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• Which still uses induction to compute the sums. May 28, 2016 at 2:11
• @Bill: Looks to me like it's plugging in known formulas, not using induction to reprove the corresponding theorem.
– user14972
May 28, 2016 at 5:18
• @BillDubuque there are other ways to prove the known sums. May 28, 2016 at 12:13
• @Hurkyl The po... | {
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A geometrical proof. A rectangle formed by a sum of gnomons.
• ...which are not a very gnomonly used shape. May 28, 2016 at 17:48
• @Micah I explain in my answer how this proof can be discovered very simply from the product rule for differences. May 28, 2016 at 22:17
• what software did you use to create the image? Ma... | {
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• The ellipses amount to using induction. May 28, 2016 at 2:06
• @Bill: No, they really don't. The places where you might be able to play that game are the relevant form of associativity and the conversion between repeated addition and multiplication.
– user14972
May 28, 2016 at 5:16
• @ClementC. I don't know whether i... | {
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We encode the sequence
\begin{align*} \left(\sum_{j=1}^{n}\left(4j-3\right)\right)_{n\geq 1}=\left(1,6,15,\cdots\right)\quad\rightarrow\quad \sum_{n=1}^\infty\left(\sum_{j=1}^n\left(4j-3\right)\right)x^n=\color{blue}{1}x+\color{blue}{6}x^2+\color{blue}{15}x^3+\cdots \end{align*} by the power series and in the same way ... | {
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• In (6) we use the formula for the geometric power series again and multiply out
• In (7) we perform the differentiation and simplify the expression
on the other hand we obtain with similar reasoning \begin{align*} \sum_{n=1}^\infty\left(2n^2-n\right)x^n &=2\sum_{n=1}^\infty n^2x^n-\sum_{n=1}^\infty nx^n\\ &=2(xD_x)... | {
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• This is a valid method to avoid induction, but at the end, it uses the same idea.
– Emre
May 29, 2016 at 22:04
• @E.Girgin Yes, telescopy is a special (but ubiquitous) form of induction. Presumably this is the same method used in the pictorial proofs (as I explain at length in my answer), though we cannot be sure sin... | {
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Hence,
$$\displaystyle\sum_{k=1}^n a_k = 2 n^2 - n$$
The most straightforward way is of course
$$\frac n2(a+\ell))=\frac n2 \big(1+(4n-3)\big)=\color{red}{2n^2-n}\qquad\blacksquare$$
Another method:
The well-known result of the sum of the first $n$ integers: $$1+2+3+4+\cdots+n=\frac{n(n+1)}2$$ Subtract $1$ from ea... | {
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So here the problem is to evaluate $\sum_{1\le i\le n}{4i-3}$. First, expand all powers of the summation variable (here, $i$) into sums of falling powers (by inspection in simple cases, or using Stirling cycle numbers). This is trivial here, since $i = i^{\underline 1}$, so $$\sum_{1\le i\le n}{4i-3} = \sum_{1\le i\le ... | {
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where $u_k = 2k^2 - 5k$.
Which converts the sum to the sum of a collapsing series.
This implies Lynn's graphical proof shown below and, though it still requires induction, I thought it shows an interesting way to compute sums. | {
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• If you've to ask whether a step requires induction, it means you haven't even proven it... Yes it does, and all the answers that rely on properties of summation also do. Worse still, your answer involves integration, and you won't be able to prove most properties of integrals without induction!! Jun 7, 2016 at 2:40
•... | {
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• In your case, your method gives no insight whatsoever because: (1) the very proof of the integral requires much more than induction. The fastest way I know that avoids Riemann sums is via anti-derivatives and yet it has to first prove the product rule. Then after that you use the property that the integral of a finit... | {
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One way using distributive laws and multiplication:
$\sum_{k=1}^n(4k-3)$
$=\sum_{k=1}^n(4k)+\sum_{k=1}^n(-3)$
$\sum_{k=1}^n(-3)=-3n$
$\sum_{k=1}^n(4k)=4\cdot\sum_{k=1}^n(k)$
Assuming n is even:
$\sum_{k=1}^n(k)=1+2+3+4...+n=(1+n)+(2+n-1)...+(\frac{n}{2}+\frac{n}{2}+1)=\frac{n(1+n)}{2}$
Assuming n is odd: $\sum_{... | {
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# [SOLVED]Prove sum of (-1)^i times n choose i equals 0
#### Jameson
Staff member
Problem: Prove that for $n>0$, $$\displaystyle \sum_{i=0}^{n} (-1)^i \binom{n}{i}=0$$
Attempt: This seems clearly like a proof based on induction.
