Search is not available for this dataset
url
string | text
string | date
timestamp[s] | meta
dict |
|---|---|---|---|
https://gmatclub.com/forum/what-is-the-value-of-229663.html | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 18 Apr 2019, 21:34
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)?
Author Message
TAGS:
### Hide Tags
Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1354
Location: Viet Nam
What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
Updated on: 28 Nov 2016, 05:08
11
00:00
Difficulty:
55% (hard)
Question Stats:
62% (02:01) correct 38% (02:08) wrong based on 183 sessions
### HideShow timer Statistics
What is the value of $$A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}$$
A. $$\frac{98}{99}$$
B. $$\frac{99}{100}$$
C. $$\frac{100}{101}$$
D. $$\frac{98}{101}$$
E. $$\frac{99}{101}$$
_________________
Originally posted by broall on 28 Nov 2016, 03:49.
Last edited by broall on 28 Nov 2016, 05:08, edited 1 time in total.
CEO
Joined: 12 Sep 2015
Posts: 3581
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
28 Nov 2016, 08:05
4
Top Contributor
3
nguyendinhtuong wrote:
What is the value of $$A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}$$
A. $$\frac{98}{99}$$
B. $$\frac{99}{100}$$
C. $$\frac{100}{101}$$
D. $$\frac{98}{101}$$
E. $$\frac{99}{101}$$
Let's find the sum of various fractions and look for a pattern
$$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}$$ = 1/2 + 1/6
= 3/6 + 1/6
= 4/6
= 2/3
$$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}$$ = 2/3 + $$\frac{1}{3 \times 4}$$
= 2/3 + 1/12
= 8/12 + 1/12
= 9/12
= 3/4
$$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\frac{1}{4 \times 5}$$ = 3/4 + $$\frac{1}{4 \times 5}$$
= 3/4 + 1/20
= 15/20 + 1/20
= 16/2
= 4/5
As we can see, the pattern is...
$$A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{(n) \times (n+1)}$$ = $$\frac{n}{n+1}$$
So, $$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}$$ = $$\frac{99}{100}$$
RELATED VIDEO
_________________
Test confidently with gmatprepnow.com
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 54371
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
28 Nov 2016, 05:06
nguyendinhtuong wrote:
What is the value of $$A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}$$
A. $$\frac{98}{99}$$
B. $$\frac{99}{100}$$
C. $$\frac{100}{101}$$
D. $$\frac{98}{101}$$
E. $$\frac{99}{101}$$
Similar question to practice: what-is-168951.html
_________________
Manager
Joined: 03 Oct 2013
Posts: 84
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
28 Nov 2016, 11:27
1
Please give Kudos+1 if you like the method
Attachments
Untitled.png [ 104.59 KiB | Viewed 10684 times ]
_________________
P.S. Don't forget to give Kudos on the left if you like the solution
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 5783
Location: United States (CA)
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
29 Nov 2016, 15:53
nguyendinhtuong wrote:
What is the value of $$A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}$$
A. $$\frac{98}{99}$$
B. $$\frac{99}{100}$$
C. $$\frac{100}{101}$$
D. $$\frac{98}{101}$$
E. $$\frac{99}{101}$$
We begin by noting that 1/(1x2) = 1/1 - 1/2, 1/(2x3) = 1/2 - 1/3, 1/(3x4) = 1/3 - 1/4, … , 1/(99x100) = 1/99 - 1/100.
If we substitute each term in the definition of A by the above, we obtain:
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + … + 1/99 - 1/100
Note that all of the terms (besides 1/1 = 1 and 1/100) cancel; therefore,
A = 1/1 - 1/100 = 99/100
_________________
# Scott Woodbury-Stewart
Founder and CEO
[email protected]
122 Reviews
5-star rated online GMAT quant
self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
Intern
Joined: 16 Aug 2016
Posts: 7
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
05 Dec 2016, 18:10
Although I can follow the explanations made in this thread I would not be able to come up with them on my own. How can we make sure that we identify the right pattern in such a question?
Manager
Joined: 03 Oct 2013
Posts: 84
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
05 Dec 2016, 18:14
VS93 wrote:
Although I can follow the explanations made in this thread I would not be able to come up with them on my own. How can we make sure that we identify the right pattern in such a question?
In general when I see a question with summation of products (numerator or denominator) I try to see if I can express the general term as a difference. This would help us cancel the terms in a summation series...
Sent from my Nexus 6 using GMAT Club Forum mobile app
_________________
P.S. Don't forget to give Kudos on the left if you like the solution
Manager
Joined: 13 Dec 2013
Posts: 151
Location: United States (NY)
Schools: Cambridge"19 (A)
GMAT 1: 710 Q46 V41
GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
15 Dec 2016, 07:18
n=1: 1/2
n=2: 1/2 + 1/2x3 = 1/2 + 1/6 = 2/3
n=3: 4/6 + 1/3x4 = 4/6 + 1/12 = 3/4
n=4: 3/4 + 1/4x5 = 3/4 + 1/20 = 4/5
So, we can deduce that: n(x) = (x-1/x) + (1/x(x+1))
Therefore, n(99) = (98/99) + (1/99*100) = 9801/9900 = 99/100
Intern
Status: One more try
Joined: 01 Feb 2015
Posts: 43
Location: India
Concentration: General Management, Economics
WE: Corporate Finance (Commercial Banking)
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
01 Jan 2017, 09:09
nguyendinhtuong wrote:
What is the value of $$A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}$$
A. $$\frac{98}{99}$$
B. $$\frac{99}{100}$$
C. $$\frac{100}{101}$$
D. $$\frac{98}{101}$$
E. $$\frac{99}{101}$$
No need to use pen /pencil
Ex :
1/(1*2) + 1/(2*3)=2/3
Similarly if u expand it you will find 3/4,4/5 going through.
Hence ans last digit 99/100
_________________
Believe you can and you are halfway there-Theodore Roosevelt
Current Student
Status: preparing
Joined: 30 Dec 2013
Posts: 39
Location: United Arab Emirates
Concentration: Technology, Entrepreneurship
GMAT 1: 660 Q45 V35
GMAT 2: 640 Q49 V28
GMAT 3: 640 Q49 V28
GMAT 4: 640 Q49 V28
GMAT 5: 640 Q49 V28
GPA: 2.84
WE: General Management (Consumer Products)
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink]
### Show Tags
26 Jul 2017, 02:00
nguyendinhtuong wrote:
What is the value of $$A=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+...+\frac{1}{99 \times 100}$$
A. $$\frac{98}{99}$$
B. $$\frac{99}{100}$$
C. $$\frac{100}{101}$$
D. $$\frac{98}{101}$$
E. $$\frac{99}{101}$$
Best way to solve these kind of questions is to simplify them to a known form.
1/2 + 1/(2x3) + 1/(3x4) + .............+ 1/(99x100)
since 1/(2x3)= 1/6 = 1/2-1/3, In a similar way all terms can be written in to this form
=1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4- 1/5+............-1/99+1/99-1/100
=1/2 + 1/2 -1/100 (all others cancelled)
=1-1/100 = 99/100
B
Re: What is the value of 1/(1*2)+1/(2*3)+...+1/(99*100)? [#permalink] 26 Jul 2017, 02:00
Display posts from previous: Sort by | 2019-04-19T04:34:24 | {
"domain": "gmatclub.com",
"url": "https://gmatclub.com/forum/what-is-the-value-of-229663.html",
"openwebmath_score": 0.8207427859306335,
"openwebmath_perplexity": 5716.801454999016,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9621075777163566,
"lm_q2_score": 0.8740772466456689,
"lm_q1q2_score": 0.8409563425072468
} |
http://math.stackexchange.com/questions/133211/complex-logarithm-my-answer-is-wrong | # Complex logarithm, my answer is wrong
I am trying to calculate $$\log(-1+i)$$
I have $$\log(-1+i) = \ln|(-1+i)| + i\operatorname{Arg}(-1+i)$$
$$= \ln\sqrt2 + i3\pi/4$$
However when I checked that in matlab and wolfram alpha they have
$$\frac{\ln\sqrt2}{2} + i3\pi/4$$
Can't see what Im doing wrong.
-
## 1 Answer
In fact Wolfram Alpha returns $\dfrac {\ln 2}2+i\dfrac{3\pi}4$ so that both are equal!
-
God damn, I am terrible for overlooking numbers and signs. Its gonna cost me in the exam :( – Jim_CS Apr 18 '12 at 0:44
@jim: fast reading is a common and shared default ;-) – Raymond Manzoni Apr 18 '12 at 6:26 | 2015-09-04T04:04:07 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/133211/complex-logarithm-my-answer-is-wrong",
"openwebmath_score": 0.5844143033027649,
"openwebmath_perplexity": 1924.7233196675788,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9621075777163566,
"lm_q2_score": 0.8740772450055545,
"lm_q1q2_score": 0.8409563409292804
} |
http://mathhelpforum.com/calculus/51173-integral-proofs.html | # Math Help - Integral proofs
1. ## Integral proofs
Actually I have 2 questions?
1) Prove that the integral of (df/dx) from a to be is equal to f(b) - f(a) using the definition of Riemann Integration and the definition of derivative.
I tried using the antiderivative proof but I dont think that is the right direction on for this one.
2) Prove that the integral of f(x)= x^2 sin(1/x) on the interval (-pi, pi) is beteen -2pi^3/3 and 2pi^3/3.
Bit clueless on this one.
2. Originally Posted by Caity
Actually I have 2 questions?
1) Prove that the integral of (df/dx) from a to be is equal to f(b) - f(a) using the definition of Riemann Integration and the definition of derivative.
I tried using the antiderivative proof but I dont think that is the right direction on for this one.
where are you stuck on this one? you've learned about partitions and all that, right?
2) Prove that the integral of f(x)= x^2 sin(1/x) on the interval (-pi, pi) is beteen -2pi^3/3 and 2pi^3/3.
Bit clueless on this one.
note that $-1 \le \sin \left( \frac 1x \right) \le 1$ for $x \ne 0$
thus, $-x^2 \le x^2 \sin \left( \frac 1x \right) \le x^2$
and so $\int_{- \pi}^{\pi}-x^2~dx \le \int_{- \pi}^{\pi} x^2 \sin \left( \frac 1x \right) ~dx \le \int_{- \pi}^{\pi} x^2~dx \Longleftrightarrow$ $2 \int_0^{\pi}-x^2~dx \le \int_{- \pi}^{\pi} x^2 \sin \left( \frac 1x \right) ~dx \le 2 \int_0^{\pi} x^2~dx$ since $x^2$ is even
3. Originally Posted by Caity
1) Prove that the integral of (df/dx) from a to be is equal to f(b) - f(a) using the definition of Riemann Integration and the definition of derivative.
$\int_a^b f'(x) dx = f(b) - f(a)$
This is because $f(x)$ is antiderivative of $f'(x)$.
4. Ok so its not the antiderivative thing. I'll go reread those definitions about partitions and all that.
5. Originally Posted by ThePerfectHacker
$\int_a^b f'(x) dx = f(b) - f(a)$
This is because $f(x)$ is antiderivative of $f'(x)$.
i might be misunderstanding something here, but aren't you using what you're supposed to be proving? namely the fundamental theorem of calculus. i would think we had to go through all that stuff about finding a partition and estimating upper and lower Riemann sums etc etc
6. I think I got it now... I used the definition of the derivative in the integral then used (b-a) for delta x. I replaced x for b. The (b-a)'s cancelled out and I ended up with f(b)-f(a)
7. If $f:[a,b]\to\mathbb R$ is an integrable function in the interval $[a,b]$ and if $\varphi:[a,b]\to\mathbb R$ is a primitive function of $f$ in $[a,b],$ then it verifies that $\int_a^b f(x)\,dx=\varphi(b)-\varphi(a).$
Proof
For any $P=\{x_0,x_1,\ldots,x_n\}$ partition of $[a,b],$ by applyin' the MVT to $\varphi$ in each one of intervals $[x_{i-1},x_i],$ we can ensure that it does exist a point $\mu_i\in[x_{i-1},x_i]$ (for each $i=1,2,\ldots,n$) such that $\varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right)=\varphi '\left( \mu _{i} \right)\left( x_{i}-x_{i-1} \right)=f\left( \mu _{i} \right)\cdot \Delta x_{i}.$ Summing member to member the previous equalities, we get $\varphi (b)-\varphi (a)=\sum\limits_{i=1}^{n}{\big( \varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right) \big)}=\sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}.$
Let $m_i$ and $M_i$ be the infimum and supremum of $f$ in $[x_{i-1},x_i]$ and according to lower and upper sums $s(P)$ and $S(P)$ of $f$ to the $P$ partition, respectively, as one wants it $m_i=f\left( \mu _{i} \right)\le M_i,$ it yields
$s(P)\le \sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}\le S(P)\implies s(P)\le \varphi (b)-\varphi (a)\le S(P).$
Hence, $\varphi (b)-\varphi (a)$ is included between the lower and upper sums of $f$ for any $P$ partition, and this property has the integral $\int_a^b f,$ from here $\int_a^b f(x)\,dx=\varphi(b)-\varphi(a).\quad\blacksquare$
8. Originally Posted by Krizalid
If $f:[a,b]\to\mathbb R$ is an integrable function in the interval $[a,b]$ and if $\varphi:[a,b]\to\mathbb R$ is a primitive function of $f$ in $[a,b],$ then it verifies that $\int_a^b f(x)\,dx=\varphi(b)-\varphi(a).$
Proof
For any $P=\{x_0,x_1,\ldots,x_n\}$ partition of $[a,b],$ by applyin' the MVT to $\varphi$ in each one of intervals $[x_{i-1},x_i],$ we can ensure that it does exist a point $\mu_i\in[x_{i-1},x_i]$ (for each $i=1,2,\ldots,n$) such that $\varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right)=\varphi '\left( \mu _{i} \right)\left( x_{i}-x_{i-1} \right)=f\left( \mu _{i} \right)\cdot \Delta x_{i}.$ Summing member to member the previous equalities, we get $\varphi (b)-\varphi (a)=\sum\limits_{i=1}^{n}{\big( \varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right) \big)}=\sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}.$
Let $m_i$ and $M_i$ be the infimum and supremum of $f$ in $[x_{i-1},x_i]$ and according to lower and upper sums $s(P)$ and $S(P)$ of $f$ to the $P$ partition, respectively, as one wants it $m_i=f\left( \mu _{i} \right)\le M_i,$ it yields
$s(P)\le \sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}\le S(P)\implies s(P)\le \varphi (b)-\varphi (a)\le S(P).$
Hence, $\varphi (b)-\varphi (a)$ is included between the lower and upper sums of $f$ for any $P$ partition, and this property has the integral $\int_a^b f,$ from here $\int_a^b f(x)\,dx=\varphi(b)-\varphi(a).\quad\blacksquare$
so you're great on the theory of integration as well i see
9. What about the definition of the derivative that I'm supposed to use along the the definition of Riemann integration??
10. Originally Posted by Jhevon
where are you stuck on this one? you've learned about partitions and all that, right?
note that $-1 \le \sin \left( \frac 1x \right) \le 1$ for $x \ne 0$
thus, $-x^2 \le x^2 \sin \left( \frac 1x \right) \le x^2$
and so $\int_{- \pi}^{\pi}-x^2~dx \le \int_{- \pi}^{\pi} x^2 \sin \left( \frac 1x \right) ~dx \le \int_{- \pi}^{\pi} x^2~dx \Longleftrightarrow$ $2 \int_0^{\pi}-x^2~dx \le \int_{- \pi}^{\pi} x^2 \sin \left( \frac 1x \right) ~dx \le 2 \int_0^{\pi} x^2~dx$ since $x^2$ is even
I see on the graph that $x^2 sin (1/x)$ is between $-x^2$ and $x^2$ but how did you know that? Also I dont quite understand the last part and how it relates to being between $-2pi^3/3$ and $2pi^3/3.$
11. Originally Posted by Caity
I see on the graph that $x^2 sin (1/x)$ is between $-x^2$ and $x^2$ but how did you know that? Also I dont quite understand the last part and how it relates to being between $-2pi^3/3$ and $2pi^3/3.$
i just multiplied my first inequality through by $x^2$. we know $\sin \left( \frac 1x \right)$ is between -1 and 1 (for $x \ne 0$). if you multiply that through by $x^2$, you get $x^2 \sin \left( \frac 1x \right)$ is between $-x^2$ and $x^2$
as far getting to the claim, just integrate each term in the last inequality i gave. | 2015-08-05T10:02:12 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/51173-integral-proofs.html",
"openwebmath_score": 0.9437620639801025,
"openwebmath_perplexity": 218.5803527328584,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9621075744568837,
"lm_q2_score": 0.8740772335247531,
"lm_q1q2_score": 0.8409563270344833
} |
https://math.stackexchange.com/questions/494244/show-that-the-diagonal-entries-of-symmetric-idempotent-matrix-must-be-in-0-1 | Show that the diagonal entries of symmetric & idempotent matrix must be in [$0,1$]
Show that the diagonal entries of symmetric & idempotent matrix must be in [$0,1$].
Let $A$ be a symmetric and idempotent $n \times n$ matrix. By the definition of eigenvectors and since $A$ is an idempotent,
$Ax=\lambda x \implies A^2x=\lambda Ax \implies Ax=\lambda Ax=\lambda^2 x.$
So $\lambda^2=\lambda$ and hence $\lambda \in \{0,1\}$. To show the part about the "diagonal matrix" I use the fact that every symmetric matrix is diagonalizable.
Is this a complete proof?
• Look at where the upper left entry in $A^2$ comes from. – Gerry Myerson Sep 15 '13 at 9:50
• @GerryMyerson I don't get it, but if you mean the typo that I made in second equation I fix it now. – James C. Sep 15 '13 at 10:00
• No, I mean what I said. When you compute $A^2$, how do you get the upper left entry? What the formula for it, in terms of the entries of $A$. – Gerry Myerson Sep 15 '13 at 10:02
• Do you mean $[0,1]$ or $\{0,1\}$? – Vishal Gupta Sep 15 '13 at 10:25
• @Vishal It means interval from $0$ to $1$ – James C. Sep 15 '13 at 10:28
Expanding my comment to an answer, as OP appears to have lost interest:
Recall the hypotheses: $A$ is $n\times n$, idempotent (so $A^2=A$), and symmetric (so $a_{ij}=a_{ji}$, if we let $a_{ij}$ be the entry in row $i$, column $j$ of $A$).
Looking at the entry in row $i$, column $i$ on both sides of $A=A^2$ we get $$a_{ii}=a_{i1}^2+a_{i2}^2+\cdots+a_{ii}^2+\cdots+a_{in}^2\ge a_{ii}^2$$ But the inequality $a_{ii}\ge a_{ii}^2$ is equivalent to $0\le a_{ii}\le1$.
Let $Q$ be a real symmetric and idempotent matrix of "dimension" $n \times n$. First, we establish the following: The eigenvalues of $Q$ are either $0$ or $1$.
proof. note that if $(\lambda,v)$is an eigenvalue- eigenvector pair of $Q$ we have
$\lambda v=Qv= Q^{2} v=Q(Qv)=Q(\lambda v) = \lambda^{2} v$. Since $v$ is nonzero then the result follows immediately.
With this result at hand the following observation gets us to the desired answer:
Let $e_{i}$ and $q_{ii}$ denote the standard unit vector and $i_{th}$ diagonal element of $Q$, respectively. Then we have
$$0= min \{\lambda_1 ,...,\lambda_{n} \} \leq q_{ii} = e_{i}' Q e_{i} \leq max \{\lambda_{1},...,\lambda_n\}=1$$
All you have shown till now (using idempotency) is that the eigenvalues are either $0$ or $1$.
Now we use symmetry to say that your matrix (let us call it $A$) is unitarily similar to the diagonal matrix consisting of eigenvalues of $A$ on its diagonal. That is
$$A = UDU^{-1}$$
for some unitary matrix $U$ and the diagonal matrix $D$ where $D$ has eigenvalues of $A$ on its diagonal. We already know that these eigenvalues are either $0$ or $1$. So now expand the above representation of $A$ to get the diagonal entries of $A$.
• I think this is the hard way. – Gerry Myerson Sep 15 '13 at 10:41
• I read your comment. That is a indeed an infinitely simpler and neat proof. Why don't you make it an answer so that we can upvote and OP can accept it? – Vishal Gupta Sep 15 '13 at 10:54
• I'm hoping that OP will understand what I'm getting at, and then I will encourage OP to post an answer. I'm willing to give OP a few days to think it over. – Gerry Myerson Sep 15 '13 at 10:57
• @GerryMyerson Nice of you! Do you think it is in general a bad idea to post answers to such questions? So did I make a mistake by posting a solution? – Vishal Gupta Sep 15 '13 at 11:03
• I take it on a case-by-case basis. If I think someone can be led to discover the answer, I usually try to do that. I don't expect everyone to take that approach. I occasionally get upset when someone robs a questioner of the pleasure of discovering an answer, but I am not upset with what you have done here. – Gerry Myerson Sep 15 '13 at 11:08 | 2019-04-20T16:49:48 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/494244/show-that-the-diagonal-entries-of-symmetric-idempotent-matrix-must-be-in-0-1",
"openwebmath_score": 0.9378826022148132,
"openwebmath_perplexity": 214.5673206098742,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9848109529751893,
"lm_q2_score": 0.8539127585282744,
"lm_q1q2_score": 0.8409426374839026
} |
https://www.physicsforums.com/threads/geometry-problem.65148/ | # Geometry Problem
1. Feb 26, 2005
### Jameson
In $$\bigtriangleup ABC , AB = BC$$ and $$CD$$ bisects angle C.
If $$y = \frac{1}{3}x$$ then z = ....
--------------------------------------------
Ok. I've tried to make a bunch of substitutions for each of the angles and I can't seem to solve any for z.
I have found these two equations:
$$y + z + c = 180$$
and
$$x + z + \frac{c}{2} = 180$$
I know I can make substitutions/variants of these, but I still am stuck.
Any help is appreciated.
Jameson
#### Attached Files:
• ###### GeometryProblem.gif
File size:
1.6 KB
Views:
81
2. Feb 26, 2005
### cepheid
Staff Emeritus
Notice that angle CDB = 180 - x. So:
(180 - x) + y + C/2 = 180 (from the upper triangle).
(180 - x) + x/3 + C/2 = 180
C/2 = 2x/3
Can you proceed from here?
3. Feb 26, 2005
### Jameson
I got the same thing. $$\frac{c}{2} = \frac{2x}{3}$$ but unfortunately I still can't seem to do the right substitution.
Substituting (2x)/3 for c/2 I get the equation:
$$x + \frac{2x}{3} + z = 180$$
This still leaves two variables. I just can't see it for some reason.
4. Feb 26, 2005
5. Feb 26, 2005
### Jameson
This is from a published SAT book from College Board, so I think there is a solution.
6. Feb 26, 2005
### cepheid
Staff Emeritus
I, too, assumed that you were meant to solve for z in terms of x, since y was given in terms of x. I don't see what else you can do with it...
7. Feb 26, 2005
### apchemstudent
no no, see how AB = BC, That's a very important clue. We know triangle ABC is an isoceles triangle with angle A = angle C. You can come up with an equation to relate x directly to z and substitute it back into one of the equations to solve for z.
Angle C = 4/3 x = Angle A = z
you should be able to take it from here.
8. Feb 27, 2005
### honestrosewater
Eh, doesn't 2z + y = 180? So knowing y = x/3 you can solve for z in terms of x.
Plus, if it's the CB's "10 Real SAT's", it has the answers. And any other SAT prep book should have the answers too.
Last edited: Feb 27, 2005
9. Feb 27, 2005
### apchemstudent
ok, that equation is essential to solve for z. However, an actual value can be determined here. It's not impossible. I solved it and i got z = 80 degrees. Let me know if you want me to show my work.
10. Feb 27, 2005
### Jameson
apchemstudent... 80 degrees is the correct answer. I would be most gracious if you could show your work.
Jameson
11. Feb 27, 2005
### apchemstudent
ok, we set angle BDC = 180 - x, so that we can solve for Angle BCD
BCD + DBC + CDB = 180 degrees
BCD = 180 -1/3 x - 180 + x
BCD = 2/3 x
BCD = 1/2 BCA so BCA = 4/3 x
Again we know that it is an isoceles triangle with angle BCA = BAC
z = 4/3 x
so BAC + ABC + BCA = 180
z + 1/3 x + 4/3 x = 180
3x = 180
x = 60 <---- you could've substituted x in terms of z and solved for z but i find it easier this way.
now z = 4/3 (60) = 80 degrees. | 2017-10-18T09:57:53 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/geometry-problem.65148/",
"openwebmath_score": 0.6686403751373291,
"openwebmath_perplexity": 1134.810441712771,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9848109489630493,
"lm_q2_score": 0.8539127566694177,
"lm_q1q2_score": 0.8409426322272626
} |
http://math.stackexchange.com/questions/17816/what-is-the-geometric-interpretation-behind-the-method-of-exact-differential-equ | # What is the geometric interpretation behind the method of exact differential equations?
Given an equation in the form $M(x)dx + N(y)dy = 0$ we test that the partial derivative of $M$ with respect to $y$ is equal to the partial derivative of $N$ with respect to $x$. If they are equal, then the equation is exact. What is the geometric interpretation of this?
Further more to solve the equation we may integrate $M(x) dx$ or $N(y)dy$, whichever we like better, and then add a constant as a function in terms of the constant variable and solve this.
e.g. If $f(x) = 3x^2$ then $F(x) = x^3 + g(y)$.
After we have our integral we set its partial differential with respect to the other variable our other given derivative and solve for $g(y)$. I have done the entire homework assignment correctly, but I have no clue why I am doing these steps. What is the geometric interpretation behind this method, and how does it work?
-
Great question. The idea is that $(M(x), N(y))$ defines a vector field, and the condition you're checking is equivalent (on $\mathbb{R}^2$) to the vector field being conservative, i.e. being the gradient of some scalar function $p$ called the potential. Common physical examples of conservative vector fields include gravitational and electric fields, where $p$ is the gravitational or electric potential.
Geometrically, being conservative is equivalent to the curl vanishing. It is also equivalent to the condition that line integrals between two points depend only on the beginning and end points and not only on the path chosen. (The connection between this and the curl is Green's theorem.)
The differential equation $M(x) \, dx + N(y) \, dy = 0$ is then equivalent to the condition that $p$ is a constant, and since this is not a differential equation it is a much easier condition to work with. The analogous one-variable statement is that $M(x) \, dx = 0$ is equivalent to $\int M(x) \, dx = \text{const}$. Geometrically, the solutions to $M(x) \, dx + N(y) \, dy = 0$ are therefore the level curves of the potential, which are always orthogonal to its gradient. The most well-known example of this is probably the diagram of the electric field and the level curves of the electrostatic potential around a dipole. This is one way to interpret the expression $M(x) \, dx + N(y) \, dy = 0$; it is precisely equivalent to the "dot product" of $(M(x), N(y)$ and $(dx, dy)$ being zero, where you should think of $(dx, dy)$ as being an infinitesimal displacement along a level curve.
(For those in the know, I am ignoring the distinction between vector fields and 1-forms and also the distinction between closed forms and exact forms.)
-
At the end of the first paragraph, what were you meaning to say? "I am being slightly biased?" – PEV Jan 17 '11 at 3:55
@Trevor: whoops. I meant to say what I said at the end. – Qiaochu Yuan Jan 17 '11 at 4:00
You might want to look up the wiki article on exact differentials and inexact differentials. The famous physical quantity which you cannot write as an exact differential is heat.
- | 2015-05-26T14:26:46 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/17816/what-is-the-geometric-interpretation-behind-the-method-of-exact-differential-equ",
"openwebmath_score": 0.8729205131530762,
"openwebmath_perplexity": 129.10496533727004,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9848109489630492,
"lm_q2_score": 0.8539127548105611,
"lm_q1q2_score": 0.8409426303966402
} |
https://math.stackexchange.com/questions/1878027/are-real-numbers-a-subset-of-the-complex-numbers | # Are real numbers a subset of the complex numbers? [duplicate]
I am having an argument with a friend. I think that in a sense, the answer is no. My reasoning is that in linear algebra, a vector $(a, b)$ is not the same as a vector $(a, b, 0)$ because the first one is in $\mathbb{R}^2$, while the second is in $\mathbb{R}^3$. However I am not sure if a similar argument can be made for real vs complex numbers.
## marked as duplicate by Matthew Towers, Bungo, user137731, GEdgar, Lee MosherAug 1 '16 at 16:55
• $\mathbb{R}$ is a subfield of $\mathbb{C}$ – reuns Aug 1 '16 at 16:39
• Recall that the set of complex numbers $\Bbb C$ is defined as $\Bbb C=\{a+ib\mid a,b\in\Bbb R\}$ where $i$ is the imaginary unit. You may see each $a+ib$ as an ordered pair $(a,b)$. What is the subset of $\Bbb C$ in which $b=0$ for every element? – learner Aug 1 '16 at 16:40
• But (a,b,0) is in $\mathbb R^2 \times {0}$. Is $\mathbb R^2 \times {0} \ne \mathbb R^2$. In what sense can that statement be meaningful and true? Does $\mathbb C = R^2$? Not really as as $a = a + 0i \in \mathbb R$ but $a + 0i \in \mathbb C$ so ... yes, $\mathbb R \subset \mathbb C$ be any definition. – fleablood Aug 1 '16 at 16:44
• To understand the natural embedding you need to look at the ring / algebra structure, not only the vector space structure. The Hamilton pair construction is a normalized version of $\Bbb C \cong R[x]/(x^2+1),\,$ i.e take $\,\Bbb R[x] =$ ring of polynomials with real coefficients, then work modulu $\,x^2+1.\,$ The embedding arises as a composition of two natural maps: $\,r\mapsto r\, x^0,\,$ which maps a real into a constant polynomial, then mapping that to its congruence class mod $\,x^2+1.\,$ – Bill Dubuque Aug 1 '16 at 17:00
• @BillDubuque Not sure you need all that. The embedding $a\to (a,0)$ works rather well. – zhw. Aug 1 '16 at 17:03
• "No imaginary part"? I think you mean "imaginary part = 0". Anyway, real numbers are not complex numbers with zero imaginary part, because complex numbers are elements of $\mathbb R^2.$ What is true is that the set of complex numbers of the form $(a,0), a \in \mathbb R,$ is isomorphic to $\mathbb R$ in every important mathematical sense you could bring to the table. – zhw. Aug 1 '16 at 16:57 | 2019-08-21T02:49:03 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1878027/are-real-numbers-a-subset-of-the-complex-numbers",
"openwebmath_score": 0.770173966884613,
"openwebmath_perplexity": 264.8669740980354,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9848109520836026,
"lm_q2_score": 0.8539127510928476,
"lm_q1q2_score": 0.8409426294000756
} |
https://math.stackexchange.com/questions/2141199/colored-balls-into-groups | # Colored balls into groups
If 3 red, 3 blue, and 3 green balls are randomly divided into three groups of three balls each, what is the probability that none of the groups have all the balls of the same color?
My attempt:
Let $A$ be the event that all groups have all balls of the same color. There is 1 way that this can occur, irrespective of inter-group ordering and considering same-colored balls to be indistinguishable.
Let $B$ be the event that exactly one group has all balls of the same color. One group of identically-colored balls can be chosen $\binom{3}{1}=3$ ways. The remaining two groups are totally determined by the initial identically-colored group, as they both must consist of two balls of one of the remaining two colors and one ball of the other remaining color. This again is irrespective of the ordering of groups or of balls within groups.
So there are $3+1=4$ ways that $A\cup B$ can occur.
There are 8 total possible outcomes: 4 outcomes belonging to $A \cup B$, 1 outcome in which all three groups contain one ball of each color, and 3 outcomes in which one group contains all differently-colored balls and the other groups contain the remaining balls.
Thus, the probability of $(A \cup B)^C$ is $\frac{1}{2}$.
Am I correct? And, is there a more generalized and efficient way to solve this sort of problem? Took me far too long...
## 1 Answer
The probability depends on how the random selection occurs.
An obvious randomization process is that we put all the balls in an urn, mix them thoroughly, and then blindly draw out one at a time. The first three balls become one group of three, the next three balls are another group of three, and the last three balls also are a group of three.
Now let's consider the probability that the first three balls drawn will all be the same color. The first ball we draw can be any color; call it color X. We now want to draw a second ball of the same color, but there are now $8$ balls in the urn with $2$ of color X, so we have a $\frac14$ chance to draw color X again. Finally, we want to draw another ball of color X. There is only $1$ such ball out of $7$ in the urn, so we have a $\frac17$ probability of drawing it. The probability that the first three balls are all the same color is therefore $\frac14 \times \frac17 = \frac{1}{28}.$
Now in order to have three same-colored groups (all red, all green, or all blue), the first three balls we draw must all be one color. We just found that has probability $\frac{1}{28}.$ But you think we have a $\frac18$ probability ($1$ case out of the $8$ you listed) that all groups have all balls the same color?
It seems you have chosen $8$ separate cases that are not equally likely under a reasonable interpretation of "randomly divided," but you are assuming they all have equal probability. You can of course construct such a probability by making a spinner with eight equal-sized sectors, label each sector with one of your cases, spin the spinner to select a case, and then arrange the balls so they match the pattern selected by the spinner. But if we resort to that kind of "randomly divided," we can make the probability of three-of-a-kind anything we want ($0,$ $1,$ or anything in between).
As a hint of how to get back on track, assuming you consider my drawing-from-the-urn scheme to be sufficiently random, I have already calculated the probability that the first group will be all one color. You can continue the process to find the probability that all three groups will be the same color. From this you can get the probability that only the first group will be one color. Now consider the probability that only the second group will be all one color, and the probability that only the third group will all be one color. Those two probabilities may seem more difficult to calculate at first, but consider the deeper symmetries of the problem: the process gives you a random permutation of the balls with all permutations equally likely.
• Thanks. Your description of the randomization process is correct, and I incorrectly assigned equal probabilities to the identified outcomes. I'm going to hold off on marking this is the answer though because I'd like a full solution (from you or someone else) to understand how I should be approaching the problem. – user1569317 Feb 12 '17 at 19:18
• I followed your hint... I came up with 27/28. The probability that all three groups are same-colored is 1/280, and the probability that only the first group (or only the second, or only the third) is same-colored is 3/280, so the total probability is (1+3+3+3)/280 = 1/28. The deeper symmetry I should have realized is that the probability of any particular group being same-colored is 1/28. Do I have this right, or am I continuing to err? – user1569317 Feb 13 '17 at 15:03
• I think the probability that only the first group is same-colored is $\frac{1}{28} - \frac{1}{280}$--the probability of all cases where the first group is same-colored, minus the cases where the first group and at least one other (which turns out to be all three groups) are same-colored. But I agree with all the rest of your reasoning; add up three times the "only one group" probability and add the "three groups" probability. – David K Feb 13 '17 at 15:14 | 2019-11-14T16:06:25 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2141199/colored-balls-into-groups",
"openwebmath_score": 0.8998207449913025,
"openwebmath_perplexity": 240.84114904767318,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9848109498546361,
"lm_q2_score": 0.8539127529517043,
"lm_q1q2_score": 0.8409426293273552
} |
https://tobydriscoll.net/fnc-julia/matrixanaly/dimreduce.html | # 7.5. Dimension reduction#
The SVD has another important property that proves very useful in a variety of applications. Let $$\mathbf{A}$$ be a real $$m\times n$$ matrix with SVD $$\mathbf{A}=\mathbf{U}\mathbf{S}\mathbf{V}^T$$ and (momentarily) $$m\ge n$$. Another way of writing the thin form of the SVD is
(7.5.1)#$\begin{split}\begin{split} \mathbf{A} = \hat{\mathbf{U}}\hat{\mathbf{S}}\mathbf{V}^T &= \begin{bmatrix} \rule[-0.3em]{0pt}{1em} \mathbf{u}_1 & \mathbf{u}_2 & \cdots & \mathbf{u}_n \end{bmatrix} \: \begin{bmatrix} \sigma_1 & & \\ & \ddots & \\ & & \sigma_n \end{bmatrix} \: \begin{bmatrix} \mathbf{v}_1^T \\ \vdots \\ \mathbf{v}_n^T \end{bmatrix}\ \\ &= \begin{bmatrix} \rule[-0.3em]{0pt}{1em} \sigma_1\mathbf{u}_1 & \cdots & \sigma_n\mathbf{u}_n \end{bmatrix}\: \begin{bmatrix} \mathbf{v}_1^T \\ \vdots \\ \mathbf{v}_n^T \end{bmatrix} \\ &= \sigma_1 \mathbf{u}_{1}\mathbf{v}_{1}^T + \cdots + \sigma_r \mathbf{u}_{r}\mathbf{v}_{r}^T = \sum_{i=1}^r \sigma_i \mathbf{u}_{i}\mathbf{v}_{i}^T, \end{split}\end{split}$
where $$r$$ is the rank of $$\mathbf{A}$$. The final formula also holds for the case $$m<n$$.
Each outer product $$\mathbf{u}_{i}\mathbf{v}_{i}^T$$ is a rank-1 matrix of unit 2-norm. Thanks to the ordering of singular values, then, Equation (7.5.1) expresses $$\mathbf{A}$$ as a sum of decreasingly important contributions. This motivates the definition, for $$1\le k \le r$$,
(7.5.2)#$\mathbf{A}_k = \sum_{i=1}^k \sigma_i \mathbf{u}_{i}\mathbf{v}_{i}^T = \mathbf{U}_k \mathbf{S}_k \mathbf{V}_k^T,$
where $$\mathbf{U}_k$$ and $$\mathbf{V}_k$$ are the first $$k$$ columns of $$\mathbf{U}$$ and $$\mathbf{V}$$, respectively, and $$\mathbf{S}_k$$ is the upper-left $$k\times k$$ submatrix of $$\mathbf{S}$$.
The rank of a sum of matrices is always less than or equal to the sum of the ranks, so $$\mathbf{A}_k$$ is a rank-$$k$$ approximation to $$\mathbf{A}$$. It turns out that $$\mathbf{A}_k$$ is the best rank-$$k$$ approximation of $$\mathbf{A}$$, as measured in the matrix 2-norm.
Theorem 7.5.1
Suppose $$\mathbf{A}$$ has rank $$r$$ and let $$\mathbf{A}=\mathbf{U}\mathbf{S}\mathbf{V}^T$$ be an SVD. Let $$\mathbf{A}_k$$ be as in (7.5.2) for $$1\le k < r$$. Then
1. $$\| \mathbf{A} - \mathbf{A}_k \|_2 = \sigma_{k+1}, \quad k=1,\ldots,r-1$$, and
2. If the rank of $$\mathbf{B}$$ is $$k$$ or less, then $$\| \mathbf{A}-\mathbf{B} \|_2\ge \sigma_{k+1}$$.
Proof
(part 1 only) Note that (7.5.2) is identical to (7.5.1) with $$\sigma_{k+1},\ldots,\sigma_r$$ all set to zero. This implies that
$\mathbf{A} - \mathbf{A}_k = \mathbf{U}(\mathbf{S}-\hat{\mathbf{S}})\mathbf{V}^T,$
where $$\hat{\mathbf{S}}$$ has those same values of $$\sigma_i$$ replaced by zero. But that makes the above an SVD of $$\mathbf{A} - \mathbf{A}_k$$, with singular values $$0,\ldots,0,\sigma_{k+1},\ldots,\sigma_r$$, the largest of which is $$\sigma_{k+1}$$. That proves the first claim.
## Compression#
If the singular values of $$\mathbf{A}$$ decrease sufficiently rapidly, then $$\mathbf{A}_{k}$$ may capture the most significant behavior of the matrix for a reasonably small value of $$k$$.
Demo 7.5.2
We make an image from some text, then reload it as a matrix.
plot(annotations=(0.5,0.5,text("Hello world",44,:center,:middle)),
grid=:none,frame=:none,size=(400,150))
savefig("hello.png")
A = @. Float64(Gray(img))
Gray.(A)
Next we show that the singular values decrease until they reach zero (more precisely, until they are about $$\epsilon_\text{mach}$$ times the norm of the matrix) at around $$k=45$$.
U,σ,V = svd(A)
scatter(σ,xaxis=(L"i"), yaxis=(:log10,L"\sigma_i"),
title="Singular values")
The rapid decrease suggests that we can get fairly good low-rank approximations.
plt = plot(layout=(2,2),frame=:none,aspect_ratio=1,titlefontsize=10)
for i in 1:4
k = 3i
Ak = U[:,1:k]*diagm(σ[1:k])*V[:,1:k]'
plot!(Gray.(Ak),subplot=i,title="rank = \$k")
end
plt
Consider how little data is needed to reconstruct these images. For rank-9, for instance, we have 9 left and right singular vectors plus 9 singular values, for a compression ratio of better than 12:1.
m,n = size(A)
compression = m*n / (9*(m+n+1))
12.099213551119178
## Exercises#
1. ✍ Suppose that $$\mathbf{A}$$ is an $$n\times n$$ matrix. Explain why $$\sigma_n$$ is the distance (in 2-norm) from $$\mathbf{A}$$ to the set of all singular matrices.
2. ✍ Suppose $$\mathbf{A}$$ is a $$7\times 4$$ matrix and the eigenvalues of $$\mathbf{A}^*\mathbf{A}$$ are 3, 4, 7, and 10. How close is $$\mathbf{A}$$ in the 2-norm to (a) a rank-3 matrix? (b) a rank-2 matrix?
3. (a) ⌨ Find the rank-1 matrix closest to
$\begin{split} \mathbf{A}=\displaystyle \begin{bmatrix} 1 & 5 \\ 5 & 1 \end{bmatrix}, \end{split}$
as measured in the 2-norm.
(b) ⌨ Repeat part (a) for
$\begin{split} \mathbf{A}=\displaystyle \begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix}. \end{split}$
4. ✍ Find the rank-1 matrix closest to
$\begin{split} \mathbf{A}=\displaystyle \begin{bmatrix} 1 & b \\ b & 1 \end{bmatrix}, \end{split}$
as measured in the 2-norm, where $$b>0$$.
5. ⌨ Following Demo 7.5.2 as a guide, load the “mandrill” test image and convert it to a matrix of floating-point pixel grayscale intensities. Using the SVD, display as images the best approximations of rank 5, 10, 15, and 20.
1
In statistics this quantity may be interpreted as the fraction of explained variance. | 2022-09-25T03:58:44 | {
"domain": "tobydriscoll.net",
"url": "https://tobydriscoll.net/fnc-julia/matrixanaly/dimreduce.html",
"openwebmath_score": 0.8358561396598816,
"openwebmath_perplexity": 425.18122406494876,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9940889311437562,
"lm_q2_score": 0.845942439250491,
"lm_q1q2_score": 0.8409420152436624
} |
https://math.stackexchange.com/questions/2317791/can-you-ignore-unused-quantifiers | # Can you ignore 'unused' quantifiers?
My question is simple, as the title says: can I safely remove (or add) a quantifier if the variable bound to it is not used in any predicate following it?
As an example: I wanted to prove $$\exists x(P(x) \implies \forall yP(y))$$ My proof went like this: $$\exists x(P(x) \implies \forall yP(y)) \iff \exists x(\neg P(x) \lor \forall yP(y)) \iff \neg \forall xP(x) \lor \exists x \forall yP(y)$$ (The last equivalence follows from distributivity of existential quantifier over disjunction).
Now, we see that in the second clause the $x$ is not used, so it could be any element of the universe of discourse. Hence we remove it and replace the bound variable $y$ to $x$ to get: $$\neg \forall xP(x) \lor \forall xP(x)$$ An obvious tautology. Q.E.D. (?)
It seems to me that all I've done is legal, at least assuming that the universe of discourse is not an empty set. Am I correct or can I not do this? Equivalently can I add quantifiers (like in the example below)? $$\forall xP(x) \land Q(y) \iff \forall xP(x) \land \forall xQ(y) \iff \forall x(P(x) \land Q(y))$$
• yes you can ignore quantifiers that don't bind anything. – Apostolos Jun 10 '17 at 21:45
Yes, this is correct. More precisely, if $P$ is any formula in which the variable $x$ does not appear, then $\exists x P$ and $\forall x P$ are each equivalent to $P$ over any nonempty universe of discourse (if you don't know the universe is nonempty, all you can say is $\exists x P \implies P$ and $P\implies\forall x P$).
(To be clear, if you're working in some specific formal deductive system rather than the informal logic that mathematicians usually work with, you would need to justify these statements using only the rules of your deductive system. Exactly how you do this depends on what your deductive system is.)
Yes, those quantifiers are called Null Quantifiers. And we have as a general equivalence principle:
Null Quantification
Where $\varphi$ does not contain $x$ as a free variable:
$\forall x \varphi \Leftrightarrow \varphi$
$\exists x \varphi \Leftrightarrow \varphi$
• How can you prove this? – Ben-ZT Feb 26 '18 at 3:34
• @Ben-ZT Are you looking for a proof based on the formal semantics of the quantifiers, or a formal derivation? – Bram28 Feb 26 '18 at 12:52 | 2021-04-12T19:27:28 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2317791/can-you-ignore-unused-quantifiers",
"openwebmath_score": 0.7007634043693542,
"openwebmath_perplexity": 248.73956690207268,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9678992923570262,
"lm_q2_score": 0.8688267813328976,
"lm_q1q2_score": 0.8409368268329444
} |
https://math.stackexchange.com/questions/3367545/a-first-course-in-mathematical-analysis-david-alexander-brannan-ch-1-1-6-exer | A First Course in Mathematical Analysis, David Alexander Brannan, ch 1, 1.6 exercises, section 1.3, problem 3
Prove the inequality $$3^{n}\ge2n^{2}+1$$
for $$n=1,2,\dots$$
This looks like a problem which might be solved using the binomial theorem. Recall \begin{align*} (1+x)^{n} & =1+nx+{n \choose 2}x^{2}+\dots\\ & =1+nx+\frac{n!}{(n-2)!2!}x^{2}+\dots\\ & =1+nx+\frac{n(n-1)}{2}x^{2}+\dots \end{align*}
Using the binomial theorem to expand the LHS of our original inequality, we have \begin{align*} 3^{n}=(1+2)^{n} & \ge2n^{2}+1\\ 1+2n+\frac{n(n-1)}{2}2^{2}+\dots & \ge1+2n^{2}\\ 1+2n+2n(n-1)+\dots & \ge1+2n^{2}\\ 1+2n+2n^{2}-2n+\dots & \ge1+2n^{2}\\ 1+2n^{2}+\dots & \ge1+2n^{2} \end{align*}
This is obviously true.
Let's try another approach: induction.
Base case
\begin{align*} 3^{0} & \ge2(0)^{2}+1\\ 1 & \ge1 \end{align*}
Now we suppose the normal case $$3^{k}\ge2k^{2}+1$$
For the inductive step, multiple both sides by 3
\begin{align*} 3\cdot3^{k} & \ge3(2k^{2}+1)\\ 3^{k+1} & \ge6k^{2}+3 \end{align*}
So it is sufficient for us to prove that \begin{align*} 6k^{2}+3 & \ge2(k+1)^{2}+1\\ & =2(k^{2}+2k+1)+1\\ & =2k^{2}+4k+3\\ 6k^{2}-2k^{2}+3-3 & \ge4k\\ 4k^{2} & \ge4k \end{align*}
which is obviously true. Thus our original inequality holds
$$3^{n}\ge2n^{2}+1.$$
Question: are there problems with either proof or style things that I could learn?
Question: For the inductive proof, did I choose the correct base case of zero? It seemed correct given the context, but I am not 100% certain.
I don't have a solution to check against. Are my proofs valid?
• your proofs look basically correct; you could choose base case $0$ to prove the proposition for $0, 1, 2, ...$ or base case $1$ to prove it for $1, 2, ...$ – J. W. Tanner Sep 24 '19 at 2:18
• How and why do you choose 0 or 1? For me, I chose 0 because there didn't seem to be any stated constraints on n, say by another inequality, and we are considering the natural numbers in induction. Is that proper reasoning? – Joe Sep 24 '19 at 2:20
• you could choose whichever you like if it works, but if you're asked to prove something for $1,2,...$, there's no need to prove it for $0$ – J. W. Tanner Sep 24 '19 at 2:21
• I forgot to consider the problem statement of domain for n. Sorry for the dumb question. – Joe Sep 24 '19 at 2:22
Your proofs look basically correct. When you did the proof by induction, you took the base case as $$0$$; you could have merely started with the base case of $$1$$, since the question asked to prove the inequality only for $$n=1,2,...$$. | 2021-01-18T19:26:46 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3367545/a-first-course-in-mathematical-analysis-david-alexander-brannan-ch-1-1-6-exer",
"openwebmath_score": 0.8785029649734497,
"openwebmath_perplexity": 516.8127895686023,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9678992923570261,
"lm_q2_score": 0.8688267796346598,
"lm_q1q2_score": 0.8409368251892211
} |
https://math.stackexchange.com/questions/2010959/monotonicity-of-quadratic-function-at-its-vertex | # Monotonicity of quadratic function at its vertex?
In my school text book, It say that the quadratic function is increasing and decreasing at some intervals based on the function, however both intervals didn't include the vertex and it was't discussed at all.
Should the vertex be constant because its slop is equal to zero? Or should I just leave it and not include it in any interval like the text book proposed? And how come I can leave it while it is from the domain of the function?
Example:
Discuss the monotonicity of the function $f(x)= {x}^{2}$:
$f(x) \text{ increases at }x \in ]0,\infty[$
$f(x) \text{ decreases at }x \in ]-\infty,0[$
As you can see they didn't include zero in any of the intervals.
Should the vertex be constant because its slop is equal to zero?
Monotonicity is generally defined on a subset of the domain which includes multiple points e.g. an interval, so (again, in general) it doesn't make sense to refer to monotonicity at one single point.
(As a side note, it does make sense for differentiable functions to associate the derivative at a point with the "rate of change" in some sense, and it's common to say that $f(x)$ is increasing at $x_0$ if $f'(x) \gt 0$ however that's more of a casual language license, and also it's a sufficient condition but not a necessary one, for example $f(x) = x^3$ is strictly increasing on $\mathbb{R}$ but $f'(0)=0$.)
Back to the question, what the book says is correct. However, it is equally correct, and in fact a stronger statement, to say that $f(x)$ is strictly decreasing on $(-\infty,0\,]$ and strictly increasing on $[\,0,\infty)$ where both intervals include $0$.
And how come I can leave it while it is from the domain of the function?
There is no requirement (or guarantee) that the intervals of monotonicity cover the entire domain.
For example $f(x) = x \cdot sin \frac{1}{x}$ for $x \ne 0$ with $f(0)=0$ is a continuous function which is not monotonic on any interval that includes $0$. For an example of a discontinuous function which is not monotonic on any interval consider the indicator function of $\mathbb{Q}$.
• Thanks, do you mind telling me why you didn't use a square bracket on the infinity and negative infinity $(-\infty,0\,]$? – Omar Ahmad Nov 13 '16 at 17:27
• Infinities are not real numbers ($\pm \infty \not \in \mathbb{R}$) so they can't belong to a real interval. The notation $x \in (-\infty,0]$ simply means $x \le 0$. – dxiv Nov 13 '16 at 17:30
If you would choose to put zero in one of the two intervals, which one would you choose?
At the vertex the slope is zero, so the function is neither increasing nor decreasing. | 2019-09-17T02:14:10 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2010959/monotonicity-of-quadratic-function-at-its-vertex",
"openwebmath_score": 0.8437479734420776,
"openwebmath_perplexity": 199.9364509911997,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9678992905050947,
"lm_q2_score": 0.8688267762381843,
"lm_q1q2_score": 0.8409368202927673
} |
https://www.jiskha.com/questions/509900/related-rates-gas-is-escaping-from-a-spherical-balloon-at-the-rate-of-2ft-3-min-how-fast | Related Rates:
Gas is escaping from a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? (A sphere of radius r has volume v=4/3 pi r^3 and surface area S=4pi r^2.)
Remember that ds/dt = ds/dr X dr/dt
Step 1: Find ds/dr
I have no clue where to start. I was going to set 2 equal to 4 pi r^2 and then take the derivative but I really have no idea.
Step 2: Find dr/dr. (HInt dv/dt= dv/dr X dr/dt
Would I set 2 equal to volume and then take derivative?
Step 3: Find ds/dt?
Step 4: Evaluate ds/dt when the radius is 12ft.
After I find my equation I would just plug in 12 correct?
1. 👍 0
2. 👎 0
3. 👁 1,134
1. S = 4pi*r^2
dS/dr = 4pi*2r = 8pi*r
V = (4/3)pi*r^3
dV/dr = 4pi*r^2
dV/dt = dV/dr * dr/dt
2 = 4pi*r^2 * dr/dt
dr/dt = 1/(2pi*r^2)
dS/dt = dS/dr * dr/dt
= 8pi*r * 1/(2pi*r^2)
= 4/r
If r = 12, just plug the value to the last equation
1. 👍 1
2. 👎 1
2. How did you get 12?
1. 👍 0
2. 👎 1
3. Nevermind I meant how did you get 4/r but I figured it out. Thank You!!
1. 👍 0
2. 👎 1
4. Is this correct for the last part?
(8)(3.14)(12) X (1/(2)(3.14)(12^2) = 4/12
1. 👍 0
2. 👎 0
5. Yes
1. 👍 0
2. 👎 0
6. Ok do I actually have to do the calculations though?
1. 👍 0
2. 👎 0
7. If you are not given the value, that means you have to write the steps I told you until the very last equation.
If you are given the value (of r), you don't have to put the value in every equation. Just follow my steps until the last equation, then plug the value after that. It's easier that way.
1. 👍 0
2. 👎 0
8. OK thank you!
1. 👍 0
2. 👎 0
9. Simplify the answer (4/12) into 1/3
1. 👍 0
2. 👎 0
## Similar Questions
1. ### calculus help thanks!
The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at
2. ### Math
A spherical balloon is to be deflated so that its radius decreases at a constant rate of 12 cm/min. At what rate must air be removed when the radius is 5 cm? Must be accurate to the 5th decimal place. It seems like an easy
3. ### math
The launch site for trigon balloon co. is 250 ft above sea level. A hot air balloon is launched from the site and begins to rise at a rate of 110 ft/min. At the same time, another balloon 2200 ft above sea level begins to descend
4. ### differential calculus
ADIABATIC EXPANSION when a certain polyatomic gas undergoes adiabatic expansion,its pressure p and volume V satisfy the equation pV^1.3 =k,where k is constant.Find the relationship between the related rates dp/dt and dV/dt. this
1. ### Calculus
A spherical balloon is inflated at a rate of 10 cubic feet per minute. How fast is the radius of the balloon changing at the instant the radius is 4 feet? And The radius of a circle is decreasing at a rate of 2 ft/minute. Find the
2. ### Calculus
a spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. how fast is the radius of the balloon increasing at the instant the radius is 30 centimeters?
3. ### PHYSICS! HELIUM BALLON
Under standard conditions, the density of air is 1.293 kg/m3 and that of helium is 0.178 kg/m3. A spherical helium balloon lifts a basket plus cargo of mass 277.0 kg. What must be the minimum diameter of the spherical balloon?
4. ### Calculus
Air is being pumped into a spherical balloon so that its volume increases at a rate of 30 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 19 cm? V= 4/3*pi*r^3 S= 4 pi r^2
1. ### ab calculus
Air is being blown into a spherical balloon at the rate of 75 cm3/s. Determine the rate at which the radius of the balloon is increasing when the radius is 28 cm.
2. ### calculus
A spherical balloon is losing air at the rate of 2 cubic inches per minute. How fast is the radius of the ballon shrinking when the radius is 8 inches.
3. ### Calculus
A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2cm/min. At what rates are the volume and surface area of the balloon increasing when the radius is 5cm?
4. ### math
a spherical ballon is inflated with gas at the rate 20cm^3 min. how fast is the radius of the ballon changing at the instant when the radius is 2cm | 2020-09-22T05:43:37 | {
"domain": "jiskha.com",
"url": "https://www.jiskha.com/questions/509900/related-rates-gas-is-escaping-from-a-spherical-balloon-at-the-rate-of-2ft-3-min-how-fast",
"openwebmath_score": 0.8398661017417908,
"openwebmath_perplexity": 1135.342167442634,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9678992951349231,
"lm_q2_score": 0.8688267694452331,
"lm_q1q2_score": 0.8409368177403934
} |
https://math.stackexchange.com/questions/2268425/proof-verification-let-x-tau-be-a-topological-space-such-that-every-subset | # Proof verification- Let $(X,\tau)$ be a topological space such that every subset is closed then prove that $(X,\tau)$ is a discrete space.
Let $(X,\tau)$ be a topological space. I need to show that if every subset is closed then it is a discrete space.
For finite $X$, let $S$ be a subset of $X$. Since $S$ is closed, $X \setminus S \in \tau$. But $X \setminus S \subseteq X$. Therefore, $X \setminus S$ is closed and $S$ is open. Since the choice of $S$ was arbitrary, $(X,\tau)$ is a discrete space.
My doubt is whether the proof holds for $X$ being infinite too.
• Your proof is okay but contains a redundancy. The obervation that $S$ is closed and consequently that its complement is open is not really used in your proof, and can be left out. Yes, the proof holds also if $X$ is infinite. – drhab May 6 '17 at 12:40
Your proof looks good to me. Actually you did not use the fact that $X$ is finite. The same logic applies to any $X$.
$\forall s \subset X$, $X \setminus s \subset X$, and thus $X \setminus s$ is closed. Thus $s \in \tau$. - we are not involving anything about the finiteness of $X$.
• Does the same logic apply to both finite and infinite subsets of $X$? – Parth May 6 '17 at 12:28
• @Parth yes, it does. On your proof, which step did you think not work for infinite set? - Are you concerning "arbitrary choice of subset s from an infinite set X"? Notice we are not involving any additional axiom here - we just say "given any set". – Yujie Zha May 6 '17 at 12:29
• math.stackexchange.com/questions/677459/… here one of the comments states that $X \setminus S$ is closed only when it is finite. – Parth May 6 '17 at 12:34
• @Parth Ah, get it. So for the link, it says every infinite subset of X is closed. That's is different, meaning finite subset is not guaranteed to be closed. However, you do not have the issue here, every subset is closed by your question. – Yujie Zha May 6 '17 at 12:35
• So shouldn't the comment state that $X \setminus S$ is closed only when it is infinite as the question guarantees that all infinite subsets are closed? – Parth May 6 '17 at 12:38
Let $x$ be a point of $X$, $X-\{x\}$ is closed so its complementary subset $\{x\}$ is open. Thus every subset is open since it is the union of its elements.
Why wouldn't the proof work for infinite sets/spaces? Take any subset $\;A\subset X\;$, then $\;X\setminus A\;$ is closed and thus $\;A\;$ is open, and thus any subset of $\;X\;$ is open and we have the discrete topology... | 2019-10-17T16:29:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2268425/proof-verification-let-x-tau-be-a-topological-space-such-that-every-subset",
"openwebmath_score": 0.9275720715522766,
"openwebmath_perplexity": 199.4164536352394,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9825575193498701,
"lm_q2_score": 0.8558511543206819,
"lm_q1q2_score": 0.8409229871220522
} |
https://marksmath.org/classes/Spring2019ComplexDynamics/text/section-13.html | ## Section2.11A few notes on computation
Many of the results in these notes have been illustrated on the computer and some of the exercises require a computational approach. Whenever using the computer, it is always wise to examine the results critically. Here's a simple numerical example where things clearly go awry. In it, we are iterating the function $f(x)=x^2-9.1x+1$ from the fixed point $x_0=0.1\text{.}$ We should generate just a constant sequence.
Uh-oh!
It must be understood that this is a simple consequence of the nature of floating point arithmetic. Part of the issue is that the decimal number $0.1$ or $1/10$ is not exactly representable in binary. In fact,
\begin{equation*} \frac{1}{10} = 0_{\dot2}0\overline{0011} = \frac{1}{2}\sum_{k=1}^{\infty}\frac{3}{16^k}. \end{equation*}
Thus, the computer must introduce round-off error in the computation. Furthermore, $0.1$ is a repelling fixed point of the function. Thus, that round-off error is magnified with each iteration. Our study of dynamics has illuminated a critical issue in numerical computation!
In the terminology of numerical analysis, computation near an attractive fixed point is stable while computation near a repelling fixed point is unstable. Generally speaking, stable computation is trustworthy while unstable computation is not. The implication for the pictures that we see here is that illustration of attractive behavior should be just fine. In figure Figure 2.3.1, for example, images (b) and (c) illustrate attraction to a fixed point that not only involves stable computation but also agrees with our theoretical development. We are happy with those figures. The cobweb plot shown in Figure 2.3.1 (d), however, should frankly be viewed with some suspicion.
The same can be said for the bifurcation diagram in figure Figure 2.6.3. In much of that image, we see a gray smear indicating chaos. How can we trust that? Well, first, theory tells us that there really is chaos. That is, there are orbits that are dense in some interval for many $c$ values. Furthermore, much of the image shows attractive regions and we can be confident in that portion.
In fact, in many of the images that we will generate later - Julia sets, the Mandelbrot set, and similar images - the stable region dominates. Thus, we can be confident in overall image because the unstable region is the complement of the stable region. We might not be confident in computations involving some particular point, but we can be confident in the overall picture. (This will, perhaps, be more clear as we move into complex dynamics.)
Nonetheless, sometimes we want to experiment with genuinely unstable dynamics. One way to improve our confidence in these kinds of computations is to use high precision numbers. Consider, for example, the cobweb plot of the doubling map shown in figure Figure 2.8.1. A naive approach to generate the first few terms of an orbit associated with the doubling map might be as follows:
We've reached the fixed point zero and now we're stuck! Even if we iterate 1000 times, we'll generate a cobweb plot that looks like figure Figure 2.11.1
The cobweb plot shown in figure Figure 2.8.1 was generated using the mpmath multi-precision library for Python with code that looked something like so:
While the truncated output looks the same, note that this was after 1000 iterates. This behavior makes perfect sense if you understand that the doubling map loses one bit of precision with every iterate. | 2019-02-17T18:21:52 | {
"domain": "marksmath.org",
"url": "https://marksmath.org/classes/Spring2019ComplexDynamics/text/section-13.html",
"openwebmath_score": 0.6882277727127075,
"openwebmath_perplexity": 443.57019126515155,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9825575142422757,
"lm_q2_score": 0.8558511488056151,
"lm_q1q2_score": 0.8409229773318412
} |
https://math.stackexchange.com/questions/375950/asymptotic-expansion-of-the-integral-int-01-exn-dx-for-n-to-infty | # Asymptotic expansion of the integral $\int_0^1 e^{x^n} dx$ for $n \to \infty$
The integrand seems extremely easy:
$$I_n=\int_0^1\exp(x^n)dx$$
I want to determine the asymptotic behavior of $I_n$ as $n\to\infty$. It's not hard to show that $\lim_{n\to\infty}I_n=1$ follows from Lebesgue's monotone convergence theorem. However, when I try to obtain more precise results, I confront difficulties. Since it's not of the canonical form of Laplace's method, I have no powerful tools to estimate $I_n$.
Is there any good approach for that? I will be pleased if we can obtain the asymptotic expansion of $I_n$. Thanks!
• The existing answers are better, but note that $e^{x^n}$ is concave and so lies above its tangent line (and obviously exceeds $1$ as well). Note that the slope of the tangent line to $y=e^{x^n}$ at $x=1$ equals $ne$. Therefore $I_n$ is larger than the integral of the piecewise linear function going through the three points $(0,1)$, $(1-1/ne,1)$, and $(1,e)$, which is $1+1/2en$. A similar argument involving secant lines would probably yield an upper bound of the same order of magnitude. Apr 29, 2013 at 6:59
• The integral seems easy, and the behavior of the expansion is expected, but the coefficients are totally not trivial. This turned out to be a very cool problem - thanks for posting! Apr 29, 2013 at 13:42
You can expand the exponential in a Taylor series quite accurately:
$$\exp{\left ( x^n \right )} = 1 + x^n + \frac12 x^{2 n} + \ldots$$
Because $x \in [0,1]$, this series converges rapidly as $n \to \infty$.
Then the integral is
$$1 + \frac{1}{n+1} + \frac12 \frac{1}{2 n+1} + \ldots = \sum_{k=0}^{\infty}\frac{1}{k!} \frac{1}{k n+1}$$
We can rewrite this as
\begin{align}I_n&=1+\frac{1}{n} \sum_{k=1}^{\infty} \frac{1}{k \cdot k!} \left ( 1+\frac{1}{k n} \right )^{-1}\\ &= 1+\frac{1}{n} \sum_{m=0}^{\infty} \frac{(-1)^m}{n^m} \: \sum_{k=1}^{\infty} \frac{1}{k^{m+1} k!}\\ &= 1+\sum_{m=1}^{\infty} (-1)^{m+1}\frac{K_m}{n^m} \end{align}
where
$$K_m = \sum_{k=1}^{\infty} \frac{1}{k^{m} k!}$$
To first order in $n$:
$$I_n \sim 1+\frac{K_1}{n} \quad (n \to \infty)$$
where
$$K_1 = \sum_{k=1}^{\infty} \frac{1}{k\, k!} = \text{Ei}(1) - \gamma \approx 1.3179$$
This checks out numerically in Mathematica.
BONUS
As a further check, I computed the following asymptotic approximation:
$$g(n) = 1+\frac{K_1}{n} -\frac{K_2}{n^2}$$
where
$$K_2 = \sum_{k=1}^{\infty} \frac{1}{k^2 k!} \approx 1.1465$$
I computed
$$\log_2{\left[\frac{\left|g\left(2^m\right)-I_{2^m}\right|}{I_{2^m}}\right]}$$
for $m \in \{1,2,\ldots,9\}$ The results are as follows
$$\left( \begin{array}{cc} 1 & -4.01731 \\ 2 & -6.56064 \\ 3 & -9.26741 \\ 4 & -12.0963 \\ 5 & -15.0028 \\ 6 & -17.9538 \\ 7 & -20.9287 \\ 8 & -23.916 \\ 9 & -26.9096 \\ \end{array} \right)$$
Note that the difference between successive elements is about $-3$; because this is a log-log table, that means that this error is $O(1/n^3)$ and that the approximation is correct.
• Very good, thanks! Apr 29, 2013 at 6:59
• Since $\exp(x^n)=\sum_0^\infty x^{nk}/k!$ converges absolutely on $[0,1]$, the interchange of summation and integration is justified. The interchange of summation is valid for the absolute convergence of the double sum. Am I right? Apr 29, 2013 at 7:18
• @FrankScience: absolutely correct. Apr 29, 2013 at 7:20
• @Ron Gordon The first few terms are numerically $\left\{\frac{1.3179}{n},-\frac{1.1465}{n^2},\frac{1.0694}{n^3},-\frac{1.03348}{n^4},\frac{1.01635}{n^5}\right\}$ Neglecting the difference from unity in the numerator and summing up the asmptotic series the asymptotic behaviour of the integral can be written in this approximation as $\frac{n+2}{n+1}$ Nov 7, 2019 at 10:05
Let $n \geq 1$. A first crude estimate can be obtained as follows. Substitute $x \leftarrow x^{1/n}$ to get
$$I_n = \frac{1}{n}\int_0^1 x^{\frac{1}{n} - 1} e^x dx.$$
Now we can estimate $e^x$ on the interval $[0,1]$ to get
$$\frac{1}{n}\int_0^1 x^{\frac{1}{n}-1}(1+x)\, dx < I_n < \frac{1}{n}\int_0^1 x^{\frac{1}{n} - 1} (1 + (e-1)x)\, dx$$ or $$1 + \frac{1}{n+1} < I_n < 1 + \frac{e-1}{n+1}.$$
• -- very nice! -- Apr 29, 2013 at 7:00
A related technique. Here is a start. Integrating by parts gives, $$I= {{\rm e}}-{\frac {{{\rm e}}\,n}{1+n}}+{\frac {{{\rm e}}\,{n}^{ 2}}{2\,{n}^{2}+3\,n+1}}-{\frac {{{\rm e}}\,{n}^{3}}{6\,{n}^{3}+11\,{ n}^{2}+6\,n+1}}$$
$$+{\frac{{{\rm e}}\,{n}^{4}}{24\,{n}^{4}+50\,{n}^{3}+ 35\,{n}^{2}+10\,n+1}}-\int _{0}^{1}\!{\frac {{n}^{5}{{\rm e}^{{ x}^{n}}}{x}^{5\,n+1}}{x \left( 4\,n+1 \right) \left( 3\,n+1 \right)\left( 1+2\,n \right) \left( 1+n \right) }}{dx}.$$
From the above, we can see that as $n\to \infty$, the integral approaches $1$
$$I = \rm e( 1 - 1 + \frac{1}{2} - \frac{1}{3!}+\frac{1}{4!} - \dots)=\rm e \rm e^{-1}=1.$$ | 2022-07-04T01:06:07 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/375950/asymptotic-expansion-of-the-integral-int-01-exn-dx-for-n-to-infty",
"openwebmath_score": 0.9875257015228271,
"openwebmath_perplexity": 304.8925525966959,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9825575137315161,
"lm_q2_score": 0.8558511451289037,
"lm_q1q2_score": 0.8409229732821266
} |
https://math.stackexchange.com/questions/2489660/prove-that-sum-i-02n-1i-frac1i2n-i-0-for-any-n2 | # Prove that $\sum_{i=0}^{2n}(-1)^i\frac{1}{i!(2n-i)!} = 0$ for any $n>2$
I've been trying to solve this problem lately, but I have been unable to do it. I want to prove that
$$\sum_{i=0}^{2n}(-1)^i\frac{1}{i!(2n-i)!} = 0$$
For any $n>2$. We can generalize the problem changing $2n$ to $a$, I don't mind, although I am focusing on the even numbers, thus the reason for using $2n$.
I have no trouble in showing that the sum is 0 for $n = \infty$, but I am interested in finding a solution for any given $n$. I have tried using induction, but since the $n$ variable is also in the denominator I cannot cleanly let the term out of the expression.
If you multiply with $(2n)!$, then with use of Binomial theorem you get
$$\sum_{i=0}^{2n}(-1)^i\frac{(2n)!}{i!(2n-i)!} = \sum_{i=0}^{2n}(-1)^i{2n\choose i} = (1-1)^{2n} =0$$
• Thanks for the answer! I was so close-minded in trying to solve it by induction that I could not imagine making use of Binomial Theorem. – Alex Martínez Ascensión Oct 25 '17 at 21:52
Surely, this question has been answered here before. I am only giving an alternative probabilistic/combinatorial proof.
Let $C$ be the collection of all subsets of $[k]:=\{1,2,\ldots,k\}$, where $k$ is a fixed positive integer. Denote by $E$ the subcollection of $C$ consisting of sets with even number of elements, and write $O$ for the complement of $E$ in $C$. Then, there exists an involution $i:C\to C$ such that $i$ maps $E$ bijectively onto $O$ and, likewise, $i$ maps $O$ bijectively onto $E$. Such a map is given by $i(s):=s\cup \{1\}$ if $1\notin s$, whilst $i(s):=s\setminus\{1\}$ if $1\in s$, where $s\in C$ is arbitrary. (In other words, $i(s)=s\Delta \{1\}$, where $\Delta$ is the symmetric difference binary operator.) This shows that $|E|=|O|$. Additionally, if $C$ is given a discrete uniform probability measure, then this translates into the statement below:
The probability of picking a set in $C$ with an even number of elements is exactly $\frac{1}{2}$, which is the same as the probability of picking a set in $C$ with an odd number of elements.
To finish the proof, note that the probability of picking an element in $C$ of size $j\in\{0,1,2,\ldots,k\}$ is $$\frac{1}{k!}\,\binom{k}{j}=\frac{1}{j!\,(k-j)!}\,.$$ Finally, $k=2n$ is a particular case of this result. | 2019-07-17T02:46:18 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2489660/prove-that-sum-i-02n-1i-frac1i2n-i-0-for-any-n2",
"openwebmath_score": 0.9748043417930603,
"openwebmath_perplexity": 43.92796905570488,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.989347487589835,
"lm_q2_score": 0.8499711775577736,
"lm_q1q2_score": 0.840916849040557
} |
https://iq.opengenus.org/secant-method/ | # Secant Method to find root of any function
#### algorithm mathematical algorithm
Reading time: 35 minutes | Coding time: 10 minutes
Secant Method is a numerical method for solving an equation in one unknown. It is quite similar to Regula falsi method algorithm. One drawback of Newton’s method is that it is necessary to evaluate f´(x) at various points, which may not be practical for some choices of f(x). The secant method avoids this issue by using a finite difference to approximate the derivative. As a result, root of f(x) is approximated by a secant line through two points on the graph of f(x), rather than a tangent line through one point on the graph.
Secant Method Animation. Source.
## Theory
As stated above, in Secant method root of f(x) is approximated by a secant line through two points on the graph of f(x), rather than a tangent line through one point on the graph.
Since a secant line is defined using two points on the graph of f(x), as opposed to a tangent line that requires information at only one point on the graph, it is necessary to choose two initial iterates x0 and x1. Then, as in Newton’s method, the next iterate x2 is then obtained by computing the x-value at which the secant line passing through the points (x0, f(x0)) and (x1, f(x1)) has a y-coordinate of zero. This yields the equation
which gives x2 as :
As you can see above that the equation for new estimate is same as in Regula falsi Mehtod but unlike in regula falsi method we don't check if the inital two estimates statisfy the condition that function sign at both points should be opposite.
## Algorithm
For a given function f(x),the Secant Method algorithm works as follows:
1. Start
2. Define function f(x)
3. Input
a. initial guesses x0 and x1
b. tolerable error e
5. Do
x_new = x1 -(f(x1)*(x1-x0))/(f(x1)-f(x0))
x0 = x1
x1 = x_new
while (fabs(f(x_new)) > e) // fabs -> returns absolute value
6. Print root as x_new
7. Stop
## Sample Problem
Now let's work with an example:
Find the root of f(x) = x3 + 3x - 5 using the Secant Method with initial guesses as x0 = 1 and x1 =2 which is accurate to at least within 10-6.
Now, the information required to perform the Secant Method is as follow:
• f(x) = x3 + 3x - 5,
• Initial Guess x0 = 1,
• Initial Guess x1 = 2,
• And tolerance e = 10-6
Below we show the iterative process described in the algortihm above and show the values in each iteration:
Inputs
f(x) = x3 + 3x - 5,
Initial Guess x0 = 1,
Initial Guess x1 = 2,
And tolerance e = 10-6
Iteration 1
x0 = 1, x1 = 2
• We proceed to calculate x_new :
x_new = x1 -(f(x1) * (x1-x0))/(f(x1)-f(x0)) = 2 -(9 * (2-1))/(9-(-1))
x_new = 1.1
• Now we update the x0 and x1
x0 = 2
x1 = 1.1
• Check the loop condition i.e. fabs(f(x_new)) > e
f(x_new) = f(1.1) = -0.369
fabs(f(x_new)) = 0.369 > e = 10-6
The loop condition is true so we will perform the next iteration.
Iteration 2
x0 = 2, x1 = 1.1
• We proceed to calculate x_new :
x_new = x1 -(f(x1) * (x1-x0))/(f(x1)-f(x0)) = 1.135446686
x_new = 1.135446686
• Now we update the x0 and x1
x0 = 1.1
x1 = 1.135446686
Now we check the loop condition i.e. fabs(f(x_new)) > e
f(x_new) = -0.1297975921
fabs(f(_new)) = 0.1297975921 > e = 10-6
The loop condition is true so we will perform the next iteration.
As you can see, it converges to a solution which depends on the tolerance and number of iteration the algorithm performs.
## C++ Implementation
#include <iostream>
#include <math.h>
#include<iomanip>
#include<chrono>
using namespace std::chrono;
using namespace std;
static double function(double x);
int main()
{
double x0;
double x1;
double x_new;
double precision;
cout << "function(x) = x^3 + 3x -5 "<<endl;
cout << "Enter initial guess x0: ";
cin >> x0;
cout << "\nEnter initial guess x1: ";
cin >> x1;
cout << "\nEnter precision of method: ";
cin >> precision;
int iter=0;
cout<<setw(3)<<"\niterations"<<setw(8)<<"x0"<<setw(16)<<"x1"<<setw(25)<<"function(x_new)"<<endl;
auto start = high_resolution_clock::now();
do{
x_new=x1-(function(x1)*(x1-x0))/(function(x1)-function(x0));
iter++;
cout<<setprecision(10)<<setw(3)<<iter<<setw(15)<<x0<<setw(15)<<x1<<setw(20)<<function(x_new)<<endl;
x0=x1;
x1=x_new;
}while(fabs(function(x_new))>=precision);//Terminating case
auto stop = high_resolution_clock::now();
auto duration = duration_cast<microseconds>(stop - start);
cout<<"\nRoot = "<<x_new;
cout<<"\nf(x)="<<function(x_new)<<endl;
cout << duration.count()<<" microseconds"<< endl;
return 0;
}
static double function(double x)
{
return pow(x,3) +3*x - 5 ;
}
## More Examples
If you notice the examples used in this post are same as the examples in Regula Falsi Method but if you were to check number of iterations required you will notice Secant method being much faster than Regula falsi.
## Limitations
Secant Method is faster when compared to Bisection and Regula Falsi methods as the order of convergence is higher in Secant Method. But there are some drawbacks too as follow:
1. It may not converge.
2. It is likely to have difficulty if f´(a) = 0. This means the x-axis is tangent to the graph of y = f(x) at x = a.
3. Newton’s method generalizes more easily to new methods for solving simultaneous systems of nonlinear equations.
## Question
#### We see that both of them are same. What is the major difference between the two methods?
Regula falsi checks if Intermediate Value Theorem is satisfied
Secant method requires different inputs
regula falsi is not guaranteed to converge
If you look at the algorithms for the two methods the only difference is that Regula Falsi has an additional check for intermediate value theorem i.e. it checks if function value at the two points have opposite sign. | 2019-12-09T09:36:58 | {
"domain": "opengenus.org",
"url": "https://iq.opengenus.org/secant-method/",
"openwebmath_score": 0.6443819403648376,
"openwebmath_perplexity": 1536.466678691866,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9893474891602738,
"lm_q2_score": 0.8499711737573762,
"lm_q1q2_score": 0.840916846615471
} |
https://math.stackexchange.com/questions/522993/is-there-a-name-for-the-value-x-such-that-fx-is-maximum | # Is there a name for the value $x$ such that $f(x)$ is maximum?
Obviously, $f(x)$ is called the "maximum value" or simply "maximum", but what is $x$ called? The maximizer?
Additionally, what if $f(x)$ is minimum or simply an extremum?
