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https://math.stackexchange.com/questions/2984453/plotting-a-system-of-linear-ode | # Plotting a system of linear ODE
I would like to plot a system of following ODE:
$$$$\mathbf{\dot{x}} = \mathbf{Ax} \text{ where } \mathbf{A} = \begin{bmatrix} -2 & 1 \\ -1 & 0 \end{bmatrix}$$$$
with general solution:
$$$$\vec{x}(t)=C_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{-t}+C_2 \left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}+\begin{bmatrix} 1 \\ 1 \end{bmatrix} t \right) e^{-t}$$$$
I know that this system is stable so arrows will be inward-pointing, also it has double eigenvalue at -1 so there will be an constant line for an eigenvalue crossing the plot but I am not sure how to fill the rest of the phase diagram. Could someone help me? Thanks in advance.
• I think you have a mistake. Shouldn't this system have a double eigenvalue of 1, not -1? – Josh B. Nov 4 '18 at 16:29
• @JoshB. I think the eigenvalue is given by $$\lambda^2+2\lambda+1$$ unless I made a mistake but I double checked the solution using Matlab – 1muflon1 Nov 4 '18 at 16:33
• If that's the case, then there is a typo in your matrix here. I just checked and the matrix you have definitely has a double eigenvalue of 1. – Josh B. Nov 4 '18 at 16:39
• @JoshB. You were correct the typo was 2 instead of -2. My bad. I am really sorry, I was copy pasting it from my latex and somehow I did not included the minus sign, dont know how did I even managed to do it – 1muflon1 Nov 4 '18 at 16:41
If you write your solution now as $$\vec{x}(t)=\left(C_1\begin{bmatrix} 1 \\ 1 \end{bmatrix}+C_2\begin{bmatrix} 0 \\ 1\end{bmatrix}\right)e^{-t}+C_2\begin{bmatrix}1 \\ 1\end{bmatrix}te^{-t}$$
it may be a little easier to see what's going on. As $$t\to\pm\infty$$, the equation is dominated by the second term, so $$\vec{x}(t)\approx C_2\begin{bmatrix} 1 \\ 1\end{bmatrix}te^{-t}\;\;\;\;\;\;\;\;\;\;\;t\to\pm\infty$$
As $$t\to\infty$$, the whole thing decays to $$0$$, so it makes sense that all solution curves begin to look like they fall along this eigenvector as they decay to $$0$$. Applying the same logic as $$t\to-\infty$$, it would seem that solutions grow along this vector as well, but this is only true on a macroscopic scale, as the solutions are growing exponentially in both directions. The first term, no longer decaying to $$0$$, shifts the solution away from the eigenvector and is growing large, but not as large as the second term. This means that when zoomed in, solutions look like they are leaving the eigenvector, but zoomed out solutions look like they are along the eigenvector (that is, until you let time grow large enough, in which you will see it move away slowly).
The last thing to consider is the direction of the curve. Suppose that $$\vec{x}$$ is in the first quadrant as $$t\to\infty$$. This means that $$C_2$$ is positive, so if $$t\to-\infty$$, then the sign on the second term is now negative, and the solution turns around to enter the third quadrant. This means that solutions sort of "spin" by turning $$180^\circ$$ around while still growing away from the origin.
The vector $$\begin{bmatrix} 0 \\ 1\end{bmatrix}$$ tells you which way the solution spins. If $$C_2$$ is positive, then that means that at $$t=0$$, the solution is above (in the xy-plane) of the line through the vector $$\begin{bmatrix} 1 \\ 1\end{bmatrix}$$ and the origin, so the solution must remain on this side of said line. The dominating term is in the first quadrant as $$t\to\infty$$, so the solution must spin clockwise to decay this way. It turns out that if solutions spin clockwise on one side of the eigenvector, they do the same on the other side as well; the same is true if they spin counterclockwise.
This type of equilibrium is called a Degenerate Node, in case you were curious. Here is an example of what one might look like.
• Thank you for very clear and detailed answer. By the way did you made the plot in matlab,? I know that it was not part of my original question and matlab inquiries dont really belong to math section, but would you mind sharing your code? I would like to learn how to do it. Or did you used another program? – 1muflon1 Nov 4 '18 at 17:36
• No, I pulled it from the encyclopedia of mathematics. Plotting this in MATLAB could be done by taking a spacing of points for t and plugging them in for x and y, then plotting the resulting curve. This would need to be done for a few different initial points to get a good diagram. – Josh B. Nov 4 '18 at 17:44 | 2019-07-16T22:22:38 | {
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https://math.stackexchange.com/questions/399394/if-left-fx-right-leq-a-fx-beta-then-f-is-a-constant-function | If $\left| f'(x) \right| \leq A |f(x)|^\beta$ then f is a constant function
Problem Let $f(x)$ be a differentiable function on $[a,b]$ satisfying $f(a)=0$. If there exist $A \ge 0$ and $\beta \ge 1$ such that the inequality
$$\left| f'(x) \right| \leq A \left| f(x) \right|^\beta$$
holds for all $x$ in $[a,b]$, then $f(x) = 0$ for any x in that interval.
My attempted solution
We shall prove by contradiction. Suppose that for some $x$ in $[a,b]$ we have $f(x) \neq 0$.
Put $S:= \left\lbrace x \right.$ such that $a \le x \le b$ and $f(x) \neq 0 \left. \right\rbrace$ and $c:=\inf S$ (the infimum value of set $S$). If $c=b$ then $f(x) = 0$ $\forall x < b$ and the continuity of $f$ implies that $f(b) = 0$, too (contradiction). Therefore $c \neq b$. If $c=a$, then $f(c) = 0$ by the hypothesis. Otherwise, when $a < c < b$ we have $f(x) = 0$ $\forall x < c$ and again, by continuity, $f(c) = 0$. Thus in all cases considered, we have $f(c) = 0$ and $a \leq c < b$
As $f$ is continuous at $c$ and $f(c) = 0$, one can choose $d>c$ and close enough to $c$ such that $|f(x)| \le 1$ and $$\tag{\star} A(x-c) \le \frac12\quad\text{for all }x\text{ with }c \le x \le d.$$ Since $f$ is continuous on $[c,d]$, it has maximum and minimum values on this closed interval, leading to the existence of $c \le t \le d$ such that $|f(t)| = \max_{c \le x \le d} |f(x)|$. As already noted, $f(c) = 0$ and $c$ is the infimum of the set of numbers whose image under $f$ is nonzero. Therefore, for any $\epsilon>0$, there is some $c<m<c+\epsilon$ satisfying $f(m) \neq 0$. This observation, together with the definition of $t$, gives us $$\tag{\star\!\star}f(t) \neq 0,$$ $t \neq c$ and so $c<t$. Applying the Lagrange theorem gives:
$$|f(t)| = |f(t) - f(c)| = |t-c||f'(u)| \leq|t-c|A|f(u)|^\beta$$ where $c<u<t\le d$. By $(\star)$: $$|t-c|A|f(u)|^\beta \le \frac12 |f(u)|^\beta \leq \frac12 |f(u)|$$
Combining the two inequalities and noting that $|f(t)| \ge |f(u)|$ (because of the definition of $t$), we have $f(u) = f(t) = 0$ and arrived at a contradiction with $(\star\star)$.