1) Base case: for $n=1$, $$\displaystyle \sum_{i=0}^{1}(-1)^i \binom{1}{i}=(-1)^0 \bino... | {
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$\chi$ $\sigma$
#### Jameson
Staff member
I would in fact use the binomial theorem here.
$\displaystyle 0=(1-1)^n=?$
The $(-1)^i$ term is the one that is throwing me off. You stated looking at 0, so maybe I can rewrite 0 in terms of the binomial theorem and show that's it's equal to the left hand side?
I'll write t... | {
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\begin{aligned} S & = \sum_{i=1}^{k+1}{k-1\choose i-1}(-1)^{i-1} + \sum_{i=0}^{k} {k-1\choose i-1} (-1)^i \\& =- \sum_{i=0}^{k}{k-1\choose i-1}(-1)^i + \sum_{i=0}^{k} {k-1\choose i-1} (-1)^i \\& = 0. \end{aligned}.
#### Deveno
##### Well-known member
MHB Math Scholar
pascal's triangle is symmetric. this takes care of... | {
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(EDIT #2: i have to say that MarkFL's and chisigma's proofs are very elegant. my proof is by comparison, "uglier", but has the advantage of using no other results ("elementary" proofs are often more difficult to understand than 'high-level ones"), except the recursive definition of the binomial coefficient (one does no... | {
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$$\sum_{i=0}^n (-1)^i \binom{n}{i} = \sum_{i \; even} \binom{n}{i} - \sum_{i \; odd} \binom{n}{i}$$
so the desired result is equivalent to showing that the number of subsets of $\{1, 2, 3, \dots ,n\}$ with an even number of elements is equal to the number of subsets with an odd number of elements. We will show this by... | {
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#### Sherlock
##### Member
i've high-lighted what i think are troublesome parts in red.
i am having some trouble deciding what:
$\displaystyle \binom{k-1}{-1}$ and $\displaystyle \binom{k-1}{k}$ should be. care to elaborate?
$\displaystyle \binom{r}{k} = \left \{\begin{array}{cc} \frac{r(r-1)\cdots(r-k+1)}{k(k-1)\cd... | {
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so far, so good, and i even agree that your change of index is kosher (the index is just a dummy variable, anyway).
so for $k = 2$, what you have above, translates to:
$\displaystyle S = \left[\binom{1}{0} - \binom{1}{1} + \binom{1}{2}\right] + \left[\binom{1}{-1} - \binom{1}{0} + \binom{1}{1}\right]$.
but we don't ... | {
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Problems
Find initial velocity when a stuntman jump from $1.25 \ m$ height
+3
−0
A stuntman jumped from $1.25 \ \text{m}$ height and, landed at distance $10 \ \text{m}$. Find velocity when he jumped. (Take $\text{g}=10 \ ms^{-2}$)
I had solved it following way.
$$h=\frac{1}{2}gt^2$$ $$=>1.25=5\cdot t^2$$ $$=>t=\fr... | {
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# Does the magnitude of a physical quantity have units or is it just a plain number?
Does the magnitude of a physical quantity have units? For example, if a velocity vector is $36\ \mathrm{m\,s^{-1}}\ \hat{u}$, is its magnitude $36\ \mathrm{m\,s^{-1}}$ or just $36$? Also why?
The magnitude has units. In your example,... | {
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• Incidentally, we can also arrive at the same result from the opposite side: After decomposing into magnitude and direction vector, the latter should have magnitude 1, but 1 is dimensionless. It wouldn't make sense to have units in your direction vector, so they have to live in the magnitude instead. – Kevin Aug 14 at... | {
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• The first being the distance $7 \,{\rm m}$ in the direction $\left(\frac{3}{7} , \frac{2}{7} , \frac{6}{7} \right)$. This is the span interpretation when one spans "x" distance along a particular line.