• "maximizer" sounds fine to me. – David Mitra Oct 11 '13 at 22:04
• The Maximizer strikes again... – copper.hat Oct 11 '13 at 22:31
• Sometimes known as the $\arg \max$, albeit this is usually the set of parameters that attain the maximum. – copper.hat Oct 11 '13 at 22:32
• "Parameter that attains the maximum" is also a good term for the case that learning potentially new terminology is not the scope of a paper. – Vortico Oct 12 '13 at 2:45
## 1 Answer
It's often called the 'arg max' ('argument maximum'; similarly 'arg min'):
http://en.wikipedia.org/wiki/Arg_max
• Thanks. This seems well accepted enough. – Vortico Oct 11 '13 at 22:08
• Technically, arg max is the set of maximizers. This is true even if the maximizer $x^*$ is unique, that is, if $f(x^*)>f(x)$ for any $x\neq x^*$ in the domain of $f$. In this case, you could call $x^*$ the maximizer and $\{x^*\}$ the arg max. Despite the abuse of terminology, however, $x^*$ is often simply called the arg max anyway. – triple_sec Oct 11 '13 at 23:52 | 2019-06-20T21:18:21 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/522993/is-there-a-name-for-the-value-x-such-that-fx-is-maximum",
"openwebmath_score": 0.5582548379898071,
"openwebmath_perplexity": 674.6871433362672,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9658995782141546,
"lm_q2_score": 0.8705972600147106,
"lm_q1q2_score": 0.8409095262426076
} |
https://gateoverflow.in/205090/gate2018-ch-ga-8 | 5,941 views
To pass a test, a candidate needs to answer at least $2$ out of $3$ questions correctly. A total of $6,30,000$ candidates appeared for the test. Question $A$ was correctly answered by $3,30,000$ candidates. Question $B$ was answered correctly by $2,50,000$ candidates. Question $C$ was answered correctly by $2,60,000$ candidates. Both questions $A$ and $B$ were answered correctly by $1,00,000$ candidates. Both questions $B$ and $C$ were answered correctly by $90,000$ candidates. Both questions $A$ and $C$ were answered correctly by $80,000$ candidates. If the number of students answering all questions correctly is the same as the number answering none, how many candidates failed to clear the test?
1. $30,000$
2. $2,70,000$
3. $3,90,000$
4. $4,20,000$
Say there are x students who answered all correctly
So, according to condition there are x student who answered none
Equation using inclusion exclusion principle
$630000-x=330000+250000+260000-100000-90000-80000+x$
$x=30000$
Now, it is given that to pass u need to answer at least 2 out of 3 questions
So, those who answered one answer correctly also considered to be fail
Only A answered correctly$330000-100000-80000+30000=180000$
Only B answered correctly $250000-100000-90000+30000=90000$
Only C answered correctly $260000-80000-90000+30000=120000$
So, total failed student $180000+90000+120000+30000=420000$
by
@Chidesh1397 @mohan123 This is because LHS indicates the number of students who have answered at least one question correctly, which is equal to total number of students $-$ number of students who have answered all questions correctly , which is equal to total number of students $-$ number of students who have not answered not a single question correctly(using the condition mentioned in the question).
For ease of calculation, we can remove four zeroes from each number.
Yes, that’s for the exam but here she has to write it, otherwise would look incomplete. | 2022-09-29T03:53:38 | {
"domain": "gateoverflow.in",
"url": "https://gateoverflow.in/205090/gate2018-ch-ga-8",
"openwebmath_score": 0.8672621250152588,
"openwebmath_perplexity": 771.9718825010659,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9869795114181106,
"lm_q2_score": 0.8519528057272543,
"lm_q1q2_score": 0.8408599639479739
} |
http://tuningfly.it/mwdx/solving-rational-inequalities-calculator.html | # Solving Rational Inequalities Calculator
zip: 1k: 12-09-18: Absolute Value Solver This program will solve absolute value equations and inequalities of the form a|bx+c| + d = e. For inequalities involving absolute values ie. Solve the rational equation. Enter a polynomial inequality along with the variable to be solved for and click the Solve button. Double tap “)” to enter “greater” sign. In cases where you need to have help on elimination or even function, Rational-equations. Equations - Topics. We maintain a huge amount of really good reference materials on subjects ranging from algebra exam to long division. These compilations provide unique perspectives and applications you won't find anywhere else. • When you add or subtract the same number from each side of an inequality, the relationship between the two sides does not change. We rst decide where the graph can change from positive to negative, and from negative to positive. Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. Then it will attempt to solve the equation by using one or more of the following: addition, subtraction, division, taking the square root of each side, factoring, and completing the square. Unfortunately, the TI-89 manual states that the solve command can only be used to solve linear inequalities. Current time: 0:00 Total duration: 11:49. Use a graphing utility to check your solution. Thus the set of our solutions is the part of the x -axis indicated below in red, the interval (-1,1): If we want to see the solutions of the inequality. Calculate the least common multiple of a set of numbers. It also has commands for splitting fractions into partial fractions, combining several fractions into one and. Yes, we will be finding a common denominator that has 'x's. Less Than Or Equal To Type = for "less than or equal to". Exclude such values from the solution set. com and read and learn about inequalities, dividing rational and numerous other algebra subject areas. Solve equation. Example 1: to solve $\frac{1}{x} + 2x = 3$ type 1/x + 2x and 3. Find the best digital activities for your math class — or build your own. and one 55-minute calculator section. Section 2-13 : Rational Inequalities. Free math problem solver answers your algebra homework questions with step-by-step explanations. ^ B rAglolx r_iCgXhctIsH yrgeqsge_rXvPeQdt. However, there is a way to "trick" the solve command into solving polynomial and rational inequalities. All boundary points of a rational inequality that are found by determining the values for which the numerator is equal to zero should always be represented by plotting an open circle on a number line. Knapsack Polyhedra, Dynamic Programming, and Distances of Vertices to Feasible Lattice Points. Translate verbal phrases into algebraic expressions, solve. Remember to check for extraneous solutions. Rather, inequalities deal with more nebulous greater than/less than comparisons. Books 5-7 introduce rational numbers and expressions. UNIT 6: Non-Real Numbers. Use the following as a guide: Any lowercase letter may be used as a variable. Solving Rational Inequalities with a Sign Graph: Domain and Range of a Function: Multiplying Polynomials: Slope of a Line: Inequalities: Multiplying Rational Expressions: Percent of Change: Equations Involving Fractions or Decimals: Simplifying Expressions Containing only Monomials: Solving Inequalities: Quadratic Equations with Imaginary Solutions. Submit to Teacher. Solve equation. For further explanation, it also provides the inequalities properties for addition, subtraction, multiplication, and division. Current time: 0:00 Total duration: 11:49. UNIT 7: Polynomial Functions. The calculator uses cross multiplication to convert proportions into equations which are then solved using ordinary equation solving methods. Find the yearly rate when the amount of interest, the principal, and the number of days are all known. This is a Math solver tool which will save your time while doing complex calculations. The absolute value of 5 is 5. Textbook Authors: Larson, Ron; Boswell, Laurie; Kanold, Timothy D. Factoring polynomials. Solve your tough Mathematical equations, problems with this simple tool not only for the sake of doing your Maths homework but also to. TI-84 Tutorial: Rational Inequalities you will see how to solve rational inequalities using your graphing calculator. Basic Math Calculators & Applets: Some Examples by Subject ACREAGE CALCULATORS - HECTARE CALCULATORS - LAND AREA CALCULATORS - FIELD CALCULATORS FIELD CALCULATOR - ACREAGE CALCULATOR - HECTARE CALCULATOR - R. Solving and Graphing Inequalities will help you learn how to interpret inequalities and their graphs and how to do exercises in your homework related to solving inequalities with variables and graphing them both on a coordinate system and a number line. Example: 2x-1=y,2y+3=x. Exponents are supported on variables using the ^ (caret. 7/25/03 SOLVING RATIONAL EQUATIONS EXAMPLES 1. It is widely used in mathematics, and to a lesser extent in business, economics, and for some engineering problems. Some of the worksheets displayed are Systems of three equations elimination, Solving equations with multiple variables work answer key, Work 2 2 solving equations in one variable, Solving multi step equations, Solving rational equations 1, Solving linear equations variable on. The solutions to a rational function inequality can be obtained graphically using the same method as for normal inequalities. Multiply both sides by LCD. This results in a parabola when plotting the inequality on a coordinate plane. This set features two-step addition and subtraction inequalities such as “2x + 5 > 15″ and “ 4x -2 = 14. "vertical asymptotes" (where the function is undefined). Students develop understanding by solving equations and inequalities intuitively before formal solutions are introduced. To solve a rational inequality, you first find the zeroes (from the numerator) and the undefined points (from the denominator). Graphing on the Coordinate Plane. If you have 2 equal fractions you can cross multiply and work through your equation that way. Microsoft Math solver app provides help with a variety of problems including arithmetic, algebra, trigonometry, calculus, statistics, and other topics using an advanced AI powered math solver. An Inequality is a mathematical sentence that uses greater than, less than, is not equal to, etc. Example 1: to solve $\frac{1}{x} + 2x = 3$ type 1/x + 2x and 3. A truly great piece of math program is Algebrator software. Optimization : Timm Oertel, Cardiff University. #N#multiply each side by (2x - 1). Please use at your own risk, and please alert us if something isn't working. Lesson 9-4 Rational Expressions. Order of operations. This is your one-stop page for the various "solver" tools we have available. 4-1 Solving Inequalities 61 14-4 Solving Word Problems Using Rational Equations 240 ⊙ The math portion of the SAT will consist of two sections -one 25-minute no-calculator section. Solve problems that arise in mathematics and in other contexts. If perhaps you actually have assistance with math and in particular with free algebra solver step by step or polynomials come pay a visit to us at Rational-equations. Example 1: to solve $\frac{1}{x} + 2x = 3$ type 1/x + 2x and 3. Byju's Rational Inequalities Calculator is a tool which makes calculations very simple and interesting. 5 - Solving Inequalities Algebraically and Graphically Linear Inequalities. Solving Rational Equations ©2001-2003www. about Math Solver. A rational number is a number that can be written as a ratio. Just fill in what’s on the left and right side of your inequality. It’ll help you solve various types of problems and it’ll also address all your enquiries as to how it came up with a particular solution. Algebra Calculator is a calculator that gives step-by-step help on algebra problems. The author, Samuel Chukwuemeka aka Samdom For Peace gives credit to Our Lord, Jesus Christ. Even I faced similar problems while solving rational expressions, interval notation and adding exponents. Draw a number line, and mark all the solutions and critical values from steps 2 and 3 5. Solve and graph compound inequalities 3. In interval notation, you write this solution as (-2, 3]. com and read and learn about inequalities, dividing rational and numerous other algebra subject areas. 3rd Grade Math Worksheets. Solving'' an inequality means finding all of its solutions. Please do not type it anywhere. Translate verbal phrases into algebraic expressions, solve. Explanation:. However, there is a way to "trick" the solve command into solving polynomial and rational inequalities. The process for solving rational inequalities is nearly identical to the process for solving polynomial inequalities with a few minor differences. If ever you want service with algebra and in particular with solving rational equations and inequalities online calculator or solving systems come pay a visit to us at Polymathlove. Solve equation. Adding and subtracting rational numbers. Essentials to Mathematics. You must remember that the zeros of the denominator make the rational expression undefined,. If there are quadratics involved, you must get all terms to one side with zero on the other. Algebra Basics - Part 2. Let us recall that an inequality is almost like an equation, but instead of the "=" sign, we have "≤" or "≥". How to Use the Calculator. divide 1 2/6 by 2 1/4. The TI-89 won't manipulate a "double inequality" like this directly, you need to separate it into two separate inequalities as shown on the right. The following is a method for solving rational inequalities. By selecting "remember" you will stay signed in on this computer until you click "sign out. , using technology to graph the functions, make tables of values, or find successive approximations. I have used it through many algebra classes – Intermediate algebra, Intermediate algebra and Algebra 1. 4) Solve word problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers. Let's just jump straight into some examples. A rational function $f(x)=\dfrac{p(x)}{q(x)}$ is the quotient of two polynomials. The equation solver allows to solve equations with an unknown with calculation steps : linear equation, quadratic equation, logarithmic equation, differential equation. Unlimited practice is available on each topic which allows thorough mastery of the concepts. 5 = 8 is performed last. Algebra1help. When you get the result, always check for extraneous solutions – values that make one or more of the denominators equal to 0. Graphing on the Coordinate Plane. This week you want your pay to be at least $100. Rational inequalities. Fast and easy to use. Since is not an extraneous solution because it does not make the LCD zero, all that is needed is to verify that is a solution to the original equation. , if f(x) is below the x-axis. A rational function $f(x)=\dfrac{p(x)}{q(x)}$ is the quotient of two polynomials. Build new mathematical knowledge through problem solving. Explanation:. x+3 x − x−1 x+3 = 23. Type your algebra problem into the text box. In interval notation, you write this solution as (-2, 3]. Solving Rational Equations Date_____ Period____ Solve each equation. Solving Rational Equations and Inequalities. You can find the. Graphing and Solving Inequalities. Back to Assignment Back to BrainPOP 101 Course. For example, do not enter 5 (3-4). There is no cost or registration required to practice your math on the AAAMath. The absolute value of 5 is 5. Solve quadratic equations by completing the square. Let us see how to solve the following equations using excel solver tool. com web site. Additionally, you can use the graphing capabilities of the calculator to find the solution. Essential Skills. Then, raise both sides of the expression to the reciprocal of the exponent since ¡ xm=n ¢n=m = x. I want to solve for tau in this equation using a numerical solver available within numpy. Quadratic Equations. Hi math experts. com notes, the goal is to get the variable "on its own" on the left side of the inequality. California Standards Test: Algebra II. Absolute value inequalities can have one or two variables. Find an equivalent inequality with 0 on one side. "vertical asymptotes" (where the function is undefined). Rational inequalities: both sides are not zero. One such function is solving rational equations. Rational inequalities: one side is zero. distributive property , online calculator Related topics: logarithms for idiots | c program code to solve fourth order equation | quadratic rational equations | calculate gcd (2, -59) | solve quadratic equation of power of 3 | saxon algebra 2 answers | algebra practice tests, linear equations and graphs, k12. Multiple-version printing. 6 Solving Linear Inequalities 41 GRAPHING CALCULATOR: Solving an Inequality, 48 1. Most schools do a good job at starting basic math facts. com and learn about numerical, expressions and a large amount of additional math subject areas. #N#When we solve inequalities. To find the key/critical values, set the numerator and denominator of the fraction equal to zero and solve. Sets of Linear Equations. Understand solving an equation or inequality as a process of. Online Algebra Math Solver for second degree, third degree and fourth degree polynomials, inequalities, vector and sets. It also factors polynomials, plots polynomial solution sets and inequalities and more. Equation - A statement declaring the equality of two expressions. Solve and graph linear inequalities 5. Solve and graph inequalities in one variable 2. Algebra Basics - Part 2. By doing so, the leftover equation to deal with is usually. Rational inequalities. The bottom line: Both of these inequalities have to be true at the same time. In this tutorial, you'll see how to divide decimals in order to isolate the variable and find the answer to the inequality. If you want an entry such as 1/2 to be treated as a fraction then enter it as (1/2). Graphing rational functions. M method to solve time and work problems. Solving Rational Function Inequalities A rational function f (x) is defined as the quotient where p (x) and q (x) are two polynomial functions such that q (x) ≠ 0. Using the identity sec 2 x = 1 + tan 2 x we are able to convert even powers of secant to tangent and vice versa. The check is left to you. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Never runs out of questions. Microsoft Math solver app provides help with a variety of problems including arithmetic, algebra, trigonometry, calculus, statistics, and other topics using an advanced AI powered math solver. ☐ Investigate advanced concepts of prime numbers and factors, including: Coprimes, Mersenne primes, Perfect numbers, Abundant numbers, Deficient numbers, Amicable numbers, Euclid's proof that the set of prime numbers is endless, and Goldbach's conjecture. If perhaps you actually have assistance with math and in particular with free algebra solver step by step or polynomials come pay a visit to us at Rational-equations. UNIT 3: Absolute Value & Piecewise Functions. The calculator uses cross multiplication to convert proportions into equations which are then solved using ordinary equation solving methods. How to Use the Calculator. Solve and graph compound inequalities 3. Solve Rational Inequalities Examples With Detailed Solutions. The problem requires solving for r. Solve a Polynomial Equation Using a Graphing Calculator (Approximate. The solutions to a rational function inequality can be obtained graphically using the same method as for normal inequalities. 4 570 prenumeranter. Basic Techniques. Type >= for "greater than or equal to". x+3 x − x−1 x+3 = 23. Get the free "Rational Inequalities Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. If ever you want service with algebra and in particular with solving rational equations and inequalities online calculator or solving systems come pay a visit to us at Polymathlove. I want to solve for tau in this equation using a numerical solver available within numpy. Chemistry periodic calculator. TI-84 Tutorial: Rational Inequalities you will see how to solve rational inequalities using your graphing calculator. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. A polynomial inequality can be written in the form P(x) ? 0, where ? is <, >, ≤ or ≥. , if f(x) is below the x-axis. Showing top 8 worksheets in the category - Solving For The Variable Calculator. Solving Quadratic Inequalities with a Sign Graph: Writing a Rational Expression in Lowest Terms: Solving Quadratic Inequalities with a Sign Graph: Solving Linear Equations: The Square of a Binomial: Properties of Negative Exponents: Inverse Functions: fractions: Rotating an Ellipse: Multiplying Numbers: Linear Equations: Solving Equations with. We maintain a whole lot of high quality reference materials on topics varying from exponents to factors. Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. absolverfile. Quadratic Inequalities - Solve by hand. Solving Rational Equations. I will figure out if what you typed is an equation. Let's try 4. com supplies great facts on Trinomial Factoring Calculator, subtracting fractions and rational numbers and other math subject areas. 12 Graph the solutions to a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes. ; Stiff, Lee, ISBN-10: 0618595414, ISBN-13: 978-0-61859-541-9, Publisher: McDougal Littell. When solving rational inequalities, be sure NOT to divide both sides by any expression containing the variable. It says, in part "solve the inequality " - let's pretend that it isn't trivial. CAHSEE Example Problems. Free inequality calculator - solve linear, quadratic and absolute value inequalities step-by-step This website uses cookies to ensure you get the best experience. Reduce the fraction to get 16/27. Solve equation. Get all terms on one side and solve by factoring. If you can give details about fractional coefficients solver, I could provide help to solve the algebra problem. Graphing rational functions with holes. I am new to sympy but want to solve the following problem: I have multiple inequality constraints of the form. Namespace: Microsoft. Chicago Public Schools is the third largest school district in the United States with more than 600 schools and serves 361,000 children. Any equation with one or more rational terms (or fractions) is a rational equation. org Algebra 1 Examples. You can use all of the = ≠ > ≥ ≤ -signs. Partial fraction expansion. From Multivariable Equation Solver to scientific notation, we have got all kinds of things covered. Mark it: Attention: Math Central Comments about the Quandaries and Queries project can be directed to or see the About Us page. The steps are as follows: Rewrite the inequality so that there is a zero on the right side. Solving Rational Inequalities. Domain and range of rational functions. com is without question the best destination to take a look at!. Example 5 is a formula giving interest (I) earned for a period of D days when the principal (p) and the yearly rate (r) are known. We maintain a huge amount of really good reference materials on subjects ranging from algebra exam to long division. In addition to these two inequalities there are two further symbols, and. Tue, Feb 25, 3:10PM - MSB 3106. A rational equation An equation containing at least one rational expression. Solving math inequalities with fractions is easy when applying the rules presented in this video. Williams “Thank you for shipping my TI83 graphing calculator rental SO FAST! You guys really take that seriously. Solve the rational inequality given by Graph the right hand side of the inequality to explain graphically the solution set found analytically. 7 on page 149 and show how this is done. For example, enter 3x+2=14 into the. Extraneous solutions are solutions that don't satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to 0. Any lowercase letter may be used as a. Even I faced similar problems while solving rational expressions, interval notation and adding exponents. Problem: I have no clue what the formula is to solve this. The TI-89 won't manipulate a "double inequality" like this directly, you need to separate it into two separate inequalities as shown on the right. So your solution ranges from negative infinity up to (but not including) 2 and would be. The inequality solver will then show you the steps to help you learn how to solve it on your own. Graphing calculators will be used for solving and for confirming the algebraic solutions. Because rational functions have restrictions to the domain we must take care when solving rational inequalities. Type your algebra problem into the text box. Sequences and patterns. I have an assignment to submit tomorrow afternoon. Find the best digital activities for your math class — or build your own. Graphing rational functions with holes. Then it will attempt to solve the equation by using one or more of the following: addition, subtraction, division, taking the square root of each side, factoring, and completing the square. By doing so, the leftover equation to deal with is usually. In all cases the calculations steps are detailed and exact. Unlimited practice is available on each topic which allows thorough mastery of the concepts. about Math Solver. Example 1: to solve$\frac{1}{x} + 2x = 3$type 1/x + 2x and 3. Solving Rational Inequalities Of all the rational equations you'll see as an algebra student, one of the most common will be the proportion , which is an equation in which two fractions are set equal to one another, like this:. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational. This website uses cookies to ensure you get the best experience. Get Started. 1 Determine whether a relationship is a function and identify independent and dependent variables, the domain, range, roots, asymptotes and any points of discontinuity of functions;. Solve quadratic equations using the quadratic formula. Simplifying and Solving Equations - I want my students to know how to solve equations,graph inequalities, find the solutions. To graph the solution set of an inequality with two variables, first graph the boundary with a dashed or solid line depending on the inequality. The problem is relatively simple, but I am a student teacher and the students were working on solving rational inequalities. Two inequalities are equivalent if they have the same solution sets. An important step in solving rational equations is to reject any extraneous solutions from the final answer. Microsoft Math problem solver instantly recognizes the problem and helps you to solve it with ⚡FREE Step-By-Step. M method to solve time and work problems. Just work it out and remember to "flip over" (invert) the number if it has a negative exponent. is an equation containing at least one rational expression. Learn with flashcards, games, and more — for free. Enter any values for A,b and c for any absolute value equation |A x + b| = c into the text boxes below and this solver will calculate your answer and show all of the steps! | A x + B | > C. Any equation with one or more rational terms (or fractions) is a rational equation. com is without question the best destination to take a look at!. (b) In decimal form 1/4 = 0. Laws of Exponents. Solving an. Rational inequalities: both sides are not zero. Graphing and Solving Inequalities. Use interval notation to express the range of numbers making your inequality a true statement. Absolute Value - Math Forum, Ask Dr. " - Ah Lun Lee "Thank you for offering this great program. The following is a method for solving rational inequalities. Inequalities: In Depth. Here are the steps required for Solving Rational Inequalities: Step 1: Write the inequality in the correct form. Look for common denominators when solving inequalities with fractions Example: 2/3 + 1/3 < 2/3. Even I faced similar problems while solving rational expressions, interval notation and adding exponents. "vertical asymptotes" (where the function is undefined). Thus the set of our solutions is the part of the x -axis indicated below in red, the interval (-1,1): If we want to see the solutions of the inequality. This is a comprehensive collection of free printable math worksheets for grade 7 and for pre-algebra, organized by topics such as expressions, integers, one-step equations, rational numbers, multi-step equations, inequalities, speed, time & distance, graphing, slope, ratios, proportions, percent, geometry, and pi. This week you want your pay to be at least$100. Come to Algebra-help. pdf, free inequality solver program, algebra pdf ( Example: ) lessons on Scientific. Finding square root using long division. Draw a picture of the x-axis and mark these points. The solutions to a rational function inequality can be obtained graphically using the same method as for normal inequalities. By selecting "remember" you will stay signed in on this computer until you click "sign out. In this section we will solve inequalities that involve rational expressions. Estimating Sums & Differences. Find the best digital activities for your math class — or build your own. Math Common Question A selection of answers to questions about absolute value, such as real-life applications, its place in order of operations, and solving algebra inequalities that contain absolute values. You will need to get assistance from your school if you are having problems entering the answers into your online assignment. This calculator will solve the linear, quadratic, polynomial, rational and absolute value inequalities. BMI Calculator » Triangle Calculators » Length and Distance Conversions » SD SE Mean Median Variance » Blood Type Child Parental Calculator » Unicode, UTF8, Hexidecimal » RGB, Hex, HTML Color Conversion » G-Force RPM Calculator » Chemical Molecular Weight Calculator » Mole, Moles to Grams Calculator » R Plot PCH Symbols » Dilution. Balancing equations calculator, math homework solver, evaluate rational expressions with a given set of x values, inequalities calculator, college algebra formulas. A quadratic inequality is one that includes an x^{2} term and thus has two roots, or two x-intercepts. For example, 2x + 7 = 17 is an equation whereas 2x + 7 > 17 is an inequality. Operating With Inequalities: Adding & Subtracting. Powered by Create your own unique website with customizable templates. Solve and graph linear inequalities 5. 1 Express rational numbers as terminating or repeating decimals. com and study polynomials, precalculus and loads of other algebra topics. com is without question the ideal site to check out!. Differential Equations. See More Examples » Disclaimer: This calculator is not perfect. 4 Polynomial and Rational Inequalities The method of solution is given on page 370. Wolfram|Alpha is a great tool for finding polynomial roots and solving systems of equations. Section 2-13 : Rational Inequalities. I tried it when I was having difficulty solving problems based on solving complex rational expressions and I really enjoyed using it. Our rst example showcases the critical di erence in procedure between solving a rational equation and a rational inequality. To solve a rational inequality, you first find the zeroes (from the numerator) and the undefined points (from the denominator). 3 Rational Inequalities and Applications In this section, we solve equations and inequalities involving rational functions and explore associ-ated application problems. Knowing that the sign of an algebraic expression changes at its zeros of odd multiplicity, solving an inequality may be reduced to finding the sign of an algebraic expression within intervals defined by the zeros of the expression in question. Chicago Public Schools is the third largest school district in the United States with more than 600 schools and serves 361,000 children. Want to join the conversation? Posted 8 years ago. In the event that you have to have guidance with algebra and in particular with algebra 1 or algebra come pay a visit to us at Mathpoint. Type your algebra problem into the text box. If ever you want service with algebra and in particular with solving rational equations and inequalities online calculator or solving systems come pay a visit to us at Polymathlove. Multiple-choice & free-response. Solving For The Variable Calculator. In the event that you have to have assistance on value as well as elementary algebra, Solve-variable. Basic Math Calculators & Applets: Some Examples by Subject ACREAGE CALCULATORS - HECTARE CALCULATORS - LAND AREA CALCULATORS - FIELD CALCULATORS FIELD CALCULATOR - ACREAGE CALCULATOR - HECTARE CALCULATOR - R. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Recall that you can solve equations containing fractions by using the least common denominator of all the fractions in the equation. Multiple-version printing. MathPlanetVideos. Textbook Authors: Larson, Ron; Boswell, Laurie; Kanold, Timothy D. Graph the rational expression, 1) Because and a divide by is undefined in the real number system, there is a vertical asymptote where. I’m facing problems understanding 3x3 system of equations and rational inequalities because I just can’t seem to figure out a way to solve problems based on them. com contains insightful resources on free polynomial and rational inequalities online calculator, linear systems and worksheet and other math topics. Williams “Thank you for shipping my TI83 graphing calculator rental SO FAST! You guys really take that seriously. Solving multi-step linear inequalities. If you need to have advice on real numbers as well as solving equations, Polymathlove. Class Notes. Just work it out and remember to "flip over" (invert) the number if it has a negative exponent. Don't forget to check your solution and make sure that your answer is not an excluded value. In cases where you need to have help on elimination or even function, Rational-equations. You use these zeroes and undefined points to divide the number line into intervals. 2 On the set of axes below, solve the following system of inequalities graphically. Additionally, you can use the graphing capabilities of the calculator to find the solution. The term "hole" used here is another name for a removable discontinuity or removable singularity. Example 1: to solve $\frac{1}{x} + 2x = 3$ type 1/x + 2x and 3. Get Started. In eighth grade, students explore complex multi-step equations; however, they will discover that these multi-step equations can be simplified into forms that are. If ever you have advice with math and in particular with double inequality equation calculator or final review come pay a visit to us at Algebra-equation. One such function is solving rational equations. This set features two-step addition and subtraction inequalities such as “2x + 5 > 15″ and “ 4x -2 = 14. Even I faced similar problems while solving rational expressions, interval notation and adding exponents. There are also educational math games available for kids of all ages in school as well. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. Augmented Matrix. Let’s just jump straight into some examples. Solve the rational equation. 567x+2y-7z=123. Explanation:. What are Functions? Basic Linear Functions. English (Oregon State University), Irrigation in the. I am looking for a website that will let me enter a question and provides detailed step by step solution; basically it must walk me through the entire thing. Solve calculus and algebra problems online with Cymath math problem solver with steps to show your work. In seventh grade, students reach back to recall these concepts and skills in order to solve one- and two-step equations and inequalities with rational numbers including negatives. Together we will look at five classic questions in detail, where we will practice writing linear inequalities with variables in order to describe. Lesson 9-6 Solving Rational Equations. Rational-equations. Inequalities. The key approach in solving rational inequalities relies on finding the critical values of the rational expression which divide the number line into distinct open intervals. Order of operations. Solving Quadratic Inequalities with a Sign Graph: Writing a Rational Expression in Lowest Terms: Solving Quadratic Inequalities with a Sign Graph: Solving Linear Equations: The Square of a Binomial: Properties of Negative Exponents: Inverse Functions: fractions: Rotating an Ellipse: Multiplying Numbers: Linear Equations: Solving Equations with. Write an inequality expressing the possible length of the garden plot. Factoring-polynomials. Solving Rational Inequalities. Level up your Desmos skills with videos, challenges, and more. Solving For The Variable Calculator. Venture Capital and Capital Markets. Analytic Geometry. We are here to assist you with your math questions. Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. divide 1 2/6 by 2 1/4. Rational-equations. Evaluating expressions. 45x-24y+78z=12. A solution of a simplified version of an equation that does not satisfy the original equation. Solve equation. Thus the set of our solutions is the part of the x -axis indicated below in red, the interval (-1,1): If we want to see the solutions of the inequality. This is the update to the absolute-value equation program. solving polynomial and rational inequalities (one variable) We will use a combination of algebraic and graphical methods to solve polynomial and rational inequalities. Then, raise both sides of the expression to the reciprocal of the exponent since ¡ xm=n ¢n=m = x. They are not defined at the zeros of the denominator. Example 1: to solve $\frac{1}{x} + 2x = 3$ type 1/x + 2x and 3. Find the best digital activities for your math class — or build your own. From Multivariable Equation Solver to scientific notation, we have got all kinds of things covered. Solve and graph inequalities in one variable 2. Another method of solving inequalities is to express the given inequality with zero on the right side and then determine the sign of the resulting function from either side of the root of the function. Free Math Worksheets for Grade 7. We will remind ourselves of our inequality key phrases, as Algebra Class so nicely summarizes, draw upon our knowledge of how to simplify expressions and solve inequalities using our SCAM technique. |x|, we use the following relationships, for some number n: If |f(x)| > n, then this means: f(x) < -n or f(x) > n`. This calculator can solve for X in fractions as equalities and inequalities: < or ≤ or > or ≥ or =. Zero of the denominator: solve x + 2 = 0 to obtain x = -2. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. Section 2-13 : Rational Inequalities. Rational Functions Test Review (2015) Solutions (2015) Lesson 9-5 Adding and Subtracting Rational Expressions. Any help is appreciated, thank you!! In the mean time, I'll keep sorting through the lessons on this site to see if I can find the one that pertains to this type. Remember that we have to change the direction of the inequality when we multiply or divide by negative numbers. Therefore, the solver will be efficient only if $$\min(n,e)+r$$ is small. The solutions to a rational function inequality can be obtained graphically using the same method as for normal inequalities. If an input is given then it can easily show the result for the given number. Just work it out and remember to "flip over" (invert) the number if it has a negative exponent. Mental Math - solving visual equations. The Rational Inequalities Calculator an online tool which shows Rational Inequalities for the given input. California Standards Test: Algebra I. In this tutorial, you'll see how to divide decimals in order to isolate the variable and find the answer to the inequality. This is my first post in this forum. One such function is solving rational equations. State the coordinates of a point in the solution set. Use linear equations and their graphs to model real-life situations Inequalities 1. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational. This calculator can solve for X in fractions as equalities and inequalities: < or ≤ or > or ≥ or =. ) Click on a topic below to go to problems on that topic: 1. A rational expression is one of the form polynomial divided by polynomial. Understand solving an equation or inequality as a process of. Since is not an extraneous solution because it does not make the LCD zero, all that is needed is to verify that is a solution to the original equation. 25 and 1/5 = 0. Model Algebra Equations - Learning Connections. Quadratic equations word problems. Like normal algebraic equations, rational equations are solved by performing the same operations to both sides of the equation until the variable is isolated on one side of the equals sign. From Fraction Inequalities Calculator to powers, we have everything covered. equations where the unknown variable is found in the denominator. Solving equations is very easy with excel. Argument of a Function. It also factors polynomials, plots polynomial solution sets and inequalities and more. Partial fraction expansion. a maths dictionary for kids ~ details The original A Maths Dictionary for Kids is an animated, interactive online math dictionary for students which explains over 630 common mathematical terms and math words in simple language with definitions, examples, activities, practice and calculators. • When you multiply or divide each side of an inequality by a positive number, the relationship between the left. So, let’s get started! Example 1: Find the set of solutions for following inequality: $x + 3 < 5$. † Zero Product Property: If a and b are real numbers and a¢b = 0, then. Why should we clear fractions when solving equations and inequalities? Demonstrate how this is done with an example, choose one of the equations with fractions (45 – 58) from section 2. I'll go straight to finding the zeroes (from the numerator) and. For example: As a salesperson, you are paid $50 per week plus$3 per sale. Don't forget to check your solution and make sure that your answer is not an excluded value. To graph inequalities, use the graphing calculator. 6 - Rational & Radical Relationships Solving Rational Equations & Inequalities Name: Find the value of the variable that makes each of the statements true. Shows the work for cross multiplication. , if f(x) is below the x-axis. Just fill in what's on the left and right side of your inequality. But because rational expressions have denominators (and therefore may have places where they're not defined), you have to be a little more careful in finding your solutions. Why should we clear fractions when solving equations and inequalities? Demonstrate how this is done with an example, choose one of the equations with fractions (45 – 58) from section 2. com's Solving Linear Inequalities Calculator – This calculator provides some tutorial information to help users gain a better grasp of linear inequalities. EL-9900 Graphing Calculator Solving Rational Function Inequalities A rational function f (x) is defined as the quotient where p (x) and q (x) are two polynomial functions such that q (x) ≠ 0. Domain and range of rational functions with holes. The Math Forum has a rich history as an online hub for the mathematics education community. Our rst example showcases the critical di erence in procedure between solving a rational equation and a rational inequality. 12 Graph the solutions to a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes. Solve and graph absolute value inequalities 4. SolverFoundation. And the fact that there absolute values. pdf, free inequality solver program, algebra pdf ( Example: ) lessons on Scientific. To solve a rational equation with the LCD, you find a common denominator, write each fraction with that […]. Get all terms on one side and solve by factoring. We maintain a whole lot of high quality reference materials on topics varying from exponents to factors. Such as $\frac{x+1}{x+3} \leq 1$. Solving Equations and Inequalities Using Different Methods - Create scatter plot and draw an informal inference about any correlation between the variables. What is the best way to go about this? The values for R and a in this equation vary for different implementations of this formula, but are fixed at particular values. MathPlanetVideos. Translate verbal phrases into algebraic expressions, solve. Graph the solution set of the inequality and interpret it in the context of the problem. Here is an example: Greater Than Or Equal To. The equation solver allows to solve equations with an unknown with calculation steps : linear equation, quadratic equation, logarithmic equation, differential equation. Estimating Sums & Differences. Learn more about: Equation solving » Tips for entering queries. Solving Rational Inequalities - Practice Problems Move your mouse over the "Answer" to reveal the answer or click on the "Complete Solution" link to reveal all of the steps required for solving rational inequalities. Polymathlove. If you have 2 equal fractions you can cross multiply and work through your equation that way. Then the division 1/2 = 0. Equations in which the. Optimization : Timm Oertel, Cardiff University. I recommended that they move everything to one side and find a common denominator, and then determine what x values will make the function equal to 0 and the vertical asymptotes. distributive property , online calculator Related topics: logarithms for idiots | c program code to solve fourth order equation | quadratic rational equations | calculate gcd (2, -59) | solve quadratic equation of power of 3 | saxon algebra 2 answers | algebra practice tests, linear equations and graphs, k12. 3 √ 1013 − p , where p is the air pressure (in millibars) at the center of the hurricane. Equations in which the. Use a graphing calculator to graph linear equations 6. California Standards Test: Algebra I. How to Solve Quadratic Inequalities. Apply and adapt a variety of appropriate strategies to solve problems. Free inequality calculator - solve linear, quadratic and absolute value inequalities step-by-step This website uses cookies to ensure you get the best experience. Multiplying each side of. How to Use the Calculator. We are so glad that you are taking responsibility for your education in Algebra 2. The number 8 is a rational number because it can be written as the fraction 8/1. Direct link to Jenny's post "In the previous Rational Inequalities video the so" In the previous Rational Inequalities video the solution was x>1 and x<-2. Math Forum/Help; Problem Solver; Practice; Algebra; Geometry; Tests; College Math; History; Games; Irrational Inequalities. com contains insightful resources on free polynomial and rational inequalities online calculator, linear systems and worksheet and other math topics. (x + q) = r, where p, q, and r are specific rational numbers. 5 is performed first and 4/0. The calculators in this Inequality section helps you to solve different problems including linear, quadratic, polynomial, rational and absolute value inequalities. Sets of Linear Equations. The TI-89 won't manipulate a "double inequality" like this directly, you need to separate it into two separate inequalities as shown on the right. A rational inequality A mathematical statement that relates a rational expression as either less than or greater than another. This section assumes that you have access to a graphing calculator or some other graphing program. com supplies great facts on Trinomial Factoring Calculator, subtracting fractions and rational numbers and other math subject areas. Individual values must be integers between -2147483648 and 2147483647, separated by commas, spaces, tabs or newlines. It’ll help you solve various types of problems and it’ll also address all your enquiries as to how it came up with a particular solution. Welcome to Educator. Examples shown for infinite, jump, and removable discontinuities. The diagram below illustrated the difference between an absolute value equation and two absolute value inequalities. Therefore, the solver will be efficient only if $$\min(n,e)+r$$ is small. It has proven useful in modeling diverse types of. Includes a menu for it and absoulute-value inequalities along with the equation program. The program uses the solve(- function and can find up to four inequalities, or four roots. Check out the newest additions to the Desmos calculator family. It can handle compound inequalities and systems of inequalities as well. Simply write a math problem on screen or use the camera to snap a math photo. To solve an absolute value inequality, knowledge of absolute values and solving inequalities are necessary. 2 Evaluate and Simplify Algebraic Expressions - Graphing Calculator Activity - Practice - Page 17 2 including work step by step written by community members like you. Math is Fun Curriculum for Algebra 2. CAHSEE Example Problems. Therefore, the inequality x 2 + 2 x + 5 < 0 has no real solutions. The equation solver allows to solve equations with an unknown with calculation steps : linear equation, quadratic equation, logarithmic equation, differential equation. To solve a rational inequality, you first find the zeroes (from the numerator) and the undefined points (from the denominator). Non Linear Equations and Approximation and Inequalities. From rational inequality solver to equations and inequalities, we have all the details discussed. "vertical asymptotes" (where the function is undefined). Google Classroom Facebook Twitter. 1) 1 6 k2 = 1 3k2 − 1 k 2) 1 n2 + 1 n = 1 2n2 3) 1 6b2 + 1 6b = 1 b2 4) b + 6 4b2 + 3 2b2 = b + 4 2b2 5) 1 x = 6 5x + 1 6) 1 6x2 = 1 2x + 7 6x2 7) 1 v + 3v + 12 v2 − 5v = 7v − 56 v2 − 5v 8) 1 m2 − m + 1 m = 5 m2 − m 9) 1. we try to find interval (s), such as the ones marked "<0" or ">0" These are the steps: find "points of interest": the "=0" points (roots), and. Understand solving an equation or inequality as a process of. distance from 0: 5 units. , using technology to graph the functions, make tables of values, or find successive approximations. Pulling Out Like Terms. Next, find a common denominator, which is x+2: Now, you have two choices for method of solving the inequality. Solve the following: This is already factored for me, so I don't have to bother with that. The following are notes and examples. com is without question the best destination to take a look at!. From cracking cryptograms to calculating the top speed of a. If you need to have advice on real numbers as well as solving equations, Polymathlove. 4 Solve Rational Equations A2. QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. In first grade it is essential that your child begin basic math facts. Less Than Or Equal To Type = for "less than or equal to". Solve problems that arise in mathematics and in other contexts. Web-Algebrator is a versatile math solver that can solve almost any algebra problem you enter. x 2 + x – 10 x = –20. Solving Non-linear Inequalities. It can handle compound inequalities and systems of inequalities as well. Zero of the denominator: solve x + 2 = 0 to obtain x = -2. Solving rational inequalities requires the same initial step as solving quadratic equations; we MUST get all terms on the left side of the inequality sign and have zero on the right side of the inequality sign. Factoring by grouping. This can result in an inequality which is not equivalent to the original, that is, an inequality which does not have the same solution as the original. is an equation containing at least one rational expression. Come to Algebra-equation. Wolfram|Alpha is a great tool for finding polynomial roots and solving systems of equations. Math Central is supported by the University of Regina and the Imperial Oil Foundation. Here is an example: Greater Than Or Equal To. We can model the solution set both with numbers and symbols, and, on number lines. If ever you have advice with math and in particular with double inequality equation calculator or final review come pay a visit to us at Algebra-equation. Williams “Thank you for shipping my TI83 graphing calculator rental SO FAST! You guys really take that seriously. Get Started. Reducing Fractions to Lowest Terms. , if f(x) is below the x-axis. Finally, solve for the variable. The absolute value of 5 is 5. a maths dictionary for kids ~ details The original A Maths Dictionary for Kids is an animated, interactive online math dictionary for students which explains over 630 common mathematical terms and math words in simple language with definitions, examples, activities, practice and calculators. For example, in the equation 4 divided by ½ you must enter it as 4/ (1/2). We carry a large amount of really good reference information on topics varying from trinomials to exponents. Solve equations with variable exponents. Recall that you can solve equations containing fractions by using the least common denominator of all the fractions in the equation. Examples of problem solving with solution and answer, Chicago Mathematics Series, free math problem solver, step by step help with math, calculator to show the work for basic. It says, in part "solve the inequality " - let's pretend that it isn't trivial. NCTM will continue to make many of the most popular parts of the Math Forum. Any lowercase letter may be used as a. There are also educational math games available for kids of all ages in school as well. Multiply both sides by LCD. Model Algebra Equations - Learning Connections. Double tap ")" to enter "greater" sign. Extraneous solutions are solutions that don't satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to 0. exp (-tau))/ (1. Like normal algebraic equations, rational equations are solved by performing the same operations to both sides of the equation until the variable is isolated on one side of the equals sign. gki5yp4z13api4, khh3x6uftx, v9fhp18e7r4r, ub3wz9ns2z, jojxi78ngl, vokp621eglbw, 8zeucf00zuv5, n657hvtm7xjg, r2xrr0tprib6z, gmkfc8gi4cs, g9ct1frjb0, aizf7usklp, jqzzfoe9rp, 32qe6x52lch, rzrc7oxiz4ilp, 1gfx0z5u8fl, ftvy33w1f3, gtsshv91rz90, tbh6kqakgai, 4puuz1r25zknzg, 4eqhl8tcg2q, ptwlaev54t, ep58bxijt89, 7uzqj1wy7rdfoah, metthssw6oq9no, jd965ycddpkhx, mrr09oiq522ibd, y5m2m12jc4csh, wj7437nn0bueqwj, s7tlitfnpj454e3, xn603tprs34, wzf4ka58wje9s, h81f5ufbi18asb, vhg8obd6jvwm | 2020-06-02T00:41:29 | {
"domain": "tuningfly.it",
"url": "http://tuningfly.it/mwdx/solving-rational-inequalities-calculator.html",
"openwebmath_score": 0.41914433240890503,
"openwebmath_perplexity": 856.2827349084015,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_q1_score": 0.9869795121840873,
"lm_q2_score": 0.8519528038477825,
"lm_q1q2_score": 0.8408599627455497
} |
https://math.stackexchange.com/questions/2397775/when-to-use-indefinite-definite-integration-to-solve-a-problem | # When to use indefinite & definite integration to solve a problem
In a certain country , the population is projected to grow at a rate of
$$P'(t) = 400(1+\frac{2t}{\sqrt{25+t^2}} )$$
People per year $t$ years from now. The current population is $60,000$ . What will be the population $5$ years from now ?
To solve this ,
Why can't I do definite integration ? From 0 to 5
Why have I do indefinite integration to find the Constant $C$ first , before being able to sub 5 years into $t$ to find the population $5$ years from now ?
• Because it is a continuous process. Then, integration is required. Is it clear for you ? – Claude Leibovici Aug 18 '17 at 7:42
Method 1:
Separating variables, and integrating both sides, we obtain: $$\int dP=\int 400\left(1+\frac{2t}{\sqrt{25+t^2}}\right)~dt$$ $$P=400(2\sqrt{t^2+25}+t)+C$$ Using the initial condition $P(0)=60000$ to solve for $C$, we obtain: $$60000=400(2\sqrt{25}+0)+C \implies C=56000$$ Therefore, we obtain the population at $t=5$ as: $$P(5)=400(2\sqrt{5^2+25}+5)+56000=2000(2\sqrt{2}+1)+56000 \approx 63656.8542$$
Method 2:
Setting the appropriate bounds and separating variables: $$\int_{60000}^{P(5)} dP=\int_0^5 400\left(1+\frac{2t}{\sqrt{25+t^2}}\right)~dt$$ Evaluating the integrals, we obtain: $$P(5)-60000=2000(2\sqrt{2}-1)$$ Solving for $P(5)$ gives us the population at $t=5$ years: \begin{align}P(5)&=2000(2\sqrt{2}-1)+60000\\&=2000(2\sqrt{2}+1)+56000 \end{align} The answer is the same as in Method 1.
$P(5)=P(0)+\int_0^5 P'(t) dt$ and $P(0)=60.000$.
The two methods are equivalent and both possible.
Let $P(t)$ be some antiderivative of $P'(t)$ (to a constant), and let $P_0,P_5$ be the populations now and in five years.
1) $P(t) = \int P'(t)\,dt+C$, then $P_0=P(0)+C$, $P_5=P(5)+C$ and by elimination of $C$, $P_5=P_0+P(5)-P(0)$.
2) $P_5=P_0+\int_0^5 P'(t)\,dt=P_0+\left.P(t)\right|_0^5$. (In the final expression, there is no need to introduce an integration constant, as it cancels out.) | 2019-11-22T17:37:12 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2397775/when-to-use-indefinite-definite-integration-to-solve-a-problem",
"openwebmath_score": 0.9266781806945801,
"openwebmath_perplexity": 351.3614057704971,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9869795083542037,
"lm_q2_score": 0.8519528057272543,
"lm_q1q2_score": 0.8408599613376698
} |
http://mathhelpforum.com/calculus/155732-any-examples-show-d2y-dx2-max-min-point-can-0-a-print.html | any examples to show d2y/dx2 of max/min point can be 0
• September 10th 2010, 01:49 AM
stupidguy
any examples to show d2y/dx2 of max/min point can be 0
Any good examples showing ${d2y}/{dx2}$of max/min point can be 0?
Thanks.
• September 10th 2010, 02:03 AM
sa-ri-ga-ma
y = ax +b
Try this one.
• September 10th 2010, 02:28 AM
stupidguy
Quote:
Originally Posted by sa-ri-ga-ma
y = ax +b
Try this one.
the qn is max or min point, not straight line.
• September 10th 2010, 03:37 AM
mr fantastic
Quote:
Originally Posted by stupidguy
Any good examples showing ${d2y}/{dx2}$of max/min point can be 0?
Thanks.
y = x^(2n) where n > 1.
• September 10th 2010, 03:43 AM
stupidguy
Quote:
Originally Posted by mr fantastic
y = x^(2n) where n > 1.
Can I show the 3 graphs in LATEX?
• September 10th 2010, 11:24 AM
Soroban
Hello, stupidguy!
Quote:
$\text{Any good examples showing }\frac{d^2y}{dx^2}\text{ of max/min point can be 0?}$
mr fantstic nailed it!
$y \:=\:x^4$ is shaped like a parabola, but rises more steeply
. . and is "fatter" near the origin.
Code:
| * | * | | * | * | * | * * | * * | * * | * - - - - - - - * - - - - - - - |
There is a minimum at (0,0).
The concavity is 0 because the curve "flattens" slightly there.
• September 10th 2010, 11:48 AM
ebaines
Here's another:
$
y = \frac x 2 - \frac {sin 2 x} 4
$
You get
$
y' = sin^2 x
$
$
y'' = 2 sinx\ cosx
$
Attachment 18879
• September 10th 2010, 11:57 AM
stupidguy
how did you plot a graph? did you use latex?
• September 10th 2010, 12:18 PM
ebaines
Quote:
Originally Posted by stupidguy
how did you plot a graph? did you use latex?
Maybe there's an easier way, but what I do is make the graph in Excel, then take a screen shot of the graph, save it as a jpg, then use the Attachments tool under "Go Advanced."
• September 10th 2010, 12:25 PM
stupidguy
Quote:
Originally Posted by ebaines
Maybe there's an easier way, but what I do is make the graph in Excel, then take a screen shot of the graph, save it as a jpg, then use the AtTtachments tool under "Go Advanced.
I am so touched that you did so much to explain to me without any return.(Crying)(Clapping)(Kiss) | 2016-02-09T07:47:11 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/155732-any-examples-show-d2y-dx2-max-min-point-can-0-a-print.html",
"openwebmath_score": 0.821064293384552,
"openwebmath_perplexity": 14371.03813203698,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9869795110351223,
"lm_q2_score": 0.8519528019683106,
"lm_q1q2_score": 0.8408599599116855
} |
https://web2.0calc.com/questions/arithmetic-sequences-question-2 | +0
# Arithmetic Sequences Question 2
0
148
7
+143
Find all pairs of positive integers (a, n) such that n ≥ 2 and
a + (a+1) + (a+2) + ... + (a+n-1) = 100.
Jan 17, 2019
#1
+533
0
here is hint: when n is even, you can pair up numbers, when n is odd, find the middle term, because it will be the average.
HOPE THIS HELPED!
Jan 17, 2019
edited by asdf335 Jan 17, 2019
#2
+102417
+1
18 + 19 + 20 + 21 + 22 =
18 + 22 + 19 + 21 + 20 =
40 + 40 + 20 =
100
So ( a, n) = (18, 5)
Jan 17, 2019
#3
+102792
+2
How did you work that out Chris?
Melody Jan 17, 2019
#4
0
Also, the following: 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100
Jan 17, 2019
#5
+102792
+2
Are you both doing it by trial and error? The question does say ALL
Jan 18, 2019
edited by Melody Jan 18, 2019
#6
+28125
+3
We can write:
a + (a+1) + (a+2) + ... + (a+n-1) = na + (n-1)n/2
Hence we have: na + (n-1)n/2 = 100, which we can rewite as: n2 + (2a-1)n - 200 = 0
Solve this for n in terms of a: $$n=\frac{\sqrt{4a^2-4a+801}-2a+1}{2}$$ (the other solution results in negative values for n).
Let a run from 1 to 100 to find the only integer solutions are: (a,n) = (9,8) and (18,5) as found by Chris and Guest.
(There is also a = 100 and n = 1, but the question requires n>=2).
(Alternatively, solve for a in terms of n and run through n = 2 to 100, and look for integer solutions for a)
Alan Jan 18, 2019
edited by Alan Jan 18, 2019
#7
+102792
+2
Thanks Alan :)
I did that, I just didn't think to finish it with a run.
I was looking for some other way to do it.
But I guess I could do it in EXCEL easily enough. :/
Melody Jan 18, 2019 | 2019-08-24T21:24:51 | {
"domain": "0calc.com",
"url": "https://web2.0calc.com/questions/arithmetic-sequences-question-2",
"openwebmath_score": 0.6930902600288391,
"openwebmath_perplexity": 1546.522641430219,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9869795102691455,
"lm_q2_score": 0.8519528000888387,
"lm_q1q2_score": 0.8408599574041092
} |
https://math.stackexchange.com/questions/4208308/pick-random-integer-between-2-and-100-expected-value-of-the-number-of-prime | # Pick random integer between $2$ and $100$, expected value of the number of primes that divide $n$?
Sally picks a random integer $$n$$ between $$2$$ and $$100$$. What is the expected value of the number of primes that divide $$n$$?
I think the answer is$$\sum_{\text{primes }p \text{ where }2 \le p \le 97} {{\text{# of those divisible by }p\text{ from }2\text{ to }100 \text{ (inclusive)}}\over{99}}$$Computing the first few terms, I get$${50\over{99}} + {{33}\over{99}} + {{20}\over{99}} + {{14}\over{99}} + \ldots$$ However, I'm lazy and I'm wondering if there is a quicker way to finish solving this problem without having to calculate every single last term and add them all up.
• Do you pick with or without replacement? Jul 27 '21 at 12:27
• Note that $\left\lfloor\frac{100}{2}\right\rfloor = 50$, $\left\lfloor\frac{100}{3}\right\rfloor = 33$, $\left\lfloor\frac{100}{5}\right\rfloor = 20$ and so on. I don't know how you were calculating $50,33,20,14,\dots$ but it shouldn't take too long. Once you get to the primes between $33$ and $50$ those contribute $2$, once you get to the primes $51$ to $99$ those contribute $1$. Jul 27 '21 at 12:34
• As for writing this as a sum like you had, that is absolutely correct and follows from the linearity of expectation. Jul 27 '21 at 12:34
• It looks like it can be approximated by $\frac{100}{99}\sum\frac1p\approx\frac{100}{99}\ln(\ln(101))$ Jul 27 '21 at 12:37
• wolfram calculation Jul 27 '21 at 12:41
The primes $$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97$$ ($$25$$ of) of them have one prime factor. But so do the powers of primes. $$4,8,16,32,64, 9, 27,81,25,49$$ so $$10$$ of those so there are $$35$$ numbers with one prime factor.
If we take those with $$1$$ prime factors and multiply by a prime that isn't a factor to get a number with $$2$$ prime factors we get.
$$2 \times 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 9,27,25,49$$ are $$18$$ such numbers. (Tally $$18$$ number with 2 prime factors. $$53$$ numbers total.)
$$4\times 3, 5, 7, 11, 13, 17, 19, 23,9, 25$$ are $$10$$ such numbers.($$28$$ with 2 factors. $$63$$ total)
$$8\times 3,5,7,11,9$$ are $$5$$ such numbers.($$33$$, $68). $$16\times 3,5$$ are $$2$$ such numbers. ($$35,70$$) $$32\times 3$$ is $$1$$.($$36,71$$) $$3\times 7, 11, 13, 17, 19, 23, 29, 31,25$$ are $$9$$ such numbers.($$45,80$$) $$9\times 7,11$$ are $$2$$ such numbers. ($$47,82$$) And $$27\times 5> 10$$ so there are no more powers of $$3$$ to consider. $$5\times 3, 5, 7, 11, 13, 17, 19$$ are$$7$$ more.($$54,89$$) $$7\times 11,13$$ are the last $$2$$. Any more would be the product of $$2$$ primes over $$10$$.($$56,91$$) So there are $$56$$ numbers with $$2$$ factors. That accounts for $$91$$ of the $$99$$ numbers. There are only $$8$$ with $$3$$ or more factors. They are $$2\times 3\times 5, 4\times 3\times 5, 2\times 9\times 5,2\times 3\times 7, 4\times 3\times 7, 2\times 3\times 11,2\times 3\times 13,2\times 5\times 7$$ And so there are $$35 + 2\times 56 + 3\times 8 = 171$$ prime factors spread among $$99$$ numbers. The expected value of the number of prime factors of $$n$$ is $$\frac {171}{99}$$. • Argh... spent to long working on that. Keith Backman did the same thing but didn't waste time calculating the just majority of those with$2\$ primes. Jul 27 '21 at 17:48
There are $$25$$ primes smaller than $$100$$. In addition, there are $$10$$ numbers that are powers of primes smaller than $$100$$: $$4,8,9,16,25,27,32,49,64,81$$. In total, there are $$35$$ numbers between $$2$$ and $$100$$ inclusive that have only one prime factor.
Since $$2\cdot 3\cdot 5\cdot7=210>100$$, there are no numbers in that range which have four or more prime factors. There are $$8$$ numbers in that range which have three prime factors: $$30,42,60,66,70,78,84,90$$. You can either count those which have two prime factors, or calculate that number as $$99-35-8=56$$.
A randomly chosen number has $$\frac{35}{99}$$ chance of having $$1$$ prime factor, $$\frac{56}{99}$$ chance of having $$2$$ prime factors, and $$\frac{8}{99}$$ chance of having $$3$$ prime factors.
The expected number of prime factors would be $$1\cdot\frac{35}{99}+2\cdot\frac{56}{99}+3\cdot\frac{8}{99}=\frac{171}{99}$$, or between $$1$$ and $$2$$, with a slight bias towards $$2$$. | 2022-01-28T09:53:23 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/4208308/pick-random-integer-between-2-and-100-expected-value-of-the-number-of-prime",
"openwebmath_score": 0.9433048963546753,
"openwebmath_perplexity": 96.41720301135736,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9869795075882268,
"lm_q2_score": 0.8519527982093666,
"lm_q1q2_score": 0.8408599532650927
} |
http://mymathforum.com/algebra/30908-venn-diagram.html | My Math Forum Venn diagram
Algebra Pre-Algebra and Basic Algebra Math Forum
October 14th, 2012, 04:27 AM #1 Senior Member Joined: Jan 2012 Posts: 739 Thanks: 7 Venn diagram In a market survey, 100 traders sell fruits. 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes and 10 sell the three fruits. Each of the 100 traders sells atleast one of the three fruits. (i) Represent the information in a venn diagram. (ii) Find the number that sells oranges and mangoes only. It difficult for me to find the number that sells oranges and mangoes only. I drew three circles to show how the traders sell the fruits: all the three fruits n(AuOuM) = 10 oranges n(O) = 46-(4+10+x) = 32 mangoes n(M). = 50-(5+10+x) = 35 apple n(A) = 40-(4+10+5) = 21 apples and oranges n(AnO) = 14-10 = 4 apples and mangoes (AnM) = 15-10 = 5 I don't know how to find the number of traders that sells oranges and mangoes only n (OnM). Please let n(OnM) be x. What step or direction of thinking should I take to find x ?
October 14th, 2012, 07:34 AM #2 Senior Member Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: Venn diagram first of all where the ovals are overlapping 3 times you know its value is 10. 10 of the 15 people who sells apples and mangoes are selling the 3 fruits so there are 5 which are only selling apples and mangoes. 10 of the 14 people who sells apples and oranges are selling the 3 fruits so there are 4 which are only selling apples and mangoes. 10+4+5 of the 40 people who sells apples are selling something else then apples so there are 21 which are only selling apples. there are in total 40+50+46-100=36 fruits which are sold in pairs. 10 of them are sold by 3 so we have to substract 10 of that 36=26. 4+5+10 of them aren't selling oranges and mangoes so there are 26-19=7 who do. 50-10-7-5=28 are selling mangoes only and 46-7-10-4=25 are selling oranges only.
October 14th, 2012, 04:34 PM #3
Math Team
Joined: Dec 2006
From: Lexington, MA
Posts: 3,267
Thanks: 408
Re: Venn diagram
Hello, Chikis!
Quote:
In a market survey: 100 traders sell fruits. [color=beige]. . [/color]40 sell apples, [color=beige]. . [/color]46 sell oranges, [color=beige]. . [/color]50 sell mangoes, [color=beige]. . [/color]14 apples and oranges, [color=beige]. . [/color]15 apples and mangoes [color=beige]. . [/color]10 sell all three fruits. Each of the 100 traders sells at least one of the three fruits. (a) Represent the information in a Venn diagram. (b) Find the number that sells oranges and mangoes only.
It looks like you were off to a good start.
Use the scroll bar to see the entire Venn diagram.
Code:
*---------* *---------*
/ Apples \/ Oranges \
/ /\ \
* 21 * * 32-x *
| | 4| |
[40] | *---+--+---* | [46]
| / |10| \ |
* / * * \ *
\ * 5 \/ x * /
\ | /\ | /
*--+--+---* *------+--*
| |
* 35-x *
\ /
\ Mangos /
*----------*
[50]
Ten sold all three fruits.
The central intersection has "10".
15 sold apples and mangos.
Hence, there are 5 in the "apples and mangos only" region.
14 sold apples and oranges.
Hence, there are 4 in the "apples and oranges only" region.
The total in the Apples circle is 40.
[color=beige]. . [/color]Hence, "apples only" = $21.$
Let $x$ = number in "oranges and mangos only".
The total in the Mangos circle is 50.
[color=beige]. . [/color]Hence, "mangos only" = $35-x$.
The total in the Oranges circle is 46.
[color=beige]. . [/color]Hence, "oranges only" = $32-x.$
If we add all the quantities in the diagram, we get $107\,-\,x$
But we are told that this total is $100.$
$\text{Therefore: }\:107\,-\,x \:=\:100 \;\;\;\Rightarrow\;\;\; \fbox{x \:=\:7}$
October 15th, 2012, 07:04 AM #4 Global Moderator Joined: Dec 2006 Posts: 20,472 Thanks: 2039 You don't need to be given the number of traders who sell all three fruits! Let N denote the "number of traders who sell" function, then 40 + 46 + 50 = N(any fruit) + N(A and M) + N(A and O) + N(M and O only) = 100 + 15 + 14 + N(M and O only). Hence N(M and O only) = 7.
October 20th, 2012, 06:45 PM #5 Senior Member Joined: Jan 2012 Posts: 739 Thanks: 7 Re: Venn diagram Thank you soroban! Thanks to all of you; I have actually seen that: (21+5+10+4)+(32- x)+x (35-x) = 100 40+35-x+32-x+x = 100 -x = 100-107 x = -7/-1 = 7 Then the number trader x that sells oranges and mangoes will be 7.
October 20th, 2012, 06:59 PM #6 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Venn diagram I have merged your duplicate topics together. As you can see duplication of effort could have been prevented had you posted the topic only once. This is one reason we ask that you pick the most appropriate forum, then post your topic once. Not only does it prevent unnecessary duplication of effort from our valued contributors, who could be helping someone else instead of laboring to give help when the problem may already have been solved elsewhere, it prevents the cluttering of our forums with redundancy.
October 21st, 2012, 07:42 AM #7 Senior Member Joined: Jan 2012 Posts: 739 Thanks: 7 Re: Venn diagram Am sorry for duplicating the post.
Tags diagram, venn
,
,
,
,
,
,
,
,
,
,
,
,
,
,
in a market survey, 100 traders sell fruits, 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes, and 10 sell the three fruits. Each of the 100 traders sells at least one of the three fruits. (i) Represent the information
Click on a term to search for related topics.
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post JPArsenal87 Algebra 8 September 6th, 2013 07:20 AM MathsLOL Algebra 14 March 3rd, 2012 10:37 AM AINtKpkC3BBNn Applied Math 0 October 2nd, 2011 08:16 PM Jungelsucks Applied Math 1 September 10th, 2011 09:39 AM herohenson Advanced Statistics 2 January 26th, 2010 09:09 PM
Contact - Home - Forums - Cryptocurrency Forum - Top | 2019-04-22T02:10:01 | {
"domain": "mymathforum.com",
"url": "http://mymathforum.com/algebra/30908-venn-diagram.html",
"openwebmath_score": 0.32317599654197693,
"openwebmath_perplexity": 6411.429415488756,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9802808759252646,
"lm_q2_score": 0.8577681104440172,
"lm_q1q2_score": 0.8408536746468203
} |
http://math.stackexchange.com/questions/15108/how-many-symmetric-relations-on-a-finite-set | # How Many Symmetric Relations on a Finite Set?
How many symmetric relations are there for an $n$-element set?
Thank you.
-
When you say relation, do you mean it has to include all the $n$ elements in it, or is the empty relation (for example) is also counted - as it satisfies symmetry. – Asaf Karagila Dec 21 '10 at 21:11
all the Symmetric relations, including the empty relation. For instance, A={1,2} then the symmetric relations are: empty set, {(1,1)},{(1,2),(2,1)} etc. – Anonymous Dec 21 '10 at 21:23
This has a homework feel to it - is it? Since I don't just want to give the answer, here's a good hint: how many total relations are there for an n-element set, and what do they correspond to? Now, what do the symmetric relations correspond to, and can you use that to find your answer? – Steven Stadnicki Dec 21 '10 at 21:46
@Qiaochu I thought we agreed that if the OP doesn't say it's homework then we can't assume it is. In the past people have been flamed for asking what Steven is asking above. Do we have a new policy on, now that the elections are over? – Yuval Filmus Dec 21 '10 at 21:51
Steven, the number of all relations could be seen as the number of all the matrices of nxn, where every entry in the matrix could be either 0 or 1 - therefore, by the multiplication principle there is a total of 2^(n^2). From here, I don't see how to find the number of all the symmetric relations - I just know that it has to be symmetric from the diagonal or on the diagonal. – Anonymous Dec 21 '10 at 21:59
This problem is very similar to Number of relations that are both symmetric and reflexive
A symmetric relation $R$ on a set $A$ is a subset $A\times A$. We can write $R$ as $B\cup C$, where $B$ is a subset of $\{(a,a)\mid a\in A\}$ and $C$ is a subset of $\{(b,c)\in A\times A\mid b\ne c\}$.
Note there are as many choices for $B$ as subsets of $A$, namely $2^n$.
To count the number of possible sets $C$, we use that $C$ is symmetric, meaning, if $(b,c)\in C$ then also $(c,b)\in C$. Now, let ${}[A]^2$ be the collection of subsets of $A$ of size 2. Note ${}[A]^2$ consists of subsets rather than ordered pairs. Given any $D$ subset of ${}[A]^2$, we can associate to it a set $C$ by setting $C=\{(b,c)\mid \{b,c\}\in D\}$. Note that $C$ is symmetric, since $\{b,c\}=\{c,b\}$.
Conversely, any $C$ symmetric corresponds to a unique $D\subseteq[A]^2$, namely $\{\{b,c\}\mid (b,c)\in C\}$. This is why in $C$ we only allow pairs $(b,c)$ with $b\ne c$, so the resulting $D$ is a subset of ${}[A]^2$ and we have a correspondence.
We have shown that there are as many $C$ as there are subsets $D$ of ${}[A]^2$. But $\displaystyle |[A]^2|={n\choose 2}=\frac{n(n-1)}2$, so the number of subsets is $2^{n\choose 2}$.
Finally, since we can pair any $B$ with any $C$, we have that the number of binary symmetric relations on a set $A$ of size $n$ is precisely $$2^{n+{n\choose 2}}=2^{{n+1}\choose 2}.$$
-
Andres, please do not provide complete answers to (what I still strongly suspect to be) homework questions. – Qiaochu Yuan Dec 21 '10 at 22:18
@Qiaochu: There is no indication to me that this is a homework problem. I expect people to act honorably, and be honest. If the OP does not explicitly say this is a homework problem, and it is not self-evident to me, I won't assume otherwise. – Andrés Caicedo Dec 21 '10 at 22:24
Hello Andres, thank you for the detailed proof, I haven't finished reading it. if A={1,2} then the number of set B we have isn't 4 but 3, if we count the subsets as sets. ({(1,1)},{(2,2)},{(1,1),(2,2)}. the fourth set is {(2,2),(1,1)} but this set is equal to {(1,1),(2,2)}. – Anonymous Dec 21 '10 at 22:25
@Anonymous: I think you are not counting the empty relation $\{\}$. It is symmetric (so you have 4 rather than 3 after all), but I am not sure whether you have some convention that excludes it. (I don't think you want to exclude it, according to the comments above.) – Andrés Caicedo Dec 21 '10 at 22:29
Andres: I think you didn't understand me. You wrote: "Note there are as many choices for B as subsets of A, namely 2^n". I think that your definition of B is not correct, so I gave an example of A={1,2} and then I wrote the sets B by your definition. – Anonymous Dec 21 '10 at 22:33
You can also think of it as a matrix $nxn$, with the elements of the matrix being $(a_i,a_j)$ with $a_i,a_j \in A$. The elements of the main diagonal can be perfectly chosen for the relation because they are symmetric. For the rest of the elements, picking a pair from the upper triangle say $(a_2,a_1)$ implies that you are also picking $(a_1,a_2)$. So from the total $n^2$ pairs you end up with only $\sum_{i=0}^{n} i = \frac{n(n+1)}{2}$ from which to choose. You can do this in $2^{\frac{n(n+1)}{2}}$ ways
- | 2015-11-25T04:13:54 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/15108/how-many-symmetric-relations-on-a-finite-set",
"openwebmath_score": 0.8764781951904297,
"openwebmath_perplexity": 182.85874773325452,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9802808724687407,
"lm_q2_score": 0.8577681068080748,
"lm_q1q2_score": 0.8408536681176796
} |
https://math.stackexchange.com/questions/1603384/calculate-inclusion-exclusion-in-birthday-paradox | Follow-on from this post
I was trying Birthday Paradox for 5-day calendar, with 3 people
The probability that NONE of them have matching birthday is
$5/5 * 4/5 * 3/5 = 0.48$
The probability there is AT LEAST one matching pair is
$1 - 0.48 = 0.52$
Question
What if I wanted to use inclusion-exclusion, i.e. the "long-cut", how to calculate this
$P = 0.52 = P(A_1,_2) + P(A_1,_3) + P(A_2,_3)$
$P = 0.52 = P(A_1,_2 \cup A_1,_3 \cup A_2,_3) - P(A_1,_2 \cap A_1,_3 \cap A_2,_3)$
This has been puzzling me for days, please assist.
• Actually probability of no match is $\frac55\frac45\frac35=0.48$ – paw88789 Jan 7 '16 at 17:15
• @paw88789 Thanks, I corrected OP – Rhonda Jan 7 '16 at 17:23
We'll let $$A = \{\text{At least one match}\}$$ and so $$\bar A = \{\text{No match}\}.$$ So, mechanically \begin{align*} P(A) &= P(A_{1,2}\cup A_{1,3}\cup A_{2,3})\\ &= P(A_{1,2})+ P(A_{1,3})+P(A_{2,3})\tag 1\\ &\qquad-[P(A_{1,2},A_{1,3})+P(A_{1,2}, A_{2,3})+P(A_{1,3}, A_{2,3})]\\ &\qquad+[P(A_{1,2},A_{1,3},A_{2,3})]\\ &= 3P(A_{1,2})-3P(A_{1,2},A_{2,3})+P(A_{1,2},A_{1,3},A_{2,3})\tag 2\\ &=3\binom{5}{1}\left(\frac{1}{5}\right)^2-3\binom{5}{1}\left(\frac{1}{5}\right)^3+\binom{5}{1}\left(\frac{1}{5}\right)^3\\ &= 0.52 \end{align*} where in $(1)$ I invoke inclusion-exclusion, and in $(2)$ I recognize that some terms are identical in probability and simply multiply by 3. There are $\binom{5}{1} = 5$ ways to choose a day in a 5-day year.
Verify, using the complement and independence, $$P(A) = 1-P(\bar A) = 1-\frac{5}{5}\cdot\frac{4}{5}\cdot\frac{3}{5} = 0.52.$$
• I'm trying to do this pictorially, i.e. three intersecting circles – Rhonda Jan 8 '16 at 19:59
• @Rhonda Does this help? – Em. Jan 8 '16 at 20:31
• Holy Cow! That's what I'm trying to do right now. I believe this shall help, let me check. – Rhonda Jan 8 '16 at 20:32
• So the $P(AB,B2,AC)$ gets subtracted three times with $-[P(AB,BC)+P(BC,AC)+P(AB,AC)]$, hence we must add it back once? Because $[P(AB,BC)+P(BC,AC)+P(AB,AC)]$ adds the intersection of $A \cap B \cap C$ three times? – Rhonda Jan 8 '16 at 20:38
• @Rhonda Yeah, notice that $P(AB,BC)$ is just saying $P(ABC)$. This why in the green box you subtract it $3$ times. Finally, inclusion-exclusion tells you to add one back. So in the end, in the green box, you only subtract 2. – Em. Jan 8 '16 at 20:43
Essentially according to the inclusion-exclusion principle, we have the following line of reasoning: The probability that the first and second people share a birthday is $p_{12} = 1/5$. Similarly, $p_{13} = p_{23} = 1/5$. If we add these three probabilities up, we get $3/5$.
However, in so doing, we have triple-counted the probability that all three share the same birthday. That probability is $(1/5)(1/5) = 1/25$, and since it is triple-counted in the sum, we should subtract twice that amount. Therefore, the overall probability that at least two people share a birthday is
$$P(\text{birthday shared}) = \frac{3}{5}-\frac{2}{25} = \frac{13}{25} = 0.52$$
• I'm trying to draw a picture of this. I understand everything up to and since it is triple-counted in the sum, we should subtract twice that amount – Rhonda Jan 8 '16 at 20:00
Since the events you are considering, $P_{1,2}, P_{1,3}, P_{2,3}$ are not disjoint, you cannot just sum their probabilities in order to get the probability of the union. You can therefore resort to the inclusion-exclusion principle which, as you know, states that for events $A_1, A_2, \dots, A_n$ $$\mbox{P}\biggl(\bigcup_{i=1}^n A_i\biggr) {} =\sum_{i=1}^n \mbox{P}(A_i) -\sum_{i<j}\mbox{P}(A_i\cap A_j) +\sum_{i<j<k}\mbox{P}(A_i\cap A_j\cap A_k)-\ \cdots\ +(-1)^{n-1}\, \mbox{P}\biggl(\bigcap_{i=1}^n A_i\biggr),$$
In a nutshell, you add the probabilities of all possible intersections of an odd number of events and subtract the probabilities of all possible intersections of an even number of events.
In your case this translates to $$\frac {3}{5} - \frac {3}{25} + \frac {1}{25} =0.52$$ | 2019-06-20T11:53:19 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1603384/calculate-inclusion-exclusion-in-birthday-paradox",
"openwebmath_score": 0.9707184433937073,
"openwebmath_perplexity": 546.6937899245298,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9802808759252645,
"lm_q2_score": 0.8577681013541613,
"lm_q1q2_score": 0.8408536657362082
} |
https://www.physicsforums.com/threads/olympiad-problem-sum-involving-many-square-roots.917270/ | # Olympiad problem -- Sum involving many square roots...
1. Jun 11, 2017
### Mathysics29
√(2-√(2^(2)-1))+√(4-√(4^(2)-1))+√(6-√(6^(2)-1))+...+√(80-√(80^(2)-1))
How the find it's value
2. Jun 11, 2017
### Nidum
Do you have any ideas yourself ?
Just out of curiosity are there any rules about how these problems have to be solved ? Do the solutions always have to be analytic ones ?
Personally if I wanted an answer to this problem for some practical purpose I would just write a program to sum all the terms numerically - it would only be a few lines of coding .
3. Jun 11, 2017
### Mathysics29
Well you are not supposed to use a calculator and the answer should be rounded of to nearest integer if needed.
4. Jun 11, 2017
### Mathysics29
Well i have an idea and here it is:
So if y=2-√(2^(2)-1)
y=2-√3
y(2+√3)=1
And this works for all the terms.
5. Jun 11, 2017
### Mathysics29
But I don't know how to connect it to the equation
6. Jun 11, 2017
### Staff: Mentor
I moved the thread to our homework section.
Or y=1/(2+√3).
If you look at larger values like 1/(78+√(782-1)) or 1/(80+√(802-1)), do you see how you could approximate that?
Does this lead to a familiar series?
7. Jun 11, 2017
### Mathysics29
Convering series? But how can I estimate the final and?
8. Jun 11, 2017
### Staff: Mentor
With an approximation for this series (well, the finite partial sum, to be more precise).
9. Jun 11, 2017
### Mathysics29
Can I say this
40
Summation (√2n-√(2n)^(2)-1))
n=1
And how can I find the partial sum for this
10. Jun 11, 2017
### Staff: Mentor
This is just your original problem written in a different way. Sure.
Find a suitable approximation, see above.
11. Jun 12, 2017
### Fred Wright
I'm posting my solution because this thread wasn't originally in homework section and doesn't appear to be homework.
After some algebra I write the problem as$$\sqrt{2}\sum_{n=1}^{40}\sqrt{n-\sqrt{n^2-.25}}$$
I multiply the sum by,
$$\frac{\sqrt{n+\sqrt{n^2-.25}}}{\sqrt{n+\sqrt{n^2-.25}}}$$
I approximate$$\sqrt{n^2-.25}\approx n$$
And after a little more algebra the sum becomes approximately$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}}$$
I approximate $\sqrt{2}\approx 1.5$ $\sqrt{3}\approx 2$ etc. I note the first two terms $\approx \frac{1.75}{2}$, the next five terms$\approx 5\frac{1}{2}\frac{1}{2}$, the next five terms $5\frac{1}{2}\frac{1}{3}$, the next ten terms $\approx 10\frac{1}{2}\frac{1}{4}$, and the last eighteen terms $\approx 18\frac{1}{2}\frac{1}{6}$. Adding it all up I get 5.2.
12. Jun 12, 2017
### Staff: Mentor
Hmm, the program (formula from post #9) ended with $5.65685 ...$
13. Jun 12, 2017
### Staff: Mentor
Quite good for an approximation. If we round we are off by 1.
Replace the sum by an integral:
$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx \frac 1 2 \int_{0.7}^{40.5} \frac{dx}{\sqrt{x}} = [\sqrt{x}]_{0.7}^{40} = \sqrt{40}-\sqrt{0.7} \approx 6.3 - 0.85 = 5.45$$
0.7 is black magic based on intuition, the square roots were estimated without calculator.
Compare it to the precise value of the sum:
$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx 5.6338$$
If we round that, we are still off by 1, but now the estimate is within 0.2 of the precise value of the original problem.
We get a better result with a numerical evaluation of the first two terms:
$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx \frac 1 2 \left( 1+\frac{1}{\sqrt 2} + \int_{2.5}^{40.5} \frac{dx}{\sqrt{x}} \right) = 0.5 + 0.35 + \sqrt{40}-\sqrt{2.5} \approx 0.85 + 6.33 - 1.59 = 5.59$$
Again without calculator (and without using the result from above...), but with a bit more thought about the square roots. | 2018-01-17T22:43:29 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/olympiad-problem-sum-involving-many-square-roots.917270/",
"openwebmath_score": 0.47044941782951355,
"openwebmath_perplexity": 1711.7250980739514,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9802808747730899,
"lm_q2_score": 0.8577681013541611,
"lm_q1q2_score": 0.8408536647479096
} |
https://mathoverflow.net/questions/261358/compact-manifolds-with-big-mapping-class-group | # Compact manifolds with big mapping class group
I was wondering if compact surfaces were the only compact manifolds with a "big" or "complicated" mapping class group.
Are there higher dimensional manifolds (which are not in some way reducible to a surface) that have an infinite mapping class group? What kind of group can arise as the mapping class group of a compact manifold?
• Is "mapping class group" defined as $\pi_0$ of diffeomorphisms or $\pi_0$ of homotopy equivalences? For surfaces these are nontrivially the same, but I'm never sure which is considered the definition. Feb 4, 2017 at 19:01
• what about the $n$-torus? its MCG is $GL_n(\mathbf{Z})$ as far as I know (whatever the definition). More generally many nilmanifolds have a big MCG.
– YCor
Feb 4, 2017 at 19:05
• $\mathbb{Z}_2^\infty \leq \pi_0(Diff(T^n))$, $n >5$. p. 5: math.cornell.edu/~hatcher/Papers/Diff%28M%292012.pdf Feb 5, 2017 at 4:30
• @TomGoodwillie Indeed, I'm not sure what is the good notion to look at here. I'd say homotopy equivalence not to add subtleties of analytic nature. Feb 5, 2017 at 9:05
Take $M^d$ to be a connected sum of $n$ copies of $S^1\times S^{d-1}$, where $d\ge 3$. Then $M$ is a closed, orientable manifold of dimension $d$ with $\pi_1(M)=F_n$, the free group of rank $n$. If $n>1$, the mapping class group $\pi_0({\rm Diff}(M))$ surjects onto the outer automorphism group ${\rm Out}(F_n)$, which is infinite, and very complicated.
