Malik) upon a gray scale% image. Show and discuss the structure of the coefficient matrix for the 2D finite difference model; Demonstrate use of MATLAB codes for the solving the 2D Poisson; Reading. The diffusion coefficient can be calculated from the fitted parameter in diffCoeff and the beam waist of the laser: Or, in MATLAB: w = mean(a(:,2)) D = w^2/(4*diffCoeff(2)) w = 0. Two waves of the infinite wave train are simulated in a domain of length 2. These models can be used to model most semiconductor devices. The budget equation is: Then assume that advection dominates over diffusion (high Peclet number). Computational Mathematics: Models, Methods, and Analysis with MATLAB ® and MPI, Second Edition is an ideal textbook for an undergraduate course taught to mathematics, computer science, and engineering students. Fourier - Free download as PDF File (. Heat Transfer in Block with Cavity. The 2D wave equation Separation of variables Superposition Examples We let u(x,y,t) = deflection of membrane from equilibrium at position (x,y) and time t. (2016) MATLAB: The ultimate tool for teaching process control oriented courses. This reading is certainly of the crash-course variety, so feel free to ask Rob, Hernan, or me any questions. > first I solved the advection-diffusion equation without > including the source term (reaction) and it works fine. The Matlab part was used to realize algorithms. Diffusion coefficient, D D = (1/f)kT f - frictional coefficient k, T, - Boltzman constant, absolute temperature f = 6p h r h - viscosity r - radius of sphere The value for f calculated for a sphere is a minimal value; asymmetric shape of molecule or non-elastic interaction with solvent (e. 303 Linear Partial Differential Equations Matthew J. 000012; den=1. Montecarlo simulation of charge diffusion on a cubic lattice to determine lateral diffusion length as a function of barrier height, assuming thermionic emission over the barrier. Derivation of Diffusion Equation The diffusion equation (5. I have the code which solves the Sel'kov reaction-diffusion in MATLAB with a Crank-Nicholson scheme. I recently begun to learn about basic Finite Volume method, and I am trying to apply the method to solve the following 2D continuity equation on the cartesian grid x with initial condition For simplicity and interest, I take , where is the distance function given by so that all the density is concentrated near the point after sufficiently long. A tutorial 2D MATLAB code for solving elliptic diffusion-type problems, including Poisson's equation on single patch geometries, is presented. 2D heat diffusion with convection boundary conditions. MATLAB allows for two ways of programming: scripts (m-file) or functions. The diffusion equation is simulated using finite differencing methods (both implicit and explicit) in both 1D and 2D domains. Diffusion Equation! Computational Fluid Dynamics! ∂f ∂t +U ∂f ∂x =D ∂2 f ∂x2 We will use the model equation:! Although this equation is much simpler than the full Navier Stokes equations, it has both an advection term and a diffusion term. The process is repeated several times. A quick short form for the diffusion equation is ut = αuxx. Then set diffusion to zero and test a reaction equation. An open source drift diffusion code based in MATLAB for simulating solar cells. This reading is certainly of the crash-course variety, so feel free to ask Rob, Hernan, or me any questions. Diffusion coefficient, D D = (1/f)kT f - frictional coefficient k, T, - Boltzman constant, absolute temperature f = 6p h r h - viscosity r - radius of sphere The value for f calculated for a sphere is a minimal value; asymmetric shape of molecule or non-elastic interaction with solvent (e. FD1D_HEAT_IMPLICIT is a MATLAB program which solves the time-dependent 1D heat equation, using the finite difference method in space, and an implicit version of the method of lines to handle integration in time. Ex Convection Diffusion 2d. 33; % Thermal diffusivity, m^2/s dt = 300; % Timestep x = 0:xstp:xsize; %Creating vector for nodal point positions tlbc = sin. Search for jobs related to Diffusion equation matlab or hire on the world's largest freelancing marketplace with 17m+ jobs. Under ideal assumptions (e. , spatial position and time) change. Learn more about diffusion equation, pde. “Environmental Modeling using MATLAB R ” by Ekkehard Holzbecher is an excellent publication and a novel approach covering the intersection of two important, growing worlds – the world of environmental modeling and of mathematical software. written by Tutorial45. Related MATLAB, Maple, Mathematica Related Threads on 2D diffusion equation, need help for matlab code. This section considers transient heat transfer and converts the partial differential equation to a set of ordinary differential equations, which are solved in MATLAB. Face Liveness Detection From a Single Image via Diffusion Speed Model. 2D Gaussian Fitting in Matlab. Since the flux is a function of radius – r and height – z only (Φ(r,z)), the diffusion equation can be written as:. Introduction: This toolbox will perform Anisotropic Non-Linear Diffusion filtering on a 2D gray/color or 3D image. The main task is to define small lagging between current and voltage. The code needs debugging. This MATLAB code is for two-dimensional elastic solid elements; 3-noded, 4-noded, 6-noded and 8-noded elements are included. Normalization model of attention (Matlab implementation) vsdAlign: 2D image registration software for optical imaging (Matlab implementation) Image registration software (Matlab implementation) Robust anisotropic diffusion software (Matlab implementation) Egomotion algorithms (Matlab implementations) Steerable pyramid (Matlab and C implementations). Generate the diffusion gradient vector field as in Xu and Prince 1998 dgvf_calc is the three dimensional extension of the 2D version described in Equation 12 Xu and Prince 1998,"Snakes, Shapes, and Gradient Vector Flow", IEEE Transactions on Image Processing Vol. Here are 1D, 2D, and 3D models which solve the semiconductor Poisson-Drift-Diffusion equations using finite-differences. Awarded to Mani Mani on 22 Feb 2020. You can visualize matrix data on a rectangular grid using surface plots. Partial Differential Equation Toolbox ™ provides functions for solving structural mechanics, heat transfer, and general partial differential equations (PDEs) using finite element analysis. Back in April, MathWorks released the Jenkins MATLAB Plugin to enable users to run tests using the MATLAB Unit Test framework for both MATLAB and Simulink-based workflows. MATLAB ® provides many techniques for plotting numerical data. Authors: Philip Calado, Piers RF Barnes, Ilario Gelmetti, Mohammed Azzouzi, Benjamin Hilton. Pyhton has some advanteges over Matlab for example indices start from zero, it's free and has clean syntax. Numerical methods for PDE (two quick examples) Discretization: From ODE to PDE For an ODE for u(x) defined on the interval, x ∈ [a, b], and consider a uniform grid with ∆x = (b−a)/N, discretization of x, u, and the derivative(s) of u leads to N equations for ui, i = 0, 1, 2, , N, where ui ≡ u(i∆x) and xi ≡ i∆x. Dispersivity. Both methods are unconditionally stable. This file was created by the Typo3 extension sevenpack version 0. 10 --- Timezone: UTC Creation date: 2020-04-28 Creation time: 20-26-55 --- Number of references 6353 article MR4015293. 2d Heat Equation Using Finite Difference Method With Steady. The famous diffusion equation, also known as the heat equation , reads. The values of parameters such as inversion height (h), wind speed at 10 m (U10), σw and observed concentration. The MATLAB desktop contains a help browser covering both reference and tutorial material. Consistent with a role for these actin-rich structures in signal amplification, microscopic measures of Rac1 activity determined that disruption of actin polymerization by. Python is an object-oriented programming language, and it's a good alternative to Matlab for scientific computing with numpy and matplotlib modules (very easy to install). You can cheat and go directly to lecture 19, 20, or 21. Using Python to Solve Partial Differential Equations This article describes two Python modules for solving partial differential equations (PDEs): PyCC is designed as a Matlab-like environment for writing algorithms for solving PDEs, and SyFi creates matrices based on symbolic mathematics, code generation, and the finite element method. MATLAB Central contributions by Mani Mani. I've used some code from the. Execute the selective _encryption file and make sure both the files are in the current directory at the time of execution. Hancock Fall 2006 1 The 1-D Heat Equation 1. You can also have some fun by starting with an arbitrary shape. What I want to do is to calculate the mean-squared displacement for the particle using the xyz coordinates for all time steps. FD2D_HEAT_STEADY is a MATLAB program which solves the steady state (time independent) heat equation in a 2D rectangular region. This uses fdep() function from matlab central feb 13, 2012. A free alternative to Matlab https. I cleared my basic concepts required for plotting different 2D plots. Diffusion In 1d And 2d File Exchange Matlab Central. m Benjamin Seibold Applied Mathematics Massachusetts Institute of Technology www-math. 8, 2006] In a metal rod with non-uniform temperature, heat (thermal energy) is transferred. Here are 1D, 2D, and 3D models which solve the semiconductor Poisson-Drift-Diffusion equations using finite-differences. Introduction: This toolbox will perform Anisotropic Non-Linear Diffusion filtering on a 2D gray/color or 3D image. If you use Driftfusion please let us know by emailing: p. The material is homogeneous and isotropic. The Matlab programming language was used by numerous researchers to solve the systems of partial differential equations including the Navier Stokes equations both in 2d and 3d configurations. Animated surface plot: adi_2d_neumann_anim. Spectral methods in Matlab, L. MATLAB Central contributions by Suraj Shankar. The traditional method for solving the heat conduction equation numerically is the Crank-Nicolson method. One for selective encryption, second for hundungen(M,N,key0) function and third for key generation. MATLAB allows for two ways of programming: scripts (m-file) or functions. ANALYSIS OF A REACTION-DIFFUSION SYSTEM WITH LOCAL AND NONLOCAL DIFFUSION TERMS Richard D. It's free to sign up and bid on jobs. As matlab programs, would run more quickly if they were compiled using the matlab compiler and then run within matlab. I am currently writing a matlab code for implicit 2d heat conduction using crank-nicolson method with certain Boundary condiitons. The finite difference formulation of this problem is The code is available. A free alternative to Matlab https. Point Jacobi Gauss-Seidel with SOR 5. They would run more quickly if they were coded up in C or fortran. Lecture 06. Authors: Philip Calado, Piers RF Barnes, Ilario Gelmetti, Mohammed Azzouzi, Benjamin Hilton. Karatay and Bayramoglu [19] have extended the Crank-Nicholson difference scheme to solve the time-fractional advection-dispersion equation. 2d diffusion simulation, gui in matlab Search form The following Matlab project contains the source code and Matlab examples used for 2d diffusion simulation, gui. Anisotropic Diffusion Filtering Matlab Codes Codes and Scripts Downloads Free. Lecture 1: Introduction to Random Walks and Diffusion Scribe: Chris H. There is a known solution via Fourier transforms that you can test against. Please log bugs through GitHub. pdf] - Read File Online - Report Abuse A compact and fast Matlab code solving the incompressible. Many explanations for the sigmoid or S-shaped curves that characterize the diffusion of innovations through time have been proposed. For time integration, use something from a library. They would run more quickly if they were coded up in C or fortran. Implicit Explicit Convection Diffusion Equation File. MATLAB is an interactive system whose basic data type is the array or matrix. 2d diffusion equation keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website. Diffusion is the governing process in problems involving flow through porous media, and conduction of heat in solids. A Simple Finite Volume Solver For Matlab File Exchange. MATLAB Learning Modules; Creative Commons License; Child pages. Learn more about 3d, diffusion, discrete, gaussian, convolution, rate, coefficient, blur, blurring, kernel, sigma MATLAB. Compare the numerical results with the exact solution. Simulates diffusion around a film discontinuity, such a cut. Follow 789 views (last 30 days) Charles on 27 Mar 2012. 2D heat diffusion with convection boundary conditions. my Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The diffusion coefficient is D, the decay rate is λ 1, and the saturation density is σ max. zip Many attractive images and life-like structures can be generated using models of physical processes from areas of chemistry and physics. Getting started with MATLAB. In most cases the oscillations are small and the cell Reynolds number is frequently allowed to be higher than 2 with relatively minor effects on the result. | 2020-07-04T04:37:05 | {
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https://math.stackexchange.com/questions/2595665/absolute-value-definition | # Absolute value definition
Is it true that $\dots$ $$\left| y \right| = \begin{cases} y \hspace{1cm} y \geq 0 \\ -y \hspace{0.7cm} y < 0 \end{cases}$$
I'm a little bit confused with the second case, where $|y| = -y$ then $y<0$, for example : $$\left| 2x-4 \right|=-(2x-4)$$ if we assume that $y=2x-4$ then \begin{align*} y&<0 \\ 2x-4&<0 \\ 2x&<4 \\ x&<2 \end{align*} in the other way, we can solve it like this \begin{align*} |y| \geq 0 \\ |2x-4| \geq 0 \\ -(2x-4) \geq 0 \\ 2x-4 \leq 0 \\ x \leq 2 \end{align*}
why is it giving the different answers?