• The second being 7 times the distances $\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \... | {
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• Why did you pick out "7 times the distances $\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$"? It's no more special than "14 times the distances $\left(\frac{3}{14}\,{\rm m} , \frac{2}{14}\,{\rm m} , \frac{6}{14}\,{\rm m} \right)$" – JiK Aug 14 at 20:31
• Stating @JiK's point more st... | {
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You would quote the speed as $36 \, \rm m\,s^{-1}$ because it is $36$ times $1 \, \rm m\,s^{-1}$.
If you wrote just $36$ what would that mean?
$36$ times what ????
Now what about $36 \, \rm m\,s^{-1}\, \hat u$?
All you have now is extra information about the direction of the velocity and there is no extra information ... | {
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$||v||=\sqrt{(v^1)^2+(v^2)^2+(v^3)^2}=\sqrt{v_x^2+v_y^2+v_z^2} \in \mathcal{R}$. (1)
$\mathcal{R}$ denotes that it is a (real) number. So once you know for example that the velocity of an object is 36 (=$||v||$). This could really mean anything. The units are arbitrary in mathematics. But of course not in physics. 36 ... | {
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That last example also leads us into a good (but false) counter point. Quantities are often given without units. In reality, even a sign which reads only "Speed Limit 65" is providing units even though it does not look like it.
Whenever we are talking, listening, reading, or writing there is practically always a conte... | {
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# Question on Axiom of replacement
This axiom comes from chapter 3 "set theory" of Tao Analysis I
Axiom $$3.6$$ (Replacement). Let $$A$$ be a set. For any object $$x \in A$$ and any object $$y$$, suppose we have a statement $$P(x, y)$$ pertaining to $$x$$ and $$y$$, such that for each $$x\in A$$ there is at most one ... | {
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# Using sum-to-product formula to solve $\sin(2\theta)+\sin(4\theta)=0$
Trying to use the sum-to-product formula to solve $$\sin(2\theta)+\sin(4\theta)=0$$ over the interval $$[0,2\pi)$$, but I'm missing solutions.
$$\sin(2\theta)+\sin(4\theta)=0$$
Apply sum-to-product formula:
$$2\sin\left(\frac{2\theta+4\theta}{2... | {
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• No solutions were missing! – Andrew Chin Jul 2 '20 at 23:05
• Wouldn't it be easier to use $$\sin(2\theta)+\sin(4\theta)= \sin(2\theta)+2\sin(2\theta)\cos(2\theta)=\sin(2\theta)\cdot(2+\cos(2\theta)?$$ – Michael Hoppe Jul 3 '20 at 11:45
This is an excellent way to proceed with this problem, and the reduction to $$\s... | {
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# Expectation of the maximum of gaussian random variables
Is there an exact or good approximate expression for the expectation, variance or other moments of the maximum of $n$ independent, identically distributed gaussian random variables where $n$ is large?
If $F$ is the cumulative distribution function for a standa... | {
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Defining, $$Z = [\max_{i} X_i]$$
By Jensen's inequality,
$$\exp \{t\mathbb{E}[ Z] \} \leq \mathbb{E} \exp \{tZ\} = \mathbb{E} \max_i \exp \{tX_i\} \leq \sum_{i = 1}^n \mathbb{E} [\exp \{tX_i\}] = n \exp \{t^2 \sigma^2/2 \}$$
where the last equality follows from the definition of the Gaussian moment generating functi... | {
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The $$\max$$-central limit theorem states that $$F_\max(x) = \left(\Phi(x)\right)^n \approx F_{\text{EV}}\left(\frac{x-\mu_n}{\sigma_n}\right)$$, where $$F_{EV} = \exp(-\exp(-x))$$ is the cumulative distribution function for the extreme value distribution, and $$\mu_n = \Phi^{-1}\left(1-\frac{1}{n} \right) \qquad \qqua... | {
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• +1. See also Section 10.5 ("The Asymptotic Distribution of the Extreme") in David and Nagaraja's Order Statistics. They explicitly discuss the normal distribution on page 302. Dec 6 '11 at 22:35
• Doesn't the inverse cdf have domain $[0,1]$? Dec 30 '12 at 7:01
• (+1) Two comments: (1) The somewhat nonstandard use of ... | {
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# How $x^4$ is strictly convex function?
