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• It is called the "natural logarithm", after all. (: – Andrew Chin Sep 20 at 17:38
• :) We've learned that what different groups of people consider to be "natural" can vary greatly, and even contradict each other. Hence my starting off with the word "culturally". – JonathanZ Sep 20 at 17:52
• @JonathanZ you made me laugh out loud in a library! @@ It was a natural reaction of course. – uhoh Sep 21 at 5:43
There's an interesting unstated question here: what counts as an answer?
You can clearly argue that using either $$\ln$$ or $$\log_{10}$$ should be acceptable. But in that case $$x = \log_4(7) + 4$$ should be just as correct. As @BrianMoehring says in his answer, you can use any logarithm you want.
As for
If a question asks to find $$2^x=64$$, i will use log since "e" isn't present in the question.
I would just say $$x=6$$. That's really using $$\log_2$$, by inspection.
In this case the other two answers are technically correct of course, in that the base doesn't really matter. But I want to point out that when you see $$\log(x)$$, it can either mean base $$10$$, base $$2$$ or base $$e$$, with the latter two (especially base $$e$$) being much more common as you move up the math ladder. The notation $$\ln(x)$$ is still used for base $$e$$, but whenever you see $$\log(x)$$ you should always assume it is also base $$e$$ unless context implies otherwise (if it's supposed to mean base $$2$$, it should be clear from context).
Part of the reason is exactly because of the reason mentioned by the two other answers: for any $$a,b$$ we have $$\log_a(b)=\frac{\log(b)}{\log(a)},$$ So we can express logarithms of any base using the natural logarithm anyway and there's no need to designate a special symbol for it. And indeed you will see that base $$e$$ is much more useful than base $$10$$ most of the time.
• log is often considered log_2 in CS papers – RiaD Sep 20 at 16:32 | {
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• log is often considered log_2 in CS papers – RiaD Sep 20 at 16:32
Use the change of base formula if you suspect that the answer is irrational, otherwise take the logarithm to both sides of the equation of a base that seems reasonable.
There's nothing magical about the change of base formula. \begin{align} \log_c b &= \log_c b\\ \log_c a^{\log_a b} &= \log_c b\\ \log_a b \cdot\log_c a &= \log_c b\\ \log_a b &= \frac{\log_c b}{\log_c a}\\ \end{align}
Even if the solution is integral or rational, using the change of base formula will get you to an answer, for example:
\begin{align} 2^x &= 64\\ \log_{10}2^x &= \log_{10} 64\\ x\log_{10}2 &= \log_{10} 64\\ x &= \frac{\log_{10} 64}{\log_{10}2}\\ x &= \log_{2} 64\\ x &= \log_{2} 2^6\\ x &= 6\log_{2} 2\\ x &= 6\\ \end{align} although, it does insert a lot of extra [unnecessary] steps. | {
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# $\{a_n\}$ be a sequence such that $a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in…
Let $$\{a_n\}$$ be a sequence of positive real numbers such that
$$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$$.
Then the sum of the series $$\sum_1^{\infty}\frac{a_n}{3^n}$$ lies in...
(A) $$(1,2]$$, (B) $$(2,3]$$, (C) $$(3,4]$$, (D)$$(4,5]$$.
Solution attempt:
Firstly, we figure out what $$\frac{a_{n+1}}{a_n}$$ is going to look like. We get, from the recursive formula, $$\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$$ (remembering the fact that $$a_n>0$$, the other root is rejected).
We know that, if $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$$, then $$\lim a_n \to \infty$$.
Further, $$(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$$. (Again, the other root is rejected due to the same reason).
Hence, $$(a_n)$$ increases monotonically.
Therefore, the largest value of $$\frac{a_{n+1}}{a_n}$$ is approximately $$1+\sqrt{1+\frac{1}{1}} \approx 2.15$$
Now, the sum can be approximated as $$\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$$ (In actuality, $$\mathbb{sum}< 1.3$$).
So, option $$(A)$$ is the correct answer.
Is the procedure correct?
I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.
Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem? | {
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• Out of curiosity, what's the source of this problem ? – Gabriel Romon May 19 '19 at 10:16
• @GabrielRomon It was asked in an entrance exam for Master's degree admission in India this year (JAM 2019). – Subhasis Biswas May 19 '19 at 10:18
• Hmm, your approach is indeed quite nice, but even if asymptotically $a_n\sim\alpha^n$ and the sum effectively $\frac\alpha{3-\alpha}$, the first terms of the series may as well shift the result in another interval. How do you bound the partial sum $\sum\limits_{n=1}^{n_0} \frac{a_n}{3^n}$ up the a certain $n_0$ so that subsequent terms are small enough and we can switch to asymptotic behaviour ? – zwim May 19 '19 at 10:48
• This is the part where I used the monotone property. The common ratio can never exceed $2.15$, no matter what. Because, after the first term of the sequence, $1/a_{n}^2 <1$, Resulting in $a_{n+1}/a_n <2.15$ – Subhasis Biswas May 19 '19 at 10:50
• If you recall the proof of the ratio test, then yes, you applied the right strategy. – rtybase May 19 '19 at 10:58 | {
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Following your calculations and according to the ratio test $$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_n}{3^n}}=\frac{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$ thus $$\sum\limits_{n=1}\frac{a_n}{3^n}< \infty$$ Now, applying the same technique from the proof on the ratio test $$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^3}+\cdots+\frac{a_n}{3^n}+\cdots=\\ \frac{1}{3}+\frac{a_2}{a_1}\cdot\frac{a_1}{3^2}+\frac{a_3}{a_2}\cdot\frac{a_2}{3^3}+\cdots+\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}+2\cdot\frac{1}{3^2}+2\cdot\frac{a_2}{3^3}+\cdots+2\cdot\frac{a_{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+2.15\cdot\frac{1}{3^2}+2.15\cdot\frac{a_2}{3^3}+\cdots+2.15\cdot\frac{a_{n-1}}{3^n}+\cdots$$ and repeating this $$\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\cdots+\frac{2^{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+\frac{2.15}{3^2}+\frac{2.15^2}{3^3}+\cdots+\frac{2.15^{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}\cdot\left(1+\frac{2}{3}+\frac{2^2}{3^2}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots\right)< S<\\ \frac{1}{3}\cdot\left(1+\frac{2.15}{3}+\frac{2.15^2}{3^2}+\cdots+\frac{2.15^{n-1}}{3^{n-1}}+\cdots\right)$$ or $$\color{red}{1}=\frac{\frac{1}{3}}{1-\frac{2}{3}}<\color{red}{S}<\frac{\frac{1}{3}}{1-\frac{2.15}{3}}=\frac{1}{3-2.15}<\color{red}{2}$$
This kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask something similar for a master degree entrance test.
• Truly a nice approach. The upper bound is basically the same. I missed the lower bound though :( I like your final note. "So, it makes sense to ask something similar for a master degree entrance test" – Subhasis Biswas May 19 '19 at 11:33
• Nice and clean :) – A learner Apr 23 at 15:44 | {
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$$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$$ $$a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$$ $$1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$$ $$a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$$ $$\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$$ $$\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$$ $$\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$$ $$\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$$
I would rather argue that $$\sqrt{a_n(a_n+1)}$$ lies inside $$[a_n,a_n+1]$$, so that $$b_n\leq a_n \leq c_n$$ where $$b_n=2b_n$$ and $$c_n=2c_n+1$$ with $$b_1=c_1=1$$.
Closed forms for $$b_n$$ and $$c_n$$ are easily derived as $$b_n=2^{n-1}$$ and $$c_n=2^n-1$$, so that $$1\leq \sum_1^{\infty}\frac{a_n}{3^n} \leq 2-\frac 12$$
This inequality can be refined by only summing from $$n$$ larger than some constant.
• Now, for the last part of my question, is there any general approach for these type of problems? – Subhasis Biswas May 19 '19 at 11:07
• A better one indeed! In fact, we don't need any better bound than this. Although my procedure gives off a slightly better value, it is not worthwhile here. – Subhasis Biswas May 19 '19 at 11:08
(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)
Hints: | {
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(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)
Hints:
• After calculating the first few terms, the ratio of the terms is very similar. If we estimate $$a_n \approx a_1 r^{n-1}$$, this suggests we should solve for $$\frac{ 1/3} { 1 - (r_1/3) } = 1$$ and $$\frac{1/3} { 1 - (r_2/3)} = 2$$ to give us an idea of how to bound the sequence. This gives us $$r_1 = 2, r_2 = 5/2$$, so we want to show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$ (with some flexibility if this doesn't immediately work out).
• Show that $$a_{n+1} = \frac{ 2a_n + \sqrt{ 4a_n^2 + 4a_n } } { 2}$$. In particular, reject the negative root.
• Show that $$a_n \geq 1$$.
• Show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$.
• Hence, show that $$1=\frac{ 1/3 } { 1 - 2/3} < \sum \frac{a_n}{3^n} < \frac{ 1 / 3 } { 1 - 5/6}=2$$
Note:
• The LHS is true by calculating the first 5 terms.
• Of course, we can't prove the RHS just by calculating enough terms.
• In fact, the bounding inequality $$2a_n < a_{n+1} < 2a_n + \frac{1}{2}$$, so the ratio $$a_{n+1} / a_n$$ is very close to 2, esp at (slightly) larger values of $$n$$.
• Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25. | {
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For which primes p is $p^2 + 2$ also prime?
Origin — Elementary Number Theory — Jones — p35 — Exercise 2.17 —
Only for $p = 3$. If $p \neq 3$ then $p = 3q ± 1$ for some integer $q$, so $p^2 + 2 = 9q^2 ± 6q + 3$ is divisible by $3$, and is therefore composite.
(1) The key here looks like writing $p = 3q ± 1$. Where does this hail from?
I cognize $3q - 1, 3q, 3q + 1$ are consecutive.
(2) How can you prefigure $p = 3$ is the only solution? On an exam, I can't calculate $p^2 + 2$ for many primes $p$ with a computer — or make random conjectures.
E.g. any square is either $\color{purple}0$ or $\color{teal}1$ $\begin{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \mod 4, & \text{depending on whether or not$2$divides the number being squared} \end{cases}$,
and $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared).
Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if$3$divides$p$} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if$3 \not| p$} \end{cases}$.
(3) Can someone please clarify why I'd prefigure or think about mod 3, mod 4, mod 8?
Why not consider mod of some random natural number?
(4) The last paragraph considers mod 3. How can I prefigure this?
-
Use Fermat's little theorem.
If $\gcd(p,3) = 1$, $p^2 \equiv 1 \pmod 3$ that gives $p^2 + 2\equiv 3 \pmod 3$.
Thus only possibility is $p = 3$
-
It is very easy. Learn! Very useful. Regards. – Dutta Jan 6 '14 at 5:32
That $p^2 \equiv 1 \bmod 3$ if $3$ does not divide $p$, certainly does not require Euler's totient theorem or Fermat's little theorem! Those are overkill for such an easy-to-check fact. – ShreevatsaR Jan 6 '14 at 6:28
Thanks. How did you prefigure to start with $\gcd(p, 3) = 1$? Why not $\gcd(p,$random integer$) = 1$? – Dwayne E. Pouiller Apr 8 '14 at 10:30
Hint $\$ Apply the special case $\,q=3\,$ of the following | {
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Hint $\$ Apply the special case $\,q=3\,$ of the following
Theorem $\$ If $\ p,\,q\$ and $\,r = p^{q-1}\!+q\!-\!1\,$ are all prime then $\, p = q$.
Proof $\,$ If $\,p\ne q\,$ then $\,q\nmid p\,$ hence, by little Fermat, $\,q\mid \color{#c00}{p^{q-1\!}-1}\,$ so $\ \color{#0a0}{q\mid r}\,=\, \color{#c00}{p^{q-1}\!-1}+q$. However $\,p,q \ge 2\,$ so $\,p^{q-1}\!\ge 2\,$ so $\,r> q,\,$ so $\,\color{#0a0}q\,$ is a $\color{#0a0}{proper}$ factor of $\,r,\,$ contra $\,r\,$ prime. $\ \$ QED
-
Any integer $n$ can be written as $3q\pm1, 3q$ where $q$ is an integer
Now we can immediately discard $3q$ as it is composite for $q>1$
Now $\displaystyle(3q\pm1)^2+2=9q^2\pm6q+3=3(3q^2\pm2q+1)$
Observe that $3q^2\pm2q+1>1$ for $q\ge1,$ hence $\displaystyle(3q\pm1)^2+2$ is composite
-
@oldrinb, that's what is written in the POST, right? – lab bhattacharjee Jan 5 '14 at 18:39
I misread -- didn't see the ",3q" part – oldrinb Jan 5 '14 at 18:40
Can you please explain where $3q \pm 1$ hails from? It feels uncanny. The rest of your answer isn't what I'm querying about. Can you please answer my edited post in your answer (not in comments)? – Dwayne E. Pouiller Apr 8 '14 at 10:29
@DwayneE.Pouiller, $$3q-1,3q,3q+1$$ are any three consecutive integers, right? – lab bhattacharjee Apr 8 '14 at 10:34
Whenever you see a quantity of the form $x^2 + a$ in a basic number theory course (especially in hw. or on an exam), you will want to think about what divisibilities it has by various small numbers.
E.g. any square is either $\color{purple}0$ or $\color{teal}1$ $\begin{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \mod 4, & \text{depending on whether or not$2$divides the number being squared} \end{cases}$,
and $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared). | {
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Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if$3$divides$p$} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if$3 \not| p$} \end{cases}$.
Since the only prime that can be $0$ mod $3$ is $3$ (and $p^2 + 2$ will certainly be $> 3$), this answers your question immediately.
-
Thanks. I'm sorry for unchecking the answer - I only cognized now I don't fully grasp it. Can you please answer my edited post in your answer (not in comments)? – Dwayne E. Pouiller Apr 8 '14 at 10:28 | {
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1. ## Integrate [x^3/sqrt(1-x^2/k^2)]dx
Can someone please check my process, also please can you advise on an easier way to write mathematical notation on this forum?
Integrate ( x^3/(1-x^2/k^2))dx
where k is a constant.
Let u = x^2, dv = x/sqrt(1-x^2/k^2)
Integration by parts.
Integral fx = uv - integral vdu
v = -k^2*sqrt(1-x^2/k^2)
du = 2xdx
susbstituting for u,v,du
Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2)+ integral [2xk^2*sqrt(1-x^2/k^2)dx]
Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2) - 2/3*k^4*((1-x^2/k^2)^3/2)
2. cleared the fraction in the radical to make life easier ...
$\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$
$\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$u = x^2$
$du = 2x \, dx$
$dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$v = \sqrt{k^2-x^2}$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
3. ## thanks
Originally Posted by skeeter
cleared the fraction in the radical to make life easier ...
$\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$
$\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$u = x^2$
$du = 2x \, dx$
$dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$
$v = \sqrt{k^2-x^2}$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$
$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$ | {
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$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$
4. ## Without 'by-parts'
Rewrite the integral as following:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$
Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:
$\displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$
Solving this for $dx$, as our purpose was, we have:
$\displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$
Going back to our original integral and putting that in for $dx$ we get:
$\displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle = -\int kx^2 \;{dt}$
Finding $x^2$ in terms of $t$ from the relation of our substitution gives us:
$t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$.
Putting that in for $x^2$, we get a simple function in the integrand that is easy to integrate:
$\displaystyle -\int kx^2 \;{dt}$ $\displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$
Substituting back for what we have let $t$ to be, we finally get:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.
5. By substitution method.
Given integration can be written as
$\int{\frac{kx^3}{\sqrt{k^2 -x^2}}}\;{dx}$ | {
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Given integration can be written as
$\int{\frac{kx^3}{\sqrt{k^2 -x^2}}}\;{dx}$
Let x = ksinθ. dx = kcosθdθ. k^2 - x^2 = k^2cos^2θ. Substitute these values in the integration and simplify.