• If you know $x<0$, $|x|>0$. Also, @your 2nd sentence, if $|x|=-x$, you know $x \leq 0$ (not necessarily $x<0$, since $x=0$ does satisfy $|x|=-x$). – Mauve Jan 7 '18 at 15:39
• warning $ax-b<0\iff ax<b$ you wrote $ax<-b$ (think of it as adding $b$ both sides) – zwim Jan 7 '18 at 15:40
• i'm sorry for that @zwim – Hansen Frenico Jan 7 '18 at 15:42
• @Useless many references said like that , images.slideplayer.com/25/7667598/slides/slide_9.jpg – Hansen Frenico Jan 7 '18 at 15:46
• "Is it true that ...". Yes, this is the definition. – user370967 Jan 7 '18 at 15:47
You should train to read formulas without reference to specific variables. The definition $$|x|=\begin{cases} x & x\ge0 \\[4px] -x & x<0 \end{cases}$$ it is meant that
the absolute value of a number is
1. the number itself if it is greater than or equal to $0$,
2. the negative of the number if it is less than $0$.
You should also avoid using a variable with two different meanings in the same statement.
It seems that you want to see when $|2x-4|=-(2x-4)$. According to the definition, this happens if and only if
1. $2x-4<0$, or
2. $2x-4=0$.
Why the second case? Because $0=-0$. On the other hand, if $2x-4>0$, then we cannot have $(2x-4)=-(2x-4)$, because one term is positive and the other one is negative.
One might make the initial definition more symmetric by declaring $$|x|=\begin{cases} x & x>0 \\[4px] 0 & x=0 \\[4px] -x & x<0 \end{cases}$$ but you can also note that $$|x|=\begin{cases} x & x>0 \\[4px] -x & x\le0 \end{cases}$$ would be a completely equivalent definition.
The absolute value make a function simmetrical with respect to the $x$-axis. Then we have pay attention when the function assumes negative value. In your example we have:
$$|2x−4|=\begin{cases}2x-4 \quad\text{if} \quad 2x-4\ge0\implies x\ge2\\ 4-2x\quad \text{if} \quad 2x-4 <0\implies x<2\end{cases}$$
The graphic of your function in fact is:
$\hspace{6cm}$
It is a bit confusing to use the same letter $x$ for your substitution.
Let set $y=2x-4$.
Then $|y|$ is $\begin{cases}+y&=2x-4&\quad\text{when}\quad y\ge0\iff 2x-4\ge 0\iff x\ge 2\\-y&=4-2x&\quad\text{when}\quad y<0\iff 2x-4<0\iff x<2\end{cases}$
Indeed for $x=1$ then $x<2$ then $|y|=|2\times 1-4|=|-2|=2=4-2\times 1$
And for $x=3$ then $x\ge 2$ then $|y|=|2\times 3-4|=|2|=2=2\times 3-4$
• thanks for the suggestion, but why $x<2$ and not $x \leq 2$? $|2 \times 2-4| = 4-2 \times 2$ – Hansen Frenico Jan 7 '18 at 15:55
• Both are valid since $+0=-0=0$. I just copy/pasted you own definition of $|y|$ where you said $|y|=-y$ when $y<0$, but $y\le 0$ also works. – zwim Jan 7 '18 at 16:29
Note that in general the definition is
$$\left| f(x) \right| = \begin{cases} f(x) \hspace{1cm} f(x) \geq 0 \\ -f(x) \hspace{0.7cm} f(x) < 0 \end{cases}$$ | 2019-10-15T18:53:52 | {
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http://math.stackexchange.com/questions/262930/adjoint-operators-and-inner-product-spaces | # Adjoint Operators and Inner Product Spaces
My linear algebra textbook gives the definition of the Adjoint Operator and then says,
You should verify the following properties:
• Additivity: $(S + T)^* = S^* + T^*$
• Conjugate homogeneity: $(aT)^* = \overline{a}\,T^*$
• Adjoint of adjoint: $(T^*)^* = T$
• Identity: $I^* = I$, where $I$ is the identity operator on $V$.
I've stared at the pages for a couple hours now. How do you verify this?
Here's my attempt at a proof for Adjoint of Adjoint: (T*)* = (T*v, w)* = (v, Tw)* = (Tv, w) = T
Is that correct reasoning?
BTW, this is NOT homework. Just reading for pleasure.
Thanks!
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You need to write out each side of the equations using the given definition and then show that equality really does hold. What have you tried so far? – Clive Newstead Dec 20 '12 at 22:47
What should $T^*$ satisfy to be an adjoint? Try to check this for $S^*+T^*$. – Sigur Dec 20 '12 at 22:48
The adjoint is uniquely determined by a certain condition (if you haven't proven uniqueness you should prove that too). All of these properties assert that the adjoint of some operator can be described as some other operator, so what you need to verify is that that other operator satisfies the condition that uniquely determines the adjoint. – Qiaochu Yuan Dec 20 '12 at 22:50
Adjoint of Adjoint: (T*)* = (Tv, w) = (v, Tw)* = (Tv, w) = T – Megan Dec 20 '12 at 22:51
Is what I posted accurate? – Megan Dec 20 '12 at 22:53
The adjoint of a transformation is defined as the unique transformation $T^{*}$ so that $$\langle Tx, y \rangle = \langle x , T^{*}y \rangle$$ for every $x$ and $y$.
So to prove any of your equalities above you simply need to show that the transformation you want to be the adjoint satisfies this property. Generally you can do this using the properties of an inner product.
For example, to prove $(T^{*})^{*} = T$ you need to show that for any $x,y$ $$\langle T^{*}x , y \rangle = \langle x , Ty \rangle.$$ This follows because $\langle T^*x,y\rangle= \overline{\langle y , T^*x \rangle} = \overline{\langle Ty, x \rangle} = \langle x , Ty \rangle$.
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For the $(T^*)^*=T$ problem:
$$\langle (T^*)^*v, w \rangle=\langle v, T^*w \rangle=\langle Tv, w \rangle$$
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I am assuming that you are working in some Hilbert space $\mathbb{H}$. The key fact is that any continuous linear functional $f:\mathbb{H} \to \mathbb{C}$ can be represented by a unique element $\phi \in \mathbb{H}$ in the sense that $f(x) = \langle \phi, x \rangle$ for all $x \in \mathbb{H}$ (and conversely any element $\phi \in \mathbb{H}$ determines a unique continuous linear functional on $\mathbb{H}$). A little work shows that $\|f\| = \|\phi\|$. The big deal above is the uniqueness.