I hear that $f(x)=x^4$ is a strictly convex function $\forall x \in \Re$. However, strict convexity condition is that the second derivative should be positive $\forall x \in \Re$. For the mentioned function second derivative is zero at $x=0$ which is in the domain of $f$. There... | {
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-
Ok. So the converse is not true... that is even if second derivative at x=0 is zero, we say that $x^4$ is strictly convex. Now my question is how to analytically see if the function is indeed strictly convex? (In case, we cannot plot the function and see if the line segment joining any two points lies above the funct... | {
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# Complicated Set Theory Problems: Sequence of Sets
Let $A_1, A_2, \ldots , A_n$ be sets such that $X = \bigcup_{i=1}^n A_i$. Prove that there exists a sequence of sets $B_1, B_2, \ldots , B_n$ such that
a) $B_i \subseteq A_i$ for each $i=1,2,\ldots ,n$.
b) $B_i \cap B_j = \varnothing$ for $i\ne j$.
c) $X = \bigcup... | {
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print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
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=... | {
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Staff - 11 months, 2 weeks ago
Thank you so much. This is enough already.
- 11 months, 1 week ago
No problem
Staff - 11 months, 1 week ago
Do you mean the following Sir?
That for (a) I will assume first that $x\in A_k$ and do all ways to prove that $x\in B_k$ given $B_i \subseteq A_i$ for each $i=1,2,\ldots ,n$?
... | {
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Series is convergent but it seems it is divergent?
I have a series:
$$\sum^\infty_{n=1}{\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}\bigg)}$$
and I thought it is a divergent series since
$$\sum{\big(f(x)-g(x)\big)} = \sum{f(x)} - \sum{g(x)}$$
and so the series equals to
$$=\sum^\infty_{n=1}{\frac{1}{\sqrt{n}}}-\... | {
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As indicated in the comments, to evaluate this series, you should telescope the series.
-
To further elaborate as to why we cannot say $\sum f(x) - g(x)$ diverges if $\sum f(x)$ and $\sum g(x)$ are individually divergent, consider the example $$\sum_{k=1}^\infty \frac{1}{k(k+1)} < \sum_{k=1}^\infty \frac{1}{k^2} < \in... | {
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Clearly, we have,
$$\lim_{n\to\infty} S_n=1+\frac{1}{\sqrt{2}}$$
-
I was wondering why the answer was $1$... and then I saw the edit. – Derek 朕會功夫 Apr 4 '14 at 5:46
@Derek朕會功夫 yeah typo. – Sabyasachi Apr 4 '14 at 5:54
Note: $$\sum_{n = 1}^\infty \left[\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 2}}\right] = 1 - \frac{... | {
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+0
# (T*(cos(45)/cos(30)))*sin(30) + (T*sin(45) - 50 = 0 What is T and how to solve?
0
948
6
(T*(cos(45)/cos(30)))*sin(30) + (T*sin(45) - 50 = 0 What is T and how to solve?
Apr 12, 2014
#1
+109766
0
$$T\times\frac{cos45}{cos30}\times sin30+Tsin45-50&=0\\\\ T\left(\frac{1}{\sqrt{2}}\div\frac{\sqrt3}{2}\right)\frac... | {
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(T*(cos(45)/cos(30)))*sin(30) + (T*sin(45) - 50 = 0 ???
When I stick "45" in for "T," we get
$$\left({\mathtt{45}}{\mathtt{\,\times\,}}\left({\frac{\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}}{\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}}\right)\right){\m... | {
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Solve $x^{2}\equiv 24 \mod 125$
Here's a congruence I'm trying to solve:
$$x^2\equiv24 \mod 125$$
What are the techniques I could use to solve it? I know about Euler's phi function, Fermat's little theorem and Chinese remainder theorem but they all seem inapplicable here. Is there something else I could use?
• Have... | {
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Remark $\$ Generally one can use Hensel's Lemma, as in lhf's answer, but that is a bit overkill in cases like this where the number is obviously congruent to a well-known power (which happens frequently for small moduli due to the law of small numbers).
• +1! that is even neater than Timbuc's nice answer - surely the ... | {
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# Finding the intersection of three sets
80 students were asked if they like math, science or humanities. 24 students did not like either of the subjects, 9 liked math only, 16 liked science only, 9 liked humanities only, 12 liked math and humanities, 7 liked math and science and 9 liked humanities and science.
a) Ho... | {
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Now, summing all of those values, we see that $$80 = 24 + 9 + 16 + 9 + 7 - x + 9 - x + 12 - x + x = 86 - 2x$$, and so $$2x = 6$$, and $$x = 3$$. Thus, the full Venn diagram looks like this:
We can now solve the questions by just reading off the diagram.