Question
I hope someone can verify if my solution is correct. Of course, other ideas, comments or solutions are welcome.
I like to post problems and my solutions to the forum because I think it's beneficial to the community, and for learners like me. First, I can hardly know if there's flaw in my own argument. Second, I may get new insights/solutions for my problem.
Thank you.
• Sorry, I am not yet clear how the two threads are related? – tom_a2 May 22 '13 at 16:43
• You are absolutely right. I misread your question. – Martin May 22 '13 at 16:45
• It looks good to me. – Julien May 22 '13 at 17:32
Let $B = \{x\in [a,b]: |f(x)|=0\}$. By continuity, this set is closed. Since $f(0) = 0$, the set is non-empty. We show that it is also open.
Let $x\in B$, i.e. $f(x) =0$. By continuity, there exists $0<r<(2A)^{-1}$, such that $|f(y)|<\frac 12$ for all $y\in B_r(x)$. Fix such a $y\in B_r(x)$. We want to show that $f(y) =0$.
By the mean value inequality, we have $$|f(y)| = |f(y) - f(x)| \le |f'(\xi_0)| |y-x| \le |f'(\xi_0)|r$$ for some $\xi_0\in B_r(x)$ lying on the segment between $x$ and $y$. By assumption, we have $|f'(\xi_0)|\le A |f(\xi_0)|^\beta$, hence $$|f(y)|\le |f'(\xi_0)|r \le A|f(\xi_0)|^\beta (2A)^{-1} \le \frac 12|f(\xi_0)|^\beta.$$ By the same argument applied to $\xi_0\in B_r(x)$ instead of $y$, we see that there exists $\xi_1\in B_r(x)$, such that $|f(\xi_0)|\le \frac 12|f(\xi_1)|^\beta$. Iterating yields a sequence $\xi_n \in B_r(x)$, such that $|f(\xi_{n})|\le \frac 12|f(\xi_{n+1})|^\beta$ for all $n$. Note that by our choice of $r$, we have $|f(\xi_{n})|\le \frac 12$ for all $n$. We conclude that $$|f(y)|\le \frac 12 |f(\xi_0)|^\beta \le \frac 1{2^2}|f(\xi_1)|^{\beta^2} \le \frac{1}{2^3}|f(\xi_2)|^{\beta^3}\le \dots \le \frac{1}{2^n}|f(\xi_{n-1})|^{\beta^n}\le \frac 1{2^{n+\beta^n}},$$ for all $n\in \mathbb N$. Letting $n\to \infty$ shows that $|f(y)| = 0$ (using also $\beta \ge 1$). This implies that $B_r(x)\subset B$, hence $B$ is open.
Since $[a,b]$ is connected, it follows that $B$ must be all of $[a,b]$, hence $f= 0$.
( Edit. The ODE analysis also explains why $\beta \geq 1$ is necessary. The condition for $|f'(x)| \leq A|H(f(x))|$ to have the property that $f$ does not change sign, when $H$ is a function such that $H(x)=0$ only at $x=0$, is that $\frac{1}{H(x)}$ an has a non-integrable singularity at $0$. Knowing that this is the answer, there is no loss in considering singularities that are powers of $f$, which is what most cases would look like and how they would be recognized.)
The differential equation $f'(x) = A(x) f(x)^\beta$ can be solved explicitly, by separation of variables. The calculation shows that $f(a)=0$ is necessary, or we could write down nowhere zero solutions, by solving the equation. This bears further analysis, because superficially the problem looks like a Lipschitz condition with $|f(x) - f(y)| \leq |x-y|^\beta$ which (as is very well known and often solved as an exercise) implies constant $f$ when $\beta > 1$.
The DE for constant $\beta > 1$, writing $y$ for $f(x)$, is $$d(\frac{1}{(\beta - 1)y^{\beta - 1}}) = A(x) dx$$
The coordinates that trivialize the equation are therefore $Y = \frac{1}{(\beta - 1)y^{\beta - 1}}$ and $X = \int A(x) dx$, in which the solutions are $Y = X + c$ for constant $c$. This brings out the idea that $A$, which appears at first to be a harmless constant removable by a linear change of variable, has a meaningful part in the problem, as velocity of the independent variable.
Under this coordinate change, $f(a)=0$ becomes $f(a)=\infty$. On a maximal interval where $f$ is nonzero, there cannot be any solutions that hit infinity at the boundary in finite "time" (here $X$ is time, so we mean a finite change in $\int A(x) dx$) while satisfying $Y - X =$ constant.
So the meaning of the problem seems to be: no singularity can be reached in finite time for this ODE.
• Let's put here the comment that for $\beta < 1$ there are counterexamples with $f(x)$ a positive power of $(x-a)$. – zyx May 26 '13 at 22:38
Supposing that f(x)=0 only for a and that f(x)>0 for the rest x in [a,b].
Then lim (f(x) - f(a))/(x-a) as x->a+ will be greater or equal to zero. If greater then f'(a)=m >0 and thats contradicting with the given inequality.Therefore f(x)=0 for all x in [a,b]. If equal, then im sorry but i need to give it some more thought. In the same manner, u can prove for f(x)<0.
I just posted this insufficient answer cuz i thought it would be simpler to understand. | 2019-07-20T16:31:31 | {
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Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
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Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 27 Jun 2018, 18:28 6 9 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64 $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
### Show Tags
29 Jun 2018, 06:58
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First, choose a nice, round number, such as 70, within the range of values.
Then calculate the average with the following formula:
Average = Nice number + Average of differences from the nice number
Average = 70 + (4 - 1 - 6 + 9 - 6 + 14 + 7)/7 = 70 + 21/7 =73
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 28 Jun 2018, 18:07 1 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
E. $85 We can determine the average using the formula: average = sum / number: average = (74 + 69 + 64 + 79 + 64 + 84 + 77)/7 = 511/7 = 73 Answer: C _________________ # Scott Woodbury-Stewart Founder and CEO [email protected] 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. VP Joined: 14 Feb 2017 Posts: 1310 Location: Australia Concentration: Technology, Strategy GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31 GMAT 6: 600 Q38 V35 GPA: 3 WE: Management Consulting (Consulting) Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink]
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Admittedly I made a stupid mistake and I organised the numbers in ascending order then selected the median (74) instead of solving Sum/# Terms
The method I used is only for consecutive integers and the punishment answer D is there for suckers like me.