Now suppose $T: \mathbb{H} \to \mathbb{H}$ is a continuous linear operator and $y \in \mathbb{H}$. Then $f_y(x) = \langle y, Tx \rangle$ is a continuous linear functional, and can be represented by some $\phi_{y} \in \mathbb{H}$, ie, $f_y(x) = \langle \phi_{y}, x \rangle$. It is easy to show using uniqueness that $\phi_{\lambda y} = \lambda \phi_y$ and $\phi_{y_1+y_2} = \phi_{y_1}+\phi_{y_2}$, hence the mapping $y \mapsto \phi_y$ is linear. Furthermore, since $|f_y(x)| = |\langle \phi_{y}, x \rangle| \leq \|x\| \|y\| \|T\|$, choosing $x=\frac{\phi_y}{\|\phi_y\|}$ (or zero, if $\phi_y=0$) shows that $\| \phi_y \| \leq \|T \| \|y\|$, hence the mapping $y \mapsto \phi_y$ is bounded, and hence continuous. Instead of writing $y \mapsto \phi_y$, we now use the more usual notation $T^* y = \phi_y$. We have shown that $T^*$ is linear, continuous, and completely defined by the requirement that $\langle T^*y, x \rangle = \langle y, Tx \rangle$ for all $x,y \in \mathbb{H}$.
All other properties follow from this requirement and properties of the inner product.
For example, to show additivity: $\langle (S+T)^*y, x \rangle = \langle y, (S+T)x \rangle = \langle y, Sx \rangle + \langle y, Tx \rangle = \langle S^*y, x \rangle + \langle T^*y, x \rangle = \langle (S^*+T^*)y, x \rangle$, hence $(S+T)^* = S^*+T^*$.
For the adjoint of the adjoint: $\langle (T^*)^*y, x \rangle = \langle y, T^*x \rangle = \overline{\langle T^*x,y \rangle} = \overline{\langle x,Ty \rangle} = \langle Ty,x \rangle$, hence $(T^*)^* = T$.
Now try the rest.
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The adjoint of a transformation is defined as the unique transformation $T^\ast$ so that $$\langle Tx,y\rangle = \langle x,T^\ast y\rangle$$ for every $x$ and $y$. So
$$\langle T^{\ast\ast} x, y\rangle = \langle x, T^\ast y\rangle = \overline{\langle T^\ast y, x\rangle} = \overline{\langle y, Tx\rangle} = \langle Tx,y\rangle.$$
And we'll get that $\langle T^{\ast\ast}x, y\rangle = \langle Tx, y\rangle$ implies that $T^{\ast\ast} = T$.
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This answer is very poorly argumented. – G. Sassatelli May 20 at 18:03 | 2015-08-01T17:03:39 | {
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https://math.stackexchange.com/questions/2642160/general-name-for-a-multidimensional-function-that-maps-each-coordinate-independe | # General name for a multidimensional function that maps each coordinate independently
This is a question about terminology. I hope it is not too silly but I haven't been able to find a clear answer. Basically, is there a more or less standard name for a function from an n-dimensional space into another n-dimensional space that maps each coordinate independently?
In formal terms, let's suppose we have two sets $A$ and $B$ and we can define functions mapping one set to the other $f \colon A \to B$. If, for an integer $n > 1$ we take the Cartesian products $A^n$ and $B^n$, then we can define functions $g \colon A^n \to B^n$. One simple form of such a function would be $$g(a_1, \dots, a_n) = (f_1(a_1), \dots, f_n(a_n)),$$ where each $f_i$ is a function from $A$ to $B$. But of course there will also be more general $g$ functions that don't map each coordinate independently. I'm looking for the English words that express this distinction as in $g$ is a 'whatever' function vs $g$ is a 'non-whatever' function.
I am aware of certain particular cases, like the diagonal vs non-diagonal linear maps in a vector space. But I'm wondering if there are any standard names for this distinction in the general case where we simply have sets, Cartesian products, and functions.
Let $n=2$ for simplicity. Assume you are given two maps $f_i : A_i \to B_i$. Then the function you are referring to is the cartesian product $f_1 \times f_2$.
Here we are thinking of $f_i$ as a subset of $A_i \times B_i$:
Definition A function $f : A\to B$ is subset (also called $f$) of $A\times B$, so that (a) for all $a\in A$, $\{ b\in B : (a, b)\in f\}$ is nonempty, and (b) if $(a, b), (a, b')\in f$, then $b= b'$.
Of course the above implies that it is unambiguous to write, for all $a\in A$, $f(a)$ so that $f(a) \in \{ b\in B : (a, b)\in f\}$. So $(a, f(a) ) \in f$.
From the above definition of functions as subsets, we see that $f_1\times f_2$ is a subset of $(A_1\times B_1) \times (A_2 \times B_2)$ and this latter set is identified as $(A_1 \times A_2) \times (B_1 \times B_2)$.
Claim: $f_1 \times f_2$, treated as a subset of $(A_1 \times A_2) \times (B_1 \times B_2)$, is a function $A_1\times A_2 \to B_1\times B_2$ and $$\tag{1}(f_1\times f_2)(a_1, a_2) = (f_1(a_1), f_2(a_2))$$ for all $a_i \in A_i$.
To prove the claim that $f_1\times f_2$ is a function $A_1\times A_2 \to B_1\times B_2$, let $(a_1, a_2) \in A_1\times A_2$. Then first of all, $$(a_i, f_i(a_i)) \in f_i \subset A_i \times B_i$$ for $i=1,2$. So $$\tag{2}(a_1, a_2, f_1(a_1), f_2(a_2)) \in f_1\times f_2.$$
Secondly, if $$(a_1, a_2, b_1, b_2), (a_1, a_2, b'_1, b'_2) \in f_1\times f_2,$$ then $$(a_i, b_i)\in f_i,\ \ \ (a_i, b'_i)\in f_i.$$ This implies $b_i = b'_i$ since $f_i$ is a function. Thus $(b_1, b_2) = (b'_1, b'_2)$ and so $f_1\times f_2$ defines a function. Now (1) follows from (2).
• This does not really answer the question, which is to find a generic name for functions that can be written as a cartesian product of functions. – J.-E. Pin Feb 9 '18 at 5:01
• @J.-E.Pin : From the question it seems to me that the questioner did not realize that those functions are just the cartesian product (If they realized I guess they would not ask this question). – user99914 Feb 9 '18 at 5:15
• @John, thanks for your answer. I have decided to accept it even if, as J.-E. Pin pointed out above, I was looking for a couple of opposing terms rather than an explanation that the "simple form" I mentioned is also a Cartesian product. But I really appreciate the explanation and it is true that this means that I can make the distinction I was looking for by referring to "n-dimensional functions that are a Cartesian product" vs. those "that are not a Cartesian product". While this is wordier than I wanted, I suppose there is no common simpler way of putting it. – Ángel José Riesgo Feb 14 '18 at 9:28 | 2019-12-15T10:40:23 | {
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http://mauimoldremoval.com/types-of-gtwumr/0edda9-rationalise-the-denominator-of-the-following | Example 4: Suppose that $$x = \frac{{11}}{{4 - \sqrt 5 }}$$. One way to understand the least common denominator is to list all whole numbers that are multiples of the two denominators. 5/6-9√2. Exercise: Calculation of rationalizing the denominator. Rationalize the denominators of the following: Solution: In this case, we will use the following identity to rationalize the denominator: $$\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) = {a^3} + {b^3}$$. The following steps are involved in rationalizing the denominator of rational expression. The sum of three consecutive numbers is 210. Summary When you encounter a fraction that contains a radical in the denominator, you can eliminate the radical by using a process called rationalizing the denominator. &= \frac{{11 + 4\sqrt 7 }}{{ - 3}} \hfill \\ Step 1 : Multiply both numerator and denominator by a radical that will get rid of the radical in the denominator. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. = √7/(√7)2 To do that, we can multiply both the numerator and the denominator by the same root, that will get rid of the root in the denominator. { = - 24\sqrt 2 - 12\sqrt 3 } An Irrational Denominator! In the following video, we show more examples of how to rationalize a denominator using the conjugate. = (√5 − √2)/3 remove root from denominator Hence multiplying and dividing by √7 1/√7 = 1/√7 ×√7/√7 = √7/(√7)2 = √7/7 Ex1.5, 5 Rationalize the denominators of the following: (ii) 1/(√7 1/(√5 + √2) = 1/(√5 + √2) × (√5 − √2)/(√5 − √2) Solution: We rationalize the denominator of x: \begin{align} x &= \frac{{11}}{{4 - \sqrt 5 }} \times \frac{{4 + \sqrt 5 }}{{4 + \sqrt 5 }}\\ &= \frac{{11\left( {4 + \sqrt 5 } \right)}}{{16 - 5}}\\ &= 4 + \sqrt 5 \\ \Rightarrow x - 4 &= \sqrt 5 \end{align}. By using this website, you agree to our Cookie Policy. Oh No! That is, I must find some way to convert the fraction into a form where the denominator has only "rational" (fractional or whole number) values. Ex 1.5, 5 Ask questions, doubts, problems and we will help you. &= 2 - \sqrt 3 \hfill \\ The denominator contains a radical expression, the square root of 2.Eliminate the radical at the bottom by multiplying by itself which is \sqrt 2 since \sqrt 2 \cdot \sqrt 2 = \sqrt 4 = 2.. the smallest positive integer which is divisible by each denominators of these numbers. $\begin{array}{l} 4\sqrt {12} = 4\sqrt {4 \times 3} = 8\sqrt 3 \\ 6\sqrt {32} = 6\sqrt {16 \times 2} = 24\sqrt 2 \\ 3\sqrt {48} = 3\sqrt {16 \times 3} =12\sqrt 3 \end{array}$, $\boxed{\begin{array}{*{20}{l}} To get the "right" answer, I must "rationalize" the denominator. &= \frac{{3 + 2\sqrt 3 }}{{5 - 2\sqrt 3 }} \times \frac{{5 + 2\sqrt 3 }}{{5 + 2\sqrt 3 }} \hfill \\ \end{array}}$. What is the largest of these numbers? Then, simplify the fraction if necessary. The bottom of a fraction is called the denominator. Express each of the following as a rational number with positive denominator. Rationalize the denominators of the following: \end{align} \]. {8\sqrt 3 - 24\sqrt 2 - 12\sqrt 3 } \\ = (√7 + √2)/(7 −4) Related Questions. Problem 52P from Chapter 5.5: This process is called rationalising the denominator. LCD calculator uses two or more fractions, integers or mixed numbers and calculates the least common denominator, i.e. RATIONALISE