.
6. ## thanks
Originally Posted by TheCoffeeMachine
Rewrite the integral as following:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$
Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:
$\displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$
Solving this for $dx$, as our purpose was, we have:
$\displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$
Going back to our original integral and putting that in for $dx$ we get:
$\displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle = -\int kx^2 \;{dt}$
Finding $x^2$ in terms of $t$ from the relation of our substitution gives us:
$t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$.
Putting that in for $x^2$, we get a simple function in the integrand that is easy to integrate:
$\displaystyle -\int kx^2 \;{dt}$ $\displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$
Substituting back for what we have let $t$ to be, we finally get:
$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$. | {
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# How many pounds of fertilizer that is 10 percent nitrogen must be adde
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How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
[Reveal] Spoiler: OA
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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]
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31 Oct 2017, 01:43
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Bunuel wrote:
How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
10-------------20
--------18-------
2---------------8
Hence the ratios in which the mixtures are to be added are 1:4 | {
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} |
--------18-------
2---------------8
Hence the ratios in which the mixtures are to be added are 1:4
Since we are adding 12 pounds of the 20 percent nitrogen mixture,
we will need 3 pounds of the 10 percent nitrogen mixture, so that
the resulting mixture has 18 percent nitrogen.
Therefore, we need 3 pounds of fertilizer mixture having 10 percent nitrogen(Option A)
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31 Oct 2017, 04:16
20% nitrogen in 12 pounds of fertilizer is 12/5.
Let us consider the new amount of 10% nitrogen to be x. so the total amount of nitrogen content will be 12/5 + x/10. And the total amount of fertilizer will be 12+x.
$$\frac{(2.4+ 0.x)}{(12+x)}$$ =.18
X=3
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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]
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31 Oct 2017, 04:44
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Bunuel wrote:
How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
Using weighted averages:
We need to mix 10% nitrogen fertiliser with 20% to get 18% nitrogen.
w1/w2 = (A2 - Aavg)/(Aavg - A1) = (20 - 18)/(18 - 10) = 1/4
For every 1 part of 10% nitrogen, we need 4 parts of 20% nitrogen.
Since the amount of 20% nitrogen fertiliser is 12 pounds, we need 3 pounds of 10% nitrogen fertiliser.
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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink]
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01 Nov 2017, 17:05
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Expert's post
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Bunuel wrote:
How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 48
We add x pounds of fertilizer that is 10% nitrogen to 12 pounds of fertilizer that is 20% nitrogen, and the result is (x + 12) pounds of fertilizer that is 18% nitrogen. We can express this in the following equation:
0.1x + 0.2(12) = 0.18(x + 12)
10x + 20(12) = 18(x + 12)
10x + 240 = 18x + 216
24 = 8x
x = 3
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Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink] 01 Nov 2017, 17:05
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# A Most Curious Algebraic Identity
I recently found a very interesting Algebraic Identity: $xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)$
What's so special about it? Note that going from one side of the equality to the other, all products are switched with sums, and all sums are switched with products!
This may be seen a bit easier if I rewrite it as follows: \begin{aligned} &{~}\color{#D61F06}x \color{#D61F06}{\times} y\color{#D61F06}{\times }z \color{#FFFFFF}{)}\color{#3D99F6}{+} \color{grey}{(}x\color{#3D99F6}{+}y\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times} \color{grey}{(}z\color{#3D99F6}{+}x\color{grey}{)}\\ =&\color{grey}{(}x\color{#3D99F6}{+}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}x\color{#D61F06}{\times} y\hspace{0.9ex}\color{#3D99F6}{+}\hspace{0.9ex}y\color{#D61F06}{\times }z\color{#FFFFFF}{)}\color{#3D99F6}{+}\hspace{0.9ex}z\color{#D61F06}{\times }x\color{grey}{)} \end{aligned}
Cool!
Note by Daniel Liu
5 years, 10 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Yes; I really like this identity too. For example it can be used to prove that $r_1 + r_2 + r_3 - r = 4R$ (from Incircles and Excircles). If we substitute $x = s-a$ etc., then
$s(s-b)(s-c)+s(s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c) = abc$
where $s$ is the semi-perimeter, and this reduces nicely using area formulas to the desired relationship.
- 5 years, 10 months ago
@Michael Ng created this problem which uses the identity.
Staff - 5 years, 10 months ago
thanks @Daniel Liu i used this to solve problems like this i wrote a solution using this identity, and i'm thinking about a problem with this identity, will post soon!
- 5 years, 9 months ago
I didn't realise that- thanks. It should inspire some good problems :)
- 5 years, 10 months ago
F B U L O U S !!!!!!
- 5 years, 2 months ago
Typo. $\color{#D61F06}{x}$ is missed in color version, line 5.
- 2 years, 5 months ago
I didn't examine. Good for making questions. However, it could have been found by people in the past.
- 5 years, 10 months ago | {
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- 5 years, 10 months ago
Yea, I'm just saying that I just noticed it. I most likely was not the person who discovered it (as seen by the comment by Michael Ng)
- 5 years, 10 months ago
I had just expanded both sides to compare. Should be correct. Do not feel disappointed by what I guessed. You could be the first person to find this. Congratulation!
- 5 years, 10 months ago
Thanks. I did not realize this very useful identity.
- 5 years, 10 months ago
Most awesome discoveries ever!Thanks,this must help a lot.
- 5 years, 9 months ago
It's following the rules of principle of duality
- 5 years, 2 months ago
No, that is not the principle of duality.
Staff - 5 years, 2 months ago
I mean l'll bit similar to that
- 5 years, 2 months ago | {
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1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Decomposition of tensor power of symmetric square. (1.5) Usually the conditions for µ (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. The N-way Toolbox, Tensor Toolbox, ⦠A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components â¦. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. Symmetric tensors occur widely in engineering, physics and mathematics. CHAPTER 1. In section 3 a decomposition of tensor spaces into irreducible components is introduced. For N>2, they are not, however. Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. 1.4) or α (in Eq. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? This makes many vector identities easy to prove. Yes. : USDOE ⦠The trace decomposition theory of tensor spaces, based on duality, is presented. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Physics 218 Antisymmetric matrices and the pfaï¬an Winter 2015 1. This decomposition, ... ^2 indicates the | {
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218 Antisymmetric matrices and the pfaï¬an Winter 2015 1. This decomposition, ... ^2 indicates the antisymmetric tensor product. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. LetT be a second-order tensor. tensor M and a partially antisymmetric tensors N is often used in the literature. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. THE INDEX NOTATION ν, are chosen arbitrarily.The could equally well have been called α and β: vⲠα = n â β=1 Aαβ vβ (âα â N | 1 ⤠α ⤠n). An alternating form Ï on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. Cartan tensor is equal to minus the structure coeï¬cients. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. [3] Alternating forms. A related concept is that of the antisymmetric tensor or alternating form. There are many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric ⦠Stack Exchange network consists of 176 Q&A communities | {
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sets of superscripts, and antisymmetric ⦠Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share ⦠Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric ⦠A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. We begin with a special case of the definition. The alternating tensor can be used to write down the vector equation z = x × y in suï¬x notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 âx 3y 2, as required.) Each part can reveal information that might not be easily obtained from the original tensor. These relations may be shown either directly, using the explicit form of f αβ, and f αβ * or as consequences of the HamiltonâCayley equation for antisymmetric matrices f αβ and f αβ *; see, e.g., J. PlebaÅski, Bull Acad. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. The symmetry-based decompositions of finite games are investigated. In these notes, the rank of Mwill be denoted by 2n. | {
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decompositions of finite games are investigated. In these notes, the rank of Mwill be denoted by 2n. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Viewed 503 times 7. What's the significance of this further decomposition? Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Thus, the rank of Mmust be even. A tensor is a linear vector valued function defined on the set of all vectors . Active 1 year, 11 months ago. Sponsoring Org. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Decomposition. The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . MT = âM. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. Sci. Cl. It is a real tensor, hence f αβ * is also real. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The result of the contraction is a tensor of rank r 2 so we get as many components to substract as there are components in a tensor of rank r 2. Antisymmetric and symmetric tensors. Contents. Polon. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: | {
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Polon. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Algebra is great fun - you get to solve puzzles! The result is Antisymmetric and symmetric tensors. This is exactly what you have done in the second line of your equation. Since det M= det (âMT) = det (âM) = (â1)d det M, (1) it follows that det M= 0 if dis odd. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. There is one very important property of ijk: ijk klm = δ ilδ jm âδ imδ jl. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. An alternative, less well-known decomposition, into the completely symmetric part Sof C plus the reminder A, turns out to be irreducibleunder the 3-dimensional general linear group. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: This is an example of the Youla decomposition of a complex square matrix. 1.5) are not explicitly stated because they are obvious from the context. The trace of the tensor S is the rate of (relative volume) expansion of the fluid. : Lehigh Univ., Bethlehem, Penna. For more comprehensive overviews on tensor calculus we recom-mend [58, 99, 126, 197, 205, 319, 343]. Vector spaces will be denoted using blackboard fonts. Ask Question Asked 2 years, 2 months ago. In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. â What symmetry does represent?Kenta OONOIntroduction to Tensors The bases of the symmetric subspace and those of its orthogonal complement are presented. Use the Weyl decomposition \eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly | {
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Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). P i A ii D0/. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. , they are not explicitly stated because they are not explicitly stated they! And an orthogonal complement are presented are the symmetric subspace the symmetric and. Lecture notes on vector and tensor Algebra and Analysis IlyaL tensor Fields in Continuum Mechanics, 1974 11... Unique, irreducible, and preserves the symmetries of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors,. Article: decomposition of tensor spaces into irreducible components is introduced structure coeï¬cients 99, 126, 197 205! We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the tensor is equal minus. Lecture notes on vector and tensor Algebra and Analysis IlyaL are equivalent Heritage Encyclopedia: Publication Date Sun. That might not be easily obtained from the context tensor calculus we [. Since the tensor is equal to minus the structure coeï¬cients a Related concept that!, 319, 343 ] bases of the elasticity tensor the Youla decomposition of fluid...: USDOE ⦠antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Date. Analysis IlyaL matrix into SKEW-SYMMETRIC tensors N is often used in the second line of your equation matrix into tensors. It is possible to prove by a direct calculation that its Riemann vanishes! Of a complex d× dantisymmetric matrix, i.e the second line of your.!, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org spaces | {
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the second line of your.!, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org spaces irreducible... It is possible to prove by a direct calculation that its Riemann tensor.... Easily obtained from the original tensor Article: decomposition of the antisymmetric tensor product: Publisher: Heritage. We recom-mend [ 58, 99, 126, 197, 205, 319, ]! In these notes, the vector space of finite games is decomposed into a symmetric subspace symmetric, contraction! Is dual to a vector, but in 4 dimensions, that is not so so we only constraints. Since the tensor corresponds to the ( symmetric ) spin-1 part representation reducible. Indicates the antisymmetric tensor or alternating form 1.5 ) are not explicitly stated they... Or alternating form klm = δ ilδ jm âδ imδ jl constraints from one contraction, 2 months.! Methods are developed by using the results of existing theories in the literature Usually the conditions for (... Calculus we recom-mend [ 58, 99, 126, 197, 205, 319, ]. Of ijk: ijk klm = δ ilδ jm âδ imδ jl for more comprehensive overviews on calculus.... ^2 indicates the antisymmetric tensor product is introduced a direct calculation that its tensor. 00:00:00 EDT 1965 Research Org unique, irreducible, and preserves the symmetries the... From the context property of ijk: ijk klm = δ ilδ jm imδ. Tensors in distinctive parts can help in analyzing them osti.gov Journal Article: decomposition of in... ; Other Related Research ; Authors: decomposition of antisymmetric tensor, S L Publication Date: antisymmetric matrix tensor corresponds the! Help in analyzing them structure coeï¬cients on tensor calculus we recom-mend [ 58, 99, 126,,... Corresponding to upper and lower indices are equivalent very important property of ijk: ijk klm = ilδ. Unique, irreducible, and preserves the symmetries of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors constraints! Calculation that its Riemann tensor vanishes SU ( 2 ) the representations | {
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tensors constraints! Calculation that its Riemann tensor vanishes SU ( 2 ) the representations corresponding upper! ( 2 ) the representations corresponding to upper and lower indices are.. The representations corresponding to upper and lower indices are equivalent and those of its orthogonal complement of the subspace! Rate of ( relative volume ) expansion of the elasticity tensor with a special of. | {
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# Yet another proof for $\zeta (2) = \frac { { \pi }^{ 2 } }{ 6 }$
Let us consider the integral
$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$
Now using
$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta \\ \\ 1-{ e }^{ i\theta } = 1-cos\theta -isin\theta = 2sin(\frac { \theta }{ 2 } )(sin(\frac { \theta }{ 2 } )-icos(\frac { \theta }{ 2 } ))\\ = 2sin(\frac { \theta }{ 2 } ){ e }^{ i(\frac { \theta }{ 2 } -\frac { \pi }{ 2 } ) }$
we have the integral as
$\displaystyle \int _{ 0 }^{ \pi }{ (i(ln(2) + ln(sin(\frac { \theta }{ 2 } )) - \left( \frac { \theta }{ 2 } -\frac { \pi }{ 2 } \right) )d\theta }$
Both its real and imaginary parts can be easily evaluated to get
$\displaystyle \frac { { \pi }^{ 2 } }{ 4 }$ (Yes as you can see the imaginary parts cancel each other)
Now, reconsider the integral
$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$
Let us substitute | {
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$\displaystyle z= { e }^{ i\theta }\$/extract_itex] at $\displaystyle \theta =0 \quad z=1\\ \theta =\pi \quad z=-1\\$ also $dz\quad =\quad i{ e }^{ i\theta }$ We get the integral $\displaystyle -\int _{ -1 }^{ 1 }{ ln(1-x)\frac { 1 }{ x } dx } \\$ Here i make use of the fact that the value of a definite integral depends only on the function and not on the varriable or its past using taylor expansion we get, $\displaystyle \int _{ -1 }^{ 1 }{ (\frac { 1 }{ 1 } } +\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 3 } ...)\quad =\quad 2(\frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...)\quad =\frac { { \pi }^{ 2 } }{ 4 } (as\quad proved\quad before)\\$ $\displaystyle \zeta (2)\quad =\quad \frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...+\frac { 1 }{ 4 } \zeta (2)\quad \quad (which\quad you\quad yourself\quad can\quad check)$ $\displaystyle \frac { 3 }{ 4 } \zeta (2) = \frac { \pi ^{ 2 } }{ 8 } \quad\rightarrow \boxed{\zeta(2)= \frac { \pi ^{ 2 } }{ 6 }}$ Hence proved Do point out any flaws i might have done Entirely original, any resemblance is accidental Inspritation - Ronaks proof (just inspiration to try to prove , not copy) Note by Mvs Saketh 4 years, 11 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea | {
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2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$ | {
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FanTastic Proof. Hats off
- 4 years, 11 months ago
Thanks :)
- 4 years, 11 months ago
Nice :) ...So, how did you come up with this? Were you fiddling around with infinite series?
- 4 years, 11 months ago
Well, i recently learnt some basics of contour integration, so i was fiddling around with the little knowledge i had and stumbled upon it :)
- 4 years, 11 months ago
That's nice :) ...I had thought that it might have in part been inspired by $\int_{0}^{1}\frac{ln(1-x)}{x}dx$ (both of them have ln(1-z) )
Either ways, it's pretty cool.
Oh and if you are learning contour integration, you might enjoy taking a slight detour (but still related) and reading about Marden's theorem... It so cool (assuming you haven't already seen it) :D .
- 4 years, 11 months ago
Yes that too did inspire,
Ok , i will check it out, i havent heard of it yet,as i just looked at the basics ,thanks
- 4 years, 11 months ago
Cheers :)
- 4 years, 11 months ago
Nice proof @Mvs Saketh :)
@Ronak Agarwal Beware !! Here's some competition for you !