For an explanation of the relation between mapping class groups and outer automorphism groups, see for instance here. For the surjectivity of the natural map $\pi_0({\rm Diff}(M)) \to {\rm Out}(F_n)$, see here.
• Why is the mapping class group in that case the outer automorphism group of the fundamental group? Feb 4, 2017 at 17:05
• Good question. I edited my answer, accordingly. Is this now better? Feb 4, 2017 at 19:39
• While I know that the isomorphism $\pi_0(\text{Diff}(M)) \to \text{Out}(F_n)$ holds for $d=3$, due to Hatcher and based on prime decomposition theory for 3-manifolds, I don't know of a reference for higher dimensions. Feb 4, 2017 at 21:09
In Infinitesimal computations in topology Sullivan shows in Theorem 13.3 that if $M$ is a simply-connected manifold of dimension $>5$, then $\pi_0(\mathrm{Diff\,} M)$ is commensurable to an arithmetic group (see p.295 for definitions of an arithmetic group).
The group of homotopy classes of homotopy self-equivalences is also commensurable to an arithmetic one (Theorem 10.3)
In Corollary 13.3 Sullivan explains how the forgetful map between the two is largely controlled by the rational Pontryagin class of the manifold (which is preserved by diffeomorphisms by not by homotopy equivalences).
• Link seems to be broken, here's one that works for ma : archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1977__47_/… Feb 6, 2017 at 6:29
• Both links appear broken to me. Feb 23, 2017 at 17:49
• Hmm, both linked initially worked. In any case this is easily searchable by title. Feb 23, 2017 at 18:52
There are situations in which surfaces are the "unique" examples with big mapping class groups. One such is closed manifolds of negative sectional curvature.
Theorem (Paulin): If $M$ is a closed $n$-dimensional manifold of negative sectional curvature and $n>2$ then $\mathrm{Out}(\pi_1M)=\mathrm{MCG}(M)$ is finite.
Proof: Paulin actually proved that if $\Gamma$ is a torsion-free word-hyperbolic group (such as the fundamental group of a closed manifold of negative curvature) and $\mathrm{Out}(\Gamma)$ is infinite then $\Gamma$ splits over a cyclic subgroup, say as $\Gamma=A*_CB$ for $C$ cyclic. (There is also the HNN-extension case, which is similar.)
Mayer--Vietoris now gives that
$\ldots\to H^{n-1}(C)\to H^n(M)\to H^n(A)\oplus H^n(B) \to H^n(C)\to\ldots$
is exact.
Since $n>2$, $H^{n-1}(C)\cong H^n(C)\cong 0$ and so $H^n(M)\cong H^n(A)\oplus H^n(B)$. But Strebel's theorem implies that $H^n(A)\cong H^n(B)\cong 0$ , a contradiction. QED
(Here, I'm taking $\mathrm{MCG}(M)$ to mean $\pi_0$ of self-homotopy-equivalences. Since negatively curved manifolds are aspherical, this coincides with $\mathrm{Out}(\pi_1M)$.)
This is actually an instance of a much more general phenomenon. In good situations, JSJ theory tells us that $\mathrm{Out}(\Gamma)$ can be 'broken up' into pieces which are essentially mapping class groups of surfaces and tori. This happens in the case of irreducible 3-manifolds, for instance.
• Where does the assumption $n>2$ come into play? Feb 5, 2017 at 1:08
• @VictorProtsak: The Mayer-Vietoris computation.
– HJRW
Feb 5, 2017 at 6:03 | 2023-02-04T03:01:48 | {
"domain": "mathoverflow.net",
"url": "https://mathoverflow.net/questions/261358/compact-manifolds-with-big-mapping-class-group",
"openwebmath_score": 0.8006851673126221,
"openwebmath_perplexity": 392.6695503418024,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9802808736209154,
"lm_q2_score": 0.8577680995361899,
"lm_q1q2_score": 0.8408536619774886
} |
https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads | # Expected Number of Coin Tosses to Get Five Consecutive Heads
A fair coin is tossed repeatedly until 5 consecutive heads occurs.
What is the expected number of coin tosses?
• Yet another copy and paste from Brilliant.org: brilliant.org/i/5rCgJ3 – Erick Wong Apr 17 '13 at 5:49
• @ErickWong: Is this a recent problem on brilliant.org? – robjohn Apr 17 '13 at 8:48
Let $e$ be the expected number of tosses. It is clear that $e$ is finite.
Start tossing. If we get a tail immediately (probability $\frac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\frac{1}{4}$), then the expected number is $e+2$. Continue $\dots$. If we get $4$ heads then a tail, the expected number is $e+5$. Finally, if our first $5$ tosses are heads, then the expected number is $5$. Thus $$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{32}(5).$$ Solve this linear equation for $e$. We get $e=62$.
• It is clear that e is finite, but how can you show it properly though ? Thanks. – Dark Jul 3 '15 at 17:39
• If one wants, let $X$ be the number of tosses. Then $\Pr(X=n)\le (1/2)^{n-5}$. So $E(X)\le \sum n (1/2)^{n-5}$, a convergent series. – André Nicolas Jul 3 '15 at 17:58
• The same method obviously generalizes to give $e_n$, the expected number of tosses to get $n$ consecutive heads ($n \ge 1$): $$e_n=\frac{1}{2}(e_n+1)+\frac{1}{4}(e_n+2)+\frac{1}{8}(e_n+3)+\frac{1}{16}(e_n+4)+\cdots +\frac{1}{2^n}(e_n+n)+\frac{1}{2^n}(n),$$ the solution of which is easily found to be $$e_n = 2(2^n - 1).$$ – r.e.s. Jul 19 '15 at 23:57
• Why are TT, TTT not considered? – Jaydev Jul 24 '17 at 1:24
• @Jaydev TT and TTT both are covered by the case "if we get a tail immediately". – David K Oct 6 '17 at 13:13
Lets calculate it for $n$ consecutive tosses the expected number of tosses needed.
Lets denote $E_n$ for $n$ consecutive heads. Now if we get one more head after $E_{n-1}$, then we have $n$ consecutive heads or if it is a tail then again we have to repeat the procedure.
So for the two scenarios:
1. $E_{n-1}+1$
2. $E_{n}{+1}$ ($1$ for a tail)
So, $E_n=\frac12(E_{n-1} +1)+\frac12(E_{n-1}+ E_n+ 1)$, so $E_n= 2E_{n-1}+2$.
We have the general recurrence relation. Define $f(n)=E_n+2$ with $f(0)=2$. So,
\begin{align} f(n)&=2f(n-1) \\ \implies f(n)&=2^{n+1} \end{align}
Therefore, $E_n = 2^{n+1}-2 = 2(2^n-1)$
For $n=5$, it will give us $2(2^5-1)=62$.
• Amazing solution, thank you – Jaydev Jul 24 '17 at 1:30
Here is a generating function approach.
Consider the following toss strings, probabilities, and terms
$$\color{#00A000}{ \begin{array}{llc} T&\frac12&\qquad\frac12x\\ HT&\frac14&\qquad\frac14x^2\\ HHT&\frac18&\qquad\frac18x^3\\ HHHT&\frac1{16}&\qquad\frac1{16}x^4\\ HHHHT&\frac1{32}&\qquad\frac1{32}x^5\\ \color{#C00000}{HHHHH}&\color{#C00000}{\frac1{32}}&\color{#C00000}{\qquad\frac1{32}x^5} \end{array} }$$ Each term has the probability as its coefficient and the length of the string as its exponent.
Possible outcomes are any combination of the green strings followed by the red string. We get the generating function of the probability of ending after $n$ tosses to be \begin{align} f(x)&=\sum_{k=0}^\infty\left(\frac12x+\frac14x^2+\frac18x^3+\frac1{16}x^4+\frac1{32}x^5\right)^k\frac1{32}x^5\\ &=\frac{\frac1{32}x^5}{1-\left(\frac12x+\frac14x^2+\frac18x^3+\frac1{16}x^4+\frac1{32}x^5\right)}\\ &=\frac{\frac1{32}x^5}{1-\frac{\frac12x-\frac1{64}x^6}{1-\frac12x}}\\ &=\frac{\frac1{32}x^5-\frac1{64}x^6}{1-x+\frac1{64}x^6} \end{align} The average duration is then \begin{align} f'(1) &=\left.\frac{\left(\frac5{32}x^4-\frac6{64}x^5\right)\left(1-x+\frac1{64}x^6\right)-\left(\frac1{32}x^5-\frac1{64}x^6\right)\left(-1+\frac6{64}x^5\right)}{\left(1-x+\frac1{64}x^6\right)^2}\right|_{\large x=1}\\ &=\frac{\frac4{64}\frac1{64}+\frac1{64}\frac{58}{64}}{\left(\frac1{64}\right)^2}\\[12pt] &=62 \end{align}
• Could you elaborate briefly on why the derivative gives the expected number of flips? – Austin Mohr Aug 20 '13 at 2:37
• @AustinMohr: If $f(x)$ is the generating function of the probability $p_n$ of the ending after $n$ tosses $$f(x)=\sum_{n=0}^\infty p_nx^n$$ then, because the probability of lasting an infinite number of tosses is $0$, we have \begin{align} f(1) &=\sum_{n=0}^\infty p_n\\ &=1 \end{align} Furthermore, \begin{align} f'(1) &=\sum_{n=0}^\infty n\,p_n\\ &=\mathrm{E}(n) \end{align} – robjohn Aug 20 '13 at 5:13
• Do you know how to find the distribution (or expectation and variance) for the number of tosses until either 5 consecutive heads or 5 consecutive tails? (Or 5 consecutive equal results from rolling dice.) Is there a question on math.se about this? – ShreevatsaR Dec 15 '15 at 14:17
• I found an answer using martingales here: quora.com/… but I'm curious if there is a generating functions way (also about the distribution, say variance or number of trials until 90% probability of seeing what we want). – ShreevatsaR Dec 15 '15 at 14:31
• @ShreevatsaR: The generating function for the probability of ending on $n$ tosses for that problem is similar: $$\frac{\frac1{16}x^5}{1-\left(\frac12x+\frac14x^2 +\frac18x^3+\frac1{16}x^4\right)}$$ From that, we can compute the expectation and variance. Looking at the coefficients of the series for the generating function we can also find out that in $66$ tosses, we will have a $90.0761\%$ chance of seeing $5$ heads or $5$ tails in a row. – robjohn Dec 15 '15 at 17:32
This problem is solvable with the next step conditioning method. Let $\mu_k$ denote the mean number of tosses until 5 consecutive heads occurs, given that $k$ consecutive heads just occured. Obviously $\mu_5=0$. Conditioning on the outcome of the next coin throw: $$\mu_k = 1 + \frac{1}{2} \mu_{k+1} + \frac{1}{2} \mu_0$$ Solving the resulting linear system:
In[28]:= Solve[Table[mu[k] == 1 + 1/2 mu[k + 1] + mu[0]/2, {k, 0, 4}],
Table[mu[k], {k, 0, 4}]] /. mu[5] -> 0
Out[28]= {{mu[0] -> 62, mu[1] -> 60, mu[2] -> 56, mu[3] -> 48,
mu[4] -> 32}}
Hence the expected number of coin flips $\mu_0$ equals 62.
• What tool did you use for solving? – pushpen.paul May 11 '15 at 17:19
• @pushpen.paul I used Mathematica – Sasha May 11 '15 at 17:20
• Can you please explain the original equation? – BOS Sep 20 '16 at 15:58
• @BOS Since $\mu_k$ is the conditional expectation, consider the next coin toss. Because a new toss was made, we add 1, in the next state, with equal probabilities we either get next head, in which case we gonna get $k+1$ heads, hence $\mu_{k+1}$, or the tail, in which case we break the streak of consecutive heads, hence $\mu_0$. – Sasha Sep 23 '16 at 3:03
• @Sasha This is the Markov way of solving it, right? This seems most intuitive to me of all methods presented here. – user3496060 Dec 22 '17 at 8:06
The question can be generalized to what is the expected number of tosses before we get x heads.Let's call this E(x). We can easily derive a recursive formula for E(x). Now, there are a total of two possibilities, first is that we fail to get the xth consecutive heads in xth attempt and second, we succeed. Probability of success is 1/(2^x) and probability of failure is 1-(1/(2^x)).
Now, if we were to fail to get xth consecutive heads in xth toss (i.e. case 1), the we will have to use a total of (E(x)+1) moves, because one move has been wasted.
On the other hand if we were to succeed in getting xth consecutive head in xth toss (i.e. case 2), the total moves is E(x-1)+1 , because we now take one move more than that was required to get x-1 consecutive heads.
So,
E(x) = P(failure) * (E(x)+1) + P(success) * (E(x-1)+1)
E(x) = [1-(1/(2^x))] * (E(x)+1) + [1/(2^x)] * (E(x-1)+1)
Also E(0) = 0 , because expected number of tosses to get 0 heads is zero, duh
now,
E(1) = (1-0.5) * (E(1)+1) + (0.5) * (E(0)+1) => E(1) = 2
E(2) = (1-0.125) * (E(1)+1) + (0.125) * (E(1)+1) => E(2) = 6
Similarly,
E(3) = 14
E(4) = 30
E(5) = 62
I would simplify the problem as follows:
Let $e$ = Expected number of flips until $5$ consecutive $H$, i.e., $E[5H]$
Let $f$ = Expected number of flips until $5$ consecutive $H$ when we have seen one $H$, i.e., $E[5H|H]$
Let $g$ = Expected number of flips until $5$ consecutive $H$ when we have seen two $H$, i.e., $E[5H|2H]$
Let $h$ = Expected number of flips until $5$ consecutive $H$ when we have seen three $H$, i.e., $E[5H|3H]$
Let $i$ = Expected number of flips until $5$ consecutive $H$ when we have seen four $H$, i.e., $E[5H|4H]$
Now Start flipping coin, there is $\frac{1}{2}$ probability of getting $H$ or $T$. So if we get $H$ then expected number of flips until 5 consecutive $H$ is $(f+1)$. Alternatively if $T$, we wasted 1 flip and expected number is still $(e+1)$ $$e=\frac12(e+1)+\frac12(f+1)\;$$
We now need $f$ to solve above to get $e$. Now we start with 1 $H$ and seeking 4 more $H$ to get total 5 $H$. Again, there is $\frac{1}{2}$ probability of getting $H$ or $T$. So if we get $H$ (total $2H$ so far) then expected number of flips until 5 consecutive $H$ is $(g+1)$. Alternatively if $T$, we wasted this flip and expected number is back to $(e+1)$
$$f=\frac12(g+1)+\frac12(e+1)\;$$
Continuing this way... $$g=\frac12(h+1)+\frac12(e+1)\;$$ $$h=\frac12(i+1)+\frac12(e+1)\;$$
Finally, Now we have 4 $H$ and seeking last $H$ to get total 5 $H$. Still, there is $\frac{1}{2}$ probability of getting $H$ or $T$. So if we get $H$ (total $5H$) then we need just $1$ flip. Alternatively if $T$ is observed, we wasted this flip and expected number is back to $(e+1)$ $$i=\frac12(1)+\frac12(e+1)\;$$
Solving these equations, $e=62$, $f=60$, $g=56$, $h=48$, $i=32$
This solution offers some insight into conditional expectations of number of flips needed till 5 consecutive $H$ given 1, 2, 3 and 4 consecutive $H$.
Expectation for getting n consecutive heads is : 2*(2^n-1). Thus for 5 heads it is = 62.
• How did you get the formula? – pushpen.paul May 11 '15 at 18:25
No one else seems to have suggested the following approach. Suppose we keep flipping a coin until we get five heads in a row. Define a "run" as either five consecutive heads or a tails flip plus the preceding streak of heads flips. (A run could be a single tails flip.) The number of coin flips is equal to the number of runs with at least four heads ($R_{4+}$), plus the number of runs with at least three heads ($R_{3+}$), and so on down to the number of runs with at least zero heads ($R_{0+}$). We can see this by expanding the terms:
$R_{0+}$ = # runs with 0 heads + # runs with 1 head + ... + # runs with 5 heads
$R_{1+}$ = # runs with 1 head + # runs with 2 heads + ... + # runs with 5 heads
...
$R_{4+}$ = # runs with 4 heads + # runs with 5 heads
# flips = # flips in runs with 0 heads + # flips in runs with 1 head + ... + # flips in runs with 5 heads
# flips in runs with 0 heads = # runs with 0 heads
# flips in runs with 1 head = 2 x # runs with 1 head
...
# flips in runs with 4 heads = 5 x # runs with 4 heads
# flips in runs with 5 heads = 5 x # runs with 5 heads
By linearity of expectation, the expected number of coin flips is $E(R_{0+}) + E(R_{1+}) + \ldots + E(R_{4+})$. $E(R_{4+})$ is $2 E(R_{5+}) = 2$, because one half of the time we flip at least four heads in a row, we go on to flip five heads in a row, i.e. the following coin flip is heads. In other words, in expectation, it takes two runs that start with four heads to achieve one run of five heads. Likewise, $E(R_{3+}) = 2 E(R_{4+}) = 4$, $E(R_{2+}) = 2 E(R_{3+}) = 8$, $E(R_{1+}) = 2 E(R_{2+}) = 16$, and $E(R_{0+}) = 2 E(R_{1+}) = 32$. The expected number of coin flips is $32 + 16 + 8 + 4 + 2 = 62$.
More generally, given a biased coin that comes up heads $p$ portion of the time, the expected number of flips to get $n$ heads in a row is $\frac{1}{p} + \frac{1}{p^2} + \ldots + \frac{1}{p^n} = \frac{1 - p^n}{p^n(1 - p)}$.
We can solve this without equations. Ask the following (auxiliary) question: how many flips you need to get either $N$ heads or $N$ tails. Then to get $N$ heads only you need twice as many flips. Start with the question of how many flips you need to get either $H$ or $T$. The answer is $$x = \frac12 (1) + \frac12 (1) = 1.$$ The reason is that there is $\frac12$ probability to get $H$, and after $1$ flip you are done. The same for $T$. To get only one $H$ you then need two flips. OK. Now we ask what it takes to get $HH$ or $TT$. The result is $$x=\frac12(1+2) + \frac12(1+2).$$ The number $2$ appears because, say you flip $H$ first, then you need on average $2$ flips to get another $H$, as we learned earlier. The same for $T$. So you need $3$ flips to get $TT$ or $HH$, and you need $6$ flips to get $HH$ only. And so on. You need $\frac12 (1+6) + \frac12(1+6) = 7$ flips to get either $HHH$ or $TTT$, and $14$ to get $HHH$ only. If you need $HHHH$ or $TTTT$, then flip $\frac12(1+14) + \frac12(1+14) = 15$ times, or $30$ times to get just $HHHH$. The sequence is $1, 3, 7, 15, \ldots$ to get either heads or tails. The formula is easy to extract: you need $2^N-1$ flips to get either $N$ heads or $N$ tails, or $2^{N+1}-2$ to get $N$ heads only. If $N=5$ we get the answer: $62$.
Use Markov chains. The nice part of Markov Chains is that they can be applied to a huge class of similar problems with relatively little thought (it's almost formulaic in application). It's also the most intuitive way to handle these problems.
(in Matlab code notation below)
%% setup full transition matrix with states from zero heads to 5 heads
T = $[ones(5,1)*.5,eye(5)*.5];$
$T = [T;zeros(1,6)]$
%%Take subset "Q" comprised of just transient states (5 heads is absorbing state) $Q = T(1:end-1,1:end-1);$
$M = inv(eye(5)-Q)$
absorbing Markov Chain has a similar example as this question BTW...
ans =
62
60
56
48
32
Where each row is the expected number of steps before being absorbed when starting in that transient state (0 through 4 heads, top to bottom).
A recursive programming solution is also possible. Below is the solution in Python
# Expected number of tosses to get n heads
def Expectation(n):
if n == 0:
return 0
return 2**n + Expectation(n-1)
>>> Expectation(1)
2
>>> Expectation(2)
6
>>> Expectation(3)
14
>>> Expectation(5)
62
>>> Expectation(10)
2046 | 2019-11-14T04:32:11 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads",
"openwebmath_score": 0.8999800086021423,
"openwebmath_perplexity": 439.89580526319247,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9802808724687407,
"lm_q2_score": 0.8577680995361899,
"lm_q1q2_score": 0.8408536609891899
} |
https://math.stackexchange.com/questions/3041656/sum-of-two-co-prime-integers | # Sum of two co-prime integers
I need some help in a proof: Prove that for any integer $$n>6$$ can be written as a sum of two co-prime integers $$a,b$$ s.t. $$\gcd(a,b)=1$$.
I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof. I mainly used arithmetic progression of $$4$$, $$(4n,4n+1,4n+2,4n+3)$$, but got not much, only to the extent of specific examples and even than sometimes $$a,b$$ weren't always co-prime (and $$n$$ was also playing a role so it wasn't $$a+b$$ it was $$an+b$$).
I would appriciate it a lot if someone could give a hand here.
• What's wrong with $n=1+(n-1)$? Perhaps you meant to require $a,b>1$ but you should specify that. – lulu Dec 15 '18 at 16:03
• Dirichlet’s theorem is useless in that kind of situation; the problem is much, much simpler. Aside from the obvious case $n=(n-1)+1$, you can try to find general patterns when $n=4m$, $n=2m+1$, $n=4m+2$. – Mindlack Dec 15 '18 at 16:05
• True i should have specified that a,b>1, sorry, my bad. – Daniel Gimpelman Dec 15 '18 at 16:07
• Daniel, you can edit your post. Also, have a look at this introduction to posting mathematical notation. – hardmath Dec 15 '18 at 16:12
• Mindlack thanks, but I don't really understand how to go from here. How do I represent a and b even in a specific case? $7=2*3+1$ where $a=2$ and $b=1$? $8=4*2$ where $a=4$ and $b=0$? (And than b isn't co-prime). $10=4*2+2$ where $a=4$ and $b=2$ (a,b aren't co-prime). Also there would always be the m, which i'm not sure if the m could be greater than 1. – Daniel Gimpelman Dec 15 '18 at 16:25
Well if $$n$$ is odd you can always do $$n-2$$ and $$2$$. Or you can do $$\frac {n-1}2$$ and $$\frac {n+1}2$$.
If $$n = 2k$$ and $$k$$ is even you can do $$k-1$$ and $$k+1$$. As $$k\pm 1$$ is odd and $$\gcd(k-1, k+1) = \gcd(k-1, k+1 -(k-1)) = \gcd(k-1,2)=1$$.
If $$n = 2k$$ and $$k$$ is odd you can do $$k-2$$ and $$k+2$$ and as $$k\pm 2$$ is odd you have $$\gcd(k-2,k+2)=\gcd(k-1, 4) = 1$$.
Just to provide an answer synthesized out of the comments already posted, your best (read as easiest) approach to this kind of problem is to toy around with general patterns until something either clicks and you can write a clever proof or until you accidentally exhaust all possible cases.
In this particular problem, we can break down cases into the residue classes $$\bmod 4$$ in order to hunt for patterns:
1) If $$n=2k+1$$ then the decomposition $$n=(k)+(k+1)$$ satisfies our criterion since consecutive numbers are always coprime and $$k\geq 3$$.
2) If $$n=4k$$ then consider the decomposition $$n=(2k-1)+(2k+1)$$. Are these numbers coprime? We can no longer rely upon the general fact that consecutive numbers are coprime, since these are not consecutive. However, if two numbers differ by exactly $$2$$, what is the only prime factor that they can share? In general, if two numbers differ by $$m$$, what prime factors can they share? Finally, are we sure that these numbers are both greater than $$1$$?
I have basically given away the entire answer, but I didn't know how to discuss this phenomenon in any other way, so I leave the final details of the second case, and the entirety of the third case, to you.
• +1 A nice combination of hint and solution. – Ethan Bolker Dec 15 '18 at 16:25
Later: the numbers between $$1$$ and $$n-1$$ that are relatively prime to $$n$$ itself come in pairs that add up to $$n$$ and are relatively prime to each other as well. If $$n=5$$ or $$n \geq 7$$ both such numbers can be chosen strictly larger than $$1.$$
Original:
A different emphasis: if Euler's totient $$\phi(n) \geq 3,$$ then there is some integer $$a$$ with $$\gcd(a,n) = 1$$ and $$1 < a < n-1.$$ If we then name $$b = n-a,$$ we find that $$\gcd(a,b) = 1$$ as well, since a prime $$p$$ that divides both $$a,n-a$$ also divides $$n,$$ and this contradicts $$\gcd(a,n) = 1.$$
So, when is $$\phi(n) \geq 3 \; ? \; \;$$ If $$n$$ is divisible by any prime $$q \geq 5,$$ then $$\phi(n)$$ is a multiple of $$\phi(q) = q-1,$$ and that is at least $$4.$$
Next, if $$n = 2^c \; 3^d \; . \;$$ When $$d=0$$ we find $$\phi(n) = 2^{c-1}$$ is at least $$3$$ when $$c \geq 3,$$ leaving $$2,4$$ out. When $$c=0$$ we find $$\phi(n) = 2 \cdot 3^{d-1}$$ is at least $$3$$ when $$d \geq 2,$$ leaving $$3$$ out. When $$c,d \geq 1,$$ we find $$\phi(n) = 2^c \cdot 3^{d-1}$$ is at least $$3$$ when either $$c \geq 2$$ or $$d \geq 2,$$ so this leaves out $$6.$$
Put it together, for $$n=5$$ or $$n \geq 7,$$ there is some $$a$$ with $$1 < a < n-1$$ and $$\gcd(a,n) = 1.$$
Here's another route you can take to solve this problem. For any $$n \ge 7$$, you want to show that there is a number $$a$$ where
1. $$gcd(a, n - a) = 1$$,
2. $$1 < a < n$$, and
3. $$1 < n - a < n$$.
One option would be to choose $$a$$ to be the smallest prime number that doesn't divide $$n$$. In that case, $$gcd(a, n - a) = 1$$ because otherwise you'd have $$gcd(a, n - a) = a$$, meaning that $$a$$ divides $$a + (n - a) = n$$, contradicting the fact that $$a$$ doesn't divide $$n$$.
What you'll need to then show is that if you pick $$n \ge 7$$ that the smallest prime number that doesn't divide $$n$$ happens to be less than $$n - 1$$. I'll leave that as an exercise to the reader. :-) | 2019-08-21T10:46:20 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3041656/sum-of-two-co-prime-integers",
"openwebmath_score": 0.8372331857681274,
"openwebmath_perplexity": 152.97029237026143,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9802808684361289,
"lm_q2_score": 0.8577680995361899,
"lm_q1q2_score": 0.8408536575301442
} |
https://mathematica.stackexchange.com/questions/184366/how-can-i-know-the-length-of-sublists-in-an-expression-without-evaluating-the-ex | How can I know the length of sublists in an expression without evaluating the expression?
I have a list called eaData which contains sublists of the form {a, b, number}, where each sublist gives the number of steps in the Euclidean algorithm (EA) for numbers $$a$$ and $$b$$.
I have the following code:
eaSteps[{a_, b_}] :=
NestWhileList[{#[[2]], Mod[#[[1]], #[[2]]]}&, {a, b}, #[[2]] != 0 &]
So for example
eaSteps[{1736, 1333}]
{{1736,1333},{1333,403},{403,124},{124,31},{31,0}}
I also have
eaData =
SortBy[
Flatten[
Table[
{a, b, Length[eaSteps[{a, b}]] - 1},
{a, 1, 100},
{b, 1, a}],
1],
Part[#,3]&]
My question is: Given that my eaData list contains 5050 sublists, how could I know there are 5050 subentries before I even run the code?
I know that for the EA, the maximum number of steps is at most 5 times the minimum number of digits of either a or b, but how does that help me here?
Further, if b <= a <= 1000, then how can I find the a and b values that require the most steps in the EA?
I'm thinking that I know a has at most 4 digits, so there are at most 20, steps. But now I'm stuck. Any help?
• shouldn't you change {a, 1, 25} to {a, 1, 100} to get 5050 sublists? – kglr Oct 22 '18 at 1:02
• yes sorry, I meant {a,1,100}, i just mean, how do I know without running the code itself that I'd get 5050. what is the logic behind it? – user130306 Oct 22 '18 at 1:05
Update 2: The functions maxEALengthPair and maxEALength can be replaced with alternatives without loops using InverseFunction[Fibonacci]:
ClearAll[fibonacciFloor, maxEALengthPair2, maxEALength2]
fibonacciFloor[n_] := Floor[InverseFunction[Fibonacci][n]]
maxEALength2[n_] := fibonacciFloor[n] - 2
maxEALengthPair2[n_] := Fibonacci[fibonacciFloor[n] + {0, -1}]
And @@ (maxEALengthPair2 @ # == maxEALengthPair @ # & /@
{50, 100, 500, 1000, 10000, 100000, 1000000, 10^7, 10^9})
True
And @@ (maxEALength2@# == maxEALength@# & /@
{50, 100, 500, 1000, 10000, 100000, 1000000, 10^7, 10^9})
True
Conjecture: For any n such that Fibonacci[k] <= n <= Fibonacci[k+1], the length of longest list eaSteps[{a, b}] (b<a<=n) is k-2 and is reached by the pair {a, b} = Fibonacci[{k, k-1}].
Update: It is also possible to find a pair of numbers {a, b} less than n with maximum length for the sequence eaSteps[{a,b}] without "running the code". It seems that Fibonacci sequence can be used to identify a pair with maximum eaSteps length (I have no idea why though):
First, consecutive triples of the Fibonacci sequence satisfy the relation implicit in eaSteps:
And @@ (# == Mod[#3, #2] & @@@ Partition[Fibonacci[Range[2, 10^5]], 3, 1])
True
Also, if we start with a consecutive Fibonacci numbers {a, b}, eaSteps traces all the Fibonacci numbers smaller than b:
Reverse@eaSteps[Fibonacci[{101, 100}]][[All, 1]] == Fibonacci[Range[2, 101]]
True
ClearAll[maxEALengthPair, maxEALength]
maxEALengthPair[n_] := Module[{i = 0},
While[Fibonacci[++i] < n]; Fibonacci[{i - 1, i - 2}]];
maxEALength[n_] := Module[{i = 0}, While[Fibonacci[++i] < n]; i - 3]
TeXForm @ Grid[{#, ## & @@ #2} & @@@ Transpose[{{"n", "pair"},
{#, maxEALengthPair /@ #} &@{50, 100, 500, 1000, 10000, 100000, 1000000, 10^7}}],
Dividers -> All]
$$\tiny\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{n} & 50 & 100 & 500 & 1000 & 10000 & 100000 & 1000000 & 10000000 \\ \hline \text{pair} & \{34,21\} & \{89,55\} & \{377,233\} & \{987,610\} & \{6765,4181\} & \{75025,46368\} & \{832040,514229\} & \{9227465,5702887\} \\ \hline \end{array}$$
Brute-force approach gives the same results for n ∈ {50, 100, 500, 1000}:
nl = {50, 100, 500, 1000} ;
{#, MaximalBy[Reverse /@ Subsets[Range[#], {2}], Length @* eaSteps][[1]] & /@ #}&@nl
Grid[{#, ## & @@ #2} & @@@ Transpose[{{"n", "pair"}, {#,
MaximalBy[Reverse /@ Subsets[Range[#], {2}], Length[eaSteps[#]] &][[1]] & /@ #} &@nl}],
Dividers -> All]
$$\tiny\begin{array}{|c|c|c|c|c|} \hline \text{n} & 50 & 100 & 500 & 1000 \\ \hline \text{pair} & \{34,21\} & \{89,55\} & \{377,233\} & \{987,610\} \\ \hline \end{array}$$
Note: For some n, there are multiple pairs with maximum eaSteps length. For example, for n=50 there are 7 pairs that give a length 7 list:
MaximalBy[Reverse /@ Subsets[Range[50], {2}], Length @* eaSteps ]
{{34, 21}, {47, 29}, {50, 29}, {49, 30}, {49, 31}, {50, 31}, {47, 34}}
Length[eaSteps @ #] - 1 &/@ %
{7, 7, 7, 7, 7, 7, 7}
The table above gives only the first of multiple pairs.
We next tabulate the pairs and lengths for some Fibonacci numbers:
Grid[{#, ## & @@ #2} & @@@ Transpose[{{"k", "F[k]", "pair", "length"},
{#, Fibonacci@#, Column[maxEALengthPair@Fibonacci@#] & /@ #,
maxEALength[Fibonacci@#] & /@ #} & @ Range[10, 60, 5]}],
Dividers -> All] // TeXForm
$$\tiny\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{k} & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50 & 55 & 60 \\ \hline \text{F[k]} & 55 & 610 & 6765 & 75025 & 832040 & 9227465 & 102334155 & 1134903170 & 12586269025 & 139583862445 & 1548008755920 \\ \hline \text{pair} & \begin{array}{l} 34 \\ 21 \\ \end{array} & \begin{array}{l} 377 \\ 233 \\ \end{array} & \begin{array}{l} 4181 \\ 2584 \\ \end{array} & \begin{array}{l} 46368 \\ 28657 \\ \end{array} & \begin{array}{l} 514229 \\ 317811 \\ \end{array} & \begin{array}{l} 5702887 \\ 3524578 \\ \end{array} & \begin{array}{l} 63245986 \\ 39088169 \\ \end{array} & \begin{array}{l} 701408733 \\ 433494437 \\ \end{array} & \begin{array}{l} 7778742049 \\ 4807526976 \\ \end{array} & \begin{array}{l} 86267571272 \\ 53316291173 \\ \end{array} & \begin{array}{l} 956722026041 \\ 591286729879 \\ \end{array} \\ \hline \text{length} & 7 & 12 & 17 & 22 & 27 & 32 & 37 & 42 & 47 & 52 & 57 \\ \hline \end{array}$$
Grid[{#, ## & @@ #2} & @@@ Transpose[{{"k", "F[k]", "pair", "length"},
{#, Fibonacci@#, Column[maxEALengthPair@Fibonacci@#] & /@ #,
maxEALength[Fibonacci@#] & /@ #} &@Range[4, 25]}],
Dividers -> All] // TeXForm
$$\tiny\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{k} & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\ \hline \text{F[k]} & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377 & 610 & 987 & 1597 & 2584 & 4181 & 6765 \\ \hline \text{pair} & \begin{array}{l} 2 \\ 1 \\ \end{array} & \begin{array}{l} 3 \\ 2 \\ \end{array} & \begin{array}{l} 5 \\ 3 \\ \end{array} & \begin{array}{l} 8 \\ 5 \\ \end{array} & \begin{array}{l} 13 \\ 8 \\ \end{array} & \begin{array}{l} 21 \\ 13 \\ \end{array} & \begin{array}{l} 34 \\ 21 \\ \end{array} & \begin{array}{l} 55 \\ 34 \\ \end{array} & \begin{array}{l} 89 \\ 55 \\ \end{array} & \begin{array}{l} 144 \\ 89 \\ \end{array} & \begin{array}{l} 233 \\ 144 \\ \end{array} & \begin{array}{l} 377 \\ 233 \\ \end{array} & \begin{array}{l} 610 \\ 377 \\ \end{array} & \begin{array}{l} 987 \\ 610 \\ \end{array} & \begin{array}{l} 1597 \\ 987 \\ \end{array} & \begin{array}{l} 2584 \\ 1597 \\ \end{array} & \begin{array}{l} 4181 \\ 2584 \\ \end{array} \\ \hline \text{length} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\ \hline \end{array}$$
You can get the length of Flatten[Table[{a, b, foo}, {a, 1, n}, {b, 1, a}], 1] using
n (n + 1)/2
(which is Sum[1, {i, 1, n}, {j, 1, i}])
or
Length[Subsets[Range[n], {1, 2}]]
For n = 100:
100 101 /2
5050
Length[Subsets[Range[100], {1, 2}]]
5050
table[n_Integer] := Flatten[Table[{a, b, foo}, {a, 1, n}, {b, 1, a}], 1];
Length[table[100]]
5050
For the second part of the question, using a brute force approach
MaximalBy[Reverse/@Subsets[Range[1000],{2}], Length[eaSteps[#]]&]
{{987, 610}}
Length[Length[eaSteps[{987, 610}]]] -1
14
• thank you! so basically the reason I would know its 5050 before I even run the code is if I use Subsets` right? – user130306 Oct 22 '18 at 1:10
• @user130306, that's correct. – kglr Oct 22 '18 at 1:22
• great, thank you again. and for the second part of the question, do the values a = 987, b = 610, n = 14 work and is that the right answer? – user130306 Oct 22 '18 at 1:23 | 2020-02-18T17:58:31 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/184366/how-can-i-know-the-length-of-sublists-in-an-expression-without-evaluating-the-ex",
"openwebmath_score": 0.18790565431118011,
"openwebmath_perplexity": 4090.3063781860264,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9546474181553803,
"lm_q2_score": 0.8807970811069351,
"lm_q1q2_score": 0.8408506593975308
} |
http://accessdtv.com/error-bound/taylor-approximation-error-bound.html | Home > Error Bound > Taylor Approximation Error Bound
# Taylor Approximation Error Bound
## Contents
Thus, we have But, it's an off-the-wall fact that Thus, we have shown that for all real numbers . solution Practice B01 Solution video by PatrickJMT Close Practice B01 like? 5 Practice B02 For $$\displaystyle{f(x)=x^{2/3}}$$ and a=1; a) Find the third degree Taylor polynomial.; b) Use Taylors Inequality to estimate The system returned: (22) Invalid argument The remote host or network may be down. Your cache administrator is webmaster. http://accessdtv.com/error-bound/taylor-series-approximation-error-bound.html
Level A - Basic Practice A01 Find the fourth order Taylor polynomial of $$f(x)=e^x$$ at x=1 and write an expression for the remainder. However, for these problems, use the techniques above for choosing z, unless otherwise instructed. Linear Motion Mean Value Theorem Graphing 1st Deriv, Critical Points 2nd Deriv, Inflection Points Related Rates Basics Related Rates Areas Related Rates Distances Related Rates Volumes Optimization Integrals Definite Integrals Integration Sometimes, we need to find the critical points and find the one that is a maximum.
## Lagrange Error Bound Formula
What is the maximum possible error of the th Taylor polynomial of centered at on the interval ? Lagrange Error Bound for We know that the th Taylor polynomial is , and we have spent a lot of time in this chapter calculating Taylor polynomials and Taylor Series. In this video, we prove the Lagrange error bound for Taylor polynomials.. So it might look something like this.
And it's going to fit the curve better the more of these terms that we actually have. Sometimes you'll see this as an error function. And that's what starts to make it a good approximation. Lagrange Error Bound Khan Academy Dr Chris Tisdell - What is a Taylor polynomial?
Let's embark on a journey to find a bound for the error of a Taylor polynomial approximation. Lagrange Error Bound Calculator I could write a N here, I could write an a here to show it's an Nth degree centered at a. So this is the x-axis, this is the y-axis. Here's the formula for the remainder term: It's important to be clear that this equation is true for one specific value of c on the interval between a and x.
If is the th Taylor polynomial for centered at , then the error is bounded by where is some value satisfying on the interval between and . Error Bound Formula Statistics The following theorem tells us how to bound this error. Note that the inequality comes from the fact that f^(6)(x) is increasing, and 0 <= z <= x <= 1/2 for all x in [0,1/2]. Clicking on them and making purchases help you support 17Calculus at no extra charge to you.
## Lagrange Error Bound Calculator
The first derivative is 2x, the second derivative is 2, the third derivative is zero. It is going to be equal to zero. Lagrange Error Bound Formula Hence, we know that the 3rd Taylor polynomial for is at least within of the actual value of on the interval . What Is Error Bound solution Practice A02 Solution video by PatrickJMT Close Practice A02 like? 10 Level B - Intermediate Practice B01 Show that $$\displaystyle{\cos(x)=\sum_{n=0}^{\infty}{(-1)^n\frac{x^{2n}}{(2n)!}}}$$ holds for all x.
ButHOWclose? http://accessdtv.com/error-bound/taylor-series-approximation-maximum-error.html What you did was you created a linear function (a line) approximating a function by taking two things into consideration: The value of the function at a point, and the value And we've seen that before. Trig Formulas Describing Plane Regions Parametric Curves Linear Algebra Review Word Problems Mathematical Logic Calculus Notation Simplifying Practice Exams 17calculus on YouTube More Math Help Tutoring Tools and Resources Academic Integrity Lagrange Error Bound Problems
But you'll see this often, this is E for error. Here is a list of the three examples used here, if you wish to jump straight into one of them. P of a is equal to f of a. have a peek here This information is provided by the Taylor remainder term: f(x) = Tn(x) + Rn(x) Notice that the addition of the remainder term Rn(x) turns the approximation into an equation.
In general, if you take an N plus oneth derivative of an Nth degree polynomial, and you could prove it for yourself, you could even prove it generally but I think Lagrange Error Ap Calculus Bc The system returned: (22) Invalid argument The remote host or network may be down. If you take the first derivative of this whole mess-- And this is actually why Taylor polynomials are so useful, is that up to and including the degree of the polynomial
## Can we bound this and if we are able to bound this, if we're able to figure out an upper bound on its magnitude-- So actually, what we want to do
Skip to main contentSubjectsMath by subjectEarly mathArithmeticAlgebraGeometryTrigonometryStatistics & probabilityCalculusDifferential equationsLinear algebraMath for fun and gloryMath by gradeK–2nd3rd4th5th6th7th8thHigh schoolScience & engineeringPhysicsChemistryOrganic ChemistryBiologyHealth & medicineElectrical engineeringCosmology & astronomyComputingComputer programmingComputer scienceHour of CodeComputer animationArts The square root of e sin(0.1) The integral, from 0 to 1/2, of exp(x^2) dx We cannot find the value of exp(x) directly, except for a very few values of x. To handle this error we write the function like this. $$\displaystyle{ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + . . . + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) }$$ where $$R_n(x)$$ is the Lagrange Error Bound Proof In short, use this site wisely by questioning and verifying everything.
What are they talking about if they're saying the error of this Nth degree polynomial centered at a when we are at x is equal to b. So how do we do that? Generated Sun, 30 Oct 2016 16:07:21 GMT by s_sg2 (squid/3.5.20) http://accessdtv.com/error-bound/taylor-error-approximation.html Essentially, the difference between the Taylor polynomial and the original function is at most .
We already know that P prime of a is equal to f prime of a. So, we have . Here's the formula for the remainder term: So substituting 1 for x gives you: At this point, you're apparently stuck, because you don't know the value of sin c. It is going to be f of a, plus f prime of a, times x minus a, plus f prime prime of a, times x minus a squared over-- Either you
So the error at a is equal to f of a minus P of a. | 2018-04-23T23:10:47 | {
"domain": "accessdtv.com",
"url": "http://accessdtv.com/error-bound/taylor-approximation-error-bound.html",
"openwebmath_score": 0.879546582698822,
"openwebmath_perplexity": 691.4229619030937,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9546474168650673,
"lm_q2_score": 0.8807970779778825,
"lm_q1q2_score": 0.8408506552738848
} |
https://math.stackexchange.com/questions/2952044/finding-the-range-of-launch-speeds-for-which-water-shot-from-a-hose-will-enter-a | # Finding the range of launch speeds for which water shot from a hose will enter a tank a distance away
A water hose is used to fill a large cylindrical storage tank of diameter $$D$$ and height $$2D$$. The hose shoots the water at $$45^\circ$$ above the horizontal from the same level as the base of the tank and is a distance $$6D$$ away. For what range of launch speeds $$v_0$$ will the water enter the tank. Ignore air resistance.
I've made a graphic representing the problem (with $$D=1$$) showing the trajectory of the water at the minimum launch speed $$v_{0\text-m}$$ (blue) and maximum launch speed $$v_{0\text-M}$$ (red) as well as the tank (light gray):
Clearly I've made a mistake in solving for the minimum launch speed - I expect the blue parabola to intersect the rectangle at its top left vertex. I'm not sure where the mistake lies, because as far as I can tell I've used the same approach to solve for $$v_{0\text-m}$$ as I had for $$v_{0\text-M}$$, which I'll show first.
The position vector for water particles escaping the hose has components
$$\begin{cases} x=v_{0\text-M}\cos45^\circ t\\ y=v_{0\text-M}\sin45^\circ t-\frac g2t^2 \end{cases}$$
With $$x=7D$$, solve for $$t$$ in the first equation, substitute into the second with $$y=2D$$ to get
$$t=\frac{7D}{v_{0\text-M}\cos45^\circ}\implies2D=v_{0\text-M}\sin45^\circ\left(\frac{7D}{v_{0\text-M}\cos45^\circ}\right)-\frac g2\left(\frac{7D}{v_{0\text-M}\cos45^\circ}\right)^2$$
$$\implies5=\frac{49Dg}{v_{0\text-M}^2}\implies v_{0\text-M}=7\sqrt{\frac{Dg}5}$$
Now to find $$v_{0\text-m}$$, the only change in the work above - I would think - would be to set $$x=6D$$.
$$t=\frac{6D}{v_{0\text-m}\cos45^\circ}\implies2D=v_{0\text-m}\sin45^\circ\left(\frac{6D}{v_{0\text-m}\cos45^\circ}\right)-\frac g2\left(\frac{6D}{v_{0\text-m}\cos45^\circ}\right)^2$$
$$\implies2=\frac{12Dg}{v_{0\text-m}^2}\implies v_{0\text-m}=\sqrt{6Dg}$$
Manipulating the code for the plot of the blue parabola, it would appear the correct answer for $$v_{0\text-m}$$ is closer to $$3\sqrt{Dg}$$. How can I salvage my attempt to obtain the right solution?
• – amd Oct 12 '18 at 1:22
• The blue trajectory is incorrect. It should reach the top left of the water tank! – hypergeometric Jan 21 '19 at 16:22
• @hypergeometric Yes, as I mentioned, "I expect the blue parabola to intersect the rectangle at its top left vertex". I found the source of the discrepancy already; see my answer below. – user170231 Jan 21 '19 at 17:23
Minor algebraic mistake...
$$2D=v_{0\text-m}\sin45^\circ\left(\frac{6D}{v_{0\text-m}\cos45^\circ}\right)-\frac g2\left(\frac{6D}{v_{0\text-m}\cos45^\circ}\right)^2$$
should reduce to
$$2D=6D-\frac g2\frac{36D^2}{v_{0\text-m}^2\cos^245^\circ}\implies4D=\frac{36D^2g}{v_{0\text-m}^2}\implies v_{0\text-m}=3\sqrt{Dg}$$
At some point I decided to divide through all terms by $$2D$$ except for one, which I had mistakenly divided by $$3D$$ instead.
The equation of trajectory of water jet is $$y = x-\dfrac{g}{v^2} x^2$$. Let $$x$$ and $$y$$ be coordinates of any point in water jet at a height of $$2D$$. So water to fall we must have $$6D < x < 7D$$. By putting $$y = 2D$$ in above equation we get quadratic in $$x$$ as $$\dfrac{g}{v^2} x^2 - x +2D = O$$. Discard negative value of $$x$$ take only positive value then put value of $$x$$ in above inequality and solve you will get
$$\sqrt{9gD} < v < \sqrt{9.8gD}$$ | 2020-07-10T13:10:00 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2952044/finding-the-range-of-launch-speeds-for-which-water-shot-from-a-hose-will-enter-a",
"openwebmath_score": 0.7169409394264221,
"openwebmath_perplexity": 302.61782883608817,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9697854103128329,
"lm_q2_score": 0.8670357735451835,
"lm_q1q2_score": 0.8408386434034203
} |
https://stats.stackexchange.com/questions/258704/does-covariance-equal-to-zero-implies-independence-for-binary-random-variables | # Does covariance equal to zero implies independence for binary random variables?
If $X$ and $Y$ are two random variables that can only take two possible states, how can I show that $Cov(X,Y) = 0$ implies independence? This kind of goes against what I learned back in the day that $Cov(X,Y) = 0$ does not imply independence...
The hint says to start with $1$ and $0$ as the possible states and generalize from there. And I can do that and show $E(XY) = E(X)E(Y)$, but this doesn't imply independence???
Kind of confused how to do this mathematically I guess.
• It is not true in general as your question's heading suggests.. – Michael R. Chernick Jan 28 '17 at 23:50
• The statement that you are trying to prove is indeed true. If $X$ and $Y$ are Bernoulli random variables wot parameters $p_1$ and $p_2$ respectively, then $E[X]=p_1$ and $E[Y]=p_2$. So, $\operatorname{cov}(X,Y)=E[XY]-E[X]E[Y]$ equals $0$ only if $E[XY]=P\{X=1,Y=1\}$ equals $p_1p_2=P\{X=1\}P\{Y=1\}$ showing that $\{X=1\}$ and $\{Y=1\}$ are independent events. It is a standard result that if $A$ and $B$ are a pair of independent events, then so are $A,B^c$, and $A^c,B$, and $A^c,B^c$ independent events, i.e. $X$ and $Y$ are independent random variables. Now generalize. – Dilip Sarwate Jan 29 '17 at 0:06
For binary variables their expected value equals the probability that they are equal to one. Therefore,
$$E(XY) = P(XY = 1) = P(X=1 \cap Y=1) \\ E(X) = P(X=1) \\ E(Y) = P(Y=1) \\$$
If the two have zero covariance this means $E(XY) = E(X)E(Y)$, which means
$$P(X=1 \cap Y=1) = P(X=1) \cdot P(Y=1)$$
It is trivial to see all other joint probabilities multiply as well, using the basic rules about independent events (i.e. if $A$ and $B$ are independent then their complements are independent, etc.), which means the joint mass function factorizes, which is the definition of two random variables being independent.
• Concise and elegant. Classy! +1 =D – Marcelo Ventura Feb 3 '17 at 22:14
Both correlation and covariance measure linear association between two given variables and it has no obligation to detect any other form of association else.
So those two variables might be associated in several other non-linear ways and covariance (and, therefore, correlation) could not distinguish from independent case.
As a very didactic, artificial and non realistic example, one can consider $X$ such that $P(X=x)=1/3$ for $x=−1,0,1$ and also consider $Y=X^2$. Notice that they are not only associated, but one is a function of the other. Nonetheless, their covariance is 0, for their association is orthogonal to the association that covariance can detect.
EDIT
Indeed, as indicated by @whuber, the above original answer was actually a comment on how the assertion is not universally true if both variables were not necessarily dichotomous. My bad!
So let's math up. (The local equivalent of Barney Stinson's "Suit up!")
### Particular Case
If both $X$ and $Y$ were dichotomous, then you can assume, without loss of generality, that both assume only the values $0$ and $1$ with arbitrary probabilities $p$, $q$ and $r$ given by \begin{align*} P(X=1) = p \in [0,1] \\ P(Y=1) = q \in [0,1] \\ P(X=1,Y=1) = r \in [0,1], \end{align*} which characterize completely the joint distribution of $X$ and $Y$. Taking on @DilipSarwate's hint, notice that those three values are enough to determine the joint distribution of $(X,Y)$, since \begin{align*} P(X=0,Y=1) &= P(Y=1) - P(X=1,Y=1) = q - r\\ P(X=1,Y=0) &= P(X=1) - P(X=1,Y=1) = p - r\\ P(X=0,Y=0) &= 1 - P(X=0,Y=1) - P(X=1,Y=0) - P(X=1,Y=1) \\ &= 1 - (q - r) - (p - r) - r = 1 - p - q - r. \end{align*} (On a side note, of course $r$ is bound to respect both $p-r\in[0,1]$, $q-r\in[0,1]$ and $1-p-q-r\in[0,1]$ beyond $r\in[0,1]$, which is to say $r\in[0,\min(p,q,1-p-q)]$.)
Notice that $r = P(X=1,Y=1)$ might be equal to the product $p\cdot q = P(X=1) P(Y=1)$, which would render $X$ and $Y$ independent, since \begin{align*} P(X=0,Y=0) &= 1 - p - q - pq = (1-p)(1-q) = P(X=0)P(Y=0)\\ P(X=1,Y=0) &= p - pq = p(1-q) = P(X=1)P(Y=0)\\ P(X=0,Y=1) &= q - pq = (1-p)q = P(X=0)P(Y=1). \end{align*}
Yes, $r$ might be equal to $pq$, BUT it can be different, as long as it respects the boundaries above.
Well, from the above joint distribution, we would have \begin{align*} E(X) &= 0\cdot P(X=0) + 1\cdot P(X=1) = P(X=1) = p \\ E(Y) &= 0\cdot P(Y=0) + 1\cdot P(Y=1) = P(Y=1) = q \\ E(XY) &= 0\cdot P(XY=0) + 1\cdot P(XY=1) \\ &= P(XY=1) = P(X=1,Y=1) = r\\ Cov(X,Y) &= E(XY) - E(X)E(Y) = r - pq \end{align*}
Now, notice then that $X$ and $Y$ are independent if and only if $Cov(X,Y)=0$. Indeed, if $X$ and $Y$ are independent, then $P(X=1,Y=1)=P(X=1)P(Y=1)$, which is to say $r=pq$. Therefore, $Cov(X,Y)=r-pq=0$; and, on the other hand, if $Cov(X,Y)=0$, then $r-pq=0$, which is to say $r=pq$. Therefore, $X$ and $Y$ are independent.
### General Case
About the without loss of generality clause above, if $X$ and $Y$ were distributed otherwise, let's say, for $a<b$ and $c<d$, \begin{align*} P(X=b)=p \\ P(Y=d)=q \\ P(X=b, Y=d)=r \end{align*} then $X'$ and $Y'$ given by $$X'=\frac{X-a}{b-a} \qquad \text{and} \qquad Y'=\frac{Y-c}{d-c}$$ would be distributed just as characterized above, since $$X=a \Leftrightarrow X'=0, \quad X=b \Leftrightarrow X'=1, \quad Y=c \Leftrightarrow Y'=0 \quad \text{and} \quad Y=d \Leftrightarrow Y'=1.$$ So $X$ and $Y$ are independent if and only if $X'$ and $Y'$ are independent.
Also, we would have \begin{align*} E(X') &= E\left(\frac{X-a}{b-a}\right) = \frac{E(X)-a}{b-a} \\ E(Y') &= E\left(\frac{Y-c}{d-c}\right) = \frac{E(Y)-c}{d-c} \\ E(X'Y') &= E\left(\frac{X-a}{b-a} \frac{Y-c}{d-c}\right) = \frac{E[(X-a)(Y-c)]}{(b-a)(d-c)} \\ &= \frac{E(XY-Xc-aY+ac)}{(b-a)(d-c)} = \frac{E(XY)-cE(X)-aE(Y)+ac}{(b-a)(d-c)} \\ Cov(X',Y') &= E(X'Y')-E(X')E(Y') \\ &= \frac{E(XY)-cE(X)-aE(Y)+ac}{(b-a)(d-c)} - \frac{E(X)-a}{b-a} \frac{E(Y)-c}{d-c} \\ &= \frac{[E(XY)-cE(X)-aE(Y)+ac] - [E(X)-a] [E(Y)-c]}{(b-a)(d-c)}\\ &= \frac{[E(XY)-cE(X)-aE(Y)+ac] - [E(X)E(Y)-cE(X)-aE(Y)+ac]}{(b-a)(d-c)}\\ &= \frac{E(XY)-E(X)E(Y)}{(b-a)(d-c)} = \frac{1}{(b-a)(d-c)} Cov(X,Y). \end{align*} So $Cov(X,Y)=0$ if and only $Cov(X',Y')=0$.
=D
• I recycled that answer from this post. – Marcelo Ventura Jan 29 '17 at 5:51
• Verbatim cut and paste from your other post. Love it. +1 – gammer Jan 29 '17 at 6:04
• How is thus an answer to the question asked? – Dilip Sarwate Feb 2 '17 at 5:18
• Your edits still don't answer the question, at least not at the level the question is asked. You write "Notice that $r~\ldots$ not necessarily equal to the product $pq$. That exceptional situation corresponds to the case of independence between $X$ and $Y$." which is a perfectly true statement but only for the cognoscenti because for the hoi polloi, independence requires not just that $$P(X=1,Y=1)=P(X=1)P(Y=1)\tag 1$$ but also $$P(X=u,Y=v)=P(X=u)P(Y=v),~u.v\in\{0,1\}.\tag 2$$ Yes, $(1) \implies(2)$ as the cognoscenti know; for lesser mortals, a proof that $(1) \implies (2)$ is helpful. – Dilip Sarwate Feb 4 '17 at 1:51
IN GENERAL:
The criterion for independence is $F(x,y) = F_X(x)F_Y(y)$. Or $$f_{X,Y}(x,y)=f_X(x)\,f_Y(y)\tag 1$$
"If two variables are independent, their covariance is $0.$ But, having a covariance of $0$ does not imply the variables are independent."
This is nicely explained by Macro here, and in the Wikipedia entry for independence.
$\text {independence} \Rightarrow \text{zero cov}$, yet
$\text{zero cov}\nRightarrow \text{independence}.$
Great example: $X \sim N(0,1)$, and $Y= X^2.$ Covariance is zero (and $\mathbb E(XY)=0$, which is the criterion for orthogonality), yet they are dependent. Credit goes to this post.
IN PARTICULAR (OP problem):
These are Bernoulli rv's, $X$ and $Y$ with probability of success $\Pr(X=1)$, and $\Pr(Y=1)$.
\begin{align}\mathrm{cov}(X,Y)&=\mathrm E[XY] - \mathrm E[X]\,\mathrm E[Y]\\[2ex] &\underset{*}{=} \Pr(X=1 \cap Y=1) - \Pr(X=1)\, \Pr(Y=1)\\[2ex] &\implies \Pr(X=1 , Y=1) = \Pr (X=1)\,\Pr(Y=1). \end{align}
This is equivalent to the condition for independence in Eq. $(1).$
$(*)$:
$$\mathrm E[XY]\quad \underset{**}{=} \quad \displaystyle \sum_{\text{domain X, Y}} \Pr(X=x\cap Y=y)\, x\,y \underset{\neq\,0\text{ iff } x \times y\neq 0}= \Pr(X=1 \cap Y=1).$$
$(**)$: by LOTUS.
As pointed out below, the argument is incomplete without what Dilip Sarwate had pointed out in his comments shortly after the OP appeared. After searching around, I found this proof of the missing part here:
If events $A$ and $B$ are independent, then events $A^c$ and $B$ are independent, and events $A^c$ and $B^c$ are also independent.
Proof By definition,
$A$ and $B$ are independent $\iff P(A\cap B) = P(A)P(B).$
But $B=(A\cap B) + ( A^c \cup B)$, so $P(B)= P(A\cap B) + P(A^c \cup B)$, which yields:
$\small P(A^c \cap B) = P(B) - P(A\cap B) = P(B) - P(A)\,P(B) = P(B) \left[1 - P(A)\right] = P(B)\,P( A^c).$
Repeat the argument for the events $A^c$ and $B^c,$ this time starting from the statement that $A^c$ and $B$ are independent and taking the complement of $B.$
Similarly. $A$ and $B^c$ are independent events.
So, we have shown already that $$\Pr(X=1 , Y=1) = \Pr (X=1)\,\Pr(Y=1)$$ and the above shows that this implies that $$\Pr(X=i , Y=j) = \Pr (X=i)\,\Pr(Y=j), ~~i, j \in \{0,1\}$$ that is, the joint pmf factors into the product of marginal pmfs everywhere, not just at $(1,1)$. Hence, uncorrelated Bernoulli random variables $X$ and $Y$ are also independent random variables.
• Actually that's not an equivalent condition to Eq (1). All you showed was that $f_{X,Y}(1,1) = f_{X}(1) f_{Y}(1)$ – gammer Jan 29 '17 at 6:08
• Please consider replacing that image with your own equations, preferably ones that don't use overbars to denote complements. The overbars in the image are very hard to see. – Dilip Sarwate Feb 3 '17 at 4:05
• @DilipSarwate No problem. Is it better, now? – Antoni Parellada Feb 3 '17 at 4:43
• Thanks. Also, note that strictly speaking, you also need to show that $A$ and $B^c$ are independent events since the factorization of the joint pdf into the product of the marginal pmts must hold at all four points. Perhaps adding the sentence "Similarly. $A$ and $B^c$ are independent events" right after the proof that $A^c$ and $B$ are independent events will work. – Dilip Sarwate Feb 3 '17 at 5:03
• @DilipSarwate Thank you very much for your help getting it right. The proof as it was before all the editing seemed self-explanatory, because of all the inherent symmetry, but it clearly couldn't be taken for granted. I am very appreciative of your assistance. – Antoni Parellada Feb 3 '17 at 5:10 | 2021-04-15T18:11:55 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/258704/does-covariance-equal-to-zero-implies-independence-for-binary-random-variables",
"openwebmath_score": 0.9997827410697937,
"openwebmath_perplexity": 485.78910135418664,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9697854103128328,
"lm_q2_score": 0.867035771827307,
"lm_q1q2_score": 0.8408386417374486
} |
https://math.stackexchange.com/questions/3793869/usamo-1989-problem-2 | # USAMO $1989$, Problem $2$
Problem $$2$$ from USAMO, $$1989$$: The $$20$$ members of a local tennis club have scheduled exactly $$14$$ two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of six games with $$12$$ distinct players.
My attempt at a solution: Since there are $$20$$ players, each of whom has played at least one game, the least number of matches which must be organised to accommodate all players would be $$10$$. Now, $$4$$ more matches are left, which can be played by at most $$8$$ of these $$20$$ players. Thus, at least $$12$$ of the players get to play no more than $$1$$ match, and hence there must exist a set of six games with $$12$$ distinct players. This, I believe, will be the guiding principle of a rigorous proof.
I have the following doubts clouding my mind:
1. Is my reasoning sound, i.e., free of any pitfalls?
2. If correct, how to frame the above reasoning in a mathematically rigorous proof?
Edit: As pointed out by Ben in the comments, the statement in italics is unjustified. I overlooked a lot of possibilities while framing this proof, it seems. Thus, I would like to get some hints to proceed with more reasonable proof.
• I think that going from "at least 12 of the players get to play only 1 match" to "there exists a set of six games with 12 distinct players" is unjustified. – Ben Grossmann Aug 17 '20 at 11:40
• @BenGrossmann Any hints on how I could fill this gap? It seems intuitively true, but again proofs in plain English are mostly shaky. – Manan Aug 17 '20 at 11:41
I am afraid there is a pitfall. If you try to prove that there are 6 games with 12 distinct players amongst those who have played exactly 1 game you will fail. Here's an example:
$$1-2\\ 3-4\\ 5-6\\ 7-8\\ 9-10\\ 11-12\\ 12-13 \\ 13-14\\ 14-15\\ 15-16\\ 16-17\\ 17-18\\ 18-19\\ 19-20$$
While there are 6 games with 12 distinct players, only 5 games are between the players with exactly one game.
So here's a hint: split the players in two groups, those with one game and those with more. Then pair players with one game,after pair players with one game with players with more and finally players with more games.
EDIT: Second approach
If every person is seen as a vertex and every game as an edge, the question asks to prove that there is a matching of size at least $$6$$. Equivalently, we can prove that the size of the maximum matching $$MM$$ is of size at least $$6$$. Note that a maximum matching partitions the vertices into two sets, one with edges in the $$MM$$, of size $$2\cdot |MM|$$, and one without, of size $$20 - 2\cdot |MM|$$ for our problem. Since all vertices are of degree at least $$1$$, we conclude that there are at least $$20 - 2\cdot |MM|$$ edges between the two partitions. Now,
$$E_{MM} + E_{\text{between}} \le |E| = 14 \implies \\ |MM| + 20 - 2\cdot |MM| \le 14 \implies \\ |MM| \ge 6$$
• What do you mean by "pair players with one game"? It sounds like you're trying to construct an arrangement of games that fits the description, whereas the goal is to prove that every arrangement of 14 games among 20 players with each player in at least one game will have these 6 games – Ben Grossmann Aug 17 '20 at 14:07
• @BenGrossmann Let's take a random player $A$ with exactly one game. If that game was another player $B$ with exactly one game then the pair(or match if you prefer) $A - B$ will be one of the 6 we are looking for. – cgss Aug 17 '20 at 14:22
• Yes, that is correct. Perhaps some pidgeonhole arguments can improve the exhausting search but in general lines that is the idea. I guess there will be other approaches as well. I am considering a graph-theoretic approach. If you have another approach feel free to share. – cgss Aug 17 '20 at 14:53
• @Maran I think I came up with one graph-theoretic one. Should I add another answer? – cgss Aug 17 '20 at 16:03
• @Maran I edited the answer. – cgss Aug 17 '20 at 17:17 | 2021-03-05T02:00:08 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3793869/usamo-1989-problem-2",
"openwebmath_score": 0.6010059118270874,
"openwebmath_perplexity": 348.67896451540287,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9697854103128328,
"lm_q2_score": 0.867035771827307,
"lm_q1q2_score": 0.8408386417374486
} |
https://engineering.stackexchange.com/questions/68/how-does-width-and-thickness-affect-the-stiffness-of-steel-plate | # How does width and thickness affect the stiffness of steel plate?
I have a 2 mm thick steel plate which is 300 mm long and 30 mm wide, supported at either end. It supports a weight-bearing wheel that can roll along the plate. It currently supports the maximum weight that I expect it to support when the wheel is in the middle, but it flexes a little bit too much. Would making it wider help to support the weight and increase its stiffness, or do I need to make it thicker?
Also is there a way to calculate how the stiffness will change with the thickness (or width if that would affect it)?
Short answer: make it thicker.
Long answer: The moment of inertia affects the beam's ability to resist flexing.
Use one of the many, free, online moment of inertia calculators (like this one) to see how increasing the height of the beam will have an exponential effect on increasing the stiffness of the beam.
And this site helps provide a pictorial view of the load(s) upon a beam depending upon differing configurations, such as where the supports are and where the load is applied. It also provides a calculator to determine the forces involved.
Wikipedia has a decent article for area moments of inertia. In your particular case, you're asking about a filled rectangular area and Ix = bh3/12. The height has an exponential factor of 3 whereas increasing the base does not have an exponential factor. So for the same amount of material, increasing the height stiffens the beam better.
To be clear, you can make the beam sag less by increasing the width of the plate. It's just more effective to make the plate thicker.
Current moment: Ix = 30 * 23 / 12 = 20 mm4
Increase width by 1mm: Ix = 31 * 23 / 12 = 20.6 mm4
Increase height by 1mm: Ix = 30 * 33 / 12 = 67.5 mm4
And if for some reason you can't easily increase the thickness of the plate, you can consider a different beam structure. Currently, your beam is a simple rectangle. You can easily use a T-beam or an I-beam in order to stiffen the plate instead.
Again, while I've provided some suggested links to online calculators feel free to search for and use others that you may prefer.
• Wow, this is amazing thanks. Despite theory/maths being way over my head, I used this to calculate that by making it 1mm thicker, the displacement will go from about 5mm down to about 1.5mm, and increasing the thickness by 2mm takes me down to 0.6mm displacement. Spot on! :) Jan 21, 2015 at 3:46
• This answer is good and probably the best advice, but I'm going to add another answer explaining exactly how making the plate wider could also help, depending on the situation.
– Rick
Jan 21, 2015 at 16:05
• This is an excellent answer. It's written in a way that concisely answers the question asked, it has links for further reading but avoids creating an "answer in another castle" situation, it clearly explains the concepts involved and shows their application through example, and it is well written and formatted in an easy to digest manner. Answers like this are the reason the stack exchange value proposition works, and it's the reason I keep coming back. Just wanted to say thanks to GlenH7 for spending the time to make this answer great. Apr 27, 2016 at 15:09
The stiffness of a rectangular cross section, be it steel, concrete, wood, or any other material, is related almost entirely to it's modulus of elasticity, $E$, and it's moment of inertia about the axis of bending, $I$.
Since you already have your material set, steel, you cannot change $E$. What you can change is your $I$.
The moment of inertia for a rectangular cross section about its neutral axis is $\frac{b \cdot d^3}{12}$. You will increase your stiffness exponentially by increasing the depth of your plate.
It's important to note that you can increase your I in other ways as well. For example, if you welded another plate to your existing plate to make a T shape in cross section, you will significantly increase your $I$ and greatly stiffen your member.
• What this answer implies but doesn't say outright is that in many cases you will find it more cost-effective to add stiffeners than to buy a thicker plate. It can even be as simple as slapping on a piece of angle iron, for small projects, with bolts or epoxy.
– Air
Mar 7, 2016 at 19:57
As mentioned in other answers, what controls the deflection is the second moment of area, and the easiest way to increase it is increase the thickness. Doubling the thickness would increase the 16 fold the second moment of area and it would increase the weight by 100%
However below I am outlining, another way to improve the second moment of area is to change the cross-section, which is more efficient. I am presenting the base line example (30 mm X 2 mm), then I am giving two additional configurations which increase the weight by 2/3 (or 66%) and I am also presenting the increase in second moment of area which is significantly greater)
More specifically
Cross-section Description $$I_{xx}$$ Increase
Flat sheet $$20 mm^4$$ -
H- section (adding 5mm flanges at either side) $$593 mm^4$$ ~ x30
C- section (adding 10mm flanges) $$1217 mm^4$$ ~60 times
From the above three configurations you can see the vast difference the relatively small addition of flanges can make on the stiffness of the beam ( I won't go through the calculation because its in most textbooks - and the question has already an accepted answer).
As a final point, it is noteworthy that the configurations H and C have the weight, yet the C section will have half the deflection. (So its really a matter of intelligently using the cross-section). | 2023-03-23T08:26:14 | {
"domain": "stackexchange.com",
"url": "https://engineering.stackexchange.com/questions/68/how-does-width-and-thickness-affect-the-stiffness-of-steel-plate",
"openwebmath_score": 0.5062732696533203,
"openwebmath_perplexity": 645.5738250016465,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9697854103128328,
"lm_q2_score": 0.8670357580842941,
"lm_q1q2_score": 0.8408386284096753
} |
https://math.stackexchange.com/questions/2924466/find-a-b-in-way-that-fa-rightarrow-b-is-bijective-when-fx-sqrtx2-1 | # find $A,B$ in way that $f:A\rightarrow B$ is bijective, when $f(x)=\sqrt{x^2-1}$
## Problem
find $$A,B$$ in way that $$f:A\rightarrow B$$ is bijective, when $$f(x)=\sqrt{x^2-1}$$.
## Attempt to solve
map $$f:A \rightarrow B$$ is bijective when it's surjective and injective simultaneously. This map is injective when:
$$\forall(x,y) \in A : x \neq y \implies f(x) \neq f(y)$$
This map is surjective when:
$$\forall y \in B \exists x \in A : f(x)=y$$
if function $$f(x)$$ is "truly" monotonic it implies that function has to be injective.
Observation:
$$\forall x \in \mathbb{R} : \frac{d}{dx}f(x) \neq 0$$
$$\frac{d}{dx}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2+1}} = f'(x)$$
It is visible that only way $$f'(x)$$ could be zero is when $$x=0$$ but $$\sqrt{-1}$$ is not defined in $$\mathbb{R}$$
which implies that $$f(x)$$ is injective if we pick $$A$$ in a way:
$$A\in \mathbb{R}\setminus[-1,1]$$
and $$B$$:
$$B \in \mathbb{R}$$
we have map: $$f:\mathbb{R}\setminus[-1,1] \rightarrow \mathbb{R}$$
Inverse function of $$f(x)$$ can be computed by solving y from following equation. Inverse function is map $$f^{-1}:B \rightarrow A$$
$$\sqrt{x^2-1}=y$$
$$f(x)$$ is defined when : $$x^2-1 \ge 0 \implies x^2\ge 1 \implies -1 \ge x \ge 1$$
$$x^2-1=y^2$$
$$x^2=y^2+1 \implies x=\pm \sqrt{x^2+1}$$
$$h^{-1}(x)=\sqrt{x^2+1}$$
meaning $$f(x)$$ is surjective. Which implies $$f(x)$$ is bijective when:
$$A\in \mathbb{R}\setminus[-1,1], B \in \mathbb{R}$$
• Your argument with the derivative does not work. Note that $f(x)=f(-x)$ when $f$ is defined in $x$. But with your reasoning you may choose $A=[1,\infty)$ and then compute the image $f(A)$. – Severin Schraven Sep 20 '18 at 20:45
• $A=B=\varnothing$. – Asaf Karagila Sep 21 '18 at 12:41
You have made a mistake here
$$\frac{d}{dx}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2-1}} = f'(x)$$
therefore $$f(x)$$ is injective for $$x>1$$ or for $$x<-1$$.
then for the surjectivity it suffices to observe that $$\lim_{x\to \pm \infty} f(x) = \infty$$
then $$f(x)$$ is bijective for the following restrictions
• $$f_1:(-\infty,a]\to [0,\infty)$$
• $$f_2:[b,\infty)\to [0,\infty)$$
with $$a\le -1$$ and $$b\ge 1$$
We can take $$A=[1,+\infty)$$
$$(\forall x>1) \; f'(x)=\frac{x}{f(x)}$$
$f$ is continuous at $A$ and strictly increasing at $(1,\infty)$ thus it is strictly increasing at $A$.
$$f(A)=[0,+\infty)=B$$
$f$ is a bijection from $A$ in $B$.
When you only have to give an $$A$$ and $$B$$ such that $$f: A\to B$$, $$f(x)=\sqrt{x^2-1}$$ is bijective, why not take simply $$A=\{1\}$$ and $$B=\{0\}$$.
If you want to finde the maximal intervalls you might go like this:
We construct $$f^{-1}$$ and while we do so, we have to note when the operations we use are defined or equivialent transformations.
$$x=\sqrt{(y-1)(y+1)}$$ First of all $$y\geq 1$$ or $$y\leq -1$$, else the square root is not defined.
We square both sides. Thus $$x\geq 0$$:
$$x^2=y^2-1\Leftrightarrow y^2=x^2+1$$
$$y_{1,2}=\pm\sqrt{x^2+1}$$
This gives us two solutions.
If $$y\leq -1$$, that means $$y\in (-\infty, -1]$$, then $$f^{-1}(x)=-\sqrt{x^2+1}$$.
If $$y\geq 1$$, that means $$y\in [1,\infty)$$, then $$f^{-1}(x)=\sqrt{x^2+1}$$
This gives us of course the intervals which $$f$$ is a bijection on. We had $$x\geq 0$$ which means $$x\in[0,\infty)$$
And the two solutions are $$f:[1,\infty)\to\mathbb{R}_{\geq 0}$$ or $$f:(-\infty,-1]\to\mathbb{R}_{\geq 0}$$
You have two good choices for $$A$$ and for each case you have a choice for $$B$$
First choice: let $$A= (-\infty,-1]$$ while $$B= [0,\infty)$$ where your function is strictly decreasing.
Second choice: Let $$A= [1,\infty)$$ while $$B= [0,\infty)$$ where your function is strictly increasing.
• There are many many choices. – hamam_Abdallah Sep 20 '18 at 20:54
• You know what I mean by these choices. – Mohammad Riazi-Kermani Sep 21 '18 at 18:24
Since $$f(x)=f(-x)$$, for $$|x|\ge1$$, your set $$A$$ can only contain one among a number and its negative.
The most natural choice is $$A=[1,\infty)$$, but obviously not the only one; also the set $$A'$$ consisting of rationals $$\ge1$$ and the irrationals $$<-1$$ would do.
Both $$A$$ and $$A'$$ are “maximal”, in the sense that you cannot add any number in the domain of $$\sqrt{x^2-1}$$ to either $$A$$ or $$A'$$ and still have an injective function. There are infinitely many choices of such “maximal domains of injectivity”, most notably $$(-\infty,-1]$$; which one to choose is a matter of preferences and simplicity.
Let's stick with $$A=[1,\infty)$$. The equation $$\sqrt{x^2-1}=y$$ can only have a solution if $$y\ge0$$. In this case it becomes $$x^2-1=y^2$$, so $$x=\sqrt{y^2+1}$$, which exists (and is $$\ge1$$) for all $$y\ge0$$.
Hence your set $$B$$ is $$[0,\infty)$$.
• @N.F.Taussig I tried to expand it – egreg Sep 21 '18 at 12:24 | 2019-07-15T18:06:56 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2924466/find-a-b-in-way-that-fa-rightarrow-b-is-bijective-when-fx-sqrtx2-1",
"openwebmath_score": 0.9752725958824158,
"openwebmath_perplexity": 185.68920861725385,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9759464478051828,
"lm_q2_score": 0.8615382165412809,
"lm_q1q2_score": 0.8408151620818756
} |
https://www.math10.com/forum/viewtopic.php?f=25&t=1712 | # Confusion over simple surds
### Confusion over simple surds
Why does $$\sqrt{180}$$ = 6$$\sqrt{5}$$
But $$\sqrt{48}$$ = 4$$\sqrt{3}$$
They are unrelated, but the way I remember doing these is sort of confusing me. This is how I've been asked to do it:
For example:
Why does $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ x $$\sqrt{3}$$ become 2 x 3 x $$\sqrt{5}$$
But then $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ become 4$$\sqrt{3}$$ ?
Am I making sense?
Guest
### Re: Confusion over simple surds
$$\sqrt{2} \times \sqrt{2} = (\sqrt{2})^2 = 2$$
$$\sqrt{2} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2} \times \sqrt{3} = (\sqrt{2} \times \sqrt{2}) \times (\sqrt{2} \times \sqrt{2}) \times \sqrt{3} =$$
$$2 \times 2 \times \sqrt{3} = 4 \times \sqrt{3}$$ or $$4\sqrt{3}$$
Guest
### Re: Confusion over simple surds
Your math question here is absurd! Please get help at https://www.math10.com/forum/viewforum.php?f=25 or try somewhere else.
Guest
### Re: Confusion over simple surds
We are basically making use of two rules:
$$\sqrt{a}\times\sqrt{a} = (\sqrt{a})^2 = a$$ (square rooting a number then squaring it, leaves you with what you started with)
$$\sqrt{a\times b} = \sqrt{a} \times \sqrt{b}$$
To see why the second rule works consider the squares of both sides.
The square of the left hand side is $$a \times b$$ because of the first rule.
The square of the right hand side is:
$$(\sqrt{a} \times \sqrt{b})^2 = (\sqrt{a} \times \sqrt{b})\times (\sqrt{a} \times \sqrt{b})$$
$$= \sqrt{a} \times \sqrt{b}\times \sqrt{a} \times \sqrt{b}$$
$$= \sqrt{a} \times \sqrt{a}\times \sqrt{b} \times \sqrt{b}$$ (the order of the terms in a multiplication is allowed to change)
$$= (\sqrt{a})^2 \times (\sqrt{b})^2$$
$$= a \times b$$ (using the first rule to get rid of square roots)
So both sides of the second rule square to the same thing so the second rule holds.
In fact the second rule generalises a bit: Consider $$\sqrt{a\times b\times c}$$ by the second rule:
$$\sqrt{a\times b\times c} = \sqrt{a\times (b\times c)}$$
$$= \sqrt{a}\times \sqrt{b\times c}$$ (by the second rule applied to $$a$$ and $$b\times c$$)
$$= \sqrt{a}\times \sqrt{b}\times\sqrt{c}$$ (by the second rule applied to $$b$$ and $$c$$)
So $$\sqrt{a\times b\times c}= \sqrt{a}\times \sqrt{b}\times\sqrt{c}$$
and similarly we can show $$\sqrt{a\times b\times c\times d}= \sqrt{a}\times \sqrt{b}\times\sqrt{c}\times \sqrt{d}$$, etc.