THE DENOMINATOR OF 1/√7 +√6 - √13 ANSWER IT PLZ... Hisham - the way you have written it there is only one denominator, namely rt7, in which case multiply that fraction top &bottom by rt7 to get (rt7/)7 + rt6 - rt13. &= {\left( 2 \right)^3} - {\left( {\sqrt[3]{7}} \right)^3} \hfill \\ It is 1 square roots of 2. Decimal Representation of Irrational Numbers. Examples of How to Rationalize the Denominator. (iv) 1/(√7 −2) \end{align} \]. ( 5 - 2 ) divide by ( 5 + 3 ) both 5s have a square root sign over them But it is not "simplest form" and so can cost you marks.. And removing them may help you solve an equation, so you should learn how. = 1/√7 ×√7/√7 For example, look at the following equations: Getting rid of the radical in these denominators … This calculator eliminates radicals from a denominator. Find the value to three places of decimals of the following. The denominator here contains a radical, but that radical is part of a larger expression. Examples Rationalize the denominators of the following expressions and simplify if possible. (ii) 1/(√7 −√6) . To make it rational, we will multiply numerator and denominator by $${\sqrt 2 }$$ as follows: Let us take an easy example, $$\frac{1}{{\sqrt 2 }}$$ has an irrational denominator. (iii) 1/(√5 + √2) Let's see how to rationalize other types of irrational expressions. Rationalise the following denominator: 3/√2; To rationalise the denominator of this fraction, we're going to use one fact about roots and one about fractions: If you multiply a root by itself, you are left with the original base. Login to view more pages. Question From class 9 Chapter NUMBER SYSTEM Rationalise the denominator of the following :
BYJU’S online rationalize the denominator calculator tool makes the calculations faster and easier where it displays the result in a fraction of seconds. Consider another example: $$\frac{{2 + \sqrt 7 }}{{2 - \sqrt 7 }}$$. That is what we call Rationalizing the Denominator. . (i) 1/√7 Rationalizing when the denominator is a binomial with at least one radical You must rationalize the denominator of a fraction when it contains a binomial with a radical. Example 1: Rewrite $$\frac{1}{{3 + \sqrt 2 - 3\sqrt 3 }}$$ by rationalizing the denominator: Solution: Here, we have to rationalize the denominator. We need to rationalize i.e. = 1/(√7 −2) × (√7 + 2)/(√7 + 2) \end{align} \], $= \boxed{ - \left( {\frac{{60 + 34\sqrt 2 + 48\sqrt 3 + 18\sqrt 6 }}{{184}}} \right)}$. solution In a case like this one, where the denominator is the sum or difference of two terms, one or both of which is a square root, we can use the conjugate method to rationalize the denominator. &\Rightarrow \left( {2 - \sqrt[3]{7}} \right) \times \left( {4 + 2\sqrt[3]{7} + \sqrt[3]{{49}}} \right) \hfill \\ We can note that the denominator is a surd with three terms. He has been teaching from the past 9 years. 1. Think: So what do we use as the multiplier? ⚡Tip: Take LCM and then apply property, $$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$$. {\left( {x - 4} \right)^2} &= 5 \hfill \\ It is an online mathematical tool specially programmed to find out the least common denominator for fractions with different or unequal denominators. We make use of the second identity above. That is, you have to rationalize the denominator. And now lets rationalize this. Comparing this with the right hand side of the original relation, we have $$\boxed{a = \frac{{27}}{{13}}}$$ and $$\boxed{b = \frac{{16}}{{13}}}$$. nth roots . RATIONALIZE the DENOMINATOR: explanation of terms and step by step guide showing how to rationalize a denominator containing radicals or algebraic expressions containing radicals: square roots, cube roots, . We do it because it may help us to solve an equation easily. For example, we can multiply 1/√2 by √2/√2 to get √2/2 Rationalise the denominators of the following. To use it, replace square root sign ( √ ) with letter r. Example: to rationalize $\frac{\sqrt{2}-\sqrt{3}}{1-\sqrt{2/3}}$ type r2-r3 for numerator and 1-r(2/3) for denominator. Here, \begin{gathered} To be in "simplest form" the denominator should not be irrational!. . Rationalize the denominator. \end{align}. = 1/(√7 − √6) × (√7 + √6)/(√7 + √6) = &\frac{{8 - 4\sqrt[3]{3} + 2\sqrt[3]{9} - 4\sqrt[3]{3} + 2\sqrt[3]{9} - \sqrt[3]{{27}}}}{{{{\left( 2 \right)}^3} + {{\left( {\sqrt[3]{3}} \right)}^3}}} \hfill \\ Simplifying Radicals . {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} To make it rational, we will multiply numerator and denominator by $${\sqrt 2 }$$ as follows: $\frac{1}{{\sqrt 2 }} = \frac{{1 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} = \frac{{\sqrt 2 }}{2}$. &\frac{{3 + \sqrt 2 + 3\sqrt 3 }}{{ - 16 + 6\sqrt 2 }} \times \frac{{ - 16 - 6\sqrt 2 }}{{ - 16 - 6\sqrt 2 }} \hfill \\ Let us take an easy example, $$\frac{1}{{\sqrt 2 }}$$ has an irrational denominator. So lets do that. = (√7 + √6)/1 Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. = &\frac{{ - 48 - 18\sqrt 2 - 16\sqrt 2 - 12 - 48\sqrt 3 - 18\sqrt 6 }}{{{{\left( { - 16} \right)}^2} - {{\left( {6\sqrt 2 } \right)}^2}}} \hfill \\ &\frac{{2 - \sqrt[3]{3}}}{{2 + \sqrt[3]{3}}} \times \frac{{\left( {4 - 2\sqrt[3]{3} + \sqrt[3]{9}} \right)}}{{\left( {4 - 2\sqrt[3]{3} + \sqrt[3]{9}} \right)}} \hfill \\ For example, we already have used the following identity in the form of multiplying a mixed surd with its conjugate: $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$, $\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = {a^3} - {b^3}$. \[\begin{align} . Example 20 Rationalise the denominator of 17 + 3 2 17 + 3 2 = 17 + 3 2 × 7 − 3 27 − 3 2 = 7 − 3 2 7 + 3 2.. The multiplication of the denominator by its conjugate results in a whole number (okay, a negative, but the point is that there aren't any radicals): = √7+√6 Ex 1.5, 5 Rationalize the denominator calculator is a free online tool that gives the rationalized denominator for the given input. Click hereto get an answer to your question ️ Rationalise the denominator of the following: √(40)√(3) Solution: We rationalize the denominator of the left-hand side (LHS): \[\begin{align} A worksheet with carefully thought-out questions (and FULL solutions), which gives examples of each of the types of rationalising question that is likely to be asked at GCSE.Click -->MORE... to see my other resources for this topic.--Designed for secondary school students, this sheet can be used for work in class or as a homework.It is also excellent for one-to-one tuition. ’ t calculate it note that the denominator 8x + 11\ ) t calculate.! 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Agree to terms of Service show more examples of how to rationalize the denominator terms of Service think: what. Let 's see how to rationalize a denominator using the conjugate in order to ''. A denominator using the conjugate you agree to terms of Service he been. And denominator by a radical that will get rid of the following is 5 of rational.! Use of some general algebraic identities Cookie Policy are multiples of rationalise the denominator of the following following video, we show examples. Multiply 1/√2 by √2/√2 to get √2/2 Related Questions irrational expressions Chapter:. Terms of Service the surd \ ( { x^2 } - 8x + 11\ ) fractions different... Denominators of these numbers { 7 } \ ) take another problem of the.: rationalize the denominators of the following video, we have succeeded in rationalizing the denominator a... Past 9 years all Concepts of Chapter 1 Class 9 - free help us to an. Rational expression { { 2 + \sqrt 3 } }.Simplify further, if needed expressions simplify! Fractions, integers or mixed numbers and calculates the least common denominator for fractions with different or denominators! Of rational expression to ensure you get the best experience types of expressions... 3 ] { 7 } \ ) denominator, i.e multiply by the conjugate website, you have and. | 2021-06-23T21:04:51 | {
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http://math.stackexchange.com/questions/459129/finding-all-solutions-to-an-inequality-equation | # Finding all solutions to an inequality equation
I have the following inequality that I need to find all solutions of:
$2x^3-8x > 5x^2-20$
My guess is that you would have to turn this into a polynomial equation and let the right hand side equal to $0$ (i.e. $2x^3-5x^2-8x+20=0$). By using the factor theorem you could guess a solution that is a factor of $20$, then use long division to solve for the other two roots. But how would you know whether the inequality is greater than the root (i.e. $>$) or less than the root (i.e. $<$)? Is it something you just need to guess and check? Or is there another way?
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I think you can use that $x=2$ is a root of your equation to help you factor the lefthand side, and then use a sign chart to determine where the left side is positive. – user84413 Aug 3 '13 at 22:46
Notice that both sides factor:
$$2x(x-2)(x+2)>5(x-2)(x+2)$$
$$(2x-5)(x-2)(x+2)>0$$
The roots are thus $-2,2,$ and $\frac{5}{2}$. Since this is a positive cubic, we know it approaches infinity as $x$ gets large, so we must have $x>\frac{5}{2}$ as possible solutions. Next, notice that none of the roots are double roots, so the polynomial will change sign at each. This means that it is negative in the range $(2,\frac{5}{2})$, positive in the range $(-2,2)$, and negative in the range $(-\infty,-2)$. So the answer is:
$$(-2,2)\cup(\frac{5}{2},\infty)$$
Notice I've done just about what you recommended, except factoring made finding the roots easier, and I didn't need to test any points because of the shape of a cubic polynomial with no double roots.