• Thank you so much!! This made much more sense!! Appreciate it!... | {
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• appreciate your input! – esker-luminous Jan 22 at 16:07
The Venn diagram doesn't seem quite correct. If you add up the numbers, you get 86, which is greater than the universe of 80 students. I think that the 12 who like math and humanities includes those who like math, humanities, and science (and likewise for the 7... | {
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775 views
A sub-sequence of a given sequence is just the given sequence with some elements (possibly none or all) left out. We are given two sequences $X[m]$ and $Y[n]$ of lengths m and n, respectively with indexes of $X$ and $Y$ starting from $0$.
We wish to find the length of the longest common sub-sequence (LCS) o... | {
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54. Answer is B. Dynamic programming is used to save the previously found LCS. So, for any index [p,q] all smaller ones should have been computed earlier. Option D is not correct as the condition given requires even L[3,2] to be computed before L[2,4] which is not a necessity if we follow row-major order.
int lcs( cha... | {
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# Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$
Can someone give me some hints on how to start solving $\sqrt{x+4}-\sqrt{x+1}=1$ for x?
Like I tried to factor it expand it, or even multiplying both sides by its conjugate but nothing comes up right.
-
$0$ of course, it's easy to guess ... square it, isolate the roots and t... | {
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Multiplying by the conjugate as you originally suggested does work here. If you multiply both sides by $\sqrt{x+4} + \sqrt{x+1}$ you get $$(x+4) - (x+1) = \sqrt{x+4} + \sqrt{x+1}$$ Which is the same as $$\sqrt{x+4} + \sqrt{x+1} = 3$$ Add this to the original equation and divide by $2$ to obtain $$\sqrt{x+4} = 2$$ Squar... | {
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Let $a = \sqrt{x+4}$ and $b=\sqrt{x+1}$. So that $a^2 = x + 4$ and $b^2 = x + 1$.
From the given equation, $$\sqrt{x+4} - \sqrt{x+1}=1 \implies a - b = 1 \ \ \ \text{and} \ \ \ a^2-b^2=3.$$ So we have that, $$a^2-b^2 =(a-b)(a+b)=1 \cdot(a+b)=3 \implies a+b=3.$$ We see that $$(a+b)+(a-b) = 2a.$$ Also, $$(a+b)+(a-b)=3+1... | {
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-
I have to wonder what the people who use a bunch of algebraic manipulation to solve these sorts of problems think about solutions like these or if my answer would have gotten accepted had I managed to post it 20 hours earlier. – Doug Spoonwood May 24 '13 at 14:48
There's also the possibility that $x$ is not a whole ... | {
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# What is the distribution of $R^2$ in linear regression under the null hypothesis? Why is its mode not at zero when $k>3$?
What is the distribution of the coefficient of determination, or R squared, $R^2$, in linear univariate multiple regression under the null hypothesis $H_0:\beta=0$?
How does it depend on the num... | {
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• Are you willing to assume error normality? Dec 23, 2014 at 1:17
• Yes, I guess one has to assume it to make this question answerable (?). Dec 23, 2014 at 1:20
• Have you checked this davegiles.blogspot.jp/2013/05/good-old-r-squared.html ? Dec 23, 2014 at 2:35
• @Khashaa: in fact, I have to admit that I did find that ... | {
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$$\{\alpha >1 , \beta \geq 1\},\;\; \text {OR}\;\; \{\alpha \geq1 , \beta > 1\}$$
Note that if $\{\alpha =1 , \beta = 1\}$, we obtain the Uniform distribution, so all the density points are modes (finite but not unique). Which creates the question: Why, if $k=3, n=5$, $R^2$ is distributed as a $U(0,1)$?