Applying the rules to $$\sqrt{180}$$ we get
$$\sqrt{180} = \sqrt{2\times 2 \times 3 \times 3 \times 5}$$ (from the prime factorization of 180)
$$= \sqrt{2}\times \sqrt{2} \times \sqrt{3} \times\sqrt{3} \times \sqrt{5}$$ (by the second rule)
$$= (\sqrt{2})^2 \times (\sqrt{3})^2 \times \sqrt{5}$$
$$= 2 \times 3 \times \sqrt{5}$$ (by the first rule applied to both 2 and 3)
$$= 6 \times \sqrt{5}$$
$$= 6\sqrt{5}$$ (the multiplication symbol is usually omitted, to keep the expressions as short as possible)
Applying the rules to $$\sqrt{48}$$ we get
$$\sqrt{48} = \sqrt{2\times 2 \times 2 \times 2 \times 3}$$ (from the prime factorization of 48)
$$= \sqrt{2}\times \sqrt{2} \times \sqrt{2} \times\sqrt{2} \times \sqrt{3}$$ (by the second rule)
$$= (\sqrt{2})^2 \times (\sqrt{2})^2 \times \sqrt{3}$$
$$= 2 \times 2 \times \sqrt{3}$$ (by the first rule)
$$= 4 \times \sqrt{3}$$
$$= 4\sqrt{3}$$ (the multiplication symbol is usually omitted, to keep the expressions as short as possible)
Hope this helped,
R. Baber.
Guest
nice post
Guest
### Re: Confusion over simple surds
Guest wrote:Why does $$\sqrt{180}$$ = 6$$\sqrt{5}$$
$$180= 9(2)(10)= 9(2)(2)(2)(5)= (3^2)(2^2)(5)$$ so $$\sqrt{180}= \sqrt{3^2}\sqrt{2^2}\sqrt{5}= 3(2)\sqrt{5}= 6\sqrt{5}$$
But $$\sqrt{48}$$ = 4$$\sqrt{3}$$[/tex]I'm not sure why you say "but". The two are not really related. $$48= 16(3)$$ so $$\sqrt{48}= \sqrt{16}\sqrt{3}= 4\sqrt{3}$$.
They are unrelated, but the way I remember doing these is sort of confusing me. This is how I've been asked to do it:
For example:
Why does $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ x $$\sqrt{3}$$ become 2 x 3 x $$\sqrt{5}$$
??? It DOESN'T! $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ x $$\sqrt{3}$$= 2(3)= 6. Did you drop a "$$\sqrt{5}$$" from the left side?
But then $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ become 4$$\sqrt{3}$$ ?
$$\sqrt{2}\sqrt{2}= 2$$ (by the definition of "$$\sqrt{}$$)
wwAm I making sense?
Yes, but you seem to be forgetting that the square root is the "inverse" of squaring: $$\sqrt{x^2}= |x|$$ (so $$\sqrt{x^2]= x$$ as long as x is positive and $$\left(\sqrt{x}\right)^2= x$$.
HallsofIvy
Posts: 72
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 27 | 2019-12-06T13:59:07 | {
"domain": "math10.com",
"url": "https://www.math10.com/forum/viewtopic.php?f=25&t=1712",
"openwebmath_score": 0.9051772952079773,
"openwebmath_perplexity": 737.2368453804386,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9759464485047916,
"lm_q2_score": 0.8615382147637196,
"lm_q1q2_score": 0.8408151609498106
} |
http://m-spitzer.de/piecewise-functions-answers.html | Identify whether or not he graph is a function. Given integral: I=∫0,05,3sinhx25+y236xdx+10yd. A piecewise function is called piecewise because it acts differently on different “pieces” of the number line. On graphing piecewise functions To graph a piecewise function, it is a good idea to follow these steps. For a piecewise function to be continuous each piecewise function must be continuous and it must be continuous at each interface between the piecewise functions. Often the unit step function u. 20 per minute. f x x 5 -4 2x 2x if if 1 x x 2 2 if x 1 3 if x 1 3. I am involved in Direct Link,which is a program for students who are too sick to actually go to school. Q: When a plane flies with the wind, it can travel 4200 miles in 6 hours. To learn more, see our tips on writing great. Piecewise functions are defined by different functions throughout the different intervals of the domain. Please be sure to answer the question. In this lesson you will learn how to graph piecewise functions by using your knowledge of graphing other functions. Define a piecewise function f(x) That has an input x, An individuals taxable income and an output the amount the person owes in federal income tax Marginal tax rate Taxable Over Not Exeeding $0. The toy sells for$17. Write f(x) = |-x + 2x + 8| as a piecewise function. Piecewise Activity You may work in groups of up to 3 people. Use MathJax to format equations. functions, with each sub function applying to a certain interval on the x-axis. Each piece behaves differently based on the input function for that interval. Lesson 1-7 – Translating Absolute Value Functions Discovery Worksheet (with answers) Name Date Graphing Absolute Value Functions Graph the following piecewise function by hand: (1) 0 0 x if x fx x if x (2) On your graphing calculator graph the function f(x) x with this WINDOW and answer the following questions. " These are called *piecewise functions*, because their rules aren't uniform, but consist of multiple pieces. I have a piecewise function that I am trying to express by means of indicator functions depending on the value of the variable z: F_Ta = @(z) F_Ta_II(z). piecewise functions practice worksheet answer key, Topic - solutions and suspension worksheet answers section a 1. Write f(x) = |2x^2 - 8x + 6| as a piecewise function. MathJax reference. Identify whether or not the graph is a function. org 2 4 On the set of axes below, graph the piecewise. Write the piecewise-defined function. Mcdougal littell biology book answers, chapter 8 resource book algebra 2 mcdougal, prentice hall answer keys, Glencoe Mathematics Algebra 1 Even Answers, matlab symbolic system linear equations. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. midpoint and distance formula worksheet. Monster Functions (Day 2) OPEN HOUSE 6:00 to 8:30 PM P. Lesson 1-7 – Translating Absolute Value Functions Discovery Worksheet (with answers) Name Date Graphing Absolute Value Functions Graph the following piecewise function by hand: (1) 0 0 x if x fx x if x (2) On your graphing calculator graph the function f(x) x with this WINDOW and answer the following questions. x 22 x 22 Date: -2, —41, -1. Express your answer as an integer or simplified fraction. lumenlearning. On graphing piecewise functions To graph a piecewise function, it is a good idea to follow these steps. A piecewise function is a function represented by two or more functions, each corresponding to a part of the domain. I have done a linear programming code. is a function that is defined by two or more. I am expecting students to struggle with writing the piecewise-defined functions but we are practicing more tomorrow so that should help students understand the worksheet. 2; Either recalculate the slope using (0,1) and (2,5) or note that the slope has not changed since the y-values of both endpoints have been increased by 1 which means that the change in y has not changed. Use MathJax to format equations. Piecewise Functions And Answers - Displaying top 8 worksheets found for this concept. Carefully graph each of the following. Piecewise Defined Functions Worksheet - Problems. function may also be referred to as a. To write a piecewise function, use the following syntax: y = {condition: value, condition: value, etc. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. The problem with your code is that when you apply an ‘if’ to an entire vector, as you have done with “if x<0”, it is not regarded as true by matlab unless all elements of that vector are true, and that of course is not the case, so the ‘if’ will fail. 3_practice_solutions. com Linear Piecewise. Worked example: graphing piecewise functions. The total cost is a function of the number of nights 𝑥 that a guest stays. Expected Learning Outcomes The students will be able to: 1) Evaluate piecewise functions. Was this article helpful? How to Solve Function Composition;. The problem with your code is that when you apply an 'if' to an entire vector, as you have done with "if x<0", it is not regarded as true by matlab unless all elements of that vector are true, and that of course is not the case, so the 'if' will fail. The domain o(the piecewise function is (- infinity, infinity). Learn more about ode45, oder23, dynamic systems, piecewise MATLAB. Piecewise Functions And Answers - Displaying top 8 worksheets found for this concept. v l [AqlQlW mrPiGgMhwtjsk Jrqe_sTeErvvreAdQ. piecewise functions project answer key, reciprocal functions. x 22 x 22 Date: -2, —41, -1. Please be sure to answer the question. Absolute Value and Reciprocal Functions Key Terms absolute value absolute value function piecewise function invariant point absolute value equation reciprocal function asymptote The relationship between the pressure and the volume of a confined gas. Carefully graph each of the following. 11th Graphing Piecewise Functions. Worksheet Piecewise Functions Author: Shai McGowan Last modified by: Jaramillo, Lianne Created Date: 9/9/2014 4:50:00 PM Company: State College Area School District Other titles: Worksheet Piecewise Functions. Left Piece When x < 0, the graph is the line given by y = x + 3. The last two problems give students a chance to graph a piecewise-defined function. Learn more about plotting, mathematics, matlab, coding, graph, equations, help, symbolic Symbolic Math Toolbox. Please be sure to answer the question. Solution for Write a piecewise function for the given graph. How to evaluate limits of Piecewise-Defined Functions explained with examples and practice problems explained step by step. Express your answer as an integer or simplified fraction. Match the piecewise function with its graph. Graph each piece individually 2. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Piecewise functions are functions that have multiple pieces, or sections. • Piecewise functions are several different functions grouped for specific domains. piecewise functions practice worksheet answer key, Topic - solutions and suspension worksheet answers section a 1. … WordPress Shortcode. In preparation for the definition of the absolute value function, it is extremely important to have a good grasp of the concept of a piecewise-defined function. Then, evaluate the graph at any specified domain value. Use MathJax to format equations. Piecewise Defined Functions Worksheet : Worksheet given in this section is much useful to the students who would like to practice problems on piecewise defined functions. • The graph is composed of part of a line. a) Write a piecewise function that represents your pay for up to 60 hours. maximum(f1_functions). Graph the piecewise function: Gimme a Hint Graph this piecewise function: Gimme a Hint = - Show Answer. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 20 per minute. com 5-2 Reteach to Build Understanding Piecewise-Defined Functions 1. This is the currently selected item. Or a piecewise function can be made up of different kinds of functions - a quadratic and a linear function on the same graph. Then, evaluate the graph at any specified domain value. Median response time is 34 minutes and may be longer for new subjects. Worked example: graphing piecewise functions. View Piecewise Problems. c) Do the same thing you did in step b above, but change the slope to the opposite sign. So, a piecewise function for the graph is f(x) = { x + 3, 2x − 1, if x < 0. Lesson 26 Applications of Piecewise Defined Functions 4 Example 3: A rental home on Airbnb rents for $100 a night for the first three nights,$90 a night for the next three nights, and $80 a night for each remaining night. Worksheet Piecewise Functions Author: Shai McGowan Last modified by: Jaramillo, Lianne Created Date: 9/9/2014 4:50:00 PM Company: State College Area School District Other titles: Worksheet Piecewise Functions. b) Graph the function below. Piecewise Functions - Explanation, Graphing by hand and calculator, and word problems Evaluating Piecewise Functions - Practice Quiz Graphing Piecewise Functions Overview. Piecewise Functions And Answers - Displaying top 8 worksheets found for this concept. Learn more about ode45, oder23, dynamic systems, piecewise MATLAB. Consider the following piecewise-defined function F*-1 if is-5 f(x) = - 4x + 6 if x > -5 Step 1 of 3: Evaluate this function at x = -5. I do you do with teacher Assignment: Graphing a Piecewise Function: Click Here (this is to be turned in for a grade) February 11-20 Lesson 2. Piecewise Functions. Graph of the Piecewise Function y = -x + 3 on the interval [-3, 0] and y = 3x + 1 on the interval [0, 3] A special example of a piecewise function is the absolute value function that states:. I have a piecewise function that I am trying to express by means of indicator functions depending on the value of the variable z: F_Ta = @(z) F_Ta_II(z). Plss I just need the true answer even if no explanation to write plss *Response times vary by subject and question complexity. Common Core Standard: F-IF. 3_practice_solutions. Lesson 1-7 – Translating Absolute Value Functions Discovery Worksheet (with answers) Name Date Graphing Absolute Value Functions Graph the following piecewise function by hand: (1) 0 0 x if x fx x if x (2) On your graphing calculator graph the function f(x) x with this WINDOW and answer the following questions. ⃣Graph piecewise functions Write equations of piecewise functions Vocabulary: piecewise function Definitions A P_____ F_____ is a function which is defined by sub-functions that each applies to a specific part of the domain. Median response time is 34 minutes and may be longer for new subjects. A graphing calculator can be used to verify that your answers "make sense" or "look right". Then find the domain and range of each piecewise function. Created Date: 4/5/2017 9:43:43 PM. f1_functions = [f1_line1, f1_line2, ] f1 = cp. The following is a piecewise function. The last two problems give students a chance to graph a piecewise-defined function. A change in the function equation occurs for different values in the domain. *Response times vary by subject and question complexity. The only way I can replicate this function is using Plot[Piecewise[{{x^2, x >= 0}, {0, x < 0}}], {x, -10, 10}] When I define the Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Making statements based on opinion; back them up with references or personal experience. If the function is undefined at the given value indicate "Undefined" < Answer 2 Points Keypad Keyboard Shortcuts Selecting a radio button will replace. com Graph this function, using time on the x-axis and miles per hour on the y-axis. Use MathJax to format equations. Writing a Piecewise Function Write a piecewise function for the graph. I want to make the following piecewise function: f(x) = 5x for 0 -5 Step 1 of 3: Evaluate this function at x = -5. Express your answer as an integer or simplified fraction. Possible Descriptions: • The graph is a function. 11th Graphing Piecewise Functions. MathJax reference. You may use your calculators to help you graph, but you must sketch it carefully on the grid! 1. The same for when x = 1 with the second line and third line. Given integral: I=∫0,05,3sinhx25+y236xdx+10yd. In this lesson you will learn how to graph piecewise functions by using your knowledge of graphing other functions. Match the piecewise function with its graph. Choose a web site to get translated content where available and see local events and offers. Look at the inequalities rst. Notes Page 6 P. Here, pupils watch as each graph is created and then pieced together to make a function. 30 per unit and the fixed costs are$98,000. 4, f(10)=96. com Features of functions from Bitwise functions worksheetwith answers, source:courses. You may use your calculators to help you graph, but you must sketch it carefully on the grid! 1. Graph each of the following piecewise functions. I have a piecewise function that I am trying to express by means of indicator functions depending on the value of the variable z: F_Ta = @(z) F_Ta_II(z). Making statements based on opinion; back them up with references or personal experience. If the function is undefined at the given value indicate "Undefined" < Answer 2 Points Keypad Keyboard Shortcuts Selecting a radio button will replace. pdf: File Size: 542 kb: Download File. Unformatted text preview: Math 2 Piecewise Functions Worksheet #2 Name: ____ Part I. You've been inactive for a while, logging you out in a few seconds. Worked example: domain & range of step function. Therefore, pieces should not intersect or overlap such that it violates the vertical line test. A graphing calculator can be used to verify that your answers "make sense" or "look right". Consider the following piecewise-defined function F*-1 if is-5 f(x) = - 4x + 6 if x > -5 Step 1 of 3: Evaluate this function at x = -5. Carefully graph each of the following. Piecewise Functions. To end today, I give the students Piecewise Functions 1. Use MathJax to format equations. Write down your solutions BEFORE looking at any answers. piecewise functions practice worksheet answer key, Topic - solutions and suspension worksheet answers section a 1. Consider the following piecewise-defined function F*-1 if is-5 f(x) = - 4x + 6 if x > -5 Step 1 of 3: Evaluate this function at x = -5. Given integral: I=∫0,05,3sinhx25+y236xdx+10yd. You've been inactive for a while, logging you out in a few seconds. AMR Piecewise Functions Name: Part I. Variables, Expressions, Functions and Equations. Write a piecewise function to describe the plan. Graph the two linear equations on the same x-y plane below y = –x – 4 y = 2x + 1 Notes Piecewise Functions If we want a function to behave differently depending on the x- values, we need a piecewise function. Q: When a plane flies with the wind, it can travel 4200 miles in 6 hours. f(x) = { x − 2, 2x + 1, if x ≤ 0 if x > 0 The expression x − 2 represents the value of f when x is less than or equal to 0. Express your answer as an integer or simplified fraction. Making statements based on opinion; back them up with references or personal experience. Some of the worksheets for this concept are Work piecewise functions, Work piecewise functions, Piecewise answers, Piecewise functions, Lesson 1 piecewise functions, Piecewise functions date period, Piecewise functions, Work piecewise functions name algebra 2. X 2 x 2 x 2 cases i. On graphing piecewise functions To graph a piecewise function, it is a good idea to follow these steps. View Piecewise Problems. If the function is undefined at the given value indicate "Undefined" < Answer 2 Points Keypad Keyboard Shortcuts Selecting a radio button will replace. v l [AqlQlW mrPiGgMhwtjsk Jrqe_sTeErvvreAdQ. Get help with your Piecewise functions homework. First of all, modifiy your preamble adding \usepackage{amsfonts} Latex piecewise function with left operator \begin{equation*} y = f(x) = \lvert x \rvert = \left\. For example, 2x-1, if x 1 f(x)= 3x+1, if x>1. Looking back at the inequalities, darken in the functions between the vertical lines. The problem with your code is that when you apply an 'if' to an entire vector, as you have done with "if x<0", it is not regarded as true by matlab unless all elements of that vector are true, and that of course is not the case, so the 'if' will fail. Worked example: domain & range of step function. Here, pupils watch as each graph is created and then pieced together to make a function. If there is no answer for f (x), write none. Let us examine where f has a discontinuity. production possibilities curve worksheet answers Redesign and Price. Piecewise Functions – Definition, Graph & Examples There are instances where the expression for the functions depends on the given interval of the input values. X 2 x 2 x 2 cases i. Show Answer. Answer: \displaystyle\lim\limits_{x. Piecewise FUNctions worksheet from mon Core Fun on from Piecewise Functions Worksheet With Answers, source:pinterest. * (z=Rd) where F_Ta_III can assume an infinite value only when z 2. Please be sure to answer the question. 2; Either recalculate the slope using (0,1) and (2,5) or note that the slope has not changed since the y-values of both endpoints have been increased by 1 which means that the change in y has not changed. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Step Functions Definition: The unit step function (or Heaviside function), is defined by ≥ < = t c t c u c t 1, 0, (), c ≥ 0. Before look at the worksheet, if you would like to know the stuff related to piecewise functions, Please click here. ALGEBRAICALLY Use the picture of the piecewise function to answer the following. The problem with your code is that when you apply an ‘if’ to an entire vector, as you have done with “if x<0”, it is not regarded as true by matlab unless all elements of that vector are true, and that of course is not the case, so the ‘if’ will fail. I'm taking a Fourier Analysis course using Churchill 's Fourier Series and Boundary Value Problems, 6th ed. Before look at the worksheet, if you would like to know the stuff related to piecewise functions, Please click here. There will be one finished product turned in per group with all the names below listed: Assign student A, B and C. Write your name next to the appropriate place. t f vMpaYdYeL YwoiBtyhe KIVnvflibnBijtmeY \PfrPe\cWaalbcVuWlwugsK. This is the “negative slope” side of the function. (a) The total cost for a. Erase around the domain 3. Worked example: domain & range of step function. Given integral: I=∫0,05,3sinhx25+y236xdx+10yd. Making statements based on opinion; back them up with references or personal experience. Unformatted text preview: Math 2 Piecewise Functions Worksheet #2 Name: ____ Part I. First of all, modifiy your preamble adding \usepackage{amsfonts} Latex piecewise function with left operator \begin{equation*} y = f(x) = \lvert x \rvert = \left\. Write a piecewise function to describe the plan. 4, f(10)=96. A piecewise function is a function that combines two or more functions. AMR Piecewise Functions Name: Part I. In this activity, students use piecewise functions to match graphs of rays and segments, interpret function values in a graphical context, and apply what they learn to a postage rates problem. MathJax reference. Piecewise Functions Evaluate the function for the given value of x. Identify whether or not the graph is a function. Piecewise Activity You may work in groups of up to 3 people. Free piecewise functions calculator - explore piecewise function domain, range, intercepts, extreme points and asymptotes step-by-step This website uses cookies to ensure you get the best experience. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Created Date: 11/4/2015 7:47:18 PM. Some of the worksheets for this concept are Work piecewise functions, Work piecewise functions, Piecewise answers, Piecewise functions, Lesson 1 piecewise functions, Piecewise functions date period, Piecewise functions, Work piecewise functions name algebra 2. Please be sure to answer the question. Before look at the worksheet, if you would like to know the stuff related to piecewise functions, Please click here. Graph of the Piecewise Function y = -x + 3 on the interval [-3, 0] and y = 3x + 1 on the interval [0, 3] A special example of a piecewise function is the absolute value function that states:. Making statements based on opinion; back them up with references or personal experience. Special care needs to be taken in for piecewise functions as explained in the exposition. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. *Response times vary by subject and question complexity. Access the answers to hundreds of Piecewise functions questions that are explained in a way that's easy for you to understand. So, a piecewise function for the graph is f(x) = { x + 3, 2x − 1, if x < 0. 4 in 1970, 96. Identify any points of discontinuity. If you don't see any interesting for you, use our search form on bottom ↓. Match the piecewise function with its graph. Then, evaluate the graph at any specified domain value. If the function is undefined at the given value indicate "Undefined" < Answer 2 Points Keypad Keyboard Shortcuts Selecting a radio button will replace. * (z=Rd) where F_Ta_III can assume an infinite value only when z 0 f(-2) f(0) Answered: Evaluate the piecewise-defined function… | bartleby menu. • The graph is composed of part of a line. midpoint and distance formula worksheet. Step Functions Definition: The unit step function (or Heaviside function), is defined by ≥ < = t c t c u c t 1, 0, (), c ≥ 0. b) Graph the function below. Show Answer. Since we have three definitions for f(x) depending on x's value, you can see that if x = -1 we would use the first line. A_____ B_____ C_____. fx 2 5 2 2 3 2 xx x x x ° ® °¯ t Function?. Greatest Integer Function. Left Piece When x < 0, the graph is the line given by y = x + 3. f( -2) = 2. Piecewise, Absolute Value and Step Functions MathBitsNotebook. Express your answer as an integer or simplified fraction. * (z=Rd) where F_Ta_III can assume an infinite value only when z=2 The final piecewise function is: f( x ) = x 2 + 1 if x < 2 = - x + 3 if x >= 2. If the function is undefined at the given value indicate "Undefined" < Answer 2 Points Keypad Keyboard Shortcuts Selecting a radio button will replace. t f vMpaYdYeL YwoiBtyhe KIVnvflibnBijtmeY \PfrPe\cWaalbcVuWlwugsK. And now the objective function of the problem will become a piece-wise function, but still linear in every part of the function. Displaying top 8 worksheets found for - Algebra 2 Piecewise Functions Answers. Some functions have simple rules, like "for every x, return x². Hi! I'm having trouble understanding how to write piecewise functions. Use MathJax to format equations. Solution for Evaluate the piecewise-defined function at the indicated values. Let x be the number of units produced and sold. Piecewise Functions – Definition, Graph & Examples There are instances where the expression for the functions depends on the given interval of the input values. This problem allows me to. " However, there can be other rules that are more elaborate. Median response time is 34 minutes and may be longer for new subjects. Hi! I'm having trouble understanding how to write piecewise functions. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. Write a piecewise function to represent individual cost of a t-shirt as function of. org 2 4 On the set of axes below, graph the piecewise. Plus each comes with an answer key. x 22 x 22 Date: -2, —41, -1. Making statements based on opinion; back them up with references or personal experience. c) Do the same thing you did in step b above, but change the slope to the opposite sign. Consider the following piecewise-defined function F*-1 if is-5 f(x) = - 4x + 6 if x > -5 Step 1 of 3: Evaluate this function at x = -5. functions, with each sub function applying to a certain interval on the x-axis. It has an infinite number of pieces: The Floor Function. Identify whether or not he graph is a function. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Piecewise Defined Functions Worksheet - Problems. Variables, Expressions, Functions and Equations. Piecewise Functions - ClassZone. Graph each piece individually 2. 2) Graph piecewise functions. To end today, I give the students Piecewise Functions 1. Problem 1 :. I have a piecewise function that I am trying to express by means of indicator functions depending on the value of the variable z: F_Ta = @(z) F_Ta_II(z). 7: Graphing Piecewise-Defined Functions www. Variables in MATLAB are by default double-precision. This quiz is incomplete! To play this quiz, please finish editing it. Step Functions Definition: The unit step function (or Heaviside function), is defined by ≥ < = t c t c u c t 1, 0, (), c ≥ 0. A teacher code is provided by your teacher and gives you free access to their assignments. function may also be referred to as a. • Piecewise functions are several different functions grouped for specific domains. Piecewise Functions. You may use your calculators to help you graph, but you must sketch it carefully on the grid! 1. Notes page 6. The Symbolic Math Toolbox extends this by allowing you to express numbers in exact symbolic form using sym and with variable-precision using vpa. We use piecewise functions to describe situations where a rule or relationship changes as the input value crosses certain “boundaries. A_____ B_____ C_____. This is the “negative slope” side of the function. The domain o(the piecewise function is (- infinity, infinity). Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. Plotting a piecewise function. MathJax reference. These functions do not share the same point at x = 0, as the first contains that point (0, 3), while the second piece contains the point (0, 1). {-x 3 + 6 x 2 - 9 x + 4 :. 7 –Piecewise Functions In real life problems, functions can be represented by a combination of equations, each corresponding to a part of the domain. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. 12th Writing Piecewise Functions. Function? Yes or No 3. Name PearsonRealize. I am involved in Direct Link,which is a program for students who are too sick to actually go to school. * (z=Rd) where F_Ta_III can assume an infinite value only when z Rational-functions-> SOLUTION: The cell phone plan is 50 per month for 450 min. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. In this lesson you will learn how to graph piecewise functions by using your knowledge of graphing other functions. Special care needs to be taken in for piecewise functions as explained in the exposition. Filesize: 634 KB. These functions do not share the same point at x = 0, as the first contains that point (0, 3), while the second piece contains the point (0, 1). Identify whether or not the graph is a function. 1 The population f(x) of one city (in thousands) was 85. Piecewise Functions - Explanation, Graphing by hand and calculator, and word problems Evaluating Piecewise Functions - Practice Quiz Graphing Piecewise Functions Overview. Variables, Expressions, Functions and Equations. " However, there can be other rules that are more elaborate. Median response time is 34 minutes and may be longer for new subjects. The last two problems give students a chance to graph a piecewise-defined function. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. Often the unit step function u. Could someone please explain how to do this problem. Piecewise Functions And Answers - Displaying top 8 worksheets found for this concept. 2: Writing a Piece-wise Function from a graph Video 1: Click Here. Get help with your Piecewise functions homework. Answer the questions below Graphs #1-4 in the spaces provided. Help would be greatly appreciated! Thanks. So this does not appear directly possible in CVXPY from the list of points. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Before look at the worksheet, if you would like to know the stuff related to piecewise functions, Please click here. Convert between absolute value functions and piecewise linear functions. fx 2 5 2 2 3 2 xx x x x ° ® °¯ t Function?. Graph the piecewise function: Gimme a Hint Graph this piecewise function: Gimme a Hint = - Show Answer. You may use your calculators to help you graph, but you must sketch it carefully on the grid! 1. Piecewise Functions. A piecewise function is a function defined by two or more equations. Access the answers to hundreds of Piecewise functions questions that are explained in a way that's easy for you to understand. Problem 1 :. They are defined piece by piece, with various functions defining each interval. f(x)={(x^2 if x<1),(x if 1 le x < 2),(2x-1 if 2 le x):}, Notice. The Absolute Value Function is a famous Piecewise Function. Common Core Standard: F-IF. Draw a dotted vertical line for each of these values. I have a piecewise function that I am trying to express by means of indicator functions depending on the value of the variable z: F_Ta = @(z) F_Ta_II(z). Solutions and suspension worksheet answer. 3 Worksheet Piecewise Function Algebra 2 Answers Afda Day 7 Block Unit 3 Absolute Values And In 2020 The Worksheets Learning Algebra. v l [AqlQlW mrPiGgMhwtjsk Jrqe_sTeErvvreAdQ. Sometimes a piecewise function consists of pieces that are not connected. Identify whether or not he graph is a function. The greatest integer function is an example of a step function, a piecewise function in which each function rule is a constant function. Use MathJax to format equations. Learn more about thrust, rocket engine, modeling and simulation, piecewise function. And now the objective function of the problem will become a piece-wise function, but still linear in every part of the function. 2: Writing a Piece-wise Function from a graph Video 1: Click Here. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. The total cost is a function of the number of nights 𝑥 that a guest stays. Function or Not A Function Worksheet with Answers Inspirational from Piecewise Functions Worksheet With Answers. This Bakpax autogradable standards-aligned Math worksheet covers Piecewise Functions. Median response time is 34 minutes and may be longer for new subjects. If the function is undefined at the given value indicate "Undefined" < Answer 2 Points Keypad Keyboard Shortcuts Selecting a radio button will replace. But if when x = -1 it also equals the second line, then it's continuous. Write f(x) = |2x^2 - 8x + 6| as a piecewise function. Piecewise FUNctions worksheet from mon Core Fun on from Piecewise Functions Worksheet With Answers, source:pinterest. If you go over 450 min. So, a piecewise function for the graph is f(x) = { x + 3, 2x − 1, if x < 0. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. To end today, I give the students Piecewise Functions 1. I ^ zAAlwlB ^rcisgShNtksW srHe[sfelrPvceldr. Absolute Value and Reciprocal Functions Key Terms absolute value absolute value function piecewise function invariant point absolute value equation reciprocal function asymptote The relationship between the pressure and the volume of a confined gas. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. ©n Q2L0`1S6\ WKUuFtTaw mSToifhtjwGaarveR VL^LwCg. You may only work a maximum of 60 hours. f1_functions = [f1_line1, f1_line2, ] f1 = cp. Write down your solutions BEFORE looking at any answers. In this lesson you will learn how to graph piecewise functions by using your knowledge of graphing other functions. Some of the worksheets for this concept are Piecewise answers, Work piecewise functions, Work piecewise functions, Piecewise functions, Piecewise functions, Work piecewise functions name algebra 2, Work piecewise functions algebra 2 answers, Piecewise functions. An example is shown below. Please be sure to answer the question. Match the piecewise function with its graph. Carefully graph each of the following. 11th Graphing Piecewise Functions. Variables, Expressions, Functions and Equations. Let x be the number of units produced and sold. However, before we jump into the fray, let’s take a look at a special type of function called a constant function. fx 2 5 2 2 3 2 xx x x x ° ® °¯ t Function?. Convert between absolute value functions and piecewise linear functions. This is the “positive slope” side of the function. Express your answer as an integer or simplified fraction. Corrective Assignment. The expression 2x + 1 represents the value of f when x is greater than 0. Evaluating a Piecewise Function. Given integral: I=∫0,05,3sinhx25+y236xdx+10yd. I have a piecewise function that I am trying to express by means of indicator functions depending on the value of the variable z: F_Ta = @(z) F_Ta_II(z). Evaluatin Greatest Inte er Ex ressions Evaluate the following: (3) (6) 1-6. Worked example: domain & range of step function. Identify whether or not he graph is a function. For example, you may have one rule for all the negative numbers, another rule for numbers bigger […]. A piecewise function is a function defi ned by two or more equations. maximum(f1_functions). Regents Exam Questions Name: _____ F. I have a piecewise function that I am trying to express by means of indicator functions depending on the value of the variable z: F_Ta = @(z) F_Ta_II(z). Help would be greatly appreciated! Thanks. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. Hi I need help with a piecewise function in excel using =IF() statements the value is 0 if x<=25, 25 if 25100 The trick is that each value may only be given out once per set of X's and must stack if not present prior to that value if you have x = 26,27,101, the values should be 25,0,150 or x = 15,16,101 the values would be 0,0,175. VERBALLY Mr. 30 per unit and the fixed costs are98,000. Problem 1 :. Please be sure to answer the question. 2) Graph piecewise functions. Possible Descriptions: • The graph is a function. Piecewise Functions Task Cards plus HW or quiz Piecewise Defined Functions This activity is designed for Algebra 2 or PreCalculus. To learn more, see our tips on writing great. Sometimes a piecewise function consists of pieces that are not connected. Get help with your Piecewise functions homework. I'm trying to do problem 3, section 24. Write f(x) = |2x^2 - 8x + 6| as a piecewise function. If there is no answer for f (x), write none. You can write absolute value functions and step functions as piecewise functions so they're easier to graph. A_____ B_____ C_____. Show Answer. 7: Graphing Piecewise-Defined Functions www. b) Graph the function below. com Graph this function, using time on the x-axis and miles per hour on the y-axis. Some of the worksheets for this concept are Piecewise answers, Work piecewise functions, Work piecewise functions, Piecewise functions, Piecewise functions, Work piecewise functions name algebra 2, Work piecewise functions algebra 2 answers, Piecewise functions. Piecewise, Absolute Value and Step Functions MathBitsNotebook. 27 Questions Show answers. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. Notes Page 6 P. Find the rule of a piecewise linear function f(x) that models this data, that is a piecewise function with f(0) = 85. If the function is undefined at the given value indicate "Undefined" < Answer 2 Points Keypad Keyboard Shortcuts Selecting a radio button will replace. Learn more about thrust, rocket engine, modeling and simulation, piecewise function. To smooth the steps out, you can use Fourier filtering (or use conv2()), or you can use a Savitzky-Golay filter. Then, evaluate the graph at any specified domain value. OK,so I am very confused and am hoping that you can help. Module 13 578 Lesson 1 DO NOT EDIT--Changes must be made through. Identify whether or not he gaph is a function. Select a Web Site. Graph the piecewise function: Gimme a Hint Graph this piecewise function: Gimme a Hint = - Show Answer. 8 Homework: Piecewise Functions Name_____ Graph each function without a calculator. 114 Chapter 2 Linear Equations and Functions Piecewise Functions Write a piecewise function that. Median response time is 34 minutes and may be longer for new subjects. I do you do with teacher Assignment: Graphing a Piecewise Function: Click Here (this is to be turned in for a grade) February 11-20 Lesson 2. function may also be referred to as a. Right Piece When x ≥ 0, the graph is the line given by y = 2x − 1. You will have to take one-sided limits separately since different formulas will apply depending on from which side you are approaching the point. Functions assign outputs to inputs. To learn more, see our tips on writing great. Writing a Piecewise Function Write a piecewise function for the graph. Free piecewise functions calculator - explore piecewise function domain, range, intercepts, extreme points and asymptotes step-by-step This website uses cookies to ensure you get the best experience. Such functions are called piecewise functions. Write a piecewise function to represent individual cost of a t-shirt as function of. GRAPH Equation of the pieces Domain for the pieces Piecewise function. com Features of functions from Bitwise functions worksheetwith answers, source:courses. Could someone please explain how to do this problem. PIECEWISE FUNCTIONS. Piecewise Functions. *Response times vary by subject and question complexity. You may use your calculators to help you graph, but you must sketch it carefully on the grid! 1. * (z=Rd) where F_Ta_III can assume an infinite value only when z 0 f(-2) f(0) Answered: Evaluate the piecewise-defined function… | bartleby menu. Each has model problems working step by step, practice problems, as well as challenging questions at the end of the sheet. Use MathJax to format equations. 20 or 25 3. ⃣Graph piecewise functions Write equations of piecewise functions Vocabulary: piecewise function Definitions A P_____ F_____ is a function which is defined by sub-functions that each applies to a specific part of the domain. PIECEWISE FUNCTIONS. A step functions is a piecewise function defined by a constant value over each part of its domain. Piecewise Functions - ClassZone. Lesson 26 Applications of Piecewise Defined Functions 4 Example 3: A rental home on Airbnb rents for $100 a night for the first three nights,$90 a night for the next three nights, and $80 a night for each remaining night. Identify whether or not the graph is a function. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. Monster Functions (Day 2) OPEN HOUSE 6:00 to 8:30 PM P. 3_practice_solutions. Displaying top 8 worksheets found for - Piecewise Word Problems With Answers. Questions; hi calculus asap. Some of the worksheets for this concept are Piecewise answers, Work piecewise functions, Work piecewise functions, Piecewise functions, Piecewise functions, Work piecewise functions name algebra 2, Work piecewise functions algebra 2 answers, Piecewise functions. To smooth the steps out, you can use Fourier filtering (or use conv2()), or you can use a Savitzky-Golay filter. Then find the domain and range of each piecewise function. Identify whether or not he graph is a function. *Response times vary by subject and question complexity. Each piece behaves differently based on the input function for that interval. For example, "If x<0, return 2x, and if x≥0, return 3x. Get help with your Piecewise functions homework. An real-world example are parking garage rates. Piecewise Functions. I have a piecewise function that I am trying to express by means of indicator functions depending on the value of the variable z: F_Ta = @(z) F_Ta_II(z). Hence the name! Graphing a Piecewise Function. pdf from MATH 100 at Richwoods High School. the shirts for the following cost. Piecewise Functions Evaluate the function for the given value of x. piecewise function, hence the name piecewise. SOLUTION Each “piece” of the function is linear. Evaluate the function for the indicated values f(x) = 3x – 2 a. Q: Show that the following line intergral is path and independent A: To show: The line integral is independent of the path. This is the “negative slope” side of the function. Piecewise Functions And Answers - Displaying top 8 worksheets found for this concept. b) Graph the function below. Express your answer as an integer or simplified fraction. To evaluate a piecewise function for a given value of x, substitute the value of x into the rule for the part of the domain that includes x. Lesson 26 Applications of Piecewise Defined Functions 4 Example 3: A rental home on Airbnb rents for$100 a night for the first three nights, $90 a night for the next three nights, and$80 a night for each remaining night. A piecewise function is called piecewise because it acts differently on different “pieces” of the number line. Write f(x) = |-x + 2x + 8| as a piecewise function. You may only work a maximum of 60 hours. This is the “positive slope” side of the function. * (z=Rd) where F_Ta_III can assume an infinite value only when z 2. Piecewise Defined Functions Worksheet - Problems. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Mathway Support October 14, 2020 22:57. Draw a dotted vertical line for each of these values. Write f(x) = |-x + 2x + 8| as a piecewise function. Q: When a plane flies with the wind, it can travel 4200 miles in 6 hours. Create the following piecewise function using relation operators and plot it in the range (-5,10). Identify any points of discontinuity. * (z=Rd) where F_Ta_III can assume an infinite value only when z=2 The final piecewise function is: f( x ) = x 2 + 1 if x < 2 = - x + 3 if x >= 2. Piecewise Activity You may work in groups of up to 3 people. Variables in MATLAB are by default double-precision. Or a piecewise function can be made up of different kinds of functions - a quadratic and a linear function on the same graph. Match each piecewise-defined function with its graph. Worked example: domain & range of step function. MathJax reference. Monster Functions (Day 2) OPEN HOUSE 6:00 to 8:30 PM P. A_____ B_____ C_____. Marking lightly, graph all the functions which are given for f. (Note: Absolute value is under. 2: Writing a Piece-wise Function from a graph Video 1: Click Here. Often the unit step function u. , they charge you an additional. If you don't see any interesting for you, use our search form on bottom ↓. Undefined function or method 'piecewise_eval' for input arguments of type 'cell', though that seems to be the syntax specified by the function. Use a graphing calculator as appropriate. When the plane flies in the. 2 - x, if x 0 f (x) =. Notes Page 6 P. " However, there can be other rules that are more elaborate. Help would be greatly appreciated! Thanks. *Response times vary by subject and question complexity. A sine can be pretty well fit by a 3rd or 5th order polynomial in the short run, so if you have the Signal Processing Toolbox, you can use sgolay() with an order 3 or 5 and it should do a pretty good job of leaving the sine wave parts along but smoothing over the abrupt steps where. 12th Writing Piecewise Functions. AMR Piecewise Functions Name: Part I. Function or Not A Function Worksheet with Answers Inspirational from Piecewise Functions Worksheet With Answers. Evaluating a Piecewise Function. Step Functions Definition: The unit step function (or Heaviside function), is defined by ≥ < = t c t c u c t 1, 0, (), c ≥ 0. Piecewise, Absolute Value and Step Functions MathBitsNotebook. x² if x < 0 f (x) = { x +l if x > 0 f(-2) f(0) Answered: Evaluate the piecewise-defined function… | bartleby menu. Piecewise Functions - ClassZone. Free piecewise functions calculator - explore piecewise function domain, range, intercepts, extreme points and asymptotes step-by-step This website uses cookies to ensure you get the best experience. Wednesday Sept. Piecewise defined functions are always difficult for students and yet are so important, especially in Calculus. Making statements based on opinion; back them up with references or personal experience. The total cost T is a function of the number of nights x that a guest stays. com Features of functions from Bitwise functions worksheetwith answers, source:courses. * (z=Rd) where F_Ta_III can assume an infinite value only when z 0 f(-2) f(0) Answered: Evaluate the piecewise-defined function… | bartleby menu. To learn more, see our tips on writing great. Draw a dotted vertical line for each of these values. Question: Graph the following piecewise function and evaluate for the given values of x. Median response time is 34 minutes and may be longer for new subjects. So, a piecewise function for the graph is f(x) = { x + 3, 2x − 1, if x < 0. A piecewise function is a function defined by two or more equations. Answers to Questions on Piecewise Functions. org 2 4 On the set of axes below, graph the piecewise. 2: Writing a Piece-wise Function from a graph Video 1: Click Here. Mcdougal littell biology book answers, chapter 8 resource book algebra 2 mcdougal, prentice hall answer keys, Glencoe Mathematics Algebra 1 Even Answers, matlab symbolic system linear equations. Subject: Piecewise Functions Name: Kait Who are you: Student. In this activity, students use piecewise functions to match graphs of rays and segments, interpret function values in a graphical context, and apply what they learn to a postage rates problem. You will also learn how they are used to solve problems. A piecewise function is a function defined by two or more equations. Identify whether or not the graph is a function. You may only work a maximum of 60 hours. Making statements based on opinion; back them up with references or personal experience. If the function is undefined at the given value indicate "Undefined" < Answer 2 Points Keypad Keyboard Shortcuts Selecting a radio button will replace. Hello everyone! I am having some serious trouble understanding this application. Variables in MATLAB are by default double-precision. Practice: Piecewise functions graphs. Common Core Algebra 9H - Piecewise Functions Practice or questions 1 - 6, graph the piecewise functions on the axes provided. Carefully graph each of the following. -5 Step 1 of 3: Evaluate this function at x = -5. 114 Chapter 2 Linear Equations and Functions Piecewise Functions Write a piecewise function that. MathJax reference. Access the answers to hundreds of Piecewise functions questions that are explained in a way that's easy for you to understand. Here, pupils watch as each graph is created and then pieced together to make a function. Define a piecewise function f(x) That has an input x, An individuals taxable income and an output the amount the person owes in federal income tax Marginal tax rate Taxable Over Not Exeeding \$0. Pieces may be single points, lines, or curves. It is a step function, and the graph is said to have "jump discontinuities " at the integers. Hence the name! Graphing a Piecewise Function. Some of the worksheets for this concept are Work piecewise functions, Work piecewise functions, Piecewise answers, Piecewise functions, Lesson 1 piecewise functions, Piecewise functions date period, Piecewise functions, Work piecewise functions name algebra 2. Consider the following piecewise-defined function F*-1 if is-5 f(x) = - 4x + 6 if x > -5 Step 1 of 3: Evaluate this function at x = -5. Graph the function. Function? Yes or No. • Evaluate on a graph by finding the x on the x-axis, move vertically until you hit the function, write the point, and the y-coordinate is the value of the function at x. Infinite Algebra 2 - 1-7 Practice (Piecewise Functions) Created Date: 20140915164014Z. I am involved in Direct Link,which is a program for students who are too sick to actually go to school. Left Piece When x < 0, the graph is the line given by y = x + 3. SOLUTION Each “piece” of the function is linear. * (z=Rd) where F_Ta_III can assume an infinite value only when z 0 f(-2) f(0) Answered: Evaluate the piecewise-defined function… | bartleby menu. There will be one finished product turned in per group with all the names below listed: Assign student A, B and C. Piecewise functions with ode45 and ode23. Write the piecewise-defined function. The slopes and the lengths of the pairs of segments are the same. You've been inactive for a while, logging you out in a few seconds. *You must graph the entire function on the given graph paper for the domain if the range is also between. fx 2 5 2 2 3 2 xx x x x ° ® °¯ t Function?. fx 2 5 2 2 3 2 xx x x x ° ® °¯ t Function?. Greatest Integer Function. Let us examine where f has a discontinuity. {-x 3 + 6 x 2 - 9 x + 4 :. | 2021-04-14T19:47:31 | {
"domain": "m-spitzer.de",
"url": "http://m-spitzer.de/piecewise-functions-answers.html",
"openwebmath_score": 0.3814734220504761,
"openwebmath_perplexity": 985.4630696164663,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9759464422083111,
"lm_q2_score": 0.8615382147637196,
"lm_q1q2_score": 0.840815155525152
} |
http://zgoe.asainfo.it/lagrange-interpolation-python.html | One easy way of obtaining such a function, is to connect the. W8V5 Python:Lagrange Interpolation 6:33. I use these data points (0,0) (1,1) (2,4) (4,16) (5,25). If f is a polynomial of degree less than or equal to , the CGL quadrature formula is exact. Let fx ign 0 be distinct real numbers and let fy ign be real. 60 gx f o xx- 1 xx- 2 xx- 3 x o - x 1 x o - x. Also the x-coordinate 230. Interpolation. ) Since , it follows that everywhere. edu 1Course G63. or create account below. 1 Weierstrass. Lagrange interpolation in python. Compiled in DEV C++ You might be also interested in : Gauss Elimination Method Lagrange interpolation Newton Divided Difference Runge Kutta method method Taylor series method Modified Euler’s method Euler’s method Waddle’s Rule method Bisection method Newton’s Backward interpolation Newton’s forward interpolation Newtons rapson. 4) for reconstructing the interpolation polynomial. In other words interpolation is the technique to estimate the value of a mathematical function, for any intermediate value of the independent variable. 5 Interpolation. 60 gx f o xx– 1 xx– 2 xx– 3 x o – x 1 x o – x. As usual, my code is available: lagrange_q. Lagrange interpolation is one of the best options. Pdf Lagrange Interpolation In Some Weighted Uniform Spaces. This "guess" at the correct subinterval can be checked. Lagrange Interpolation The computations in this small article show the Lagrange interpolation. The interpolants Pn(x) oscillated a great deal, whereas the function f(x) was nonoscillatory. Author content. Polynomial Interpolation. ; In standard output format, only the domain element of an InterpolatingFunction object is printed explicitly. The formula can be derived from the Vandermonds determinant but a much simpler way of deriving this is from Newton's divided difference formula. But we get a different straight line depending on our coordinate system. Commented: KSSV on 4 May 2017 My Code is missing two values and I need help: function y = lagrange (X, Y, x) n = length(X); if n ~= length (Y). For Python: % timeit lagrange(x_int, y_int, x_new) with result. from matplotlib import pyplot as plt. consider linear interpolation. Python supports multiple ways to format text strings and these includes %-formatting, sys. Visit the online book containing this material. The interpolation calculator will return the function that best approximates the given points according to the method chosen. The difference is that I will change the sampling, that is, I will use non-uniform sampling. The Extensions regions defines a few extensions to allows for matrix manipulations. The code above uses a single header file , and there are no user defined functions. Linear Interpolation Method Using C Programming. Points are divided into a hier-. Método que permite encontrar un polinomio que interpola un conjunto de puntos mediante un sistema de ecuaciones. Interpolation is the process of deriving a simple function from a set of discrete data points so that the function passes through all the given data points (i. Interpolation par la méthode de Lagrange Le programme en C. When graphical data contains a gap, but data is available on either side of the gap or at a few specific points within the gap, an estimate of values within the gap can be made by interpolation. The Whittaker Shannon interpolation is equivalent to convolution with the impulse response of an ideal low pass filter. It's a whole a lot easier than Newton's divided differences interpolation polynomial, because there is no divided difference part that need a recursive function. By construction, on. It is useful at least in data analy-sis (interpolation is a form of regression), industrial design, signal processing (digital-to-analog conversion) and in numerical analysis. The remaining elements are indicated by <>. 93 KB #!/usr/bin/env python. Lagrange Polynomial Interpolation on Python. Nominators and denominators fo the base-polynomials are calculated and used to build ab the interpolation polynomial. from matplotlib import pyplot as plt. Aubin The University of Wisconsin-Milwaukee, 2019 Under the Supervision of Professor Lei Wang In this thesis, a treecode implementing Hermite interpolation is constructed to approximate a summation of pairwise interactions on large data sets. An overview of numerical methods and their application to problems in physics and astronomy. The double prime notation in the summation indicates that the first and last terms are halved. This webinar will review the interpolation modules available in SciPy and in the larger Python community and provide instruction on their use via example. An interpolation on two points, (x0, y0) and (x1, y1), results in a linear equation or a straight line. 1 Interpolation and the Lagrange Polynomial SolutionsbyJonLoKimLin-Fall2014 MATH 104A HW 05 SOLUTION KEY For those who need a quick primer on programming, I highly recommend the python course by. Lagrange Interpolation Calculator. abedkime 13 août 2013 à 3:49:33. Bonjour, je sollicite de l'aide pour pour pouvoir réaliser un programme en C qui fait l'interpolation polynomiale par la méthode de Lagrange. Interpolation, a fundamental topic in numerical analysis, is the problem of constructing a function. Create a new le named Newton interpolant. Lagrange showed that this polynomial function is given by, $I(x) = \sum\limits_{k=1}^{N}y_k\prod\limits_{i=1\,(i e k)}^{N}\frac{x-x_i}{x_k-x_i}. It is necessary because in science and engineering we often need to deal with. Looking up Lagrange interpolation on Wikipedia, I found something new to me: the barycentric form of Lagrange interpolation. Algorithms Android problem Assembly Bangla Love Poem Books C CPP Database HSC Html JAVA JavaScript Others Perl Php Presentation Project Prolog Prolog2 Python Saturday, October 7, 2017 Others Perl Inverse lagrange interpolation formula theory, algorithm and flowchart with a lot of example. We see that they indeed pass through all node points at , , and. For a given set of distinct points and numbers. Colour and Normal Interpolation As it applies to triangles and quadrilaterals in the rendering of 3D surfaces Written by Paul Bourke September 2002 It is frequently desirable to estimate the colour or normal at a point in the interior of a 3 or 4 vertex planar polygon given only the colour and normal at each of the vertices. My teacher recommended to use poly and conv function. The value of x at which y is desired, xdesired = 0. For a given set of points with no two values equal, the Lagrange polynomial is the polynomial of lowest degree that assumes at each value the corresponding value, so that the functions coincide at each point. Polynomial Interpolation Polynomials 𝑃𝑃 𝑛𝑛 𝑥𝑥= 𝑎𝑎 𝑛𝑛 𝑥𝑥 𝑛𝑛 +⋯ +𝑎𝑎 2 𝑥𝑥 2 +𝑎𝑎 1 𝑥𝑥+𝑎𝑎 0 are commonly used for interpolation. 93 KB #!/usr/bin/env python. Click here to do the Environment Modeling topic if you haven't already. ) \endgroup - Michael E2 7 hours ago. 4A For polynomial interpolation at the zeros of the. I have found a python code to plot these approximation as a graph, but how can I use these to find the approximated Langrange polynomium in the interval x in(0,3)? Here is the code: numerical-methods python lagrange-interpolation. Polynomial Interpolation; Piece-wise Interpolation; Spoiler: Natural Cubic Spline is under Piece-wise Interpolation. The function being estimated is the same as in previous sections:. Furthermore, I am not specifying the exact meaning of. edu 1Course G63. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. You will see updates in your activity feed. # Save the plot fig. interpolate packages wraps the netlib FITPACK routines (Dierckx) for calculating smoothing splines for various kinds of data and geometries. We do it in the following way: •Let. 223144 fx = lnx i x i f i g 0. Create a new le named Newton interpolant. Interpolation of an N-D curve¶ The scipy. 1 Interpolation and the Lagrange Polynomial SolutionsbyJonLoKimLin-Fall2014 MATH 104A HW 05 SOLUTION KEY For those who need a quick primer on programming, I highly recommend the python course by. Also, the weighted basis polynomials of each of the three methods are. Lagrange interpolation (or Lagrangian interpolation) method is one of the most basic and common methods to apply the interpolation polynomials. All practical interpolation methods will also involve a low pass filter. If the number did not appear in the table, then the two numbers above and below x were used, and interpolation provided the solution. Consequently y = f(x). Today we have Lagrange interpolation, again. If f is a polynomial of degree less than or equal to , the CGL quadrature formula is exact. Lagrange & Newton interpolation In this section, we shall study the polynomial interpolation in the form of Lagrange and Newton. NEAREST_INTERP_1D, a Python library which interpolates a set of data using a piecewise constant interpolant defined by the nearest neighbor criterion. y array of data: ydata = 1 1 3 LINEAR INTERPOLATION: x data chosen: x1 = 1, x2 = 0 , x3 = -2. Looking up Lagrange interpolation on Wikipedia, I found something new to me: the barycentric form of Lagrange interpolation. In numerical analysis, Lagrange polynomials are used for polynomial interpolation. It is useful at least in data analy-sis (interpolation is a form of regression), industrial design, signal processing (digital-to-analog conversion) and in numerical analysis. Objectives of Lagrange Interpolation The first goal of this section is to convert any set of tabulated data such as that found in Abramowitz_Stegun into. But let us explain both of them to appreciate the method later. How global polynomial interpolation works. Find the Lagrange Interpolation Formula given below, Solved Examples. In many real world applications of science and engineering, it is required to find the value of dependent variable corresponding to some value of independent variable by analyzing data which are obtained from some observation. Interpolation is the process of deriving a simple function from a set of discrete data points so that the function passes through all the given data points (i. interpolate. Earlier in Linear Interpolation Method Algorithm article we discussed about interpolation and we developed an algorithm for interpolation using Linear interpolation Method. # lagrange interpolation example from __future__ import print_function import math import numpy as np import matplotlib. data_fname = 'knot_points. ) Since , it follows that everywhere. lagrange is a Python package implementing likelihood models for geographic range evolution on phylogenetic trees, with methods for inferring rates of dispersal and local extinction and ancestral ranges. In this section, we shall study the interpolation polynomial in the Lagrange form. interpolate. Lagrange Interpolation Formula. This program help improve student basic fandament and logics. It is necessary because in science and engineering we often need to deal with. In this section, we shall study the polynomial interpolation in the form of Newton. Consequently y = f(x). Chapter 3 Interpolation Interpolation is the problem of tting a smooth curve through a given set of points, generally as the graph of a function. It is indeed equal to a constant that is '1'. We have from (2. Matlab Function for Lagrange Interpolation. Interpolation is going in the opposite direction, that is, estimating a value for the independent variable x, from the function, x = inverse( f(x) ). where is the barycentric weight, and the Lagrange interpolation can be written as: ( 24 ) We see that the complexity for calculating for each of the samples of is (both for and the summation), and the total complexity for all samples is. Computer Engineering. The remaining elements are indicated by <>. Hermite interpolation constructs an interpolant based not. Original data (dark) and interpolated data (light), interpolated using (top) forward filling, (middle) backward filling and (bottom) interpolation. If the trends, seasonality and longer term cycles are known then interpolation is easy. pyplot as plt from numpy. Compiled in DEV C++ You might be also interested in : Gauss Elimination Method Lagrange interpolation Newton Divided Difference Runge Kutta method method Taylor series method Modified Euler's method Euler's method Waddle's Rule method Bisection method Newton's Backward interpolation Newton's forward interpolation Newtons rapson. 60 gx f o xx– 1 xx– 2 xx– 3 x o – x 1 x o – x. There is a unique straight line passing through these points. McClarren, in Computational Nuclear Engineering and Radiological Science Using Python, 2018. All practical interpolation methods will also involve a low pass filter. Survey: Interpolation Methods in Medical Image Processing Thomas M. In the mathematical field of numerical analysis, a Newton polynomial, named after its inventor Isaac Newton, is the interpolation polynomial for a given set of data points in the Newton form. Khan Academy is a 501(c)(3) nonprofit organization. By voting up you can indicate which examples are most useful and appropriate. Sincepolynomialsaretypicallyrepresentedintheirexpandedformwithcoefficientsoneachof. rv variable stands for return value. Click here to review slope-intercept form of a line. can be arbitrary real or complex numbers, and in 1D can be arbitrary symbolic expressions. """ import numpy as np import matplotlib. Given a set of data-points , the Lagrange Interpolating Polynomial is a polynomial of degree , such that it passes through all the given data-points. The Lagrange form of polynomial interpolation is useful in some theoretical contexts and is easier to understand than other methods, however, it has some serious drawbacks that prevent it from being a useful method of interpolation. Visit Stack Exchange. Testing You can test the code by cloning the directory, entering it, and typing make test. edu 1Course G63. In this topic. This technique is the most easily implemented by humans (at least for linear and quadratic interpolating polynomials), but is also the. Given a set of data-points , the Lagrange Interpolating Polynomial is a polynomial of degree , such that it passes through all the given data-points. 2 Lagrange Polynomials. Lagrange interpolation, multivariate interpolation. Given two 1-D arrays x and w, returns the Lagrange interpolating polynomial through the points (x, w). b) Make a module Lagrange_poly2 containing the p_L, L_k, test_p_L, and graph functions. He is the author of the asciitable, cosmocalc, and deproject packages. 25) x: {0,1,2,3,4,5,6} f(x): {0,1,8,27,64,125,216} Output: Result after Lagrange interpolation f(3. He uses Python for Chandra spacecraft operations analysis as well as research on several X-ray survey projects. Lagrange Interpolation Method: Algorithm, Computation and Plot | Numerical Computing with Python - Duration: 18:28. You can define a function that does the job (here an older one from me used in lectures 2015), this goes over the Lagrange base polynomials. Corollary 6. Lagrange Method of Interpolation *****Input Data***** x array of data: xdata = 1 0 -2. Original data (dark) and interpolated data (light), interpolated using (top) forward filling, (middle) backward filling and (bottom) interpolation. Hermite interpolation constructs an interpolant based not. Shannon Hughes author of LAGRANGE'S INTERPOLATION METHOD FOR FINDING f(X) is from London, United Kingdom. Example: Approximate function by a polynomial of degree , based on the following points:. Return a Lagrange interpolating polynomial. Let's first explain the Lagrange polynomial, then we will proceed to the algorithm and the implementation. String interpolation is a process substituting values of variables into placeholders in a string. Polynomial Interpolation is the simplest and the most common type of interpolation. or create account below. Le but pour moi est de trouver une fonction polynôme passant par les points d'interpolations que j'aurai à me fixer. In the mathematical field of numerical analysis, a Newton polynomial, named after its inventor Isaac Newton, is the interpolation polynomial for a given set of data points in the Newton form. Given a set of n data points (xi,yi) where no two xi are the same,. McClarren, in Computational Nuclear Engineering and Radiological Science Using Python, 2018. Survey: Interpolation Methods in Medical Image Processing Thomas M. Testing You can test the code by cloning the directory, entering it, and typing make test. pyplot as plt # globals to control some behavior func_type = "tanh" # can be sine or tanh points = "variable" # can be variable or fixed npts = 15 def fun_exact(x): """ the exact function that we sample to get the points to interpolate through """ if func_type == "sine. NEAREST_INTERP_1D, a Python library which interpolates a set of data using a piecewise constant interpolant defined by the nearest neighbor criterion. Journal of Physics: Conference Series 1447 , 012002. W8V5 Python:Lagrange Interpolation 6:33. consider linear interpolation. Put this code in a file called lagrange. abedkime 13 août 2013 à 3:49:33. It is necessary because in science and engineering we often need to deal with. Lagrange interpolation is one of the best options. Lagrange's interpolation is also an degree polynomial approximation to f ( x ). The code is broken into five regions. Lagrange Interpolation Method Algorithm. Polynomial Interpolation Polynomials 𝑃𝑃 𝑛𝑛 𝑥𝑥= 𝑎𝑎 𝑛𝑛 𝑥𝑥 𝑛𝑛 +⋯ +𝑎𝑎 2 𝑥𝑥 2 +𝑎𝑎 1 𝑥𝑥+𝑎𝑎 0 are commonly used for interpolation. Named after Joseph Louis Lagrange, Lagrange Interpolation is a popular technique of numerical analysis for interpolation of polynomials. For any distinct complex numbers and any complex numbers , there exists a unique polynomial of degree less than or equal to such that for all integers, , and this polynomial is. The Newton polynomial is sometimes called Newton's divided differences interpolation polynomial because the coefficients of the polynomial are calculated. Corollary 6. It is necessary because in science and engineering we often need to deal with. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Although the data is evenly spaced in this example, it need not be so to use this routine. In standard output format, only the domain element of an InterpolatingFunction object is printed explicitly. Interpolation (scipy. This is an interpolation problem that is solved here using the Lagrange interpolating polynomial. Suppose we want yield rate for a four-years maturity bond, what shall we do? Solution: Draw a smooth curve passing through these data points (interpolation). Native implementation of the Lagrange interpolation algorithm over finite fields. Best of luck! (Oh, let me just say that polynomial interpolation of 90 points usually has numerical problems. Begin with n + 1 interpolation points evenly spaced in [0; 2pi]. For example, Figure 1 shows 4 points and a polynomial which passes through them. Create a new le named Newton interpolant. Purpose Native implementation of the Lagrange interpolation algorithm over finite fields. W8V5 Python:Lagrange Interpolation 6:33. where is the barycentric weight, and the Lagrange interpolation can be written as: ( 24 ) We see that the complexity for calculating for each of the samples of is (both for and the summation), and the total complexity for all samples is. %L is the function which will be used to find the approximating function. from matplotlib import pyplot as plt. So, it may be po. (Suggestion: Look at a Python code provided with this homework. By voting up you can indicate which examples are most useful and appropriate. The following code takes in a single value, x, and a list of points, X, and determines the value of the Lagrange polynomial through the list of points at the given x value. (5) This property makes it possibly to determine the interpolation polynomial without solving a linear system of equations. Let fx ign 0 be distinct real numbers and let fy ign be real. The interpolation problem attempts to nd a function p(x) with the property p(x i) = y i for all i. In summary, the Lagrange form of the interpolating polynomial is useful theoretically because it does not require solving a linear system explicitly shows how each data value f. Efficient calculation of the barycentric polynomial interpolant requires that the function to be interpolated be sampled at points from a known. This technique is the most easily implemented by humans (at least for linear and quadratic interpolating polynomials), but is also the. Testing You can test the code by cloning the directory, entering it, and typing make test. So, I am trying create a stand-alone program with netcdf4 python module to extract multiple point data. The Foundation region is where the parent Interpolation class is defined. mechtutor com 568 views. PolynomialInterpolationPolynomial Interpolation Thepolynomialinterpolationproblemistheproblemofconstructingapolynomialthatpassesthroughor interpolatesn+1datapoints(x0. 1000 loops, best of 3: 1. From Process Model Formulation and Solution: 3E4 < Software tutorial. This presents a problem in most \real" applications, in which functions are used to model relationships between quantities,. Lagrange Cubic Interpolation Using Basis Functions • For Cubic Lagrange interpolation, N=3 Example • Consider the following table of functional values (generated with ) • Find as: 0 0. He uses Python for Chandra spacecraft operations analysis as well as research on several X-ray survey projects. In that sense, in Section 2 we consider the construction of the unique Lagrange interpolating polynomial on a set of interpolating nodes on several radial rays. 223144 fx = lnx i x i f i g 0. 3 Newton's Form of the Interpolation Polynomial One good thing about the proof of Theorem 2. There are different method, for example Lagrangian interpolation or Barycentric Lagrange Interpolation. pyplot as plt from numpy. %It is a matrix and is filled with 1s for multiplication purposes. Technically, I am not posting about a different method, but just using the same algorithm for interpolation. Polynomial interpolation¶ This example demonstrates how to approximate a function with a polynomial of degree n_degree by using ridge regression. The interpolation problem attempts to nd a function p(x) with the property p(x i) = y i for all i. I hope the question makes sense. The problem with having lots of data, especially if it's roughly equally spaced apart, is that polynomial interpolation suffers from Runge's Phen. Given a set of (n+1) data points and a function f, the aim is to determine a polynomial of degree n which interpolates f at the points in question. There is a unique straight line passing through these points. Nominators and denominators fo the base-polynomials are calculated and used to build ab the interpolation polynomial. lagrange taken from open source projects. 93 KB #!/usr/bin/env python. It's a whole a lot easier than Newton's divided differences interpolation polynomial , because there is no divided difference part that need a recursive function. Polynomial Interpolation. """ Demonstration module for quadratic interpolation. We can calculate the interpolated values directly with the interpolation functions:. I believe your interpolation example is in fact a prediction example and not interpolation. 5 is repeated the data above; I assume it's a typo, otherwise you won't be able to do Lagrange interpolation, which requires distinct x-coordinates. When i extract data, result values are all the same! All values are -9. 5 Interpolation. The biggest drawback was implementing my own polynomial class. All content in this area was uploaded by Xue-Zhang Liang on May 26, 2017. Langrange polynomial interpolation. Python library with a basic native implementation of Lagrange interpolation over finite fields. The Foundation region is where the parent Interpolation class is defined. Python String Interpolation In this article we will learn about the python string interpolation. to implement scilab program for lagrange interpolation. Numerical Methods I Polynomial Interpolation Aleksandar Donev Courant Institute, NYU1 [email protected] Interpolation. Re: Polynomial interpolation On Mon, Apr 28, 2008 at 09:20:48AM -0700, Ed Rahn wrote: > The group of people who use scipy is much greater than you and your > colleagues. The ℓk(x) are known as Lagrange polynomials. from_derivatives. The purpose of this paper is to give a local tricubic interpolation scheme in three dimensions that is both C1 and isotropic. 4) x k+1 = x k 1 1 2 (x k 1 x. y array of data: ydata = 1 1 3 LINEAR INTERPOLATION: x data chosen: x1 = 1, x2 = 0 , x3 = -2. InterpolatingFunction works like Function. where is the barycentric weight, and the Lagrange interpolation can be written as: ( 24 ) We see that the complexity for calculating for each of the samples of is (both for and the summation), and the total complexity for all samples is. In the special case of the first-kind Chebyshev polynomials, the preceding lemma gives the following specific result. An interpolation on two points, (x0, y0) and (x1, y1), results in a linear equation or a straight line. For simple Lagrange interpolation, the main task is to evaluate the. In this section, we shall study the polynomial interpolation in the form of Newton. C code to implement Lagrange interpolation method. 1 Introduction. Given a set of points ( xi, yi ) for i = 0, 1, 2, , n, we want to find a function (usually a polynomial) which passes through all of the points. Straight forward interpolating polynomials. Polynomial interpolation is a method of estimating values between known data points. interpolate. In numerical analysis, Lagrange polynomials are used for polynomial interpolation. Compute the coefficients of the polynomial interpolating the points (xi[i],yi[i]) for i = 0,1,2. Returns the same object type as the caller, interpolated at some or all NaN values. Advantages of Lagrange's Interpolation Method. There are several approaches to polynomial interpolation, of which one of the most well known is the Lagrangian method. Lagrange Interpolation The computations in this small article show the Lagrange interpolation. Implementing Linear and Cubic Spline Interpolation in C#. The two pictures below were generated using this python code to compare the Lagrange interpolating polynomial and Spline Interpolation using 5 data points. interpolate packages wraps the netlib FITPACK routines (Dierckx) for calculating smoothing splines for various kinds of data and geometries. In that sense, in Section 2 we consider the construction of the unique Lagrange interpolating polynomial on a set of interpolating nodes on several radial rays. Lagrange Method of Interpolation - More Examples. An interpolation on two points, (x0, y0) and (x1, y1), results in a linear equation or a straight line. Best of luck! (Oh, let me just say that polynomial interpolation of 90 points usually has numerical problems. Lagrange Polynomial Interpolation is useful in Newton-Cotes Method of numerical integration. Let fx ign 0 be distinct real numbers and let fy ign be real. A second order polynomial interpolation. Installing ActivePython is the easiest way to run your project. y data chosen: y1 = 1, y2 = 1 , y3 = 3. Lagrange Interpolation (curvilinear interpolation) The computations in this small article show the Lagrange interpolation. Python String Interpolation In this article we will learn about the python string interpolation. The concept of interpolation can be shown in series analysis and regression analysis in statistics. Specifically, it gives a constructive proof of the theorem below. or create account below. The straight line we get using linear X and Y (blue in the charts below) is not the same as the straight line we get when our X axis is logarithmic (orange). We shall resort to the notion of divided differences. InterpolatingFunction […] [x] finds the value of an approximate function with a particular argument x. The formula can be derived from the Vandermonds determinant but a much simpler way of deriving this is from Newton's divided difference formula. Piecewise polynomial in the Bernstein basis. He did not,. In this method, one of the variables is forced to be constant and, with another variable, the Lagrange polynomials can be written by using the given data. In this situation, g(x, y, z) = 2x + 3y - 5z. lagrange's inverse interpolation method Basic GAUSS ELIMINATION METHOD, GAUSS ELIMINATION WITH PIVOTING, GAUSS JACOBI METHOD, GAUSS SEIDEL METHOD Program to construct Newton's Forward Difference Interpolation Formula from the given distinct equally spaced data points. We follow the procedure given by (2. A matrix of the form of $$A$$ is called Vandermonde matrix. format (), string. All practical interpolation methods will also involve a low pass filter. Repeat the previous problem using linear, quadratic, third order, and fourth Lagrange polynomial interpolation. Polynomial interpolation will always be of an order one less than the number of points used; it will always go through the basis points you use to create the interpolation. L=ones(m,length(x));. Consequently y = f(x). It follows from (5) that the interpolation polynomial. Lagrange Interpolation. It has the drawback that exceptional cases may cause overflow or underflow. Let ; and for. The interpolation nodes are given as: is the ith Lagrange base polynomial of degree N. Also the x-coordinate 230. Polynomials of degree 3 are cubic functions. for plotting $$p_L(x)$$ in Exercise 25: Implement Lagrange's interpolation formula, based on interpolation points taken from some mathematical function $$f(x)$$ represented by the argument f. The Lagrange interpolation formula is a way to find a polynomial which takes on certain values at arbitrary points. Installing ActivePython is the easiest way to run your project. 3 Newton’s Form of the Interpolation Polynomial One good thing about the proof of Theorem 2. Lagrange interpolation: Runge phenomenon. savefig ('polynomial_interpolation_Python.$ The app below calculates the polynomial fit through the series of points given in the text box. First, Lagrange interpolation is O(n2) where other. The algorithms use their respective interpolation/basis functions, so are capable of producing curves of any order. %It is a matrix and is filled with 1s for multiplication purposes. We discuss the remedies for this, including: optimal distribution of. The Lagrange polynomial, displayed in red, has been calculated using this class. For example, if we have two data points, then we can fit a polynomial of degree 1 (i. Suppose we want yield rate for a four-years maturity bond, what shall we do? Solution: Draw a smooth curve passing through these data points (interpolation). Hence this 'constraint function' is generally denoted by g(x, y, z). Template and f-strings. Lagrange interpolation (or Lagrangian interpolation) method is one of the most basic and common methods to apply the interpolation polynomials. • Same format as all other interpolants • Function diff finds difference of elements in a vector • Find appropriate sub-interval •Evaluate • Jargon: x is called a “knot” for the linear spline interpolant. For instance, if you have a template for saying. In this method, one of the variables is forced to be constant and, with another variable, the Lagrange polynomials can be written by using the given data. Content uploaded by Xue-Zhang Liang. Lagrange polynomials are the simplest way to interpolate a set of points. Like Like Reply. This software implements methods described in Ree, R H and S A Smith. You may receive emails, depending on your notification preferences. There are several approaches to polynomial interpolation, of which one of the most well known is the Lagrangian method. In summary, the Lagrange form of the interpolating polynomial is useful theoretically because it does not require solving a linear system explicitly shows how each data value f. Here are the examples of the python api scipy. Interpolation is a useful mathematical and statistical tool used to estimate values between two points. Testing You can test the code by cloning the directory, entering it, and typing make test. When graphical data contains a gap, but data is available on either side of the gap or at a few specific points within the gap, an estimate of values within the gap can be made by interpolation. The code computes y -coordinates of points on a curve given their x -coordinates. It is the process of finding a value between two points on a line or a curve. (b) Plot your interpolating polynomial and include the four points clearly on the plot. NEXT Polynomial functions and derivative (1): Linear functions. Even linear interpolation can be interpreted as a low pass filter, it's just a very bad one. or create account below. Polynomial interpolation¶ This example demonstrates how to approximate a function with a polynomial of degree n_degree by using ridge regression. We introduce the fundamentals of the spectral-element method developing a. C/C++ program to Lagrange's Interpolationwe are provide a C/C++ program tutorial with example. The algorithm is based on a specific 64×64 matrix that gives the relationship between the derivatives at the corners of the elements and the coefficients of the tricubic interpolant for this element. solve problems using Lagrangian method of interpolation, and 3. Lagrange polynomials are used for polynomial interpolation. In this article, I am using C# for coding. There are different method, for example Lagrangian interpolation or Barycentric Lagrange Interpolation. In the special case of the first-kind Chebyshev polynomials, the preceding lemma gives the following specific result. interpolate packages wraps the netlib FITPACK routines (Dierckx) for calculating smoothing splines for various kinds of data and geometries. Here are the examples of the python api scipy. Le but pour. This piece of code is a Matlab/GNU Octave function to perform Lagrange interpolation. Looking for the full power of Python 3? Check out our Python 3 Trinket. Survey: Interpolation Methods in Medical Image Processing Thomas M. The derivative of a lineal function is a constant function. Learning a. (2020) Barycentric rational interpolation method for numerical investigation of magnetohydrodynamics nanofluid flow and heat transfer in nonparallel plates with thermal radiation. Piecewise cubic polynomials (Akima interpolator). But let us explain both of them to appreciate the method later. Given a set of (n+1) data points and a function f, the aim is to determine a polynomial of degree n which interpolates f at the points in question. A matrix of the form of $$A$$ is called Vandermonde matrix. Use MathJax to format equations. We can consider the polynomial function that passes through a series of points of the plane. and a function lagrange_polynomial(z,x,y) with: Input: z, point (or array of points) where we will evaluate the polynomial and the coordinates of the nodes x and y. In the special case of the first-kind Chebyshev polynomials, the preceding lemma gives the following specific result. We can write the formula for a straight line as P1(x)=a0 + a1x In fact, there are other more convenient ways. Lagrange Interpolation Formula. interpolation type of this function is the typical interpolation of the old logarithms table. Interpolation Calculator. The interpolation problem is to construct a function Q(x) that passes through these points, i. 1 is that it is constructive. The Newton polynomial is sometimes called Newton's divided differences interpolation polynomial because the coefficients of the polynomial are calculated. W8V5 Python:Lagrange Interpolation 6:33. Polynomial Interpolation is the simplest and the most common type of interpolation. ; domain specifies the domain of the data from which the InterpolatingFunction was constructed. where are the data-points. This implies that $$\displaystyle p(x) = \sum_{i=0}^n y_i \cdot L_i(x)$$ is an interpolation of our data points. The function utilizes the rSymPy library to build the interpolating polynomial and approximate the value of the function f for a given value of x. Tridiagonal Matrix region defines a Tridiagonal class to solve a system of linear equations. In this topic. The concept of interpolation can be shown in series analysis and regression analysis in statistics. Mathematical interpolation theory considers a function f, defined on a regular grid N. The interpolation problem attempts to nd a function p(x) with the property p(x i) = y i for all i. For the Lagrange interpolation, we have to follow this equation. Featured Examples — click an image to try it out! Want to use this to teach? Sign up for trinket! Log in with Edmodo. * Regression: Here we try to fit a specific form of curve to the given data points. Cubic Spline Interpolation Sky McKinley and Megan Levine Math 45: Linear Algebra Abstract. Let's first explain the Lagrange polynomial, then we will proceed to the algorithm and the implementation. METHOD OF QUADRATIC INTERPOLATION 3 The minimizer of qis easily found to be 0b=2aby setting q(x) = 0. W8V6 Numerical Integration 7:31. But before applying Lagrange Multiplier method we should make sure that g(x, y, z) = c where 'c' is a constant. It represents the text that will be interpolated into the document when our snippet is. ndarrays so I could do easy plotting. I found that SciPy implements a polynomial class. String interpolation is a process substituting values of variables into placeholders in a string. It is called multivariate since the data points are supposed to be sampled from a function of several variables. Dictionary meaning of interpolation is the estimation of an unknown quantity between two known quantities. • It is also possible to set up specialized Hermite interpolation functions which do not include all functional and/or derivative values at all nodes • There may be some missing functional or derivative values at certain nodes. Mathematical interpolation theory considers a function f, defined on a regular grid N. The lagrange_interp_ND series of functions are global interpolators and should be used only if your grid points are stable for high-order interpolation. Lagrange Interpolation Polynomial – C PROGRAM. for plotting $$p_L(x)$$ in Exercise 25: Implement Lagrange's interpolation formula, based on interpolation points taken from some mathematical function $$f(x)$$ represented by the argument f. This is a bit out-of-date; we'll try to update it when we can. from matplotlib import pyplot as plt. One easy way of obtaining such a function, is to connect the. In other words, we can use the proof to write down a formula for the interpolation polynomial. 25) x: {0,1,2,3,4,5,6} f(x): {0,1,8,27,64,125,216} Output: Result after Lagrange interpolation f(3. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia. interpolate. LINEAR INTERPOLATION The simplest form of interpolation is probably the straight line, connecting two points by a straight line. Sopasakis: FMN050/FMNF01-2015 86. Let's first explain the Lagrange polynomial, then we will proceed to the algorithm and the implementation. By voting up you can indicate which examples are most useful and appropriate. Lagrange I n terpolat io. The remaining elements are indicated by <>. Lagrange interpolation, multivariate interpolation. PREVIOUS Polynomial Functions (3): Cubic functions. """ Demonstration module for quadratic interpolation. I used scipy. (2020) Barycentric Lagrange interpolation for solving Volterra integral equations of the second kind. The Lagrange polynomial, displayed in red, has been calculated using this class. You must implement a interpolation what you do by hand when interpolate. abedkime 13 août 2013 à 3:49:33. There are several approaches to polynomial interpolation, of which one of the most well known is the Lagrangian method. In this article, I am using C# for coding. 6 Open Newton-Cotes Formula See Figure 4. Interpolation Calculator. Sahni (Computer algorithms in C++) has an understandable implementation of the algorithm. The Lagrange interpolation formula is a way to find a polynomial which takes on certain values at arbitrary points. interpolation type of this function is the typical interpolation of the old logarithms table. Thanks for contributing an answer to Code Review Stack Exchange! Please be sure to answer the question. You may receive emails, depending on your notification preferences. Khan Academy is a 501(c)(3) nonprofit organization. 1) are satisfied (see Figure 3. So the function delivers all the Lagrange base-polynomials. This technique is the most easily implemented by humans (at least for linear and quadratic interpolating polynomials), but is also the. 25) x: {0,1,2,3,4,5,6} f(x): {0,1,8,27,64,125,216} Output: Result after Lagrange interpolation f(3. It is one of those. Lecture 3: The Runge Phenomenon and Piecewise Polynomial Interpolation (Compiled 16 August 2017) In this lecture we consider the dangers of high degree polynomial interpolation and the spurious oscillations that can occur - as is illustrated by Runge's classic example. Try changing a data point in the data to see how the interpolation function changes. Python script to interpolate with Lagrange method. Or copy & paste this link into an email or IM:. First we'll use the slope intercept form of a line to define each frame along a straight line. I have tried this code. The double prime notation in the summation indicates that the first and last terms are halved. In this method, one of the variables is forced to be constant and, with another variable, the Lagrange polynomials can be written by using the given data. For instance, if you. And in another article Linear Interpolation Method Pseudocode, we developed pseudocode for this method. Advantages for using polynomial: efficient, simple mathematical operation such as differentiation and integration. The interpolants Pn(x) oscillated a great deal, whereas the function f(x) was nonoscillatory. 93 KB #!/usr/bin/env python. When the data points $$x_i$$ are mutually different, it is known that the Vandermonde matrix is invertible (). The function being estimated is the same as in previous sections:. The people who use scipy do so because it uses python. In this article, I am using C# for coding. Named after Joseph Louis Lagrange, Lagrange Interpolation is a popular technique of numerical analysis for interpolation of polynomials. However, first we need to convert the read dates to datetime format and set them as the index of our dataframe: df = df0. I found that SciPy implements a polynomial class. cpp that contains two functions: Matrix Newton_coefficients(Matrix& x, Matrix& y);. interpolate. InterpolatingFunction works like Function. ndarrays so I could do easy plotting. Create a new le named Newton interpolant. pyplot as plt # globals to control some behavior func_type = "tanh" # can be sine or tanh points = "variable" # can be variable or fixed npts = 15 def fun_exact(x): """ the exact function that we sample to get the points to interpolate through """ if func_type == "sine. Scilab Program / Source Code: The following is the source code of scilab program for polynomial interpolation by numerical method known as lagrange interpolation. If f is a polynomial of degree less than or equal to , the CGL quadrature formula is exact. He is the author of the asciitable, cosmocalc, and deproject packages. But let us explain both of them to appreciate the method later. a guest May 15th, 2014 2,132 Never Not a member of Pastebin yet? it unlocks many cool features! raw download clone embed report print diff Python 0. Given some data points {xi, yi}, the aim is to find a polynomial which goes exactly through these points. It is useful at least in data analy-sis (interpolation is a form of regression), industrial design, signal processing (digital-to-analog conversion) and in numerical analysis. So the solution exists and is unique $$\blacksquare$$. 2d Lagrange Interpolation : Directly to the input form: Polynomial interpolation in higher dimensions is in the case of complete rectangular grids as easy as in one dimension, especially if one uses the representation by Lagrange's basis polynomials. GitHub Gist: instantly share code, notes, and snippets. So, it may be po. Lagrange Interpolation. The polynomial interpolations generated by the power series method, the Lagrange and Newton interpolations are exactly the same, , confirming the uniqueness of the polynomial interpolation, as plotted in the top panel below, together with the original function. In numerical analysis, Lagrange polynomials are used for polynomial interpolation. Lagrange Interpolation python Search and download Lagrange Interpolation python open source project / source codes from CodeForge. Python scripts can be embedded inside UltiSnips snippets using !p. He did not,. Corollary 6. Lagrange Interpolation Calculator. Or copy & paste this link into an email or IM:. ; InterpolatingFunction […] [x] finds the value of an approximate function with a particular argument x. ) Since , it follows that everywhere. format(), string. The main difference between these two is that in interpolation we need to exactly fit all the data points whereas it's not the case in regression. format (), string. Regarding number of lines we have: 34 in Python and 37 in Julia. 60 gx f o xx– 1 xx– 2 xx– 3 x o – x 1 x o – x. Let's have a look how to implement Lagrange polynomials and interpolation with Lagrange polynomials on the computer using Python. The Lagrange polynomial, displayed in red, has been calculated using this class. Lagrange Method of Interpolation - More Examples. Clearly there. Working C C++ Source code program for Lagrange's interpolation /***** Lagrange's interpolation *****/ #include< Object tracking in Java - detect position of colored spot in image Red spot in image - position to be detected later Object Tracking plays important role in Image Processing research projects. LINEAR INTERPOLATION The simplest form of interpolation is probably the straight line, connecting two points by a straight line. Python scripts can be embedded inside UltiSnips snippets using !p. Interpolation (scipy. linalg import solve def quad_interp (xi, yi): """ Quadratic interpolation. Newton Interpolation: A C++ function Lagrange() for evaluating the polynomial interpolant of a set of data points using the Lagrange basis has been provided on the course web page in the le Lagrange. PIECEWISE POLYNOMIAL INTERPOLATION exploit the systematic "migration" of the evaluation point as it moves left to right across the subintervals. Given a set of n data points (xi,yi) where no two xi are the same,. Follow 725 views (last 30 days) Muhammed Ahkbar on 4 May 2017. Polynomial interpolation is the interpolation of a given data set by a polynomial. Matlab Code for Lagrange Interpolation. Nominators and denominators fo the base-polynomials are calculated and used to build ab the interpolation polynomial. We introduce the fundamentals of the spectral-element method developing a. data_fname = 'knot_points. We see that they indeed pass through all node points at , , and. By construction, on. Lagrange I n terpolat io. I don't think you can say splines are always better, but for a lot of data sets it can be beneficial. the latter only guarantees continuity of the zeroeth derivative (the interpolated function itself). Oblivious you will use the pit hon data structures and so on, but is possible. (Suggestion: Look at a Python code provided with this homework. Interpolation gives us the coordinates of a point on a straight line between two known points. This leads us to consider the function , the sum of the absolute values of the Lagrange basis polynomials. Then, this value can be complicated for the nal form of 2D-Lagrange interpolation. METHOD OF QUADRATIC INTERPOLATION 3 The minimizer of qis easily found to be 0b=2aby setting q(x) = 0. You will use Lagrange's interpolation formula to interpolate sin x over the range [0; 2pi]. Find more on LAGRANGE'S INTERPOLATION METHOD FOR FINDING f(X) Or get search suggestion and latest updates. Shamir's Secret Sharing is an algorithm in cryptography created by Adi Shamir. lagrange(x, w)¶. 3 Newton’s Form of the Interpolation Polynomial One good thing about the proof of Theorem 2. Original data (dark) and interpolated data (light), interpolated using (top) forward filling, (middle) backward filling and (bottom) interpolation. Lagrange interpolation consists of computing the Lagrange basis functions then combining themwiththey-values. Applicable for unequally spaced values of x, this program for Lagrange interpolation in C language is short and simple to understand. from matplotlib import pyplot as plt. There are different method, for example Lagrangian interpolation or Barycentric Lagrange Interpolation. 5) may be written in the form i(x)= φ n+1(x) (x−x i)φ n+1 (x i), (6. Testing You can test the code by cloning the directory, entering it, and typing make test. The problem is to find f in a continuum that includes N. They are of degree n−1. The purpose of this paper is to give a local tricubic interpolation scheme in three dimensions that is both C1 and isotropic. Find the Lagrange Interpolation Formula given below, Solved Examples. This is an argument cover in all the books of numerical analysis for university level. No installation required. Template and f-strings. y data chosen: y1 = 1, y2 = 1 , y3 = 3. An interpolation on two points, (x0, y0) and (x1, y1), results in a linear equation or a straight line. The code is broken into five regions. reproduces the data points exactly) and can be used to estimate data points in-between the given ones. The Foundation region is where the parent Interpolation class is defined. poly1d([0]). data_fname = 'knot_points. the former guarantees continuity of both the zeroeth and first derivative. Thanks for contributing an answer to Code Review Stack Exchange! Please be sure to answer the question. Mathematical interpolation theory considers a function f, defined on a regular grid N. Lagrange Interpolation on a Sphere. Video created by Ludwig-Maximilians-Universität München (LMU) for the course "Computers, Waves, Simulations: A Practical Introduction to Numerical Methods using Python". This leads us to consider the function , the sum of the absolute values of the Lagrange basis polynomials. In this context, UltiSnips predefines a few Python objects and variables for us. I want to describe a visual tool to help you investigate this question yourself. Interpolation par la méthode de Lagrange Liste des forums; Rechercher dans le forum. Akima1DInterpolator. Polynomial Interpolation; Piece-wise Interpolation; Spoiler: Natural Cubic Spline is under Piece-wise Interpolation. 1 Chapter 05. ; domain specifies the domain of the data from which the InterpolatingFunction was constructed. i think the bicubic interpolation is more likely 3rd-order Hermite polynomial than 3rd-order Lagrange polynomial interpolation. The function values and sample points , etc. derive Lagrangian method of interpolation, 2. Returns the same object type as the caller, interpolated at some or all NaN values. String interpolation is a process substituting values of variables into placeholders in a string. Purpose Native implementation of the Lagrange interpolation algorithm over finite fields. Shannon Hughes author of LAGRANGE'S INTERPOLATION METHOD FOR FINDING f(X) is from London, United Kingdom. rv variable stands for return value. The function utilizes the rSymPy library to build the interpolating polynomial and approximate the value of the function f for a given value of x. Try changing a data point in the data to see how the interpolation function changes. The call to test_p_L described in Exercise 25: Implement Lagrange's interpolation formula and the call to graph described above should appear in the module's test block. By construction, on. Interpolation is the process of deriving a simple function from a set of discrete data points so that the function passes through all the given data points (i. The Lagrange’s Interpolation formula: If, y = f(x) takes the values y0, y1, …, yn corresponding to x = x0, x1, …, xn then, This method is preferred over its counterparts like Newton’s method because it is applicable even for unequally spaced values of x. In the mathematical field of numerical analysis, a Newton polynomial, named after its inventor Isaac Newton, is the interpolation polynomial for a given set of data points in the Newton form. | 2020-07-09T09:05:15 | {
"domain": "asainfo.it",
"url": "http://zgoe.asainfo.it/lagrange-interpolation-python.html",
"openwebmath_score": 0.5778627395629883,
"openwebmath_perplexity": 897.424159526933,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9759464513032269,
"lm_q2_score": 0.8615382058759129,
"lm_q1q2_score": 0.8408151546867461
} |
https://math.stackexchange.com/questions/754827/does-a-15-puzzle-always-have-a-solution | # Does a 15-puzzle always have a solution
For those that are not familiar with (this type of) sliding puzzles, basically you have a number of tiles on a board equal to (n * m) - 1 (possibly more holes if you want). The goal is to re-arrange the tiles in such a way that solves the puzzle.