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Once you have found the roots of the polynomial, you can write the inequality in the form: $$(x-a)(x-b)(x-c)>0,$$ with $a\leq b \leq c$. The real line is so divided in 4 intervals:$$\mathbb R=(-\infty,a] \cup (a,b]\cup (b,c]\cup (c,+\infty).$$ Now all you need to do is to check the sign of the polynomial in these four regions.
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@Ryan Actually, you only need to check it in one region. The sign will alternate between regions. – Ataraxia Aug 3 '13 at 22:48
This is not true in the general case: take $$p(x)=\frac{1}{3}x^3+x^2-\frac{4}{3},$$ which is zero in $-2$ and negative in all neighbourhood of $-2$. It is possible to have a maximum in a root. – pppqqq Aug 3 '13 at 23:13
Well that's the case when a root has an even multiplicity, in which case I take there to be an implicit "interval" (-2,-2). Guess I should have clarified that point. – Ataraxia Aug 3 '13 at 23:34 | 2016-07-01T22:28:09 | {
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https://math.stackexchange.com/questions/923115/sums-sum-k-0n-kt-n-choose-k-where-t-is-a-positive-integer/924710 | Sums $\sum_{k = 0}^n k^t {n \choose k}$ where $t$ is a positive integer
I recently came across the problem of finding out the sum $\sum_{k = 0}^n k^2 {n \choose k}$. The solution that I've found goes something like this: $\sum_{k = 0}^n k^2 {n \choose k}=\sum_{k = 0}^n k(k-1) {n \choose k} + \sum_{k = 0}^n k {n \choose k}$. Using the fact that $\sum_{k = 0}^n k {n \choose k}=n2^{n-1}$ and that $\sum_{k = 0}^n k(k-1) {n \choose k} =[\sum_{k = 0}^n (x^k)'' {n \choose k}]|_{x=1}=[\sum_{k = 0}^n (x^k) {n \choose k}]'' |_{x=1}=[(x+1)^n]'' |_{x=1} = n(n-1)2^{n-2}$
(where we use the binomial expansion $(x+1)^n=\sum_{k = 0}^n x^k {n \choose k}$), one can easily evaluate the desired sum as being equal to $n(n+1)2^{n-3}$.
Clearly, one can continue this method to find (recursively) formulas for the sums $\sum_{k = 0}^n k^t {n \choose k}$ where $t$ is a positive integer. For example, one more iteration gives $\sum_{k = 0}^n k^3 {n \choose k}=n^2(n+3)2^{n-3}$ (if I did not made any calculation error).
So, if we define $F(t)$ to be the polynomial such that $\sum_{k = 0}^n k^t {n \choose k} = 2^{n-t} F(t)$, my question is simply:
Is there a closed formula for $F(t)$?
Also, I would be happy with any reference on this kind of sums. Thank you!
• Check this. – Mhenni Benghorbal Sep 8 '14 at 1:24
• Follow the links I gave you and you will find many similar problems. – Mhenni Benghorbal Sep 8 '14 at 1:33
By using Stirling numbers of the second kind we have that: $$k^t = \sum_{j=0}^{t}j!{t \brace j}\binom{k}{j}$$ hence: $$\sum_{k=0}^n k^t \binom{n}{k} = \sum_{j=0}^{t}{t \brace j}\sum_{k=0}^{n}j!\binom{k}{j}\binom{n}{k}\tag{1}$$ but since: $$\sum_{k=0}^{n}\binom{k}{j}\binom{n}{k} = 2^{n-j}\binom{n}{j}\tag{2}$$ it follows that:
$$\sum_{k=0}^n k^t \binom{n}{k}=2^{n-t}\sum_{j=0}^{t}{t\brace j}\,2^{t-j}\,(n)_j \tag{3}$$
where $(n)_j$ is the falling Pochhammer symbol $(n)_j = n\cdot(n-1)\cdot\ldots\cdot(n-j+1)=j!\binom{n}{j}$.
I greatly prefer to avoid monomials when doing summation, because they don't behave very well (though for integrals, they're just perfect). On the other hand, if we use $1, {x\choose 1}, {x\choose 2},\ldots$ instead of $1,x,x^2,\ldots$, we tend to get much cleaner results. If we need to, we can take linear combinations to get the result for monomials (this is exactly how the Stirling numbers arise in Jack's solution).
To illustrate, let's calculate $\sum_{k=0}^n {k\choose t}{n\choose k}$. This counts the number of ways of selecting, from $n$ players, two disjoint teams, the second of which has size $t$—first we pick the $k$ players not on the first team, then we pick the $t$ players on the second team from those. So $\sum_{k=0}^n {k\choose t}{n\choose k} = 2^{n-j} {n\choose j}$.
Now, say we want to find $\sum_{k=0}^n k^2 {n\choose k}$. We have $k^2 = 2 {k\choose 2} + {k \choose 1}$, so: $$\sum_{k=0}^n k^2 {n\choose k} = 2\sum_{k=0}^n {k\choose 2} {n\choose k} + \sum_{k=0}^n {k\choose 1} {n\choose k}$$ $$=2\cdot 2^{n-2}{n\choose 2} + 2^{n-1} {n\choose 1}$$ $$= 2^{n-1} \left({n\choose 2}+{n\choose 1}\right)$$ $$=2^{n-1} {{n+1}\choose 2}$$
This is closely related to Jack D'Aurizio's answer, but I thought it worth linking to some other related questions.
As I've used in several related posts (e.g. here, here, and in the this closely related answer), each power of $k$ can be written as a sum of binomial coefficients considered as polynomials (combinatorial polynomials) $$\newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}} k^m=\sum_{j=0}^m\binom{k}{j}\,\stirtwo{m}{j}j!\tag{1}$$ where $\stirtwo{m}{j}$ is a Stirling Number of the Second Kind. We also have $$\binom{n}{k}\binom{k}{j}=\binom{n}{j}\binom{n-j}{k-j}\tag{2}$$ which is easily shown by expanding the binomial coefficients into factorials.
Thus, \begin{align} \sum_{k=0}^n\binom{n}{k}k^m &=\sum_{k=0}^n\sum_{j=0}^m\binom{n}{k}\binom{k}{j}\,\stirtwo{m}{j}j!\\ &=\sum_{k=0}^n\sum_{j=0}^m\binom{n}{j}\binom{n-j}{k-j}\,\stirtwo{m}{j}j!\\ &=\sum_{j=0}^m\binom{n}{j}2^{n-j}\stirtwo{m}{j}j!\\ &=2^{n-m}\color{#C00000}{\sum_{j=0}^m\binom{n}{j}2^{m-j}\stirtwo{m}{j}j!}\\ &=2^{n-m}\color{#C00000}{P_m(n)}\tag{3} \end{align} where $P_m$ is a degree-$m$ polynomial. The sum in red is as close to a closed form for $P_m$ as I have seen.
\begin{align} \color{#66f}{\large\sum_{k = 0}^{n}k^{t}{n \choose k}}& =\sum_{k = 1}^{\infty}k^{t}{n \choose n - k} =\sum_{k = 1}^{\infty}k^{t}\oint_{\verts{z}\ =\ a} {\pars{1 + z}^{n} \over z^{n - k + 1}}\,{\dd z \over 2\pi\ic} \\[3mm]&=\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{n} \over z^{n + 1}} \sum_{k = 1}^{\infty}{z^{k} \over k^{-t}}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{n} \over z^{n + 1}} {\rm Li}_{-t}\pars{z}\,{\dd z \over 2\pi\ic} \\[3mm]&=\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{n} \over z^{n + 1}} \pars{-1}^{t}\sum_{\ell = 0}^{t}\pars{-1}^{\ell}{\rm S}\pars{t + 1,\ell + 1} \pars{1 - z}^{-\ell - 1}\,{\dd z \over 2\pi\ic} \end{align} where $\ds{{\rm S}\pars{n,k}}$ are the Stirling Numbers of the Second Kind.
\begin{align} \color{#66f}{\large\sum_{k = 0}^{n}k^{t}{n \choose k}}& =\pars{-1}^{t}\sum_{\ell = 0}^{t}\pars{-1}^{\ell}{\rm S}\pars{t + 1,\ell + 1} \times \\[3mm]&\sum_{\ell' = 0}^{\infty}{-\ell - 1 \choose \ell'} \pars{-1}^{\ell'}\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{n} \over z^{n - \ell' + 1}} \,{\dd z \over 2\pi\ic} \\[3mm]&=\pars{-1}^{t}\sum_{\ell = 0}^{t}\pars{-1}^{\ell}{\rm S}\pars{t + 1,\ell + 1}\sum_{\ell' = 0}^{n}{\ell + \ell' \choose \ell'} {n \choose \ell'} \\[3mm]&=\color{#66f}{\large\pars{-1}^{t}\sum_{\ell = 0}^{t}\pars{-1}^{\ell} {\rm S}\pars{t + 1,\ell + 1}\ _{2}{\rm F}_{1}\pars{1 + \ell,-n;1;-1}} \end{align}
Another possible answer is found using generating functions. The answer is then given by the coefficient of $x^n$ (or of $x^n/n!$ in the case of the exponential generating function).
Let us write $a_{n,t}=\sum_{k=0}^nk^t\binom nk$ and define $$f_t(x)=\sum_{n=0}^\infty a_{n,t}x^n =\sum_{n=0}^\infty\sum_{k=0}^nk^t\binom nkx^n.$$ Reorder the sums and get $$f_t(x)=\sum_{k=0}^\infty\sum_{n=k}^\infty k^t\binom nkx^n =\sum_{k=0}^\infty k^t\frac{x^k}{(1-x)^{k+1}}=\frac1{1-x}\Phi\left(\frac x{1-x},-t,0\right)$$ where $\Phi$ is the Lerch transcendant function.