IMPLICATIONS
... | {
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To respond to the new issue regarding the mode of the $R^2$ distribution, I can offer the following line of thought (not geometrical), which links it to the "spurious fit" phenomenon: when we run least-squares on a data set, we essentially solve a system of $n$ linear equations with $k$ unknowns (the only difference fr... | {
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• Its mathematical. For $k=2$ the first parameter of the beta distribution (the "$\alpha$" in standard notation) becomes smaller than unity. In that case the Beta distribution has no finite mode, play around with keisan.casio.com/exec/system/1180573226 to see how the shapes change. Dec 23, 2014 at 9:53
• @Alecos Excell... | {
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I won't rederive the $\mathrm{Beta}(\frac{k-1}{2}, \, \frac{n-k}{2})$ distribution in @Alecos's excellent answer (it's a standard result, see here for another nice discussion) but I want to fill in more details about the consequences! Firstly, what does the null distribution of $R^2$ look like for a range of values of ... | {
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The algebra is straightforward for $\alpha$: we have $\frac{k-1}{2}=1$ so $k=3$. This is indeed the only column of the facet plot that's filled blue on the left. Similarly $\alpha < 1$ for $k<3$ (the $k=2$ column is red on the left) and $\alpha > 1$ for $k>3$ (from the $k=4$ column onwards, the left side is green).
Fo... | {
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The Beta distribution itself is symmetric if $\alpha = \beta$. For us this occurs if $n = 2k-1$ which correctly identifies the panels $(k=2, n=3)$, $(k=3, n=5)$, $(k=4, n=7)$ and $(k=5, n=9)$. The extent to which the distribution is symmetric across $R^2 = 0.5$ depends on how many regressor variables we include in the ... | {
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Special distributions
When $k=n$ we have $\beta=0$, which isn't a valid parameter. However, as $\beta \to 0$ the distribution becomes degenerate with a spike such that $\mathsf{P}(R^2 = 1)=1$. This is consistent with what we know about a model with as many parameters as data points - it achieves perfect fit. I haven't... | {
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You may also have noticed the triangular distributions at $(k=5,n=7)$ and its reflection $(k=3,n=7)$. We can recognise from their $\alpha$ and $\beta$ that these are just special cases of the power-law and reflected power-law distributions where the power is $2-1=1$.
Mode
If $\alpha>1$ and $\beta>1$, all green in the... | {
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Mean
The question asked about the mode, but the mean of $R^2$ under the null is also interesting - it has the remarkably simple form $\frac{k-1}{n-1}$. For a fixed sample size it increases in arithmetic progression as more regressors are added to the model, until the mean value is 1 when $k=n$. The mean of a Beta dist... | {
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df2 <- mutate(df2,
kname = paste("k =", k),
nname = paste("n =", n),
a = (k-1)/2,
b = (n-k)/2,
meanR2 = ifelse(k > n, NaN, a/(a+b)),
modeR2 = ifelse((a>1 & b>=1) | (a>=1 & b>1), (a-1)/(a+b-2),
ifelse(a<1 & b>=1 & n>=k, 0, ifelse(a>=1 & b<1 & n>=k, 1, NaN))),
sdR2 = ifelse(k > n, NaN, sqrt(a*b/((a+b)^2 * (a+b+1)))),
mea... | {
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• Really illuminating visualization. +1 Dec 24, 2014 at 8:16
• Great addition, +1, thanks. I noticed that you call $0$ a mode when the distribution goes to $+\infty$ when $x\to 0$ (and nowhere else) -- something @Alecos above (in the comments) did not want to do. I agree with you: it is convenient. Dec 24, 2014 at 10:2... | {
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# Does the coin flipping game terminate?
Imagine I were to toss a coin, so that
1. if it is tails, I will toss again.
2. if it is heads, I stop the game.
It is clear that if I would to play the game forever the probability of me tossing tails over and over again tends to $0$, because $\lim\limits_{n \rightarrow \inf... | {
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Note that in practice you cannot perform this experiment: whatever resources you may have (millions of people tossing billions of coins for many many years) you are not guaranteed to ever get a Head. The above therefore relates to a thought experiment that is allowed to go on "forever".
How about the followings? The p... | {
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The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. — It is a matter of security to change such things every now and then, to keep the enemy in the dark. — But look, I have chosen my number 1033 ... | {
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... |
## Sample
input:
1 2 3 4 3 1033 8179 1373 8017 1033 1033
output:
1 2 3 6 7 0
## Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 #in... | {
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one-to-one correspondence
My textbook states the following proposition: Let $f:R \rightarrow S$ be a ring homomorphism and let $s$ be in the image of $f$. Then $\{r \in R \mid f(r) = s\}$ is in one-to-one correspondence with $\ker(f)$.