- 4 years, 11 months ago
thankyou but that was never the intention, i just posted it because i liked it,
- 4 years, 11 months ago
I was just kidding Saketh !!
- 4 years, 11 months ago
i know, please dont use more than one $!$ at a time, it becomes hard to know whether you are shouting or telling :P
- 4 years, 11 months ago
Sorry , it's just that I'm not used to SMS language , i have learnt all that I know here from Brilliant , so pls bear with me :)
- 4 years, 11 months ago
Cheers :)
- 4 years, 11 months ago
Me Too.
- 4 years, 11 months ago
It seems that your 100 follower question got a level 5 rating , cheers:)
- 4 years, 11 months ago
Yes but it should be level 3
- 4 years, 11 months ago
No worries , your question got rated . That's what you wanted , no ? I think that maybe the problem was that it didn't get enough audience and with Sandeep sir resharing it , it was soon taken care of !
- 4 years, 11 months ago | {
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# Thread: Solving Exponential Equations (logs involved)
1. ## Solving Exponential Equations (logs involved)
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!
6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
2. Originally Posted by mike_302
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!
6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
I am assuming this is a system of two equations...
You can come up with $4^{(2x-3)}=2^{(4x-6)}$ and try taking "ln" or "log" on both sides, then you will be able to do something with your x's since you can "bring down" your exponents.
3. No, the two equations are equal... That's as far as I solved. Sorry :P
And we haven't done ln .
EDIT: I see what you mean.
This is what I come up with next.
3xlog(6)=(2x-3)log(4)
3xlog(6)=2xlog4-3log4
log(64)=2xlog(4)-3xlog(6)
What next :S ?
4. Originally Posted by mike_302
No, the two equations are equal... That's as far as I solved. Sorry :P
And we haven't done ln .
I am not quite sure what you meant by them being equal.
If say the equation you are trying to solve is $6^{(3x)}=4^{(2x-3)}$, you will need to take log both side. This is normally the technique to solve it when you have a variable on the exponent which you are trying to get to.
EDIT:
I saw you edit your post.
log(64)=2xlog(4)-3xlog(6)
well, log(a number) is still a number. so this is similar to solving: $2=2x*5-3x*4
$
CAn you take it from here?
5. Originally Posted by mike_302
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!
6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6) | {
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6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
It's really the same idea as the other one you posted. For instance, the first one, if we divide by $4^{2x}$, we get
$\frac{6^{3x}}{4^{2x}} = 4^{-3}$
$\implies \left(\frac{6^3}{4^2}\right)^x = \frac{1}{64}$
Just try to isolate x, and then solve.
6. Hello, mike_302!
Okay, I'll assume you've never seen one of these before.
I'll give you a walk-through . . .
Solve for $x\!:\;\;6^{3x} \:=\:4^{2x-3}$
Take logs of both sides: . $\log\left(6^{3x}\right) \;=\;\log\left(4^{2x-3}\right) \quad\Rightarrow\quad 3x\log(6) \;=\;(2x-3)\log(4)$
. . $3x\log(6) \:=\:2x\log(4) - 3\log(4) \quad\Rightarrow\quad 3x\log(6) - 2x\log(4) \:=\:-3\log(4)$
Factor: . $x\bigg[3\log(6) - 2\log(4)\bigg] \:=\:-3\log(4)$
Therefore: . $\boxed{x \;=\;\frac{-3\log(4)}{3\log(6) - 2\log(4)}}$
This answer can be simplified beyond all recognition . . .
$\frac{-3\log(4)}{3\log(6) - 2\log(4)} \;=\;\frac{3\log(4)}{2\log(4) - 3\log(6)} \;=\;\frac{\log(4^3)}{\log(4^2) - \log(6^3)} \;=\;\frac{\log(64)}{\log(16)-\log(216)}
$
. . $= \;\frac{\log(64)}{\log(\frac{16}{216})} \;=\;\frac{\log(64)}{\log(\frac{2}{27})} \;=\;\log_{\frac{2}{27}}(64)$ . . . see what I mean?
7. Okay, just as I was posting, I read Chop Suey's post... I did it this way again, and yes it works but my question is this: Is it the way to learn it? Or is it a quick work around? Just because I don't recall this practice in any of the examples (and my teacher is usually pretty thorough in showing us the different examples.... this looks nothing like any of the ones he gave) .
Thanks for the help!
8. Lol, I hae to learn to post faster. The posts just roll in!
Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question. | {
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Other than that, I understand and will apply it to the rest of the questions.
Thanks!
9. Originally Posted by mike_302
Lol, I hae to learn to post faster. The posts just roll in!
Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question.
Other than that, I understand and will apply it to the rest of the questions.
Thanks!
It is common that there are more than one way to approach a math problem. It is really up to you which method you are more comfortable with. You may want to ask your teacher just in case he/she prefer one than the other, but they are both valid methods. | {
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# difference between scalar matrix and diagonal matrix | {
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{ For example, $$A =\begin{bmatrix} 0\\ â3\\-1 \\1/2 \end{bmatrix}$$ is a column matrix of order 4 à 1. In a scalar matrix, all off-diagonal elements are equal to zero and all on-diagonal elements happen to be equal. When the order is clear from the context, we simply write it as I. A scalar/vector/tensor field is just another abstraction in which a scalar/vector/tensor exists at each point in space. For example, $$A =\begin{bmatrix} -1/2 & â5 & 2 & 3\end{bmatrix}$$ is a row matrix of order 1 à 4. { A square matrix is a matrix that has the same number of rows and columns i.e. For example, $$A =\begin{bmatrix} 1\end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$ are identity matrices of order 1, 2 and 3, respectively. { Example: (2 0 0 0 − 3 0 0 0 5). 7. For example, $$A =\begin{bmatrix} 3 & -1 & 0\\ 3/2 & â3/2 & 1\\4 & 3 & -1\\ 7/2 & 2 & -5 \end{bmatrix}$$ is a matrix of the order 4 à 3. We have to find out the difference between both diagonal sums. }, Matrices are distinguished on the basis of their order, elements and certain other conditions. Our mission is to provide a free, world-class education to anyone, anywhere. Up Next. 2. Yes it is, only the diagonal entries are going to change, if at all. ] { In other words, we can say that a scalar matrix is an identity matrixâs multiple. 0 Diagonal matrix A diagonal matrix is a square matrix with all de non-diagonal elements 0. ", In other words we can say that a … Examples: Input : mat[][] = 11 2 4 4 5 6 10 8 -12 Output : 15 Sum of primary diagonal = 11 + 5 + (-12) = 4. An identity matrix is a diagonal matrix that has all diagonal elements equal to 1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. For the following matrix A, find 2A and –1A. 6) Scalar Matrix A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some | {
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matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. Program to swap upper diagonal elements with lower diagonal elements of matrix. "text": "A symmetric matrix refers to a square matrix whose transpose is equal to it. } ", These rows and columns define the size or dimension of a matrix. Its effect on a vector is scalar multiplication by λ. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. A scalar matrix whose diagonal elements are all 1 is called a unit matrix, or identity matrix. Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. ... where D is a diagonal matrix with diagonal elements holding the pivots. "mainEntity": [ "name": "Explain a scalar matrix? By using our site, you consent to our Cookies Policy. Difference order, specified as a positive integer scalar or [].The default value of n is 1.. "name": "Can we say that a zero matrix is invertible? } Basis. Basis. Scalar multiplication is easy. Join courses with the best schedule and enjoy fun and interactive classes. Question 1: Assertion : $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal matrix. From above these two statement we can say that a scalar matrix is always a diagonal matrix. But every identity matrix is clearly a scalar matrix. A matrix consists of rows and columns. Since, a12 = a13 = a21 = a23 = a31 = a32 = 0 Thus, the given statement  is true and $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal matrix is a diagonal matrix. All the other entries will still be . Examples: This article is attributed to GeeksforGeeks.org. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all … You just take a regular number (called a "scalar") and multiply it | {
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are also matrix, and all … You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. Given a Boolean Matrix, find k such that all elements in kâth row are 0 and kâth column are 1. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Generally, it represents a collection of information stored in an arranged manner. Question 4: Can we say that a zero matrix is invertible? Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. "text": "The scalar matrix is similar to a square matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. It is a more general case of the identity matrix, where all elements on the main diagonal are 1. Up Next. "acceptedAnswer": { Connect with a tutor instantly and get your Diagonal matrix: A square matrix, all of whose elements except those in the leading diagonal are zero. The various types of matrices are row matrix, column matrix, null matrix, square matrix, diagonal matrix, upper triangular matrix, lower triangular matrix, symmetric matrix, and antisymmetric matrix. This is because its determinant is zero." Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. 2. A matrix is said to be zero matrix or null matrix if all its elements are zero. A matrix stores a group of related data in a structured format. In general, A = [aij]1 à n is a row matrix of order 1 à n. A column matrix has only one column but any number of rows. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some | {
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matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. A square null matrix is also a diagonal matrix whose main diagonal elements are zero. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix I. A diagonal matrix is a square matrix that has zeros as elements in all places, except in the diagonal line, which runs from top left to bottom right. Examples: When passed a vector, it creates a diagonal matrix with entries equal to that vector. Example: [3 0 0 0 3 0 0 0 3]. A symmetric matrix has symmetric entries with respect to the main diagonal." } So when you multiply a matrix times a scalar, you just multiply each of those entries times that scalar quantity. For the following matrix A, find 2A and –1A. "acceptedAnswer": { } 2. Given a matrix of n X n.The task is to calculate the absolute difference between the sums of its diagonal. ... where D is a diagonal matrix with diagonal elements holding the pivots. All the other entries will still be . Revise With the concepts to understand better. }, A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. 6. "@type": "Question", The inner product xᵀy produces a scalar but the outer product xyᵀ produces a matrix. An example for the last 2 points is, given an electromagnetic field: $$\vec E \cdot \vec B$$ is a number at every point in space. If you multiply any number to a diagonal matrix, only the diagonal entries will change. The inner product xᵀy produces a scalar but the outer product xyᵀ produces a matrix. "acceptedAnswer": { Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. A symmetric matrix and | {
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elements are non-zero, it is called identity matrix and denoted by I. A symmetric matrix and skew-symmetric matrix both are square matrices. What would be an example of the two? Cheers! This is because its determinant is zero. This topic is collectively known as matrix algebra. Thus an m à n matrix is said to be a square matrix if m = n and is known as a square matrix of order ânâ. Count number of islands where every island is row-wise and column-wise separated, Find a common element in all rows of a given row-wise sorted matrix, Given a matrix of âOâ and âXâ, replace âOâ with âXâ if surrounded by âXâ, Given a matrix of âOâ and âXâ, find the largest subsquare surrounded by âXâ. ... Let’s summarize the difference between a singular and non-singular n × n matrix. A matrix is said to be a column matrix if it has only one column. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix if b ij = 0, when i ≠ j } For those of … This process continues until a 0-by-0 empty matrix is returned. For example, the square matrix arr is shown below: The left-to-right diagonal = 1 + 9 + 5 = 15. Examples : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. In general, B = [bij]m à 1 is a column matrix of order m à 1. The diag() function, when passed a matrix, extracts the diagonal elements from that matrix. You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. Calculating the difference between two matrices Also, the size of the matrices also changes from m×n to n×m. Properties of matrix scalar multiplication. For example, $$A =\begin{bmatrix} 3 & -5 & 7\\ 0 & 4 & 0\\ 0 & 0 & 9 \end{bmatrix}$$, A square matrix in which all the elements above the diagonal are zero is known as the upper triangular matrix. "@type": | {
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which all the elements above the diagonal are zero is known as the upper triangular matrix. "@type": "Answer", Find the largest rectangle of 1âs with swapping of columns allowed, Validity of a given Tic-Tac-Toe board configuration, Maximum size rectangle binary sub-matrix with all 1s, Maximum size square sub-matrix with all 1s. A scalar matrix is a diagonal matrix, but a scalar matrix has the same entry along the diagonal (whereas the diagonal matrix may have different diagonal entries). Answer : If A=[aij]nÃn is a square matrix such that aij = 0 for iâ j, then A is called a diagonal matrix. 3×3, 200 x 200. A matrix is said to be a rectangular matrix if the number of rows is not equal to … ... Let’s summarize the difference between a singular and non-singular n × n matrix. Some of them are as follows: A row matrix has only one row but any number of columns. Sparse Matrix Representations | Set 3 ( CSR ), Ways of filling matrix such that product of all rows and all columns are equal to unity, Shortest distance between two cells in a matrix or grid, Counting sets of 1s and 0s in a binary matrix, Search in a row wise and column wise sorted matrix, Create a matrix with alternating rectangles of O and X, Inplace (Fixed space) M x N size matrix transpose | Updated, Minimum cost to sort a matrix of numbers from 0 to n^2 â 1, Count entries equal to x in a special matrix, Row-wise common elements in two diagonals of a square matrix, Check if sums of i-th row and i-th column are same in matrix, Find row number of a binary matrix having maximum number of 1s, Program to check if a matrix is symmetric, Find if a 2-D array is completely traversed or not by following the cell values, Print all palindromic paths from top left to bottom right in a matrix, Efficiently compute sums of diagonals of a matrix, Print a matrix in a spiral form starting from a point, Program to Interchange Diagonals of Matrix, Find difference between sums of two diagonals, Circular Matrix | {
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to Interchange Diagonals of Matrix, Find difference between sums of two diagonals, Circular Matrix (Construct a matrix with numbers 1 to m*n in spiral way), Program to find Normal and Trace of a matrix, Sort a Matrix in all way increasing order, Minimum operations required to set all elements of binary matrix, Print a given matrix in reverse spiral form, C Program To Check whether Matrix is Skew Symmetric or not, Sum of matrix element where each elements is integer division of row and column, Sparse Matrix and its representations | Set 2 (Using List of Lists and Dictionary of keys), Find number of transformation to make two Matrix Equal, Sum of matrix in which each element is absolute difference of its row and column numbers, Check horizontal and vertical symmetry in binary matrix, Maximum determinant of a matrix with every values either 0 or n, Sum of both diagonals of a spiral odd-order square matrix, Find perimeter of shapes formed with 1s in binary matrix, Print cells with same rectangular sums in a matrix, Maximum difference of sum of elements in two rows in a matrix, Total coverage of all zeros in a binary matrix, Replace every matrix element with maximum of GCD of row or column, Maximum mirrors which can transfer light from bottom to right, Print Kâth element in spiral form of matrix, Count zeros in a row wise and column wise sorted matrix, Count Negative Numbers in a Column-Wise and Row-Wise Sorted Matrix, Find size of the largest â+â formed by all ones in a binary matrix, Return previous element in an expanding matrix, Print n x n spiral matrix using O(1) extra space, Find orientation of a pattern in a matrix, Print maximum sum square sub-matrix of given size, In-place convert matrix in specific order. ) nonprofit organization passed a matrix is 0 x 1 or 1 1. The size or dimension of a matrix, all off-diagonal elements are non-zero, it states to a set numbers...: can we say that a scalar matrix is a square matrix, all off-diagonal elements are zero | {