The puzzle could be anything, from number games to images.
While writing a small app for this, I found that if I were to initialize the puzzle by randomly shuffle all of the pieces, I could end up in a situation where there is no solution if my puzzle was 2x2.
So the problem I have is: given a sliding puzzle with n-by-m dimensions, is there always a solution if there are a sufficient number of tiles (eg: a 3x3 board)? How would I even begin to prove this, or simply convince myself that it is the case?
If it is possible that a random shuffle could result in no-solution, then I'd have to figure out how to verify that there exists a solution, which is a completely different beast.
• If it started in a solved position, simply doing the reverse of all the random moves would bring you back to it.
– VBCPP
Apr 14, 2014 at 16:53
• Wikipedia -> 15 puzzle -> Solvability: "Johnson & Story (1879) used a parity argument to show that half of the starting positions for the n-puzzle are impossible to resolve, no matter how many moves are made..."
– gnat
Apr 14, 2014 at 16:56
• Thanks, I didn't know there was a specific name for this class of puzzles. Apr 14, 2014 at 17:36
• If I'm understanding the conclusions correctly, if I just randomly swap two tiles at a time, for an even number of times, I should have a solvable configuration. Apr 14, 2014 at 18:07
• This question appears to be off-topic because it is about puzzles not programming. While algorithms are used to find (optimal) solutions to puzzles, this question is asking if a solution always exists.
– GlenH7
Apr 14, 2014 at 22:40
Martin Gardner had a very good writeup on the 14-15 puzzle in one of his Mathematical Games books.
Sam Loyd invented the puzzle. He periodically posted rewards for solutions to certain starting configurations. None of those rewards were claimed.
Much analysis was expended, and it was finally determined, through a parity argument (as mentioned in the comment above), that half of the possible starting configurations were unsolvable. Interestingly, ALL of Loyd's reward configurations were unsolvable.
SO: No, every possible configuration is not solvable. If you START with a solved puzzle, and apply only legal transformations (moves) to it, you always wind up at a solvable configuration.
For the GENERAL nxm question, you'd probably have to expand the parity argument.
• You don't have to expand the parity argument at all. The puzzle is solvable precisely when the parity of the configuration is even. This applies to general $n \times m$ as well, as long as $n,m \ge 3$ so you have enough room. Apr 15, 2014 at 14:24
Just a hint. Once I had to show with basic algebra (permutation groups) that a standard $15$ puzzle had no solution. The idea there was the following:
to rearrange the puzzle, you have to perform a permutation of the $15$ tiles.
Now, notice that once you write any permutation allowed, it is written as a product of an even number of $2$-cycles (you always move the "empty" tile, it starts in the corner and it has to be still there at the end of your moves). Hence permutation with sign $-1$ are not allowed.
In my case it was enough to conclude the puzzle had no solution (I had to perform an exchange between two tiles, so it had sign $-1$). Maybe it can help you to exclude some configurations.
The solvability of an n puzzle can be tested after shuffling by computing the permutations of the puzzle. "While odd permutations of the puzzle are impossible to solve, all even permutations are solvable."
For the math behind this, please see http://mathworld.wolfram.com/15Puzzle.html
• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review Oct 3, 2016 at 3:14
• The answer is useful and is summarized in "While odd permutations of the puzzle are impossible to solve, all even permutations are solvable." The provided link is a reliable one. Oct 3, 2016 at 15:00
Yes, it will always have a solution as long as you start with a good solution and then make legal moves to randomize the tiles.
BUT if you, for example, pop two pieces out and switch them, then you can get an insolvable puzzle. ;) As you discovered.
Here, we consider the following problem: For a given position on the board to say whether there is a sequence of moves leading to a decision or not.
Let there be given a position on the board:
+------------------+.
| 11 | 15 | 2 | 13 |.
|----|----|---|----|.
| 14 | 1 | 8 | 3 |.
|----|----|---|----|.
| 7 | 6 | 0 | 10 |.
|----|----|---|----|.
| 4 | 12 | 9 | 5 |.
+------------------+.
wherein one of the elements is zero and indicates an empty cell. Consider the numbers in the matrix serially (rowwise):
11 12 2 13 14 1 8 3 7 6 0 10 4 12 9 5
Denote N - the number of inversions in the permutation (i.e. the number of such elements a[i] and a[j] that i < j but a[i] > a[j]).
Next, let K- line number in which there is an empty element (i.e. in our notation K = 3).
Then, a solution exists if and only if N + K is even.
• I tried to edit your post to make the math formatting work, but I failed (partly because I could not understand some of it). Have a look at the introduction to posting mathematical expressions. May 31, 2017 at 2:08
• Thanks @hardmath. Fixed the formatting. Aug 2, 2017 at 12:23 | 2023-03-30T10:38:32 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/754827/does-a-15-puzzle-always-have-a-solution",
"openwebmath_score": 0.625257670879364,
"openwebmath_perplexity": 449.4043882564316,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9759464471055738,
"lm_q2_score": 0.8615382076534743,
"lm_q1q2_score": 0.8408151528051123
} |
https://math.stackexchange.com/questions/3990081/number-of-ways-to-group-students-with-certain-conditions | # Number of ways to group students with certain conditions
So i have this specific question and i really can't seem to wrap my head around it. The question is as below;
There are 4 girls in a class of 16 students where 2 of the girls are twins. If we want to split them into two equal groups how many different ways can we do so for
a) when there is no restriction,
In this case i think its just 16C8
b) all girls to be in the same group
for this i think its 4C4 x 12C4
c) not wanting the twins in the same group,
This one im not too sure how to go about it.
I've always had trouble with understanding permutations and combinations and this one really got me thinking, i would appreciate any explanation i could get the answer doesn't really concern me but if i can finally understand what's happening here i would be more than grateful.
• I think you have (a) and (b) correct. For (c), build the two groups by placing the twins first (one in each group), and then arrange the remaining 14 people to complete the groups. Jan 18 at 13:32
• @JaapScherphuis The groups are not labeled, so the answer to (a) is not correct. Jan 18 at 13:39
Your a) is nearly correct, $$\binom {16}{8}$$ chooses the children to go in one group but the remaining children can also be selected in the same way; so you can get the same division twice. You can adjust for this by dividing by two (correct), $$\frac 12 \binom {16}{8}$$, or by picking an arbitrary child as being on one team and choosing the other $$7$$ chlldren to join her, $$\binom {15}7$$
Your b) is correct but a little more complicated than it needs to be: $$\binom 44 = 1$$ ("the girls are all on one team"), so the answer calculates as expected to be just $$\binom {12}4$$ ("choose the $$4$$ boys that are on the same team with the girls"). Here the group with the girls is thus labelled and there is no ambiguity about the choice as previously.
Solution to c) can start by allocating the twins one to each group (which are now therefore distinct eg. "Brenda's team" and "Glenda's team") and picking the other $$7$$ children to be on one of these.
There are $$4$$ girls in a class of $$16$$ students where $$2$$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when there is no restriction?
You have counted each group twice, once when you select the group and once when you select its complement since choosing $$8$$ of the $$16$$ students to be in one group also determines which $$8$$ students are in the other group. The answer should be $$\frac{1}{2}\binom{16}{8}$$
Another way to see this is to suppose James is one of the members. We must choose which $$7$$ of the other $$15$$ students must be in the same group as James. Then the other eight students must be in the other group. Hence, there are $$\binom{15}{7}$$ ways to divide the students into two groups of equal size.
As you can verify, $$\frac{1}{2}\binom{16}{8} = \binom{15}{7}$$
There are $$4$$ girls in a class of $$16$$ students where $$2$$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when all the girls are placed in the same group?
There are $$4$$ girls in a class of $$16$$ students where $$2$$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when the twins are not placed in the same group.
Suppose Katharine is one of the twins. Since her twin cannot be in the same group, we must choose $$7$$ of the other $$14$$ students to be in the same group as Katharine, which can be done in $$\binom{14}{7}$$ ways. The remaining students must in the group with Katharine's twin. | 2021-12-02T00:31:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3990081/number-of-ways-to-group-students-with-certain-conditions",
"openwebmath_score": 0.660674512386322,
"openwebmath_perplexity": 180.67617193293165,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9759464443071381,
"lm_q2_score": 0.8615382040983515,
"lm_q1q2_score": 0.8408151469245435
} |
http://mathhelpforum.com/number-theory/52508-fermats-little-theorem-2-a-print.html | # Fermat's Little Theorem 2
• October 7th 2008, 05:06 PM
bigb
Fermat's Little Theorem 2
Use Fermat's Little Theorem to compute 29^202 mod 13
• October 7th 2008, 06:53 PM
o_O
By Fermat's theorem: $a^{12} \equiv 1 \ (\text{mod } 13)$
Note that $29 \equiv 3 \ (\text{mod } 13)$.
So: $\left(3^{12}\right)^{17} \equiv 1^{17} \ (\text{mod } 13)$
$\iff 3^{204} \equiv 1 \ (\text{mod } 13)$
$\iff 3^{2} 3^{202} \equiv 1 \ (\text{mod } 13)$
...
• October 7th 2008, 07:01 PM
bigb
Quote:
Originally Posted by o_O
By Fermat's theorem: $a^{12} \equiv 1 \ (\text{mod } 13)$
Note that $29 \equiv 3 \ (\text{mod } 13)$.
So: $\left(3^{12}\right)^{17} \equiv 1^{17} \ (\text{mod } 13)$
$\iff 3^{204} \equiv 1 \ (\text{mod } 13)$
$\iff 3^{2} 3^{202} \equiv 1 \ (\text{mod } 13)$
...
The answer is 81=3mod13..I am messing around with what you did but still cant get the answer. This problem is in my lecture notes but I just cant seem to understand what i wrote.
• October 7th 2008, 07:15 PM
icemanfan
Following after o_O:
$3^23^{202} \equiv 1 (\mod 13)$
$9 \times 3^{202} \equiv 1 (\mod 13)$
$27 \times 3^{202} \equiv 3 (\mod 13)$
$(13 \times 2 + 1) \times 3^{202} \equiv 3 (\mod 13)$
$3^{202} \equiv 3(\mod 13)$
• October 7th 2008, 07:20 PM
bigb
Quote:
Originally Posted by icemanfan
Following after o_O:
$3^23^{202} \equiv 1 (\mod 13)$
$9 \times 3^{202} \equiv 1 (\mod 13)$
$27 \times 3^{202} \equiv 3 (\mod 13)$
$(13 \times 2 + 1) \times 3^{202} \equiv 3 (\mod 13)$
$3^{202} \equiv 3(\mod 13)$
• October 7th 2008, 07:23 PM
icemanfan
Quote:
Originally Posted by bigb
It became $(13 \times 2 + 1) \equiv 1$.
• October 7th 2008, 07:38 PM
bigb
Quote:
Originally Posted by icemanfan
It became $(13 \times 2 + 1) \equiv 1$.
COmpute 3^302 mod 5
Would this be right??
a^4=1mod5
3=8mod5
8^4=1mod5
8^4*76=1^76mod5
8^2 * 8^302=1mod5
8^2 * 8^302=-64mod5
8^302=-1mod5
8^302=4mod5
or
3^4 = 1 (mod 5) by Fermat theorems.
So (raise to the 75) both sides,
3^300 = 1 (mod 5)
Multiply 3^2 both sides,
3^302 = 3^2 = 9 = 4 (mod 5)
• October 7th 2008, 07:40 PM
o_O
(Yes) I would do it the second way
• October 7th 2008, 07:43 PM
bigb
Quote:
Originally Posted by o_O
(Yes) I would do it the second way
It can be done the first way right thou?? If not can the second method be applied to the problem 29^202mod13
sry for all the dumb questions. I am just trying to learn the material better
• October 7th 2008, 07:47 PM
o_O
Well it really is the same work but you just did the unnecessary work of showing that 3 is congruent to 8 mod 5.
And this method WAS used for your problem. Go through it again. The principle is the same. From Fermat's theorem, you have your number to the power of p - 1 is congruent to 1. Raise it as close to the desired power as possible and manipulate it to get what you need. | 2013-12-08T00:46:15 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/number-theory/52508-fermats-little-theorem-2-a-print.html",
"openwebmath_score": 0.8408783674240112,
"openwebmath_perplexity": 2276.167496855653,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9915543740571682,
"lm_q2_score": 0.8479677622198947,
"lm_q1q2_score": 0.8408061436886053
} |
http://merganser.math.gvsu.edu/david/linear.algebra/ula/ula/knowl/example-24.html | ###### Example4.3.4
We will try to find a diagonalization of $$A = \left[\begin{array}{rr} -5 \amp 6 \\ -3 \amp 4 \\ \end{array}\right] \text{.}$$
First, we find the eigenvalues of $$A$$ by solving the characteristic equation
\begin{equation*} \det(A-\lambda I) = (-5-\lambda)(4-\lambda)+18 = (-2-\lambda)(1-\lambda) = 0\text{.} \end{equation*}
This shows that the eigenvalues of $$A$$ are $$\lambda_1 = -2$$ and $$\lambda_2 = 1\text{.}$$
By constructing $$\nul(A-(-2)I)\text{,}$$ we find a basis for $$E_{-2}$$ consisting of the vector $$\vvec_1 = \twovec{2}{1}\text{.}$$ Similarly, a basis for $$E_1$$ consists of the vector $$\vvec_2 = \twovec{1}{1}\text{.}$$ This shows that we can construct a basis $$\bcal=\{\vvec_1,\vvec_2\}$$ of $$\real^2$$ consisting of eigenvectors of $$A\text{.}$$
We now form the matrices
\begin{equation*} D = \left[\begin{array}{rr} -2 \amp 0 \\ 0 \amp 1 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_1 \amp \vvec_2 \end{array}\right] = \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 1 \\ \end{array}\right] \end{equation*}
and verify that
\begin{equation*} PDP^{-1} = \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 1 \\ \end{array}\right] \left[\begin{array}{rr} -2 \amp 0 \\ 0 \amp 1 \\ \end{array}\right] \left[\begin{array}{rr} 1 \amp -1 \\ -1 \amp 2 \\ \end{array}\right] = \left[\begin{array}{rr} -5 \amp 6 \\ -3 \amp 4 \\ \end{array}\right] = A\text{.} \end{equation*}
There are, of course, many ways to diagonalize $$A\text{.}$$ For instance, we could change the order of the eigenvalues and eigenvectors and write
\begin{equation*} D = \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp -2 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_2 \amp \vvec_1 \end{array}\right] = \left[\begin{array}{rr} 1 \amp 2 \\ 1 \amp 1 \\ \end{array}\right]\text{.} \end{equation*}
If we choose a different basis for the eigenspaces, we will also find a different matrix $$P$$ that diagonalizes $$A\text{.}$$ The point is that there are many ways in which $$A$$ can be written in the form $$A=PDP^{-1}\text{.}$$
in-context | 2019-02-23T16:06:48 | {
"domain": "gvsu.edu",
"url": "http://merganser.math.gvsu.edu/david/linear.algebra/ula/ula/knowl/example-24.html",
"openwebmath_score": 1.0000100135803223,
"openwebmath_perplexity": 1003.5425916342662,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9915543728093005,
"lm_q2_score": 0.8479677583778258,
"lm_q1q2_score": 0.8408061388208335
} |
http://mathhelpforum.com/pre-calculus/158108-proof-definition-limit-e-exists-please-help.html | Hi,
I'm reviewing some basic calculus concepts, and I have the following questions that will probably be very easy for you guys to answer, although I'm stuck since a long time on them.
Given the usual limit that is used in the definition of "e": lim (n->infinity) (1+1/n), how do we prove that the limit exists and it is finite?
Can we use the binomial theorem expansion for a finite n, and prove that the coefficients tend to be 1/k! for n that goes to infinity, so we get the Taylor expansion? I'd like to see a rigorous proof of this derivation of e's expression as a power series, we shouldn't forget to prove that the coefficients of the binomial expansion don't diverge (in fact, i guess they should converge) as n goes to infinity...
Or without expanding it, can we prove the following statements:
let a(n)=(1+1/n)^n
1-prove that a(n) is strictly increasing
2-prove that a(n) is limited above by a finite positive number L (for instance, we could try to prove it for L=3 or L=4): a(n)<L for all n.
Final questions:
A-if we manage to prove the two statements above, then did we successfully prove that the limit exists?
B-is there a better way than this to prove that the limit exists?
Thanks! And please excuse me for the length of the post...
2. Using the notation $e^x=\exp(x)$ we show it.
It is easy to use Napier’s inequality to show that $\dfrac{1}{n+1}\le\ln\left(1+\frac{1}{n}\right)\le\ dfrac{1}{n}.$
Raise e to those powers $\exp\left(\frac{1}{n+1}\right)\le\left(1+\frac{1}{ n}\right)\le\exp\left(\frac{1}{n}\right).$.
Raise each of those to the nth power
$\exp\left(\frac{n}{n+1}\right)\le\left(1+\frac{1}{ n}\right)^n\le e.$.
It is squeezed between e & e.
3. Thank you for answering so soon, the proof is very simple, but how do you define ln(x) without proving the limit of e first? I mean, wouldn't it be circular logic?
We should find a way to prove it without using the natural logarithm...
Alternatively, we could define ln(x) as the integral of 1/t from 1 to x, and then napier's inequality should be simple to infer, but we need to use a integral definition...
there should be an elementary method to prove it...
4. There are so many different ways to do this.
Notice that for $n\ge 2$ we have $\[
\left( {1 - \frac{1}{{n^2 }}} \right)^n \geqslant \left( {1 - \frac{1}
{n}} \right)\; \Rightarrow \;\left( {1 + \frac{1}{n}} \right)^n \geqslant \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1}$
.
So the sequence $\left( {1 + \frac{1}{n}} \right)^n$ is increasing and bounded above. That means it converges.
Call that limit $e$.
EDIT: If you need to show it is bounded.
$\left( {1 + \frac{1}
{n}} \right)^n = \sum\limits_{k = 0}^n {C(n,k)\left( {\frac{1}
{n}} \right)^k } < 1 + \left( {1 + \frac{1}
{2} + \cdots + \frac{1}
{{2^{n - 1} }}} \right) < 3$
5. Ok thank you very much, I understood how you deduced the facts from those inequalities, I will work to prove them.
You have been very kind, thanks again.
6. Here is a couple of links from PlanetMath.
7. Great, thanks again!
8. By definition, $e$ is the base of the exponential function that is its own derivative.
Therefore $\frac{d}{dx}(e^x) = e^x$.
If we wish to evaluate $e$, we need to use this definition.
$\frac{d}{dx}(e^x) = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}$
$= \lim_{h \to 0}\frac{e^xe^h - e^x}{h}$
$= \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$
$= e^x\lim_{h \to 0}\frac{e^h - 1}{h}$.
Since we know by definition that $\frac{d}{dx}(e^x) = e^x$, that means
$e^x\lim_{h \to 0}\frac{e^h - 1}{h} = e^x$
$\lim_{h \to 0}\frac{e^h - 1}{h}= 1$
$\lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h$
$\lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)$
$\lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$
$e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$.
Now if we let $h = \frac{1}{n}$, then as $h \to 0, n \to \infty$.
Thus $e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$.
9. Another useful way to define ln(x) is simply:
$ln(x)= \int_1^x \frac{1}{t}dt$
Since $\frac{1}{t}$ is defined for all non-zero t, this defines ln(x) for all positive x. It is clearly continuous and differentiable for all x. By the fundamental theorem of Calculus, its derivative is 1/x. Since that is positive for all positive x, ln(x) is an increasing function and so a one-to-one function.
It is not difficulty to prove, by substitutions in the integral, that
1) For any positive x, $ln(1/x)= - ln(x)$.
2) For any positive x and y, $ln(xy)= ln(x)+ ln(y)$.
3) For any positive x and y any real number, $ln(x^y)= y ln(x)$.
For example, for the last one, $ln(x^y)$ is, by this definition, $\int_1^{x^y}\frac{1}{t}dt$.
If $y\ne 0$, define $u= t^{1/y}$ so that $t= u^y$, $du= y u^{y- 1}du$ so that $\frac{1}{t}dt= \frac{1}{u^y}(y u^{y-1}du)= y \frac{1}{u}du$. When t= 1, $u= 1^{1/y}= 1$ and when $t= x^y$, $u= (x^y)^{1/y}= x$ so the integral becomes
$ln(x^y)= \int_1^x y \frac{1}{u}du= y\int_1^x \frac{1}{u}du= y ln(x)$
(If y= 0 then $x^y= x^0= 1$ and $ln(x^y)= ln(1)= \int_1^1 \frac{1}{t}dt= 0= 0 ln(x)= y ln(x)$ still.)
It follows then that ln(x) is a one-to-one function from the positive real numbers onto the real numbers and so has an inverse function, exp(x), from the real numbers to the positive real numbers.
Finally, if y= exp(x), then x= ln(y). If x is not 0, divide both sides by x to get $1= \frac{1}{x}ln(y)= ln(y^{1/x})$. Going back to the exponential form, $y^{1/x}= exp(1)$ so $y= (exp(1))^x$.
(If x= 0, then y= 1 because $ln(1)= \int_1^1 \frac{1}{t}dt= 0$. So we still have $exp(x)= exp(0)= 1= (exp(1))^0$. | 2015-10-07T21:38:26 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/pre-calculus/158108-proof-definition-limit-e-exists-please-help.html",
"openwebmath_score": 0.972873866558075,
"openwebmath_perplexity": 292.972672417335,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9780517494838938,
"lm_q2_score": 0.8596637559030338,
"lm_q1q2_score": 0.8407956404288572
} |
https://math.stackexchange.com/questions/4107957/is-ax2-bx-0-considered-a-quadratic-equation-or-is-it-linear-since-it-si | # Is $ax^2 + bx = 0$ considered a quadratic equation? Or is it linear, since it simplifies to $ax+b=0$?
I know that a quadratic equation can be represented in the form $$ax^2 + bx + c = 0$$ where $$a$$ is not equal to $$0$$, and $$a$$, $$b$$, and $$c$$ are real numbers. However, if there is an equation in the form $$ax^2 + bx = 0$$ would it be classified as a quadratic equation since the conditions are satisfied, or would it be a linear equation since it can be simplified into $$ax + b = 0$$?
• We define a polynomial by the highest power present Apr 19, 2021 at 8:30
• It can't be simplified into $ax+b=0$, which has one solution. It can be factorised, like other quadratics, as $x(ax+b)=0$; and this gives two solutions, in the normal way. Apr 19, 2021 at 8:50
It is a quadratic equation as it satisfies the definition.
Notice that $$ax^2+bx=0$$ and $$ax+b=0$$ are not equivalent, the first one has $$0$$ as a solution for sure and $$\frac{-b}a$$ as a root as well.
• Oh I see. Thanks! Apr 19, 2021 at 8:31
A quadratic equation is an equation that can be rearranged as $$ax^2+bx+c=0$$ where $$a$$ is not equal to $$0$$ and $$b$$ and $$c$$ are real numbers. If $$a=0$$ then the equation is linear not quadratic since the $$x^2$$ has no influence .
Hints
1. If you draw the graph of $$y=ax^2+bx$$ what shape is it? (Plug in some non-zero values for $$a$$ and $$b$$)
2. If you factorize $$ax^2+bx=0$$ and then apply null factor law, how many solutions are there?
You need only $$a≠0$$ for $$ax^2+bx+c=0.$$
If the degree of your polynomial is equal to $$2$$, then you have a quadratic polynomial. This means , your equation $$ax^2+bx=0$$ is still a quadratic equation, if $$a≠0.$$
But, the degree of the polynomial $$ax+b$$ is equal to $$1$$. This implies, $$ax+b=0$$ is not a quadratic.
Small Supplement:
$$ax^2+bx=0$$ is not equivalent to $$ax+b=0$$. Because, $$x=0$$ is not always a root of $$ax+b=0.$$
• I see. But upon multiplying x to both sides of ax + b = 0, we get ax^2 + bx = 0. Would this be a quadratic equation (since a is not equal to 0)? Or would it be linear since it is the same as ax + b = 0? Apr 19, 2021 at 8:30
• @Twilight I added a small supplement. Apr 19, 2021 at 8:41
• @Twilight: Note that you cannot multiply by something on both sides of an equation until you're sure that that something is not zero. Apr 19, 2021 at 8:44 | 2022-08-16T10:51:08 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/4107957/is-ax2-bx-0-considered-a-quadratic-equation-or-is-it-linear-since-it-si",
"openwebmath_score": 0.8013526201248169,
"openwebmath_perplexity": 122.64424090234667,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9780517469248845,
"lm_q2_score": 0.8596637577007393,
"lm_q1q2_score": 0.8407956399872187
} |
https://mathsgee.com/36733/what-difference-between-collinear-vectors-parallel-vectors | 0 like 0 dislike
241 views
What is the difference between collinear vectors and parallel vectors?
| 241 views
0 like 0 dislike
Suppose that $\mathbf{v}$ and $\mathbf{w}$ are vectors in 2 -space or 3 -space with a common initial point. If one of the vectors is a scalar multiple of the other, then the vectors lie on a common line, so it is reasonable to say that they are collinear.
However, if we translate one of the vectors, then the vectors are parallel but no longer collinear.
This creates a linguistic problem because translating a vector does not change it. The only way to resolve this problem is to agree that the terms parallel and collinear mean the same thing when applied to vectors.
Although the vector $\mathbf{0}$ has no clearly defined direction, we will regard it as parallel to all vectors when convenient.
by Platinum (164,290 points)
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike | 2023-02-05T02:23:36 | {
"domain": "mathsgee.com",
"url": "https://mathsgee.com/36733/what-difference-between-collinear-vectors-parallel-vectors",
"openwebmath_score": 0.7685940265655518,
"openwebmath_perplexity": 868.3133537087781,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9780517514031506,
"lm_q2_score": 0.8596637523076225,
"lm_q1q2_score": 0.8407956385622745
} |
http://openstudy.com/updates/5569d3d8e4b050a18e836654 | ## Kainui one year ago Here's a fun problem I came across. During an hour two independent events can happen at any time. What's the probability that the events are at least 10 minutes apart?
1. rational
|dw:1432999492312:dw|
2. Kainui
Hahaha yeah you got it.
3. rational
P(|E1-E2| > 10 minutes) = (Area of shaded region) / (Total area) = 5^2/6^2 = 25/36 ?
4. Kainui
Yep, exactly. =)
5. rational
these are really fun xD i got some practice a couple of months ago when @amistre64 was preparing for some probability theory exam
6. Kainui
Something that just occurred to me that I don't know the answer to is what is the probability of having 3 events 10 minutes apart? Or, is there a good reason why the probability is 5^2/6^2? If the events were separated by 20 minutes would the answer then become 4^2/6^2 ? I guess this would be easy to check for the general answer.
7. rational
very interesting, 3 events requires a triple integral is it
8. rational
second part of the question is easy yeah we always get squares in top and bottom because of symmetry : |E1-E2| > 20
9. Kainui
for the 2D case in general the answer is: $\frac{(60-t)^2}{60^2}$ which is quite nice!
10. ParthKohli
Wow, that's beautiful.
11. ParthKohli
Looks like my book has this... is this what is called "infinitistic probability"?
12. Kainui
Yup, I guess? I guess continuous probability distribution maybe? It's important for quantum mechanics!
13. ikram002p
:)
14. dan815
|dw:1433020038430:dw|
15. dan815
oh greater so 5/6
16. dan815
|dw:1433020148264:dw|
17. dan815
is that it?
18. dan815
i see that for one case its, 5/6 then therefore for every other point its always 5/6 of the total there so 5/6 all the time :)
19. dan815
i solved it by looking at a number line
20. dan815
|dw:1433020285038:dw|
21. dan815
so i thought about summing all these individual events up (kind of like an integral) however its constant here its always 5/6 of the smaller case, that event 1 happens at the one single point
22. dan815
Hence 5/6 for the complete thing
23. dan815
here is also something interesting... prolly not so much but kinda cool to visualze lol
24. dan815
cause its not good to think ofa number line but a circular number like for this or like a clock
25. dan815
|dw:1433020557004:dw|
26. dan815
|dw:1433020604712:dw|
27. Kainui
the problem is your answer seems to be wrong, the answer is 25/36 not 5/6 for the first question
28. dan815
ya i got that i change it xD
29. dan815
for a sec i thought u were asking less than 10
30. dan815
oh dang really
31. dan815
why is it 5^/6^2?
32. dan815
lemmee think
33. Kainui
this picture: |dw:1433020730099:dw| imagine each point is when two events happen, so all the points on the diagonal happen simultaneously. so for instance when one event happens, (10,20) at that point, that will fall in the black, so it's greater or equal to 10 min right? So the ratio of the black region to the entire square is the probability.
34. dan815
oh :O i get it
35. dan815
theres a problem here with assuming the circular thing too, because once event 1 occurs later
36. dan815
the prob is greater for 10 mins
37. dan815
:D
38. dan815
or wait no, the even can happen back in time too
39. Kainui
This is pretty similar to the Buffon needle problem which is a cool way to approximate $$\pi$$.
40. dan815
i dont think i understand the graph
41. Kainui
Honestly I don't think it gets much better than this general case: $\frac{(60-t)^2}{60^2}$ this tells you the probability that two random events happen with at least a separation of time t within an hour. Like if t=10 in the original then we get: $\frac{(60-10)^2}{60^2} = \frac{25}{36}$ and of course the probability that they're 0 minutes apart will give us 100% and probability they are 60 min apart will give us 0% just like we expect, it's just from calculating it based off of a variable triangle size inside cut out basically lol.
42. Kainui
|dw:1433021207310:dw| Here are some random points picked in there. The first represents some time that one event happened and the other represents the time the second event happened. So look along the diagonal, (0,0) means both events happened at hte beginning of the our. (30,30) means both events happened at the 30 minute mark. Makes sense? Now look at the point (10,50) that means the first event happened at 10 minutes and the other event happened at 50 minutes. This is one of the points that falls within "both events happening 10 minutes apart or greater" since these are obviously separated by 40 minutes yeah?
43. dan815
ah i gotcha now xD
44. dan815
the graph labelling didnt quite make sense to me, now its clear thanks
45. Kainui
cool so what if we have 3 events separated by 10 minutes, what does that cube look like?
46. dan815
|dw:1433021555908:dw|
47. Kainui
You would think so, that was my first guess too, but it doesn't seem to be right if you think of evaluating say, this point: |dw:1433021616607:dw|
48. dan815
ok lemme see more algrebrically first a+b+c<60*3 and diff between a,b,c is greater than 10
49. dan815
(5/6)^3?
50. dan815
^ thats a guess
51. dan815
wait what is the question?is it all 3 have to be apart by 10? or as long as all 3 dont happen in 10
52. Kainui
|dw:1433021690423:dw|
53. dan815
ok wat u drew there looks like its as long all 3 dont happen within 10, so u can have 2 with in 10 and other out
54. Kainui
So one example of an event can be like (10, 20, 30) since they are all separated by 10 min.
55. dan815
|dw:1433021882772:dw|
56. Kainui
yeah what I drew is wrong I think idk haha
57. Kainui
Yeah I just didn't wanna draw it with those lines in there cause it looked tacky
58. dan815
|dw:1433021915541:dw|
59. dan815
lol
60. dan815
oh u can think of the 2by 2 case as 1 big event again
61. dan815
|dw:1433022504943:dw|
62. dan815
so u solve for cases, when event 1 and 2 are close, and when they are not
63. dan815
eh nvm i feel like there is something easier to compute off the differences
64. dan815
did u already solve it?
65. Kainui
Nahhhh
66. dan815
ok :) im on it then sir
67. Kainui
some useful things: $$20 \le x+y+z \le 40$$ $$10 \le x+y \le 50$$ $$10 \le x+z \le 50$$ $$10 \le y+z \le 50$$
68. dan815
that what i was lookign at
69. dan815
like can we write out system of equationa for lines
70. dan815
and break it down into 4 or 6 or 9 cases or something
71. Kainui
I think this is enough to make up an integral, those are our bounds
72. dan815
hey btw the first lines ganeshew drew was
73. dan815
y=x+10 and x=y+10 right
74. Kainui
yeah
75. Kainui
notice they're symmetric so we only needed to find one part of that, then multiply by 2
76. dan815
ugh its gonna into these inequality cases | 2017-01-17T21:42:33 | {
"domain": "openstudy.com",
"url": "http://openstudy.com/updates/5569d3d8e4b050a18e836654",
"openwebmath_score": 0.6549059152603149,
"openwebmath_perplexity": 1866.9834815748513,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9780517450056273,
"lm_q2_score": 0.8596637541053281,
"lm_q1q2_score": 0.8407956348208047
} |
http://quiz.geeksforgeeks.org/graph-theory/ | Question 1
Consider an undirected random graph of eight vertices. The probability that there is an edge between a pair of vertices is 1/2. What is the expected number of unordered cycles of length three?
A 1/8 B 1 C 7 D 8
GATE CS 2013 Graph Theory
Discuss it
Question 1 Explanation:
Question 2
Which of the following statements is/are TRUE for undirected graphs?
P: Number of odd degree vertices is even.
Q: Sum of degrees of all vertices is even.
A P only B Q only C Both P and Q D Neither P nor Q
GATE CS 2013 Graph Theory
Discuss it
Question 2 Explanation:
Question 3
The line graph L(G) of a simple graph G is defined as follows: · There is exactly one vertex v(e) in L(G) for each edge e in G. · For any two edges e and e' in G, L(G) has an edge between v(e) and v(e'), if and only if e and e'are incident with the same vertex in G. Which of the following statements is/are TRUE?
(P) The line graph of a cycle is a cycle.
(Q) The line graph of a clique is a clique.
(R) The line graph of a planar graph is planar.
(S) The line graph of a tree is a tree.
A P only B P and R only C R only D P, Q and S only
GATE CS 2013 Graph Theory
Discuss it
Question 3 Explanation:
Question 4
Let G be a simple undirected planar graph on 10 vertices with 15 edges. If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to
A 3 B 4 C 5 D 6
GATE CS 2012 Graph Theory
Discuss it
Question 4 Explanation:
If the graph is planar, then it must follow below Euler's Formula for planar graphs
v - e + f = 2
v is number of vertices
e is number of edges
f is number of faces including bounded and unbounded
10 - 15 + f = 2
f = 7
There is always one unbounded face, so the number of bounded faces = 6
Question 5
Which of the following graphs is isomorphic to
A A B B C C D D
GATE CS 2012 Graph Theory
Discuss it
Question 5 Explanation:
Question 6
Let G be a complete undirected graph on 6 vertices. If vertices of G are labeled, then the number of distinct cycles of length 4 in G is equal to
A 15 B 30 C 45 D 360
GATE CS 2012 Graph Theory
Discuss it
Question 6 Explanation:
There can be total 6C4 ways to pick 4 vertices from 6. The value of 6C4 is 15. Note that the given graph is complete so any 4 vertices can form a cycle. There can be 6 different cycle with 4 vertices. For example, consider 4 vertices as a, b, c and d. The three distinct cycles are cycles should be like this (a, b, c, d,a) (a, b, d, c,a) (a, c, b, d,a) (a, c, d, b,a) (a, d, b, c,a) (a, d, c, b,a) and (a, b, c, d,a) and (a, d, c, b,a) (a, b, d, c,a) and (a, c, d, b,a) (a, c, b, d,a) and (a, d, b, c,a) are same cycles. So total number of distinct cycles is (15*3) = 45. **NOTE**: In original GATE question paper 45 was not an option. In place of 45, there was 90.
Question 7
A K4 is planar while Q3 is not B Both K4 and Q3 are planar C Q3 is planar while K4 is not D Neither K4 nor Q3 are planar
GATE CS 2011 Graph Theory
Discuss it
Question 7 Explanation:
A Graph is said to be planar if it can be drawn in a plane without any edges crossing each other. Following are planar embedding of the given two graphs (Source: http://clweb.csa.iisc.ernet.in/rahulsharma/gate2011key.html)
Question 8
Let G = (V,E) be a graph. Define ξ(G) = Σd id x d, where id is the number of vertices of degree d in G. If S and T are two different trees with ξ(S) = ξ(T),then
A |S| = 2|T| B |S| = |T|-1 C |S| = |T| D |S| = |T|+1
GATE CS 2010 Graph Theory
Discuss it
Question 8 Explanation:
The expression ξ(G) is basically sum of all degrees in a tree. For example, in the following tree, the sum is 3 + 1 + 1 + 1.
a
/ | \
b c d
Now the questions is, if sum of degrees in trees are same, then what is the relationship between number of vertices present in both trees? The answer is, ξ(G) and ξ(T) is same for two trees, then the trees have same number of vertices. It can be proved by induction. Let it be true for n vertices. If we add a vertex, then the new vertex (if it is not the first node) increases degree by 2, it doesn't matter where we add it. For example, try to add a new vertex say 'e' at different places in above example tee.
Question 9
The degree sequence of a simple graph is the sequence of the degrees of the nodes in the graph in decreasing order. Which of the following sequences can not be the degree sequence of any graph?
I. 7, 6, 5, 4, 4, 3, 2, 1
II. 6, 6, 6, 6, 3, 3, 2, 2
III. 7, 6, 6, 4, 4, 3, 2, 2
IV. 8, 7, 7, 6, 4, 2, 1, 1
A I and II B III and IV C IV only D II and IV
GATE CS 2010 Graph Theory
Discuss it
Question 9 Explanation:
A generic algorithm or method to solve this question is 1: procedure isV alidDegreeSequence(L) 2: for n in list L do 3: if L doesn’t have n elements next to the current one then return false 4: decrement next n elements of the list by 1 5: arrange it back as a degree sequence, i.e. in descending order 6: if any element of the list becomes negative then return false 7: return true Rationale behind this method comes from the properties of simple graph. Enumerating the f alse returns, 1) if L doesn’t have enough elements after the current one or 2) if any element of the list becomes negative, then it means that there aren’t enough nodes to accommodate edges in a simple graph fashion, which will lead to violation of either of the two conditions of the simple graph (no self-loops and no multiple-edges between two nodes), if not others. See http://www.geeksforgeeks.org/data-structures-and-algorithms-set-25/ This solution is contributed by Vineet Purswani. Another one: A degree sequence d1,d2,d2. . . dn of non negative integer is graphical if it is a degree sequence of a graph. We now introduce a powerful tool to determine whether a particular sequence is graphical due to Havel and Hakimi Havel–Hakimi Theorem : → According to this theorem, Let D be sequence the d1,d2,d2. . . dn with d1 ≥ d2 ≥ d2 ≥ . . . dn for n≥ 2 and di ≥ 0. → Then D0 be the sequence obtained by: → Discarding d1, and → Subtracting 1 from each of the next d1 entries of D. → That is Degree sequence D0 would be : d2-1, d2-1, d3-1 . . . , dd1+1 -1 . . . , dn → Then, D is graphical if and only if D0 is graphical. Now, we apply this theorem to given sequences: option I) 7,6,5,4,4,3,2,1 → 5,4,3,3,2,1,0 → 3,2,2,1,0,0 → 1,1,0,0,0 → 0,0,0,0 so its graphical. Option II) 6,6,6,6,3,3,2,2 → 5,5,5,2,2,1,2 ( arrange in ascending order) → 5,5,5,2,2,2,1 → 4,4,1,1,1,0 → 3,0,0,0,0 → 2,-1,-1,-1,0 but d (degree of a vertex) is non negative so its not a graphical. Option III) 7,6,6,4,4,3,2,2 → 5,5,3,3,2,1,1 → 4,2,2,1,1,0 → 1,1,0,0,0 → 0,0,0,0 so its graphical. Option IV) 8,7,7,6,4,2,1,1 , here degree of a vertex is 8 and total number of vertices are 8 , so it’s impossible, hence it’s not graphical. Hence only option I) and III) are graphic sequence and answer is option-D This solution is contributed by Nirmal Bharadwaj.
Question 10
What is the chromatic number of an n-vertex simple connected graph which does not contain any odd length cycle? Assume n >= 2.
A 2 B 3 C n-1 D n
GATE-CS-2009 Graph Theory
Discuss it
Question 10 Explanation:
The chromatic number of a graph is the smallest number of colours needed to colour the vertices of so that no two adjacent vertices share the same colour. These types of questions can be solved by substitution with different values of n. 1) n = 2 This simple graph can be coloured with 2 colours. 2) n = 3 Here, in this graph let us suppose vertex A is coloured with C1 and vertices B, C can be coloured with colour C2 => chromatic number is 2 In the same way, you can check with other values, Chromatic number is equals to 2 This solution contributed by Anil Saikrishna Devarasetty //A simple graph with no odd cycles is bipartite graph and a Bipartite graph can be colored using 2 colors (See this)
Question 11
Which one of the following is TRUE for any simple connected undirected graph with more than 2 vertices?
A No two vertices have the same degree. B At least two vertices have the same degree. C At least three vertices have the same degree. D All vertices have the same degree.
GATE-CS-2009 Graph Theory
Discuss it
Question 11 Explanation:
Since the graph is simple, there must not be any self loop and parallel edges. Since the graph is connected, the degree of any vertex cannot be 0. Therefore, degree of all vertices should be be from 1 to n-1. So the degree of at least two vertices must be same.
Question 12
Which of the following statements is true for every planar graph on n vertices?
A The graph is connected B The graph is Eulerian C The graph has a vertex-cover of size at most 3n/4 D The graph has an independent set of size at least n/3
Graph Theory GATE CS 2008
Discuss it
Question 12 Explanation:
A planar graph is a graph which can drawn on a plan without any pair of edges crossing each other. A) FALSE: A disconnected graph can be planar as it can be drawn on a plane without crossing edges. B) FALSE: An Eulerian Graph may or may not be planar. An undirected graph is eulerian if all vertices have even degree. For example, the following graph is Eulerian, but not planar C) TRUE: D) FALSE:
Question 13
G is a graph on n vertices and 2n - 2 edges. The edges of G can be partitioned into two edge-disjoint spanning trees. Which of the following is NOT true for G?
A For every subset of k vertices, the induced subgraph has at most 2k-2 edges B The minimum cut in G has at least two edges C There are two edge-disjoint paths between every pair to vertices D There are two vertex-disjoint paths between every pair of vertices
Graph Theory GATE CS 2008
Discuss it
Question 13 Explanation:
Counter for option D is as follows. Take two copies of K4(complete graph on 4 vertices), G1 and G2. Let V(G1)={1,2,3,4} and V(G2)={5,6,7,8}. Construct a new graph G3 by using these two graphs G1 and G2 by merging at a vertex, say merge (4,5). The resultant graph is two edge connected, and of minimum degree 2 but there exist a cut vertex, the merged vertex. Thanks to Renjith P for providing above explanation.
Question 14
Let G be the non-planar graph with the minimum possible number of edges. Then G has
A 9 edges and 5 vertices B 9 edges and 6 vertices C 10 edges and 5 vertices D 10 edges and 6 vertices
Graph Theory GATE-CS-2007
Discuss it
Question 14 Explanation:
For a simple, connected, planar graph with v vertices and e edges, the following simple conditions hold: If v ≥ 3 then e ≤ 3v − 6; Note that the question is about non-planar graph G. Only option C doesn't follow above. Alternate Explanation: We know that K5K5 (which has 10 edges and 5 vertices) and K3,3K3,3 (which has 9 edges and 6 vertices) are non-planar graphs. Since we are asked about minimum number of edges, answer should be k3,3k3,3 i.e. option (B) is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2007.html
Question 15
Which of the following graphs has an Eulerian circuit?
A Any k-regular graph where kis an even number. B A complete graph on 90 vertices C The complement of a cycle on 25 vertices D None of the above
Graph Theory GATE-CS-2007
Discuss it
Question 15 Explanation:
A graph has Eulerian Circuit if following conditions are true. ….a) All vertices with non-zero degree are connected. We don’t care about vertices with zero degree because they don’t belong to Eulerian Cycle or Path (we only consider all edges). ….b) All vertices have even degree. Let us analyze all options. A) Any k-regular graph where k is an even number. is not Eulerian as a k regular graph may not be connected (property b is true, but a may not) B) A complete graph on 90 vertices is not Eulerian because all vertices have degree as 89 (property b is false) C) The complement of a cycle on 25 vertices is Eulerian. In a cycle of 25 vertices, all vertices have degree as 2. In complement graph, all vertices would have degree as 22 and graph would be connected.
Question 16
Let G=(V,E) be a directed graph where V is the set of vertices and E the set of edges. Then which one of the following graphs has the same strongly connected components as G ?
A A B B C C D D
Graph Theory GATE-CS-2014-(Set-1)
Discuss it
Question 16 Explanation:
If we reverse directions of all arcs in a graph, the new graph has same set of strongly connected components as the original graph. Se http://www.geeksforgeeks.org/strongly-connected-components/ for more details.
Question 17
Consider an undirected graph G where self-loops are not allowed. The vertex set of G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}. There is an edge between (a, b) and (c, d) if |a − c| <= 1 and |b − d| <= 1. The number of edges in this graph is __________.
A 500 B 502 C 506 D 510
Graph Theory GATE-CS-2014-(Set-1)
Discuss it
Question 17 Explanation:
Given:
The vertex set of G is {(i, j): 1 <= i <= 12, 1 <= j <= 12}.
There is an edge between (a, b) and (c, d) if |a − c| <= 1 and
|b − d| <= 1.
There can be total 12*12 possible vertices. The vertices are (1, 1),
(1, 2) ....(1, 12) (2, 1), (2, 2), ....
The number of edges in this graph?
Number of edges is equal to number of pairs of vertices that satisfy
above conditions. For example, vertex pair {(1, 1), (1, 2)} satisfy
above condition.
For (1, 1), there can be an edge to (1, 2), (2, 1), (2, 2). Note that
there can be self-loop as mentioned in the question.
Same is count for (12, 12), (1, 12) and (12, 1)
For (1, 2), there can be an edge to (1, 1), (2, 1), (2, 2), (2, 3),
(1, 3)
Same is count for (1, 3), (1, 4)....(1, 11), (12, 2), ....(12, 11)
For (2, 2), there can be an edge to (1, 1), (1, 2), (1, 3), (2, 1),
(2, 3), (3, 1), (3, 2), (3, 3)
Same is count for remaining vertices.
For all pairs (i, j) there can total 8 vertices connected to them if
i and j are not in {1, 12}
There are total 100 vertices without a 1 or 12. So total 800 edges.
For vertices with 1, total edges = (Edges where 1 is first part) +
(Edges where 1 is second part and not first part)
= (3 + 5*10 + 3) + (5*10) edges
Same is count for vertices with 12
Total number of edges:
= 800 + [(3 + 5*10 + 3) + 5*10] + [(3 + 5*10 + 3) + 5*10]
= 800 + 106 + 106
= 1012
Since graph is undirected, two edges from v1 to v2 and v2 to v1
should be counted as one.
So total number of undirected edges = 1012/2 = 506.
Question 18
An ordered n-tuple (d1, d2, … , dn) with d1 >= d2 >= ⋯ >= dn is called graphic if there exists a simple undirected graph with n vertices having degrees d1, d2, … , dn respectively. Which of the following 6-tuples is NOT graphic?
A (1, 1, 1, 1, 1, 1) B (2, 2, 2, 2, 2, 2) C (3, 3, 3, 1, 0, 0) D (3, 2, 1, 1, 1, 0)
Graph Theory GATE-CS-2014-(Set-1)
Discuss it
Question 18 Explanation:
The required graph is not possible with the given degree set of (3, 3, 3, 1, 0, 0). Using this 6-tuple the graph formed will be a Disjoint undirected graph, where the two vertices of the graph should not be connected to any other vertex ( i.e. degree will be 0 for both the vertices ) of the graph. And for the remaining 4 vertices the graph need to satisfy the degrees of (3, 3, 3, 1).
Let's see this with the help of a logical structure of the graph :
Let's say vertices labelled as <ABCDEF> should have their degree as <3, 3, 3, 1, 0, 0> respectively.
Now E and F should not be connected to any vertex in the graph. And A, B, C and D should have their degree as <3, 3, 3, 1> respectively. Now to fulfill the requirement of A, B and C, the node D will never be able to get its degree as 1. It's degree will also become as 3. This is shown in the above diagram.
Hence tuple <3, 3, 3, 1, 0, 0> is not graphic.
Question 19
The maximum number of edges in a bipartite graph on 12 vertices is __________________________.
A 36
Graph Theory GATE-CS-2014-(Set-2)
Discuss it
Question 19 Explanation:
Number of edges would be maximum when there are 6 edges on each side and every vertex is connected to all 6 vertices of the other side.
Question 20
A cycle on n vertices is isomorphic to its complement. The value of n is _____.
A 2 B 4 C 6 D 5
Graph Theory GATE-CS-2014-(Set-2)
Discuss it
Question 20 Explanation:
Below is a cyclic graph with 5 vertices and its complement graph. The complement graph is also isomorphic (same number of vertices connected in same way) to given graph.
Question 21
If G is a forest with n vertices and k connected components, how many edges does G have?
A floor(n/k) B ceil(n/k) C n-k D n-k+1
Graph Theory GATE-CS-2014-(Set-3)
Discuss it
Question 21 Explanation:
Each component will have n/k vertices (pigeonhole principle). Hence, for each component there will be (n/k)-1 edges. Since there are k components, total number of edges= k*((n/k)-1) = n-k.
Question 22
Let d denote the minimum degree of a vertex in a graph. For all planar graphs on n vertices with d ≥ 3, which one of the following is TRUE?
A In any planar embedding, the number of faces is at least n/2 + 2 B In any planar embedding, the number of faces is less than n/2 + 2 C There is a planar embedding in which the number of faces is less than n/2 + 2 D There is a planar embedding in which the number of faces is at most n/(d+1)
Graph Theory GATE-CS-2014-(Set-3)
Discuss it
Question 22 Explanation:
Euler's formula for planar graphs:
v − e + f = 2.
v => Number of vertices
e => Number of edges
f => Number of faces
Since degree of every vertex is at least 3,
below is true from handshaking lemma (Sum of
degrees is twice the number of edges)
3v >= 2e
3v/2 >= e
Putting these values in Euler's formula.
v - 3v/2 + f >= 2
f >= v/2 + 2
Question 23
The 2n vertices of a graph G corresponds to all subsets of a set of size n, for n >= 6 . Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements. The number of vertices of degree zero in G is:
A 1 B n C n+1 D 2n
Graph Theory GATE-CS-2006
Discuss it
Question 23 Explanation:
There are n nodes which are single and 1 node which belong to empty set. And since they are not having 2 or more elements so they won’t be connected to anyone hence total number of nodes with degree 0 are n+1 hence answer should be none. Thanks to roger for the explanation.
Question 24
The 2n vertices of a graph G corresponds to all subsets of a set of size n, for n >= 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements. The number of connected components in G is:
A n B n+2 C 2n/2 D 2n / n
Graph Theory GATE-CS-2006
Discuss it
Question 24 Explanation:
n+1 nodes of the graph not connected to anyone as explained in question 70 while others are connected so total number of connected components are n+2 (n+1 connected components by each of the n+1 vertices plus 1 connected component by remaining vertices).
Question 25
Let G be a simple connected planar graph with 13 vertices and 19 edges. Then, the number of faces in the planar embedding of the graph is
A 6 B 8 C 9 D 13
Graph Theory GATE-CS-2005
Discuss it
Question 25 Explanation:
An undirected graph is called a planar graph if it can be drawn on a paper without having two edges cross and such a drawing is called Planar Embedding. We say that a graph can be embedded in the plane, if it planar. A planar graph divides the plane into regions (bounded by the edges), called faces. Graph K4 is palanar graph, because it has a planar embedding as shown in figure below. Euler's Formula : For any polyhedron that doesn't intersect itself (Connected Planar Graph),the • Number of Faces(F) • plus the Number of Vertices (corner points) (V) • minus the Number of Edges(E) , always equals 2. This can be written: F + V − E = 2. Solution : Here as given, F=?,V=13 and E=19 -> F+13-19=2 -> F=8 So Answer is (B). This solution is contributed by Nirmal Bharadwaj We can apply Euler's Formula of planar graphs. The formula is v − e + f = 2.
Question 26
Let G be a simple graph with 20 vertices and 100 edges. The size of the minimum vertex cover of G is 8. Then, the size of the maximum independent set of G is
A 12 B 8 C Less than 8 D More than 12
Graph Theory GATE-CS-2005
Discuss it
Question 26 Explanation:
Background Explanation: Vertex cover is a set S of vertices of a graph such that each edge of the graph is incident to at least one vertex of S. Independent set of a graph is a set of vertices such that none of the vertices in this set have an edge connecting them i.e. no two are adjacent. A single vertex is an independent set, but we are interested in maximum independent set, that is largest set which is independent set. Relation between Independent Set and Vertex Cover : An interesting fact is, the number of vertices of a graph is equal to its minimum vertex cover number plus the size of a maximum independent set. How? removing all vertices of minimum vertex cover leads to maximum independent set. So if S is the size of minimum vertex cover of G(V,E) then the size of maximum independent set of G is |V| - S. Solution: size of minimum vertex cover = 8 size of maximum independent set = 20 - 8 =12 Therefore, correct answer is (A). References : vertex cover maximum independent set. This solution is contributed by Nitika Bansal.
Question 27
Which one of the following graphs is NOT planar?
A G1 B G2 C G3 D G4
Graph Theory GATE-CS-2005
Discuss it
Question 27 Explanation:
A graph is planar if it can be redrawn in a plane without any crossing edges. G1 is a typical example of nonplanar graphs.
Question 28
Let s and t be two vertices in a undirected graph G + (V, E) having distinct positive edge weights. Let [X, Y] be a partition of V such that s ∈ X and t ∈ Y. Consider the edge e having the minimum weight amongst all those edges that have one vertex in X and one vertex in Y. Let the weight of an edge e denote the congestion on that edge. The congestion on a path is defined to be the maximum of the congestions on the edges of the path. We wish to find the path from s to t having minimum congestion. Which one of the following paths is always such a path of minimum congestion?
A a path from s to t in the minimum weighted spanning tree B a weighted shortest path from s to t C an Euler walk from s to t D a Hamiltonian path from s to t
Graph Theory GATE-CS-2005
Discuss it
Question 28 Explanation:
Suppose shortest path from A->B is 6, but in MST, we have A->C->B (A->C = 4, C->B = 3), then along the path in MST, we have minimum congestion, i.e 4
Question 29
The minimum number of colours required to colour the following graph, such that no two adjacent vertices are assigned the same colour, is
A 2 B 3 C 4 D 5
Graph Theory GATE-CS-2004
Discuss it
Question 29 Explanation:
Two vertices are said to be adjacent if they are directly connected, i.e., there is a direct edge between them. So, here, we can assign same color to 1 & 2 (red), 3 & 4 (grey), 5 & 7 (blue) and 6 & 8 (brown). Therefore, we need a total of 4 distinct colors. Thus, C is the correct choice.
Please comment below if you find anything wrong in the above post.
Question 30
Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between
A k and n B k - 1 and k + 1 C k - 1 and n - 1 D k + 1 and n - k
Graph Theory GATE-CS-2003
Discuss it
Question 30 Explanation:
Minimum: The removed vertex itself is a separate connected component. So removal of a vertex creates k-1 components. Maximum: It may be possible that the removed vertex disconnects all components. For example the removed vertex is center of a star. So removal creates n-1 components.
Question 31
How many perfect matchings are there in a complete graph of 6 vertices ?
A 15 B 24 C 30 D 60
Graph Theory GATE-CS-2003
Discuss it
Question 31 Explanation:
A perfect matching, every vertex of the graph is incident to exactly one edge of the matching. A perfect matching is therefore a matching of a graph containing n/2 edges, the largest possible, meaning perfect matchings are only possible on graphs with an even number of vertices. (Source http://mathworld.wolfram.com/PerfectMatching.html)
Question 32
A graph G = (V, E) satisfies |E| ≤ 3 |V| - 6. The min-degree of G is defined as . Therefore, min-degree of G cannot be
A 3 B 4 C 5 D 6
Graph Theory GATE-CS-2003
Discuss it
Question 32 Explanation:
Let the min-degree of G be x, then G has at least |v| *x/2 edges. |v|*x/2 <= 3* |v| -6 for x=6, we get 0 <= -6, Therefore, min degree of G cannot be 6. Hence answer is (D).
Question 33
The minimum number of colours required to colour the vertices of a cycle with η nodes in such a way that no two adjacent nodes have the same colour is
A 2 B 3 C 4 D n - 2⌊n/2⌋ + 2
Graph Theory GATE-CS-2002
Discuss it
Question 33 Explanation:
We need 3 colors to color a odd cycle and 2 colors to color an even cycle.
Question 34
Maximum number of edges in a n - node undirected graph without self loops is
A n2 B n(n - 1)/2 C n - 1 D (n + 1) (n)/2
Graph Theory GATE-CS-2002
Discuss it
Question 34 Explanation:
Background required - Basic Combinatorics Since the given graph is undirected, that means the order of edges doesn't matter. Since we have to insert an edge between all possible pair of vertices, therefore problem reduces to finding the count of the number of subsets of size 2 chosen from the set of vertices. Since the set of vertices has size n, the number of such subsets is given by the binomial coefficient C(n,2) (also known as "n choose 2"). Using the formula for binomial coefficients, C(n,2) = n(n-1)/2.e This explanation has been contributed by Pranjul Ahuja.
Question 35
Let G be a connected planar graph with 10 vertices. If the number of edges on each face is three, then the number of edges in G is _______________.
A 24 B 20 C 32 D 64
Graph Theory GATE-CS-2015 (Set 1)
Discuss it
Question 35 Explanation:
Euler's formula states that if a finite, connected, planar graph is drawn in the plane without any edge intersections, then
v − e + f = 2.
v -> Number of vertices
e -> Number of edges
f -> Number of faces
As per the question
v = 10
And number of edges on each face is three
Therefore, 2e = 3f [Note that every edge is
shared by 2 faces]
Putting above values in v − e + f = 2
10 - e + 2e/3 = 2
e = 3*10 - 6 = 24
Question 36
A graph is self-complementary if it is isomorphic to its complement. For all self-complementary graphs on n vertices, n is
A A multiple of 4 B Even C Odd D Congruent to 0 mod 4, or 1 mod 4
Graph Theory GATE-CS-2015 (Set 2)
Discuss it
Question 36 Explanation:
An n-vertex self-complementary graph has exactly half number of edges of the complete graph, i.e., n(n − 1)/4 edges, and (if there is more than one vertex) it must have diameter either 2 or 3. Since n(n −1) must be divisible by 4, n must be congruent to 0 or 1 mod 4; for instance, a 6-vertex graph cannot be self-complementary. Source: http://en.wikipedia.org/wiki/Self-complementary_graph
Question 37
In a connected graph, a bridge is an edge whose removal disconnects a graph. Which one of the following statements is True?
A A tree has no bridge B A bridge cannot be part of a simple cycle C Every edge of a clique with size ≥ 3 is a bridge (A clique is any complete subgraph of a graph) D A graph with bridges cannot have a cycle
Graph Theory GATE-CS-2015 (Set 2)
Discuss it
Question 37 Explanation:
A bridge in a graph cannot be a part of cycle as removing it will not create a disconnected graph if there is a cycle.
Question 38
What is the number of vertices in an undirected connected graph with 27 edges, 6 vertices of degree 2, 3 vertices of degree 4 and remaining of degree 3?
A 10 B 11 C 18 D 19
Graph Theory GATE-IT-2004
Discuss it
Question 38 Explanation:
The idea is to use Handshaking Lemma :- In any graph, the sum of all the vertex-degree is equal to twice the number of edges.
Let x = Number of vertices of degree 3.
By Handshaking Lemma
6*2 + 3*4 + (x-9)*3 = 27*2
24 + (x-9)*3 = 54
x-9 = 10
x = 19
Question 39
The minimum number of colours that is sufficient to vertex-colour any planar graph is _______________ [This Question was originally a Fill-in-the-blanks Question]
A 1 B 2 C 3 D 4
Graph Theory GATE-CS-2016 (Set 2)
Discuss it
Question 39 Explanation:
A planar graph is a graph on a plane where no two edges are crossing each other. The set of regions of a map can be represented more abstractly as an undirected graph that has a vertex for each region and an edge for every pair of regions that share a boundary segment. Hence the four color theorem is applied here. Here is a property of a planar graph that a planar graph does not require more than 4 colors to color its vertices such that no two vertices have same color. This is known four color theorem.
Question 40
If all the edge weights of an undirected graph are positive, then any subset of edges that connects all the vertices and has minimum total weight is a
A Hamiltonian cycle B grid C hypercube D tree
Tree Traversals Graph Theory GATE IT 2006
Discuss it
Question 40 Explanation:
As here we want subset of edges that connects all the vertices and has minimum total weight i.e. Minimum Spanning Tree Option A - includes cycle, so may or may not connect all edges. Option B - has no relevance to this question. Option C - includes cycle, so may or may not connect all edges. Related: http://www.geeksforgeeks.org/greedy-algorithms-set-2-kruskals-minimum-spanning-tree-mst/ http://www.geeksforgeeks.org/greedy-algorithms-set-5-prims-minimum-spanning-tree-mst-2/ This solution is contributed by Mohit Gupta.
Question 41
Consider a weighted undirected graph with positive edge weights and let uv be an edge in the graph. It is known that the shortest path from the source vertex s to u has weight 53 and the shortest path from s to v has weight 65. Which one of the following statements is always true?
A weight (u, v) < 12 B weight (u, v) ≤ 12 C weight (u, v) > 12 D weight (u, v) ≥ 12
Graph Shortest Paths Graph Theory Gate IT 2007
Discuss it
Question 41 Explanation:
The minimum weight happens when (S,U) + (U,V) = (S,V) Else (S,U) + (U,V) >= (S,V) Given (S,U) = 53, (S,V) = 65 53 + (U,V) >= 63 (U,V) >= 12. This solution is contributed by Anil Saikrishna Devarasetty
Question 42
What is the chromatic number of the following graph?
A 2 B 3 C 4 D 5
Graph Theory Gate IT 2008
Discuss it
Question 42 Explanation:
The chromatic number of a graph is the smallest number of colors needed to color the vertices so that no two adjacent vertices have the same color. In this graph, minimum number of colors needed to color given graph would be equal to 3.
Question 43
G is a simple undirected graph. Some vertices of G are of odd degree. Add a node v to G and make it adjacent to each odd degree vertex of G. The resultant graph is sure to be
A regular B Complete C Hamiltonian D Euler
Graph Theory Gate IT 2008
Discuss it
Question 43 Explanation:
For a graph to be Euler graph all the degrees must be Even for all nodes. In any graph all the Odd degree nodes are connected with a node.
And number of Odd degree vertices should be even.
So degree of this new node will be Even and as a new edge is formed between this new node and all other nodes of Odd degree hence here is not a single node exists with degree Odd
=> Euler Graph
There are 43 questions to complete.
## GATE CS Corner
See Placement Course for placement preparation, GATE Corner for GATE CS Preparation and Quiz Corner for all Quizzes on GeeksQuiz. | 2017-02-26T16:48:42 | {
"domain": "geeksforgeeks.org",
"url": "http://quiz.geeksforgeeks.org/graph-theory/",
"openwebmath_score": 0.564095675945282,
"openwebmath_perplexity": 504.4346073515372,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9780517469248845,
"lm_q2_score": 0.8596637505099167,
"lm_q1q2_score": 0.8407956329542221
} |
https://math.stackexchange.com/questions/2573638/domain-of-fx-arctan-left-fracx22-x2-right | # Domain of $f(x)=\arctan\left(\frac{x^2}{2-x^2}\right)$
I need to find the domain of the function $f(x)=\arctan\left(\frac{x^2}{2-x^2}\right)$. I used a graphing calculator and found that the domain is $(-\infty,-\sqrt 2) ∪ (-\sqrt 2, \sqrt 2)∪(\sqrt 2, \infty)$ but I'm unsure as to how to do it without a calculator.
My assumption is that, using what we know about the calculation of the domain of $\sin$ and $\cos$, we make the denominator equal to $0$. So $$2-x^2=0$$ $$2=x^2$$ $$\pm\sqrt 2=x$$
But how do I get from there to the domain stated above?
• How would you describe the set of real numbers not equal to $\pm\sqrt2$? – Lord Shark the Unknown Dec 19 '17 at 20:17
• $\arctan$ is defined everywhere, and $\frac{x^2}{2-x^2}$ is only undefined where you showed. What can't you put into $\arctan\left(\frac{x^2}{2-x^2}\right)$? – Austin Weaver Dec 19 '17 at 20:20
• But in the answer $(-\sqrt 2, \sqrt 2)$ is also a possible solution to the question. – Ski Mask Dec 19 '17 at 20:22
The function $\arctan$ is defined everywhere in $\mathbb{R}$, but in mathematics you mustn’t divide by zero. This means that the only condition you have to apply is: $2-x^2\neq 0$. And this means $x\neq \pm \sqrt 2$.
Then the domain is $(-\infty,-\sqrt 2) ∪ (-\sqrt 2, \sqrt 2)∪(\sqrt 2, \infty)$ that is the same of $\mathbb{R}-\{-\sqrt 2,\sqrt 2\}$.
As you can see from the graphic of the function, it as the asymptotes in $x=\pm \sqrt 2$:
The function $arctan(x)$ is continuous, so this on its own would have no discontinuities. The only place then that this function will be discontinuous is when there is a division by $0$, which you correctly found to be $\sqrt2$ and $-\sqrt2$, therefore the domain is all reals except these two values.
To be clear, you don't have a function unless you have its domain to begin with. Asking for the domain of a function is like asking for the age of a man whose age is given to be $35.$
What you have with $\arctan (x^2/(2-x^2))$ is an expression. The set of all $x$ for which an expression is defined is often called the "natural domain" of the expression.
Assuming the natural domain is what we're after here leads to two questions: For which $x$ is $x^2/(2-x^2)$ defined, and for which $y$ is $\arctan y$ defined. The answers are: for the first question, it's all $x\ne \pm \sqrt 2;$ for the second question, it's all $y.$ Thus the natural domain for the expression $\arctan (x^2/(2-x^2))$ is $\{x\in \mathbb R: x\ne \pm \sqrt 2\}.$ | 2019-09-21T11:04:16 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2573638/domain-of-fx-arctan-left-fracx22-x2-right",
"openwebmath_score": 0.8765677809715271,
"openwebmath_perplexity": 103.5849120984816,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.97805174980377,
"lm_q2_score": 0.8596637451167995,
"lm_q1q2_score": 0.8407956301543479
} |
https://math.stackexchange.com/questions/3829606/number-of-ways-to-arrange-books | # number of ways to arrange books
I am solving number of arrangements for following question:
Eight books are placed on a shelf. Three of them form a 3-volume series, two form a 2-volume series, and 3 stand on their own. In how many ways can the eight books be arranged so that the books in the 3-volume series are placed together according to their correct order, and so are the books in the 2-volume series? Noted that there is only one correct order for each series.
I analyzed this problem in following way: 3 volume book as A, 2 volume book as B, and remaining books as C. So three can be arranged as 3!, and C in turn can be arranged as 3! which by product rule we can have 3! * 3! = 36. Why this analysis or approach to this problem is wrong. Correct answer is 120 (i.e. 5!). Kindly help. thanks
• The C books do not need to be next to each other. Sep 17, 2020 at 9:58
You can think of the three books who make a volume as a single big book (their relative positions can't change) and so for the other two volumes books, they can be considered as a single one too since their relative position can't change, therefore you have the equivalent of 5 books:
(3 books), (2 books), (1 book), (1 book), (1 book)
And the number of ways you can arrange them is $$5!$$ indeed.
Since the $$3$$-volume series are placed together according to their correct order, we have $$A_{1}A_{2}A_{3}$$ and for the $$2$$ volume series we also have $$B_{1}B_{2}.$$
So we can consider the $$3$$ volume books to be a single book denoted by $$A$$ and the $$2$$ volume books to be a single book denoted by $$B$$. Thus we need to arrange $$A,B$$ and the remaining $$3$$ books. This is the same as the number of ways to rearrange $$5$$ books.
Since the number of ways in which $$5$$ books can be arranged is $$5!=120,$$ we have $$120$$ ways.
• thanks for explanation. Can you point out what is wrong in my analysis Sep 17, 2020 at 11:24
• As mentioned in the comments, you calculated the $C$ books to be next to each other, but they do not need to be next to each other. Sep 17, 2020 at 11:26 | 2022-07-06T02:14:04 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3829606/number-of-ways-to-arrange-books",
"openwebmath_score": 0.726427435874939,
"openwebmath_perplexity": 277.66685596464566,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9780517450056274,
"lm_q2_score": 0.8596637487122111,
"lm_q1q2_score": 0.8407956295460572
} |
http://planetmath.org/310whenarepropositionstruncated | # 3.10 When are propositions truncated?
At first glance, it may seem that the truncated versions of $+$ and $\Sigma$ are actually closer to the informal mathematical meaning of “or” and “there exists” than the untruncated ones. Certainly, they are closer to the precise meaning of “or” and “there exists” in the first-order logic which underlies formal set theory, since the latter makes no attempt to remember any witnesses to the truth of propositions. However, it may come as a surprise to realize that the practice of informal mathematics is often more accurately described by the untruncated forms.