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 25 Jun 2018, 05:10 Solution Given: • Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77
To find:
• Smith’s average weekly grocery bill over the 7-week period
Approach and Working:
• Smith’s total bill amount over the 7-week period = (74 + 69 + 64 + 79 + 64 + 84 + 77) = 511
• Therefore, Smith’s average bill amount over the same period = $$\frac{511}{7}$$ = 73
Hence, the correct answer is option C.
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28 Aug 2018, 03:42
Bunuel wrote:
Over the past 7 weeks, the Smith family had weekly grocery bills of $74,$69, $64,$79, $64,$84, and $77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period? A.$64
B. $70 C.$73
D. $74 E.$85
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Easy way out is consider avg 70 -
so diff for all terms is 4, -1 , -6, 9 , -6,14, 7 = sum is 21 = divided by 7 = equal to 3 so 70 +3 = 73 avg
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Re: Over the past 7 weeks, the Smith family had weekly grocery bills of $7 [#permalink] ### Show Tags 10 Sep 2018, 07:34 Bunuel wrote: Over the past 7 weeks, the Smith family had weekly grocery bills of$74, $69,$64, $79,$64, $84, and$77. What was the Smiths' average (arithmetic mean) weekly grocery bill over the 7-week period?
A. $64 B.$70
C. $73 D.$74
E. $85 NEW question from GMAT® Official Guide 2019 (PS07369) $$\frac{( 70 + 4 ) + ( 70 - 1 ) + ( 60 + 4 ) + ( 80 - 1 ) + ( 60 + 4 ) + ( 80 + 4 ) + ( 70 + 7 )}{7}$$ = $$\frac{( 70 + 70 + 60 + 80 + 60 + 80 + 70 ) + ( 4 - 1 + 4 - 1 + 4 + 4 + 7 )}{7}$$ = $$\frac{490 + 21}{7}$$ = $$\frac{490}{7} + \frac{21}{7}$$ = $$70 + 3$$ = $$73$$, Answer must be (C) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Intern Joined: 02 May 2018 Posts: 12 GMAT 1: 620 Q46 V29 Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink]
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10 Sep 2018, 07:46
We can rule out
64, 74 and 85 as they are the extreme numbers.
That leaves just two choices to choose from 70 and 73.
Only 3 values in the 60s and therefore less likely for 70.
73 is the most likely choice. Confirm 73 by taking difference and summing it up
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Is |x| = |y|? (1) x = -y (2) x^2 = y^2
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Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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Is |x| = |y|?
(1) x = -y
(2) x^2 = y^2
[Reveal] Spoiler:
I found this question in GMAT Prep and selected ST2 as the answer but that's wrong.
Normally, x^2 would equal to lxl since it will contain both signs "positive as well as negative". How come st 1 is true?
Does it assume that since +x = -y, it must be true that -x is also equal to -y or?
What am I missing?
[Reveal] Spoiler: OA
Last edited by Bunuel on 11 Nov 2017, 01:25, edited 3 times in total.
Formatted the question.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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HarveyKlaus wrote:
Is lxl = lyl ?
A) x=-y
b) x^2 = y^2
[Reveal] Spoiler:
I found this question in GMAT Prep and selected ST2 as the answer but that's wrong.
Normally, x^2 would equal to lxl since it will contain both signs "positive as well as negative". How come st 1 is true?
Does it assume that since +x = -y, it must be true that -x is also equal to -y or?
What am I missing?
You are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.
Per statement 1:$$x=-y$$ . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.
Per statement 2: $$x^2=y^2$$ ---> $$x= \pm y$$. Again both the cases, x=y and x=-y give you a "yes" for the question asked as is hence sufficient.
Both statements are sufficient ---> D is the correct answer.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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17 Dec 2015, 06:54
statement 1 is sufficient as absolute value for -ve should give +ve
statement 2 is sufficient as
it will give either x and y both are +ve or both are -ve
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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10 Nov 2017, 15:24
Is |x|=|y|?
1) x=-y
2) x^2=y^2
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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10 Nov 2017, 17:20
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You are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.
Per statement 1:x=−yx=−y . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.
Per statement 2: x2=y2x2=y2 ---> x=±yx=±y. Again both the cases, x=y and x=-y give you a "yes" for the question asked as is hence sufficient.
Both statements are sufficient ---> D is the correct answer.
Please give me kudos. I need them to unlock gmatclub tests.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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11 Nov 2017, 01:25
SamDGold wrote:
Is |x|=|y|?
1) x=-y
2) x^2=y^2
Merging topics. Please check the discussion above.
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Kudos [?]: 133106 [0], given: 12411
Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink] 11 Nov 2017, 01:25
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http://math.stackexchange.com/questions/94359/need-help-deriving-recurrence-relation-for-even-valued-fibonacci-numbers | # Need help deriving recurrence relation for even-valued Fibonacci numbers.
That would be every third Fibonacci number, e.g. $0, 2, 8, 34, 144, 610, 2584, 10946,...$
Empirically one can check that:
$a(n) = 4a(n-1) + a(n-2)$ where $a(-1) = 2, a(0) = 0$.
If $f(n)$ is $\operatorname{Fibonacci}(n)$ (to make it short), then it must be true that $f(3n) = 4f(3n - 3) + f(3n - 6)$.
I have tried the obvious expansion:
$f(3n) = f(3n - 1) + f(3n - 2) = f(3n - 3) + 2f(3n - 2) = 3f(3n - 3) + 2f(3n - 4)$ $= 3f(3n - 3) + 2f(3n - 5) + 2f(3n - 6) = 3f(3n - 3) + 4f(3n - 6) + 2f(3n - 7)$ ... and now I am stuck with the term I did not want. If I do add and subtract another $f(n - 3)$, and expand the $-f(n-3)$ part, then everything would magically work out ... but how should I know to do that? I can prove the formula by induction, but how would one systematically derive it in the first place?
I suppose one could write a program that tries to find the coefficients x and y such that $a(n) = xa(n-1) + ya(n-2)$ is true for a bunch of consecutive values of the sequence (then prove the formula by induction), and this is not hard to do, but is there a way that does not involve some sort of "Reverse Engineering" or "Magic Trick"?
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Why isn't $f_{n+2}=3f_n-f_{n-2}$ suitable? – Guess who it is. Dec 27 '11 at 4:31
@J.M., sorry I do not understand. If I were to expand $f_{n+2}$ the way I do it, I would end up with $2f_n + f_{n - 2} + f_{n - 3}$. Again, I can make this work if I know what I am trying to get. I wonder if you have read the question correctly - I am looking for even-valued (not even-indexed) fib numbers. If my assumption is mistaken, then sorry. However, I am not sure how to systematically arrive at a relation you have given and how to use it to help me simplify things. – Job Dec 27 '11 at 4:45
It seems I did misunderstand you (and I apologize for this); have you looked at the references here by any chance? – Guess who it is. Dec 27 '11 at 4:50
## 8 Answers
The definition of $F_n$ is given:
• $F_0 = 0$
• $F_1 = 1$
• $F_{n+1} = F_{n-1} + F_{n}$ (for $n \ge 1$)
Now we define $G_n = F_{3n}$ and wish to find a recurrence relation for it.