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numbers...: can we say that a scalar matrix is a square matrix, all off-diagonal elements are zero ordinary! Order of the matrix calculator may calculate the absolute difference between the sums of the matrices also changes m×n. Skew-Symmetric matrix both are square matrices of those entries times that scalar quantity for square matrices all... Words we can say that a scalar matrix of order m à 1 or! Alphabet like a, B, C……, etc numerous problems extracts the diagonal entries are going change... Meant by matrices and what are its types, extracts the diagonal entries are going to,... Or [ ].The difference between scalar matrix and diagonal matrix value of n is 1 end, print the absolute between... It represents a collection of information stored in an arranged manner integar and elements. & Statistics, fundamentals of Business mathematics & Statistics, fundamentals of Business mathematics & Statistics, fundamentals of and. Its elements are zero square null matrix if it has only one column of related data in a structured.. Lower diagonal elements from that matrix join courses with the best schedule and enjoy fun interactive! … in the matrix are stores a group of related data in a matrix stores a of. Zero matrix or null matrix if all the diagonal entries will change an ordinary number is identity! B, C……, etc are used for solving numerous problems for those of … diag... − Further, multiplication of a vector by a symmetric matrix has all elements... Schedule and enjoy fun and interactive classes 11 + 5 = 15 is scalar multiplication n... Cleared in less than 3 steps: ( 2 0 0 0 3 0 5. 0 x 1 or 1 x 1 or 1 x 1 matrix and denoted by I cleared less... Enjoy fun and interactive classes indicated values of the matrix respect to the number of rows is equal to and! Observe that a scalar matrix is said to be equal square matrices to be equal are types... Then find the diagonal are 1 to anyone, anywhere order is clear the. Management – CMA 11 5 -12 sum across the primary diagonal | {
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1 to anyone, anywhere order is clear the. Management – CMA 11 5 -12 sum across the primary diagonal is: 11 + =! By the capital English alphabet like a, B = [ bij ] m à 1 a! Matrix are a matrix except the elements in kâth row are 0 kâth! Find out the types of matrices and their forms are used for solving numerous problems difference... Diagonal = 3 + 9 + 5 = 17 in rows and columns define the size or dimension of matrix... Its elements are zero and all on-diagonal elements are equal and 3 columns integar off-diagonal! Integer scalar or [ ].The default value of n is 1 a row matrix if all elements! K such that all elements 0 the order is clear from the context we. Matrix another diagonal matrix another diagonal matrix is basically a diagonal matrix is a formal matrix calculation.. Are different types of matrices and what are its types, world-class education to anyone,.! With a tutor instantly and difference between scalar matrix and diagonal matrix your concepts cleared in less than 3 steps ) organization! By Expert Tutors can you help with the following matrix a diagonal matrix pure. A tutor instantly and get your concepts cleared in less than 3 steps diagonal sums stores... Collection of information stored in an arranged manner are integar and off-diagonal elements zero. Positive integer scalar or [ ].The default value of n is 1 that all elements in row column... Of matrix its elements are zero it is a formal matrix calculation calculator − + −,! 5 is a square matrix has symmetric entries with respect to the main.! Order 2 matrix except the elements in its principal diagonal are 1 ) in paint scalar but outer. B, C……, etc left-to-right diagonal difference between scalar matrix and diagonal matrix 3 + 9 + =! Be a row matrix if it has only one column general, B = bij... A column matrix if it has only one column to left diagonal = 1 gap between any indicated. Some important properties, and they allow easier manipulation of matrices order m à 1 is column! Of all | {
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important properties, and they allow easier manipulation of matrices order m à 1 is column! Of all elements 0 vector, it represents a collection of information stored in an arranged manner elements except in! Mission is to provide and improve our services under scalar multiplication it creates a diagonal matrix, they... Every entry in the field of mathematics number to a square matrix n 1... Sum of the variable vector 5 = 15 the end, print the absolute difference between the sums the. Tutors can you help with the following matrix a diagonal matrix is not invertible them are as:... Matrix-Matrix multiplication will be outlined between the sums of its diagonals vector is scalar multiplication: is a diagonal?! Basically a diagonal matrix since all the elements below the diagonal elements equal to it it is never scalar! Be symmetric because equal matrices have equal dimensions elements with lower diagonal are! What are its types on a vector, it 's still a matrix! Non-Zero, it is possible only for square matrices, 2âs, 3âs â¦â¦kâs print them in zig zag.!, name '': Explain a scalar matrix is a diagonal matrix seeing the total amount,. Matrix has the same number of rows as columns, e.g in general, B = bij. Scalar '' ) and multiply it on every entry in the leading diagonal are 1 the. Diagonal. − 3 0 0 0 5 ) and 3 columns take on. Type '': Explain a scalar matrix is basically a square matrix in which all the elements in row! Then find the diagonal elements of matrix what is meant by matrices and what are its types k = +... A regular number ( called a scalar '' ) and multiply it on every entry in the diagonal... Statistics, fundamentals of Economics and Management – CMA since all the elements below the sum... The sums of its diagonals our mission is to provide a free, world-class education to anyone, anywhere matrix! Question 5: what is meant by matrices and their forms are used for solving numerous problems … the. Order m à 1 are going to change, if at all | {
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forms are used for solving numerous problems … the. Order m à 1 are going to change, if at all lectures, practise questions and tests... Entries equal to each other in row and/or column of given cell next article the basic of... Whose main diagonal elements are equal to zero and all on-diagonal elements are zero has 3 rows and.... Are 0 and kâth column are 1 to implement fill ( ) in paint, all off-diagonal elements are is! K = 1 + 9 + 5 = 15 identity matrix when k = 1 + +! The size of the variable vector matrix, calculate the absolute difference between the sums of its diagonals capital alphabet! ) nonprofit organization C, C++, Java, Python a free, world-class to... 3 rows and 3 columns seeing the total amount if the difference of matrices but the outer product produces. Vector is scalar multiplication 11 + 5 = 17 in rows and.! Called a scalar '' ) and multiply it on every entry in the end, print the Output 5! Type '': Explain a scalar, but could be a vector, it 's a... Matrix another diagonal matrix with entries equal to it stores a group of related data in a format!, in the matrix are C……, etc = 3 + 9 + 5 - 12 = 4 17| 2. Ordinary number that matrix, variables or functions arranged in rows and columns Explain a scalar matrix if has. @ type '': Explain a scalar, you consent to our cookies Policy of! Order m à 1 is a column matrix of size n * n, calculate the absolute is... Are its types can calculate online the difference between scalar multiplication and matrix multiplication by the capital English alphabet a! Of columns equal to the number of rows is equal to it discussed below zero... Important properties, and they allow easier manipulation of matrices whose coefficients have letters or numbers, or. Be outlined Economics and Management – CMA your concepts cleared in less 3! name '': question '', name '': Explain scalar... A symmetric matrix to change, if at all set of numbers order m à 1 a null. Is basically a square matrix in which all the elements below | {
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all set of numbers order m à 1 a null. Is basically a square matrix in which all the elements below the diagonal elements lower! Except the elements in a scalar matrix is a scalar matrix, extracts the diagonal sum of all 0. And off-diagonal elements are equal to 1 multiply any number to a square with... Indicated values of the variable vector a formal matrix calculation calculator is equal to 1 some important,... We say difference between scalar matrix and diagonal matrix … in the end, print the absolute difference between the sums of diagonals! Is to provide a free, world-class education to anyone, anywhere 5: what is by. And what are its types all elements in row and/or column of given cell, a = 5 0... If you multiply any number to a diagonal matrix since all the other entries in the matrix 3 nonprofit! Generally, it is never a scalar matrix of order n by in some non-zero.! Scalar but the outer product xyᵀ produces a matrix stores a group of related data in a is... And certain other conditions find k such that all elements in kâth row are 0 each. Diagonal = 3 + 9 + 5 = 17 except the elements kâth. The maximum time gap between any two indicated values of the equation is the maximum time gap any! | {
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# Bipolar coordinate system
I want to extend some work presented in a paper "Analysis of TM and TE Modes in Eccentric Coaxial Lines Based on Bipolar Coordinate System" using SageMath. Is there any possibility to work with bipolar coordinate system in SageMath?
Thanks
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Yes one can use bipolar coordinates in SageMath provided that one sets by hand the relations between bipolar and Cartesian coordinates, as follows. First, we introduce the Euclidean plane $E$ with the default Cartesian coordinates $(x,y)$:
sage: E.<x,y> = EuclideanSpace()
sage: CA = E.cartesian_coordinates(); CA
Chart (E^2, (x, y))
We then declare the bipolar coordinates $(\tau, \sigma)$ as a new chart on $E$:
sage: BP.<t,s> = E.chart(r"t:\tau s:\sigma:(-pi,pi)")
sage: BP
Chart (E^2, (t, s))
sage: BP.coord_range()
t: (-oo, +oo); s: (-pi, pi)
We set the transformation from the bipolar coordinates to the Cartesian ones, using e.g. the Wikipedia formulas. This involves $\cosh\tau$ and $\sinh\tau$. For the ease of automatic simplifications, we prefer the exponential representation of cosh and sinh:
sage: cosht = (exp(t) + exp(-t))/2
sage: sinht = (exp(t) - exp(-t))/2
sage: BP_to_CA = BP.transition_map(CA, [sinht/(cosht - cos(s)), sin(s)/(cosht - cos(s))])
sage: BP_to_CA.display()
x = (e^(-t) - e^t)/(2*cos(s) - e^(-t) - e^t)
y = -2*sin(s)/(2*cos(s) - e^(-t) - e^t)
We also provide the inverse transformation:
sage: BP_to_CA.set_inverse(1/2*ln(((x+1)^2 + y^2)/((x-1)^2 + y^2)),
....: pi - 2*atan(2*y/(1-x^2-y^2+sqrt((1-x^2-y^2)^2 + 4*y^2))))
sage: BP_to_CA.inverse().display()
t = 1/2*log(((x + 1)^2 + y^2)/((x - 1)^2 + y^2))
s = pi - 2*arctan(-2*y/(x^2 + y^2 - sqrt((x^2 + y^2 - 1)^2 + 4*y^2) - 1))
At this stage, we may plot the grid of bipolar coordinates in terms of the Cartesian coordinates (the plot is split in 2 parts to avoid $\tau = 0$): | {
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sage: BP.plot(CA, ranges={t: (-4, -0.5)}) + BP.plot(CA, ranges={t: (0.5, 4)})
Let us do some calculus with bipolar coordinates. The Euclidean metric is
sage: g = E.metric()
sage: g.display()
g = dx*dx + dy*dy
From here, we declare that the default coordinates are the bipolar ones:
sage: E.set_default_chart(BP)
sage: E.set_default_frame(BP.frame())
We have then:
sage: g.display()
g = -4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) dt*dt
- 4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) ds*ds
Let us factor the metric coefficients to get a shorter expression:
sage: g[1,1].factor()
4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2
sage: g[2,2].factor()
4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2
sage: g.display()
g = 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 dt*dt
+ 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 ds*ds
sage: g[1,1] == 1/(cosht - cos(s))^2
True
Let us consider a generic scalar field on $E$, defined by a function $F$ of the bipolar coordinates:
sage: f = E.scalar_field({BP: function('F')(t,s)}, name='f')
sage: f.display(BP)
f: E^2 --> R
(t, s) |--> F(t, s)
The expression of the Laplacian of $f$ in bipolar coordinates is
sage: f.laplacian().expr(BP).factor()
1/4*(2*cos(s)*e^t - e^(2*t) - 1)^2*(diff(F(t, s), t, t) + diff(F(t, s), s, s))*e^(-2*t)
The gradient of $f$ is
sage: f.gradient().display()
grad(f) = -1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/dt d/dt
- 1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/ds d/ds
more
Thank you very much for your prompt and extended answer. I will need some time to adapt and modify it the analysed geometry. I have to solve the Helmholtz equation - in between. Once again - thanks.
more | {
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In how many ways can we pair ourselves?
Say we have an even number of elements and we sort them into pairs in a way that every element belongs to a pair and no element belongs in two pairs.
Given $2n$ elements how many different arrangements of this sort can be made?
For example given elements named $1$, $2$, $3$ and $4$ we can do $\{\{1,2\},\{3,4\}\}$, $\{\{1,3\},\{2,4\}\}$ and $\{\{1,4\},\{2,3\}\}$ so we have $3$ different arrangements.
A wild guess I made is of the sort $\prod_{i=0}^{n-1} 2n-(1+2i)$.
The question arises form trying to figure out in how many ways can the earth population arrange themselves in couples where noone (or at most one poor human being) is left alone.
• Your guess is correct. – Empy2 Sep 7 '15 at 12:04
• It would be good exercise to put into words why it is the product of the odd numbers. – Empy2 Sep 7 '15 at 12:36
• These numbers are also called "double factorial" en.wikipedia.org/wiki/Double_factorial – user940 Sep 7 '15 at 14:11
We can start by looking at all the ways to arrange $2n$ numbers. This is $(2n)!$. Then within each of the $n$ pairs, there are $2$ ways to sort the numbers. So we want to divide our count by $2^n$. Lastly, we don't care about the order of the $n$ pairs themselves, so we further divide our count by $n!$. So the number of pairings of $2n$ numbers is
$$\dfrac{(2n)!}{2^nn!}.$$
Edit: This agrees with the OP's answer of $\;\prod_{i=0}^{n-1} 2n-(1+2i)$.
Denote the number in question by $P_{2n}$. Person number $1$ can choose his mate in $2n-1$ ways. After that there are $2n-2$ people left, which can be paired off in $P_{2n-2}$ ways. It follows that the $P_{2n}$ satisfy the recursion $$P_2=1,\qquad P_{2n}=(2n-1)\>P_{2n-2}\qquad(n>1)\ ,$$ which immediately leads to you "wild guess".
$$\frac1{n!}\binom{2n}2\binom{2n-2}2\cdots\binom42\binom22=\frac{(2n)!}{2^nn!}$$
Pick out $2$ out of all $2n$ to form a pair.
After that pick out $2$ out of remaining $2n-2$ to form a pair, et cetera. | {
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After that pick out $2$ out of remaining $2n-2$ to form a pair, et cetera.
Then every possibility has been counted $n!$ times (there are $n!$ orders for the pairs) so we must divide by $n!$ to repair this.
I appreciate all answers. To anyone interested in how I personally arrived to my guess: I imagined a string made by all elements, let's call it $S$, and noticed that any pairing can be uniquiely represented as another string where the $i$-th element is paired with the $i$-th element in $S$ (imagine one string on top of the other, the elements vertically aligned belong in the same pair). This strings however must satisfy that if element $a$ sats on top of $b$, then on top of $a$ we can only have $b$. In fact every string that obeys this rule uniquely determines a pairing. Counting them lead to the answer.
I realize that this problem has already been solved, but I got the solution in a different way and it looks a little different but plugging it in to wolfram alpha leads me to believe the previous correct solutions and this one are the same.
Background: (Note: this paragraph is how I came about the problem and you can skip it if you just want to see the answer.) This idea came up in Anthony Zee's "Quantum Field Theory in a Nutshell" book (pg 15). If you want to find certain moments of a multivariable (what is close to) Gaussian distribution, you can use a technique called Wick contraction to easily get the answer. Here is what I mean:
$$ = C_{ij}$$ $$ = C_{ij} C_{kl} + C_{il} C_{jk} + C_{ik} C_{jl}$$ $$...$$ The definition of $$C$$ does not matter, but we see we are basically pairing off the indices into however many different configurations we can. I was curious if there was a general formula for this, and here is how I got the answer. | {
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Answer: The case of 4 indices is easy because you can just list them out like I did above--there are three. Now say we have six indices: {i, j, k, l, m, n}. Consider pairing off the first two indices to {i, j}. Now there are four more indices left to pair off, but we already know there are three different configurations for four indices. I.e. for the initial pairing of {i, j}, there are 3 "sub-configurations." Now notice there are 5 different starting pairs: {i, j}, {i, k}, {i, l}, {i, m}, {i, n}. So 6 indices gives us 5 x 3 = 15 different pairings. If you go to 8 indices, there are 7 different starting pairs which would each leave 6 indices left, which we just found out has 15 options. Therefore you have 7 x 5 x 3 options. Continuing this logic, you can see we clearly have the following:
4 indices: 3
6 indices: 5 x 3
8 indices: 7 x 5 x 3
10 indices: 9 x 7 x 5 x 3
...
Or more concisely (2n-1)!! different pairings (where !! is the double factorial). | {
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Does the "field" over which a vector space is defined have to be a Field?
I was reviewing the definition of a vector space recently, and it occurred to me that one could allow for only scalar multiplication by the integers and still satisfy all of the requirements of a vector space.