For example, consider a statement like “every prime number is either $2$ or odd.” The working mathematician feels no compunction about using this fact not only to prove theorems about prime numbers, but also to perform constructions on prime numbers, perhaps doing one thing in the case of $2$ and another in the case of an odd prime. The end result of the construction is not merely the truth of some statement, but a piece of data which may depend on the parity of the prime number. Thus, from a type-theoretic perspective, such a construction is naturally phrased using the induction principle for the coproduct type “$(p=2)+(p\text{ is odd})$”, not its propositional truncation.
Admittedly, this is not an ideal example, since “$p=2$” and “$p$ is odd” are mutually exclusive, so that $(p=2)+(p\text{ is odd})$ is in fact already a mere proposition and hence equivalent to its truncation (see http://planetmath.org/node/87851Exercise 3.7). More compelling examples come from the existential quantifier. It is not uncommon to prove a theorem of the form “there exists an $x$ such that …” and then refer later on to “the $x$ constructed in Theorem Y” (note the definite article). Moreover, when deriving further properties of this $x$, one may use phrases such as “by the construction of $x$ in the proof of Theorem Y”.
A very common example is “$A$ is isomorphic to $B$”, which strictly speaking means only that there exists some isomorphism between $A$ and $B$. But almost invariably, when proving such a statement, one exhibits a specific isomorphism or proves that some previously known map is an isomorphism, and it often matters later on what particular isomorphism was given.
Set-theoretically trained mathematicians often feel a twinge of guilt at such “abuses of language”. We may attempt to apologize for them, expunge them from final drafts, or weasel out of them with vague words like “canonical”. The problem is exacerbated by the fact that in formalized set theory, there is technically no way to “construct” objects at all — we can only prove that an object with certain properties exists. Untruncated logic in type theory thus captures some common practices of informal mathematics that the set theoretic reconstruction obscures. (This is similar to how the univalence axiom validates the common, but formally unjustified, practice of identifying isomorphic objects.)
On the other hand, sometimes truncated logic is essential. We have seen this in the statements of $\mathsf{LEM}$ and $\mathsf{AC}$; some other examples will appear later on in the book. Thus, we are faced with the problem: when writing informal type theory, what should we mean by the words “or” and “there exists” (along with common synonyms such as “there is” and “we have”)?
A universal consensus may not be possible. Perhaps depending on the sort of mathematics being done, one convention or the other may be more useful — or, perhaps, the choice of convention may be irrelevant. In this case, a remark at the beginning of a mathematical paper may suffice to inform the reader of the linguistic conventions in use therein. However, even after one overall convention is chosen, the other sort of logic will usually arise at least occasionally, so we need a way to refer to it. More generally, one may consider replacing the propositional truncation with another operation on types that behaves similarly, such as the double negation operation $A\mapsto\neg\neg A$, or the $n$-truncations to be considered in http://planetmath.org/node/87580Chapter 7. As an experiment in exposition, in what follows we will occasionally use adverbs to denote the application of such “modalities” as propositional truncation.
For instance, if untruncated logic is the default convention, we may use the adverb merely to denote propositional truncation. Thus the phrase
“there merely exists an $x:A$ such that $P(x)$
indicates the type $\mathopen{}\left\|\mathchoice{\sum_{x:A}\,}{\mathchoice{{\textstyle\sum_{(x:A)% }}}{\sum_{(x:A)}}{\sum_{(x:A)}}{\sum_{(x:A)}}}{\mathchoice{{\textstyle\sum_{(x% :A)}}}{\sum_{(x:A)}}{\sum_{(x:A)}}{\sum_{(x:A)}}}{\mathchoice{{\textstyle\sum_% {(x:A)}}}{\sum_{(x:A)}}{\sum_{(x:A)}}{\sum_{(x:A)}}}P(x)\right\|\mathclose{}$. Similarly, we will say that a type $A$ is merely inhabited to mean that its propositional truncation $\mathopen{}\left\|A\right\|\mathclose{}$ is inhabited (i.e. that we have an unnamed element of it). Note that this is a definition of the adverb “merely” as it is to be used in our informal mathematical English, in the same way that we define nouns like “group” and “ring”, and adjectives like “regular” and “normal”, to have precise mathematical meanings. We are not claiming that the dictionary definition of “merely” refers to propositional truncation; the choice of word is meant only to remind the mathematician reader that a mere proposition contains “merely” the information of a truth value and nothing more.
On the other hand, if truncated logic is the current default convention, we may use an adverb such as purely or constructively to indicate its absence, so that
“there purely exists an $x:A$ such that $P(x)$
would denote the type $\mathchoice{\sum_{x:A}\,}{\mathchoice{{\textstyle\sum_{(x:A)}}}{\sum_{(x:A)}}{% \sum_{(x:A)}}{\sum_{(x:A)}}}{\mathchoice{{\textstyle\sum_{(x:A)}}}{\sum_{(x:A)% }}{\sum_{(x:A)}}{\sum_{(x:A)}}}{\mathchoice{{\textstyle\sum_{(x:A)}}}{\sum_{(x% :A)}}{\sum_{(x:A)}}{\sum_{(x:A)}}}P(x)$. We may also use “purely” or “actually” just to emphasize the absence of truncation, even when that is the default convention.
In this book we will continue using untruncated logic as the default convention, for a number of reasons.
1. 1.
We want to encourage the newcomer to experiment with it, rather than sticking to truncated logic simply because it is more familiar.
2. 2.
Using truncated logic as the default in type theory suffers from the same sort of “abuse of language” problems as set-theoretic foundations, which untruncated logic avoids. For instance, our definition of “$A\simeq B$” as the type of equivalences between $A$ and $B$, rather than its propositional truncation, means that to prove a theorem of the form “$A\simeq B$” is literally to construct a particular such equivalence. This specific equivalence can then be referred to later on.
3. 3.
We want to emphasize that the notion of “mere proposition” is not a fundamental part of type theory. As we will see in http://planetmath.org/node/87580Chapter 7, mere propositions are just the second rung on an infinite ladder, and there are also many other modalities not lying on this ladder at all.
4. 4.
Many statements that classically are mere propositions are no longer so in homotopy type theory. Of course, foremost among these is equality.
5. 5.
On the other hand, one of the most interesting observations of homotopy type theory is that a surprising number of types are automatically mere propositions, or can be slightly modified to become so, without the need for any truncation. (See Lemma 3.3.5 (http://planetmath.org/33merepropositions#Thmprelem4),http://planetmath.org/node/87577Chapter 4,http://planetmath.org/node/87580Chapter 7,http://planetmath.org/node/87583Chapter 9,http://planetmath.org/node/87584Chapter 10.) Thus, although these types contain no data beyond a truth value, we can nevertheless use them to construct untruncated objects, since there is no need to use the induction principle of propositional truncation. This useful fact is more clumsy to express if propositional truncation is applied to all statements by default.
6. 6.
Finally, truncations are not very useful for most of the mathematics we will be doing in this book, so it is simpler to notate them explicitly when they occur.
Title 3.10 When are propositions truncated? \metatable | 2018-08-20T16:06:32 | {
"domain": "planetmath.org",
"url": "http://planetmath.org/310whenarepropositionstruncated",
"openwebmath_score": 0.9029503464698792,
"openwebmath_perplexity": 430.16087788279094,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9780517450056273,
"lm_q2_score": 0.8596637451167997,
"lm_q1q2_score": 0.8407956260295587
} |
https://www.freemathhelp.com/forum/threads/110313-Quadratic-Simultaneous-Eqns-quot-quot-J-and-T-share-money-If-I-square-J-s-money-and-add-quot?goto=nextoldest | # Thread: Enough conditions for this question? smallest number if exactly 93.6% answered
1. ## Enough conditions for this question? smallest number if exactly 93.6% answered
For the following question, do we have enough conditions to get the solution?
-----
What is the fewest number of people surveyed if exactly 93.6% of the people surveyed actually completed the whole survey? Explain your answer or show your work.
-----
thanks,
2. Originally Posted by mathdaughter
For the following question, do we have enough conditions to get the solution?
-----
What is the fewest number of people surveyed if exactly 93.6% of the people surveyed actually completed the whole survey? Explain your answer or show your work.
-----
thanks,
Yes you do.
First set up an equation.
$x = \text { number of persons surveyed.}$
$y = \text { number of persons who completed survey.}$
$0.936x = y.$
Can you give a survey to a fraction of a person?
Can you have a fraction of a person complete a survey?
So x and y must be positive whole numbers. Now what?
3. Originally Posted by mathdaughter
For the following question, do we have enough conditions to get the solution?
-----
What is the fewest number of people surveyed if exactly 93.6% of the people surveyed actually completed the whole survey? Explain your answer or show your work.
-----
thanks,
78% means 78/100. 78.5%=78.5/100 or 785/1000
Do the same for 93.6% and then reduce and you'll have your answer. Post back with your work so it can be verified as correct or to show where you went wrong.
4. 0.936x = y;
The fewest x should be 105 to make y a whole number. Because 6*5 is ended with 0, so 93.6*5 will a whole number, then 105 must be the smallest. Am I correct?
5. Originally Posted by mathdaughter
0.936x = y;
The fewest x should be 105 to make y a whole number. Because 6*5 is ended with 0, so 93.6*5 will a whole number, then 105 must be the smallest. Am I correct?
If 105 people were surveyed, then 0.936*105 =98.28 people completed the survey?
That's a fractional number of people. So no, that is not the right answer.
0.936x = y
divide both sides of the equation by x:
0.936 = y/x
You want the smallest whole numbers y and x that make this equation true. As Jomo hinted, y = 936 and x = 1000 certainly make the equation true, but they're not the smallest numbers that do. So you need to reduce the fraction 936/1000 to an equivalent one with the smallest possible values.
6. Originally Posted by mathdaughter
0.936x = y;
The fewest x should be 105 to make y a whole number. Because 6*5 is ended with 0, so 93.6*5 will a whole number, then 105 must be the smallest. Am I correct?
You are thinking in the correct direction, BUT
$0.936 * 105 = 98.28.$ NOT A WHOLE NUMBER.
The 5 helps with the 0.006, but not with 0.93.
Let's think like this:
$0.936x = y \implies 0.936 = \dfrac{y}{x} \implies 5 * 0.936 = 4.68 = 5 * \dfrac{y}{x} \implies$
$5 * 4.68 = 23.4 = 5 * 5 * \dfrac{y}{x} = 25 * \dfrac{y}{x} \implies$
$5 * 23.4 = 117 = 5 * 25 * \dfrac{y}{x} = 125 * \dfrac{y}{x} \implies WHAT?$
EDIT: Jomo's way is perhaps easier arithmetically, but perhaps less intuitive relative to your personal thought process. There are frequently several ways to solve a problem.
7. I was wrong.
0.936x = y
y/x = 0.936 = 936/1000 = 468/500 = 234/250 = 117/125
so x = 125 is the fewest?
8. Originally Posted by mathdaughter
I was wrong.
0.936x = y
y/x = 0.936 = 936/1000 = 468/500 = 234/250 = 117/125
so x = 125 is the fewest?
Yes. Well done.
9. Originally Posted by mathdaughter
I was wrong.
0.936x = y
y/x = 0.936 = 936/1000 = 468/500 = 234/250 = 117/125
so x = 125 is the fewest? | 2018-12-12T07:16:02 | {
"domain": "freemathhelp.com",
"url": "https://www.freemathhelp.com/forum/threads/110313-Quadratic-Simultaneous-Eqns-quot-quot-J-and-T-share-money-If-I-square-J-s-money-and-add-quot?goto=nextoldest",
"openwebmath_score": 0.7517880201339722,
"openwebmath_perplexity": 1370.5013401287154,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9891815487957215,
"lm_q2_score": 0.849971181358171,
"lm_q1q2_score": 0.8407758096076047
} |
http://mathhelpforum.com/trigonometry/95176-deduce-inequality.html | # Math Help - Deduce to inequality
1. ## Deduce to inequality
By using the formulae expressing $sin\theta$ and $cos\theta$ in terms of t, where $(t\equiv tan\frac{\theta}{2})$ or otherwise, show that $\frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^ 2}{9+t^2}$.
Deduce that $0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10} {9}$
I have done the first part using the identities $sin\theta\equiv\frac{2t}{1+t^2}$ and $cos\theta\equiv\frac{1-t^2}{1+t^2}$ where $t\equiv tan\frac{\theta}{2}$, but I don't know how to do the second part.
Thanks
2. Originally Posted by arze
By using the formulae expressing $sin\theta$ and $cos\theta$ in terms of t, where $(t\equiv tan\frac{\theta}{2})$ or otherwise, show that $\frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^ 2}{9+t^2}$.
Deduce that $0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10} {9}$
I have done the first part using the identities $sin\theta\equiv\frac{2t}{1+t^2}$ and $cos\theta\equiv\frac{1-t^2}{1+t^2}$ where $t\equiv tan\frac{\theta}{2}$, but I don't know how to do the second part.
Thanks
the left side of the inequality is clear. to prove the right side we have: $\frac{(1+t)^2}{9+t^2} -\frac{10}{9}=\frac{9(1+t)^2 - 10(9+t^2)}{9(9+t^2)}=\frac{-t^2+18t-81}{9(9+t^2)}=\frac{-(t-9)^2}{9(9+t^2)} \leq 0.$
but what if the question was this: find the maximum value of $f(t)=\frac{(1+t)^2}{9+t^2}, \ t \in \mathbb{R}$? (no calculus allowed!)
3. We have to prove that $0\leq\frac{(1+t)^2}{9+t^2}\leq\frac{10}{9}$
The first inequality is obviously true.
For the second:
$\frac{(1+t)^2}{9+t^2}\leq\frac{10}{9}\Leftrightarr ow 9+18t+9t^2\leq 90+10t^2\Leftrightarrow t^2-18t+81\geq 0\Leftrightarrow(t-9)^2\geq 0$
4. $
\frac{1 + \sin\theta}{5 + 4\cos{\theta}}
$
$= \frac{ (\sin{\theta/2} + \cos{\theta/2})^2}{1 + 4(2)(\cos^2{\theta/2})}$
$= \cos^2{\theta/2} \frac{ (1 + \tan{\theta/2})^2}{ \sin^2{\theta/2} + 9\cos^2{\theta/2}}$
$= \frac{ (1+\tan{\theta/2})^2}{ 9 + \tan^2{\theta/2}}$
for part 2 , besides using $b^2 - 4ac \geq 0$
here is another method :
$\frac{(1+t)^2 }{t^2 + 9}$
$= \frac{ t^2 + 9 + 2t - 8}{(t-4)^2 + 8(t-4) + 25}$
$= 1 + \frac{2}{ (t-4) + 8 + \frac{25}{t-4}}$
$\leq 1 + \frac{2}{2\sqrt{25} + 8}$
$= \frac{ 10}{9}$ | 2015-08-05T10:38:23 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/trigonometry/95176-deduce-inequality.html",
"openwebmath_score": 0.9338953495025635,
"openwebmath_perplexity": 116.15778347435547,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9891815487957216,
"lm_q2_score": 0.8499711794579723,
"lm_q1q2_score": 0.8407758077279632
} |
https://math.stackexchange.com/questions/3493871/a-collection-of-sets-that-cover-all-edges-in-kn | # A collection of sets that cover all edges in Kn?
The problem is the following:
Let $$\mathcal{F}$$ be a family of distinct proper subsets of {1,2,...,n}. Suppose that for every $$1\leq i\neq j\leq n$$ there is a unique member of $$\mathcal{F}$$ that contains both $$i$$ and $$j$$. Prove that $$\mathcal{F}$$ has at least $$n$$ elements.
I realized that each member $$A_i$$ in $$\mathcal{F}$$ can be regarded as a collection of edges in a clique induced by the element in $$A_i$$, and at the end all the members in $$\mathcal{F}$$ exactly cover all the edges in $$K_n$$ and every edge must live in exactly one member in $$\mathcal{F}$$. So if we write $$\mathcal{F}=\{A_1,...,A_m\}$$ with sizes $$a_1,...,a_m$$, respectively, then we obtain the following equation:
$$a_1\choose 2$$+...+$$a_m\choose 2$$ = $$n\choose 2$$ =$$\frac{n(n-1)}{2}$$.
I got stuck from here. Even if I assumed $$m, I cannot obtain a contradiction. Is my approach correct at all? If so, how should I continue? I appreciate any help and insights.
• Interesting question. You want to assume $n\ge2$ because of the silly counterexample $n=1$, $\mathcal F=\emptyset$. The statement is true vacuously for $n=2$ as there is no such $\mathcal F$, non-vacuously for $n=3$. Maybe there is a clever proof by induction, but I don't see it at the moment. An interesting example of the structure you're talking about is a finite projective plane, where $\{1,2,\dots,n\}$ is the set of all points and $\mathcal F$ the set of all lines, or vice versa. Is that where this came up?
– bof
Jan 1, 2020 at 9:55
• @bof interesting..., and thanks for the new perspective. I didn't expect it would be related to this kind of questions. In fact it's an exercise from my course in algebraic combinatorics Jan 5, 2020 at 20:36
This result is known as the de Bruijn–Erdős theorem. It is usually stated in terms of incidence geometries. An incidence geometry has:
• A set of objects called points.
• A collection of sets of points called lines. If a line $$\ell$$ contains a point $$P$$, we also say that $$P$$ lies on $$\ell$$, or $$\ell$$ passes through $$P$$.
We require that for any two points, there is a unique line passing through both.
We also make some nontriviality assumptions. Each line contains two points (though this doesn't matter for this question, since single-point lines only help us). There is not a line containing all the points (which does matter for your question: the subsets in $$\mathcal F$$ must be proper subsets, or the conclusion is false).
In your case, $$\{1,2,\dots,n\}$$ is the set of points and $$\mathcal F$$ is the set of lines.
The de Bruijn–Erdős theorem simply says that in any incidence geometry, there are at least as many lines as points.
If you've gone ahead and read the Wikipedia article I linked to earlier, then you've seen the proof there, which assumes a Euclidean incidence geometry: one in which the points are some finite set of points in the plane, and the lines are actual lines drawn through those points. This is not the proof that de Bruijn and Erdős gave, for which you can read their original paper; their proof applies to any abstract incidence geometry, and so it solves your problem.
A few years ago, I also wrote an alternative writeup of their proof, which you can find here.
A parital solution:
Let $$n \in \mathbb{N}$$ and let $$\mathcal{F}$$ such that $$\forall i,j \in [n]\exists S\in \mathcal{F}:i,j\in S$$ and that $$S$$ is unique.
Denote by $$t$$ the size of the biggest set in $$\mathcal{F}$$ and assume WLOG that the set is $$S=\{1,...,t\}$$.
Case 1: $$t\ge \frac{n}{2}$$:
For each $$i \in S$$, we must have a set containing the pair $$(t+1,i)$$, and as we can't have $$2$$ elements from $$S$$ together in another set, we must have $$t$$ new sets.
As $$t+2$$ can be in at most $$1$$ of the above sets, we need at least $$t-1$$ new sets for $$t+2$$ (to have the edges $$(t+2,i): i\in S$$). We can continue similarly until $$n=t+(n-t)$$ intoduces $$t-(n-t-1)$$ new sets.
We got a total of $$1+t(n-t)-(0+1+...+n-t-1)=1+t(n-t)-\frac{(n-t)(n-t-1)}{2}=$$ $$1+\frac{n-t}{2}(3t-n+1)$$ sets.
Treating $$t$$ as a variable and $$n$$ fixed, we can see that the function has a global maxima at $$t \sim \frac{2}{3}n$$ so it is enough to check for $$t=n-1, \frac{n}{2}$$.
For $$t=n-1$$ we get $$n$$ subsets, which is exactly enough.
For $$t=\frac{n}{2}$$ we get $$\ge \frac{n}{4}(\frac{n}{2}+1)$$. We want it to be $$\ge n$$ so we need $$\frac{n}{2} + 1 \ge 4$$ or $$n\ge 6$$.
So we solved for this case for every $$n$$ above $$5$$, and for $$n=2,3,4,5$$ it can be verified manually.
Case 2: $$t < \frac{n}{2}$$
First we can observe that if there are less than $$n$$ subsets in $$\mathcal{F}$$, then $$t\ge \sqrt n$$:
Each set contributes at most $$t \choose 2$$ pairs and we have at most $$n-1$$ subsets, and so at most $$(n-1)\binom{t}{2}$$. On the other hand, we know there are exactly $$n \choose 2$$ pairs. So, we got that $$\binom{n}{2}\le (n-1)\binom{t}{2}\rightarrow n\le t(t-1)\le t^2 \rightarrow t\ge \sqrt n$$
We can use a similar method to the previous case to deduce that there are at least $$1+2+...+t$$ other subsets, so there are at least $$\binom{\sqrt{n} + 1}{2}\ge \frac{n}{2}$$ subsets.
I was not able to improve the bound in this case.
Hope it gives you an idea that can generalize it to a complete solution. | 2022-05-23T11:00:42 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3493871/a-collection-of-sets-that-cover-all-edges-in-kn",
"openwebmath_score": 0.9934759736061096,
"openwebmath_perplexity": 730.46779631572,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9891815526228442,
"lm_q2_score": 0.8499711718571775,
"lm_q1q2_score": 0.8407758034623413
} |
https://cs.stackexchange.com/questions/140049/douglas-peucker-line-simplification-algorithm-time-complexity | # Douglas-Peucker line simplification algorithm time complexity
I am analyzing the time complexity of the Douglas-Peucker line simplification algorithm. Reading online I've found that it has a worst-case running time of $$O(n^2)$$ where $$n$$ is the number of points on the line. However, I haven't been able to provide complete proof. Here is the pseudo-code I am working with:
Here is my analysis:
Line 1: $$O(1)$$
Line 2: We can say that it goes through all points in the polyline formed by the first point and the last point in $$PL$$ and return the point with the maximum distance to the straight line between the mentioned points. We can safely say this is $$O(n)$$
Lines 4-6: If we are here we are doing two recursive calls. The number of elements used in each call could be different (not always in half). Depends on the point found in Line 2. Leaving implementation details aside, we can say that splitting is $$O(1)$$. Also, we have a $$JOIN$$ operation which would take $$O(n)$$ to join both lists.
Line 8: This is also a $$JOIN$$ operation but this only takes $$O(1)$$.
After looking at these steps I think we have the following recurrence relation:
$$T(n) = T(index + 1) + T(n - index) + 2n$$ where $$index \in \{1, \dots, n-1\}$$
Am I right? If so, how could I prove this is $$O(n^2)$$? Master Theorem?
Thank you very much in advance
EDIT: Changed description on Lines 4-6. Thus, also changed recurrence formula.
Your analysis seems correct, except for that you need $$T(n)=T(index)+T(n-index+1)+2n$$ instead because when $$index$$ is either $$0$$ or $$n-1$$ you won't get in the right hand side another factor of $$T(n)$$, which is not allowed.
In fact, it will yield something very close to the recurrence of the running time in the quicksort algorithm.
## Worst case scenario
When index is either $$0$$ or $$n-1$$, the recurrence will boil down to $$T(n)=T(1)+T(n-1)+2n = T(n-1) + (2n + T(1))$$. You can easily show by induction that $$T(n)=\Theta(n^2)$$ in this case.
## Best case scenario
We assume the input in the recursion will always be cut down by half. That is, the recursion is $$T(n)=2T(n/2)+2n$$.
Define $$b=a=2$$ and $$f(n)=2n$$. Then, $$T(n)=aT(n/b)+f(n)$$.
We can calculate $$c_{crit}=\log_{b}(a)=\log_2(2)=1$$ and thus $$f(n)=2n=\Theta(n)=\Theta(n^{c_{crit}})$$.
Now we can directly use the second case to get that $$T(n)=\Theta(n^{c_{crit}}\log(n))=\Theta(n\log(n))=O(n^2)$$
## Average case scenario
If we assume that the input is a random variable, such that the choice of $$index$$ is uniformly distributed in $$\{0,\dots,n-1\}$$, then the algorithm takes expected time $$\Theta(n\log(n))$$. Take a look at the proof for quicksort to see how its done.
• Thanks for answering, does this means that $O(n^2)$ is a loose upper bound for this, right? And the tighter one would be $O(n\lg n)$? May 8 at 0:26
• @nir_shahar. I changed the information on the question because I was assuming the information was cut in half and it is not. It depends on the farthest point. How would that change the proof? May 8 at 1:04
• @DarK_FirefoX I updated my answer according to it. May 8 at 9:16
• Thank you very much! I am only interested in the worst-case scenario, but your explanation of the other cases help me understand better the algortihm. May 8 at 16:30 | 2021-10-21T14:39:33 | {
"domain": "stackexchange.com",
"url": "https://cs.stackexchange.com/questions/140049/douglas-peucker-line-simplification-algorithm-time-complexity",
"openwebmath_score": 0.7504863142967224,
"openwebmath_perplexity": 218.56562225967426,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9891815538985516,
"lm_q2_score": 0.8499711680567799,
"lm_q1q2_score": 0.8407758007873725
} |
https://math.stackexchange.com/questions/3981907/how-does-subtracting-an-exponent-from-an-exponent-return-a-greater-value | # How does subtracting an exponent from an exponent return a greater value?
I'm just stuck on a problem where I have to simplify an expression. This is the expression:
$$\sqrt{x^2\:-\:\left(\frac{x}{2}\right)^2}$$
The textbook has the answer as:
$$\frac{\sqrt{3}x}{2}$$
I have no idea how to get to that. I've tried to figure it out for the past hour and I have tried online solvers but I can't understand what the steps are. Any help would be very much appreciated.
Thanks!
• What have you tried so far? – perpetuallyconfused Jan 12 at 2:21
• $x^2-\left(\dfrac x2\right)^2=\dfrac{3x^2}4$ – J. W. Tanner Jan 12 at 2:28
• A little bit picky, but the text-book is wrong. The expression actually simplifies to $\frac {|x|\sqrt3}{2}$ – Doug M Jan 12 at 2:36
• @DougM Not picky at all, I would say. – saulspatz Jan 12 at 2:40
• J.W. I understand that but where do you get the 3 from? – Scratch Cat Jan 12 at 2:49
You solution assumes that $$x\ge 0$$.
\begin{align} x^2-(\frac{x}{2})^2 &=x^2-\frac{x^2}{4}\\ &=1\cdot x^2-\frac{1}{4}\cdot x^2\\ &=(1-\frac{1}{4})x^2= \frac34 x^2 \end{align}
So $$\sqrt{x^2-(\frac{x}{2})^2}=\sqrt{\frac34 x^2} =\frac{\sqrt{3}}{\sqrt{4}}\sqrt{x^2}=\frac{\sqrt{3}}{2}x\quad x\ge 0$$ | 2021-05-12T18:00:26 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3981907/how-does-subtracting-an-exponent-from-an-exponent-return-a-greater-value",
"openwebmath_score": 0.7426757216453552,
"openwebmath_perplexity": 839.807830321249,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9891815491146485,
"lm_q2_score": 0.8499711699569787,
"lm_q1q2_score": 0.8407757986008344
} |
https://www.physicsforums.com/threads/a-quadratic-polynomial.604349/ | Hello, I just wanna clear up a confusion.
Is f(x,y) = a + bx + cy + dxy
where x and y are variables and a,b,c,d are constants.
Mark44
Mentor
Hello, I just wanna clear up a confusion.
Is f(x,y) = a + bx + cy + dxy
where x and y are variables and a,b,c,d are constants.
No. A quadratic polynomial in two variables looks like this:
f(x, y) = ax2 + bxy + cy2 + dx + ey + f. See http://en.wikipedia.org/wiki/Quadratic_polynomial (two variables case).
HallsofIvy
Homework Helper
Mark44, I don't understand your response. You say "no" but then you give the general form for a quadratic in two variables which fits the one given by n0ya.
$a+ bx+ cy+ dxy$
is the same as $ax^2+ bxy+ cy^2+ dx+ ey+ f$.
Those are the same with your a= 0, b equal to nOya's d, c= 0, d equal to n0ya's b, e equal to n0ya's c, and your f equal to n0ya's a.
mathman
Hello, I just wanna clear up a confusion.
Is f(x,y) = a + bx + cy + dxy
where x and y are variables and a,b,c,d are constants.
The usual terminology would call this bilinear.
Mark44
Mentor
I was under the impression that a and c could not both be zero, but that b could be zero. Clearly, we can't have a = b = c = 0, or the function wouldn't be quadratic, so there need to be some restrictions on a, b, and c (but not on d, e, and f). Perhaps the restriction is that not all of a, b, c can be zero.
Hello, I just wanna clear up a confusion.
Is f(x,y) = a + bx + cy + dxy
where x and y are variables and a,b,c,d are constants.
To reduce this problem to its essence, take f(x,y) = xy. On the one hand there is no x2 or y2 term; but there is an xy term. So is this a quadratic or not?
chiro
Hey n0ya and welcome to the forums.
As a guiding rule for further questions like this, look at the highest term (in terms of the order) and use that as a basis to figure out what kind of equation/polynomial/blah something is on top of other things that need to be considered (for example polynomials only have integer powers so a x^(1/2) in the expression would make it a non-polynomial instantly).
My problem is really whether f(x,y) = xy is a quadratic polynomial, as SteveL27 points out.
I know that this polynomial is of degree 2, and according to Wikipedia, "a quadratic polynomial or quadratic is a polynomial of degree two", it is a quadratic polynomial. However, some argue that a x^2 or a y^2 term is needed in order to call it 'quadratic', so I just wanted to clear up this confusion.
HallsofIvy
Homework Helper
Yes, it is a quadratic polynomial. If you were to rotate the coordinate system through 45 degrees, calling the new coordinates x' and y', f(x,y)= xy would become $x'^2- y'^2= 1$. Another way of looking at it is that the graph is a conic section, a hyperbola.
HallsofIvy
Homework Helper
Hello, I just wanna clear up a confusion.
Is f(x,y) = a + bx + cy + dxy
where x and y are variables and a,b,c,d are constants.
The usual terminology would call this bilinear.
No, it would not. "xy" is not linear.
No, it would not. "xy" is not linear.
It's bilinear, meaning that
x(y1 + y2) = xy1 + xy2
and x(ay) = a(xy) for a constant a;
and likewise in y. It's linear in each variable.
mathman
In my experience there is a distinction between polynomials in one variable and polynomials in two variables. Highest order term x2y2 is called biquadratic and x3y3 is called bicubic.
Thanks for the replies.
Then, as far as I understand it, if you for example have: x = t; y = t, then f(x,y)=xy will be a parabola (t^2).
Anyway, I think both quadratic and bilinear are both correct terms here. However, it might be more intuative to call it a bilinear mapping, instead of a quadratic polynomial, since we have no x^2 or y^2 terms.
coolul007
Gold Member
Thanks for the replies.
Then, as far as I understand it, if you for example have: x = t; y = t, then f(x,y)=xy will be a parabola (t^2).
Anyway, I think both quadratic and bilinear are both correct terms here. However, it might be more intuative to call it a bilinear mapping, instead of a quadratic polynomial, since we have no x^2 or y^2 terms.
most of the time when you have an xy term the curve is a hyperbola
Mark44
Mentor
Then, as far as I understand it, if you for example have: x = t; y = t, then f(x,y)=xy will be a parabola (t^2).
most of the time when you have an xy term the curve is a hyperbola
f(x, y) = xy represents a surface in three dimensions, not a curve in the plane, as are parabolas and hyperbolas. According to wolframalpha, the surface is a hyperbolic paraboloid.
coolul007
Gold Member
f(x, y) = xy represents a surface in three dimensions, not a curve in the plane, as are parabolas and hyperbolas. According to wolframalpha, the surface is a hyperbolic paraboloid.
I think you found a bug, if you plot the old fashioned way n=xy, you get all of the real factors of n and it's a hyperbola. BTW, where's the third dimension variable?
Mentallic
Homework Helper
BTW, where's the third dimension variable?
f(x,y) IS the 3rd dimension variable. It's basically equivalent to z=xy. For any fixed value z, then yes, you're dealing with a hyperbola, but if you consider the graph intersected by the plane y=mx, then you're dealing with a parabola z=mx2, hence why it's called a hyperbolic paraboloid.
coolul007
Gold Member
f(x,y) IS the 3rd dimension variable. It's basically equivalent to z=xy. For any fixed value z, then yes, you're dealing with a hyperbola, but if you consider the graph intersected by the plane y=mx, then you're dealing with a parabola z=mx2, hence why it's called a hyperbolic paraboloid.
It must be too early, I was thinking f(x) not f(x,y)...so what is the answer to the original question? did we settle on bi-linear?
Mentallic
Homework Helper
I don't know. There seems to be quite a few different proposals and arguments for each term. I don't think we should settle on anything yet :tongue: | 2021-01-26T12:41:11 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/a-quadratic-polynomial.604349/",
"openwebmath_score": 0.7496936321258545,
"openwebmath_perplexity": 561.7730328490368,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9637799462157138,
"lm_q2_score": 0.8723473846343394,
"lm_q1q2_score": 0.8407509154443022
} |
https://math.stackexchange.com/questions/1942983/find-int-02-pi-frac15-4-cos-x-dx | # Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$
$$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$ How do I compute this integral? An online integral calculator gives an antiderivative as $$\frac{2\arctan(3\tan\frac x2)}3$$ but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative vanishes at $x=0$ and $x=2\pi$.
• Another way would be to use contour integration =). – Jacky Chong Sep 27 '16 at 0:52
Using complex analysis your integral can be expressed as $$\int \limits^{2\pi }_{0}\frac{1}{5-4\cos \left( x\right) } dx=\oint \frac{1}{5-2\left( z+\frac{1}{z} \right) iz} dz$$ Now we have two singularites at $z=2$and at $z=\frac {1}{2}$ Now apply Residue theorem $$\oint \frac{1}{5-2\left( z+\frac{1}{z} \right) iz} dz = 2\pi i \ \text {Res}\sum \left( f\left( z\right) , z\right)$$ We have only one singularity in the contour.
So the Residue will be $$2 \pi i \ \text {Res} \sum(f (z), z)=2 \pi i \left[\lim \limits_{z\to \frac{1}{2} }\left( 2z-1\right) \frac{1}{2i\left( -z+2\right) \left( 2z-1\right) } =\frac{1}{3i} \right]=\frac {2 \pi}{3}$$ So $$\color{Brown }{\int \limits^{2\pi }_{0}\frac{1}{5-4\cos \left( x\right) } dx=\boxed {\frac{2\pi }{3}} }$$
• I came across this integral by reducing to it from the integral of $\frac{1}{2z^2 - 5z + 2}$ about the counter clockwise unit circle. Haven't done the residue theorem yet. – Vik78 Sep 27 '16 at 6:29
Your example shows that a computer algebra system may break the Fundamental theorem of calculus. That is, it may give a different result for $\int_a^bf(x)\;\mathrm dx$ and $F(b)-F(a)$, where $F$ is an antiderivative of $f$, even if $f$ has no singularity. This is a very bad thing.
Since the fundamental theorem of calculus is true, it simply means the antiderivative must be incorrect, or the definite integral, or both.
Here the CAS may be doing what one does by hand: a change of variable $u=\tan \frac x2$ (see Tangent half-angle substitution on Wikipedia).
$$\int \frac{\mathrm dx}{5-4\cos x}= \int \frac{1}{5-4\dfrac{1-u^2}{1+u^2}}\frac{2\mathrm du}{1+u^2} =\int \frac{2\mathrm du}{1+(3u)^2}=\frac23\arctan(3u)\\=\frac23\arctan(3\tan \frac x2)$$
But you can do that only on an interval where $u=\tan \frac x2$ is a continuous and differentiable bijection, so not on $[0,2\pi]$ (it's discontinuous at $\pi$). However, you can do that on $[0,\pi[$, and
$$\int_0^{2\pi} \frac{\mathrm dx}{5-4\cos x}=2\int_0^\pi \frac{\mathrm dx}{5-4\cos x}=\frac43[\arctan(3\tan\frac x2)]_0^{\to\pi^-}=\frac23\pi$$
So what happens is this: when asked for an antiderivative, it gives one that is only valid on an interval, though the function is continuous everywhere, so there should be an antiderivative valid on the whole of $\Bbb R$. This is wrong. However, the CAS is clever enough to handle correctly the definite integral. I tried with Maxima, and it does the same.
The correct antiderivative is not given by a single expression, but a disjunction of cases:
• if $x\in]2k\pi-\pi,2k\pi+\pi[$, $F(x)=\frac23\arctan(3\tan\frac x2)+\frac23k\pi+C$
• if $x=(2k+1)\pi$, $F(x)=\frac13(2k+1)\pi+C$
Where $C$ is the same constant for all values of $x$.
Here is a plot showing the difference. The correct antiderivative (with $C=0$) is in blue, and the expression given by the CAS in red.
Notice that on all intervals $]2k\pi-\pi,2k\pi+\pi[$, the antiderivative given by the CAS is correct. It just has to be shifted properly on each interval to get a function with only removable singularities at $(2k+1)\pi$, and after defining the values at $(2k+1)\pi$, you get a continuous antiderivative on $\Bbb R$. That's how $F$ is defined above.
• Nice answer! Here's an interesting article by D. J. Jeffrey about such matters: jstor.org/stable/2690852 – Hans Lundmark Sep 27 '16 at 8:21
• @HansLundmark Very interesting article! Actually, I was wrong when I stated that the antiderivative is not given by a single expression: this article gives $$\frac13x+\frac23\arctan \frac{\sin x}{2-cos x}$$ I'll have a look at the method he develops in this article. There is the detailed derivation for the same example, $\displaystyle{\int \frac{3\mathrm dx}{5-4\cos x}}$. – Jean-Claude Arbaut Sep 28 '16 at 6:15
A simple way: $$I=\int_{0}^{2\pi}\frac{dx}{5-4\cos x}=\int_{0}^{\pi}\left(\frac{1}{5-4\cos x}+\frac{1}{5+4\cos x}\right)\,dx\tag{1}$$ $$I = 10\int_{0}^{\pi}\frac{dx}{25-16\cos^2 x}=20\int_{0}^{\pi/2}\frac{dx}{25-16\cos^2 x} \tag{2}$$ and now, by setting $x=\arctan t$, $$I = 20 \int_{0}^{+\infty}\frac{dt}{25(1+t^2)-16} = 20 \int_{0}^{+\infty}\frac{dt}{9+25 t^2}\tag{3}$$ so by setting $t=\frac{3}{5}u$, $$I = 20\cdot\frac{3}{5}\cdot\frac{1}{9}\int_{0}^{+\infty}\frac{du}{1+u^2}=\frac{4}{3}\cdot\frac{\pi}{2}=\color{red}{\frac{2\pi}{3}}.\tag{4}$$
• Really like this approach. Love the trick you use to get the first equality. – Vik78 Sep 28 '16 at 2:41
• The first line here is a beautiful observation on the graphs for those two functions. – complexmanifold May 24 '18 at 11:12 | 2019-07-15T20:17:04 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1942983/find-int-02-pi-frac15-4-cos-x-dx",
"openwebmath_score": 0.9415411949157715,
"openwebmath_perplexity": 230.4848508550575,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9637799482964025,
"lm_q2_score": 0.8723473763375644,
"lm_q1q2_score": 0.8407509092631201
} |
https://stats.stackexchange.com/questions/175271/can-i-get-a-cholesky-decomposition-from-the-inverse-of-a-matrix/268534 | # Can I get a Cholesky decomposition from the inverse of a matrix?
I have the inverse of a giant covariance matrix from which I'd like to draw random instances.
The way I know how to do this is to do a Cholesky decomposition on the covariance matrix and use it to transform a vector of independent Gaussians. So the straightforward process is to invert the giant matrix and then do the Cholesky decomposition.
Since Cholesky decomposition can be used to get inverses in the first place, it seems like there might be a way to get what I need without the inversion step (which takes a long time).
Question: Is there a way to use Cholesky decomposition to do this without inverting the matrix?
You can avoid inverting the matrix by generating draws by means of the eigendecomposition method. According to this method, the draws are generated by doing this product: $$(V D)^\top X^\top \,,$$
where $V$ is the eigenvectors of the matrix, $D$ is a diagonal matrix containing the square roots of the eigenvalues and $X$ is a matrix containing draws from the standard univariate $N(0,1)$ distribution.
It is straightforward to adapt this method and avoid recovering the original covariance matrix by using these results: 1) the eigenvalues of a matrix $A$ are the reciprocal of the eigenvalues of its inverse $A^{-1}$; 2) the eigenvectors of a matrix A are also eigenvectors of its inverse $A^{-1}$.
Example:
Let's say that the original covariance matrix is the following:
$$A = \left[ \begin{array}{rrr} 1 & 0.8 & -0.4 \\ 0.8 & 2 & 0.3 \\ -0.4 & 0.3 & 3 \end{array} \right] \,.$$ But you have the inverse of this matrix, $B=A^{-1}$: $$A^{-1}= B = \left[ \begin{array}{rrr} 1.699 & -0.725 & 0.299 \\ -0.725 & 0.817 & -0.178 \\ 0.299 & -0.178 & 0.391 \end{array} \right] \,.$$
The eigendecomposition method based on the original matrix $A$ yields the following covariance matrix for a sample of draws:
A <- rbind(c(1,0.8,-0.4), c(0.8,2,0.3), c(-0.4,0.3,3))
e1 <- eigen(A, symmetric=TRUE)
set.seed(1)
X <- matrix(rnorm(5000*ncol(A)), ncol=ncol(A))
draws1 <- t(e1$vectors %*% sqrt(diag(e1$values)) %*% t(X))
draws1.cov <- cov(draws1)
draws1.cov
# [,1] [,2] [,3]
#[1,] 0.9765023 0.8030752 -0.3970233
#[2,] 0.8030752 1.9941052 0.3229827
#[3,] -0.3970233 0.3229827 3.1689348
Using the matrix that you have (the inverse of A), you just need to invert the eigenvalues:
B <- solve(A)
e2 <- eigen(B, symmetric=TRUE)
e2$values <- 1/e2$values
draws2 <- t(e2$vectors %*% sqrt(diag(e2$values)) %*% t(X))
draws2.cov <- cov(draws2)
draws2.cov
# [,1] [,2] [,3]
#[1,] 0.9765023 0.8030752 -0.3970233
#[2,] 0.8030752 1.9941052 0.3229827
#[3,] -0.3970233 0.3229827 3.1689348
A covariance matrix closely matching the original one is obtained and we didn't need to invert B in order to recover the original covariance matrix A.
A small simulation to check the validity of this approach:
set.seed(3)
niter <- 1000
m <- matrix(0, nrow=ncol(A), ncol=ncol(A))
for (i in seq_len(niter))
{
X <- matrix(rnorm(5000*ncol(A)), ncol=ncol(A))
draws2 <- t(e2$vectors %*% sqrt(diag(e2$values)) %*% t(X))
m <- m + cov(draws2)
}
m/niter
# average covariance matrix
# [,1] [,2] [,3]
#[1,] 1.0005129 0.7995872 -0.4005644
#[2,] 0.7995872 1.9993231 0.2990850
#[3,] -0.4005644 0.2990850 2.9957277
# original covariance matrix 'A'
# [,1] [,2] [,3]
#[1,] 1.0 0.8 -0.4
#[2,] 0.8 2.0 0.3
#[3,] -0.4 0.3 3.0
We can see that the covariance matrix of the draws are on average very close to the original covariance matrix A.
• Unrelated request: I have proposed to make arma a synonym of arima and advertised this on Meta here a while ago (without much success). As one of the top scorers under the arima tag, you may vote on the proposal here. – Richard Hardy Mar 21 '17 at 12:56 | 2019-08-26T03:44:26 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/175271/can-i-get-a-cholesky-decomposition-from-the-inverse-of-a-matrix/268534",
"openwebmath_score": 0.7065805196762085,
"openwebmath_perplexity": 747.9065985529569,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9637799462157138,
"lm_q2_score": 0.8723473713594991,
"lm_q1q2_score": 0.8407509026502774
} |
https://math.stackexchange.com/questions/4459712/why-is-the-dot-product-of-a-vector-with-itself-not-a-linear-function | # Why is the dot product of a vector with itself not a linear function? [closed]
If an inner product is linear by definition, i.e., $$\langle\mathbf{v+w},\mathbf{u}\rangle=\langle\mathbf{v},\mathbf{u}\rangle+\langle\mathbf{w},\mathbf{u}\rangle$$ and $$\langle a\mathbf{v},\mathbf{w}\rangle=a\langle\mathbf{v},\mathbf{w}\rangle$$, and the dot product is an inner product, then why is $$f(\mathbf{x})=\mathbf{x}\cdot\mathbf{x }$$ not a linear function?
Thanks
• Can you tell me what is $f(2\mathbf{x})$? More specifically, with $\mathbf{x}$ a non-zero vector. May 27 at 9:02
• because it is linear in both components seperately, so if you try to manipulate both sides at the same time (which you are doing by throwing in x in both) you get something that behaves more like a sqare. May 27 at 9:04
• Question looks like: $x\cdot y$ is linear function of both $x$ and $y$, so why $x\cdot x$ is not linear function of $x$. May 27 at 10:10
• Expanding on the comment by @IvanKaznacheyeu (which at a glance might still appear to be referring to dot product!): Take the ordinary multiplication $xy$. For a given $y$ (not varying with $x$), $f(x)=xy$ is linear in $x$, and vice versa. ($f(x,y)$ is bilinear.) But $f(x,x)$ is not linear in $x$, because the second argument is varying with $x$. May 27 at 18:19
If an inner product is linear by definition ...
No, an inner product $$\langle ,\cdot,\rangle$$ of a (real) vector space $$V$$, is bilinear, not linear. It is linear in each component.
Moreover, the domain of an inner product on $$V$$ is $$V\times V$$, while that of a linear map on $$V$$ is $$V$$. These two objects have completely different domains.
A linear map is a function between two vector spaces that preserves vector addition and scalar multiplication. Given a vector space $$V$$ over $$K$$ (where $$K$$ equals $$\mathbb R$$ or $$\mathbb C$$), an inner product on $$V$$ is a function $$V\times V\to K$$ that satisfies certain axioms: conjugate symmetry, linearity in the first argument, and positive-definiteness. These axioms only "view" $$V\times V$$ as a set, and $$K$$ as a field; neither $$V\times V$$ nor $$K$$ are assumed to be equipped with a vector space structure. Therefore, it doesn't really make sense to assert that the inner product is linear.
What we can say is that the inner product is linear in the first argument, i.e. $$\langle\mathbf{v+w},\mathbf{u}\rangle=\langle\mathbf{v},\mathbf{u}\rangle+\langle\mathbf{w},\mathbf{u}\rangle$$ for all $$\mathbf{u},\mathbf{v},\mathbf{w}\in V$$, but this is not the same thing. And, as you have discovered in your question, this does not give the inner product the same properties as linear maps.
Finally, the function $$f:\mathbb R^n\to\mathbb R$$ given by $$f(\mathbf x)=\mathbf x\cdot\mathbf x$$ is not the same as the dot product$$\langle.,.\rangle:\mathbb R^n\times\mathbb R^n\to\mathbb R$$ given by $$\langle\mathbf{x},\mathbf{y}\rangle=\mathbf x\cdot\mathbf y$$. If we equip $$\mathbb R^n$$ and $$\mathbb R$$ with the usual vector space structure, then we see that $$f$$ is not linear, e.g. $$8=f(2,2)=f(2(1,1))\neq2f(1,1)=2\cdot 2 \, .$$
• "K is just a set and not assumed to be equipped with a vector space structure" - K is field (and hence defines a K-vector space) isn't it? May 27 at 12:53
• @Filippo: It is true that we can consider $K$ to be a vector space over $K$. What I mean is: none of the axioms that define an inner product "view" $K$ as a vector space. (They do view $K$ as field, however—that is a good point.)
– Joe
May 27 at 14:26
Yes, the function \begin{align} A_v:V&\to F\\ w&\mapsto\langle w,v\rangle \end{align} is linear for all $$v\in V$$. This does not imply that \begin{align} B:V&\to F\\ w&\mapsto\langle w,w\rangle \end{align} is linear since there is no $$v\in V$$ s.t. $$B=A_v$$, so there is no contradiction.
The definition of a linear function requires $$f(ax) = af(x)$$.
But for $$f(ax) = x \cdot x$$, $$f(ax) = (ax) \cdot (ax) = a^2f(x)$$
• "But for..." - Well, a function can satisfy both properties at the same time :) May 27 at 16:43
• @Filippo: True, if $a \in \{-1, 0, 1\}$.
– Dan
May 27 at 16:44
• I meant that a function $f$ can satisfy $$\forall x:\forall a: f(ax)=af(x)=a^2f(x)$$I.e. the fact that$$\forall x:\forall a:f(ax)=a^2f(x)$$does not exclude the possibility that $f$ is linear. May 27 at 16:56 | 2022-06-26T12:16:40 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/4459712/why-is-the-dot-product-of-a-vector-with-itself-not-a-linear-function",
"openwebmath_score": 0.9474844336509705,
"openwebmath_perplexity": 169.1276334082125,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9637799399736476,
"lm_q2_score": 0.8723473746782093,
"lm_q1q2_score": 0.8407509004035337
} |
https://math.stackexchange.com/questions/3038605/flirting-sequences-real-analysis | # Flirting Sequences (Real Analysis)
I need help with a homework problem and am pretty sure my real analysis teacher made the following definition up:
In a metric space, a sequence $$\{P_n\}_n$$ $$\$$ flirts with $$p$$ iff for each $$\epsilon > 0$$, there is a $$n \in \mathbb{N}$$ and $$m > n$$ such that $$0 < d(p_n,p) < \epsilon$$ and $$d(p_n,p) < d(p_m,p)$$. The sequence $$\{P_n\}_n$$ $$\$$ is a flirting sequence if there is a point $$p$$ such that $$\{P_n\}$$ flirts with $$p$$.
Give an example of a sequence that flirts with $$1$$.
I am trying to $$(1)$$ understand the concept of "flirting" and $$(2)$$ figure out an example of a sequence that flirts with $$1$$. Actually, I would be happy to see an example of a any sequence that flirts with something.
I translated the definition like this:
In a metric space, the sequence $$\{P_n\}$$ flirts with $$p$$ iff $$\forall \epsilon > 0 \quad \exists(m,n \in \mathbb{N}, m >n): 0 < d(p_n,p) < \epsilon \quad \text{and} \quad d(p_n,p) < d(p_m,p).$$
I have concluded that in $$\mathbb{R}$$ with the usual metric, the sequences {$$2-\frac1n$$} and {$$1-\frac1n$$} do not flirt with $$1$$.
Can anyone help me understand this concept and/or provide an example of any sequence that flirts to some point?
• It looks like you just need the sequence to come near $1$ a lot, but then move away. If the definition didn't have the requirement $d(p_n,1)>0$ you could just take the sequence $\{p_n\}$ with $p_{2n}=1,p_{2n+1}=2$. Can you modify that sequence to meet the requirements? – lulu Dec 13 '18 at 21:50
• Hi lulu, thanks for the suggestion. I'm wondering... wouldn't that sequence fail the part where $\forall \epsilon \quad d(p_n,1) = |p_n - 1| < \epsilon$? Because if we consider $\epsilon = 0.5$, then either $|p_n - 1| = 0$ or $|p_n - 1| = 1$ and neither is less than 0.5. Or did I misunderstand the sequence? – mcmath27 Dec 13 '18 at 21:56
• @mcmath27 No, my sequence passes that test. Indeed, every $p_n$ with $n$ even is a distance $0$ from $1$ so certainly less than $\epsilon$. My sequence fails because it actually hits $1$ a lot, which is not desired. But you can easily modify the sequence to work. – lulu Dec 13 '18 at 21:58
• Just to stress: the test does not require that for all sufficiently large $n$ we have $|p_n-1|<\epsilon$. Rather it just says that we can always find some $n$ that works. The idea, informally, is that the sequence should come near the target infinitely often but then veer off infinitely often as well. – lulu Dec 13 '18 at 22:00
• @lulu Ohhh I see what you mean! So for example, with $\epsilon = 0.5$, we just pick $n =2$ and $m = 4$ and that first part works. But then wouldn't the next part fail, where $d(p_n,1) < d(p_m,1)$? Is that what you mean by the sequence fails because it hits 1 a lot? – mcmath27 Dec 13 '18 at 22:04
Consider the following real-valued sequence $$(p_n)_n$$ which flirts with $$0$$: $$p_n = \begin{cases} 1 & \text{ if } n \text{ is even}\\ \frac{1}{n} &\text{ if } n \text{ is odd} \end{cases}$$ can you see why this is the case?
Note that
• a sequence can flirt with $$p$$ but not converge to $$p$$: as in the above (it will have a subsequence converging to $$p$$, though)
• a sequence can flirt with $$p$$ and converge to $$p$$: $$p_n = \begin{cases} \frac{1}{2^n} & \text{ if } n \text{ is even}\\ \frac{1}{n} &\text{ if } n \text{ is odd} \end{cases}$$
• Hello and thank you. I can see why the sequence flirts with 0! I was starting to believe no sequences flirt with anything, but you have convinced me otherwise. – mcmath27 Dec 13 '18 at 22:15
• @mcmath27 Glad this helped! – Clement C. Dec 13 '18 at 22:16
Remember that a easy example of a flirting sequence is $$~-1, ~1/2,~ -1/3,~ 1/4, ~-1/5, ~1/6,~$$ etc.
Mainly if it is even it is positive, if it is odd it is negative, just like $$~(-1)^n~.~$$
I think the formula is $$-1/n,~~ +1/(n+1)$$
Even better, $$-1/(2*(n)),~~ +1/(2*n+1)$$. | 2019-12-07T14:21:48 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3038605/flirting-sequences-real-analysis",
"openwebmath_score": 0.9178101420402527,
"openwebmath_perplexity": 300.1908867868779,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9637799451753696,
"lm_q2_score": 0.8723473697001441,
"lm_q1q2_score": 0.8407509001434827
} |
https://math.stackexchange.com/questions/851650/dot-product-notation | # Dot product notation
Let $\mathbf{A=(a_1,a_2,\ldots, a_n)}$ and $\mathbf{B=(b_1,b_2,\ldots,b_n)}$. Many linear algebra books and texts define the dot product as $$\mathbf{A\cdot B^T=a_1b_1+a_2b_2+\cdots+a_nb_n}$$ where $\mathbf{B^T}$ is the transpose of row vector $\mathbf{B}$. But Serge Lang in Linear Algebra define as $$\mathbf{A\cdot B=a_1b_1+a_2b_2+\cdots+a_nb_n}$$ There is no diference between the results. But which is more correct and why?
EDIT
As Cameron Williams said, $\mathbf{A\cdot B^T}$ is implied by matrix multiplication. So I correct a part of my question:
Many linear algebra books and texts define the dot product as matrix multiplication $$\mathbf{A B^T=a_1b_1+a_2b_2+\cdots+a_nb_n}$$
• I would prefer to write $AB^T$ or $A\cdot B$ but I would not put a dot in between $A$ and $B^T$ since it is implied by matrix multiplication. – Cameron Williams Jun 29 '14 at 21:26
• Additionally, I think the preferred notation is to treat vectors implicitly as column vectors. Treating them as row vectors is a little strange to me though there's no real reason to choose one over the other than to make things later on a little bit nicer. – Cameron Williams Jun 29 '14 at 21:32
• I think the first one is better because it hints how to generalize from $\mathbb R$ to $\mathbb C$ and the second one doesn't. – Git Gud Jun 29 '14 at 21:35
Really, there isn't a notation that is more correct. It is just a matter of convention. All of them mean the operation $\sum_{i = 1}^n a_ib_i$. The important thing is that you understand what you must do. Like you said yourself, in $\mathbf{A \cdot B^T}$, we see $\mathbf{A}$ and $\mathbf{B}$ as row vectors. The $\mathbf{^T}$ serves just to remind you that you can see the dot product as a matrix multiplication, after all, we will have a $1 \times n$ matrix times a $n \times 1$, which is well defined, and gives as result a $1 \times 1$ matrix, i.e., a number.
The notation $\mathbf{A \cdot B}$ doesn't sugest any of these things, and you can think directly of the termwise multiplication, then sum.
In Linear Algebra, we often talk about inner products in arbitrary vector spaces, a sort of generalization of the dot product. Given vectors $\mathbf{A}$ and $\mathbf{B}$, a widely used notation is $\langle \mathbf{A}, \mathbf{B} \rangle$. An inner product (in a real vector space), put simply, is a symmetric bilinear form (form means that the result is a number), which is positive definite. That means:
i) $\langle \mathbf{A}, \mathbf{B} \rangle=\langle \mathbf{B}, \mathbf{A} \rangle$;
ii) $\langle \mathbf{A} + \lambda \mathbf{B}, \mathbf{C} \rangle = \langle \mathbf{A}, \mathbf{C} \rangle + \lambda \langle \mathbf{B}, \mathbf{C} \rangle$ ;
iii) $\langle \mathbf{A}, \mathbf{A} \rangle > 0$ if $\mathbf{A} \neq \mathbf{0}$
I, particularly, don't like the notation $\mathbf{A \cdot B^T}$, because when working in more general spaces than $\Bbb R^n$, we don't always have a finite dimension, so matrices don't work so well. I never saw a notation different from those three I talked about. But I enforce what I said at the beginning: there isn't a correct notation, but you should be used to all of them, as possible.
To phrase CameronWilliams's comments another way, you can simply define the dot product as a matrix multiplication:
$$\mathbf a \cdot \mathbf b := \mathbf a^T \mathbf b$$
where $\mathbf a = [a_1, \ldots, a_n]^T$ and $\mathbf b = [b_1, \ldots, b_n]^T$ are column vectors. (If they are row vectors, the dot product would be $\mathbf a \cdot \mathbf b^T$, but as Cameron says, this is less standard.) To combine the dot product and transpose in the definition itself is confusing. If I saw $\mathbf a \cdot \mathbf b^T$, I would assume that $\mathbf a$ was a column vector and $\mathbf b$ was a row vector.
Incidentally, in terms of notation, I think it is nicer to reserve bold type ($\mathbf a$) for matrices and vectors, and italic type ($a$) for their entries. This provides a visual advantage to the reader, so that it is easy to see at a glance that, for example, $\mathbf A = [\mathbf a_1, \ldots, \mathbf a_n]$ is a matrix consisting of column vectors, whereas $\mathbf a = [a_1, \ldots, a_n]^T$ is a column vector with (say) real entries. | 2019-12-06T06:03:37 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/851650/dot-product-notation",
"openwebmath_score": 0.9278148412704468,
"openwebmath_perplexity": 131.4371516498472,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9637799472560582,
"lm_q2_score": 0.8723473663814338,
"lm_q1q2_score": 0.8407508987600596
} |
http://mathhelpforum.com/algebra/222339-factorial-print.html | # Factorial
• Sep 27th 2013, 01:45 PM
Natasha1
Factorial
So here goes my problem. I am not allowed a calculator through this problem...
4! = 4 x 3 x 2 x 1
5! = 5 x 4 x 3 x 2 x 1 etc
How many zeros are there at the end of 50! ?
It would take me hours to do 50! = 50 x 49 x....x 1 by hand so how can I show the number of zeros at the end of 50! ?
All I have done so far is to try and see a pattern but I am stuck... Please help
1! = 1
2! = 2 x 1
3! = 3 x 2 x 1
4! = 4 x 3 x 2 x 1
5! = 5 x 4 x 3 x 2 x 1
6! = 6 x 5 x 4 x 3 x 2 x 1
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
• Sep 27th 2013, 01:48 PM
Shakarri
Re: Factorial
An extra zero is added every time you multiply by a number which is a multiple of 5
• Sep 27th 2013, 01:56 PM
Natasha1
Re: Factorial
So 50! = 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
So as there are 10 multiples of 5 in 50! there are 10 zeros at the end of 50!
Is this correct?
• Sep 27th 2013, 02:00 PM
Plato
Re: Factorial
Quote:
Originally Posted by Natasha1
So here goes my problem. I am not allowed a calculator through this problem...
How many zeros are there at the end of 50! ?[/B]
For a trailing zero (zero at the end) of a product requires a factor of five and a factor two.
$15!$ is a product containing three fives and many twos. So $15!$ as three trailing zeros
$24!$ as four trailing zeros because there four factors of five and many twos.
BUT $25!$ as six trailing zeros because there six factors of five and many twos. Where did the extra five come from?
So what is the answer for $50!$?
Here is a routine to do what you want: $126!$ as thirty-one trailing zeros.
You can 'play with this change 126 to 119 to 121, see what happens.
• Sep 27th 2013, 02:14 PM
Natasha1
Re: Factorial
Is this correct?
• Sep 27th 2013, 02:17 PM
Natasha1
Re: Factorial
50! has 10 factors of five and 25 factors of two so it has 10 trailing zeros. Right?
• Sep 27th 2013, 02:19 PM
Plato
Re: Factorial
Quote:
Originally Posted by Natasha1
Is this correct?
Is what correct? Use that webpage to see about $50!$.
Change the 126 to 50 and enter.
• Sep 27th 2013, 02:23 PM
Natasha1
Re: Factorial
I am sorry but I do not get that writing. Does that mean there are 12 factors from that link you sent me?
• Sep 27th 2013, 02:31 PM
Plato
Re: Factorial
Quote:
Originally Posted by Natasha1
I am sorry but I do not get that writing. Does that mean there are 12 factors from that link you sent me?
YES it does mean exactly that.
$50!$ has twelve factors of five $\frac{50}{5}+\frac{50}{25}=10+2=12$.
EVERY PAIR of factors of $\{5,2\}$ gives a zero.
• Sep 27th 2013, 02:38 PM
Natasha1
Re: Factorial
Quote:
Originally Posted by Plato
YES it does mean exactly that.
$50!$ has twelve factors of five $\frac{50}{5}+\frac{50}{25}=10+2=12$.
EVERY PAIR of factors of $\{5,2\}$ gives a zero.
Many thanks Plato
• Sep 27th 2013, 11:25 PM
thevinh
Re: Factorial
Here is an interesting question, how many trailing zeroes does 1,000,000! have? Can you find a general formula for any non-negative integer?
Spoiler:
$\text{Let}~ \lambda=\text{the number of trailing zeroes,}~n=\text{any non-negative integer}$
$\lambda=\left\lfloor \sum_{k=1}^{k_{max}} \frac{n}{5^k} \right\rfloor \quad \text{where} \quad k_{max}=\left \lfloor \frac{\ln n}{\ln5} \right \rfloor$
• Sep 28th 2013, 01:44 PM
Natasha1
Re: Factorial
I have a question...
Just one thing... 50! has 10 factors of five right. What does the 50/25 mean exactly in your sum when you say 50/5 + 50/25? That 25 is a factor of 50, right? But isn't it already accounted for in the 10 factors of five?
• Sep 28th 2013, 02:04 PM
Plato
Re: Factorial
Quote:
Originally Posted by Natasha1
I have a question...
Just one thing... 50! has 10 factors of five right. What does the 50/25 mean exactly in your sum when you say 50/5 + 50/25? That 25 is a factor of 50, right? But isn't it already accounted for in the 10 factors of five?
Here is a list of multiples of five in 50 written in factored form.
$5,\;2 \cdot 5,\;3 \cdot 5,\;4 \cdot 5,\;5 \cdot 5,\;6 \cdot 5,\;7 \cdot 5,\;8 \cdot 5,\;9 \cdot 5,\;2 \cdot 5 \cdot 5$
those are
$5,~10,~15,~20,~25,~30,~35,~40,~45,~50$.
COUNT the number of fives in the first list. There are twelve factors of five.
• Sep 28th 2013, 11:06 PM
thevinh
Re: Factorial
Quote:
Originally Posted by Natasha1
I have a question...
Just one thing... 50! has 10 factors of five right. What does the 50/25 mean exactly in your sum when you say 50/5 + 50/25? That 25 is a factor of 50, right? But isn't it already accounted for in the 10 factors of five?
To find out the number of trailing zeroes upon factorial expansion we need to count how many 5's there are in the product. Let's do a simple example, 16!. In this case you can see that there are three 5's in this product (1.2.3...5...10...15...16), hence you will have 3 trailing zeroes. Let's complicate thing a little bit, what about 47!?
Well, 1...5...10...15...20...25...30...35...40...45.46.4 7, here you have nine 5's + 1 extra from 25. Thus you will have total of 10 trailing zero.
So you can see that for every power of 5, you will have one extra zero (such as 25=5^2, 125=5^3, 625=5^4 ....). Let's do 128!
5.10.15.20.25.30.35.40.45.50.55.60.65.70.75.80.85. 90.95.100.105.110.120.125.126.127.128 =>
5.10.15.20.5.5.30.35.40.45.2.5.5.55.60.65.70.3.5.5.80.85.90.95.4.5.5.105.110.120.5.5.5.126.127.128
In general, you need to n/5 + n/25 + n/125 + n/625 + .... n/(5^kmax).
I hope this help. Let me know if you need more explanation! | 2017-01-20T16:45:12 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/algebra/222339-factorial-print.html",
"openwebmath_score": 0.7295138239860535,
"openwebmath_perplexity": 440.6088152328344,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9637799451753696,
"lm_q2_score": 0.8723473614033683,
"lm_q1q2_score": 0.8407508921472167
} |
https://math.stackexchange.com/questions/2584096/maximum-possible-order-of-an-element-in-s-7-text-and-s-10 | # Maximum possible order of an element in $S_7 \text{ and } S_{10}$
Exercise :
Find the maximum possible order of an element of the group of permutations $S_7$. Do the same thing for $S_{10}$.
Discussion :
Recalling that any permutation can be written as a product of disjoint cycles :
$$c=c_1 c_2\dots c_r$$
the order of $|σ|=\text{lcm}(|σ_1||σ_2|\dots|c_r|)$ and if $c_i$ has length $k_i$ then it will be $|c_i|=k_i$.
So what I have to do is find all the possible products of disjoint cycles, which will be :
$$(1,2) \space \text{ of order } \space 2$$ $$(1,2,3) \space \text{ of order } \space 3$$ $$(1,2,3,4) \space \text{ of order } \space 4$$ $$(1,2)(3,4) \space \text{ of order } \space 2$$ $$(1,2,3,4,5) \space \text{ of order } \space 5$$ $$(1,2,3)(4,5) \space \text{ of order } \space 6$$ $$(1,2,3,4,5,6) \space \text{ of order } \space 6$$ $$(1,2,3,4)(5,6) \space \text{ of order } \space 4$$ $$(1,2)(3,4)(5,6) \space \text{ of order } \space 2$$ $$(1,2,3)(4,5,6) \space \text{ of order } \space 3$$ $$(1,2,3)(4,5)(6,7) \space \text{ of order } \space 6$$ $$(1,2,3,4,5)(6,7) \space \text{ of order } \space 10$$ $$(1,2,3,4)(5,6,7) \space \text{ of order } \space 12$$ $$(1,2,3,4,5,6,7) \space \text{ of order } \space 7$$
So, the maximum possible order of an element in $S_7$ is $12$.
Question :
What I wanted to ask is $(1)$ am I correct, first of all?
And $(2)$ how am I supposed to find the maximum order of an element in $S_{10}$ ? Judging from all the possible cycle products in $S_7$ it will be a pretty big list for $S_{10}$. Am I missing a trick or some clever way to reach the desired result faster ?
P.S. : I know that Landau's function calculates exactly that thing, but we haven't been taught about it yet.
• Why $(1,2,3)(4,5)(6,7)$ has order $2$? – Skills Dec 29 '17 at 11:55
• @Skills Sorry, was typo. – Rebellos Dec 29 '17 at 11:57
• Case it by the order of the largest cycle, from high to low, so just 10 cases, and most will be already dominated. Also, no need to write out the actual cycles, just write their orders (e.g. 4-3-3 => 12). – quasi Dec 29 '17 at 11:57
• Besides the typo pointed out by Skills, yes, it looks right. You can write a computer program to do it; you are effectively trying to maximize least common multiple of a bunch of cycles such that their lengths add up to 10. It should not be that hard to do it by hand. (For instance, do you think you will have a cycle of length 9?) – E-A Dec 29 '17 at 11:58
• @E-A The problem is that this is an exercise that should be hand-written/solved. This is why I'm concerned about larger $n$. – Rebellos Dec 29 '17 at 12:03
First of all yes, your answer for $S_7$ is correct. And indeed listing all cycle types quickly becomes unmanageable as $n$ grows.
It is quite helpful and not difficult to prove that the maximum is always assumed for a permutation that is the product of disjoint cycles of prime power lengths. For increasing values of $n$ this eliminates an increasing proportion of permutations; for $n=7$ and $n=10$ it does not eliminate much.
It is also quite helpful and not difficult to prove that the maximum is always assumed for a permutation that has at most one fixed point. This eliminates a large proportion of permutations for any $n$.
With these two facts in mind, for $n=7$ we are left with permutations of cycle types $$(7),\ (5,2),\ (4,3),\ (4,2,1),\ (3,3,1),\ (3,2,2)\quad \text{and}\quad (2,2,2,1).$$ For $n=10$ we are left with permutations of cycle types $$(9,1),\ (8,2),\ (7,3),\ (7,2,1),\ (5,5),\ (5,4,1),\ (5,3,2),\ (4,4,2),\ (4,3,3),\ (4,3,2,1),\ (3,3,3,1),\ (3,3,2,2),\quad\text{and}\quad (2,2,2,2,2).$$ Of course one can already see some recursion in this problem; given that the maxima for $n=3$ and $n=5$ are assumed for cycle types $(3)$ and $(3,2)$, respectively, for $n=10$ we can eliminate the cycle types $$(7,2,1),\ (5,5)\quad\text{and}\quad(5,4,1).$$ There are a lot of simple facts to note that each reduce the number of cycle types to consider by a little bit. For your problem there aren't that many to begin with, and the facts above are enough to make the problem manageable by hand. But it can just as well be done by hand for $n=100$ and $n=1000$, though this requires more careful analysis of the problem which I won't go into here. I'll leave you with this vague intuition though:
Heuristically, for large $n$ the maximum is assumed when writing $n$ as a sum of prime powers close to $\sqrt{n}$, and padding the remainder with small primes. It is not easy to compute the maximum precisely for large values of $n$, though asymptotically it is $\exp((1+o(1))\sqrt{n\ln{n}})$.
• For the observation on prime power lengths: Let $\sigma\in S_n$ be a cycle of length $m=\prod_{p\mid m}p^{m_p}>1$. We want to show that there exists a $\tau\in S_n$ of order $m$ that is a product of disjoint cycles of prime power lengths. If $\sum_{p\mid m}p^{m_p}\leq n$ we can simply take disjoint cycles of lengths $p^{m_p}$ for every prime $p$ dividing $m$. Then the order of the product is $\operatorname{lcm}_{p\mid m}p^{m_p}=\prod_{p\mid m}p^{m_p}=m$. As $n\geq m$ we can do this whenever $$\sum_{p\mid m}p^{m_p}\leq m=\prod_{p\mid m}p^{m_p}.$$ This holds for all $m>1$. (Continued) – Servaes Dec 29 '17 at 17:43
• This follows quickly from the fact that $p^{m_p}\geq2$ for all $p$. Now for a general $\sigma\in S_n$ (that is not necessarily a cycle), write $\sigma$ as a product of disjoint cycles and apply the above to each cycle. – Servaes Dec 29 '17 at 17:44
• Hello and thanks a lot for your detailed and amazing answer. I checked it fast yesterday (had to attend a Christmas table later on) and I had the time to sit down, write and prove everything. I have to say that your answer was really helpful and written in such a good way with details and mathematical formality that gave me a great insight and understanding ! Also the comments were really helpful regarding the proofs and leading me to a general understanding. Thanks a really lot, I would upvote double times if I could ! – Rebellos Dec 30 '17 at 12:20
You see that in $S_7$ the maximum possible order is given by $$(1,2,3,4)(5,6,7)$$ so used element $s_1,s_2$ such that $o(s_1)+o(s_2)=7$ that maximize $l.c.m.$ So you have to choose a similar case in $S_{10}$. The solution is $s_1,s_2,s_3$ such that $o(s_1)=5,o(s_2)=3,o(s_3)=2$ so: $$(1,2,3,4,5)(6,7,8)(9,10).$$ I don't know if it's possible to generalize it for $S_n$. | 2019-10-17T08:24:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2584096/maximum-possible-order-of-an-element-in-s-7-text-and-s-10",
"openwebmath_score": 0.8945714831352234,
"openwebmath_perplexity": 138.4150866922279,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9845754492759498,
"lm_q2_score": 0.8539127548105611,
"lm_q1q2_score": 0.840741534210072
} |
http://math.stackexchange.com/questions/210787/where-is-the-calculation-hiding-in-this-proof-about-how-p-splits-in-mathbbq | # Where is the calculation hiding in this proof about how $p$ splits in $\mathbb{Q}(\zeta_n)$?
I just worked through a proof in Daniel Marcus' book Number Fields that if $p\nmid n$, the inertial degree of any prime ideal of $\mathbb{Q}(\zeta_n)$ lying over $p$ is equal to the order of $p$ in $(\mathbb{Z}/n\mathbb{Z})^\times$ (p.77). The argument rests on the equation $(1-\zeta_n)(1-\zeta_n^2)\dots(1-\zeta_n^{n-1})=n$. I tried to reorganize the proof and was surprised to end up with what seems to me to be a complete proof that doesn't make explicit use of this fact. My question is, where did I hide it? (Or, is my proof missing something? Or, is this fact actually extrinsic to the result?)
The setup: We have a prime $p$ not dividing $n$, and the discriminant of $\mathcal{O}_{\mathbb{Q}(\zeta_n)}=\mathbb{Z}[\zeta_n]$ divides a power of $n$ so is prime to $p$, so $(p)$ splits into distinct prime ideals in $\mathbb{Z}[\zeta_n]$:
$$p\mathbb{Z}[\zeta_n] = P_1\dots P_r$$
The field extension is normal and degree $\varphi(n)$, so we have $fr=\varphi(n)$ where $f$ is the inertial degree of the $P_i$. By definition, $f$ is the degree of the field extension $[\mathbb{Z}[\zeta_n]/P_i:\mathbb{Z}/(p)]$. This field extension is cyclic, with Galois group generated by the Frobenius automorphism $\sigma$, which therefore has order $f$. Meanwhile, $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong (\mathbb{Z}/n\mathbb{Z})^\times$ has an automorphism $\tau$ mapping $\zeta_n\mapsto \zeta_n^p$, since $p$ is prime to $n$. The order of $\tau$ is the order of $p$ in $(\mathbb{Z}/n\mathbb{Z})^\times$. The problem is to show that $\tau$ and $\sigma$ have the same order.
Marcus' argument: Consider the action of $\tau$ on the field generator $\zeta_n$ and of $\sigma$ on its image in $\mathbb{Z}[\zeta_n]/P_i$. $\tau^m=\mathrm{id.}$ if and only if $\tau^m(\zeta_n)=\zeta_n$ because it is a generator. Likewise, $\sigma^m=\mathrm{id.}$ if and only if $\sigma^m(\bar\zeta_n)=\bar\zeta_n$, with calculation taking place in $\mathbb{Z}[\zeta_n]/P_i$, i.e. mod $P_i$. The problem is thus to show that $\zeta_n^{p^m}=\zeta_n$ if and only if $\zeta_n^{p^m}=\zeta_n\mod P_i$. The "only if" is trivial. If $\zeta_n^{p^m}=\zeta_n\mod P_i$ then $\zeta_n^{p^m-1}=1\mod P_i$ because $\zeta_n$ is a unit. So $1-\zeta_n^{p^m-1}\in P_i$. However,
$$(1-\zeta_n)(1-\zeta_n^2)\dots (1-\zeta_n^{n-1}) = n\notin (p)\subset P_i$$
Therefore, $p^m-1$ can't be equal to $1,2,\dots,n-1$ mod $n$. The only possibility is $p^m-1=0\mod n$, and then $\zeta_n^{p^m-1}=1$, so we have our "if" statement.
My argument: $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ acts transitively on the $P_i$, of which there are $r$, so by the orbit-stabilizer theorem, $f=|\operatorname{Stab}P_i|$. Let $\pi:\mathbb{Z}[\zeta_n]\rightarrow\mathbb{Z}[\zeta_n]/P_i$ be the canonical homomorphism. $\tau$ acts trivially on $\mathbb{Z}$ and sends $\zeta_n$ to its $p$th power; $\sigma$ sends the image of $\zeta_n$ to its $p$th power (since it is the $p$th power map on everything) and acts trivially on the image of $\mathbb{Z}$. By consideration of their action on $\mathbb{Z},\zeta_n$ which generate the ring $\mathbb{Z}[\zeta_n]$, we conclude that $\pi\circ\tau$ and $\sigma\circ\pi$ are equal. It follows that $\tau\in\operatorname{Stab}P_i$ and that $\tau\mapsto \sigma$ under the homomorphism $\operatorname{Stab}P_i\rightarrow Gal((\mathbb{Z}[\zeta_n]/P_i)/(\mathbb{Z}/(p)))$ induced by $\pi$. It is immediate that this map is surjective because $\sigma$ generates the image. Since domain and image both have order $f$, this means it is an isomorphism, and it follows that $\sigma,\tau$ have the same order.
To reiterate my question: where did I hide Marcus' calculation $(1-\zeta_n)\dots(1-\zeta_n^{n-1})=n?$ Or, is my proof wrong? Or, is the calculation actually extrinsic to the conclusion?
-
seems right to me. – user29743 Oct 11 '12 at 1:58
Actually I prefer your proof ! – user18119 Apr 15 '13 at 22:18
Your proof is correct. The crucial part in both proofs consist in proving the map $$\mathrm{Stab}(P_i)\to \mathrm{Gal}((\mathbb Z[\zeta_n]/P_i)/\mathbb F_p)$$ is injective. This map is known to be always surjective. The difference between the two proofs:
• You use implicitely the fact that the inertia group at $P_i$ is trivial (when you say the stablizer has order $f$).
• Marcus is proving directly the injectivity, using the identity on the product of the $(1-\zeta_n^k)$'s. The same identity can be used to show the extension is unramified above $p$ (hence the inertia at $P_i$ trivial).