Clearly
• $G_0 = F_0 = 0$
• $G_1 = F_3 = 2$
Now we can repeatedly use the definition of $F_{n+1}$ to try to find an expression for $G_{n+1}$ in terms of $G_n$ and $G_{n-1}$.
\begin{align*} G_{n+1}&= F_{3n+3}\\ &= F_{3n+1} + F_{3n+2}\\ &= F_{3n-1} + F_{3n} + F_{3n} + F_{3n+1}\\ &= F_{3n-3} + F_{3n-2} + F_{3n} + F_{3n} + F_{3n-1} + F_{3n}\\ &= G_{n-1} + F_{3n-2} + F_{3n-1} + 3 G_{n}\\ &= G_{n-1} + 4 G_{n} \end{align*}
so this proves that $G$ is a recurrence relation.
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At André's request, I've decided to write an answer. I've also arbitrarily decided to be ambitious and greedy, and I will thus derive a recurrence for the $k$-th increment Fibonacci number $f_{kn}$. (For OP's specific case, $k=3$)
Like André, I shall also start with Binet:
$$f_{kn}=\frac{\phi^{kn}-(-\phi)^{-kn}}{\sqrt 5}$$
Letting $u=\phi^k$ and $v=\left(-\dfrac1\phi\right)^k$, the formula takes the form
$$f_{kn}=pu^n+qv^n$$
This means that the characteristic polynomial for the recurrence satisfied by $f_{kn}$ takes the form
\begin{align*} x^2-(u+v)x+uv&=x^2-\left(\phi^k+\left(-\frac1\phi\right)^k\right)x+\left(\phi^k\left(-\frac1\phi\right)^k\right)\\ &=x^2-\left(\phi^k+\left(-\frac1\phi\right)^k\right)x+(-1)^k \end{align*}
and the recurrence itself goes like
$$f_{k(n+1)}=\left(\phi^k+\left(-\frac1\phi\right)^k\right)f_{kn}-(-1)^k f_{k(n-1)}$$
You might say that the form $\ell_k=\phi^k+\left(-\dfrac1\phi\right)^k$ is a bit unwieldy, and I agree. There are two ways to go about (slightly) simplifying this. One way makes use of the Newton-Girard formulae. These formulae express $\ell_k$ in terms of $\phi-\dfrac1\phi=1$ and $\phi\left(-\dfrac1\phi\right)=-1$. To use $k=3$ as an example:
$$\alpha^3+\beta^3=(\alpha+\beta)^3-3(\alpha+\beta)(\alpha\beta)$$
Making the replacement $\alpha+\beta=1$ and $\alpha\beta=-1$, we have
$$\ell_3=(1)^3-3(1)(-1)=4$$
The slick way is to recognize that since $\ell_k$ is itself a linear combination of $\phi^k$ and $\left(-\dfrac1\phi\right)^k$, it also satisfies the Fibonacci recurrence:
$$\ell_{k+1}=\ell_k+\ell_{k-1}$$
The $\ell_k$ are in fact the (not-so-famous) Lucas numbers. With $\ell_0=2$ and $\ell_1=1$, we have the sequence $2, 1, 3, 4, 7, 11,\dots$
In short, the recurrence is of the form
$$f_{k(n+1)}=\ell_k f_{kn}-(-1)^k f_{k(n-1)}$$
For $k=3$, we have $f_{3(n+1)}=\ell_3 f_{3n}-(-1)^3 f_{3(n-1)}$ or $f_{3(n+1)}=4 f_{3n}+f_{3(n-1)}$.
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Very nice post. This is the kind of solution where each piece falls nicely at its right place. – Did Dec 27 '11 at 9:35
It turns out that the identity derived here is listed in the Wolfram Functions site. – Guess who it is. Dec 28 '11 at 0:21
@JM: this identity is a special case of the identity $$F_{n+2k}=L_kF_{n+k}-(-1)^kF_n$$ However, the indices do not need to be multiples of $k$, they just need to differ by $k$. – robjohn Jan 16 '12 at 19:43
Let $\alpha$ and $\beta$ be the two roots of the equation $x^2-x-1=0$. Then the $n$-th Fibonacci number is equal to $$\frac{\alpha^n-\beta^n}{\sqrt{5}}.$$
We are interested in the recurrence satisfied by the numbers $$\frac{\alpha^{3n}-\beta^{3n}}{\sqrt{5}}.$$
If $x$ is either of $\alpha$ or $\beta$, then $x^2=x+1$. Multiply by $x$. We get $x^3=x^2+x=2x+1$. It follows that $x^4=2x^2+x=3x+2$. But then $x^5=3x^2+2x=5x+3$, and then $x^6=5x^2+3x=8x+5$.
We want $x^6=Ax^3+B$, where $A$ and $B$ are rational, indeed integers. So we want $8x+5=A(2x+1)+B$. Reading off $A$ and then $B$ is obvious: we need $A=4$ and $B=1$.
So the numbers $\alpha^{3n}$ and $\beta^{3n}$ satisfy the recurrence $y_n=4y_{n-1}+y_{n-2}$. By linearity, so do the numbers $\frac{\alpha^{3n}-\beta^{3n}}{\sqrt{5}}$.
Comment: Note that using the same basic strategy, we can write down the recurrence satisfied by $\frac{\alpha^{kn}-\beta^{kn}}{\sqrt{5}}$. The coefficients that we painfully computed by hand, step by step, can be expressed simply in terms of Fibonacci numbers, and therefore so can the recurrence for the numbers $\frac{\alpha^{kn}-\beta^{kn}}{\sqrt{5}}$.
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Nice! I think Newton-Girard might also yield a useful route here. – Guess who it is. Dec 27 '11 at 5:11
I thought I would take an "unfancy" route through facts likely to be well-known. – André Nicolas Dec 27 '11 at 5:18
Here's the Newton-Girard route for completeness: one wants the characteristic polynomial $x^2-(\alpha^3+\beta^3)x+\alpha^3 \beta^3$ without knowing $\alpha$ or $\beta$, but knowing that $\alpha+\beta=1$ and $\alpha\beta=-1$ (Vieta). It is easily seen that $\alpha^3 \beta^3=-1$. Using Newton-Girard, we have the symmetric polynomial expansion $\alpha^3+\beta^3=(\alpha+\beta)^3-3(\alpha+\beta)(\alpha\beta)$, and thus $\alpha^3+\beta^3=(1)^3-3(1)(-1)=4$. The characteristic polynomial is thus $x^2-4x-1$. – Guess who it is. Dec 27 '11 at 5:25
@J.M.: I think that the comment above should be made into an answer. – André Nicolas Dec 27 '11 at 5:29
By inspection $f(3n+3)=4f(3n)+f(3n-3)$, as you’ve already noticed. This is easily verified:
\begin{align*} f(3n+3)&=f(3n+2)+f(3n+1)\\ &=2f(3n+1)+f(3n)\\ &=3f(3n)+2f(3n-1)\\ &=3f(3n)+\big(f(3n)-f(3n-2)\big)+f(3n-1)\\ &=4f(3n)+f(3n-1)-f(3n-2)\\ &=4f(3n)+f(3n-3)\;. \end{align*}
However, I didn’t arrive at this systematically; it just ‘popped out’ as I worked at eliminating terms with unwanted indices.