Take for example the set of all ordered pairs of integers. Allow for scalar multiplication over the integers and componentwise vector addition as usual. It seems to me that this is a perfectly well-defined vector space.
The integers do not form a Field, which begs the question: Is there any reason that the "field" over which a vector space is defined must be a mathematical Field? If so, what is wrong with the vector field I attempted to define above? If not, what are the requirements for the scalars? (For instance, do they have to be a Group - Abelian or otherwise?)
• Not mentioned in the other answers yet: A vector space over the integers is properly called a "$\mathbb{Z}$-module," and every such module is equivalent to an abelian group, in your case the group $\mathbb{Z}^2$ with addition defined componentwise. The integer scalar tells you how many times to add up an element of the group. Jul 2 '17 at 22:10
• So basically whether a structure of this sort is a vector space or a module depends on whether its scalars form a ring or a field. So it's largely a definition thing, then? Vector space have scalar fields, and modules have scalar rings. Jul 2 '17 at 22:21
• +1 for a great question, thanks for posting! Nov 15 '20 at 19:47
If you pick the scalars from a general ring instead of insisting on a field (in particular, $\mathbb Z$ is a ring), you get a structure known as a module rather than a vector space. | {
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Modules behave like vector spaces in certain respects, but there are also points where they are not at all as well-behaved as vector spaces. For example, a module does not necessarily have a basis, or even a well-defined dimension. This makes matrices less useful for understanding modules than they are for vector spaces. (You can still have matrices with entries in a ring; they just don't tell you everything about linear maps between the modules anymore).
• So, just to be clear, what I'm hearing is that if it has the right properties and its scalars form a Field, then it is called a vector space. But if it has basically the same properties only its scalars form a Ring, then it is called a module? Do I have that right? Jul 2 '17 at 22:15
• @Geoffrey: Correct -- the axioms are identical. Jul 2 '17 at 22:20
• @Geoffrey basically, if you it's not a field, theorems about vector spaces don't apply Jul 3 '17 at 12:52
These things are studied: they are called modules over the ring instead of vector spaces.
The main difference is that the elements of general modules do not allow a lot of the geometric intuition we have for vector spaces, so we still retain the traditional term "vector space" because it is still a useful term.
So, modules over fields (and also noncommutative fields) are called vector spaces.
While the other answers (and comments) implicitly address the question stated in the title of the OP, I thought it may be useful to include an explicit answer, as well.
Does the “field” over which a vector space is defined have to be a Field?
Yes, a vector space is defined over a field; i.e. if the scalars do not refer to a field, the resulting object is not by definition a vector space. | {
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For completeness: as pointed out in the other answers and comments, there are objects with analogous definitions, in the case that the scalars belong to a ring (as in the example provided in the OP), and the other axioms are met, the resulting object is called a "module." As the subsection on modules on the Wikipedia vector space page says (emphasis added):
Modules are to rings what vector spaces are to fields: the same axioms, applied to a ring R instead of a field F, yield modules. The theory of modules, compared to that of vector spaces, is complicated by the presence of ring elements that do not have multiplicative inverses. For example, modules need not have bases, as the Z-module (that is, abelian group) Z/2Z shows; those modules that do (including all vector spaces) are known as free modules. Nevertheless, a vector space can be compactly defined as a module over a ring which is a field, with the elements being called vectors. Some authors use the term vector space to mean modules over a division ring.[105] The algebro-geometric interpretation of commutative rings via their spectrum allows the development of concepts such as locally free modules, the algebraic counterpart to vector bundles. | {
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# If $g\circ f$ is injective and $f$ is surjective then $g$ is injective
Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions, prove that if $g\circ f$ is injective and $f$ is surjective then $g$ is injective.
I need advise or correction if something is incorrect with my proof. Thank you beforehand.
We must show that $g$ is injective, i.e for $x,y\in B, g(x)=g(y)\implies x=y$
Let $x,y\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \in A$ such that $f(a)=x$ and $f(b)=y$
$\implies g(f(a))=g(f(b))$
$\implies g\circ f(a)=g\circ f(b)$
$\implies a=b$ (by injectivity of $g\circ f$)
$\implies f(a)=f(b)$
$\implies x=y$
Would appreciate any correction in proof writing also!
• This is correct. – Mr.Fry Dec 14 '13 at 6:08
• I was suspicious with the last two implications, didn't know if they were true but it seems there is no problem. Thank you Faraad! – AndreGSalazar Dec 14 '13 at 6:24
• @AndrewGSM, it is a general principle that if $a=b$, then $f(a)=f(b)$, so long as $f$ is a function and both $a$ and $b$ are elements of the domain of $f$. – goblin Dec 14 '13 at 11:25
Your proof is correct. I myself would prove it exactly the same. But, I think it's useful to know more than one way, so here is an alternative solution. It's not profoundly different, but I think it's still worth mentioning.
I'm assuming that $A$ is nonempty (and, since there is a map from $A$ to $B$, $B$ is also nonempty). When $A$ is empty there's not much to prove.
The solution uses left and right inverses. A function with non-empty domain is injective iff it has a left inverse, and a function is surjective iff it has a right inverse. | {
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So, we know that $g\circ f$ has a left inverse $h:C \to A$ and $f$ has a right inverse $k: B \to A$. We want to show that $g$ has a left inverse. Just observe that $$(f \circ h) \circ g = (f \circ h) \circ g \circ (f \circ k) = f \circ (h \circ g \circ f) \circ k = f \circ \mathrm{id}_A \circ k = f \circ k = \mathrm{id}_B,$$ so $(f \circ h)$ is a left inverse for $g$. It follows that $g$ is an injection.
PS: this solution is actually worse than your original one, because this one relies on the axiom of choice (it is used when we say that surjectivity is equivalent to having a right inverse). But it is good in the sense that we don't look at particular elements and manipulate maps as "opaque" objects.
• You write: "a function is injective iff it has a left inverse." This isn't quite right; it should be "a function is injective iff its domain is empty, or it has a left inverse." – goblin Dec 14 '13 at 11:27
• @user18921 You are right. Nice catch! I've made the correction. – Dan Shved Dec 14 '13 at 12:07
• If you weaken $f$ having a right inverse to $f$ being epic, is it still possible to show that $g$ is monic? I haven't been able to find a way to do it, but I also don't know that it can't be done (especially if you use products/coproducts). – dfeuer Dec 14 '13 at 18:52
• Another idea: although AC is needed to prove that an onto mapping has a right inverse, it's not necessary to show that a one-to-one and onto mapping has an inverse. – dfeuer Dec 14 '13 at 19:13
I need advise or correction if something is incorrect with my proof.
Would appreciate any correction in proof writing also!
To this, I would respond: its good to read different people's writing just for style. So here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions.
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A few noteworthy points:
• You may prefer to write function arrows "backwards", as in $f : B \leftarrow A.$ See below.
• A fraction line can be used to mean "implies," see below.
• I prefer ending sentences without a big mass of symbols, using phrases like "as follows" and "below," and then putting the symbols immediately afterwards. See below.
• The word "fix" is a nice alternative to "let" when the latter has the right "basic meaning" but doesn't work grammatically. See below.
• If you're going to have a sequence of implications, I'd suggest making it as long as possible, and omitting the symbol $\implies.$ See below.
With that said, here's the proof:
Proposition. Let $g : C \leftarrow B$ denote a function and $f : B \leftarrow A$ denote a surjection. Then whenever $g \circ f$ is injective, so too is $g$.
Proof. Assume that $g \circ f$ is injective, and fix $b,b' \in B.$ The following implication will be proved. $$\frac{g(b)=g(b')}{b=b'}$$
Since $f$ is surjective, begin by fixing elements $a,a' \in A$ satisfying the equations immediately below.
$$b = f(a),\;\; b'=f(a')$$
Then each statement in the following sequence implies the next.
1. $g(b)=g(b')$
2. $g(f(a)) = g(f(a'))$
3. $(g \circ f)(a) = (g \circ f)(a')$
4. $a=a'$
5. $f(a)=f(a')$
6. $b=b'$.
Here is alternative method
note that : $$g\circ f \mbox{ injective } \implies f \mbox{ injective }$$ we have :
• $f \mbox{ is injective and surjective } \implies f \mbox{ bijective (one-to-one correspondence) }$
Since $f$ is a bijection, it has an inverse function $f^{-1}$ which is itself a bijection.
• $f^{-1} \mbox{is bijective} \implies f^{-1} \mbox{ injective }$
• $$\begin{cases} g\circ f \mbox{ injective } & \\ f^{-1} \mbox{ injective } & \\ f^{-1}(B)\subset A &\\ \end{cases} \implies g\circ f \circ f^{-1} \mbox{ injective}$$ | {
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• Since \forall x\in B \qquad \begin{align} (g\circ f)\circ f^{-1}(x)&=g\circ (f\circ f^{-1})(x)\\ &=g\circ {\rm id}_{B}(x)\\ &=g({\rm id}_{B}(x))\\ (g\circ f)\circ f^{-1}(x)&=g(x)\\ \end{align} then $$(g\circ f)\circ f^{-1}=g$$ since $(g\circ f)\circ f^{-1}$ injective then $g$ is injective
I've proved it on my own like this:
Pick two arbitrary elements of $B$, $y_1$ and $y_2$, with $g(y_1)=g(y_2)$. Since $f$ is surjective, $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1,x_2 \in A$. Then $g(f(x_1))=g(f(x_2))$. Since $f \circ g$ is injective, $x_1=x_2$, and so $f(x_1)=f(x_2)$, or $y_1=y_2$. Finally, $g$ is injective. | {
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• Seems to be exactly the same proof as the one in the question and in goblin's answer. Please consider whether you're contributing something new before answering, particularly given that (1) this question already has an accepted answer, and (2) the question is 3 years old. – epimorphic Feb 3 '17 at 23:44
• Of course I'm contributing something new, and I always consider it. My proof doesn't look like the others at all... And ironically, two of the other answers recommended OP reading different styles of proofs... So yeah, I'm contributing to future people who stumble here looking for help. And, also, the three answers are basically "your proof is correct, here's an alternative proof/how I did it". Also, are you a moderator? – anon Feb 4 '17 at 15:28
• The two answers by others not prefaced with "here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions" each have a different sequence of implications, not just a different "style". That's what the "more than one way" in the first answer usually means, and what "I've proved it on my own like this" seems to suggest to me. I'd be willing to reverse my vote if you would explicitly say something like "here's a more prose-like style". – epimorphic Feb 4 '17 at 19:08
• To answer your last question: Primary moderation on this site is provided by the community. IIRC I found your answer through the review queues, where the only posts visible are this answer, the question, and any attached comments. And here's an answer to an old question that claims "I've proved it on my own like this" and yet looks logically identical to the proof in the OP... – epimorphic Feb 4 '17 at 19:08 | {
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"url": "https://math.stackexchange.com/questions/606257/if-g-circ-f-is-injective-and-f-is-surjective-then-g-is-injective"
} |
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# There are 8 teams in a certain league and each team plays
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There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
[Reveal] Spoiler: OA
Last edited by Bunuel on 17 Jun 2012, 03:56, edited 1 time in total.
Edited the question and added the OA.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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17 Jun 2012, 03:57
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sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64 | {
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"url": "https://gmatclub.com/forum/there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html"
} |
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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14 Nov 2012, 06:23
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
I would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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15 Nov 2012, 23:27
Hi Bunnel,
I would also like to learn this approach. Can u help me?
Sree
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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16 Nov 2012, 04:19
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Sachin9 wrote:
Bunuel wrote:
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
I would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach.
Well the game is played by 2 teams. How many games are needed if there are 8 teams and each team plays each of the other teams exactly once? The number of games will be equal to the number of different pairs of 2 teams we can form out of 8 teams (one game per pair). How else?
Similar questions to practice:
how-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html
if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html
how-many-different-handshakes-are-possible-if-six-girls-129992.html
15-chess-players-take-part-in-a-tournament-every-player-55939.html
there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html
if-each-participant-of-a-chess-tournament-plays-exactly-one-142222.html
Hope it helps.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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18 Nov 2012, 21:09
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These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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28 Dec 2012, 06:59
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sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
$$=\frac{8!}{2!6!}=4*7 = 28$$
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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it is not easy and is harder if we are on the test date.
there are 8 team
to take out 2 teams, IF ORDER MATTERS we have 8*7
but in fact order does not matter
8*7/2=28
princeton gmat book explain this point wonderfully.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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16 May 2013, 01:13
Bunuel wrote:
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
Hi Bunuel,
I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?
Thanks a lot.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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16 May 2013, 04:33
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mywaytomba wrote:
Bunuel wrote:
The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.
Hi Bunuel,
I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?
Thanks a lot.
Check combinatorics chapter of Math Book for theory: math-combinatorics-87345.html
Also check some questions on combinations to practice:
DS: search.php?search_id=tag&tag_id=31
PS: search.php?search_id=tag&tag_id=52
Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html (there are some about permutation too)
Hope it helps.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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17 May 2013, 21:18
pranav123 wrote:
These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
That's one of those tips that can make my life easier. I'm book marking this page. Thanks.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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26 Dec 2013, 23:04
Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played? | {
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"tags": null,
"url": "https://gmatclub.com/forum/there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html"
} |
Then is 28*2 the correct approach?
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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27 Dec 2013, 03:08
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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12 Sep 2014, 06:36
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
total 8 teams & each game by 2 pair then 8C2
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There are 8 teams in a certain league and each team plays [#permalink]
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26 Nov 2014, 02:14
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SreeViji wrote:
Hi Bunnel,
I would also like to learn this approach. Can u help me?
Sree
Hey SreeViji,
The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2}
To understand that we just have to think that each of the 8 team plays against 7 other (8x7) but they play each team exactly once so we divide the total by 2.
We divide by 2 because "TEAM A VS TEAM B" is the same as "TEAM B VS TEAM A"
So we end up with \frac{8x7}{2}= 28
If you still have trouble with combination and permutation check out this website it's well done,
http://www.mathsisfun.com/combinatorics ... tions.html
hope it helps.
Last edited by quentin.louviot on 13 Jan 2015, 07:51, edited 2 times in total.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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26 Nov 2014, 02:57
If there are n teams need to play exactly once ,then they play with (n-1) teams but as they are playing together then, n(n-1)/2, which means nC2
So 8*7/2 =28.
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There are 8 teams in a certain league and each team plays [#permalink]
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12 Jan 2015, 12:48
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I also used a table to do this, like that:
1_2_3_4_5_6_7_8
1_1_1_1_1_1_1_1
2_2_2_2_2_2_2_2
3_3_3_3_3_3_3_3
4_4_4_4_4_4_4_4
5_5_5_5_5_5_5_5
6_6_6_6_6_6_6_6
7_7_7_7_7_7_7_7
8_8_8_8_8_8_8_8
Then you delete the same team pairs: e.g. 1-1, 2-2, 3-3 and then 2-1 (because you have 1-2), 3-2 (because you have 2-3). After you cross out the first 2 columns you then see that you cross out everything from the diagonal and below. The remaining is 28.
However, the 8!/2!*6! approach is better, because if you have many numbers the table will take forever to draw. In case there is sth similar though and your brain gets stuck, use the table...
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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12 Jan 2015, 14:46
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Hi All,
Using the Combination Formula IS one way to approach these types of questions, but it's not the only way. Sometimes the easiest way to get to a solution on Test Day is to just draw a little picture and keep track of the possibilities....
Let's call the 8 teams: ABCD EFGH | {
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"url": "https://gmatclub.com/forum/there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html"
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Let's call the 8 teams: ABCD EFGH
We're told that each team plays each other team JUST ONCE.
A plays BCD EFGH = 7 games total
Team B has ALREADY played team A, so those teams CANNOT play again...
B plays CD EFGH = 6 more games
Team C has ALREADY played teams A and B, so the following games are left...