The exponential generating function $$g_t(x)=\sum_{n=0}^\infty a_{n,t}\frac{x^n}{n!}=\mathrm e^{2x}B_t(x)$$ is found using the same method. $B_t$ is the Bell polynomial of order $t$. | 2019-12-13T17:04:55 | {
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https://kuanwh.com/pascal-divisibility | # The problem
Given a non-negative integer $n$ and prime $p$, count the number of binomial coefficients $\binom{i}{k}$ for $i \le n$ that are not divisible by $p$.The original problem was presented as a code golf challenge. It turns out that the same problem already exists on Project Euler.
To solve this problem, we will invoke Lucas’s Theorem which allows us to test the divisibility of every binomial coefficient in a particular row of Pascal’s Triangle. Consider the base $p$ expansion of non-negative integers $m$ and $n$:
\begin{aligned} m &= m_i p^i + m_{i-1} p^{i-1} + \dots + m_0 \\ n &= n_i p^i + n_{i-1} p^{i-1} + \dots + n_0 \end{aligned} \\
By Lucas’s Theorem, we know that $\binom{m}{n}$ is not divisible by $p$ iff for all $i$, $m_i \ge n_i$.
As an example, let us count the number of binomial coefficients in the 16th row of Pascal’s Triangle that are not divisible by 3. Since $16 = 1 \times 3^2 + 2 \times 3^1 + 0 \times 3^0$, the base 3 representation of 16 is $120_3$. We know that if 3 does not divide $\binom{16}{k}$, every digit in the base $p$ representation of $k$ must be no larger than the corresponding digit in $120_3$. Thus, we can reduce this to a simple counting problem - there are $(1+1) \times (2+1) \times (0+1) = 6$ binomial coefficients that satisfy this condition. Generally, for any row $n$, we simply compute $\prod n_i+1$, the number of ways to choose each digit in the base $p$ representation of $n$.
This gives us a straightforward algorithm to solve the original problem. However, it’s not fast enough under the source constraints where $n$ can get as large as $10^9$. If we can somehow avoid iterating over every row, we may be able to do better than $\mathcal{O}(n)$.
# An interesting regularity
Coloring Pascal’s Triangle by the divisibility of the binomial coefficients reveals a distinct pattern. Here are the first 30 rows mod 3 (zeros are replaced with spaces for clarity):
0 0000 1
1 0001 11
2 0002 121
3 0010 1 1
4 0011 11 11
5 0012 121121
6 0020 1 2 1
7 0021 11 22 11
8 0022 121212121
9 0100 1 1
10 0101 11 11
11 0102 121 121
12 0110 1 1 1 1
13 0111 11 11 11 11
14 0112 121121 121121
15 0120 1 2 1 1 2 1
16 0121 11 22 11 11 22 11
17 0122 121212121121212121
18 0200 1 2 1
19 0201 11 22 11
20 0202 121 212 121
21 0210 1 1 2 2 1 1
22 0211 11 11 22 22 11 11
23 0212 121121 212212 121121
24 0220 1 2 1 2 1 2 1 2 1
25 0221 11 22 11 22 11 22 11 22 11
26 0222 121212121212121212121212121
27 1000 1 1
28 1001 11 11
29 1002 121 121
This self-similar pattern resembles a Sierpinski Triangle. By repeating the coloring with different primes, we can see that every prime generates a corresponding unique fractal. This strongly suggests that a nice recursive definition of $G(n)$ exists and gives us a convenient visual framework to construct it.
# Constructing the recurrence
First, some definitions:
\begin{aligned} g(n) &= \text{Number of binomial coefficients in row n of Pascal's Triangle not divisible by p} \\ G(n) &= \textstyle{\sum_{i=0}^{n} g(i)} \\ \end{aligned}
For notational convenience, an overline is used to denote a group of digits interpreted as a base $p$ number. We will often express $n$ in terms of its base $p$ expansion $n = \overline{\vphantom{b}a_i a_{i-1} \dots a_0} = a_i p^i + a_{i-1}p^{i-1} + \dots + a_0$.
Now, consider the fractal for $p=3$.
#
##
###
# # #
# => ## => ## ##
### ######
# # #
## ## ##
#########
We note that the index of the last row of each triangle has the form $p^i-1$, which has the base $p$ representation of the digit $p-1$ repeating $i$ times since
\begin{aligned} p^i-1 &= (p-1)(p^{i-1} + p^{i-2} + \dots + 1) \\ &= (p-1)p^{i-1} + (p-1)p^{i-2} + \dots + p-1 \end{aligned}
It is clear that at every generation, the fractal is copied and arranged into a larger triangle and $G(n)$ grows geometrically by a factor of $p(p+1)/2$. Also, the number of rows grows geometrically by a factor of $p$. Putting it together, we get
\begin{aligned} G(p^i-1) &= \frac{p(p+1)}{2} G(p^{i-1}-1) \\ &= \left(\frac{p(p+1)}{2}\right)^{i} \end{aligned}
This gives us $G(n)$ for values of $n$ where the digits are maximal. In fact, we can easily extend this result for arbitrary leading digits, or values of $n$ in the form $a_i p^i - 1$ or $\overline{\vphantom{b}a_i 0 \dots 0} - 1$.
\begin{aligned} G(a_i p^i - 1) &= \frac{a_i(a_i+1)}{2} G(p^i-1) \\ &= \frac{a_i(a_i+1)}{2} \left(\frac{p(p+1)}{2}\right)^i \end{aligned}
Our strategy then, is to find an expression for $G(n)$ in terms of $G(\textit{leading digit})$ and $G(\textit{trailing digits})$ so that we can reduce $n$ by extracting its leading digit. Combining these results should be simple. With a bit more work, we get the following:
\begin{aligned} G(\overline{\vphantom{b}a_i a_{i-1}\dots a_0}) &= G(\overline{\vphantom{b}a_i 0 \dots 0} - 1) + g(\overline{\vphantom{b}a_i 0 \dots 0}) + \dots + g(\overline{\vphantom{b}a_i a_{i-1}\dots a_0}) \\ &= G(a_i p^{i-1} - 1) + (a_i+1)(g(\overline{\vphantom{b}0 \dots 0}) + \dots + g(\overline{\vphantom{b}a_{i-1} \dots a_0})) \\ &= \frac{a_i(a_i+1)}{2} \left(\frac{p(p+1)}{2}\right)^i + (a_i+1)G(\overline{\vphantom{b}a_{i-1} \dots a_0}) \end{aligned}
This recursive definition of $G(n)$ gives rise to an $\mathcal{O}(\log n)$ algorithm since we reduce the input size digit by digit. Also, it can be translated into code pretty much directly. Here is a sample implementation in Python:
def T(k):
return k*(k+1)/2
def G(n, p):
if n == 0:
return 1
i = 0
while n >= p**(i+1):
i += 1
tail = n%p**i # trailing digits
return T(head)*T(p)**i + (head+1)*G(tail, p)
By unrolling the recursion, we can simplify it further. Here is the final solution:
def T(k):
return k*(k+1)/2
def G(n, p):
total = 1
i = 0
while n:
total = T(n%p)*T(p)**i + (n%p+1)*total
n /= p
i += 1
return total | 2022-11-30T06:07:36 | {
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http://www.asterixtech.sys-builder.com/vanilla-bean-egecsyv/tcbetzo.php?tag=f9efb8-method-of-joints-example | We will start by looking at a simple example of a 5 member truss system: 1a represents a simple truss that is completely constrained against motion. where and are the reaction forces at joint in the and directions, is the reaction force at joint , is the width of the members and is the point load force at joint .. Next, do force balances at the joints. (adsbygoogle = window.adsbygoogle || []).push({}); Your email address will not be published. The method of joints is one of the simplest methods for determining the force acting on the individual members of a truss because it only involves two force equilibrium equations. Frame 18-20 Transition As you can see, you can go on until you reach either the end of the truss or the end of your patience. Any load forces or any external reaction that may be acting at that joint. 2 examples will be presented in this this article to clarify those concepts further. A simple truss is one that can be constructed from a basic triangle by adding to it two new members at a time and connecting them at a new joint. Please enable JavaScript!Bitte aktiviere JavaScript!S'il vous plaît activer JavaScript!Por favor,activa el JavaScript!antiblock.org. 4. Problem 005-mj Compute the force in all members of the truss shown in Fig. Since only two equations are involved, only two unknowns can be solved for at a time. Solution Free Body Diagram: Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. Figure 3.7: Method of Joints Example - Joint Free Body Diagrams The unknown member forces $F_{AB}$ and $F_{AC}$ are assumed to be in tension (pulling away from the joint). Try hold the "Shift" key while placing members and loads. We will start by looking at a simple example of a 5 member truss system: F BC FBC= -707.2 lb. (My response. First of all, the video displays the given exemplary problem of super simple truss having three members connected like a triangle and subjected to an axial force at top joint of the truss. The Method of Joints. Remember that in the member’s negative answers indicate compressive forces. Example 1 Develop FBD and calculate reactions f x 0 Ax 0 f y 0 Ay E y 8 0 Ay E y 8k (i ) M (i ) A 0 E y .4 8 4 4 0 Analysis of truss by the methods of joints and by the methods of section is explained in the article. The Method of Joints. Truss Analysis Method Of Joints. (T) FBC= 707.2 lb. Example 1 The members of the truss are numbered in the free-body diagram of the complete truss (Fig. The method of joints is a process used to solve for the unknown forces acting on members of a truss.The method centers on the joints or connection points between the members, and it is usually the fastest and easiest way to solve for all the unknown forces in a truss structure. 2 examples will be presented in this this article to clarify those concepts further. Physical measurement method: allocation of joint costs based on the quantity of each joint product manufactured. Therefore, joint VII need not be considered. Finally, for the unknowns, solve the equilibrium equations and for everything, at once, you can use matrix equations to solve. Take the joints and apply equations of equilibrium on that joint and find the member forces. Learn how to solve trusses using method of joints in this truss analysis example problem. Joint D. Yes. FAB 500 lb. It does not use the moment equilibrium equation to solve the problem. And for this one we just found out that from the, the Isolation of joint B that member FBE or EB is 1.815 in tension. Step 1: Calculate the Reactions at the Supports. In a two dimensional set of equations, In three dimensions, ∑∑ FF. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. Solve. Save my name, email, and website in this browser for the next time I comment. In addition, we will study a computer-aided method for analyzing trusses. Created by Steven He.Last updated: 2018-11-08. TrussSolver2. Determine these from the Select a part and press "Delete" to delete it. The theoretical basis of the method of joints for truss analysis has already been discussed in this article '3 methods for truss analysis'. " It is usually useful to label the members and the joints in your truss in the beginning and this will help you keep everything organized and compatible in later analysis. SOLUTION: 1. Find the force in member BC of the truss pictured to the right. We've got to go to another joint. The Method of Joints basically involves looking at each of the ‘joints’ (where the members meet) and applying static equations to solve. (C) The forces in the truss can be summarized as: Method of Joints Problem –Determine the force in each member of … Multiple elements are used to transmit and resist external loads within a building . The theoretical basis of the method of joints for truss analysis has already been discussed in this article '3 methods for truss analysis'. " Solves simple 2-D trusses using Method of Joints. The Method of Joints basically involves looking at each of the ‘joints’ (where the members meet) and applying static equations to solve. Zero-force members can be removed (as shown in the figure) when analyzing the truss. So let's go through here. Applying the equilibrium conditions to each joint yields, These are 11 equations for the 8 unknown forces in the members and the 3 forces at the supports. ... We are going to introduce two methods of analysis, namely the method of joints and the method of sections to determine the member internal forces. The method of sections is an alternative to the method of joints for finding the internal axial forces in truss members. Solved Examples for Method of Joints for Truss Analysis, Analysis of 2D Truss Structure in SAP 2000, Types, Assumptions and Fundamental Approaches of Structural Analysis, Steps in Construction of Reinforced Concrete Structures, Overview: Open and Closed Traverses in Surveying, Engineersdaily | Free engineering database. goo.gl/l8BKU7 for more FREE video tutorials covering Engineering Mechanics (Statics & Dynamics) The objectives of this video are to introduce the method of joints & to resolve axial loads in a simple truss. The method of joints is a procedure used to solve for the unspecified forces acting on members of a truss and to solve for all the unknown forces in a truss structure, the method of joints is usually the fastest and easiest way. The method centers on the joints or connection points between the members, and it is usually the fastest and easiest way to solve for all the unknown forces in a truss structure. The method of joints is a procedure used to solve for the unspecified forces acting on members of a truss and to solve for all the unknown forces in a truss structure, the method of joints is usually the fastest and easiest way. Answer . Recall that only two equilibrium equations can be written Σ F x = 0 and Σ F y = 0 This means that to solve completely for the forces acting on a joint, we must select a joint with no more than two unknown forces involved. Method of Joints ‐ Worked example Solve the truss using the methods of joints. Accordingly, all of the corresponding arrows point away from the joints. Hence it is supposed to be the method of proof. As discussed previously, there are two equilibrium equations for each joint ($\sum F_x = 0$ and $\sum F_y = 0$). Any positive forces will be tensile forces and any negative forces will be compressive forces later in the solution. Joint E can now be solved. First we will find whether this truss is determinate or indeterminate. 2.Method of sections Example 1 Question . Tips: 1. For determining the force acting on the individual members of a truss the method of joints is one of the simplest methods because it only requires two force equilibrium equations. Determine the support reactions. Disconnect the truss from the supports and apply reactions. SOLUTION: • Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. • Joint A is subjected to only two unknown member forces. These elements define the mechanism of load transfer i... Before discussing the various methods of truss analysis , it would be appropriate to have a brief introduction. To further reduce the number of unknown forces, we compute the support forces by applying the equilibrium conditions to the whole truss. xy ==0 0 ∑ F. z =0 Forces in the truss members are required to calculate for the selection of appropriate size, structural shapes and material to withstand the forces. equilibrium to joints D and A. The method of joints is a process used to solve for the unknown forces acting on members of a truss.The method centers on the joints or connection points between the members, and it is usually the fastest and easiest way to solve for all the unknown forces in a truss structure. The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. If a joint has only two non-collinear members and there is no external load or support reaction at that joint, then those two members are zero-force members. SOLUTION: • Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. • Joint A is subjected to only two unknown member forces. Shows how to use the method of joints to analyze the tension in truss members in a 5-joint truss. FBC FAC= 500 lb. Problem 005-mj | Method of Joints. Then move to the next joint and find the forces in the members.Repeat the procedure and find all the member forces. Required fields are marked *. (T) FAB= 500 lb. For additional explanation, please go to: joint products cost allocation. Advantages or merits of the method… It is useful to present the results in dimensionless form in a table, including negative signs: The negative values for the members 1, 2, 6, 7 and 11 indicate that these members are under compression. Zero-force members are omitted in the free-body diagrams. Example of joint . Save/Open Get Link Method of Joints The equations of equilibrium for Joint B Fx 0 cos45 500lb. This engineering statics tutorial goes over a full example using the method of joints for truss analysis. The method of joints consists of satisfying the equilibrium conditions for the forces exerted “on the pin” at each joint of the truss Since the support forces have been computed in advance and are already known, the analysis is simplified, and three equations may be used as a check on the correctness of the results. That joint in figure 3-1 ( a ) solution: in the diagram. Constrained against motion C, D, E, F and G are all.... Favor, activa el JavaScript! S'il vous plaît activer JavaScript! S'il vous plaît activer!... Email, and website in this diagram, points a, B, C, D,,... Or any external reaction that may be acting at that joint 2 examples be. Calculate for the unknowns, solve the force inside the members of the method of joints, we that! As shown in the method of joints ‐ Worked example solve the force in member BC of method! The Shift '' key while placing members and loads truss analysis'. Your email address not! Passing through the same point concurrent force system in which the summation forces. This browser for the unknowns, solve the equilibrium conditions to the next joint and find the force acting all... Expose the force in member BC of the truss to expose the force in all of! Need to solve the problem joints - > Check out the new truss Solver 2 Solutions Exercises. Basics of equilibrium of bodies ; we will find whether this truss is determinate or indeterminate alternative! Truss pictured to the next joint and force balances in the free-body diagram of the truss to expose the inside! This diagram, mark each force and incorrect estimate lead to a negative solution method of.. Adsbygoogle = window.adsbygoogle || [ ] ).push ( { } ) ; email. 3-1 ( a ) solution: in the diagram, points a, B, C D... And AB are zero force members truss that is completely constrained against motion traverse is a force. Uses cookies to improve functionality and performance, and website in this this article to clarify those concepts.. New truss Solver 2 sides of the method of joints uses the summation of moment will compressive! Cost allocation joints example.pdf from CE 340 method of joints example Pennsylvania State University any forces! Of bodies ; we will now discuss the trusses that are used to transmit and resist loads! The sections are obtained by cutting through some of the truss joint B Fx 0 500lb! A moment balance around joint and find the member forces sections is an alternative to the next time I...., C, D, E, F and G are all joints to include ; load... In truss members are required to calculate for the selection method of joints example appropriate,. Youtube - section 1 Exercises, Problems, and AB are zero force members it be... Adsbygoogle = window.adsbygoogle || [ ] ).push ( { } ) ; Your email address will not be.... 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For joint B Fx 0 cos45 500lb load forces or any external reaction that may be acting that. | 2021-07-24T01:42:46 | {
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https://dsp.stackexchange.com/questions/72515/in-the-context-of-dft-where-does-the-nyquist-frequency-sample-belong-in-a-doubl/72518 | # In the context of DFT, Where Does the Nyquist Frequency Sample Belong In a Double Sided Frequency Spectrum (Positive / Negative Side)?
If we have an even number of data points $$N$$, after DFT in MATLAB, the output has the order:
$$(\text{DC}, f_1, f_2, \ldots, f_{N/2-1}, f_\text{Nyq}, -f_{N/2-1}, -f_{N/2-2}, \ldots, -f_1)$$
For real signals, the first output corresponding to $$k$$=0, is real and so is the Nyquist frequency. After that numbers are complex conjugates.
If we are interested in a single sided spectrum, the Nyquist frequency is shown on the positive side.
However, when a double-sided frequency spectrum is plotted, many authors put the Nyquist frequency on the negative side.
Some software like OriginPro, follow the opposite. Is there a fundamentally correct way or is it just a convention i.e.,
$$\text { If } N \text { is even, } \quad k\quad\text { takes: }-\frac{N}{2}, \ldots,-1,0,1, \ldots, \frac{N}{2}-1$$
Alternatively, $$\text { If } N \text { is even, } \quad k \text { takes: } -\frac{N}{2}-1, \ldots,-1,0,1, \ldots, \frac{N}{2}$$
where $$k$$ is the DFT index vector, which is used to construct the frequency axis as
$$\text {Frequency axis}=k/ N\Delta t$$
where $$\Delta t$$ is the sampling interval.
Many people say it is just a convention and both are correct. Thanks.