Added: Here’s a systematic approach, but I worked it out after the fact.
The generating function for the Fibonacci numbers is $$g(x)=\frac{x}{1-x-x^2}=\frac1{\sqrt5}\left(\frac1{1-\varphi x}-\frac1{1-\hat\varphi x}\right)\;,$$ where $\varphi = \frac12(1+\sqrt5)$ and $\hat\varphi=\frac12(1-\sqrt5)$, so that $f(n)=\frac1{\sqrt5}(\varphi^n-\hat\varphi^n)$. Thus, $f(3n)=\frac1{\sqrt5}(\varphi^{3n}-\hat\varphi^{3n})$. Thus, we want
\begin{align*} h(x)&=\frac1{\sqrt5}\sum_{n\ge 0}(\varphi^{3n}-\hat\varphi^{3n})x^n\\ &=\frac1{\sqrt5}\left(\sum_{n\ge 0}\varphi^{3n}x^n-\sum_{n\ge 0}\hat\varphi^{3n}x^n\right)\\ &=\frac1{\sqrt5}\left(\frac1{1-\varphi^3 x}-\frac1{1-\hat\varphi^3 x}\right)\\ &=\frac1{\sqrt5}\cdot\frac{(\varphi^3-\hat\varphi^3)x}{1-(\varphi^3+\hat\varphi^3)x+(\varphi\hat\varphi)^3x^2}\;. \end{align*}
Now $\varphi+\hat\varphi=1$, $\varphi-\hat\varphi=\sqrt5$, $\varphi\hat\varphi=-1$, $\varphi^2=\varphi+1$, and $\hat\varphi^2=\hat\varphi+1$, so
\begin{align*} h(x)&=\frac1{\sqrt5}\cdot\frac{(\varphi^3-\hat\varphi^3)x}{1-(\varphi^3+\hat\varphi^3)x+(\varphi\hat\varphi)^3x^2}\\ &=\frac{(\varphi^2+\varphi\hat\varphi+\hat\varphi^2)x}{1-(\varphi^2-\varphi\hat\varphi)x-x^2}\\ &=\frac{(\varphi^2-1+\hat\varphi^2)x}{1-(\varphi^2+1+\hat\varphi^2)x-x^2}\\ &=\frac{(\varphi+\hat\varphi+1)x}{1-(\varphi+3+\hat\varphi)x-x^2}\\ &=\frac{2x}{1-4x-x^2}\;. \end{align*}
It follows that $(1-4x-x^2)h(x)=2x$ and hence that $h(x)=4xh(x)+x^2h(x)+2x$. Since the coefficient of $x^n$ in $h(x)$ is $f(3n)$, this tells me that
\begin{align*} \sum_{n\ge 0}f(3n)x^n&=h(x)=4xh(x)+x^2h(x)+2x\\ &=\sum_{n\ge 0}4f(3n)x^{n+1}+\sum_{n\ge 0}f(3n)x^{n+2}+2x\\ &=\sum_{n\ge 1}4f(3n-3)x^n+\sum_{n\ge 2}f(3n-6)x^n+2x\;, \end{align*}
which by equating coefficients immediately implies that $f(3n)=4f(3n-3)+f(3n-6)$ for $n\ge 2$. It also gets the initial conditions right: the constant term on the righthand side is $0$, and indeed $f(3\cdot 0)=0$, and the coefficient of $x$ is $4f(0)+2=2=f(3)$, as it should be.
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Actually the "Magic Trick" or "Reverse Engineering" idea works nicely.
First, as André pointed
$$f_{3n}=\frac{\alpha^{3n}+\beta^{3n}}{\sqrt{5}} \,.$$
This means that if $(x-\alpha^{3})(x-\beta^3)=x^2-Ax-B$ then $f_{3n}$ is the recurrence satisfying
$$x_{n+2}=Ax_{n+1}+Bx_{n} \, x_{0}=f_0, x_1=f_{3} \,.$$
Thus,
$$f_6=Af_3+Bf_0$$ $$f_9=Af_6+Bf_3$$
Solve it and get $A,B$.
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Let $S$ be the shift operator on sequences (as in Bill Dubuque's answer). Note that the Fibonacci sequence is killed by $S^2-S-1$. The Fibonacci sequence will then be killed by any "polynomial" multiple of $S^2-S-1$. To get a recurrence for every $k^{\rm{th}}$ term, all we need to do is find a multiple of $S^2-S-1$ that only involves powers of $S^k$.
First note that $S^2-S-1=(S-a)(S-b)$ where $a=\phi$ (the golden ratio) and $b=-1/\phi$. Consider the operator $(S^k-a^k)(S^k-b^k)=S^{2k}-(a^k+b^k)S^k+(ab)^k$. It is a polynomial multiple of $S^2-S-1$, so it kills the Fibonacci sequence. It only involves powers of $S^k$.
Recall that one formula for the $k^{\rm{th}}$ Lucas number is $L_k=a^k+b^k$, and note that $ab=-1$. Thus, we get that $S^{2k}-L_kS^k+(-1)^k$ kills the Fibonacci sequence.
Therefore, in summary, we get $$F_{n+2k}=L_kF_{n+k}-(-1)^kF_n\tag{1}$$ For example, $$F_{n+2}=F_{n+1}+F_n\tag{k=1}$$ $$F_{n+4}=3F_{n+2}-F_n\tag{k=2}$$ $$F_{n+6}=4F_{n+3}+F_n\tag{k=3}$$ $$F_{n+8}=7F_{n+4}-F_n\tag{k=4}$$ $$F_{n+10}=11F_{n+5}+F_n\tag{k=5}$$
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We can write linear recurrence relations in terms of matrix multiplication like so:
$$F_{n+2} = \left[ \begin{array}{cc} 1 & 1 \end{array} \right] \left[ \begin{array}{cc} F_n \\ F_{n+1} \end{array} \right] = 1 \cdot F_n + 1 \cdot F_{n+1}.$$
Now if the sequence 0,2,8,34,144,610,2584,10946,... is called $G_n$, let's make the unjustified assumption that it is also a second order recurrence relation, then not only would we have
$$G_{n+2} = \left[ \begin{array}{cc} c_1 & c_2 \end{array} \right] \left[ \begin{array}{cc} G_n \\ G_{n+1} \end{array} \right]$$
for some unknowns $c_1$ and $c_2$, but also we can collect several instances of the above identity together into one e.g.
$$\left[ \begin{array}{cc} 34 & 144 \end{array} \right] = \left[ \begin{array}{cc} c_1 & c_2 \end{array} \right] \left[ \begin{array}{cc} 2 & 8 \\ 8 & 34 \end{array} \right]$$
and we can solve this using PARI/GP like so:
? [34,144]/[2,8;8,34]
% = [1, 4]
Therefore $G_{n+2} = 1 \cdot G_n + 4 \cdot G_{n+1}$.