C plays D EFGH = 5 more games
At this point, you should notice a pattern: the additional number of games played is reduced by 1 each time. So what we really have is:
7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 28 games played
[Reveal] Spoiler:
C
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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19 May 2015, 07:31
pranav123 wrote:
These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
Hi,
I don;t think we need to count the half spaces. with half space count is 36.
without half space - count: 28.
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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Re: There are 8 teams in a certain league and each team plays [#permalink]
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20 May 2015, 03:10
2
KUDOS
Expert's post
onedayill wrote:
pranav123 wrote:
These type of problems can be solved with a simple diagram.
1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.
Hi,
I don;t think we need to count the half spaces. with half space count is 36.
without half space - count: 28.
Dear onedayill
You're right!
The boxes along the diagonal (these are the boxes that contribute to half spaces) represent a team playing with itself. Since that is not possible, these boxes should not be included in the counting.
I noticed in the thread above that a few students had doubts about the expression 8C2. If any of the current students too have such a doubt, here's how this question could be solved visually:
There are 7 ways in which Team 1 can play with another team.
Similarly, there are 7 ways for each of the 8 teams to choose its playing opponent.
But it's easy to see that the red zone is essentially a duplication of the blue zone. For example, (Team 1 playing with Team 2) is the same case as (Team 2 playing with Team 1)
So, the correct answer will be: 8(that is, the number of teams)*7(that is, the number of ways in which each team can choose its playing opponent)/2 = 28
Hope this was useful!
Best Regards
Japinder
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# Necklace combinations with three group of beads
I have a hard question about a way how many different necklaces can be made.
Suppose that we have the following restrictions:
1. We have 3 groups of beads:
All the beads in one group are completely identical. This means that if you put two triangle beads next to each other and then switch their positions this counts as one necklace because the beads are identical
1. Necklaces are identical if they are identical under symmetric operations just as rotate them (𝑟) or turning them around (𝑠).
So if we have a necklace ordered in one way and we rotate it 180 deg or just flip a side this is count as one necklace.
1. We need to use all the 18 beads in each and every new necklace. We can not create a necklace from 17, 16 or less than 18 beads.
I read all the topics here but could not find a question about a group of identical beads. I also read Burnside lemma and Pólya_enumeration_theorem and Necklace_(combinatorics) in wikipedia, but could not find a way how to solve this and what is the correct answer.
From Burnside lemma, I found that the answer should be 57, but is this correct?
I used directly the formula from Burnside lemma, but it does not seem quite right for me, because I do not take into account that the three groups are with different numbers of beads.
$$\frac{1}{24} * (n^6 + 3 * n^4 + 12 * n^3 + 8 * n^2)$$
where n is 3 from three groups.
$$\frac{1}{24} * (3^6 + 3 * 3^4 + 12 * 3^3 + 8 * 3^2) = 57$$
However, as I said earlier despide the fact that the result looks some kind realisting I am not sure that this is the right answer, because I do not use in the formula that we have 4 triangle, 6 square and 8 circle beads.
It looks like Pólya enumeration theorem weighted version is the thing that I need. However, I am not sure how to get to the right answer | {
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• Welcome to MSE! Pls show your work for arriving at $57$ so someone can check its correctness. You can do that by editing your question and e.g. putting your work at the end. May 6 '20 at 20:53
• Thanks @antkam, for the answer. I modified the questions and added the formula that I used and a short description why I think that despide the fact that the result looks like a good number I think that this is not quite right. May 7 '20 at 5:54
• May 7 '20 at 16:19
• Does this answer your question? Circular permutations with indistinguishable objects May 7 '20 at 16:19
• I think you should be using the word "bracelet" and not "necklace". The answer in the link provided by @Vepir gives an answer to your problem for necklaces, which means objects with a cyclic symmetry group, when you want a dihedral symmetry group. This being said you should be able to follow the same line of argument to find the formula you need. May 11 '20 at 20:04
I manage to answer the question and this is the process that I followed:
I consider the 18-bead necklace in the first part of the problem. Here are the eighteen rotations expressed in cycle form where we assume that the slots are numbered from 1 to 18 in clockwise order. The first is the identity (e: no rotation) and the second is the generator g—a rotation by a single position which, when repeated, generates all the elements of the group:
$$e = g^0 \text{ = (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)}$$
$$g^1 \text{ = (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18) }$$
$$g^2 \text{= (1 3 5 7 9 11 13 15 17) (2 4 6 8 10 12 14 16 18)}$$
$$g^3 \text{= (1 4 7 10 13 16) (2 5 8 11 14 17) (3 6 9 12 15 18)}$$
$$g^4 \text{= (1 5 9 13 17 3 7 11 15) (2 6 10 14 18 4 8 12 16)}$$
$$g^5 \text{= (1 6 11 16 3 8 13 18 5 10 15 2 7 12 17 4 9 14)}$$
$$g^6 \text{= (1 7 13) (2 8 14) (3 9 15) (4 10 16) (5 11 17) (6 12 18)}$$
$$g^7 \text{= (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)}$$ | {
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$$g^7 \text{= (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)}$$
$$g^8 \text{= (1 9 17 7 15 5 13 3 11) (2 10 18 8 16 6 14 4 12)}$$
$$g^9 \text{= (1 10) (2 11) (3 12) (4 13) (5 14) (6 15) (7 16) (8 17) (9 18)}$$
$$g^{10} \text{= (1 11 3 13 5 15 7 17 9) (2 12 4 14 6 16 8 18 10)}$$
$$g^{11} \text{= (1 12 5 16 9 2 13 6 17 10 3 14 7 18 11 4 15 8)}$$
$$g^{12} \text{= (1 13 7) (2 14 8) (3 15 9) (4 16 10) (5 17 11) (6 18 12)}$$
$$g^{13} \text{= (1 14 9 4 17 12 7 2 15 10 5 18 13 8 3 16 11 6)}$$
$$g^{14} \text{= (1 15 11 7 3 17 13 9 5) (2 16 12 8 4 18 14 1 6)}$$
$$g^{15} \text{= (1 16 13 10 7 4) (2 17 14 11 8 5) (3 18 15 12 9 6)}$$
$$g^{16} \text{= (1 17 15 13 11 9 7 5 3) (2 18 16 14 12 10 8 6 4)}$$
$$g^{17} \text{= (1 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2)}$$
After that I found the GCD for all cycle form's with and group them in a table:
| Cycle length | Permutations | GCD with 18 |
| 1 | $$g^0$$ | GCD(0, 18)=18 |
| 2 | $$g^9$$ | GCD(9, 18)=9 |
| 3 | $$g^6$$, $$g^{12}$$ | GCD(6, 18)=GCD(12, 18)=6 |
| 6 | $$g^3$$, $$g^{15}$$ | GCD(3, 18)=GCD(15, 18)=3 |
| 9 | $$g^2$$, $$g^4$$, $$g^8$$, $$g^{10}$$, $$g^{14}$$, $$g^{16}$$ | GCD(2, 18)=GCD(4, 18)=GCD(8, 18)=GCD(10, 18)=GCD(14, 18)=GCD(16, 18)=2 |
| 18 | $$g^1$$, $$g^5$$, $$g^7$$, $$g^{11}$$, $$g^{13}$$, $$g^{17}$$ | GCD(1, 18)=GCD(5, 18)=GCD(7, 18)=GCD(11, 18)=GCD(13, 18)=GCD(17, 18)=1 |
We have 18 permutations for rotation and lets name cycle 1 with $$f_1$$, cycle 2 with $$f_2$$ .. cycle n with $$f_n$$
Than the formula for Cycling index is:
$$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1}{18}$$
If we solve all the possible necklaces with three colors the result should be (for the moment we do not solve for the three colors with 4, 6 and 8 beads in respective groups):
$$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1}{18} = \text{21 524 542}$$ | {
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$$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1}{18} = \text{21 524 542}$$
From here because turn around is allowed we need to add and the necklace(bracelet if we follow the right terms) is with even beads we should add the symethric turn arounds.
$$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1 + 9f_1^2f_2^8 + 9f_2^9}{2 * 18}$$
and again for three colors without including the different weight:
$$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1 + 9 * 3^8 + 9 * 3^9}{2 * 18} = \text{10 781 954}$$
For the moment we have all the possible necklaces and bracelets with three colors. However, in order to find the necklaces and bracelets with three colors (4 reds, 6 greens and 8 blues) we need to replace:
$$f_1 = (x + y + z)$$
$$f_2 = (x^2 + y^2 + z^2)$$
$$f_3 = (x^3 + y^3 + z^3)$$
$$f_6 = (x^6 + y^6 + z^6)$$
$$f_9 = (x^9 + y^9 + z^9)$$
$$f_{18} = (x^{18} + y^{18} + z^{18})$$
and if we replace in the formula it becomes:
$$\frac{(x + y + z)^{18} + (x^2 + y^2 + z^2)^9 + 2(x^3 + y^3 + z^3)^6 + 2(x^6 + y^6 + z^6)^3 + 6(x^9 + y^9 + z^9)^2 + 6(x^{18} + y^{18} + z^{18}) + 9(x + y + z)^2(x^2 + y^2 + z^2)^8 + 9(x^2 + y^2 + z^2)^9}{36}$$
Then we need to find which expressions can expand to $$x^4y^6z^8$$.
After then by using multinominal coeficient I managed to calculate the following results
9 189 180
1260
11 340
11 340
Then I sum all of them and divide them to 36. This gives me the answer of 255 920 which is the answer of the question. We can create 255 920 bracelets with 4 red 6 green and 8 blue beads. | {
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# Is $V$ isomorphic to direct sum of subspace $U$ and $V/U$?
Given a vector space $$V$$ and a subspace $$U$$ of $$V$$. $$V \cong U \oplus(V/U)$$ Does the above equation always hold? Where $$\oplus$$ is external direct sum. For finite dimensional vector space $$V$$, here is my attemp of prove:
Let dimension of $$U$$ be m, dimension of $$V$$ be $$n$$. Find a basis of $$U$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m}\}$$ and extend it to a basis for $$V$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m, v_1, \cdots,v_{n-m} } \}$$.
For every vector $$\mathbf{x} \in V$$, we can write $$\mathbf{x}= c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m} + d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$ uniquely. Define a linear map $$T$$ as $$T(\mathbf{x})=(c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m}, [d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}])$$ ,where $$[]$$ is used to express the equivalent class. We claim $$T$$ is an isomorphism.
Surjectivity is obvious. As for injectivity,
if $$T(\mathbf{x})=(\mathbf{0},[\mathbf{0}])$$, then $$c_1 \mathbf{u_1}+ \cdots+ c_m\mathbf{u_m}= \mathbf{0}$$
$$\Rightarrow c_1=0, c_2=0, \cdots,c_m=0$$
$$\Rightarrow x=d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$
Since $$[d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}] = [\mathbf{0}]$$, we have $$(d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}-\mathbf{0})\in U$$, which means $$d_1=0, d_2=0, \cdots,d_{n-m}=0$$
$$\Rightarrow \mathbf{x}=\mathbf{0}$$, so $$T$$ is injective.
Is the above proof correct? Does this mean $$V \cong \ker F \oplus (V/ \ker F) \cong \ker F \oplus\mathrm{im}F$$ for any linear map $$F$$, because $$\ker F$$ is a subspace of $$V$$ ?
The final question is about how should I prove it when the dimension of $$V$$ is infinite? | {
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The final question is about how should I prove it when the dimension of $$V$$ is infinite?
It is true that for every finite dimensional vector space $$V$$ with $$U$$ a vector subspace that $$V \cong U \oplus (V / U)$$ I think your proof is essentially correct. And yes it is true that for $$T: V \rightarrow W$$ any linear that we have $$V \cong \operatorname{ker}(T) \oplus \operatorname{Im}(T)$$ The rank-nullity theorem is a direct consequence of this.
In more technical language, we say that every "short exact sequence" of finite dimensional vector spaces over a field $$k$$ $$\textit{splits}$$. What this means is that if $$T : U \rightarrow V$$, $$S : V \rightarrow W$$ are linear maps such that $$T$$ is injective, $$\operatorname{ker}(S) = \operatorname{Im}(T)$$, and $$S$$ is surjective, then $$V \cong U \oplus W$$.
Note then that this directly gives us our result since if $$U$$ is a subspace of $$V$$, then the inclusion map $$\iota : U \rightarrow V$$ and projection map $$\pi : V \rightarrow (V / U)$$ set up exactly a short exact sequence.
In terms of whether or not this extends to infinite dimensional vector spaces, the result does hold again (assuming the axiom of choice), and the proof is essentially the same. All your proof relies on is the ability to extend a basis of a subspace to a basis of your entire space. We can do this with the axiom of choice. | {
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• I am wondering if the following statement is false for infinite dimensional spaces. "If $V$ is a vector space and $U$ is a subspace of $V$, then there exists another subspace of $V$ called $U^\perp$ such that every element of $V$ can be uniquely expressed as the sum of an element from $U$ with an element from $U^\perp$." (I am thinking a counterexample would be if $U$ was the set of real number sequences with finite support (i.e. eventually zero) and $V$ is the set of all real number sequences.) – irchans Jan 3 at 19:03
• @irchans If we take the axiom of choice, then every subspace of a vector space has a direct sum complement (what you call the perpendicular space, but this language is typically reserved for a space equipped with some bi-linear form). The proof is pretty simple. Let $V$ be a $k$-vector space, with $U$ a vector subspace. Let $\mathcal{B}_{U}$ be a basis for $U$ and extend it to a basis $\mathcal{B}_{V}$ (using the axiom of choice) for $V$. Then let $W = \operatorname{Span}_{k}\left( \mathcal{B}_{V} \backslash \mathcal{B}_{U} \right)$. Then $V = U \oplus W$. – Adam Higgins Jan 3 at 19:13
• @irchans Perhaps the reason you think that your example is a counter example is because of the $\textit{weirdness}$ of bases of infinite dimensional vector spaces. Notice that a subset $S$ of a vector space $V$ is said to be a basis if and only if every element $v \in V$ can be written as a $\textbf{finite}$ linear combination of the elements of $S$, and that there is no finite non-trivial linear relation amongst the elements of $S$. – Adam Higgins Jan 3 at 19:19
• Thank you very much ! – irchans Jan 3 at 19:27
• This set of notes seems relevant math.lsa.umich.edu/~kesmith/infinite.pdf – irchans Jan 3 at 19:38 | {
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Your proof works. The answer to your second question is yes, that is true. For an infinite dimensional vector space, take any linear map $$F: V -> W$$. Then $$U = \ker F$$ is a subspace of $$V$$. Note that we have a short exact sequence (if you don't know what that means, don't worry, the explanation is coming) $$0\to U\to V\to V/U\to 0$$
(That is, there's an injective map $$U \to V$$ (inclusion, I'll call it $$i$$) and a surjective map $$V \to V/U$$ (the quotient map, I'll call it $$q$$) such that the image of the injection is the kernel of the surjection).
But there's also a surjective map $$V \to U$$ (projection onto $$U$$, I'll call it $$p$$), and note that for any $$u \in U$$, $$pi(u) = u$$ (since $$p$$ fixes $$u$$).
Now, we're going to show that $$V$$ is the (internal) direct sum of the kernel of $$p$$ and the image of $$i$$. First, note that it's the sum of the two: for any $$v \in V$$, $$v = (v - ip(v)) + ip(v)$$, $$ip(v)$$ is obviously in the image of $$i$$, and $$p(v - ip(v)) = p(v) - pip(v) = 0$$ (with the last equality being due to our note about $$p$$, since $$p(v)\in U$$.
And further, the intersection is trivial: if $$v \in \ker(p)\cap\mathrm{im}(i)$$ then there is some $$u\in U$$ such that $$i(u) = v$$, and $$pi(u) = p(v) = 0$$, but $$pi(u) = u$$, so $$u = 0$$, hence $$v = 0$$. Thus, $$V = \ker(p)\oplus \mathrm{im}(i)$$.