What does it mean to have one to one correspondence with $\ker(f)$? Does it mean t... | {
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-
@Bill cool, it's actually the proof! I see that you are saying each $r \in R$ corresponds to (r' + k) for each $k \in ker f$, which implies the same cardinality. but what about the cases where (r'+2k), (r'+3k), and so on? since f maps them to s as well, wouldn't that make the set R have more elements than ker f? I wo... | {
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@Bill well, the same thing applies to your space shifting as well. one (newbie like me) might ask, after shifting r1 to the homogeneous solutions to obtain a set of non-homogeneous solutions, why not then shift the homogeneous solutions by r2 to obtain even more solutions? well, let's say f(r2) = s, and r2 = r1 + r0. t... | {
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# Can someone clarify this doubt of mine, regarding calculation of average for a probability distribution?
Suppose there is a question on probability which asks us to find the expected number of successful attempts or average number of successful attempts, with the probability of success given. On computation using th... | {
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it is more like
"the average of several values I will come up in my experiment".
• Thank you. That sure helped. Just for a clarification, could you refer to this question: Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of... | {
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# Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$.
I can show that $$x^2+y^2=\frac12(1+\log2)$$ is the equation of the circle of largest area inscribed in $$y=\pm e^{-x^2}$$:
The minimum distance $$r$$ (which will be the... | {
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We can assume by symmetry and without loss of generality that the ellipse can be parametrized by $$(x,y) = (a \cos \theta, b \sin \theta), \quad a, b > 0, \quad \theta \in [0,2\pi).$$ We require tangency to the curve $$y = e^{-x^2}$$ as well as a single point of intersection in the first quadrant. That is to say, $$b \... | {
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• Does the max area ellipse contact the Bell curve at its inflection point, by any chance? – Narasimham Jun 3 at 8:05
• @Narasimham Yes, it does. It is easy to show that for the maximal area, the point of tangency of the ellipse to the curve in the first quadrant is at $(2^{-1/2}, e^{-1/2})$. – heropup Jun 3 at 8:29
• ... | {
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I think what you say holds good if the Bell curve has a single constant in its parametrization. But there are two constants in: $$y= y_{max} e^{- x^2/(2 \sigma^2)}$$
The touching curve is an ellipse, not a circle.
This is vague/intuitive at present,trying to prove that ellipse area is maximized when constrained on an... | {
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You are currently browsing the tag archive for the ‘school’ tag.
Wikipedia calls this a classic paradox of elementary probability theory.
Taken from Wikipedia.org:
Card version
Suppose you have three cards:
• black card that is black on both sides,
• white card that is white on both sides, and
• mixed card that is... | {
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In our question, however, you have already selected a card from the hat and it shows a black face. At first glance it appears that there is a 50/50 chance (ie. probability 1/2) that the other side of the card is black, since there are two cards it might be: the black and the mixed. However, this reasoning fails to expl... | {
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Bayes’ theorem
Given that the shown face is black, the other face is black if and only if the card is the black card. If the black card is drawn, a black face is shown with probability 1. The total probability of seeing a black face is 1/2; the total probability of drawing the black card is 1/3. By Bayes’ theorem,[3] ... | {
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Note the logical fact that the B/B card contributes significantly more (in fact twice) to the number of times “B” is on top. With the card B/W there is always a 50% chance W being on top, thus in 50% of the cases card B/W is drawn, card B/W virtually does not count. Conclusively, the cards B/B and B/W are not of equal ... | {
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# Prove that the square root of a positive integer is either an integer or irrational
Is my proof that the square root of a positive integer is either an integer or an irrational number correct?
The proof goes like this:
Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt... | {
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Therefore, this contradicts the fact that $\sqrt n$ is rational. Therefore, $\sqrt n$ must be irrational.
Is this sufficient? Or is there any parts I did not explain well?
• I think its correct and very well explained. – Shobhit Jan 30 '17 at 14:25
• It's a bit wordy, but logically you've got a solid proof, the even-... | {
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Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.