About the assumption, it can be proved in general based on the ideas of characteristic function which you have seen in most of the answers. Given that, there is no need for any induction proofs or anything. Just computing the vector completes the proof of $G_{n+2} = 1 \cdot G_n + 4 \cdot G_{n+1}$.
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Here's an interesting identity involving Lucas and Fibonacci numbers: $$\begin{pmatrix}\ell_k&(-1)^{k+1}\\1&0\end{pmatrix}=\begin{pmatrix}f_k&f_{k-1}\\0&1\end{pmatrix}\cdot\begin{pmatrix}1&1\\1&0\end{pmatrix}^k \cdot\begin{pmatrix} f_k&f_{k-1}\\0&1\end{pmatrix}^{-1}$$ – Guess who it is. Dec 28 '11 at 15:21
It is easy by operator algebra. Let the Shift, Triple $\mathbb C$-linear operators $\rm\ S\:n\: :=\: n+1,\ \ T\:n\: :=\: 3\:n\:$ act on fibonacci's numbers by $\rm\ S\:f(a\:n+b) = f(a\:(n+1)+b)\$ and $\rm\ T\:f(a\:n+b) = f(3\:a\:n+b)\:.$ Below I show a general method that works for any Lucas sequence $\rm\:f(n)\:$ that involves only simple high-school polynomial arithmetic (albeit noncommutative). Namely, one employs a commutation rule $\rm\: TS\:\to\: (a\ S + b)\ T\$ to shift $\rm\:T\:$ past powers of $\rm\:S\:,\:$ in order to transmute the known recurrence $\rm\ q(S)\ f(n)\: =\: 0\$ into $\rm\ \bar{q}(S)\:T\:f(n)\:=\:0\:,\$ the sought recurrence for $\rm\ T\:f(n)\: =\: f(3\:n)\:.\:$
We know $\rm\ q(S)\ f(n) := (S^2 - S - 1)\ f(n)\: =\: f(n+2) - f(n+1) - f(n)\: =\: 0\:.\:$ We seek an analogous recurrence $\rm\ \bar{q}(S)\ T\: f(n)\ =\ 0\$ for $\rm\ T\:f(n) = f(3\:n)\:,\:$ and some polynomial $\rm\:\bar{q}(S)\:.\:$ Since clearly we have that $\rm\ T\:q(S)\ f(n)\: =\: 0\:,\:$ it suffices to somehow transmute this equation by shifting $\rm\:T\:$ past $\rm\:q(S)\:$ to yield $\rm\:\bar{q}(S)\:T\:f(n)\:=\:0\:.\:$ To do this, it suffices to find some commutation identity $\rm T\:S\: =\: r(S)\: T\$ to enable us to shift $\rm\:T\:$ past $\rm\:S$'s in each monomial $\rm\ S^{\:i}\: f(n)\: =\: f(n+i)\:$ from $\rm\:q(S)\:.\:$ The sought commutation identity arises very simply: iterate the recurrence for $\rm\:f(n)\:$ so to rewrite
$\rm\ ST\ f(n)\ =\ f(3\:n+3)\$ as a linear combination of $\rm\ f(3\:n+1) = TS\ f(n)\:,\:$ $\rm\ f(3\:n) = T\ f(n)\:,\:$ viz.
$\rm\ \ \ ST\ f(n+i)\ =\ f(3n+3+i)\ =\ f(3n+2+i) + f(3n+1+i)\ =\ 2\ f(3n+1+i) + f(3n+i)$
$\rm\ \ \ \phantom{ST\ f(n+i)}\ =\ (2\:TS+T)\ f(n+i)\quad$ for all $\rm\:i\in \mathbb Z\:$
$\rm\ 2\:TS\ f(n+i)\ =\ (S-1)\:T\ f(n+i)\:,\$ i.e. $\rm\ 2\:TS\ =\ (S-1)\:T\:,\$ the sought commutation identity.
Thus $\rm\qquad\quad\:\ 0\ =\ 4\: T\: (S^2 - S - 1)\ f(n)\$
$\rm\qquad\qquad\qquad\quad\ \ =\ (2\:(2TS)S - 2\:(2TS) - 4\:T)\ f(n)$
$\rm\qquad\qquad\qquad\quad\ \ =\ ((S-1)\:2TS - 2\:(S-1)\:T - 4\:T)\ f(n)$
$\rm\qquad\qquad\qquad\quad\ \ =\ ((S-1)^2 - 2\:(S-1)\: - 4)\ T\: f(n)$
$\rm\qquad\qquad\qquad\quad\ \ =\ (S^2 - 4\ S - 1)\ T\: f(n)$
$\quad$ i.e. $\rm\qquad\quad\: 0\ =\ f(3(n+2)) - 4\ f(3(n+1)) - f(3\:n)\qquad$ QED
NOTE $\$ Precisely the same method works for any Lucas sequence $\rm\:f(n)\:,\:$ i.e. any solution of $\rm\ 0\ =\ (S^2 + b\ S + c)\ f(n)\ =\ f(n+2) + b\ f(n+1) + c\ f(n)\$ for constants $\rm\:b,\:c\:,\:$ and for any multiplication operator $\rm\:T\:n = k\ n\:$ for $\rm\:k\in \mathbb N\:.\:$ As above, we obtain a commutation identity by iterating the recurrence (or powering its companion matrix), in order to rewrite
$\rm\ ST\ f(n)\ =\ f(k\:n+k)\$ as a $\rm\:\mathbb C$-linear combination of $\rm\ f(kn+1) = TS\ f(n)\$ and $\rm\ f(kn) = T\ f(n)\:$
say $\rm\ \ ST\ f(n)\ =\ f(k\:n+k)\ =\ a\ f(k\:n+1) + d\ f(k\:n)\ =\ (a\ TS + d\ T)\ f(n)\ \$ for some $\rm\:a,d\in \mathbb C$
$\rm\:\Rightarrow\ a\ TS\ f(n) =\ (S-d)\ T\ f(n)\ \Rightarrow\ a\ TS\ =\ (S-d)\ T\$ on $\rm\ S^{\:i}\: f(n)\$ as above.
Again, this enables us to transmute the recurrence for $\rm\:f(n)\:$ into one for $\rm\:T\:f(n) = f(k\:n)\:$ by simply commuting $\rm\:T\:$ past all $\rm\:S^i\:$ terms. Hence the solution involves only simple polynomial arithmetic (but, alas, the notation obscures the utter simplicity of the method).