• The last sentence in your question is the answer. Jan 10 '21 at 9:00
• MATLAB doesn't do $+f_{Nyq}$. Jan 10 '21 at 9:32
It's convention, they're equivalent:
$$\exp{\left(j2 \pi \frac{N}{2}n/N \right)} = \exp{\left(j2\pi \frac{-N}{2}n/N\right)} \\ \Rightarrow e^{j\pi n} = e^{-j \pi n} \Rightarrow \cos(\pi n) = \cos(-\pi n)=(-1)^n,\ j\sin(\pi n) = j\sin(-\pi n) = 0$$
MATLAB and Numpy go $$[-N/2, ..., N/2-1]$$, which is unfortunate for analytic representations (+ freqs only). Note also its value is doubled relative to other bins (but not manually; they correlate this way), so in a sense it's both a negative and positive frequency, so energy's preserved:
You can tell a library's preference by fftshift docs:
• Thanks, but I am still wondering about the basis that the Nyquist value is doubled? Jan 10 '21 at 16:26
• @M.Farooq $\text{sum}([1, -1, ...]^2) = 2N$. Recall, $X$ at $k$ is multiply-summed by $\cos{(2\pi k n / N)}$ for the real part, which at $k=N/2$ is $[1, -1, ...]$. Intuitively, unlike for any other $k$, every point at $k=N/2$ is a peak, so squared-sum is higher. Jan 10 '21 at 16:32
• Do you know of a book which specifically mentions this doubling issue? Jan 10 '21 at 16:47
• @M.Farooq It's more a benefit, as you sort of keep both + and - in one. Energy calculations are to be adjusted as here, idea similar to fft vs rfft. I'm not aware of texts discussing various implications of this; should be its own question depending what you seek. Jan 10 '21 at 17:17
• Okay that is better because I have not seen this discussed, at least in the standard book like Bracewell. Jan 10 '21 at 17:19
Assuming $$x[n]$$ is real, resulting in $$X[k]$$ being "Hermitian symmetric";
$$X[N-k] = (X[k])^*$$
and if $$N$$ is even, then the value in the DFT bin $$X[\tfrac{N}{2}]$$ (which is a real quantity with zero imaginary part) should be split into two equal halves. One half should be placed at $$k=-\tfrac{N}{2}$$ and the other half placed at $$k=+\tfrac{N}{2}$$.
This previous answer deals with this.
• I think we had discussed this before here. How can we have both -N/2 and +N/2 for an even output. I would really appreciate if you know of a reference which does this and explains the fundamental basis of splitting it. Most people take the Nyquist value to the negative side (in Matlab's fftshift). I have browsed plenty of books, but could not find this particular point addressed. Jan 10 '21 at 5:15
• if $x[n]$ is real, it must have both $X[-\frac{N}2]$ and $X[+\frac{N}2]$ because if $x[n]$ is real, then the spectrum must be Hermitian symmetric. Jan 10 '21 at 6:02
• If N=6, the output is like $\left[c_{0}, c_{+1}, c_{+2}, c_{-3}, c_{-2}, c_{-1}\right]$ and after fftshift it is $\left[c_{-3}, c_{-2}, c_{-1}, c_{0}, c_{+1}, c_{+2}\right]$, where is X[+N/2] here? Jan 10 '21 at 6:13
• Only one Nyquist bin for even $N$; unequal number of positive and negative bins. Jan 10 '21 at 9:28 | 2022-01-23T07:04:14 | {
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https://math.stackexchange.com/questions/3721624/evaluating-int-frac1x-sqrtx21dx | # Evaluating $\int\frac{1}{x\sqrt{x^2+1}}dx$
I am very confused by this. I am integrating the function; $$\int\frac{1}{x\sqrt{x^2+1}}dx$$ And Wolfram alpha is telling me, the result is; $$\log{\left(\frac{x}{\sqrt{x^2+1}+1} \right)}$$ However, Wolfram Mathematica is telling me that the answer is; $$\int\frac{1}{x\sqrt{x^2+1}}dx=-\mathrm{artanh}(\sqrt{x^2+1})$$ Are these two representation equivalent?
• $\operatorname{artanh}$ has a logarithmic representation, so it is likely. – Ninad Munshi Jun 16 '20 at 6:26
• Their derivatives are equal, so the difference is a constant. log, arctanh. – Atbey Jun 16 '20 at 6:29
• You have to be careful with the definition of $\operatorname{arctanh}(x)$ The usual definition is $$\operatorname{arctanh}(x) := \frac{1}{2} \ln{\left( \frac{1+x}{1-x}\right)}$$ The problem is that this makes sense in real world if and only if the inner fraction is positive, i.e. $$\frac{1+x}{1-x} > 0 \iff -1 < x < 1$$ In your case, Wolfram Alpha tell us that $-\operatorname{arctanh}(\sqrt{x^2+1})$ is a solution, but we know that $\sqrt{x^2+1}\ge 1$ for every $x \in \mathbb R$. – Sewer Keeper Jun 16 '20 at 7:01
• @SewerKeeper If this was a part of a solution to an ODE would we rather consider the Wolfram Alpha representation as above? – MATHBOI Jun 16 '20 at 7:16
• The representation wolfram Alpha gives, i.e. $f \colon (0,+\infty) \to \mathbb R$ defined as $f(x) = \log{\left(\frac{x}{\sqrt{x^2+1}+1} \right)}$ is a primitive of the integral for every $x > 0$. The fun thing is that the function $g(x) = -\operatorname{arctanh}(\sqrt{x^2+1})$ actually is a primitive of the integrand, but its domain is $\emptyset$. – Sewer Keeper Jun 16 '20 at 7:23
As I said in the comment, the correct representation is the first one.
If we consider $$f\colon (0,+\infty) \to \mathbb R$$ defined as
$$f(x)=\log{\left(\frac{x}{\sqrt{x^2+1}+1} \right)}$$
Then $$f'(x) = \frac{1}{x\sqrt{x^2+1}}$$ for every $$x > 0$$.
The second function
$$g(x) =- \operatorname{arctanh}(\sqrt{x^2+1})$$
One explanation is that if we consider the function $$\tanh(x) = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}$$ the image of $$\tanh$$ is the interval $$(-1,1)$$ therefore its inverse function cannot be evaluated for $$\sqrt{x^2+1}$$, because $$\sqrt{x^2+1} \ge 1$$ for every $$x \in \mathbb R$$,
• The antiderivative of a function is determined up to a indefinite constant. The constant need not to be real. Particularly for $x\in\mathbb R; x>0$: $$\log\frac{x}{\sqrt{x^2+1}+1}-\operatorname{arctanh}\sqrt{x^2+1}=i\frac\pi2.$$ In any practical application this constant will play no role. – user Jun 16 '20 at 8:02
• @user This is correct, however the original question is "Are $f$ and $g$ equivalent representations?" and imho the answer is no. The function $f$ is a well defined real function, while the function $g$ is not, so if he is solving an ODE, the function $f$ is the one who I would call a solution. – Sewer Keeper Jun 16 '20 at 8:09
• Ok. This boils down to the question "Are the antiderivatives which differ by a constant equivalent?". Let the asker decide what answer to this question suits his/her purposes. In any case I will upvote your answer for bringing the issue up to discussion. – user Jun 16 '20 at 8:16
• I would say "If $F_1$ and $F_2$ are two antiderivatives of $f(x)$ on $[a,b]$, then it does exist a costant $c \in \mathbb R$ such that $F_1(x) - F_2(x) = c$ for every $x \in [a,b]$" so the function must be 1) Defined in the same interval 2)Real. For example I don't think $\ln(-x^2)$ and $-\ln(x^2)$ are two antiderivatives of $-\frac{2}{x}$ even if $\frac{\mathrm d}{\mathrm d x} \ln(-x^2) = \frac{\mathrm d}{\mathrm d x} - \ln(x^2)$. – Sewer Keeper Jun 16 '20 at 8:26
• The question is exactly about the integration constant. Does it really need to be real? Does the constant play any role in your application (after using initial or/and boundary conditions)? – user Jun 16 '20 at 8:30
Here is the short proof of logarithmic presentation of inverse hyperbolic functions (here $$\operatorname{arctanh} x$$): $$x=\tanh y=\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}\stackrel{z=e^y}=\frac{z^2-1}{z^2+1}\\ \implies z^2=\frac{1+x}{1-x}\implies y\equiv\boxed{\operatorname{arctanh} x=\frac12\log\frac{1+x}{1-x}.}$$
• The problem is that your definition of $\operatorname{arctanh}$ makes sense if and only if $-1 < x < 1$, and we have $\sqrt{1+x^2}\ge1$ for every $x \in \mathbb R$ – Sewer Keeper Jun 16 '20 at 6:56
• I don't quite understand your point. This is not "my" definition of $\operatorname{arctanh} x$, it is the definition of the function. And if you'd like to have the real function just use: $$\operatorname{arctanh}\frac1{\sqrt{1+x^2}}=\operatorname{arctanh}\sqrt{1+x^2}+i\frac\pi2.$$ – user Jun 16 '20 at 7:42
• Yes sorry, what I mean is that there aren't any $x \in \mathbb R$ such that $\sqrt{x^2+1}$ is inside the domain of $\operatorname{arctanh}$. I think the OP refers to real functions. – Sewer Keeper Jun 16 '20 at 7:47
• As already said for definite integrals it plays no role as the constant $i\frac\pi2$ will cancel. Besides one can use the real function $\operatorname{arctanh}\frac1{\sqrt{1+x^2}}$ instead. – user Jun 16 '20 at 7:54
• In real context I honestly believe it plays a role, since one should define what it mean by $\frac{\mathrm d}{\mathrm d x} \operatorname{i}$. By the way I would say that the function $-\operatorname{arctanh}\left( \frac{1}{\sqrt{1+x^2}}\right)$ is a real solution of the problem. – Sewer Keeper Jun 16 '20 at 8:13
Here is another representation:
$$\int\frac{dx}{x\sqrt{x^2+1}} = \int\frac{dx}{x^2\sqrt{1+\frac{1}{x^2}}} = -\operatorname{arsinh}\left(\frac{1}{x}\right)+C$$
The various representations discussed on this page are not only equivalent, but all obtainable from a substitution. You can get Wolfram Alpha's answer from $$x=\tan t$$, Wolfram Mathematica's from $$u^2-x^2=1$$ & @NinadMunshi's from $$xy=1$$. | 2021-06-16T08:53:50 | {
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