Thanks! What's the idea for using the identity to show that $p$ is unramified? – Ben Blum-Smith Apr 16 '13 at 22:32
The identity implies that mod $p$, $\Phi_n(X)$ is separable because the difference of any two distinct roots $\zeta_n^i-\zeta_n^j=\zeta_n^i(1-\zeta_n^{i-j})$ is non-zero mod any prime ideal lying over $p$. Hence the cyclotomic extension is unramified over $p$. But I notice that in Marcus the non-ramification is proved earlier and differently. – user18119 Apr 17 '13 at 20:17 | 2015-09-02T22:06:48 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/210787/where-is-the-calculation-hiding-in-this-proof-about-how-p-splits-in-mathbbq",
"openwebmath_score": 0.9648061394691467,
"openwebmath_perplexity": 104.38016924399712,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9845754497285467,
"lm_q2_score": 0.8539127492339909,
"lm_q1q2_score": 0.8407415291059963
} |
https://tbc-python.fossee.in/convert-notebook/Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter4_2.ipynb | # Chapter 4 : Truncation Errors and the Taylor Series¶
## Example: 4.1 Page No:79¶
In [13]:
from math import factorial
from scipy.misc import derivative
def f(x):
y=-0.1*x**4-0.15*x**3-0.5*x**2-0.25*x+1.2#
return y
xi=0#
xf=1#
h=xf-xi#
fi=f(xi)##function value at xi
ffa=f(xf)##actual function value at xf
#for n=0, i.e, zero order approximation
ff=fi#
Et_1=ffa-ff##truncation error at x=1
print "The value of f at x=0 :",fi
print "The value of f at x=1 due to zero order approximation :",ff
print "Truncation error :",Et_1
print "----------------------------------------------"
#for n=1, i.e, first order approximation
def f1(x):
y=derivative(f,x)
return y
f1i=f1(xi)##value of first derivative of function at xi
f1f=fi+f1i*h##value of first derivative of function at xf
Et_2=ffa-f1f##truncation error at x=1
print "The value of first derivative of f at x=0 :",f1i
print "The value of f at x=1 due to first order approximation :",f1f
print "Truncation error :",Et_2
print "----------------------------------------------"
#for n=2, i.e, second order approximation
def f2(x):
y=derivative(f1,x)
return y
f2i=f2(xi)##value of second derivative of function at xi
f2f=f1f+f2i*(h**2)/factorial(2)##value of second derivative of function at xf
Et_3=ffa-f2f##truncation error at x=1
print "The value of second derivative of f at x=0 :",f2i
print "The value of f at x=1 due to second order approximation :",f2f
print "Truncation error :",Et_3
print "----------------------------------------------"
#for n=3, i.e, third order approximation
def f3(x):
y=derivative(f2,x)
return y
f3i=f3(xi)##value of third derivative of function at xi
f3f=f2f+f3i*(h**3)/factorial(3)##value of third derivative of function at xf
Et_4=ffa-f3f##truncation error at x=1
print "The value of third derivative of f at x=0 :",f3i
print "The value of f at x=1 due to third order approximation :",f3f
print "Truncation error :", Et_4
print "----------------------------------------------"
#for n=4, i.e, fourth order approximation
def f4(x):
y=derivative(f3,x)
return y
f4i=f4(xi)##value of fourth derivative of function at xi
f4f=f3f+f4i*(h**4)/factorial(4)##value of fourth derivative of function at xf
Et_5=ffa-f4f##truncation error at x=1
print "The value of fourth derivative of f at x=0 :",f4i
print "The value of f at x=1 due to fourth order approximation :",f4f
print "Truncation error :",Et_5
print "----------------------------------------------"
The value of f at x=0 : 1.2
The value of f at x=1 due to zero order approximation : 1.2
Truncation error : -1.0
----------------------------------------------
The value of first derivative of f at x=0 : -0.4
The value of f at x=1 due to first order approximation : 0.8
Truncation error : -0.6
----------------------------------------------
The value of second derivative of f at x=0 : -1.8
The value of f at x=1 due to second order approximation : -0.1
Truncation error : 0.3
----------------------------------------------
The value of third derivative of f at x=0 : -0.9
The value of f at x=1 due to third order approximation : -0.25
Truncation error : 0.45
----------------------------------------------
The value of fourth derivative of f at x=0 : -2.4
The value of f at x=1 due to fourth order approximation : -0.35
Truncation error : 0.55
----------------------------------------------
## Example 4.2: Page No:82¶
In [2]:
from math import pi,cos,factorial
from scipy.misc import derivative
def f(x):
y=cos(x)
return y
xi=pi/4#
xf=pi/3#
h=xf-xi#
fi=f(xi)##function value at xi
ffa=f(xf)##actual function value at xf
#for n=0, i.e, zero order approximation
ff=fi#
et1=(ffa-ff)*100/ffa##percent relative error at x=1
print "The value of f at x=1 due to zero order approximation :",ff
print "% relative error :",et1
print "----------------------------------------------"
#for n=1, i.e, first order approximation
def f1(x):
y=derivative(f,x)
return y
f1i=f1(xi)##value of first derivative of function at xi
f1f=fi+f1i*h##value of first derivative of function at xf
et2=(ffa-f1f)*100/ffa##% relative error at x=1
print "The value of f at x=1 due to first order approximation :",f1f
print "% relative error :",et2
print "----------------------------------------------"
#for n=2, i.e, second order approximation
def f2(x):
y=derivative(f1,x)
return y
f2i=f2(xi)##value of second derivative of function at xi
f2f=f1f+f2i*(h**2)/factorial(2)##value of second derivative of function at xf
et3=(ffa-f2f)*100/ffa##% relative error at x=1
print "The value of f at x=1 due to second order approximation :",f2f
print "% relative error :",et3
print "----------------------------------------------"
#for n=3, i.e, third order approximation
def f3(x):
y=derivative(f2,x)
return y
f3i=f3(xi)##value of third derivative of function at xi
f3f=f2f+f3i*(h**3)/factorial(3)##value of third derivative of function at xf
et4=(ffa-f3f)*100/ffa##% relative error at x=1
print "The value of f at x=1 due to third order approximation :",f3f
print "% relative error :",et4
print "----------------------------------------------"
#for n=4, i.e, fourth order approximation
def f4(x):
y=derivative(f3,x)
return y
f4i=f4(xi)##value of fourth derivative of function at xi
f4f=f3f+f4i*(h**4)/factorial(4)##value of fourth derivative of function at xf
et5=(ffa-f4f)*100/ffa##% relative error at x=1
print "The value of f at x=1 due to fourth order approximation :",f4f
print "% relative error :",et5
print "----------------------------------------------"
#for n=5, i.e, fifth order approximation
f5i=(f4(1.1*xi)-f4(0.9*xi))/(2*0.1)##value of fifth derivative of function at xi (central difference method)
f5f=f4f+f5i*(h**5)/factorial(5)##value of fifth derivative of function at xf
et6=(ffa-f5f)*100/ffa##% relative error at x=1
print "The value of f at x=1 due to fifth order approximation :",f5f
print "% relative error :",et6
print "----------------------------------------------"
#for n=6, i.e, sixth order approximation
def f6(x):
y=derivative(f5,x)
return y
f6i=(f4(1.1*xi)-2*f4(xi)+f4(0.9*xi))/(0.1**2)##value of sixth derivative of function at xi (central difference method)
f6f=f5f+f6i*(h**6)/factorial(6)##value of sixth derivative of function at xf
et6=(ffa-f6f)*100/ffa##% relative error at x=1
print "The value of f at x=1 due to sixth order approximation :",f6f
print "% relative error :", et6
print "----------------------------------------------"
The value of f at x=1 due to zero order approximation : 0.707106781187
% relative error : -41.4213562373
----------------------------------------------
The value of f at x=1 due to first order approximation : 0.551333569463
% relative error : -10.2667138927
----------------------------------------------
The value of f at x=1 due to second order approximation : 0.534175415889
% relative error : -6.83508317772
----------------------------------------------
The value of f at x=1 due to third order approximation : 0.535435376789
% relative error : -7.08707535775
----------------------------------------------
The value of f at x=1 due to fourth order approximation : 0.535504768061
% relative error : -7.10095361216
----------------------------------------------
The value of f at x=1 due to fifth order approximation : 0.535501917392
% relative error : -7.10038347839
----------------------------------------------
The value of f at x=1 due to sixth order approximation : 0.535501819651
% relative error : -7.10036393016
----------------------------------------------
## Example 4.3 : Page No:85¶
In [2]:
from math import pi,cos,factorial
m=input("Input value of m:")
h=input("Input value of h:")
def f(x):
y=x**m
return y
x1=1#
x2=x1+h#
fx1=f(x1)#
fx2=fx1+m*(fx1**(m-1))*h#
if m==1:
R=0#
elif m==2 :
R=2*(h**2)/factorial(2)#
elif m==3:
R=(6*(x1)*(h**2)/factorial(2))+(6*(h**3)/factorial(3))#
elif m==4:
R=(12*(x1**2)*(h**2)/factorial(2))+(24*(x1)*(h**3)/factorial(3))+(24*(h**4)/factorial(4))
print "\nRemainder:",fx2,"\nThe value by first order approximation:",R
print "True Value at x2:",f(x2)
Input value of m:4
Input value of h:5
Remainder: 21
The value by first order approximation: 1275
True Value at x2: 1296
## Example 4.4: Page No:92¶
In [3]:
from scipy.misc import derivative
def f(x):
y=-0.1*(x**4)-0.15*(x**3)-0.5*(x**2)-0.25*(x)+1.2
return y
x=0.5#
h=input("Input h:")
x1=x-h#
x2=x+h#
#forward difference method
fdx1=(f(x2)-f(x))/h##derivative at x
et1=abs((fdx1-derivative(f,x))/derivative(f,x))*100#
#backward difference method
fdx2=(f(x)-f(x1))/h##derivative at x
et2=abs((fdx2-derivative(f,x))/derivative(f,x))*100#
#central difference method
fdx3=(f(x2)-f(x1))/(2*h)##derivative at x
et3=abs((fdx3-derivative(f,x))/derivative(f,x))*100#
print "For h=",h
print "and percent error=",fdx1,"Derivative at x by forward difference method=",et1
print "and percent error=",fdx2,"Derivative at x by backward difference method=",et2
print "and percent error=",fdx3,"Derivative at x by central difference method=",et3
Input h:1.232323
For h= 1.232323
and percent error= -2.70944264922 Derivative at x by forward difference method= 114.60931875
and percent error= -0.178591334206 Derivative at x by backward difference method= 85.854151746
and percent error= -1.44401699172 Derivative at x by central difference method= 14.3775835022
## Example 4.5: Page No: 95¶
In [4]:
from scipy.misc import derivative
def f(x):
y=x**3
return y
x=2.5#
delta=0.01#
deltafx=abs(derivative(f,x))*delta#
fx=f(x)#
print "true value is between : ",fx-deltafx,"and",fx+deltafx
true value is between : 15.4275 and 15.8225
## Example 4.6: Page No: 96¶
In [5]:
from scipy.misc import derivative
def f(F,L,E,I):
y=(F*(L**4))/(8*E*I)
return y
Fbar=50##lb/ft
Lbar=30##ft
Ebar=1.5*(10**8)##lb/ft**2
Ibar=0.06##ft**4
deltaF=2##lb/ft
deltaL=0.1##ft
deltaE=0.01*(10**8)##lb/ft**2
deltaI=0.0006##ft**4
ybar=(Fbar*(Lbar**4))/(8*Ebar*Ibar)#
def f1(F):
y=(F*(Lbar**4))/(8*Ebar*Ibar)
return y
def f2(L):
y=(Fbar*(L**4))/(8*Ebar*Ibar)
return y
def f3(E):
y=(Fbar*(Lbar**4))/(8*E*Ibar)
return y
def f4(I):
y=(Fbar*(Lbar**4))/(8*Ebar*I)
return y
deltay=abs(derivative(f1,Fbar))*deltaF+abs(derivative(f2,Lbar))*deltaL+abs(derivative(f3,Ebar))*deltaE+abs(derivative(f4,Ibar))*deltaI#
print "The value of y is between:",ybar-deltay,"and",ybar+deltay
ymin=((Fbar-deltaF)*((Lbar-deltaL)**4))/(8*(Ebar+deltaE)*(Ibar+deltaI))#
ymax=((Fbar+deltaF)*((Lbar+deltaL)**4))/(8*(Ebar-deltaE)*(Ibar-deltaI))#
print "ymin is calculated at lower extremes of F, L, E, I values as =",ymin
print "ymax is calculated at higher extremes of F, L, E, I values as =",ymax
The value of y is between: 0.528721343471 and 0.596278656529
ymin is calculated at lower extremes of F, L, E, I values as = 0.524066539965
ymax is calculated at higher extremes of F, L, E, I values as = 0.602846335915
## Example 4.7 : Page No:98¶
In [6]:
from math import pi,tan
from scipy.misc import derivative
def f(x):
y=tan(x)
return y
x1bar=(pi/2)+0.1*(pi/2)#
x2bar=(pi/2)+0.01*(pi/2)#
#computing condition number for x1bar
condnum1=x1bar*derivative(f,x1bar)/f(x1bar)#
print "The condition number of function for x =",condnum1,"is :",x1bar
if abs(condnum1)>1:
print "Function is ill-conditioned for x =",x1bar
#computing condition number for x2bar
condnum2=x2bar*derivative(f,x2bar)/f(x2bar)#
print "The condition number of function for x =",condnum2,"is :",x2bar
if abs(condnum2)>1:
print "Function is ill-conditioned for x =",x2bar
The condition number of function for x = 0.18201112073 is : 1.72787595947
The condition number of function for x = 0.0160083243793 is : 1.58650429006 | 2020-05-27T06:22:56 | {
"domain": "fossee.in",
"url": "https://tbc-python.fossee.in/convert-notebook/Numerical_Methods_For_Engineers_by_S._C._Chapra_And_R._P._Canale/Chapter4_2.ipynb",
"openwebmath_score": 0.8045533299446106,
"openwebmath_perplexity": 7739.287048737271,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9845754461077708,
"lm_q2_score": 0.8539127492339909,
"lm_q1q2_score": 0.8407415260141696
} |
https://math.stackexchange.com/questions/2349232/what-can-be-the-maximum-and-minimum-number-of-prime-factors | # What can be the maximum and minimum number of prime factors?
A number is having exactly $72$ factors or $72$ composite factors. what can be the maximum and minimum number of prime factors of this number? I search on internet and I found a solution of that question, solution like:
factorise $72$ as $2\times2\times2\times3\times3$ so total no of maximum prime factors is $5$ and minimum is $1$.
But I could not understand why is this and what is the correct way to find maximum and minimum prime factors from total no of factors or total composite factors. So please help me out. Thanks in advance.
First, note that the positive integer $n = p_1^{\alpha_1} \cdots p_n^{\alpha_n}$ has
$$\prod_{k=1}^n (\alpha_k + 1) = (\alpha_1 + 1) \cdot (\alpha_2 +1 ) \cdots (\alpha_k + 1)$$
factors.
So if your number $n$ has 72 positive divisors, then
$$72 = (\alpha_1 + 1) \cdot (\alpha_2 +1 ) \cdots (\alpha_k + 1)$$
where $k$ is the number of distinct primes in the prime factorization for $n$. You correctly noted that $72 = 2\cdot 2\cdot 2\cdot 3\cdot 3$. To get as many prime factors as possible, set $k = 5$ and $\alpha_1 = \alpha_2 = \alpha_3 = 1$ and $\alpha_4 = \alpha_5 = 2$. So for example your $n$ could be equal to $2^2 3^2 5^2 7^1 11^1$. Check that this has $72$ factors. For the minimum, just set $k=1$ and $\alpha_1 = 71$. So for example your $n$ could be $2^{71}$.
If you're just looking at composite factors, the process is similar. Since $n$ has $\prod_{k=1}^n (\alpha_k + 1)$ total factors, $k$ prime factors and $1$ (which is not composite) as factor, the number of composite factors is
$$\left( \prod_{k=1}^n (\alpha_k + 1)\right) - (k+1).$$
The minimum is still $1$ (set $\alpha_1 = 73$). For the maximum, you might need to do some casework. You want the number $72 + k + 1$ to have exactly $k$ prime factors. So $k$ cannot get too large. (See below for precise reasoning.) I tried the first $10$ values of $k$ and $k = 2$ (yielding $\prod_{k=1}^n (\alpha_k + 1) = 75$) was the only match. So the maximum number of prime factors for the case when you only want composite factors is $2$. An example number would be $2^{24} 3^2$. Check that this has $75$ total factors, three of which ($1$, $2$, $3$) are not composite.
Here is why $k$ cannot get too large in the composite case. Let $\{ p_1, \cdots, p_k, \cdots \}$ be the set of primes in order. So $p_1 = 2$, etc. Let the smallest number with exactly $k$ prime factors be $S_k = p_1 \cdots p_k$. So $S_1, S_2, S_3, S_4 = 2,6,30,210$. Note that $S_k \geq 2^k$.
We want $72 + k+1$ to have exactly $k$ prime factors, which means $72 + k + 1 \geq S_k$. But this means $72 + k + 1 \geq 2^k$, which is false for all $k \geq 7$. (Polynomials grow more slowly than exponentials.) So we can stop our search in the composite case for all $k \geq 7$.
For any $k$ you can find a number with exactly $k$ factors but only $1$ prime factor, e.g. $2^{k-1}$ has factors $2^0,2^1,...,2^{k-1}$.
In general, if a number $n$ has prime factors $p_1,...,p_r$ you can write it as $p_1^{a_1}\times\cdots\times p_r^{a_r}$, where each $a_i\geq 1$. The factors of this are all the numbers expressible as $p_1^{b_1}\times\cdots\times p_r^{b_r}$, where for each $b_i$ we have $0\leq b_i\leq a_i$. Thus there are $a_i+1$ choices for $b_i$, and $(a_1+1)\times\cdots\times(a_r+1)$ choices overall. So if a number has $r$ prime factors and $k$ factors, we must have $k$ being a product of $r$ numbers, each greater than $1$.
The largest product which gives $k=72$ is $2\times2\times2\times3\times3$, so you can have at most $5$ prime factors (and for exactly $5$, you must have $n=p_1p_2p_3p_4^2p_5^2$ for some primes $p_1,...,p_5$). | 2019-06-24T08:48:06 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2349232/what-can-be-the-maximum-and-minimum-number-of-prime-factors",
"openwebmath_score": 0.9469801187515259,
"openwebmath_perplexity": 102.54189938307915,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9914225130656699,
"lm_q2_score": 0.8479677622198947,
"lm_q1q2_score": 0.8406943298187204
} |
https://stats.stackexchange.com/questions/398822/finding-the-joint-cdf-using-the-joint-pdf-why-cant-i-do-this | # Finding the joint CDF using the joint PDF; why can't I do this?
Find the joint CDF of the independent random variables $$X$$ and $$Y$$, where
$$f_X(x)=x/2, 0\le x \le 2,$$ and
$$f_Y(y)=2y, 0 \le y \le 1$$.
To do this, we can find the CDF separately for each of the marginal PDFs, and then multiply them together to get the joint CDF (since the variables are independent).
However, why doesn't it work to first multiply the marginal PDFs together, to get a joint PDF of $$f_{X,Y}(x,y)=xy$$, and then take the double integral (with respect to $$x$$ and then $$y$$) to get the joint CDF?
This is the way I thought to do it first but it seems like it wouldn't give the same answer; why not?
Thank you!
If $$X$$ and $$Y$$ are independent random variables, then $$F_{X,Y}(x,y)=\int_{-\infty}^x \int_{-\infty}^y f_{X,Y}(w,v)\,dv\,dw = \int_{-\infty}^x \int_{- \infty}^{y} f_X(w)f_Y(v)\,dv\,dw$$ $$=\int_{-\infty}^x f_X(w)\,dw\int_{-\infty}^{y}f_Y(v)\,dv = F_X(x)F_Y(y).$$
Method 1 (joint pdf approach) gives: $$f_{X,Y}(x,y)=f_X(x)f_Y(y)=xy,$$ if $$0\leq x \leq 2, 0\leq y \leq 1$$ and zero otherwise. Then $$F_{X,Y}(x,y)=\int_0^x \int_0^yf_{X,Y}(w,v)\,dv\,dw=\int_0^x \int_0^y wv \, dv \, dw=\frac{x^2y^2}{4}.$$
Method 2 (marginal cdf approach) gives: $$F_X(x) = \int_0^x \frac{w}{2} \, dw =\frac{x^2}{4}$$ $$F_Y(y) = \int_0^y 2v \, dv =y^2.$$ It follows $$F_{X,Y}(x,y)=F_X(x)F_Y(y)=\frac{x^2y^2}{4}.$$
I should note the cdf's are also piecewise functions: $$F_X(x)$$ and $$F_Y(y)$$ take a value of zero if $$x \leq 0$$ and $$y \leq 0$$, respectively, and take a value of 1 if $$2 \leq x$$ and $$1 \leq y$$, respectively. $$F_{X,Y}(x,y)$$ takes a value of zero if either $$x \leq 0$$ or $$y \leq 0$$, and takes a value of $$1$$ if $$2 \leq x$$ and $$1 \leq y$$.
• Thank you so much!! This is super helpful. Why did you decide to set the bounds for the integral from $0$ to $x$ and $0$ to $y$, as opposed to $0$ to $2$ for $x$, and $0$ to $1$ for $y$, because aren't we given that those are the bounds? Thank you! – Sarina Mar 22 at 19:32 | 2019-10-14T21:30:16 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/398822/finding-the-joint-cdf-using-the-joint-pdf-why-cant-i-do-this",
"openwebmath_score": 0.9312463998794556,
"openwebmath_perplexity": 140.13430496969784,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9822876976669629,
"lm_q2_score": 0.8558511543206819,
"lm_q1q2_score": 0.8406920599232752
} |
https://math.stackexchange.com/questions/3150114/two-inner-products-different-by-a-scalar | # Two inner products different by a scalar
Let $$\langle \cdot, \cdot \rangle_{1}$$ and $$\langle \cdot,\cdot \rangle_{2}$$ be inner product on a finite-dimensional vector space with the property that \begin{align*} \langle u,v\rangle_{1}=0 \Leftrightarrow \langle u,v\rangle_{2}=0 \end{align*} for all $$u,v\in V$$. Show that there is a scalar $$a\in F$$ for which \begin{align*} \langle u,v\rangle_{2}=a\langle u,v\rangle_{1} \end{align*}
I saw a solution wrote like this:
Presumably, $$F=\mathbb {R}$$ or $$\mathbb {C}$$. For every nonzero vector $$v\in V$$, define $$c_{v}=\frac {\langle v,v\rangle_{2}}{\langle v,v\rangle_{1}}$$. Now, for any $$v,w\in V$$, let $$x=w-\frac {\langle w,v\rangle_{1}}{\langle v,v\rangle_{1}}v$$. Then $$\langle x,v\rangle_{1}=0$$. Hence, $$\langle x,v\rangle_{2}=0$$ and in turn \begin{align*} \langle w,v\rangle_{2}&=\frac {\langle w,v\rangle_{1}}{\langle v,v\rangle_{1}}\langle v,v\rangle_{2}\\ &=\frac {\langle v,v\rangle_{2}}{\langle v,v\rangle_{1}}\langle w,v\rangle_{1}\\ &=c_{v}\langle w,v\rangle_{1} \end{align*} for all nonzero $$v,w\in V$$. Therefore, \begin{align*} c_{v}\langle w,v\rangle_{1}&=\langle w,v\rangle_{2}\\ &=\overline {\langle v,w\rangle_{2}}\\ &=\overline {c_{w}\langle v,w\rangle_{1}}\\ &=c_{w}\langle w,v\rangle_{1} \end{align*} for all $$v,w\not=0$$. Now, for any $$v,w\not =0$$, there exists some $$y\in \{w+tv:t\in \mathbb {R}\}$$ such that $$\langle w,y\rangle_{1}$$, $$\langle y,v\rangle_{1}\not =0$$. Hence, we have $$c_{v}\langle y,v\rangle_{1}=c_{y}\langle y,v\rangle_{1}$$ and $$c_{y}\langle w,y\rangle_{1}=c_{w}\langle w,y\rangle_{1}$$. Hence $$c_{v}=c_{y}=c_{w}$$ for all $$v,w\not=0$$. In other word, all the $$c_{v}$$'s are equal to some common constant $$c>0$$. Thus, the proof is done. But I don't know what if $$\langle w,v\rangle_{1}=0$$ and why do we care bout the vectors in $$\{w+tv:t\in \mathbb {R}\}$$.
• If $\langle w,v\rangle_1=0$, then $\langle w,v\rangle_2=0$ and $0=c\times 0$ holds for any $c$, so I don't see any problem in that case. – learner Mar 16 at 5:53
• @learner Then there's no way to find the value of c. – Jiexiong687691 Mar 16 at 6:18
• You don't really need to find a particular value of $c$, you have to show that there exists at least one such value that works. In the above case, any value of $c$ works, so we're done. – learner Mar 16 at 6:32
What if $$\langle w,v\rangle_1 = 0$$?
The idea is this : if $$\langle w,v\rangle \neq 0$$ (and we know for sure that such $$w$$ exists, simply taking $$w = v$$), then we can get $$c_v$$ from there. Now, does that $$c_v$$ work when $$\langle w,v\rangle_1 = 0$$, is the question.
But, if $$\langle w,v \rangle_1 = 0$$ then $$\langle w,v\rangle_2 = 0$$ by the given conditions, so in fact $$\langle w,v\rangle_1 = c_v \langle w,v\rangle_2$$ (obviously because $$0 \times c_v = 0$$).
In short, it will still hold that for all non-zero $$v$$ and $$w$$ that $$\langle w,v\rangle_2 = c_v \langle w,v\rangle_1$$,because it will hold through the given argument if $$\langle w,v\rangle_1 \neq 0$$, and both sides are zero if $$\langle w,v \rangle_1 = 0$$.
Why do we care about the vectors $$w+tv$$?
The best way to put it is this : if we show for any $$v,w$$ that $$c_v = c_w$$, then we are done. If $$w=cv$$ for some scalar $$c$$ then we can easily show that $$c_w = c_v$$.
If $$w$$ and $$v$$ are linearly independent, then consider the two dimensional subspace generated by $$v$$ and $$w$$, and restrict the inner products $$1$$ and $$2$$ to this space. The point is , that the conditions for the inner products still hold in this two dimensional subspace. So the conclusion must hold for this subspace. Therefore, using only the vectors in this two dimensional subspace, we should be able to conclude that $$c_v = c_w$$. So we care about these vectors, because the proof should require only the properties of these vectors to conclude.
Which is why we restrict our attention to the vectors $$\{w+tv : t \in \mathbb R\}$$, which (up to scaling) covers every element of the span of $$v$$ and $$w$$. Then the rest should be straightforward. | 2019-04-24T12:19:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3150114/two-inner-products-different-by-a-scalar",
"openwebmath_score": 0.9794548153877258,
"openwebmath_perplexity": 148.06630511316456,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.982287698185481,
"lm_q2_score": 0.8558511524823263,
"lm_q1q2_score": 0.8406920585612555
} |
https://scicomp.stackexchange.com/questions/10024/numerical-iterative-method-estimating-error | # Numerical iterative method, estimating error
Given iterative method: $x_{n+1}=0.7\sin x_n +5 = \phi(x_n)$ for finding solution for $x=0.7\sin x +5$, I want to estimate $|e_6|=|x_6-r|$ as good as possible, with $x_0=5$, where $r$ is exact solution. This method obviously converges, because $\phi$ is contraction mapping, so $r=\phi(r)$ is a fixed point. So, with mean value theorem:
$|e_{n+1}|=|x_{n+1}-r|=|\phi(x_n)-\phi(r)|\le \max_{c\in\mathbb{R}}|\phi'(c)|\cdot |x_n-r|$
and we have:
$|e_n|\le 0.7^n \cdot |e_0|$
But I don't know how can I estimate $|e_0|$ without a computer? I suppose there is some simple way to finish it and with clever observation $|e_0|\le 0.7$. Can anybody help?
• If you knew the error exactly, that implies you know the solution, which makes the whole exercise trivial. I think your observation of e_0 < 0.7 is a fairly good one. And cosidering the true error is about 0.66, you likely won't get much better. – Godric Seer Nov 12 '13 at 22:34
• Yes, but that $e_0\le 0.7$ was only my guess. I don't know that. I'm looking for some trick that will help me with derivation of some decent upper bound for $|e_0|$ with given $x_0$. Without use of a computer. – xan Nov 12 '13 at 22:40
• Only thing I have so far is that $|e_n|\le 0.7^n \cdot |e_0|$ but this is useless without any approximation of $|e_0|$ which can be very big for many $x_0$. – xan Nov 12 '13 at 22:41
• Ok, I think substituting $x=4.3$ into $f(x)=x-0.7\sin x -5$ will suffice. It's easy to see that $f(x)$ is negative and for $x=5$ positive, so $|e_0|\le 0.7$ in this case. – xan Nov 12 '13 at 22:47
Given some $x_0$, you are looking for a bound on the error of the root for the equation:
$0 = 0.7 \mathrm{sin}(x) - x + 5$
Note that the function is bounded above by $5.7 - x$ and below by $4.3 - x$
Each of these have a trivial root at $5.7$ and $4.3$ respectively. You know that any roots of your function then fall in $[ 4.3, 5.7]$. Therefore, for any $x_0$
$e_0 = max\{|x_0 - 5.7|,|x_0 - 4.3|\}$
• @xan : $\max\{|x_0-5.7|,|x_0-4.3|\} = 0.7 + |x_0-5|$. – Stefan Smith Nov 28 '13 at 1:01 | 2019-10-16T22:56:17 | {
"domain": "stackexchange.com",
"url": "https://scicomp.stackexchange.com/questions/10024/numerical-iterative-method-estimating-error",
"openwebmath_score": 0.8589611649513245,
"openwebmath_perplexity": 185.92147668167803,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9822877038891779,
"lm_q2_score": 0.8558511396138366,
"lm_q1q2_score": 0.8406920508022118
} |
https://math.stackexchange.com/questions/904033/probability-of-bingo | # probability of bingo
It is the first time I heard about bingo game and I would like to learn more on this game by mathematical analysis. To make it simple, I consider the American BINGO with 75 balls used and each game will at maximum draw 50 balls. The winning pattern on each card is the center cross (the column pass the free and the row pass the free) and two diagonals. I want to know what's the probability for not winning the game within 50 draws for a single board. Here is my math
the probability of winning a single pattern is $P_1=C(50,4)/C(75,4)=0.189477$, since there are 4 patterns to win, the total winning probability is $P_4=4P_1=0.757909$. The probability of nothing winning should be $1-P_4=0.242901$. However, by computer simulation, I've got that the probability of not winning the game is 0.414359. I think there might be something wrong in my math. Any comment will be appreciated.
## 1 Answer
Define event $W_i$ = "Line $i$ wins" for $i = 1,2,3,4$.
Then the probability of none of the $4$ lines winning is $1 - P(W_1 \cup W_2 \cup W_3 \cup W_4)$.
Because more than one of the $4$ lines could win on any given draw, $P(W_1 \cup W_2 \cup W_3 \cup W_4) \neq P(W_1) + P(W_2) + P(W_3) + P(W_4)$.
This is why, while $P_1 = C(50,4)/C(75,4)$ is correct, multiplying that by $4$ is not.
Instead, we use the inclusion-exclusion principle:
\begin{eqnarray*} P(\mbox{no winning lines}) &=& 1 - P(W_1 \cup W_2 \cup W_3 \cup W_4) \\ && \\ &=& 1 - \sum_{i=1}^{4}{P(W_i)} + \sum_{i \lt j}{P(W_i \cap W_j)} - \sum_{i \lt j \lt k}{P(W_i \cap W_j \cap W_k)} \\ && \qquad + P(W_1 \cap W_2 \cap W_3 \cap W_4) \\ && \\ &=& 1 - \binom{4}{1} \binom{50}{4} \bigg/ \binom{75}{4} + \binom{4}{2} \binom{50}{8} \bigg/ \binom{75}{8} \\ && \qquad - \binom{4}{3} \binom{50}{12} \bigg/ \binom{75}{12} + \binom{4}{4} \binom{50}{16} \bigg/ \binom{75}{16} \\ && \\ &\approx& 1 - 0.7579086 + 0.1909348 - 0.0185883 + 0.0005759 \\ && \\ &=& 0.4150138 \end{eqnarray*}
• Thanks Mick. I see why my math is wrong now :) Aug 24 '14 at 19:51 | 2021-11-28T09:27:24 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/904033/probability-of-bingo",
"openwebmath_score": 0.9999856948852539,
"openwebmath_perplexity": 846.0013424024413,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9822876997410348,
"lm_q2_score": 0.8558511414521923,
"lm_q1q2_score": 0.840692049057813
} |
https://mathematica.stackexchange.com/questions/60053/approximation-to-the-prime-counting-function | # Approximation to the prime counting function
Is there a function similar to PrimePi that gives approximate value for large numbers?
In particular, I need a reasonably good approximation for $$\pi(x)$$, where $$x\approx10^{1000}$$.
More or less a function that gives $$\int_2^x\frac{dt}{\ln t}$$ or better.
Edit: I settled for the definite integral approximation but if someone knows of a built-in function that performs better, I'm definitely interested.
• LogIntegral is this integral. Try this: Show[{Plot[LogIntegral[x], {x, 1, 2000}], ListPlot[Table[PrimePi[n], {n, 1, 2000}]]}] Sep 18, 2014 at 22:36
The best analytic built-in approximation is the Riemann Prime Counting Function; it is implemented in Mathematica as RiemannR. So far we know exact values of $\pi$ prime counting function for n < 10^25, however in Mathematica its counterpart PrimePi[n] can be computed exactly to much lower values i.e. up to 25 10^13 -1, see e.g. What is so special about Prime? for more details.
RiemannR[10^1000] // N
4.344832576401197453*10^996
See e.g. Prime-counting function, it claims that the best estimator is $\operatorname{R}(x) - \frac1{\ln x} + \frac1\pi \arctan \frac\pi{\ln x}$
let's define it as APi:
APi[x_] := RiemannR[x] - 1/Log[x] + 1/Pi ArcTan[ Pi/Log[x]]
it provides the same approximation as RiemannR, at least IntegerPart does, in fact
Grid[ Table[ IntegerPart @ { 5 10^k, RiemannR[5 10^k] - PrimePi[5 10^k],
APi[5 10^k] - PrimePi[5 10^k]}, {k, 3, 13}],
Frame -> All, Alignment -> Left]
We can realize how good RiemanR is plotting related errors of interesting approximations where we know exact values of the prime counting function (see e.g. "Mathematica in Action" by Stan Wagon):
Plot[{ LogIntegral[x] - PrimePi[x], RiemannR[x] - PrimePi[x],
x/(Log[x] - 1) - PrimePi[x]}, {x, 2, 3 10^5}, MaxRecursion -> 3,
Frame -> True, PlotStyle -> {{Thick, Red}, {Thick, Darker @ Green},
{Thick, Darker @ Cyan}},
PlotLegends -> Placed["Expressions", {Left, Bottom}],
ImageSize -> 800, AxesStyle -> Thick]
Plot[{ LogIntegral[x] - PrimePi[x], RiemannR[x] - PrimePi[x],
x/(Log[x] - 1) - PrimePi[x]}, {x, 10^6, 10^7}, MaxRecursion -> 3,
Frame -> True, PlotStyle -> {{Thick, Red}, {Thick, Darker @ Green},
{Thick, Darker @ Cyan}},
PlotLegends -> Placed["Expressions", {Left, Bottom}],
Axes -> {True, False}, AxesStyle -> Thick, ImageSize -> 800]
• Thank you. I wish I could accept both replies as the answer; however, I'm obliged to accept Karsten's correct reply because it was posted first. However, let me stress that I thoroughly enjoyed reading your reply and, particularly, the content of the links you provided.
– user11356
Sep 19, 2014 at 0:35
• @user2943324 You can accept whatever you want however you don't have to choose the first answer, moreover accepting too quickly you discourage others (including me) to post another answers possibly better. As far as I can say I could elaborate my answer improving it, even though so far I've provided more details than the other answer. Sep 19, 2014 at 0:48
• Agreed. It seems that I am too quick to mark a reply as accepted. You have to forgive me. I don't frequent these forums often enough and when I do, it seems, I inadvertently get too excited.
– user11356
Sep 19, 2014 at 1:01
RiemannR seems to be a better choice than LogIntegral based on this plot:
Plot[{PrimePi[n], LogIntegral[n], RiemannR[n]}, {n, 1, 5000}, PlotStyle -> {Black, Blue, Red}]
RiemannR[1.*10^1000]
4.344832576401197453*10^996
• That gets my vote (as does the other RiemannR response). I'll note it is somewhat slower to compute, although still fast enough for whatever purposes. Sep 18, 2014 at 23:43 | 2022-07-07T05:08:37 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/60053/approximation-to-the-prime-counting-function",
"openwebmath_score": 0.46831458806991577,
"openwebmath_perplexity": 3065.218831675757,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.982287697666963,
"lm_q2_score": 0.8558511414521923,
"lm_q1q2_score": 0.8406920472827163
} |
https://math.stackexchange.com/questions/1310490/finding-the-jordan-canonical-form-of-a-6-times-6-matrix | Finding the Jordan Canonical form of a $6 \times 6$ matrix
Find the Jordan Canonical Form of the following matrix $$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 & 1 & 1\\ \end{bmatrix}$$
My try: I go about finding the Jordan Basis for this matrix. It is clear that $1$ is the only eigenvalue of this matrix. So $$A-I=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1 & 1 & 0\\ \end{bmatrix}$$
Moreover Rank$(A-I)^2=1$. Moreover Rank$(A-I)^3=0$ . So We don't need to go further on evaluating the powers of matrices in our search for generalized eigenvectors. $$(A-I)^2=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 4 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$$
Also $(A-I)^3=0$.
It is seen that the generalized eigenspace (say $U$) consists of
$U$=span$\{v_1=(0,1,0,0,0,0)^t,v_2=(0,0,1,0,0,0)^t,v_3=(0,0,0,1,0,0)^t,v_4=(0,0,0,0,1,0)^t,v_5=(0,0,0,0,0,1)^t\}$
Here I run into a little problem. Since $(A-I)v_i=v_5$ for each $i=1,2,3,4$, I have only five vectors with me for the Jordan Canonical Basis. Moreover I can't choose any other arbitrary vector linearly independent to these four simply because if $v$ were such a vector , then there is no $i \gt 0$ (and integer) such that $(A-I)v^{i}=0$
I have also another question here. Since $(A-I)^3=0$, why should I not choose general eigen vectors corresponding to $(A-I)^3$??
I am a little stuck here.
Thanks for the help!!!
• $A-I$ is nilpotent, it will most certainly not have the same rank as its square! You'll find that $(A-I)^2$ has all its coefficients zero, except for its last line which equals $$(\;4\quad 0\quad 0\quad 0\quad 0\quad 0\;)$$ – Olivier Bégassat Jun 3 '15 at 11:00
• @OlivierBégassat Thanks. Edited – tattwamasi amrutam Jun 3 '15 at 11:14
From rank computations you find: there are four linearly independent eigenvectors ($rank(I-A)=2$), hence four Jordan blocks. The nilpotence index of $I-A$ is three, hence the largest Jordan block has size $3\times 3$. Thus, the other blocks are $1\times 1$ blocks.
You need a vector $v_6$ that satisfies $$(I-A)^3 v_6=0, \ (I-A)^2v_6\ne0.$$ Take $$v_6=\pmatrix{1&0&0&0&0&0}.$$ Then $(I-A)^2v_6, (I-A)v_6$ are in the Jordan basis too, which are $$(I-A)^2v_6 = \pmatrix{0&0&0&0&0&1}^T=v_5, \\ (I-A)v_6 = \pmatrix{0&1&1&1&1&1}^T=:v_7.$$ Now you have three linearly independent vectors. Complement this set with three eigenvectors to form a basis, for instance $$\pmatrix{0&1&-1&0&0&0}^T, \pmatrix{0&1&0&-1&0&0}^T, \pmatrix{0&1&0&0&-1&0}^T.$$ Arrange them in the right order to obtain the Jordan basis.
• what happens if I take $e_2$, $e_3$ and $e_4$ instead of the eigen vectors?? Can I take them and form a basis as well??? – tattwamasi amrutam Jun 3 '15 at 11:50
• No, these are no eigenvectors, and all of them satisfy $(I-A)e_i=v_5$. The resulting matrix will not be in Jordan form (it would contain three $1$'s in a row). – daw Jun 3 '15 at 11:52 | 2019-06-17T01:04:48 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1310490/finding-the-jordan-canonical-form-of-a-6-times-6-matrix",
"openwebmath_score": 0.9249415993690491,
"openwebmath_perplexity": 276.1623377752475,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9867771770811147,
"lm_q2_score": 0.8519528094861981,
"lm_q1q2_score": 0.8406875883511152
} |
http://raisingthebar.nl/2018/04/24/solution-to-question-1-from-uol-exam-2016-zone-b/ | 24
Apr 18
## Solution to Question 1 from UoL exam 2016, Zone B
This problem is a good preparation for Question 2 from UoL exam 2015, Zone A (FN3142), which is more difficult.
## Problem statement
Two corporations each have a 4% chance of going bankrupt and the event that one of the two companies will go bankrupt is independent of the event that the other company will go bankrupt. Each company has outstanding bonds. A bond from any of the two companies will return $R=0$% if the corporation does not go bankrupt, and if it goes bankrupt you loose the face value of the investment, i.e., $R=-100$%. Suppose an investor buys $1000 worth of bonds of the first corporation, which is then called portfolio $P_1$, and similarly, an investor buys$1000 worth of bonds of the second corporation, which is then called portfolio $P_2$.
(a) [40 marks] Calculate the VaR at $\alpha=5$% critical level for each portfolio and for the joint portfolio $P=P_1+P_2$.
(b) [30 marks] Is VaR sub-additive in this example? Explain why the absence of sub-additivity may be a concern for risk managers.
(c) [30 marks] The expected shortfall $ES^\alpha$ at the $\alpha=5$% critical level can be defined as $ES^\alpha=E_t[R|R\le VaR^\alpha_{t+1}]$. Calculate the expected shortfall for the portfolios $P_1,\ P_2$ and $P$. Is this risk measure sub-additive?
## Solution
There are a couple of general ideas to understand before embarking on calculations. The return on the bond of one company is a binary variable taking values 0% and -100%. All calculations involving it are similar to the ones for the coin. After doing calculations the return figures can be translated to dollar amounts by multiplying by $1000. While the use of the notions of the distribution function and generalized inverse can be avoided, I prefer to use them to show the general approach. (a) The return on one bond is described by the table Table 1. Probability table for return on one bond Return values Probability 0 0.96 -100 0.04 Therefore its distribution function can be found in the same way as for the coin: Figure 1. Distribution function for return on one bond The distribution function is shown in red. It is zero for $R<-100$, 0.04 for $-100\le R<0$ and 1 for $0\le R<\infty$. The definition of the VaR requires inversion of this function. The graph of this function has flat pieces and its usual inverse does not exist. We have to use the generalized inverse defined by $F^{-1}(y)=\inf\{x:F(x)\ge y\}$, see the definition of the infimum here. In our case $y=0.05$ and the verbal procedure is: 1) find those returns for which $F(R)\ge 0.05$ (it's the half-axis $[0,\infty)$) and 2) among them find the least return. The answer is $VaR^\alpha=0$%. This is the Value at Risk for each of $P_1,P_2$. What we do next is very similar to the derivation of the sampling distribution for two coins. Table 2. Joint probability table for returns on two portfolios First portfolio 0 -100 Second portfolio 0 $0.96^2=0.9216$$0.96^2=0.9216$ $0.04\times 0.96=0.0384$$0.04\times 0.96=0.0384$ -100 $0.04\times 0.96=0.0384$$0.04\times 0.96=0.0384$ $0.04^2=0.0016$$0.04^2=0.0016$ The main body of the table contains probabilities of pairs $(R_1,R_2)$ of the two returns. For the total portfolio the possible return values are 0 (none of the companies goes bankrupt), -50 (one goes bankrupt and the other does not) and -100 (both go bankrupt). The corresponding probabilities follow from Table 2 and we get Table 3. Probability table for return on the total portfolio Total return Probabilities 0 0.9216 -50 $2\times 0.0384=0.0768$$2\times 0.0384=0.0768$ -100 0.0016 This table results in the following distribution function: Figure 2. Distribution function for return on two bonds Since 0.0784>0.05, the Value at Risk is -50% (use the generalized inverse). (b) Translating the percentages to dollars, at 5% the risk for each of the bonds is$0 and for the total portfolio it is $1000 (50% of$2000, I am passing from negative percentages to positive loss figures).
We say that Value at Risk is sub-additive if $VaR_P^\alpha\le Var^\alpha_{P_1}+Var^\alpha_{P_2}$. Our calculations show that Value at Risk is not sub-additive in case of independent returns. This has an important practical implication. Suppose that a financial institution has several branches and each of them keeps their Value at Risk, say, at zero. Nevertheless, the Value at Risk for the whole institution may well be large and threaten its stability.
(c) Here we have to apply the definition of the conditional expectation: $ES^\alpha_{P_i}=\frac{E_t[R1_{\{R\le VaR^\alpha_{t+1}\}}]}{P(R\le VaR^\alpha_{t+1})}$. Since $P(R\le 0)=0.96+0.04=1$, this is the same as
$ES^\alpha_{P_i}=E_t[R1_{\{R\le 0\}}]=0\times 0.96+(-100)\times 0.04=-4$%, $i=1,2$.
For the total portfolio we get
$ES^\alpha_P=\frac{E_t[R1_{\{R\le -50\}}]}{P(R\le -50)}=\frac{(-50)\times 0.0768+(-100)\times 0.0016}{0.0768+0.0016}=\frac{-3.84-0.16}{0.0784}=-51.02$%.
In monetary terms, this translates (again passing to positive values) to $40 for each bond and to$1020.40 for the total portfolio. The conclusion is that expected shortfall is not sub-additive. | 2018-07-20T18:25:57 | {
"domain": "raisingthebar.nl",
"url": "http://raisingthebar.nl/2018/04/24/solution-to-question-1-from-uol-exam-2016-zone-b/",
"openwebmath_score": 0.422188937664032,
"openwebmath_perplexity": 3010.7697244119145,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9867771766922545,
"lm_q2_score": 0.8519528076067262,
"lm_q1q2_score": 0.8406875861652048
} |
https://math.stackexchange.com/questions/2341163/does-a-convex-polynomial-always-reach-its-minimum-value | # does a convex polynomial always reach its minimum value?
Consider a convex polynomial $p$ such that $p(x_1,~x_2,\dots x_n)\geq 0~\forall x_1,~x_2,\dots x_n\in \mathbb{R}^n$. Does the polynomial reach its minimum value?
This is not true for non-convex polynomials like $(1-x_1x_2)^2+x_1^2$, see the response of J.P. McCarthy on a similar question on general polynomials. This is not true for general functions either. Function $e^x$ is infinitely differentiable, convex and bounded from below, but there is no $x$ that does reach the minimum value 0 (thanks to zhw for this simpler example).
• There are no bounded convex functions on $\mathbb R$ except constants. Your example $\to \infty$ at $-\infty.$ So might as well use something simpler here, like $h(x)=e^x.$ – zhw. Jul 2 '17 at 19:30
The restriction of $p$ to any straight line is a polynomial in one variable that is bounded below, therefore is either constant or goes to $\infty$ in both directions. If there is a line $L$ on which $p$ is constant, then using convexity it is easy to see that $p$ must be constant on all lines parallel to $L$, and by taking a cross-section we reduce the dimension by $1$. So we can assume wlog there is no line on which $p$ is constant.
Now consider $A = \{x \in \mathbb R^n: p(x) < C\}$ where $C > p(0)$.
This is a convex set. The restriction of $p$ to any ray through $0$ is a nonconstant polynomial in one variable and bounded below, therefore goes to $+\infty$ in both directions. Thus $A$ contains no ray through $0$. For each $s$ in the unit sphere $\mathbb S^{n-1}$, there is some $t > 0$ such that $p(ts) > C$ and by continuity this holds (with the same $t$) in some neighbourhood of $s$. Note that by convexity, $p(t' s) > C$ for all $t' > t$. Using compactness, we conclude that $A$ is bounded. And then the infimum of $p$ is the infimum of $p$ on the compact set $\overline{A}$, which is attained.
EDIT: As requested, I'll expand on "using compactness". For each $s \in \mathbb S^{n-1}$, there is $t > 0$ such that $p(ts) > C$. Thus the open sets $\{s \in \mathbb S^{n-1}: p(t s) > C\}$ for $t > 0$ form an open covering of $\mathbb S^{n-1}$. Because $\mathbb S^{n-1}$ is compact, this has a finite subcovering, i.e. $t_1, \ldots, t_k$ such that for every $s \in \mathbb S^{n-1}$, some $p(t_j s) > C$. But that says $p(x) > C$ for all $x$ with $\|x\| \ge \max(t_1, \ldots, t_k)$, i.e. $\|x\| < \max(t_1, \ldots, t_k)$ for all $x \in A$.
• Thanks for the answer. Could you please develop a bit "Using compactness, we conclude that $A$ is bounded."? Anyways, I agree $A$ is bounded. For any $s$ in the unit ball $\mathbb S$, denote $t_s$ the minimum value s.t. $p(t_s s)\geq C$ and $p(t's)>C~\forall t'>t_s$. We can not have a sequence $(s_i)$ with $s_i\in \mathbb S$ and $\lim_{i\to \infty} t_{s_i}=\infty$, since the Bolzano–Weierstrass theorem states $\lim_{i\to\infty} s_{n_i}=s$ for some sub-sequence $(s_{n_i})$. Using what you said on $s$, there is a neighborhood/ball $N_s$ around $s$ such that $s'\in N_s\implies t_{s'}<t_s+1$. – Daniel Porumbel Jun 30 '17 at 13:04
• It could be useful to expand on "$p$ must be constant on all lines parallel to $L$". My solution is as follows. Consider $p(x+tL)=c~\forall t\in \mathbb{R}$. We investigate the value of $p(y+tL)~\forall t\in\mathbb{R}$. By convexity, we have $p(y+\frac 12 tL) \leq \frac 12\left(p(x+tL)+p(2y-x)\right)=\frac12 \left(c+p(2y-x)\right)~\forall t\in\mathbb{R}$, and so, $p$ can not go to $\infty$ in either direction along $y+tL$ with $t\in\mathbb{R}$. As such, it needs to be constant on any line parallel to $L$. – Daniel Porumbel Jul 3 '17 at 21:03 | 2020-02-23T18:13:55 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2341163/does-a-convex-polynomial-always-reach-its-minimum-value",
"openwebmath_score": 0.9579554200172424,
"openwebmath_perplexity": 75.1630678328784,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9867771813585751,
"lm_q2_score": 0.8519527982093666,
"lm_q1q2_score": 0.8406875808675897
} |
https://math.stackexchange.com/questions/1868697/how-to-prove-if-this-sum-l-0n-binomnl-2n-is-valid-for-all-n-in | # How to prove if this $\sum_{l=0}^{n}\binom{n}{l}=2^{n}$ is valid for all $n\in \mathbb{N}$? [duplicate]
Prove for for all $n\in \mathbb{N}$: $\sum_{l=0}^{n}\binom{n}{l}=2^{n}$
I know the steps of induction but i have no idea how to prove this equation with binomial coefficient.
1) For the induction base i need to take one number and it should be $n=1$. So, do i use this value $n$ for both $n$ and $l$ in the equation? In that case i believe that equation is not valid.
2) Should i for inductive claim increase both $n+1$ and $l+1$?
• Are you allowed to use the binomial theorem? – carmichael561 Jul 23 '16 at 17:34
• Technically, for the induction you may want to use $n=0$ as base case. – Clement C. Jul 23 '16 at 17:49
• Also, see this page 5/6 of this document for three proofs: Link – FraGrechi Jul 23 '16 at 17:57
Since the meat of your question seems to be about how to prove this claim with induction (even though the other answers provide much shorter and easier proofs), that is how I will choose to approach.
Checking base case: Here, we check that the statement is true for some starting value(s) of $n$. The statement's truth will depend only upon $n$ as $l$ is only used as a shorthand here to make the sum easier to write ($l$ will range between all values from $0$ to $n$ and will not be a specific single number in general)
If you consider $0\in\Bbb N$, we start with the base case of checking if $\sum\limits_{l=0}^0\binom{0}{l}=\binom{0}{0}=\frac{0!}{0!0!}=1=2^0$, so the base case is valid.
Otherwise, for $n=1$ we check that $\sum\limits_{l=0}^1\binom{1}{l}=\binom{1}{0}+\binom{1}{1}=1+1=2=2^1$, so the base case is again valid.
Inductive step: Again, $l$ will be all possible values in the range, so we do not change how $l$ appears in the inductive step. We assume that the statement $2^n = \sum\limits_{l=0}^n \binom{n}{l}$ is true for some $n$ and we wish to prove that from this it follows that it is true for the next value, $n+1$.
Here, we rely on something known as pascal's identity: $\binom{n+1}{r} = \binom{n}{r}+\binom{n}{r-1}$.
Examining the case of $n+1$, we have then:
$$\begin{array}{rlr}\sum\limits_{l=0}^{n+1}\binom{n+1}{l}&=\sum\limits_{l=0}^{n+1}\left(\binom{n}{l}+\binom{n}{l-1}\right)&\text{via pascal's identity}\\ &=\left(\sum\limits_{l=0}^{n+1}\binom{n}{l}\right)+\left(\sum\limits_{l=0}^{n+1}\binom{n}{l-1}\right)&\text{by splitting into separate summations}\\ &=\left(\binom{n}{n+1}+\sum\limits_{l=0}^n\binom{n}{l}\right) + \left(\binom{n}{-1}+\sum\limits_{l=1}^{n+1}\binom{n}{l-1}\right)&\text{by breaking off last and first pieces respectively}\\ &=\left(\sum\limits_{l=0}^{n}\binom{n}{l}\right)+\left(\sum\limits_{l=0}^{n}\binom{n}{l}\right)&\text{broken off terms equalled zero and reindexing of right}\\ &=2^n+2^n&\text{by induction hypothesis}\\ &=2\cdot (2^n)=2^{n+1}&\text{by simplification}\end{array}$$
Hence, it follows that it is true for $n+1$ as well, therefore by induction the statement is true for all $n\in\Bbb N$.
• And can you give me any hint how could I prove it easier, without induction? – Tars Nolan Jul 25 '16 at 11:31
If $S$ is a set with $n$ elements, then $\binom{n}{l}$ is the number of subsets of $S$ with $l$ elements.
$\sum_{l=0}^{n}\binom{n}{l}$ is then the number of all subsets of $S$ and this is equal to $2^{n}$ because the subsets of $S$ can be put into bijection with the set of $n$-bit strings.
For a prove by induction:
You can start the induction at $n=0$, which is trivial. So let $n$ be greater than $0$
This is actually needed in the calculations that follow, since we will use the induction hypothesis on $n-1$ which should not be smaller than $0$
\begin{align} \sum_{k=0}^{n} \binom{n}{k} &= 1+ \sum_{k=1}^{n-1} \binom{n}{k} + 1 \\ &= 1+ \sum_{k=1}^{n-1} \left[ \binom{n-1}{k-1} + \binom{n-1}{k} \right] + 1 \\ &= 1+ \sum_{k=1}^{n-1} \binom{n-1}{k-1} + \sum_{k=1}^{n-1}\binom{n-1}{k} + 1 \\ &= \sum_{k=1}^{n} \binom{n-1}{k-1} + \sum_{k=0}^{n-1}\binom{n-1}{k} \\ &=\sum_{k=0}^{n-1} \binom{n-1}{k} + \sum_{k=0}^{n-1}\binom{n-1}{k} \\ &= 2 \cdot \sum_{k=0}^{n-1} \binom{n-1}{k} = 2\cdot 2^{n-1} \end{align} In the last step the induction hypothesis was used.
The binomial coefficient $n\choose i$ is by definition, the coefficient of $x^i$ in the expanded binomial $(1+x)^n$, i.e., $(1+x)^n=\sum_{i=0}^n{n\choose i}x^i$. Use this definition with $x=1$.
• I don't think I've ever seen $\binom{n}{i}$ defined that way. The definition $\binom{n}{i} = |\{ A \subseteq \{ 1, 2, \dots, n \} : |A|=i \}|$ is far more common, in which case there is more work to be done. The definition $\binom{n}{i} = \frac{n!}{i!(n-i)!}$ is (abhorrent but) also common (when $i \le n$). – Clive Newstead Jul 23 '16 at 18:10
Yeah if you use binomial theorem ..then $$\sum_{l=0}^{n} nC_l=^nC_0+^nC_1+^nC_2+......$$ Which is equal to $$1+ ^nC_1+^nC_2+....+^nC_n$$ That is $(1+1)^n$ ...taking x=1and a=1 in the expansion of $(x+a)^n$ | 2020-01-18T23:46:57 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1868697/how-to-prove-if-this-sum-l-0n-binomnl-2n-is-valid-for-all-n-in",
"openwebmath_score": 0.9982945322990417,
"openwebmath_perplexity": 166.8502027226624,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9867771778588348,
"lm_q2_score": 0.8519527982093666,
"lm_q1q2_score": 0.8406875778859761
} |
https://math.stackexchange.com/questions/2711633/finding-the-point-on-6x2y2-262090-that-is-nearest-to-the-point-1045-0 | Finding the point on $6x^2+y^2=262090$ that is nearest to the point $(1045,0)$
I know that to find the point on $6x^2+y^2=262090$ that is nearest to the point $(1045,0)$, we can try to minimize the squared distance $S=(x-1045)^2+262090-6x^2$. However, calculus tells us that this function does not have a minimum point (instead only a maximum point exists).
But if we try to minimize the distance function (without squaring) then we can find the minimum.
So, When exactly can we actually square the distance function to find the max/min point for distance problems?
• It does not have a minimum on $\Bbb R$, but it does in the relevant range of $x$ - specifically, the set $\{x\in\Bbb R\,:\, -\sqrt{262090}\le x\le\sqrt{262090}\}$. – user228113 Mar 28 '18 at 11:22
• Thanks a lot for that, that is exactly what I thought. But it is widely thought that to find the minimum distance, it suffices to just consider the squared distance function. This example, however, shows that it is in fact NOT SUFFICIENT. – Probability is wonderful Mar 28 '18 at 11:26
• It is sufficient. The problem is that you must apply calculus correctly. – user228113 Mar 28 '18 at 11:27
• So is it true that whenever we consider the squared distance function, we need to explicitly specify the range of every variable? – Probability is wonderful Mar 28 '18 at 11:29
• Just of the variable you are considering (in this case $x$). And of course, looking for zeros of the derivative is not sufficient. You need to evaluate the appropriate extremal points of the subintervals where the derivative is $> 0$ (which may include the boundary of the domain). – user228113 Mar 28 '18 at 11:30
There is nothing wrong with using the squared distance.
You converted the problem into a one-parameter minimization problem. You are looking for the minimum of a smooth function on the interval $[-\sqrt{262090/6},\sqrt{262090/6}]$. The minimum value is obtained at a zero of the derivative (a critical point) or at one of the endpoints. The function is a downward opening parabola, so you know that any critical point is a local maximum. Therefore you have to look at the endpoints.
Alternatively, you could have converted the problem into minimization over $y$. You can solve that $x=\pm\sqrt{(262090-y^2)/6}$. If you draw a picture, it becomes clear that the closest point must be in the right half of the ellipse. (If you don't believe in pictures, you can treat the two halves separately.) In this half $x>0$.
This leads to the squared distance being $$\begin{split} &(x-1045)^2+(y-0)^2 \\=& \frac16(262090-y^2)-2090\sqrt{(262090-y^2)/6}+1045^2+y^2 \\=& \frac56y^2-2090\sqrt{(262090-y^2)/6}+1045+262090/6. \end{split}$$ Now each term is at its smallest when $y=0$, so the minimum is at $y=0$. The corresponding value of $x$ is $\sqrt{262090/6}$ — the endpoint of the $x$-interval!
• I'm afraid the result you gave $x=\sqrt{262090}$ does not satisfy the equation given: $6x^2+y^2 = 6\cdot(\sqrt{262090})^2+0^2 = 1572540$ which does not equal the RHS... – CiaPan Mar 28 '18 at 11:53
• @CiaPan Good catch, thanks! I forgot to divide by some sixes. Does it make more sense now? – Joonas Ilmavirta Mar 28 '18 at 11:55
• Perfect answer! I just wonder why in so many textbooks when they calculate the min/max distances by considering the squared distance, they do not check the boundary points? – Probability is wonderful Mar 28 '18 at 12:25
• @Probabilityiswonderful Thanks! Why many sources are as sloppy as they are, focusing on computation instead of understanding, is beyond my understanding. Too many things are left implicit. (Did you know you have the privilege to vote up any questions and answers you like?) – Joonas Ilmavirta Mar 28 '18 at 12:28
• @Probabilityiswonderful Probably because in most textbooks' problems the domain is an open interval, bounded or not, hence no edge-cases exist to be tested. – CiaPan Mar 28 '18 at 12:29 | 2019-10-15T06:53:12 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2711633/finding-the-point-on-6x2y2-262090-that-is-nearest-to-the-point-1045-0",
"openwebmath_score": 0.8453709483146667,
"openwebmath_perplexity": 243.66428164768274,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9867771766922545,
"lm_q2_score": 0.8519527963298946,
"lm_q1q2_score": 0.8406875750374847
} |
https://math.stackexchange.com/questions/1044766/proving-floor-and-ceiling-of-a-rational-number | # Proving Floor and Ceiling of a Rational Number
Suppose x,y $\in \mathbb{Z}^+$
Prove $\lceil x/y \rceil = \lfloor (x-1)/y \rfloor + 1$
I was considering using the definition of floor and ceiling to prove this. But this does not seem like a valid proof to me as I assume the right hand side is already equal to the left hand side.
This is what I have so far:
Let n$\in \mathbb{Z}$
$\lceil x/y \rceil = n \iff n-1 < x/y \leq n$
$\lfloor (x-1)/y \rfloor = n -1 \iff n-1 \leq (x-1)/y < n$
Then $\lfloor (x-1)/y \rfloor +1 = n \iff n-1 \leq (x-1)/y < n + 1$
Then both sides are equivalent.
Would a better way be to prove using a contradiction by assuming both sides are not equal?
• What does $n$ have to do with all this? Nov 30, 2014 at 9:31
• n is the result of the floor and ceiling function. eg. $\lceil 5/2\rceil = 3$ So n = 3. Nov 30, 2014 at 9:32
• I don't see it anywhere in the statement to prove. Nov 30, 2014 at 9:33
• I guess I should put it in my proof part. edited. Nov 30, 2014 at 9:34
If $y$ divides $x$ then:
• $\lceil\frac{x}{y}\rceil=\frac{x}{y}$
• $\lfloor\frac{x-1}{y}\rfloor=\frac{x}{y}-1$
Hence $\lceil\frac{x}{y}\rceil=\lfloor\frac{x-1}{y}\rfloor+1$
If $y$ does not divide $x$ then:
• $\lceil\frac{x}{y}\rceil=\lfloor\frac{x}{y}\rfloor+1$
• $\lfloor\frac{x-1}{y}\rfloor=\lfloor\frac{x}{y}\rfloor$
Hence $\lceil\frac{x}{y}\rceil=\lfloor\frac{x-1}{y}\rfloor+1$
• I'm confused as to how you obtained the y does not x case. Did you use the definition of floor and ceiling? Also, wouldn't any positive integer divide another positive integer? Nov 30, 2014 at 9:50
• @onesevenfour: 1. There are two statements there, which one are you confused about? 2. When I say "divide", I mean "divide without a remainder" (what else could it mean? the only number that does not divide other numbers is $0$). Nov 30, 2014 at 9:52
starting from line 2, the inequality can be rewritten as
\begin{equation*} n-1+1/y \leq x/y < n+1/y \end{equation*} since x,y are integers, the left inequality is equivalent to
\begin{equation*} n-1 < x/y \end{equation*}
and the right inequality is equivalent to \begin{equation*} x/y \leq n \end{equation*}
• I get this proof, but does the case $\lfloor (x-1)/y \rfloor \neq n -1$ need to be considered? Nov 30, 2014 at 9:54
• I edited it to be more clear. this shows that the right hand sides of the first two lines are equivalent. therefore the left hand sides are as well Nov 30, 2014 at 10:14
I might want to use a property of integer arithmetic $a \lt b \iff a+1 \le b$, as in:
$\qquad\;\;\lceil x/y \rceil = n$
$\iff n-1 \lt x/y \leq n$
$\iff (n-1)y \lt x \leq ny$
$\iff (n-1)y \le x-1 \lt ny$
$\iff n-1 \le (x-1)/y \lt n$
$\iff \lfloor(x-1)/y\rfloor =n-1$
$\iff \lfloor(x-1)/y\rfloor +1=n$
so $\lceil x/y \rceil = \lfloor(x-1)/y\rfloor$ + 1 | 2022-06-26T06:26:31 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1044766/proving-floor-and-ceiling-of-a-rational-number",
"openwebmath_score": 0.9076536893844604,
"openwebmath_perplexity": 393.60474615260176,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9736446456243805,
"lm_q2_score": 0.863391617003942,
"lm_q1q2_score": 0.8406366249728638
} |
https://math.stackexchange.com/questions/3832935/in-how-many-ways-can-4-cards-be-drawn-randomly-from-a-pack-of-52-cards-such-that | # In how many ways can 4 cards be drawn randomly from a pack of 52 cards such that there are at least 2 kings and at least 1 queen among them?
So i tried this question in two ways
(i)In my first method I made different possible arrangements and then find the number of ways
So, the different possibilities are:
2 kings and 1 queen and 1 other card
Or, 3 kings and 1 queen
Or, 2 kings and 2 queens
Total ways possible = $${4\choose2}{4\choose1}{44\choose1}+{4\choose3}{4\choose1}+{4\choose2}{4\choose2} = 1056 + 16 + 36 = 1108$$
Total ways possible = 1108 And this is the correct answer.
(ii) In order to shorten the above method, I did this
We need at least 2 kings and 1 queen, so total ways possible = $${4\choose2}{4\choose1}{49\choose1}=1176$$(49 because I subtracted the 3 cards from the deck of 52 cards).
So what's the problem with second method? Why I'm getting additional 68 ways (1176 - 1108= 68)? And Is there any way to solve this question without making cases?
Thanks and stay safe.
Since in your second method, for example you have the case 3 kings 1 queen counted three times and 2 kings 2 queens counted twice.
Since assume that $${4\choose2}$$ picked up the king of spades and the king of hearts, and then the $${49\choose1}$$ picked up the king of diamonds. $$k_s,k_h,q,k_d$$ this is one way of choosing your cards. But this method counts $$k_d,k_h,q,k_s$$ as another way and this is wrong they should only be counted once.
On the other hand, in your first method when you use $${4\choose3}$$ for the case where there are 3 kings, you're being sure that all the equivalent ways are counted once.
Now why they are 68?
$${4\choose3}×2×4 + {4\choose 2}×{4\choose 2}=32+36=68$$
$${4\choose 3}$$ for the case we have 3 kings, $$2$$, because we want to cancel 2 ways (i.e. if we have $$k_1,k_2,q,k_3$$ we want to cancel $$k_1,k_3,q,k_2$$ and $$k_3,k_2,q,k_1$$ since they must be counted only once) and the $$4$$ because we have 4 possibilities for the queen.
Simlarly for the queens, $${4\choose 2}$$ for the case we have 2 queens, and another $${4\choose 2}$$ because we have $${4\choose 2}=6$$ possibilities of the first 2 kings. (Note that here we don't multiply by $$2$$ here since if we have $$k,k,q_1,q_2$$ we only want to cancel only 1 equivalent case and that is $$k,k,q_2,q_1$$)
• You count three kings one queen three times, so need to subtract them twice. Sep 20 '20 at 4:45
• Yes thats right. Sep 20 '20 at 4:59
You're overcounting in the second method.
Label the four kings $$K_1, K_2, K_3, K_4$$. In the second method, you will count the case in which you pick an arbitrary queen $$Q$$ followed by $$K_i$$ followed by $$K_j$$ separate from the case in which you pick a queen $$Q$$ followed by $$K_j$$ followed by $$K_i$$. However, these two outcomes should be treated as the same outcome. | 2021-09-17T15:23:25 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3832935/in-how-many-ways-can-4-cards-be-drawn-randomly-from-a-pack-of-52-cards-such-that",
"openwebmath_score": 0.6315335035324097,
"openwebmath_perplexity": 247.09401755711824,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9736446463891304,
"lm_q2_score": 0.8633916152464017,
"lm_q1q2_score": 0.840636623921923
} |
https://math.stackexchange.com/questions/3330549/how-to-solve-this-equation-by-hand-4-68-4-50-cos-alpha-1-23-alpha-0 | # How to solve this equation by hand? $4.68-4.50\cos\alpha-1.23\alpha=0$
I'm trying to solve the next equation
$$4.68-4.50\cos\alpha-1.23\alpha=0$$
But when using a calculator it gives $$\alpha$$ as $$0.226$$ but it s supposed to be $$4.483$$.
As far as I know it has no closed form solution so its necessary a numeric method, so which is the right number, aside isn't a way to at least manually approximate the solution?
• There is indeed no closed form, but have you tried graphing the equation in DESMOS? – imranfat Aug 22 '19 at 1:52
• This is a little baffling. The possible zero(es) near $\alpha = 0$ is/are in the interval $[0.2768, 0.2769]$, which is not close to your $0.226$ and the straightforward zero at $4.5166\dots$ is not close to your $4.483$. Are you sure these numbers and the coefficients in your equation are copied correctly? – Eric Towers Aug 22 '19 at 1:58
• I imagine the error between $4.483$ and the true zero might be a result of the intended approximation method, whatever that is - it's close enough to feel that way to me in any event. Like imagine if the exact answer involved an infinite summation, and the exercise intended you to approximate this by taking some finite number of terms in said sum. This is purely a guess though. – Eevee Trainer Aug 22 '19 at 2:00
• @EricTowers A closer look shows there is no zero in the interval $[0.2768, 0.2769]$: there is a local minimum at $a = \arcsin(4.50/1.23) \approx 0.2768566278$, where the function value is approximately $0.0108291488$. – Robert Israel Aug 22 '19 at 2:01
• @RobertIsrael : Agreed. But the discrepancy between the reported location and the observed location leads me to wonder if there is a transcription error. (I could certainly imagine a calculator's rootfinder getting stuck in the local minimum, but then how does it miss that minimum by so much?) – Eric Towers Aug 22 '19 at 2:02
I don't know how the calculator got its value, $$0.226$$, or where the value $$4.483$$ comes from. Plotting $$f(\alpha) = 4.68 - 4.50 \cos \alpha - 1.23 \alpha$$ gives
We expect this graph to be a cosine with midline given by $$4.68 - 1.23 \alpha$$, so as soon as a local minimum is above zero, we need not proceed further in that direction (to the left), and as soon as a local maximum is below zero, we need not proceed further in that direction (to the right). The graph above indicates those features, so all possible zeroes are in the interval shown, $$\alpha \in [-6,10]$$.
Zooming in on the potential zero(es) near $$\alpha = 0$$, we see that there is a local minimum above the $$\alpha$$ axis, so there is/are no actual zero(es) there.
(This plot is an example of why it is useful to include "round" numbers on axes when your interval of interest is very close to that round number. If we only zoom in to show the minimum, we have to actually look at the tick labels to determine whether there is a zero. For instance, whether the above plot shows a zero is immediately clear. Whether the following one does takes a little more inspection.
)
Going back to the overview plot, there should be a zero of $$f$$ in the interval $$[4,5]$$. So let's see if we can find it. (Spoiler: It's at $$\alpha = 4.516\,602\,526\,340\,883\,204\,3\dots$$.)
Let's separate the polynomial (actually, linear in this problem) and non-polynomial parts and plot them separately. We want $$4.68 - 1.23 \alpha = 4.50 \cos \alpha$$ Since $$-1 \leq \cos \alpha \leq 1$$, the right-hand side is in the interval $$[-4.50, 4.50]$$. The left-and side is a line and we can solve the the interval of $$\alpha$$s where the height of the line is in $$[-4.50,4.50]$$. \begin{align*} -4.50 &\leq 4.68 - 1.23 \alpha \\ -9.18 &\leq -1.23 \alpha \\ 7.463\,414\,634\,146\,341\,463\,4\dots =\frac{9.18}{1.23} &\geq \alpha \\ \end{align*} and \begin{align*} 4.68 - 1.23 \alpha &\leq 4.50 \\ -1.23 \alpha &\leq -0.18 \\ \alpha &\geq \frac{0.18}{1.23} = 0.146\,341\,463\,414\,634\,146\,34\dots \end{align*} (Notice that this possible interval for a zero of $$f$$ is smaller than the one we got above from looking at the graph.) Plotting the two sides over this interval, we have
We know from the above that there is/are no zero(es) of $$f$$ (intersections of the line from the left-hand side and the cosine from the right-hand side of the above) near $$\alpha = 0$$, but there is clearly one near $$\alpha = 4.5$$.
## Binary Search
So we can proceed by binary search. We know that the $$f$$ is positive on one side of the zero and negative on the other, so we can cut the interval containing the zero in half for each evaluation of $$f$$. You seem to be using thousandths as precision, so we need only evaluate about ten times. We indicate the calculation of $$f$$ at the midpoint of the current interval containing the zero, then write down the new (smaller) interval which we may infer contains the zero. \begin{align*} f(4) &= 2.7013\dots \\ f(5) &= -2.7464\dots & (4,5) \\ f(4.5) &= 0.09358\dots & (4.5,5) \\ f(4.75) &= -1.3317\dots & (4.5, 4.75) \\ f(4.6) &= -0.4733\dots & (4.5, 4.6) \\ f(4.55) &= -0.1889\dots & (4.5, 4.55) \\ f(4.52) &= -0.01918\dots & (4.5, 4.52) \\ f(4.51) &= 0.03724\dots & (4.51, 4.52) \\ f(4.515) &= 0.009043\dots & (4.515, 4.52) \\ f(4.517) &= -0.002243\dots & (4.515, 4.517) \\ f(4.516) &= 0.003400\dots & (4.516, 4.517) \end{align*} The decimal expansion of every point in that last interval begins "$$4.516$$", agreeing to three decimal places, so the zero is about $$\alpha = 4.516$$.
(Note that all we really care about is the sign of these values of $$f$$. If we know the sign, we know which endpoint to replace with the trial $$\alpha$$ in that step, so we do not have to evaluate these very carefully (especially in the first few steps). Also, we didn't use the exact midpoints. If we had, the interval containing the zero would be shorter than the interval we found for the same number of evaluations of $$f$$. It would also need a lot of calculator button pushing since the sequence of midpoints is $$4.5$$, $$4.75$$, $$4.625$$, $$4.5625$$, $$4.53125$$, $$4.515625$$, $$4.5234375$$, $$4.51953125$$, $$4.517578125$$, $$4.5166015625$$, $$4.51708984375$$, and $$4.516845703125$$ to get both endpoints to agree to $$3$$ decimals.)
## Taylor Series
There are other methods we could try. We could replace the cosine with leading segments of its Taylor expansion. Then we are seeking roots of the resulting polynomial. This works if we can center the series close to the root. The Taylor series of cosine centered at $$0$$ is $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots + \frac{x^{2k}}{(2k)!} + \cdots \text{,}$$ but we want to find a zero near $$4.5$$, so we should center close to $$4.5$$. Picking standard angles that are around $$4.5 \leq 1.5 \cdot \pi$$, $$\frac{4}{3}\pi = 4.188\dots$$ and $$\frac{3}{2}\pi = 4.712\dots$$, so $$\frac{3}{2}\pi$$ is closer to $$4.5$$. The Taylor series of cosine centered at $$\frac{3}{2} \pi$$ is $$\cos x = \left( x - \frac{3}{2}\pi \right) - \frac{1}{3!}\left( x - \frac{3}{2}\pi \right)^3 + \frac{1}{5!}\left( x - \frac{3}{2}\pi \right)^5 + \cdots + \frac{1}{(2k+1)!}\left( x - \frac{3}{2}\pi \right)^{2k+1} + \cdots \text{.}$$ We want to use low degrees since polynomials of high degree can be hard to factor. Let's go up to degree $$5$$ (by $$2$$s since this series for cosine only has odd degree terms). We list the degree of the initial part of the cosine series we are keeping in that row, the resulting approximation for $$f$$, and the real roots of that polynomial. \begin{align*} 1& :& -5.73 \alpha + 25.885\dots &= 0 :& \{& 4.517\dots \} \\ 3& :& 0.75 \alpha ^3-10.602\dots \alpha ^2+44.234\dots \alpha -52.598\dots &= 0 :& \{& 2.051\dots, 4.516\dots, 7.569\dots \} \\ 5& :& -0.0375 \alpha ^5+0.883\dots \alpha ^4-7.577\dots \alpha ^3+28.639\dots \alpha ^2-48.227\dots \alpha +34.545 &= 0 :& \{& 4.516\dots\} \end{align*} We get our zero to three decimals almost immediately.
## Newton's method
The last method I'll show is Newton's Method. We have prior information that $$f$$ has a zero near $$\alpha = 4$$, so we use a linear approximation to $$f$$ at that point, find that approximation's $$\alpha$$-intercept, and report that as an improved location of the zero. We need to know that $$\alpha - \frac{f(\alpha)}{f'(\alpha)} = \frac{150 \alpha \sin \alpha + 150 \cos \alpha - 156}{150 \sin \alpha - 41} \text{.}$$ \begin{align*} 4& :& 4&.5827\dots \\ 4&.5827 :& 4&.5168\dots \\ 4&.5168 :& 4&.5166\dots \\ 4&.5166 :& 4&.5166\dots \end{align*} Since we have found an approximate fixed point of the iteration for Newton's method, and we have graphical evidence that there is a simple zero of $$f$$ near this last $$\alpha$$, we have found that $$\alpha = 4.516$$ is an approximate zero of $$f$$.
• Whoah!!!, And I thinked this has nothing left to do about! – riccs_0x Aug 23 '19 at 17:25
• And the binary search its new for me! – riccs_0x Aug 23 '19 at 18:50
• I think this answer must be locked/protected its highly informative – riccs_0x Aug 23 '19 at 19:03
• Binary search in the numerical context is commonly known as bisection method. Conversely, sometimes the binary search in ordered lists is also called bisection. – Lutz Lehmann Aug 24 '19 at 8:32
If you know an approximate of the solution, you can use the fixed-point method to find a less approximate (!) solution. Define $$\alpha_{n+1}=\sqrt[k+1]{{4.68-4.50\cos \alpha_{n}\over 1.23}\cdot \alpha_n^{k}}$$and find the limit. I tried $$k=3$$ and obtained $$\approx 4.5166$$. All you have to do is that using a scientific calculator press an approximate of solution e.g. $$4$$ and then press $$Ans$$ button. Then input the expression $$\sqrt[k+1]{{4.68-4.50\cos Ans\over 1.23}\cdot Ans^{k}}$$for some $$k\in \Bbb R$$ and press $$"="$$ repetitively to obtain a rather precise solution.
P.S.
Such equations are generally impossible to be solved by hand. | 2021-04-21T18:06:28 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3330549/how-to-solve-this-equation-by-hand-4-68-4-50-cos-alpha-1-23-alpha-0",
"openwebmath_score": 0.9997368454933167,
"openwebmath_perplexity": 499.5031089071328,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.973644652507129,
"lm_q2_score": 0.863391602943619,
"lm_q1q2_score": 0.840636617225613
} |
https://math.stackexchange.com/questions/2924137/two-vectors-with-the-same-normal-surface-projection-and-the-same-normal-surface | # Two vectors with the same normal surface projection and the same normal surface cross product, are equal?
I have two vectors, $$\mathbf a$$ and $$\mathbf b$$, that fulfill the following conditions:
$$(\mathbf a-\mathbf b)\cdot \mathbf n= 0$$
$$(\mathbf a-\mathbf b)\times \mathbf n=\mathbf 0$$
being $$\mathbf n$$ a unit surface normal.
My question is, is $$\mathbf a = \mathbf b$$ ?
I have confirmed this by doing the cross product in a reference frame for which its first direction is coincident with the surface normal. Since both the dot and cross products are invariant under reference frame transformations, the results should be confirmed for all coordinate systems. Is this reasoning ok?