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https://math.stackexchange.com/questions/2876194/proof-verification-if-a-subset-b-and-b-subset-c-then-a-cup-b-subset/2876237 | # Proof Verification: If $A \subset B$ and $B \subset C$, then $A \cup B \subset C$
I am trying to prove that:
If $A \subset B$ and $B \subset C$, then $A \cup B \subset C$
My proof is : Given some $x \in A \cup B$, it is true that either $x \in A$ and/or $x \in B$. IN the case that $x \in A$ it is true that $x \in B$, as $A \subset B$, and that $x \in C$ , as $B \subset C$. In the case that $x \in B$ it is true that $x \in C$, as $B \subset C$. Therefore, $A\cup B \subset C$
Is this correct?. Any tips to improve this would be appreciated as I am self taught and new to proof writing.
• Yes, it is correct. You could also prove that $A \cup B = B$, and then the fact follows from $B = A \cup B \subset C$. – астон вілла олоф мэллбэрг Aug 8 '18 at 15:14
• Your proof is totally correct. Just be careful about the use of English: it is better to say "Given some $x \in A \cup B$, it is true that $x \in A$ or $x \in B$". Do not use "either" and "and" unnecessarily as it may make your statement confusing. In maths, the word "or" means : either the first case only, either the second only, either both cases at the same time. – Suzet Aug 8 '18 at 15:16
• @астонвіллаолофмэллбэрг Thank you – Ewan Miller Aug 8 '18 at 15:16
• @Suzet Ok thank you – Ewan Miller Aug 8 '18 at 15:17
• It's totally correct, although there's an alternative way. Since $A \subset B$, $A \cup B= B$. Now from $B \subset C$, the result follows. – Anik Bhowmick Aug 8 '18 at 15:32
Just to answer : yes, the approach is correct.
You could also prove that $A∪B=B$, and then the fact follows from $B=A∪B\subset C$.
For example, if $x \in B$ is true, then of course $x$ is in $A$ or $x$ is in $B$ is true, so $B \subset A \cup B$. If $x \in A$, then $x \in B$ because $A \subset B$, and if $x \in B$, then of course $x \in B$, so $A \cup B \subset B$, hence $A \cup B = B$.
Another approach is by using Venn diagrams. Draw circles $A$, $B$ and $C$ for three sets such that $A$ is contained in $B$ and $B$ is contained in $C$ (according to given set inclusions). So you have $A$ as the innermost, $B$ in the middle and $C$ as the outermost of them. Now $A\cup B$ is given by the middle circle which is offcourse contained in $C$ (the outer circle).
Is this correct?
I believe that your proof regarding this Question is correct.
Any tips to improve this would be appreciated as I am self-taught and new to proof writing.
Set Theory which you are using (Elementary-Set Theory) was mostly Contributed for its Development by George Cantor. But, in Later Stage, there are some famous Loopholes found in this Theory, the most popular of them is Russel's Paradox. After Russel's paradox, numerous other Paradox came after which ZFC Set Theory (Zermelo Franklin Set Theory) came which is based on some of the Basic Set Axioms (Related to Mathematical Logic)
## Proofs in Set Theory
Most of the Proofs in Elementary Set Theory is based on Mathematical Logic. Some of the Basic Theorems and Properties are below -
1. De Morgan's Theorem -https://en.wikipedia.org/wiki/De_Morgan%27s_laws
2. Complement - https://en.wikipedia.org/wiki/Complement_(set_theory) $A^` = U - A$
3. Intersection and Union -
## Alternate Methods to Proof Set Theory Question
Your Question Can be Proved Using Venn Diagram too like here -
• Using The Laws and Properties of Sets (Intersection, Complement and Union)
The most natural way seems to be:
1. (Transitivity) From $A \subset B$ and $B \subset C$, show $A \subset C$.
If $x \in A$, then $x \in B$.
If $x \in B$, then $x \in C$.
Thus if $x \in A$, then $x \in C$.
1. From $A \subset C$ and $B \subset C$, show $(A \cup B) \subset C$.
If $x \in (A \cup B)$, then $x \in A$ or $x \in B$.
But if $x \in A$, then $x \in C$, and if $x \in B$, then $x \in C$.
Thus, either way, $x \in C$.
You implicitly used this method in your (correct) proof, but you didn't separate out the ideas. It can be especially useful to organise larger proofs in terms of simpler definitions, lemmas and theorems. | 2020-04-08T06:58:43 | {
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http://mathhelpforum.com/algebra/49692-simultaneous-equations.html | Math Help - Simultaneous Equations
1. Simultaneous Equations
I don't usually have any problem with simultaneous equations but whilst revising for a test, I came across two types that I haven't seen before and the book I have only gives answers but no explanations of how it got those answers. I've tried in various ways... but my answers never seem to match up.
Could anyone solve these two simultaneous equations step-by-step so I could see how it's done?
Q1)
$y = 1 - 4x$
$y = -4x^2$
Q2)
$xy = 9$
$x - 2y = 3$
If anyone could I would really appreciate it, thanks. And Hi to everyone on the forum - just found this forum and it looks great.
2. Hello, PaulSelb!
Welcome aboard!
These are not linear equations.
. . The recommended method is Substitution.
$1)\;\;\begin{array}{cccc}y &=& 1 - 4x &{\color{blue}[1]} \\
y &= &-4x^2& {\color{blue}[2]}\end{array}$
Equation [1] is already solved for $y\!:\;\;y \:=\:1-4x$
Substitute into [2]: . $1 - 4x \:=\:-4x^2 \quad\Rightarrow\quad 4x^2 - 4x + 1 \:=\:0$
. . $(2x-1)^2 \:=\:0 \quad\Rightarrow\quad 2x-1 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{1}{2}$
Substitute into [1]: . $y \:=\:1 - 4\left(\frac{1}{2}\right) \:=\:-1$
Solution: . $x \:=\:\frac{1}{2},\;y \:=\:-1$
$2);\;\begin{array}{cccc}xy &=& 9 & {\color{blue}[1]} \\
x - 2y &=& 3 & {\color{blue}[2]}\end{array}$
Solve [2] for $x\!:\;\;x \:=\:2y+3\;\;{\color{blue}[3]}$
Substitute into [1]: . $(2y+3)y \:=\:9 \quad\Rightarrow\quad 2y^2 + 3y - 9 \:=\:0$
Factor: . $(2y - 3)(y + 3)\:=\:0\quad\Rightarrow\quad y \:=\:\frac{3}{2},\;-3$
Substitute into [3]: . $\begin{Bmatrix}x &=& 2\left(\frac{3}{2}\right) + 3 & = & 6 \\ \\[-3mm] x &=&2(-3) + 3 &=&-3 \end{Bmatrix}$
Solutions: . $(x,y) \;=\;\left(6,\:\frac{3}{2}\right),\;(-3,\:-3)$
3. Thanks, great explanation . Got my test in a few hours so hopefully if one of these questions turns up I shouldn't have any problems.