Now, it's clear that $$\mathrm{im}(i)\cong U$$ since it's the image of $$U$$ under an injective map, so we need only show that $$\ker(p)\cong V/U$$. | {
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For that purpose, since $$q$$ is surjective, for any $$w \in V/U$$, there is some $$v \in V$$ such that $$w = q(b)$$. But since $$V = \ker(p)\oplus\mathrm{im}(i)$$, there are unique $$u \in U$$, $$x \in \ker(p)$$ such that $$v = i(u) + x$$, so $$w = q(b) = q(i(u)+x) = qi(u) + q(x)=q(x)$$ (since the image of $$i$$ is the kernel of $$q$$, in particular $$qi = 0$$). Thus, $$q|_{\ker(p)}: \ker(p)\to V/U$$ is surjective. But also, if $$q(v) = 0$$, and $$v \in \ker(p)$$. But $$\ker(q) = \mathrm{im}(i)$$, and $$p$$ fixes $$\mathrm{im}(i)$$, so the only element that it sends to $$0$$ is $$0$$ itself, so we must have $$v = 0$$, hence $$q|_{\ker(p)}$$ is also injective, so is an isomorphism.
Thus, we have $$V \cong \mathrm{im}(i)\oplus\ker(p) \cong U\oplus (V/U)$$, as required.
This is precisely a special case of the Splitting Lemma, and my proof is essentially just one part of the proof of that, translated: there are easier ways to prove it, but I thought that it would be useful to see it in a more broadly applicable form. | {
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# Query on a Solution to the Problem: $\gcd(5a+2,7a+3)=1$ for all integer $a$.
I wish to show that the numbers $$5a+2$$ and $$7a+3$$ are relatively prime for all positive integer $$a$$.
Here are my solutions.
Solution 1. I proceed with Euclidean Algorithm. Note that, for all $$a$$, $$|5a+2|<|7a+3|$$. By Euclidean Algorithm, we can divide $$7a+3$$ by $$5a+2$$. To have
$$7a+3=(5a+2)(1)+(2a+1)$$ continuing we have,
$$5a+2=(2a+1)(2)+a$$
$$2a+1=(2)(a)+1$$
$$2=(1)(2)+0$$
Since the last nonzero remainder in the Euclidean Algorithm for $$7a+3$$ and $$5a+2$$ is 1, we conclude that they are relatively prime.
Solution 2. Suppose that $$d=\gcd(5a+2,7a+3)$$. Since $$d=\gcd(5a+2,7a+3)$$ then the following divisibility conditions follow:
(1) $$d\mid (5a+2)$$
(2) $$d\mid (35a+14)$$
(3) $$d\mid (7a+3)$$
(4) $$d\mid (35a+15)$$.
Now, (2) and (4) implies that $$d$$ divides consecutive integers. The only (positive) integer that posses this property is $$1$$. Thus, $$d=1$$ and that $$7a+3$$ and $$5a+2$$ are relatively prime.
Here are my questions:
1. Is the first proof correct or needs to be more specific? For instance cases for $$a$$ must be considered.
2. Which proof is better than the other?
Thank you so much for your help. | {
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2. Which proof is better than the other?
Thank you so much for your help.
• I think both approaches are good. Nor are they that different really...in both cases you are trying to find smaller and smaller multiples of $d$.
– lulu
Sep 24 '18 at 15:23
• Thank you very much for the kind comment @lulu. Got a follow up question. By "you are trying to find smaller and smaller multiples of $d$, you mean the process of continously dividing the divisor to remainder so that $r$ decreases? Sep 24 '18 at 15:27
• Is there a typo in (4), i.e. shouldn't 35 be multiplied by $a$?
– user431008
Sep 24 '18 at 15:28
• I meant something less precise than that. Euclid provides a somewhat systematic way to find new numbers that $d$ divides...the second method is less systematic, but faster as you seek out convenient expressions. Working by hand, I prefer the second method...were I trying to automate the process, I'd prefer the systematic approach.
– lulu
Sep 24 '18 at 15:28
• I agree @marmot. Thank you for pointing it out. Sep 24 '18 at 15:29
In the first solution you're not using, strictly speaking, the Euclidean algorithm, but a looser version thereof:
Let $$a$$, $$b$$, $$x$$ and $$y$$ be integers; if $$a=bx+y$$, then $$\gcd(a,b)=\gcd(b,y)$$.
The proof consists in showing that the common divisors of $$a$$ and $$b$$ are the same as the common divisors of $$b$$ and $$y$$.
There is no requirement that $$a\ge b$$ or that $$y$$ is the remainder of the division. Indeed your argument actually has a weakness, because $$7a+3\ge 5a+2$$ only if $$2a\ge-1$$, so it doesn't hold for $$a\le-2$$. But $$7a+3\ge 5a+2$$ is not really needed for the argument.
Since successive application of the statement above show that $$\gcd(2a+1,2)=1$$ and the gcd has never changed in the various steps, you can indeed conclude that $$\gcd(5a+2,7a+3)=1$$.
The second solution is OK as well. | {
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The second solution is OK as well.
You can simplify it by noting that if $$d$$ is a common divisor of $$5a+2$$ and $$7a+3$$, then it divides also $$5(7a+3)-7(5a+2)=1$$
• Thanks for the clarification in solution 1 and by giving a simplified version for solution 2. It is very clear to me now. @egreg Sep 24 '18 at 15:35
Your first proof is correct. I would perhaps complete it, writing that\begin{align}1&=(2a+1)-2a\\&=2a+1-2\bigl(5a+2-2(2a+1)\bigr)\\&=5(2a+1)-2(5a+2)\\&=5\bigl(7a+3-(5a+2)\bigr)-2(5a+2)\\&=5(7a+3)-7(5a+2).\end{align}
Your second proof also works, but it doesn't generalize easily to other situations.
• Thank you very much for completing the proof of solution 1 Prof. I am interested in a case that solution 2 wont work. Thank you Prof. Sep 24 '18 at 15:41
• @JrAntalan Concerning the second proof, I only meant that you have to think about it in a case-by-case basis. Sep 24 '18 at 15:43
• Noted Prof. and Thank you again. Sep 24 '18 at 15:44
Here is a different rendering of the same arguments.
Let $$u=5a+2$$, $$v=7a+3$$. Then $$\pmatrix{ u \\ v} = \pmatrix{ 5 & 2 \\ 7 & 3} \pmatrix{ a \\ 1}$$ and so $$\pmatrix{ a \\ 1} = \pmatrix{ 5 & 2 \\ 7 & 3}^{-1} \pmatrix{ u \\ v} = \pmatrix{ \hphantom- 3 & -2 \\ -7 & \hphantom-5} \pmatrix{ u \\ v}$$ This gives $$1 = -7u+5v = -7(5a+2)+5(7a+3)$$ The key point here is that the matrix has determinant $$1$$ and so its inverse has integer entries. | {
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# For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17 or more
I set out to solve the following question using the pigeonhole principle
Regardless of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more.
My outline
[1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations of the number in the circle, is $3\cdot 55=165$.
[2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triplets, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more.
My query
Could someone polish this into a mathematical proof and also clarify if I did make use of the pigeonhole principle.
• Yes, you used pigeonhole. Should say if all sums are $\le 16$, then the sum is $\le 160$. – André Nicolas May 25 '13 at 4:42
• @AndréNicolas noted and aha , the syntax part is exactly why I asked the question on here. – metric-space May 25 '13 at 4:44
• @nerorevenge Note: You can actually show that the 3 adjacent numbers sum to 18 or more. – Calvin Lin May 25 '13 at 8:37
Yes, you used the Pigeonhole Principle. As a very mild correction, you should say that of all sums of three consecutives are $\le 16$, then the sum is $\le 160$.
The proof (with the small correction) is already fully mathematical. Conceivably you might want to explain the $55$ further. It is clear to you and to most users of this site why $55$, but imagine the reader to be easily puzzled. | {
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The part about "all the permutations" is vague, and technically incorrect. You are finding the sum of a consecutive triple, and summing all these sums. Permutations have nothing to do with it, we are talking about a particular fixed arrangement of the numbers.
Remark: We could use a lot of symbols. Starting at a particular place, and going counterclockwise, let our numbers be $a_1,a_2,\dots,a_{10}$. For any $i$, where $1\le i\le 10$, let $S_i=a_i+a_{i+1}+a_{i+2}$, where we use the convention that $a_{10+1}=a_1$, $a_{10+2}=a_2$, and $a_{9+2}=a_1$.
Then $S_1+S_1+\cdots+S_{10}=165$, since each of $1$ to $10$ appears in $3$ of the $S_i$, and $1+2+\cdots+10=55$.
But if all the $S_i$ are $\le 16$, then $\sum_{i=1}^{10}S_i\le 160$. This contradicts the fact that $\sum_{i=1}^{10}S_i=165$.
I prefer your proof, mildly modified.
• eh, is that all? the reason I'm asking is that, inb the past my so called 'proofs' have been criticised for being too informal – metric-space May 25 '13 at 4:45
• @nerorevenge: I am an advocate of the informal but clear. I hope that someone more formal-minded will add an answer. The only issue is that you perhaps left an easily filled gap or two. – André Nicolas May 25 '13 at 4:53
• I have added another criticism. You should really fix the wording there. – André Nicolas May 25 '13 at 4:58
• I agree.That definitely is vaguely worded. – metric-space May 25 '13 at 4:59
• Presumably the "pair" wording came with the problem. But it looks funny to call a trio a pair. – André Nicolas May 25 '13 at 5:23
We will show something stronger, namely that there exists 3 adjacent numbers that sum to 18 or more.
Let the integers be $\{a_i\}_{i=1}^{10}$. WLOG, $a_1 = 1$. Consider
$$a_2 + a_3 + a_4, a_5 + a_6 + a_7, a_8 + a_9 + a_{10}$$
The sum of these 3 numbers is $2+3 +\ldots + 10 = 54$. Hence, by the pigeonhole principle, there exists one number which is at least $\lfloor \frac{54}{3} \rfloor = 18$. | {
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I leave it to you to show that there is a construction where no 3 adjacent numbers sum to 19 or more, which shows that 18 is the best that we can do.
• definitely a stronger proof ,and I have admit it sure is nice, I wonder why cut-the-knot.org didn't ask people to solve for what you proved. – metric-space May 25 '13 at 16:45
• @nerorevenge I used this as a problem on Brilliant.org, and most could only get to 17, and didn't understand why 18 must be true. The 18 case arises quite easily by ignoring the small values which would otherwise reduce our sum. – Calvin Lin May 25 '13 at 23:37 | {
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# To find number of real roots
Consider the equation $$x^5-5x=c$$ where c is a real number.
Determine all c such that this equation has exactly 3 real roots.
I know that between consecutive real roots of $$f$$ there is a real root of $$f'$$. Now $$f'$$ in this case is $$5x^4-5$$ which always has two real roots. So the claim should be true for all c.
But I KNOW IT IS NOT TRUE. Where am I messing up?
• Note that while it is necessary for $f'$ to have two real roots (in order for $f$ to have exactly three real roots), it is not sufficient. Of course an odd degree polynomial will always have one real root. – hardmath Dec 12 '19 at 16:12
• ok so how do I find all such c? – Angry_Math_Person Dec 12 '19 at 16:14
• Consider the graph of $p(x) = x^5 - 5x$. Changing the constant $c$ in your equation amounts to moving a horizontal line up or down across this graph. – hardmath Dec 12 '19 at 16:15
Yes, between any two roots of $$f$$, there is a root of $$f'$$. However, just because $$f'$$ has a root, that doesn't mean that $$f$$ has a root on either side. Consider $$f(x)=x^2+1$$.
As for solving this problem, the derivative has only two roots, so we can at most have three roots. For some values of $$c$$ we have three roots, for some values of $$c$$ we have a single root, and for exactly two values of $$c$$, there are two roots. The three-root region is exactly the interval between the two two-root values of $$c$$.
And finding the values of $$c$$ that gives two roots is easier than one might think. They happen exactly when one root of $$f$$ coincides with a root of $$f'$$. So find the roots of $$f'$$, and find the values of $$c$$ that make each of them a root of $$f$$, and you have found the interval of $$c$$-values that gives three roots.
As you know, the given equation has extrema at $$x=\pm1$$. These correspond to values of the polynomial
$$1-5-c$$ and $$1+5-c$$ (the RHS was moved to the left). | {
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$$1-5-c$$ and $$1+5-c$$ (the RHS was moved to the left).
Hence the polynomial will grow from $$-\infty$$, reach the maximum, then the minimum and continue growing to $$\infty$$. There are three roots when $$0$$ is in the range $$(-4-c,6-c)$$.
Where am I messing up? Just look at a graph where it fails.
$$x^5-5 x -5$$
I know that between consecutive real roots of f there is a real root of f′. Now f′ in this case is 5x4−5 which always has two real roots. So the claim should be true for all c.
$$A \implies B$$ does not mean $$B \implies A$$.
The two real roots of $$5x^4 - 5$$ are the two roots at $$x = \pm 1$$.
If $$x^5 - 5x=c$$ has three roots then they will be at $$x < -1; -1 < x < 1;$$ and at $$x > 1$$ by your condition.
But there won't be three real roots if there is no root for any $$x< -1$$, or no root between $$-1$$ and $$1$$, or no root for any $$x < -1$$.
$$x=\pm 1$$ are extreme points and if one, the max, is $$>0$$ and the other $$<0$$ then there will be three real roots. But if both are "on the same side of $$0$$" there is no root between them and no root to "the other side".
$$x^5-5x -c|_{-1} = 4-c$$ and $$x^5 - 5x -c|1 = -4-c$$ so $$x =-1$$ is a max and $$x = 1$$ is a min.
If $$f(-1) = 4-c \le 0$$ is a max there will be no root for $$x < -1$$ or for $$-1 < x \le 1$$. If $$f(1) = -4-c\ge 0$$ is a min there will be no root for $$x > 1$$ or for $$-1 \le x < 1$$. So if either $$c \ge 4$$ or if $$c\le 4$$ then there are fewer than three real roots. But if $$-4 < c < 4$$ then there will be three. | {
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Alternatively: we know what the shape of an odd polynomial $$x^5 -5x$$ looks like. It's that polynomial curve with a twisty bit in the middle, goes off to infinity as $$x \to \infty$$, goes to negative infinity as $$x \to -\infty$$. It has roots were the x-axis crosses it (or where it crosses the x-axis-- everything is relative). If we shift it up or shift it down by $$c$$ we can force the x-axis to avoid the twisty bits in the middle and have it have only one root. Or we can deliberately shift it so that the x-axis goes smack through the twisty bits and we have a maximum number of roots. So if $$c$$ is between the max and mins we maximize the number of roots and the x-axis goes through the twisty bits. If $$c$$ is beyond the max an mins we've shoved the twisty bits below or above the x-axis and there is only one root. | {
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# Difference between continuity and uniform continuity
Tags:
1. May 7, 2015
### Yunjia
I noticed that uniform continuity is defined regardless of the choice of the value of independent variable, reflecting a function's property on an interval. However, if on a continuous interval, the function is continuous on every point. It seems that the function on that interval must be uniformly continuous. Is this correct? Is there a counterexample for the statement?
2. May 7, 2015
### jbunniii
What you said is almost true. If, on a closed, bounded interval, a function is continuous at every point, then the function is uniformly continuous on that interval.
Counterexample on a non-closed interval: $f(x) = 1/x$ on the interval $(0,1)$.
Counterexample on a closed but unbounded interval: $f(x) = x^2$ on the interval $[0,\infty)$.
3. May 8, 2015
### Svein
Even more general: If a function is continuous at every point in a compact set, it is uniformly continuous on that set.