• A slick one-liner proof: $$\mathbf{a}=(\mathbf{a}\cdot\mathbf{n})\mathbf{n}-(\mathbf{a}\times\mathbf{n})\times\mathbf{n}=\dots=\mathbf{b}$$ – user10354138 Sep 20 at 15:49
• This is certainly excellent! – nodarkside Sep 20 at 16:10
Yes it is correct, indeed we have that
• $(\vec a-\vec b)\times \vec n=\vec 0 \implies \vec a-\vec b$ is a multiple of $\vec n$ that is $\vec a-\vec b=k\vec n$
• $(\vec a-\vec b)\cdot \vec n=0 \implies \vec a-\vec b$ is orthogonal to $\vec n$ that is $k\vec n\cdot \vec n=0 \implies k=0$
therefore
$$\vec a-\vec b=\vec 0 \implies \vec a=\vec b$$
• Thanks very much @gimusi ! – nodarkside Sep 20 at 15:24
• @nodarkside You are welcome! Bye – gimusi Sep 20 at 15:25
You could also do this using the fact that $$\vec{v} \cdot \vec{w} = \vert\vec{v}\rvert \lvert \vec{w} \rvert \cos\theta$$ and $$\lvert\vec{v} \times \vec{w}\rvert = \vert\vec{v}\rvert \lvert \vec{w} \rvert \sin\theta$$. In your case this leads to \begin{align*} (\vec{a} - \vec{b})\ \cdot \ \vec{n} = \lvert\vec{a} - \vec{b}\rvert \lvert\vec{n}\rvert \cos\theta &= 0 \\ \lvert(\vec{a} - \vec{b})\ \times \ \vec{n}\rvert = \lvert\vec{a} - \vec{b}\rvert \lvert\vec{n}\rvert \sin\theta &= 0 \end{align*} From here, it is safe to cancel the $$\lvert\vec{n}\rvert$$, as it is unit normal. We cannot safely cancel out the $$\cos\theta$$ or $$\sin\theta$$ because these may be $$0$$, but notice that if $$\sin\theta$$ = 0, then $$\cos\theta \neq 0$$ and vice versa. Thus, we are left to the conclusion that $$\lvert \vec{a} - \vec{b} \rvert = 0$$, so $$\vec{a} = \vec{b}$$.
• Thanks very much @Alerra! This proof is also very useful. – nodarkside Sep 20 at 16:02
• @Alerra you might want to correct your writing: $|\vec v \times \vec w|,$ similarly bellow. – user376343 Sep 20 at 21:07
• Just fixed it. Although I did it from my phone so it shows up as Community edit I guess? – Alerra Sep 20 at 21:15
One definition of the cross product $$v , w$$ is the unique element $$v \times w$$ that satisfies $$\langle x , v \times w \rangle = \det \begin{bmatrix} x & v & w \end{bmatrix}$$.
Let $$d=a-b$$. If $$d \times n = 0$$, then $$d,n$$ must lie on the same line (otherwise we could find an $$x$$ such that the above determinant is non zero). Hence we can write $$d = \lambda n$$ for some $$\lambda$$.
Then $$\langle d,n \rangle = \lambda = 0$$. Hence $$a=b$$. | 2018-12-13T19:41:08 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2924137/two-vectors-with-the-same-normal-surface-projection-and-the-same-normal-surface",
"openwebmath_score": 0.9958087801933289,
"openwebmath_perplexity": 434.60856353542977,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9736446433301306,
"lm_q2_score": 0.8633916082162403,
"lm_q1q2_score": 0.8406366144359291
} |
https://en.khanacademy.org/math/get-ready-for-ap-calc/xa350bf684c056c5c:get-ready-for-applications-of-integration/xa350bf684c056c5c:extraneous-solutions/a/analyzing-extraneous-solutions-of-square-root-equations | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## Get ready for AP® Calculus
### Course: Get ready for AP® Calculus>Unit 7
Lesson 3: Extraneous solutions
# Extraneous solutions of radical equations
Practice some problems before going into the exercise.
## Introduction
### Practice question 1
Caleb is solving the following equation for x.
x, equals, square root of, x, plus, 2, end square root, plus, 7
His first few steps are given below.
\begin{aligned}x-7&=\sqrt{x+2}\\ \\ (x-7)^2&=(\sqrt{x+2})^2\\ \\ x^2-14x+49&=x+2 \end{aligned}
Is it necessary for Caleb to check his answers for extraneous solutions?
### Practice question 2
Elena is solving the following equation for x.
cube root of, 3, x, minus, 5, end cube root, plus, 2, equals, 7
Her first few steps are given below.
\begin{aligned}\sqrt[3]{3x-5}&=5\\ \\ \left(\sqrt[3]{3x-5}\right)^3&=(5)^3\\ \\ 3x-5&=125 \end{aligned}
Is it necessary for Elena to check her answers for extraneous solutions?
### Practice question 3
Addison solves the equation below by first squaring both sides of the equation.
2, x, minus, 1, equals, square root of, 8, minus, x, end square root
What extraneous solution does Addison obtain?
x, equals
### Practice question 4
Which value for the constant d makes x, equals, minus, 1 an extraneous solution in the following equation?
square root of, 8, minus, x, end square root, equals, 2, x, plus, d
d, equals
## Want to join the conversation?
• is 0 an extraneous solution
• It doesn't have to be. For example, x^2+4x=0 has solutions 0 and -4 (try them)
• So take the following as an example.
sqrt(4x + 41) = x+5
Originally, x must be greater than -41/4 to be valid. When we solve this equation algebraically, we get the following two solutions: X = -8 and X = 2.
When you plug X = -8 back into the equation, you end up with sqrt(9) = -3.
This is extraneous because we are supposed to use the principle root.
I have three questions.
#1: Why do we have to use the principle root? It seems logical that the sqrt(9) should be equal to both +3 and -3.
#2: Why is this extraneous solution still within the original domain of the function, as it is > -41/4.
#3: What happened mathematically to produce this extraneous solution. If you graph the two equations separately, they only intersect at x = 2, therefore x = 8 is not a solution. But why is this? What happened to produce this extra answer?
• 1. In your example, we use the principal square root because the original problem statement says so. It says √(4x + 41), and not -√(4x + 41). By definition, the radical notation without a minus or ± sign in front of it means “the principal root”, which is always positive. The solutions to Power Equations and the solutions to Radical Equations are different things. For example, x^2 = 4 has two solutions: x = ±2, while x = √4 has only one solution: x = 2, and x = -√4 also has one solution: x = -2. We define the square root as a function, so it must have only one output for each input. If we define the function y = √x as having two solutions, then it is no longer a function.
2. The solution x = -8 is extraneous to the original equation √(4x + 41) = x + 5. However, it is the solution to the equation -√(4x + 41) = x + 5. The expression under the radical is same in both equations, so in terms of keeping the radicand non-negative, the value -8 is OK. If we take the function y = √(4x + 41), then -8 would be a valid input value for x. However, for the equation √(4x + 41) = x + 5, the value x = -8 is not a solution, because it leads to an invalid statement 3 = -3, which is not true.
3. As mentioned above, x = -8 is the solution to the equation -√(4x + 41) = x + 5. So, if we graph -√(4x + 41) instead of √(4x + 41), it will intersect with x + 5 at x = -8.
• In the last situation, why did they plug -1 in the equation? was it a random number?
• This got me at first too. In the original question, it specifically asks you to use x=-1
• I still don't understand why I should care about extraneous solution. It's outside the domain, not a solution, a wrong answer. Or is there any use of finding it?
• Extraneous solutions are not necessarily outside the domain. But they can appear as extra solutions when we square both sides of an equation, because when we square an equation, we would get the same result whether the original equation was positive or negative. So one solution corresponds to the positive equation, and the other to the negative equation. But both of them will fall out of the algebra. We need to be able to tell which solution is extraneous and which works.
• square root 9 should be either 3 or -3.......why should -3 be extraneous then?
• I am having a really hard time with this unit. I have watched all the videos several times and I am still very confused. I would really appreciate some help.
• The key idea for solving square root equations is to isolate a radical on one side and square both sides. If there's still a radical in the equation, then this process would need to be performed a second time. (By the way, don't forget to include the middle term when squaring a binomial. Many students forget to do this.)
After you have solved the resulting linear or quadratic equation for x, remember that you're not finished yet! Because every positive number has a positive and a negative square root, but the radical symbol denotes only the positive (principal) square root, the act of squaring both sides can create invalid (extraneous) solutions! So plug in your solutions to the original equation to determine which solutions work and which solutions must be discarded.
Have a blessed, wonderful day!
• Hi I don't understand why √9 would not equal -3 when it equals 3. In practice question 3, why wouldn't x=-1 be a correct solution??
• I think they're looking for the principal sqrt of 9. | 2023-03-26T01:45:56 | {
"domain": "khanacademy.org",
"url": "https://en.khanacademy.org/math/get-ready-for-ap-calc/xa350bf684c056c5c:get-ready-for-applications-of-integration/xa350bf684c056c5c:extraneous-solutions/a/analyzing-extraneous-solutions-of-square-root-equations",
"openwebmath_score": 0.8496931195259094,
"openwebmath_perplexity": 589.92899920974,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9736446456243804,
"lm_q2_score": 0.8633916047011594,
"lm_q1q2_score": 0.8406366129943255
} |
https://math.stackexchange.com/questions/4256774/why-multiplication-of-142857-with-2-3-4-5-6-gives-the-same-digits-shifted | Why multiplication of 142857 with 2,3,4,5,6 gives the same digits shifted?
I was reading about a not so practical way to determine the divisibility of a number by $$7$$.
At some point the following number is mentioned: $$142857$$ (which is the result of $$\frac{999999}{7}$$) and apparently this number as I have verified if multiplied by $$2,3,4,5, 6$$ it gives the same digits in different order e.g. $$142857 \cdot 3 = 428571$$
Why does this number have this property? I see that the numbers $$2,3,4,5,6$$ are all remainders if we divide $$100, 10, 10000, 100000, 1000$$ by $$7$$ respectively but I am not sure if there is any correlation with the property.
I'd like to understand the intuition behind this "trick"
Note: I have found a similar post but I don't see any explanation on this
• $$\frac{142857}{999999}=\frac17.$$ So $$\frac17=0.\overline{142857}.$$ Then use what you noticed. Sep 21 at 21:54
• More generally if $10$ is a generator modulo prime $p,$ you will get something like this. For example, $p=17$ gives the same result for $058235294117647,$ multiplying by $1,\dots,16.$ It is not just a different order, but a rotated order - some number of digits are cut off the end of $142857$ and placed in the front. Sep 21 at 22:04
• – lhf
Sep 21 at 22:12
• Does this answer your question? Why $142857$ when multiplied by $1$ to $6$ gives the same digits? Sep 21 at 23:57
• @JeanMarie Same question yes , but the answers are so useless. I think we can get a better dupe target. Sep 22 at 8:13
Well, one can note that $$142857$$ is a bit of a special number in that $$\frac{1}{7}=0.\overline{142857}$$, which means that, by long division, one obtains $$\overline{142857}=142857142857142857...$$ ad infinitum via the following algorithm:
Start with $$1$$.
$$1=\color{green}{0}\times7+\color{red}{1}$$ so $$\frac{1}{7}=0.\text{something}$$
To get this something, do:
$$\color{red}{1}\times10 = 10 = \color{green}{1}\times7+\color{red}{3}$$ (*)
$$\color{red}{3}\times10 = 30 = \color{green}{4}\times7+\color{red}{2}$$ (**)
$$\color{red}{2}\times10 = 20 = \color{green}{2}\times7+\color{red}{6}$$
$$\color{red}{6}\times10 = 60 = \color{green}{8}\times7+\color{red}{4}$$
$$\color{red}{4}\times10 = 40 = \color{green}{5}\times7+\color{red}{5}$$
$$\color{red}{5}\times10 = 50 = \color{green}{7}\times7+\color{red}{1}$$
Now, the remainder is $$\color{red}{1}$$, which is what we started with at equation (*), so we now have a loop which yields $$\frac{1}{7}=0.\color{green}{\overline{142857}}$$. In group-theoretic parlance, the fact that the loop closes after $$6$$ iterations means that $$10$$ has order $$6$$ in $$(\mathbb{Z}/7\mathbb{Z})^*$$, i.e. $$10^6\equiv1 \pmod{7}$$ and $$10^k\neq1\pmod7$$ for $$1 \leq k \leq 5$$.
To understand why the digits are shifted when e.g. multiplying by $$3$$, just note that multiplying $$\frac{1}{7}$$ by $$3$$ means considering $$\frac{3}{7}$$ instead of $$\frac{1}{7}$$, and so the algorithm would start at equation (**) instead of (*), thus yielding $$\color{green}{\overline{428571}}$$ instead of $$\color{green}{\overline{142857}}$$.
So $$\frac{3}{7}=\frac{3\times 142857}{999999} = \frac{428571}{999999} \Rightarrow 3\times142857 = 428571$$
This explains the shift.
• I don't follow the logic from the To get this something, do and on. What exactly does the remainder with 10,30,20,60,40,50 show and what is the idea behind doing this to determine the decimal part of the result of $\frac{1}{7}$?
– Jim
Sep 22 at 16:36
Good question!
Let's see if we can noodle it out.
$$\frac {1,000,000 -1}7 = 142857$$.
And $$2\times \frac {1,000,000 -1}7 =$$ what... now you point out that $$100 = 7k + 2$$ where $$7 =14$$ so $$2 = 100-7k$$
$$2\times \frac {100,000 - 1}7= (100-7k)\frac {1,000,000-1}7 =$$
$$100\frac {1,000,000 - 1}7 - k(1,000,000 -1) =$$
$$14,285,700 - \overline{k000000} +k$$.
Hmm.... that we know $$k = 14$$ seems to be a really nice coincidence.
$$14,285,700 - 14,000,000 + 14 = 285714$$ and of course the numbers are reversed.
If we can assume that if $$10^{m_j} = (\text{first }m_j\text{ digits of }142857)\times 7 + j$$ for $$j=2,3,4,5,6$$ we'd be done.
After all we'd then have
$$j \times 142,857=$$
$$(10^{m_j} - (\text{first }m_j\text{ digits of }142857)7)\frac {1,000,000-1}7 =$$
$$10^{m-j}142,857 - (\text{first }m_j\text{ digits of }142857)7)1,000,000 + (\text{first }m_j\text{ digits of }142857)7)=$$
$$[\overline{(\text{first }m_j\text{ digits of }142857)(\text{last }6-m_j\text{ digits of }142857)\underbrace{0...0}_{m-j}}]- [\overline{(\text{first }m_j\text{ digits of }142857)000000}]+(\text{first }m_j\text{ digits of }142857)=$$
$$\overline{(\text{last }6-m_j\text{ digits of }142857)(\text{first }m_j\text{ digits of }142857)}$$
......
So how can we show that for all $$m_j \le 6$$ that $$10^{m_j} = (\text{first }m_j\text{ digits of }142857)\times 7 + j$$
Well.....
we know that $$\frac {1,000,000 -1}7 = 142,857$$.
This means $$\frac 17= \frac 1{1000000}\frac {1000000}7 = \frac 1{1000000}[142,857 + \frac 17]$$. That means $$\frac 17 = 0.142857\overline{142,857}$$.
Okay....
So if we multiply $$\frac 17$$ times $$10^{m_j}$$ we get $$10^{m_j}\frac 17 = (\text{first }m_j\text{ digits of }142857) + (\text{some decimal part less than } 1)$$.
So $$10^{m_j} = 7 \times (\text{first }m_j\text{ digits of }142857) + 7\times(\text{some decimal part less than } 1)=$$
$$7 \times (\text{first }m_j\text{ digits of }142857) + (\text{some value less than } 7)$$
but as $$10^{m_j}$$ is a natural number (not divisible by $$7$$) we have
$$10^{m_j} = 7 \times (\text{first }m_j\text{ digits of }142857) +j$$ for some $$1 \le j \le 7$$
And that's it.
This will actually be true of any $$n$$ so that $$\gcd(n,10) = 1$$
Take, say $$\frac 1{17}= 0.\overline{0588235294117647}$$
We should get $$0588235294117647 \times 2..... 16$$ should be the same digits shifted. Try it.
• $\frac{999,999}{7}=142857$. You used $1,000,000-1$ instead. But I don't understand how from $2\cdot 999,999 = 2\times \frac {1,000,000 -1}7 = 1,999,998$ we go to $285714$ via replacing $2$ with the formula for $100$.
– Jim
Sep 22 at 19:55
• If $100 = 7\times 14 + 2$ then $2 = 100 - 7\times 14$. So $2\times \frac {999,999}7 = (100-7\times 14)\times \frac {999999}7 = (100 - 7\times 14)\times \frac {10^6 -1}7$ Now just expand $(100-7\times 14)\times \frac{10^6 - 1}7 = 100\times \frac {10^6-1}7 - 7\times 14\times \frac {10^6-1}7= 100\times \frac {10^6-1}7 +-14\times (10^6 -1) = 100\times \frac {10^6-1}7 - 14\times 10^6 + 14$. ... Now $100\times {10^6-1}7$ will add two zeros to $142857$ to get $14285700$. And $-14\times 10^6=-14000000$ removes the leading $14$ to get $285700$ and $+14$ adds it back to the other side: $285714$. Sep 22 at 20:06
• $1999998 = 2\cdot 999,999 \ne 2\cdot \frac {999999}7=285714$. Why did you do anything with $2\times 999999$ that was not part of the question and never part of my solution? Sep 22 at 20:08
• It $2 = 100 - 7\times 14$ (which it does) then $$2 \times ANYTHING = (100 -7\times 14)ANYTHING = 100\times ANYTHING - 7\times 14\times ANYTHING$$. And if $$ANYTHING = 142857 = \frac {1000000 - 1}7$$ then $$2\times 142857 = 100\times 142857 - 7\times 14\times\frac {1000000 -1}7 = 100\times 142857 - 14\times 1000000 + 14$$. $$=\color{green}{14}\color{red}{2857}\color{purple}{00} - \color{green}{14}\color{red}{0000}\color{purple}{00} + \color{purple}{14}= \color{red}{2857}\color{purple}{00}$$That's all. Sep 22 at 20:11
• Yes I got confused, sorry about that
– Jim
Sep 23 at 20:50 | 2021-11-27T02:27:44 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/4256774/why-multiplication-of-142857-with-2-3-4-5-6-gives-the-same-digits-shifted",
"openwebmath_score": 0.8414235711097717,
"openwebmath_perplexity": 196.69623841647777,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9890130596362787,
"lm_q2_score": 0.8499711775577736,
"lm_q1q2_score": 0.8406325949190644
} |
https://mathforums.com/threads/alternating-sums-of-powers-of-2.18518/ | # Alternating Sums of powers of 2
#### Rutzer
Hello,
Here's an interesting problem that I have solved, but would like someone else's input simply to see if people come up with the same solution or possibly find a more elegant one. I hope anyone who tries this has fun with the problem!
Which numbers can be written by selecting a subset of the powers of two and alternating them (positive-negative,..., negative-positive,...) and list them in increasing order to form a sum. *A sum cannot be made by simply one power of two, namely n=n. SO, can ALL positive integers be written this way? how many ways can a number 'n' be written as an alternating sum of powers of two?
Rutzer
Last edited by a moderator:
#### CRGreathouse
Forum Staff
I think that all numbers can be written in this form, yes. I think there's a nice bijection with the binary digits of the number by taking the consecutive 1 bits and representing them as the difference of two powers of two.
#### Rutzer
Excellent, it seems we think alike! I took the binary approach as well. but I also saw the possibility for an induction... thanks for your response.
rutzer
Last edited by a moderator:
#### xingzheli
I agree that all numbers seem to be expressible in this way and this can be shown with binary. I also think there are 2 ways to express all non-powers of 2, 1 ways to express powers of 2.
Last edited by a moderator: | 2020-04-03T21:00:06 | {
"domain": "mathforums.com",
"url": "https://mathforums.com/threads/alternating-sums-of-powers-of-2.18518/",
"openwebmath_score": 0.910398006439209,
"openwebmath_perplexity": 400.34128965141906,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.989013056721729,
"lm_q2_score": 0.8499711775577735,
"lm_q1q2_score": 0.840632592441781
} |
http://mathhelpforum.com/calculus/219579-sum-complex-series-print.html | # sum of the complex series
• Jun 4th 2013, 12:39 PM
alteraus
sum of the complex series
Hello everybody, I am wondering how can I find the sum of the series $\sum_{n=1}^{\infty}\frac{ni^n2^n}{(z+i)^{n+1}}$ ? I found out that the series converges for |z+i|>2, but I am not able to find the sum of the series , thank you very much
• Jun 4th 2013, 01:25 PM
HallsofIvy
Re: sum of the complex series
If it weren't for that "n" at the beginning, that would be a geometric series. But that n makes me think of a derivative. If we set $f(z)= -(2i)^n (z+ i)^{-n}$ then $f'(z)= n(2i)^n (z+ i)^{-n- 1}= \frac{n 2^n i^n}{(z+ i)^{n+1}}$. Does that give you any ideas?
• Jun 5th 2013, 01:43 AM
alteraus
Re: sum of the complex series
So you're saying that I could at first find the sum
$f(z)=\sum_{n=1}^{\infty}\Big(\frac{2i}{z+i}\Big)^n$ and I get the function f(z)
and just count the derivative of f'(z) ?
this series looks not as difficult as the first one, but I still can't find the sum ..
would you please give me another hint?
• Jun 5th 2013, 02:34 AM
Prove It
Re: sum of the complex series
Quote:
Originally Posted by alteraus
So you're saying that I could at first find the sum
$f(z)=\sum_{n=1}^{\infty}\Big(\frac{2i}{z+i}\Big)^n$ and I get the function f(z)
and just count the derivative of f'(z) ?
this series looks not as difficult as the first one, but I still can't find the sum ..
would you please give me another hint?
It's a geometric series...
• Jun 5th 2013, 03:05 AM
alteraus
Re: sum of the complex series
okey,
can you please explain how does the geometric series work for complex numbers?
I dealt only with real geometric series and the sum is then easy
is there any sum formula for complex terms?
in my series number $z \in \mathbb{C}$
thank you
• Jun 9th 2013, 08:24 AM
alteraus
Re: sum of the complex series
no more suggestions?
• Jun 9th 2013, 08:49 AM
Prove It
Re: sum of the complex series
It works exactly the same way as a real geometric series.
• Jun 9th 2013, 09:04 AM
alteraus
Re: sum of the complex series
thank you very much, I found the result which looks to be right
• Jun 9th 2013, 08:31 PM
Soroban
Re: sum of the complex series
Hello, alteraus!
I think I found the sum.
But check my work . . . please!
Quote:
$\sum_{n=1}^{\infty}\frac{n(2i)^n}{(z+i)^{n+1}}$
I found out that the series converges for $|z+i|>2$,
but I am not able to find the sum of the series.
$\begin{array}{cccccc} \text{We are given:} & S &=& \dfrac{1\cdot(2i)}{(z+i)^2} + \dfrac{2\cdot(2i)^2}{(z+1)^3} + \dfrac{3\cdot (2i)^3}{(z+i)^4} + \cdots \\ \text{Multiply by }\frac{2i}{z+i}\!: & \dfrac{2i}{z+i}S &=& \qquad\qquad\;\; \dfrac{1\cdot (2i)^2}{(z+i)^3} + \dfrac{2\cdot (2i)^3}{(z+i)^4} + \cdots \end{array}$
$\text{Subtract: }\:\left(1 - \frac{2i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2} + \frac{(2i)^2}{(z+i)^3} + \frac{(2i)^3}{(z+i)^4} + \cdots$
. . . . . . . . . . $\left(\frac{z-i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2}\underbrace{\left[1 + \frac{2i}{z+i} + \frac{(2i)^2}{(z+i)^2} + \cdots \right]}_{\text{geometric series}}$
The geometric series has first term $a = 1$ and common ratio $r = \frac{2i}{z+i}$
. . Its sum is: . $\frac{1}{1-\frac{2i}{z+i}} \:=\:\frac{z+i}{z-i}$
We have: . $\left(\frac{z-i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2}\cdot\frac{z+i}{z-i}$
. . . . . . . . . . . . . . $S \;=\;\frac{2i}{(z+i)^2}\cdot\frac{z+i}{z-i}\cdot\frac{z+i}{z-i}$
. . . . . . . . . . . . . . $S \;=\;\frac{2i}{(z-i)^2}$
• Jun 10th 2013, 01:53 AM
alteraus
Re: sum of the complex series
Thank you very much Soroban,
nice way how to avoid the term-by-term differentiation,
by which I got the same sum of the series, so it looks to be right :) | 2017-12-17T17:55:48 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/219579-sum-complex-series-print.html",
"openwebmath_score": 0.9206525087356567,
"openwebmath_perplexity": 589.9391870214656,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9890130586647623,
"lm_q2_score": 0.8499711718571775,
"lm_q1q2_score": 0.8406325884553395
} |
https://cs.stackexchange.com/questions/14376/count-elements-of-a-sorted-matrix-that-fall-into-a-given-interval/14377 | # Count elements of a sorted matrix that fall into a given interval
I have a $n\times n$ matrix called $M$, and two integers $k_\min$ and $k_\max$. Each row and each column of M is sorted in the increasing order.
I would like to know if there is way I can count the number of its elements which are inside $[k_\min, k_\max]$, using a $O(n)$ algorithm.
• What have you tried? Which algorithms do you know? Why is this problem interesting to you? – Raphael Sep 17 '13 at 12:08
First, notice that you only need to know how to count the number of elements bigger than $k_{min}$, because then, by a very simple transformation, you can count the number of element smaller than $k_{max}$, and compute the desired result : since the set $A$ of element bigger than $k_{min}$ is of the form $[a;max]$, and the set $B$ of element smaller than $k_{max}$ is $[min;b]$, you have $|A \cap B| = |A| + |B| - n^2$
Now, find the last element smaller than $k_{min}$ on the last row, at column $c_0$. On this last row, exactly $n - c_0$ elements are bigger than $k_{min}$. On the row above, the last element smaller than $k_{min}$ has to be right to $c_0$, because otherwise, since $M$ is sorted on row and column, you would have a contradiction. Hence, in order to find the last element smaller than $k_{min}$ on this row, you can start at column $c_0$ and go right to find $c_1$. Then again, the number of elements bigger than $k_{min}$ on this row is $n - c_1$.
Repeating this process, you then sum the number of elements on each row to compute the result, and you'll scan only at most $2n$ elements since the path followed by the algorithm is a path from the point $(0,n-1)$ to the point $(n-1,0)$, going only up and left, and these kind of path have length $2n$. This is hence indeed an $\mathcal{O}(n)$ algorithm.
• I don't get your first paragraph. Knowing $|A|$ and $|B|$, nothing is known about $|A \cap B|$ without further knowledge. Your algorithmic idea is nice, though, and can be adapted to solve the original problem in one pass by moving two points upwards in the matrix and summing over the difference in their positions. – Raphael Sep 17 '13 at 12:11
• IMO. Your answer seems to be correct, but, I think you missed to mention two things : 1. Should also find biggest element which is smaller than $k_{max}$. 2. For counting (in O(n)) you should say subtract then index of largest and smallest element of each row +1, instead of simply saying count them (because may be someone iterate through them to count them, which is $O(n^2)$). – user742 Sep 17 '13 at 12:16
• Is this not $\mathcal{O}(nlogn)$? If $k_{min},k_{max}$ are outside the matrix range.. Each find call in each row will be $\mathcal{O}(logn)$ by binary search, and there are $n$ rows. Am I missing anything? – Ioannis Sep 17 '13 at 15:50
• @Raphael & Saeed: in this case, we can recover $|A \cap B|$ since $A$ is of the form $[a;\max]$ and $B$ is $[\min;b]$, hence $|A \cap B| = |A| + |B| - n^2$ – Tpecatte Sep 17 '13 at 17:06 | 2020-11-30T17:52:53 | {
"domain": "stackexchange.com",
"url": "https://cs.stackexchange.com/questions/14376/count-elements-of-a-sorted-matrix-that-fall-into-a-given-interval/14377",
"openwebmath_score": 0.7881278395652771,
"openwebmath_perplexity": 174.19838802394975,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9890130554263734,
"lm_q2_score": 0.8499711718571775,
"lm_q1q2_score": 0.8406325857028023
} |
https://math.stackexchange.com/questions/2034613/distance-covered-by-falling-particle-p | # Distance covered by falling particle P
Particle P released from rest at O. Falls freely under gravity until reaching point A which is $1.25$m below O.
(i) Find speed of P at A and time taken for P to reach A.
P continues to fall, but now its downward acceleration $t$ seconds after passing through A is $(10-0.3t)$ metres per second square.
(ii) Find the total distance P has fallen, $3$ s after being released from O.
I have solved (i) and the speed is $5$ m/s and time is $0.5$ s. (ii) is easy, I think, but the given answer is $44.2$ m, while my answer is coming $43.7$ m.
• What is your assumption of g for part i)? – Doug M Nov 28 '16 at 17:56
• 10, as given to be used. otherwise I would have used 9.81 – Hammad Shariq Nov 28 '16 at 17:59
• the answers I have found are given just like part (ii). – Hammad Shariq Nov 28 '16 at 17:59
• It looked like you had used 10. But that is vital information to the problem. In order to help. we need to to be clear on the assumptions. – Doug M Nov 28 '16 at 18:13
• You are right, I forgot to mention that. Will take care in the future. :) – Hammad Shariq Nov 28 '16 at 18:14
conventionally, down is negative... It would probably be easier to do this all in terms of positive numbers, but I am going to stick with convention.
if $g = -10$ then the time to $A = 0.5 s$ and the velocity at $A$ is $-5 m/s$
$3s$ from $O$ is $2.5 s$ after passing though $A$.
$a = \frac {d^2x}{dt^2}= -10+0.3t\\ v = \frac {dx}{dt}= -5 - 10t + 0.15t^2\\ x = -1.25 - 5t -5t^2 + 0.05 t^3$
$t=2.5$ solve for $x.$ Again $x$ will be negative, (distance below $O$). $|x|$ is the distance from $O.$
I get $44.2$
• OK, so I was not using the constants from part (i) in the integrals of part (ii), but the rest was easy. Thanks a lot. – Hammad Shariq Nov 28 '16 at 18:13 | 2019-08-25T19:44:56 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2034613/distance-covered-by-falling-particle-p",
"openwebmath_score": 0.656004786491394,
"openwebmath_perplexity": 472.08665170410467,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9715639644850629,
"lm_q2_score": 0.865224084314688,
"lm_q1q2_score": 0.8406205415247365
} |
https://math.stackexchange.com/questions/207910/prove-convergence-of-the-sequence-z-1z-2-cdots-z-n-n-of-cesaro-means | # Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means [duplicate]
Prove that if $\lim_{n \to \infty}z_{n}=A$ then: $$\lim_{n \to \infty}\frac{z_{1}+z_{2}+\cdots + z_{n}}{n}=A$$ I was thinking spliting it in: $$(z_{1}+z_{2}+\cdots+z_{N-1})+(z_{N}+z_{N+1}+\cdots+z_{n})$$ where $N$ is value of $n$ for which $|A-z_{n}|<\epsilon$ then taking the limit of this sum devided by $n$ , and noting that the second sum is as close as you wish to $nA$ while the first is as close as you wish to $0$. Not sure if this helps....
• Possibly easier to first show for $A=0$ – Thomas Andrews Oct 5 '12 at 19:19
• @Thomas Andrews I edited the question, adding my idea. Could you tell me what you think, please? – Mykolas Oct 5 '12 at 19:29
• Changed the title. There is no series here. – Did Oct 5 '12 at 20:54
• Maybe this question was asked here before (I did not search), but at least I remember that we had questions such that this can be obtained as a consequence of the results from those questions, see, e.g., If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\limsup\sigma_n \leq \limsup s_n$ and limit of quotient of two series. – Martin Sleziak Oct 6 '12 at 9:28
• It is also worth mentioning that this is often called Cesaro mean. – Martin Sleziak Oct 27 '12 at 6:53
It seems like Homework problem, hence I'll just give hint: $$\frac{z_1+z_2+\cdots +z_n}{n}-A=\frac {(z_1-A)+(z_2-A)+\cdots +(z_n-A)}{n}$$ Now use the defn of limit that for every $\epsilon > 0$ there exists $N_0 \in \mathbb N$ such that $|z_m-A| < \epsilon \ \forall m \geq N_0$
Also remember triangle inequality : $|a_1+a_2+\cdots +a_n| \leq |a_1| + |a_2| +\cdots +|a_n|$
Can you find proper $a_i$ in terms of say $z_i$'s??
• thankyou for help. It's not homework. Would $a_{i}$ be $\frac{z_{i}}{n}$, and since for $n \to\infty$all sums go to $0$, it would be proved? @TheJoker – Mykolas Oct 5 '12 at 19:40
• and would my first idea be wrong? – Mykolas Oct 5 '12 at 19:44
• You can take $a_i=\frac{z_1-A}{n}$. Now use the fact that limit of $z_i$ is $A$. about your approach,it is true,just write rigorously :-).also there is no dofference between mine and your approach as your next step would be same. – TheJoker Oct 5 '12 at 19:51
Let $$\epsilon >0$$
We have that $$\lim_{n \rightarrow \infty}z_n=A$$ thus $$\exists n_1 \in \mathbb{N}$$ such that $$|z_n-A|< \epsilon, \forall n \geqslant n_1$$
$$|\frac{z_1+...+z_n}{n}-A|=|\frac{(z_1-A)+...(z_{n_1-1}-A)}{n}+\frac{(z_{n_1}-A)+...+(z_n-A)}{n}| \leqslant \frac{|z_1-A|+...+|z_{n_1-1}-A|}{n}+ \frac{|z_{n_1}-A|+...+|z_n-A|}{n}$$
Exists $$n_2 \in \mathbb{N}$$, by Archimedean property, such that $$\frac{|z_1-A|+...+|z_{n_1-1}-A|}{n}< \epsilon, \forall n \geqslant n_2$$
Now for $$n \geqslant n_0= \max\{n_1,n_2\}$$
$$|\frac{z_1+...+z_n}{n}-A| \leqslant \epsilon+ \frac{(n-n_1) }{n}\epsilon<2 \epsilon$$
• Thanks for your answer; could you please explain why 'there exists $n_2$ such that...' – Daniele1234 May 20 '18 at 21:58
• Because the numerator is a positive constant number and we use the Archimedean property of the real numbers(or you can see it as the convergence of $1/n$ to zero) – Marios Gretsas May 21 '18 at 1:01
This can be an easy consequence of a more general statement which is from Polya's Problems and Theorems in Analysis:
Let $\{a_n\}_{n=1}^{\infty}$ be a real sequence such that $\lim_{n\to\infty}a_n=a$. And we have a family of finite sequences $\{\{b_{nm}\}_{m=1}^{m=n}\}_{n=1}^{\infty}$: $$b_{11}\\ b_{21},b_{22}\\ b_{31},b_{32},b_{33}\\ \cdots$$ such that $$b_{mn}\geq 0$$ for all $m,n$, and $\sum_{m}b_{nm}=1$ for each $n=1,2,\cdots$. Let $\{c_n\}_{n=1}^{\infty}$ be such that $$c_n=\sum_{m=1}^na_mb_{nm}$$ Then $\lim_{n\to\infty}c_n=a$ if and only if $\lim_{n\to\infty}b_{nm}=0$ for each $m$.
The question in OP is a special case of the statement by letting $$b_{nm}=\frac{1}{n},\quad m=1,2,\cdots.$$ | 2021-04-18T12:27:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/207910/prove-convergence-of-the-sequence-z-1z-2-cdots-z-n-n-of-cesaro-means",
"openwebmath_score": 0.9024229645729065,
"openwebmath_perplexity": 203.4709374608049,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.971563966131786,
"lm_q2_score": 0.8652240808393984,
"lm_q1q2_score": 0.8406205395730549
} |
http://mathhelpforum.com/algebra/136477-proof-ellipse.html | # Math Help - Proof of an ellipse
1. ## Proof of an ellipse
The question I am stuck on is (iii):
This part concerns the conic with equation
$3x^2+5y^2=75$
(i) By rearranging the equation of the conic, classify it as an ellipse, parabola or hyperbola in standard position, and sketch the curve.
I have re-arranged the equation and discovered it is an ellipse in standard position:
$(x^2)/5^3 + (y^2)/(\sqrt{15})^2 = 1$
a=5, b= $\sqrt{15}$
The points of the ellipse are (-5,0), (0, $\sqrt{15}$), (5,0) and (0, $-\sqrt{15}$)
(ii) Find exact values for the eccentricity, foci and directrices of this conic, and mark the foci and directrices on your sketch.
eccentricity (e) = $\sqrt{1-b^2/a^2}$= $\sqrt{10}/5$
foci = +-ae = $(-\sqrt{10}, 0)$ and $(\sqrt{10}, 0)$
directrices = +-a/e = $+-25/\sqrt{10}$
(iii) Check your answers to part (a)(ii) by verifying that the equation PF = ePd holds at each of the two points P where the conic intersects the x-axis, where F is the focus with negative x-coordinate, d is the corresponding directrix and e is the eccentricity.
point, P where the conic intersect the x-axis ... etc. (-5,0)
focus with negative x co-ordinate = $(-\sqrt{10},0)$
corresponding directrix = $-25/\sqrt{10}$
eccentricity = $\sqrt{10}/5$
I know that PF = a-ae = $-5+\sqrt{10}$
and ePd = e(a-ae)/e = $\sqrt{10}/5\times(-5+\sqrt{10}/\sqrt{10}/5$
= $-5+\sqrt{10}$
so PF = ePd, but I'm not sure what they mean when they say the two points, P where the conic intersects the x-axis ... etc.
I hope what I have done so far is correct? Thanks again for your help. It is most appreciated
2. Originally Posted by cozza
The question I am stuck on is (iii):
This part concerns the conic with equation
$3x^2+5y^2=75$
(i) By rearranging the equation of the conic, classify it as an ellipse, parabola or hyperbola in standard position, and sketch the curve.
I have re-arranged the equation and discovered it is an ellipse in standard position:
$(x^2)/5^3 + (y^2)/(\sqrt{15})^2 = 1$
a=5, b= $\sqrt{15}$
The points of the ellipse are (-5,0), (0, $\sqrt{15}$), (5,0) and (0, $-\sqrt{15}$)
Surely you meant "the points where the ellipse meets the axis are..."
(ii) Find exact values for the eccentricity, foci and directrices of this conic, and mark the foci and directrices on your sketch.
eccentricity (e) = $\sqrt{1-b^2/a^2}$= $\sqrt{10}/5$
foci = +-ae = $(-\sqrt{10}, 0)$ and $(\sqrt{10}, 0)$
directrices = +-a/e = $+-25/\sqrt{10}$
(iii) Check your answers to part (a)(ii) by verifying that the equation PF = ePd holds at each of the two points P where the conic intersects the x-axis, where F is the focus with negative x-coordinate, d is the corresponding directrix and e is the eccentricity.
point, P where the conic intersect the x-axis ... etc. (-5,0)
focus with negative x co-ordinate = $(-\sqrt{10},0)$
corresponding directrix = $-25/\sqrt{10}$
eccentricity = $\sqrt{10}/5$
I know that PF = a-ae = $-5+\sqrt{10}$
and ePd = e(a-ae)/e = $\sqrt{10}/5\times(-5+\sqrt{10}/\sqrt{10}/5$
= $-5+\sqrt{10}$
so PF = ePd, but I'm not sure what they mean when they say the two points, P where the conic intersects the x-axis ... etc.
Any function intersects the x-axis when y=0, and the y-axis when x=0...
Tonio
I hope what I have done so far is correct? Thanks again for your help. It is most appreciated
.
3. so PF = ePd, but I'm not sure what they mean when they say the two points, P where the conic intersects the x-axis ... etc.
The other point of intersection is (5, 0).
For this point, distances to the directrix with negative x co-ordinate is PF = (a + ae) , and PD is (2a + a/e)
4. Originally Posted by sa-ri-ga-ma
so PF = ePd, but I'm not sure what they mean when they say the two points, P where the conic intersects the x-axis ... etc.
Originally Posted by sa-ri-ga-ma
The other point of intersection is (5, 0).
For this point, PF = (a + ae) , and PD is (2a + a/e)
Thank you for the quick reply. I understand that the other point of intersection is (5,0), but the question specifically says:
Check your answers to part (a)(ii) by verifying that the equation PF = ePd holds at each of the two points P where the conic intersects the x-axis, where F is the focus with negative x-coordinate, d is the corresponding directrix and e is the eccentricity.
Surely there is only one point where the conic intersects the x-axis, where F is the focus with negative x-coordinate, d is the corresponding directrix and e is the eccentricity?
Am I missing something here? At the point (5,0) the f has a positive x-coordinate, as does the corresponding directrix, doesn't it?
Thanks again
5. At the point (5,0) the f has a positive x-coordinate, as does the corresponding directrix, doesn't it?
Here you have to take the distance of the point P (5,0) from negative focus and the corresponding diretrix. Then prove the relation PF = e*PD. | 2014-09-17T18:33:09 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/algebra/136477-proof-ellipse.html",
"openwebmath_score": 0.8332107067108154,
"openwebmath_perplexity": 994.4736582724797,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9715639718953155,
"lm_q2_score": 0.8652240756264639,
"lm_q1q2_score": 0.8406205394951001
} |
http://stat545.com/block011_write-your-own-function-02.html | ### Where were we? Where are we going?
In part 1 we wrote our first R function to compute the difference between the max and min of a numeric vector. We checked the validity of the function’s only argument and, informally, we verified that it worked pretty well.
In this part, we generalize this function, learn more technical details about R functions, and set default values for some arguments.
As usual, load the Gapminder excerpt.
library(gapminder)
### Restore our max minus min function
Let’s keep our previous function around as a baseline.
mmm <- function(x) {
stopifnot(is.numeric(x))
max(x) - min(x)
}
### Generalize our function to other quantiles
The max and the min are special cases of a quantile. Here are other special cases you may have heard of:
• median = 0.5 quantile
• 1st quartile = 0.25 quantile
• 3rd quartile = 0.75 quantile
If you’re familiar with box plots, the rectangle typically runs from the 1st quartile to the 3rd quartile, with a line at the median.
If $$q$$ is the $$p$$-th quantile of a set of $$n$$ observations, what does that mean? Approximately $$pn$$ of the observations are less than $$q$$ and $$(1 - p)n$$ are greater than $$q$$. Yeah, you need to worry about rounding to an integer and less/greater than or equal to, but these details aren’t critical here.
Let’s generalize our function to take the difference between any two quantiles. We can still consider the max and min, if we like, but we’re not limited to that.
### Get something that works, again
The eventual inputs to our new function will be the data x and two probabilities.
First, play around with the quantile() function. Convince yourself you know how to use it, for example, by cross-checking your results with other built-in functions.
quantile(gapminder$lifeExp) ## 0% 25% 50% 75% 100% ## 23.5990 48.1980 60.7125 70.8455 82.6030 quantile(gapminder$lifeExp, probs = 0.5)
## 50%
## 60.7125
median(gapminder$lifeExp) ## [1] 60.7125 quantile(gapminder$lifeExp, probs = c(0.25, 0.75))
## 25% 75%
## 48.1980 70.8455
boxplot(gapminder$lifeExp, plot = FALSE)$stats
## [,1]
## [1,] 23.5990
## [2,] 48.1850
## [3,] 60.7125
## [4,] 70.8460
## [5,] 82.6030
Now write a code snippet that takes the difference between two quantiles.
the_probs <- c(0.25, 0.75)
the_quantiles <- quantile(gapminder$lifeExp, probs = the_probs) max(the_quantiles) - min(the_quantiles) ## [1] 22.6475 IQR(gapminder$lifeExp) # hey, we've reinvented IQR
## [1] 22.6475
### Turn the working interactive code into a function, again
I’ll use qdiff as the base of our function’s name. I copy the overall structure from our previous “max minus min” work but replace the guts of the function with the more general code we just developed.
qdiff1 <- function(x, probs) {
stopifnot(is.numeric(x))
the_quantiles <- quantile(x = x, probs = probs)
max(the_quantiles) - min(the_quantiles)
}
qdiff1(gapminder$lifeExp, probs = c(0.25, 0.75)) ## [1] 22.6475 qdiff1(gapminder$lifeExp, probs = c(0, 1))
## [1] 59.004
mmm(gapminder$lifeExp) ## [1] 59.004 Again we do some informal tests against familiar results. ### Argument names: freedom and conventions I want you to understand the importance of argument names. I can name my arguments almost anything I like. Proof: qdiff2 <- function(zeus, hera) { stopifnot(is.numeric(zeus)) the_quantiles <- quantile(x = zeus, probs = hera) return(max(the_quantiles) - min(the_quantiles)) } qdiff2(zeus = gapminder$lifeExp, hera = 0:1)
## [1] 59.004
While I can name my arguments after Greek gods, it’s usually a bad idea. Take all opportunities to make things more self-explanatory via meaningful names.
This is better:
qdiff3 <- function(my_x, my_probs) {
stopifnot(is.numeric(my_x))
the_quantiles <- quantile(x = my_x, probs = my_probs)
return(max(the_quantiles) - min(the_quantiles))
}
qdiff3(my_x = gapminder$lifeExp, my_probs = 0:1) ## [1] 59.004 If you are going to pass the arguments of your function as arguments of a built-in function, consider copying the argument names. Again, the reason is to reduce your cognitive load. This is what I’ve been doing all along and now you know why: qdiff1 ## function(x, probs) { ## stopifnot(is.numeric(x)) ## the_quantiles <- quantile(x = x, probs = probs) ## max(the_quantiles) - min(the_quantiles) ## } We took this detour so you could see there is no structural relationship between my arguments (x and probs) and those of quantile() (also x and probs). The similarity or equivalence of the names accomplishes nothing as far as R is concerned; it is solely for the benefit of humans reading, writing, and using the code. Which is very important! ### What a function returns By this point, I expect someone will have asked about the last line in my function’s body. Look above for a reminder of the function’s definition. By default, a function returns the result of the last line of the body. I am just letting that happen with the line max(the_quantiles) - min(the_quantiles). However, there is an explicit function for this: return(). I could just as easily make this the last line of my function’s body: return(max(the_quantiles) - min(the_quantiles)) You absolutely must use return() if you want to return early based on some condition, i.e. before execution gets to the last line of the body. Otherwise, you can decide your own conventions about when you use return() and when you don’t. ### Default values: freedom to NOT specify the arguments What happens if we call our function but neglect to specify the probabilities? qdiff1(gapminder$lifeExp)
## Error in quantile.default(x = x, probs = probs): argument "probs" is missing, with no default
Oops! At the moment, this causes a fatal error. It can be nice to provide some reasonable default values for certain arguments. In our case, it would be crazy to specify a default value for the primary input x, but very kind to specify a default for probs.
We started by focusing on the max and the min, so I think those make reasonable defaults. Here’s how to specify that in a function definition.
qdiff4 <- function(x, probs = c(0, 1)) {
stopifnot(is.numeric(x))
the_quantiles <- quantile(x, probs)
return(max(the_quantiles) - min(the_quantiles))
}
Again we check how the function works, in old examples and new, specifying the probs argument and not.
qdiff4(gapminder$lifeExp) ## [1] 59.004 mmm(gapminder$lifeExp)
## [1] 59.004
qdiff4(gapminder\$lifeExp, c(0.1, 0.9))
## [1] 33.5862
### Check the validity of arguments, again
EXERCISE FOR THE READER: upgrade our argument validity checks in light of the new argument probs
## problems identified during class
## we're not checking that probs is numeric
## we're not checking that probs is length 2
## we're not checking that probs are in [0,1]
### Wrap-up and what’s next?
Here’s the function we’ve written so far:
qdiff4
## function(x, probs = c(0, 1)) {
## stopifnot(is.numeric(x))
## the_quantiles <- quantile(x, probs)
## return(max(the_quantiles) - min(the_quantiles))
## }
What we’ve accomplished:
• we’ve generalized our first function to take a difference between arbitrary quantiles
• we’ve specified default values for the probabilities that set the quantiles
Where to next? In Part 3, we tackle NAs, the special ... argument, and formal unit testing. | 2017-01-23T02:17:44 | {
"domain": "stat545.com",
"url": "http://stat545.com/block011_write-your-own-function-02.html",
"openwebmath_score": 0.498393714427948,
"openwebmath_perplexity": 3628.986077166162,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9715639653084245,
"lm_q2_score": 0.8652240791017536,
"lm_q1q2_score": 0.8406205371724297
} |
https://www.physicsforums.com/threads/epsilon-delta-proof-problem.389633/ | # Epsilon-Delta Proof problem
Hey guys heres the problem,
## Homework Statement
lim (4x^2+9) / (3x^2 +5) = 4/3
x->infinity
find k, such that x> k/sqrt(epsilon) guarantees abs((4x^2+9) / (3x^2 +5) - 4/3) < epsilon
## The Attempt at a Solution
By removing the absolute sign and making the denominator common, we get
7 / (9x^2 + 15) < epsilon
keep solving we get
x > 1/3 sqrt(7/epsilon - 15)
Here is where I got stuck. How do we find k as a constant? 15 is inside the square root and if we want to make the whole thing as a fraction we would get x > 1/3 sqrt((7-15*epsilon)/epsilon). I can't just get rid of 15 either because that would screw up the inequality and I am meant to find the MINIMUM value of k. Any help would be appreciated.
jbunniii
Homework Helper
Gold Member
As you said,
$$\frac{4x^2 + 9}{3x^2 + 5} - \frac{4}{3} = \frac{7}{9x^2 + 15}$$
This is positive for all $x$ so we only need to worry about the upper bound. We have
$$\frac{7}{9x^2 + 15} < \epsilon$$
iff
$$7 < 9x^2 \epsilon + 15 \epsilon$$
iff
$$9x^2 > \frac{7 - 15\epsilon}{\epsilon}$$
iff
$$9 x^2 > \frac{7}{\epsilon} - 15$$
Therefore the following is certainly sufficient:
$$9x^2 > \frac{7}{\epsilon}$$
i.e. if $x$ satisfies this inequality, then it satisfies all the ones above. It should be easy to find a suitable $k$ now. (Note that the problem didn't ask you to find the smallest possible $k$, just any $k$ that works.)
Yeah I get your working. Just a side question, is it then impossible to obtain a minimum value of k?
Cheers.
jbunniii
Homework Helper
Gold Member
Yeah I get your working. Just a side question, is it then impossible to obtain a minimum value of k?
Cheers.
OK, go back to
$$9x^2 > \frac{7}{\epsilon} - 15$$
If this were an equality instead of an inequality, then we would have a "boundary case" where $x$ just barely fails, but any larger $x$ would pass. Let's investigate this boundary case:
$$9x^2 = \frac{7}{\epsilon} - 15$$
Let's also set $x$ to the minimum allowed:
$$x = \frac{k}{\sqrt{\epsilon}}$$
and substitute this into the boundary case:
$$\frac{9k^2}{\epsilon} = \frac{7}{\epsilon} - 15$$
or equivalently
$$9k^2 = 7 - 15\epsilon$$
Then we have
$$k = \frac{1}{3}\sqrt{7 - 15\epsilon}$$
assuming $\epsilon$ is small enough that we can take the square root.
I think this $k$ is the smallest possible, if $k$ is allowed to depend on $\epsilon$.
From this we can see that the smallest $k$ that works for ALL $\epsilon$ is
$$k = \frac{1}{3}\sqrt{7}$$
which is the same answer as before.
Last edited:
Hey man thanks again. However I just realised if 9x^2 > 7/epsilon -15, then how can we assume 9x^2 > 7/epsilon? if a > b -c we can't just say a > b can we? grr this is annoying lol...
jbunniii | 2022-06-28T08:46:57 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/epsilon-delta-proof-problem.389633/",
"openwebmath_score": 0.8796982169151306,
"openwebmath_perplexity": 537.3409573243597,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9715639677785088,
"lm_q2_score": 0.8652240756264638,
"lm_q1q2_score": 0.8406205359331397
} |
http://math.stackexchange.com/questions/499544/expressing-the-values-of-a-matrix-at-pow-n | # Expressing the values of a matrix at pow N
I have a square matrix (that comes from a Markov Chain) that looks like that:
$$Q = \begin{bmatrix} 0 & 1& 0 & 0 & .. & 0 & 0\\ 0 & a & 1-a & 0 & .. & 0 & 0\\ 0 & 0 & b & 1 - b & .. & 0 & 0\\ .. & .. & .. & .. & .. & .. & .. \\ 0 & 0 & 0 & 0 & .. & 0 & 1 \end{bmatrix}$$
with $a, b, c,$ etc. real numbers between $0$ and $1$ included.
I am interested in the values of the first line of the matrix $Q^N$. Currently, I am using a library (numpy) that allows me to compute $Q^N$ and then I read the first line of this matrix. But with large matrices ($> 500 \times 500$) and large values of $N$ (~ 10000), it is a bit too slow for my usage.
By curiosity, I have plotted $Q^N_{0,j}$ for $N$ between 1 to 1000 and I found that they follow something that looks like a Poisson distribution or similar (but I don't know if it's only due to my specific input matrix $Q$ or not).
My question is, given such a matrix Q, is there a way to get the values of $Q^N_{0,j}$ without having to compute $Q^N$?
Edit: the terms on the diagonal are such as: 0 <= a <= b <= c <= ... < 1
Edit2:
It appears that if I can diagonalize $Q$, I can use $Q^n = P D^n P^{-1}$ which is faster for large values of $n$ than an exponentiation by squaring (as used by numpy). Problem is that I am not sure it is possible for any matrix $Q$. And if it's not possible, I'd accept a slight modification of $Q$, $P$ or $D$ if the result is close enough.
Edit3:
The values of the first line of $Q^N$ for $N \in [1,100]$:
Edit4:
it's $a <= b$ and not $a < b$, sorry!
-
If ${\bf R}^n$ has a basis consisting of eigenvectors for $Q$, then $Q^N=PD^NP^{-1}$, where $P$ is the matrix whose columns are the eigenvectors of $Q$, and $D$ is the diagonal matrix whose diagonal entries are the eigenvalues of $Q$. The point is, it is very easy to compute $D^N$, $D$ being diagonal. – Gerry Myerson Sep 20 '13 at 12:47
@GerryMyerson After writing my question, I wondered how the matrix power can be done so I'd tend to agree with your comment. But I'm wondering what is so long when I ask numpy to compute $Q^n$, could it be the time to compute the eigenvectors of Q and copying values? Maybe I could save time by doing it myself and save the $P$ and $D$ matrices (since I am computing $Q^N$ for thousands of $N$ values). Am I right to say that the eigenvalues of $Q$ are $(0, a, b, ..., 1)$ (because the matrix is triangular)? – Maxime Sep 20 '13 at 13:29
I don't know what "numpy" is, much less how it works, but, yes, you have the eigenvalues right. – Gerry Myerson Sep 20 '13 at 13:39
@GerryMyerson, Thanks. numpy is a Python library that allows to do linear algebra (and not only). numpy.org I assume that it needs to compute the matrices $P$ and $D$ in order to compute $Q^N$. Now that I ask him to compute $P$, $D$ and $P^{-1}$ before, it's 8 times faster to compute thousands of powers. – Maxime Sep 20 '13 at 14:31
@GerryMyerson One other question: Am I sure that my matrix Q will not be defective? For any dimension and values of a, b, c, etc. ? – Maxime Sep 20 '13 at 15:12
A general bidiagonal matrix is not always diagonalizable; for instance if $a=b=\ldots=\frac{1}{2}$ you will get a Jordan block of eigenvalues $\frac{1}{2}$.
But, since all of your entries are distinct, you're in luck. I'm going to change your notation a bit:
$$M = \left[\begin{array}{ccccc} a_1 & 1-a_1 & & &\\ & a_2 & 1-a_2 & &\\ & & \ddots & &\\ & & & a_{n-1} & 1-a_{n-1}\\ & & & & 1\end{array}\right].$$ (In your case, $a_1 = 0$.)
Then the eigenvalues of $M$ are simply $a_1, \ldots, a_{n-1},1$. The eigenvector corresponding to $1$ is obviously the constant 1 vector. The eigenvector for $a_i$ is the vector $v^i$, where $$v^i_j = \begin{cases}0, &j > i\\ \frac{\prod_{k=j}^{i-1} (a_k-1)}{\prod_{k=j}^{i-1} (a_k-a_i)}, & j \leq i\end{cases}$$ and $v^i_i = 1$ by the empty product convention. You can check this formula by looking at the $j$th row of $Mv^i$:
$$a_j \frac{\prod_{k=j}^{i-1} (a_k-1)}{\prod_{k=j}^{i-1} (a_k-a_i)} + (1-a_j+1)\frac{\prod_{k=j}^{i-1} (a_k-1)}{\prod_{k=j+1}^{i-1} (a_k-a_i)}=\left(a_j+\frac{(1-a_j)(a_j-a_i)}{a_j-1}\right)v^i_j = a_i v^i_j.$$
So now that you have the eigenvectors and eigenvalues, you can decompose $M = BDB^{-1}$ and compute $M^n = BD^nB^{-1}$.
EDIT: Here is the formula for $B^{-1}$. Write $$B = \left[\begin{array}{cccc} v^1 & v^2 & \ldots & v^n\end{array}\right]$$ where $v^i$ is as above and $v^n$ is the one vector. Then $$B^{-1} = \left[\begin{array}{cccc} w^1 & w^2 & \ldots & w^n\end{array}\right]$$ with $$w^i_j = \begin{cases}0, &j > i\\ \frac{\prod_{k=j}^{i-1} (a_k-1)}{\prod_{k=j+1}^{i} (a_k-a_j)}, & j \leq i\end{cases}$$ for $i<n$ and $$w^n_j = \begin{cases}1, &j = n\\ -\frac{\prod_{k=j+1}^{n-1} (a_k-1)}{\prod_{k=j}^{n-2} (a_k-a_{n-1})}, & j < n.\end{cases}$$
-
I am sorry, I've made a mistake, my diagonal entries are not distinct. But I guess I can slightly change them so they become different. Would it change a lot the result if I add something like 0.0000001 when necessary? – Maxime Sep 20 '13 at 20:18
The entries of $M^n$ are continuous functions of the entries of $M$. Maybe someone else can give more precise bounds on the error introduced by perturbing the equal values. – user7530 Sep 20 '13 at 20:23
Notice that the entries of $B$ and $B^{-1}$ become ill conditioned when two diagonal values approach each other, though. Be careful. – user7530 Sep 20 '13 at 20:25
Yes, I was also afraid of that... – Maxime Sep 20 '13 at 20:28 | 2016-02-06T01:13:40 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/499544/expressing-the-values-of-a-matrix-at-pow-n",
"openwebmath_score": 0.9275763034820557,
"openwebmath_perplexity": 157.19637011718174,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9715639653084245,
"lm_q2_score": 0.8652240686758841,
"lm_q1q2_score": 0.8406205270430305
} |
https://math.stackexchange.com/questions/1567310/prove-that-t-is-linear-one-to-one-and-onto | Prove that $T$ is linear, one to one and onto.
Consider the function $T:\mathbb R^2 \to \mathbb R^2$ given by $$T \begin{pmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \end{pmatrix} = \begin{bmatrix} 2x+y \\ -3x \end{bmatrix}$$ Prove that $T$ is linear, one-to-one, and onto. Here is what I 've got so far:
Let $u,v \in \mathbb R^2$ such that $u = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}$ and $v = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ then, $$T(u + v) = T\begin{pmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \end{pmatrix} =\ldots \text{ (skipping tedious steps)} \ldots = T(u) + T(v) \to$$ preserves addition and,
$$T(ru) = T\begin{pmatrix} r \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \end{pmatrix}= \ldots = rT(u) \to$$ preserves scalar multiplication $\therefore T$ is linear.
To prove it is one-to-one I believe I need to show that it is impossible for $u=v$ right? on this i am not positive. Also to prove that it is onto I am pretty sure I need to show that $sp(T) = sp(\mathbb R^2)$ again not really sure so I don't know where to start.
Any help will be greatly appreciated.
Edit 1:
please correct me if im wrong here:
Let $T(v)=0$, then $0=T\begin{pmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \end{pmatrix} = \begin{bmatrix} 2v_1 + v_2 \\ -3v_2 \end{bmatrix}$, so $2v_1 + v_2=0$ and $-3v_1=0$
$\therefore v=\begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0$ and $T$ is one-to-one
• It is enough to show the kernel is trivial: $T(x)=0\Rightarrow x=0$ – Ranc Dec 9 '15 at 10:11
• @Ranc that proves one to one correct? – justin shores Dec 9 '15 at 10:13
• Yes, in the finite dimensional case; If $T\colon E\rightarrow E$ is a linear map, then if $T$ is $1-1$ then it must be onto. This is due to the relation: $\dim E = \dim \mathrm {Im} T +\dim \ker T$ – Ranc Dec 9 '15 at 12:39
The matrix of $T$ is $$A = \begin{bmatrix} &2 &1 \\ &-3 &0 \end{bmatrix}$$ The kernel of $T$ is $\bf0$, because the only solution to $A * [x,y]^T = \bf 0$ is $x=y=0$. Because $$\operatorname{dim}\,\operatorname{ker}\,A + \operatorname{dim}\,\operatorname{rank}\,A = \text{# of columns of A},$$ it follows that $\operatorname{dim}\,\operatorname{rank}\,A = 2$. Thus $A$ is invertible, so $T$ has an inverse — that is, $T$ is an injection and a bijection.
Solve $2x+y = w, -3x=z$ for $x,y$ to get $x=-{1 \over 3} z$ and $y={2 \over 3} z+w$.
Hence for any $(w,z)$ there is a unique $(x,y)$ such that $T(x,y) = (w,z)$.
It follows that $T$ is injective and surjective.
Hint:
1. One-to-one: Let $u, v$ as you have them. Show that $T(u)=T(v) \implies u=v$. You will need to solve a two-by-two system of equations.
2. Onto: Show that $\dim \Im(T)=2$. How? By finding two vectors in the image of $T$ that are linearly independent. For example: what is the image of the vector $e_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ under $T$? Can you find another one, say $z$ so that $T(e_2)$ and $T(z)$ are linearly independent (and therefore span $\mathbb R^2$)?
Of course as @kevinzakka comments, it suffices to show that $T$ is one-to-one. This is equivalent to $\dim Ker (T)=0$ and hence by the Rank - nullity Theorem you can conclude that $\dim \Im (T)=2$.
• Since the image and the co-image are the same, all you have to show is one or the other. Usually proving it is one-to-one is easier. – Kevin Zakka Dec 9 '15 at 15:14 | 2020-10-25T14:16:29 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1567310/prove-that-t-is-linear-one-to-one-and-onto",
"openwebmath_score": 0.9305991530418396,
"openwebmath_perplexity": 99.98469886694613,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9799765622193215,
"lm_q2_score": 0.8577681104440172,
"lm_q1q2_score": 0.8405926440542913
} |
https://math.stackexchange.com/questions/1320137/defining-change-in-an-interval/1320153 | # Defining Change in an interval
I don't know an appropriate topic for this. Here is my question though.
How does one define change in something? For example, regarding Riemann sums one defines change in the width of the rectangles to be $$\Delta x_i = x_i - x_{i-1}~~~,~~~i=1,2,\ldots, n$$
Would I be wrong in defining change as $$\Delta x_i = x_{i+1} - x_i~~?$$ In other words, if I have an interval $[x_0, x_1]$, would I write $\Delta x_0 = x_1 - x_0$ or $\Delta x_1 = x_1 - x_0$ and why?
• math isn't an opinion, but notation sure is. – Exodd Jun 10 '15 at 16:01
• Both are fine to use. – muaddib Jun 10 '15 at 16:04
For example, a common definition for an approximation to the derivative could be $f'(x) = (f(x+h) - f(x))/h = (f(x_{i+1})-f(x_i))/h$. This is known as a 'forward difference' and has first order accuracy in the size of your interval $h$. So if you decrease the size of your $h$, the error in your derivative approximation will correspondingly decrease by that amount.
On the other hand if we define the approximation to the derivative be the 'centered difference' : $f'(x) = (f(x+h) - f(x-h))/(2h) = (f(x_{i+1})-f(x_{i-1}))/(2h)$ we can show that this results in $O(h^2)$ accuracy - i.e. if your $h$ is twice as small, the error in your derivative approximation will be 4 times less.
• I think it's $2h$ in the central difference – Exodd Jun 10 '15 at 16:14 | 2020-07-08T14:46:25 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1320137/defining-change-in-an-interval/1320153",
"openwebmath_score": 0.9483143091201782,
"openwebmath_perplexity": 182.66959490304475,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.979976559880137,
"lm_q2_score": 0.8577681104440171,
"lm_q1q2_score": 0.8405926420478133
} |
http://mathhelpforum.com/calculus/123912-series.html | # Math Help - series
1. ## series
Is there a better way to do this problem?
For the following convergent series, at least how many terms do you have to sum so that the partial sum is within 0.001 of the sum of the series?
$1 - \frac{2}{3}+ \frac{3}{9} - \frac{4}{27}+...+(-1)^{k-1} \frac{k}{3^{k-1}}+...$
If you add the first k terms, the remainder is
$(-1)^{k} \frac{k+1}{3^{k}} + (-1)^{k+1} \frac{k+2}{3^{k+1}} + (-1)^{k+2} \frac{k+3}{3^{k+2}}+... = \ (-1)^{k} \frac{1}{3^{k}} \Big((k+1)-\frac{(k+2)}{3}$ $+ \frac{k+3}{9} +...\Big)$
Using the fact that
$x^{k+1} - \frac{x^{k+2}}{3} +\frac{x^{k+3}}{9}+... = \frac{x^{k+1}}{1+\frac{x}{3}}\ , \ |\frac{x}{3}| < 1$
take the derivative of both sides
$(k+1)x^{k}- \frac{k+2}{3}x^{k+1} + \frac{k+3}{9}x^{k+2} + ... = \frac{(k+1)x^{k}(1+\frac{x}{3})-\frac{1}{3}x^{k+1}}{(1+\frac{x}{3})^{2}}$
let x=1
$(k+1) - \frac{k+2}{3} + \frac{k+3}{9} + ... = \frac{12k+9}{16}$
so we want
$\Big|(-1)^{k}\frac{1}{3^{k}}\frac{12k+9}{16}\Big|<\frac{1 }{1000}$
$\frac{12k+9}{3^{k}} < \frac{2}{125}$
but that can't be solved explicitly for k
although it appears to be true for k >8
2. See attachment
3. Hello, Random Variable!
I can help get you started . . .
A least how many terms do you have to sum so that the partial sum
is within 0.001 of the sum of the series?
$1 - \frac{2}{3}+ \frac{3}{9} - \frac{4}{27}+ \hdots + \frac{(\text{-}1)^{k-1}k}{3^{k-1}}+ \hdots$
$\begin{array}{ccccc}\text{We have:} & S &=& 1 - \dfrac{2}{3} - \dfrac{3}{9} + \dfrac{4}{27} - \dfrac{5}{81} + \hdots \\ \\[-3mm]
\text{Multiply by }\frac{1}{3}: & \dfrac{1}{3}S &=& \quad\;\;\; \dfrac{1}{3} - \dfrac{2}{9} + \dfrac{3}{3^3} - \dfrac{4}{3^4} + \hdots\end{array}$
$\text{Subtract: }\qquad\quad\frac{2}{3}S \;\:=\;\;\underbrace{1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} - \frac{1}{81} + \hdots}_{\text{geometric series}}$
The geometric series has its sum: . $\frac{1}{1-\left(\text{-}\frac{1}{3}\right)} \:=\:\frac{1}{\frac{4}{3}} \:=\:\frac{3}{4}$
Hence, we have: . $\frac{2}{3}S \:=\:\frac{3}{4} \quad\Rightarrow\quad S \:=\:\frac{9}{8} \quad\Leftarrow\text{ sum of the series}$
4. Originally Posted by Calculus26
See attachment
I might be wrong, but it seems like what you did is find the first term whose absolute value is less than 0.001.
5. Yes that is from the theorem on the sum of an alternating series:
|S-Sn|<a(n+1)
the difference between the actual sum S of an alternating series and the nth partial sum Sn is bounded by the n+1st term of the sequence an.
6. If interested I have the proof of this Theorem on my web site
Infinie Series
7. Originally Posted by Calculus26
Yes that is from the theorem on the sum of an alternating series:
|S-Sn|<a(n+1)
the difference between the actual sum S of an alternating series and the nth partial sum Sn is bounded by the n+1st term of the sequence an.
But what if you couldn't use that fact?
8. .
9. But what if you couldn't use that fact?
Well its a handy tool and I always try to look classify a series and determine the best approach.
Soroban cleverly worked it into a geometric series-- something I wish I had thought of.
10. Originally Posted by Soroban
Hello, Random Variable!
I can help get you started . . .
$\begin{array}{ccccc}\text{We have:} & S &=& 1 - \dfrac{2}{3} - \dfrac{3}{9} + \dfrac{4}{27} - \dfrac{5}{81} + \hdots \\ \\[-3mm]$ $
\text{Multiply by }\frac{1}{3}: & \dfrac{1}{3}S &=& \quad\;\;\; \dfrac{1}{3} - \dfrac{2}{9} + \dfrac{3}{3^3} - \dfrac{4}{3^4} + \hdots\end{array}" alt="
\text{Multiply by }\frac{1}{3}: & \dfrac{1}{3}S &=& \quad\;\;\; \dfrac{1}{3} - \dfrac{2}{9} + \dfrac{3}{3^3} - \dfrac{4}{3^4} + \hdots\end{array}" />
$\text{Subtract: }\qquad\quad\frac{2}{3}S \;\:=\;\;\underbrace{1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} - \frac{1}{81} + \hdots}_{\text{geometric series}}$
The geometric series has its sum: . $\frac{1}{1-\left(\text{-}\frac{1}{3}\right)} \:=\:\frac{1}{\frac{4}{3}} \:=\:\frac{3}{4}$
Hence, we have: . $\frac{2}{3}S \:=\:\frac{3}{4} \quad\Rightarrow\quad S \:=\:\frac{9}{8} \quad\Leftarrow\text{ sum of the series}$
It looks like you made a sign error. I think the sum is 9/16.
This is what I did:
$x -\frac{x^{2}}{3} + \frac{x^{3}}{9} - \frac{x^{4}}{27} + ... = \frac{x}{1+\frac{x}{3}} \ , \ \Big|\frac{x}{3}\Big|<1$
differentiate
$1 - \frac{2x}{3} + \frac{3x^{2}}{9} - \frac{4x^{3}}{27} + ...= \frac{9}{(3+x)^{2}}$
let x = 1
$1 -\frac{2}{3} +\frac{3}{9}-\frac{4}{27} + ... = \frac{9}{16}$
so know I need to solve $\Big|\frac{9}{16} - (1-\frac{2}{3} + \frac{3}{9} - \frac{4}{27}+ ...(-1)^{k}\frac{k}{3^{k-1}})\Big| < 0.001$ for k ?
That difference is the remainder. | 2014-09-17T06:27:50 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/123912-series.html",
"openwebmath_score": 0.9027306437492371,
"openwebmath_perplexity": 1119.3627832047905,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9799765587105447,
"lm_q2_score": 0.8577681086260461,
"lm_q1q2_score": 0.8405926392630053
} |
https://dsp.stackexchange.com/questions/55212/periodicity-of-constant-discrete-time-signals | # periodicity of constant discrete time signals
are constant discrete time signals periodic?
example $$$$e^{i10\pi n}$$$$ my proffesor says that this signal is aperiodic, in the discrete sense. but it seems wrong, because unlike in the continuous case, i can calculate the smallest time period , which is 1.
• Welcome to SE.SP! As you say, $e^{\imath 10 \pi n}$ at integer $n$ is a constant. Are constants periodic? See this question and answer on SE.math. – Peter K. Jan 31 at 19:28
• @PeterK. Since this is a discrete signal, I'd say it is periodic with period 1, as abhishek suspects. fourier.eng.hmc.edu/e101/lectures/Fundamental_Frequency/… – MBaz Jan 31 at 19:47
• @MBaz Yes, it looks like you're correct, it just doesn't have a fundamental or minimal period. – Peter K. Jan 31 at 20:00
• @PeterK. the document you gave a reference to says that a constant continuous time signal has no fundamental or minimal period. The document says nothing about a constant discrete time signal. – abhishek Feb 1 at 16:15
• @abhishek Surely that’s true in discrete time too? If the fundamental period must be greater than zero? – Peter K. Feb 1 at 20:56
normally, "$$n$$" is the symbol we use here for discrete-time. if your professor said that:
\begin{align} x[n] &= e^{i10 \pi n} \\ &= e^{i 2 \pi (5n)} \\ \end{align}
is not periodic with a period of $$1$$ (assuming $$n \in \mathbb{Z}$$) or a period of $$\frac15$$ (assuming $$n \in \mathbb{R}$$), then your professor is mistaken.
• More generally, if the frequency is a rational multiple of $pi$, then the sequence is periodic. $10pi$ is a rational multiple of $pi$. – Juancho Feb 1 at 14:46 | 2019-04-20T17:15:45 | {
"domain": "stackexchange.com",
"url": "https://dsp.stackexchange.com/questions/55212/periodicity-of-constant-discrete-time-signals",
"openwebmath_score": 0.9972156882286072,
"openwebmath_perplexity": 879.5118918785192,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9799765581257487,
"lm_q2_score": 0.857768108626046,
"lm_q1q2_score": 0.8405926387613859
} |
http://math.stackexchange.com/questions/90526/proof-of-product-rule-for-derivatives-using-proof-by-induction/90531 | # Proof of Product Rule for Derivatives using Proof by Induction
I am trying to understand the proof of the General Result for the Product Rule for Derivatives by reading this.
Relevant parts are as follows:
Basis for the induction $$D_x \left({f_1 \left({x}\right) f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) f_2 \left({x}\right) + f_1 \left({x}\right) D_x \left({f_2 \left({x}\right)}\right)$$
Induction Hypothesis $$D_x \left({\prod_{i=1}^k f_i \left({x}\right)}\right) = \sum_{i=1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_i \left({x}\right)}\right)$$
Induction Step \begin{align} \tag{1} \kern-30pt D_x \left({\textstyle\prod\limits_{i=1}^{k+1} f_i \left({x}\right)}\right) &= D_x \left({\left({\textstyle\prod\limits_{i=1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)}\right) \\ &= \tag{2} D_x \left({f_{k+1} \left({x}\right)}\right) \left({\textstyle\prod\limits_{i=1}^k f_i \left({x}\right)}\right) + D_x \left({\textstyle\prod\limits_{i=1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right) \\ &=\tag{3} D_x \left({f_{k+1} \left({x}\right)}\right) \left({\textstyle\prod\limits_{i=1}^k f_i \left({x}\right)}\right) + \left({\sum_{i=1}^k \left({D_x \left({f_i \left({x}\right)}\right) \textstyle\prod\limits_{j \ne i} f_i \left({x}\right)}\right)}\right) f_{k+1} \left({x}\right) \\ &= \tag{4} \sum_{i=1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right)\textstyle \prod\limits_{j \ne i} f_i \left({x}\right)}\right) \end{align}
Question
I am stuck at (3). How do I go from (3) to (4)? Specifically, what sort of algebraic manipulations need to be done and what are the motivations for doing those algebraic manipulations in order to arrive at (4)? To put it in another way, I would like to know that is the thought process that one goes through when simplifying (3) to (4).
Note: I hope someone can correct my LaTeX typesetting. I was under the impression that the align environment would automatically number the formulas I write. Thanks in advance.
-
How's that (I had to make the products small to make things fit)? – David Mitra Dec 11 '11 at 18:54
@DavidMitra: Thank you. – Sara Dec 15 '11 at 3:19
Note that $$D_x (f_{k+1}(x))\left( \prod_{i=1}^k f_i(x))\right) = \sum_{i=k+1}^{k+1} D_x(f_i(x)) \left( \prod_{j \neq i}^{k+1} f_i(x) \right)$$
and
$$\left( \sum_{i=1}^k \left( D_x(f_i(x)) \prod_{j \neq i}^k f_i(x)\right) \right) f_{k+1}(x) = \sum_{i=1}^{k} D_x(f_i(x)) \left( \prod_{j \neq i}^{k+1} f_i(x) \right)$$, bringing the $f_{k+1}(x)$ term under the product. Adding these up, the result follows.
If still unclear, you can always try writing it out for small values of $k$.
-
HINT $\$ It is much clearer upon scaling, where it becomes additivity of logarithmic derivatives.
$$\rm\begin{eqnarray} D(f_{n+1}\cdots f_1)\ &=&\rm\ (D\:f_{n+1})\ f_n\cdots f_1\ +\ f_{n+1}\:D(f_n\cdots f_1) \\ \\ \Rightarrow\quad\rm\ \dfrac{D(f_{n+1}\cdots f_1)} {f_{n+1}\cdots f_1}\ &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D(f_{n}\cdots f_1)} {f_{n}\cdots f_1} \\ \\ &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D\:f_{n}}{f_{n}}\ +\ \dfrac{D(f_{n-1}\cdots f_1)} {f_{n-1}\cdots f_1} \\ \\ \\ &\cdots& \\ \\ \Rightarrow\qquad\rm\dfrac{D(f_{n+1}\cdots f_1)} {f_{n+1}\cdots f_1}\ &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D\:f_{n}}{f_{n}}\ +\ \cdots\ +\ \dfrac{D\:f_1}{f_1} \end{eqnarray}$$
Multipying both sides above by $\rm\ f_{n+1}\cdots f_1\$ yields the sought result.
Key is this. With $\rm\ L\: f\: :=\: D\:f/f\$ we have $\rm\ L(f\:g)\ =\ L(f) + L(g)\:.\:$ The above proof is simply the inductive extension to a product of $\rm\:n+1\:$ terms, i.e. $\rm\ L(f_{n+1}\cdots f_1)\:=\: L(f_{n+1})+\:\cdots\:+L(f_1)\:.$ Multiplying this through by $\rm\:f_{n+1}\cdots f_1\:$ yields the sought derivative product rule (but, alas, obfuscates said key homomorphic property of the logarithmic derivative).
Perhaps you might also find helpful this hint from one of my prior posts.
logarithmic differentiation makes the n-ary generalization obvious:
$$\rm (abc)'\:=\ abc \; log(abc)'\: =\ abc \;(log\; a + log\; b + log\; c)'\: =\ abc \; \bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c}\bigg)$$
Obviously the same proof works for arbitrary length products yielding
$$\rm (abc\: \cdots\: f)'\: =\ \: abc\:\cdots f\:\ \bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c} +\:\cdots\:+ \frac{f\:'}{f}\bigg)$$
-
:The answer looks interesting even if I am yet able to comprehend it, but thank you for answering anyway. – Sara Dec 15 '11 at 3:20
@Sara I've expanded the answer. Please let me know if anything is still not clear. – Bill Dubuque Dec 15 '11 at 4:18
My main problem is that my syllabus have not covered logarithmic differentiation yet. Sorry for being dense but there are also certain parts of the wikipedia link that I don't understand. I will get back to you once I have more time to study it in more detail. – Sara Dec 19 '11 at 5:29
@Sara All you need to know is the product rule for derivatives - that's all I use in the first half of the post. While it is convenient to view it in terms of logs, as I explain in the second half, you can safely ignore that for now. The logarithmic derivative of $f$ is simply $f\:'/f$. That's equal to $(\log f)'$ but you don't need to employ that to see the additivity of logarithmic derivatives. – Bill Dubuque Dec 19 '11 at 6:15
Thank you for not giving up on me. I can see the additivity of the logathrithmic derivatives now. Now I just need to understand why the derivative of $f$ is $f^\prime /f$. Anyway, thank you very much for your time. – Sara Dec 25 '11 at 11:27 | 2014-12-18T19:45:59 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/90526/proof-of-product-rule-for-derivatives-using-proof-by-induction/90531",
"openwebmath_score": 0.993211567401886,
"openwebmath_perplexity": 426.5295709643267,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.979976559295341,
"lm_q2_score": 0.8577681068080749,
"lm_q1q2_score": 0.8405926379830558
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.