This can all be said more simply and to argue that if $\sqrt{n}$ is an integer we can conclude $\sqrt{n}$ is rational or that $n$ is therefore a perfect square, is a little heavy handed. Those are de... | {
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Bill Dubuque in the comments noted what you meant to say was "each prime factor will be raised to any even power".
.
However, we can also express n as a product of primes. Since n is not a square number, then there exist at least one prime number that has an odd number of primes. Therefore, there exists at least one ... | {
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But if I want to be completely accurate, you (and I) have actually only proven that positive integer $n$ either has an integer square root or it has no rational square root at all. Which is the same thing as saying if positive integer $n$ has a square root, the root is either integer or irrational. But we have not actu... | {
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• Thank you for the critique! It definitely helped me a lot! – Icycarus Jan 31 '17 at 11:00
• Can you explain or state a case where we cannot find the square root of a positive integer $n$ (as you state in the last line)? – Astrobleme Apr 15 '17 at 12:56
• I didn't say there are positive integers without square roots. ... | {
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We know that $\sqrt{4} = 2$ and $\sqrt{2} = 1.414...$ are rational and irrational respectively, so all we have to do is to show that if $n\in \mathbb{Z+}$ such that $\sqrt{n} = \dfrac{a}{b}$ where $a$ and $b$ are positive integers and the expression $\dfrac{a}{b}$ is in its simplest form then $\sqrt{n}$ is integral. Sq... | {
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• How is that a critique of the OP's proof? – fleablood Jan 30 '17 at 18:51
• The problem lies in the first paragraph, the op argues that if $\sqrt{n}$ is an integer then it is rational which is very true but the op does not show that all the rational square roots to be integers in fact the op shows that the square roo... | {
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• I believe if you reread the op's post you especially the areas I pointed out you will understand where my problem arises but as you said before I was only simplifying the post and corrected the few areas which I thought needed to be repaired. – Tom Carter Jan 31 '17 at 6:34 | {
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Finding real $(x,y)$ solutions that satisfies a system of equation.
I was given:
$$x + y^2 = y^3 ...(i) \\ y + x^2 = x^3...(ii)$$
And was asked to find real $$(x,y)$$ solutions that satisfy the equation.
I substracted $$(i)$$ by $$(ii)$$:
$$x^3 - y^3 + y^2 - x^2 + x - y = 0$$
Then factored it out so I have:
$$(x... | {
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"openwebmath_score": 0.8389489054679871,
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You can begin by noting that the $$2$$ functions are inverses of each other (and only involve odd non-zero exponents). Using the fact that inverse functions are reflections in the line $$y=x$$, we can now see that the intersection points must be along the line $$y=x$$. Substituting $$y$$ into $$x$$ or vice-versa, we ob... | {
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"openwebmath_score": 0.8389489054679871,
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# Metric space consisting two elements.
Suppose that $S$ be a set consisting exactly $2$ elements. Suppose we define a function $\displaystyle d:S \times S \to [0,\infty)$ by $\displaystyle d(x,y)=\begin{cases}1 &\text{ , if }x\not=y\\0 &\text{ , if} x=y\end{cases}$
How I can show that $d$ defines a metric on $S$ ?
... | {
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"openwebmath_score": 0.9288551807403564,
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• A lawyer might write the Triangle Inequality as " For any point, hereinafter called x, and any point hereinafter called y, and notwithstanding that x and y are different names, the point called x and the point called y may or may not be the same point, and any point hereinafter called z..... (etc)" ...or something li... | {
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"lm_q1_score": 0.9838471647042429,
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"lm_q2_score": 0.861538211208597,
"openwebmath_perplexity": 174.5951011550481,
"openwebmath_score": 0.9288551807403564,
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# Solution to $y'=y^2-4$
I recognize this as a separable differential equation and receive the expression:
$\frac{dy}{y^2-4}=dx$
The issue comes about when evaluating the left hand side integral:
$\frac{dy}{y^2-4}$
I attempt to do this integral through partial fraction decomposition using the following logic:
$\f... | {
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• two constant solutions, $y=2$ and $y=-2.$ The other solutions split into three types, $y < -2,$ very similar $y > 2,$ rather different $-2 < y < 2$ – Will Jagy May 29 '15 at 22:03
• Ok but how does this help me in understanding why my answer is wrong? – Filip May 29 '15 at 22:09
• That is up to you. – Will Jagy May 2... | {
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