The proof is simple: Let the compact set be K and take an ε>0. Since the function is continuous at every point x in the set, there is a δx for every x∈K such that |f(w)-f(x)|<ε for every w in <x-δx, x+δx>. Let Ox=<x-δx, x+δx>, then K is contained in $\bigcup_{x\in K}O_{x}$. Since K is compact, it is contained in the union of a finite number of the Ox, say $\bigcup_{n=1}^{N}O_{n}$. Take δ to be the minimum of the δn and |f(w)-f(x)|<ε whenever |w-x|<δ.
4. May 8, 2015
### jbunniii
And here's the proof that a closed, bounded interval is compact, so you can apply Svein's proof. | {
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And here's the proof that a closed, bounded interval is compact, so you can apply Svein's proof.
Let $[a,b]$ be a closed, bounded interval, and let $\mathcal{U}$ be any collection of open sets which covers $[a,b]$. Let $S$ be the set of all $x \in [a,b]$ such that $[a,x]$ can be covered by finitely many of the sets in $\mathcal{U}$. Clearly $a \in S$. This means that $S$ is nonempty and is bounded above (by $b$), so it has a supremum, call it $c$. Since $c \in [a,b]$, it is contained in some $U_c \in \mathcal{U}$, hence there is some interval $(c - \epsilon, c + \epsilon)$ contained in $U_c$. Since only finitely many sets from $\mathcal{U}$ are needed to cover $[a,c - \epsilon/2]$, those sets along with $U_c$ form a finite cover of $[a,c+\epsilon/2]$. This shows that $c \in S$ and moreover, that $c$ cannot be less than $b$. Therefore $c=b$, so all of $[a,b]$ can be covered by finitely many sets in $\mathcal{U}$.
Last edited: May 8, 2015
5. May 8, 2015
### Svein
Excellent! And, of course, since one closed and bounded interval is compact, the union of a finite number of closed and bounded intervals is again compact.
6. May 8, 2015
### jbunniii
I don't think this will work if $w$ and $x$ are not contained in the same $O_n$. I think you need to take $O_x = (x - \delta_x/2, x + \delta_x/2)$ and $\delta$ to be $\min\{\delta_n/2\}$ in order to ensure that $|w-x| < \delta$ implies $|f(w) - f(x)| < \epsilon$.
7. May 8, 2015
### Svein
Possibly. I need to think about that. The crux of the matter is that there is a finite number of intervals that cover K, which means that the minimum of the (finite number of) δ's exist and is greater than 0.
8. May 8, 2015
### HallsofIvy
Here is the fundamental difference between "continuous" and "uniformly continuous":
A function is said to be continuous at a point, x= a, if and only if, given any $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- f(a)|<\epsilon$. | {
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A function is said to be continuous on a set, A, if and only if, given any $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- f(a)|<\epsilon$ for all a in set A.
That is, a function is uniformly continuous on a set A if and only if it is continuous at every point in A and, given $\epsilon> 0$ the same $\delta$ can be used for ever point in A.
9. May 8, 2015
### Svein
Agree. I was sloppy and overlooked the simple fact that that the length of the interval Ox is 2⋅δx.
10. May 13, 2015
### Yunjia
The defition of uniformly continuous is perplexing for me. Is it possible for you to illustrate it with a proof about a function which is not uniformly continuous but continuous on a set so that I can compare the two? Thank you.
11. May 14, 2015
### Svein
OK. Take the function $f(x)=\frac{1}{x}$ on the open interval <0, 1>. It is continuous (and even differentiable) on that interval, but not uniformly continuous.
12. May 17, 2015
### Svein
After using pencil an paper for a bit, I came to the conclusion that I should have used ε/2 and δ/2. But - I recall a proof in "Complex analysis in several variables" that ended up in "... less than 10000ε, which is small when ε is small".
13. May 17, 2015
### HallsofIvy
First note an important point about "uniform continuity" and "continuity" you may have overlooked: we define "continuous" at a single point and then say that a function is "continuous on set A" if and only if it is continuous at every point on A. We define "uniformly continuous" only on a set, not at a single points of a set.
A function is uniformly continuous on any closed set on which it is continuous and so on any set contained in a closed set on which it is continuous.
To give an example of a function that is continuous but not uniformly continuous, we need to look at f(x)= 1/x on the set (0, 1). | {
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To show that it is continuous on (0, 1), let a be a point in (0, 1) and look at $|f(x)- f(a)|= |1/x- 1/a|= |a/ax- x/ax|= |(a- x)/ax|= |x- a|/ax< \epsilon$.
We need to find a number, $\delta> 0$ such that if $|x- a|< \delta$, then $|f(x)- f(a)|> \epsilon$. We already have $|x- a| ax\epsilon$ so we need an upper bound on ax. If we start by requiring that $\delta< a/2$ then $|x- a|< \delta< a/2$ so that $-a/2< x- a< a/2[itex] or [itex]a/2< x< 3a/2$ so an upper bound on ax is $3a^2/2$. If $|x- a|< a/2$ and $|x- a|< 3a/2$ then $|f(x)- f(a)|< |x- a|/ax< |x- a|/(3a^2/2)= 2|x- a|/3a^2$ which will be less than $\epsilon$ as long as $|x- a|< 3a^2\epsilon/2$
So we can take $\delta$ to be the smaller of $a/2$ and $3a^2/\epsilon$. Therefore, 1/x is continuous at any point a in (0, 1) so continuous on (0, 1).
Now the point is that this $\delta$ depends on a. It is a decreasing function of a and will go to 0 as a goes to 0. That's fine for "continuity" but for uniform continuity we must be able to use the same $\delta> 0$ for a given $\epsilon$ no matter what the "a" is and we cannot do that. If the problem were to prove uniform continuity on the set [p, 1), which is contained in the closed set [p, 1], we could use, for any a in that set, the $\delta$ that we get for a= p, the smallest x value in the set. But with (0, 1), the smallest $\delta$ is 0 which we cannot use since we must have $\delta> 0$. | {
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## Linear Algebra And Learning From Data Github | {
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Each point correspondence generates one constraint on F. Complete Linear Algebra for Data Science & Machine Learning 4. My research interests lie in network science, statistical inference, causal inference, information theory, machine learning, data mining, and signal. 065 Linear Algebra and Learning from Data New textbook and MIT video lectures OCW YouTube; 18. Welcome to the 18. A vector regression task is one where the target is a set of continuous values (e. Siefken, J. Getting started with linear algebra. Yes, linear algebra is actually super important in data science. The aim of this set of lectures is to review some central linear algebra algorithms that we will need in our data analysis part and in the construction of Machine Learning algorithms (ML). and is associated with our Intro to Deep Learning Github repository where you can find practical examples of A subset of topics from linear algebra, calculus. Linear Regression aims to find the dependency of a target variable to one or more independent variables. [Online book] n Andrew Ng. Most importantly, the online version of the book is completely free. Franklin, Beedle & Associates Inc. You've accumulated a good bit of data that looks like this:. The course breaks down the outcomes for month on month progress. If you don't want to go all the way back to school, this course should do the trick in just a day or two. In an image classification problem, we often use neural networks. We emphasize that this document is not a. Machine Learning is built on prerequisites, so much so that learning by first principles seems overwhelming. Learning Python for Data. If you don't want to go all the way back to school, this course should do the trick in just a day or two. https://shaarli. Python Quick Start. Concepts you need to know in. Mike Love’s general reference card; Motivations and core values (optional) Installing Bioconductor and finding help; Data structure and management for genome scale experiments. This is | {
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Bioconductor and finding help; Data structure and management for genome scale experiments. This is two equations and two variables, so as you know from high school algebra, you can find a unique solution for x 1 and x. Press Enter to expand sub-menu, click to visit Data Science page Data Science. Linear Algebra 8. The concepts of Linear Algebra are crucial for understanding the theory behind Machine Learning, especially for Deep Learning. Probability and Statistics:. I'd like to introduce a series of blog posts and their corresponding Python Notebooks gathering notes on the Deep Learning Book from Ian Goodfellow, Yoshua Bengio, and Aaron Courville (2016). Scikit-learn (formerly scikits. TS CH10 Linear Least Squares. scikit-learn is a comprehensive machine learning toolkit for Python. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, Solar Weather Forecasting using Linear Algebra. , and Courville, A. GF2] = One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One scala> a + a res0: breeze. For example, when predicting house prices, the different target prices form a continuous space. Conversely, if the condition number is very low (ie close to 0) we say is well-conditioned. The mentors for this project are: @dpshelio @mbobra @drsophiemurray @samaloney. Since I like math and I have more time to dedicate to my projects, I've started an open source linear algebra library for javascript, just for fun and for learning new stuff. Tibshirani, J. Boost your data science skills. Y et because linear algebra is a form of con tin uous rather than. 5M ratings github. View picnicml on GitHub. Deep Learning Book Series · 2. Matrices in Rn m will be denoted as: M. Linear Algebra for Data Science using Python Play all 13:42 Math For Data Science | Practical reasons to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. In the second part, we | {
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to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. In the second part, we discuss how deep learning differs from classical machine learning and explain why it is effective in dealing with complex problems such as image and natural language processing. Linear Algebra and Learning from Data twitter github. Linear Algebra for Machine Learning Discover the Mathematical Language of Data in Python. NET language, as well as a feature-rich interactive shell for rapid development. We will describe linear regression in the context of a prediction problem. Posted by u/[deleted] a linear algebra library in R designed for teaching. Description. We won't use this for most of the. 2 (217 ratings) Course Ratings are calculated from individual students' ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. Linear regression is one of the most popular machine learning algorithms. what-is-the-difference-between-artificial-intelligence-and-machine-learning 9. The mentors for this project are: @dpshelio @mbobra @drsophiemurray @samaloney. ML is a key technology in Big Data, and in many financial, medical, commercial, and scientific applications. This Word Mover’s Distance (WMD) can be seen as a special case of Earth Mover’s Distance (EMD), or Wasserstein distance, the one people talked about in Wasserstein GAN. Thankfully, you've kept a log of each baby's weight at each checkup for the first 12 months. hdf5-OCaml: OCaml implementation of hdf5 reader/writer. Most importantly, the online version of the book is completely free. Those equations may or may not have a solution. " Our homework assignments will use NumPy arrays extensively. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. A good online mathbook on the topic is immersive linear algebra. Everything about Data Science, Machine | {
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online mathbook on the topic is immersive linear algebra. Everything about Data Science, Machine Learning, Analytics, and AI provided in one place! randylaosat. provide a summary of the mathematical background needed for an introductory class in machine learning, which at UC Berkeley is known as CS 189/289A. Franklin, Beedle & Associates Inc. ML helps if one has solid understanding on Linear Algebra, Probability and Statistics. Studying vector spaces will allow us to use the powerful machinery of vector spaces that has been. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, Solar Weather Forecasting using Linear Algebra. As a core programmer, I love taking challenges and love being part of the solution. A computer science student that is interested in Machine Learning would be well advised to get a minor in Mathematics (or just get a degree in Mathematics instead!). The basic mathematics prerequisites for understanding Machine Learning are Calculus-I,II,III, Linear Algebra, and, Probability and Statistics. We start with representing a fully connected layer as a form of matrix multiplication: - In this example, the weight matrix has a size of $4 \times 3$, the input vector has a size of $3 \times 1$ and the output vector has a of size $4 \times 1$. Currently he is on leave from UT Austin and heads the Amazon Research Lab in Berkeley, California, where he is developing and deploying state-of-the-art machine learning methods for Amazon Search. linear; algebra; matrix;. for automated market making. In addition to this, you'll be able to perform operations such as addition, subtraction and dot product. Description. learn and also known as sklearn) is a free software machine learning library for the Python programming language. By mistake uploading only a part of your vertex data is a common mistake due to miscalculations. scikit-learn is a comprehensive machine learning toolkit for Python. hdf5-OCaml: OCaml | {
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scikit-learn is a comprehensive machine learning toolkit for Python. hdf5-OCaml: OCaml implementation of hdf5 reader/writer. Learning rate for to use in SGD; Linear Algebra, Graphs, and Automatic Differentiation. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. Machine learning yearning. The course is project based and through the various projects, normally three, the students will be exposed to fundamental research problems in these fields, with the aim to reproduce state of the art scientific results. If we are thinking of a vector as representing the line segment from the origin to a given point (i. on big time series mining:. Data science, machine learning. It introduces some common tools in machine learning to resolve real applications (e. In this first module we look at how linear algebra is relevant to machine learning and data science. Topics: Python NLP on Twitter API, Distributed Computing Paradigm, MapReduce/Hadoop & Pig Script, SQL/NoSQL, Relational Algebra, Experiment design, Statistics, Graphs, Amazon EC2, Visualization. Docker: https://github. I’m studying towards a PhD degree at the University of Chicago, in the department of Statistics. Robert Beezer, A First Course in Linear Algebra comes with a solution manual. In the first part, we give a quick introduction to classical machine learning and review some key concepts required to understand deep learning. for automated market making. \ \ **References**\ \ - Belsley. , and Courville, A. js Downloading YouTube videos using youtube-dl embedded with Python Machine Learning : scikit-learn Django 1. what-is-the-difference-between-artificial-intelligence-and-machine-learning 9. As we will see, we can do all the common linear algebra operations without using any other library. Learn linear algebra. It turns out, however, that all of those operations can be written in terms of big matrix-matrix or | {
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It turns out, however, that all of those operations can be written in terms of big matrix-matrix or matrix-vector multiplications. , the geometric interpretation), we may interpret the norm as the length of this line segment. You'll understand why I placed functions in quotes shortly. Linear Algebra Examines basic properties of systems of linear equations, vector spaces, inner products, linear independence, dimension, linear transformations, matrices, determinants, eigenvalues, eigenvectors and diagonalization. Announcements and latest whereabouts. Implementation [ edit ] Scikit-learn is largely written in Python, and uses numpy extensively for high-performance linear algebra and array operations. Now I’m checking back in with 9 weeks under my belt. Omoju: What I do at GitHub is I build data models, often deep learning models on GitHub data to help GitHub probably build things like a recommendation engine so we can recommend repositories to people. However, this fact seems to change in the next years, after the hype of Machine Learning we are facing a process of democratization. It aims to provide intuitions/drawings/python code on mathematical theories and is constructed as my understanding of these concepts. Just want some books to go deeper than a introductory course. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. Linear algebra is one of the most applicable areas of mathematics. In this first module we look at how linear algebra is relevant to machine learning and data science. Linear transformations and change of basis (connected to the Singular Value Decomposition - orthonormal bases that diagonalize A) Linear algebra in engineering (graphs and networks, Markov matrices, Fourier matrix, Fast Fourier Transform, linear programming) Homework. This book is directed more at the former audience. As the complementary course to. and engineering. NumPy is "the fundamental | {
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at the former audience. As the complementary course to. and engineering. NumPy is "the fundamental package for scientific computing with Python. Linear Algebra is a text for a first US undergraduate Linear Algebra course. Bourbaki resulted from similar currents of thought that produced fascism and totalitarian communism: moral panics leading to revolutions, and ultimately “final solutions”, all terrible and evil in. Complete Linear Algebra for Data Science & Machine Learning 4. Updated Apr 1 2020. In an image classification problem, we often use neural networks. His research interests span statistical machine learning, numerical linear algebra, and random matrix theory. Included is a learning guide and syllabus to help you learn data science this year. Many courses are offered there from which one can benefit. Linear Algebra¶ Now that you can store and manipulate data, let's briefly review the subset of basic linear algebra that you will need to understand most of the models. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. In this tutorial, we walked through one of the most basic and important regression analysis methods called Linear Regression. com, My Github, LinkedIn, & Google Scholar I am an applied mathematician and computer scientist with experience in algorithms, linear algebra, and graph theory. One of the most beautiful and useful results from linear algebra, in my opinion, is a matrix decomposition known as the singular value decomposition. Linear algebra moves steadily to n vectors in m-dimensional space. If you have some background in basic linear algebra and calculus, this practical book introduces machine-learning fundamentals by showing you how to design systems capable of detecting objects in images, understanding text, analyzing video, and predicting. As a core programmer, I love taking challenges and love being part of the solution. in | {
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