4. I had a slightly different type in the exam the other day (possibly solved in the same way)... but I got it wrong
If anyone reads this thread again could you let me know how to solve this one?
$k(k - m) = 12$
$k(k + m) = 60$
Sorry for the easy stupid questions, I really enjoy math and want to understand bits more!
EDIT: I think I did it may have done it actually...
$k^2 - km = 12$
$k^2 + km = 60$
$k+m = \frac{60}{k}$
$m = \frac{60}{k} - k$
$k^2 - k(\frac{60}{k} - k) = 12$
$2k^2 - 60 = 12$
$2k^2 - 72 = 0$
$(2k - 12)(k + 6) = 0$
$k = \pm 6$
$36 + \pm6(m) = 60$
$m = \frac{24}{\pm6}$
$m = \pm 4$
I think that's about right anyway... | 2014-09-16T16:14:45 | {
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https://stats.stackexchange.com/questions/241652/simulating-order-statistics | # Simulating order statistics
I am having a problem with the density of the first order statistic of a series of n random variables iid with common distribution (standard normal).
I am using Arnold's book as a reference for such a density function for the k'th order stat: $$f_{X_{k:n}}(x) = \frac{n!}{k-1! (n-k)!} F(x)^{k-1} (1-F(x))^{n-k} f(x).$$
The I simulate data in R store the min and compare the obtained empirical distribution against the theoretical function. These curves diverge substantially.
What am I doing wrong? or missing?
# Distribution
Fx <- function(x) pnorm(x)
fx <- function(x) dnorm(x)
n <- 10
# ATTENTION: 6M simulation may take some time
z <- replicate(1e6, min(rnorm(n)))
# Probabiltiy z > y
prob <- function(x) {
sapply(x, function(u) mean(z>u))
}
# Density
fos_k <- function(x, n, k) {
fact <- factorial(n) / (factorial(k-1) * factorial(n-k))
fact * fx(x) * Fx(x)^(k-1) *(1-Fx(x))^(n-k)
}
curve(fos_k(x, n=n, k=1), from=-1, to=1)
curve(prob, add=T, col=2, from=-1, to=1)
• The "observed empirical distribution" is not the density. One way to plot the observed density is to approximate it with a KDE and plot that. – whuber Oct 21 '16 at 19:29
• @Xi'an could you please clarify – mrb Oct 21 '16 at 20:44
• Compare your equation to the second line of fos_k: your equation omits the term you have coded as fx(x). But that's only an error of exposition; the error in your code lies in prob, which is not a density function. – whuber Oct 21 '16 at 21:40
• For what it's worth, here's the computation done in a single line in Mathematica: Show[Histogram[ParallelTable[Min[RandomVariate[NormalDistribution[0, 1], 10]], {10^6}], {0.1}, "PDF"], Plot[Evaluate[PDF[OrderDistribution[{NormalDistribution[0, 1], 10}, 1], z]], {z, -4, 1}]] On my machine, the runtime is less than 1 second. – heropup Oct 21 '16 at 22:08
• Thanks to everyone. @whuber is right. @whuber would you like to go on and reply to this question with an example of your R code? – mrb Oct 22 '16 at 1:54
Although the problem is primarily with R code, it raises issues we would have to confront in any statistical computing environment. This reply focuses on those general issues.
A correct formula for the density of the $k^\text{th}$ smallest of $n$ independent identically distributed (iid) values from a distribution $F$ with density $f$ is presented in my answer at https://stats.stackexchange.com/a/225990/919. It is
$$f_{[k]}(x) = \frac{n!}{(k-1)!(1)!(n-k)!} F(x)^{k-1} (1-F(x))^{n-k} f(x).\tag{1}$$
Because factorials and potentially high powers are combined, it is numerically better to compute its logarithm as
\eqalign{ \log f_{[k]}(x) = &\log n! - \log(k-1)! - \log(n-k)! + \\&(k-1)\log F(x) + (n-k) \log(1-F(x)) + \log f(x).\tag{2} }
Furthermore, because order statistics provide a way to peer far out into the tails of the distribution, where values grow close to $1$ and $0$, it is best to compute $1-F(x)$ directly rather than subtracting $F(x)$ from $1$.
With these caveats in mind, the other issue in this thread concerns graphing an empirical density. The commonest way to do so is with a histogram: it looks like a barplot in which the bar areas (not heights!) are proportional to the relative frequencies of the data.
To illustrate these two main points, I wrote a quick solution in R that simulates many iid samples, extracts specified order statistics from each, plots the histogram of each order statistic, and overplots the density function $(1)$ computed by exponentiating the logarithm $(2)$. Here is an example of its output applied to samples of size $10$ from a standard Normal distribution and of size $25$ from a Gamma$(3/2,5)$ distribution.
Clearly the agreement between the empirical results (the histogram bars) and the theoretical formula (the colored curves) is good.
This code applies the preceding suggestions about numerical computation by exploiting the log.p, log, and lower.tail arguments that are standard in the families of R distribution functions.
f <- function(n.sim, n, k=1:n, p=pnorm, d=dnorm, r=rnorm, name, ...) {
if (missing(name)) name <- ""
k <- sort(unique(k))[1 <= k & k <= n]
# Perform the simulation.
sim <- apply(matrix(r(n.sim*n, ...), nrow=n), 2, sort)[k, , drop=FALSE]
# Plot the requested order statistics.
for (i in 1:length(k)) {
# Define a function to plot the density of an order statistic.
dord <- function(x, k, n, ...) {
z <- lfactorial(n) - lfactorial(k-1) - lfactorial(n-k) +
(k-1)*p(x, log.p=TRUE, ...) +
(n-k)*p(x, log.p=TRUE, lower.tail=FALSE, ...) +
d(x, log=TRUE, ...)
return(exp(z))
}
# Plot the empirical distribution.
hist(sim[i, ], freq=FALSE,
xlab="Value",
sub=paste(n.sim, "iterations with sample size", n),
main=paste(name, "order statistic", k[i]))
# Overplot the theoretical density.
curve(dord(x, k[i], n, ...), add=TRUE, col=hsv(runif(1), 0.8, 0.7), lwd=2)
}
}
#
# Set up to simulate and display the results.
#
par(mfrow=c(2,4))
n.sim <- 1e4
set.seed(17)
# Study Normal order statistics.
f(n.sim, 10, c(1,3,5,9), name="Normal(0,1)")
# Study Gamma order statistics.
# This illustrates how to use f for general distributions.
f(n.sim, 25, c(2, 4, 16, 24), name="Gamma(1.5,5)",
p=pgamma, d=dgamma, r=rgamma, shape=1.5, scale=5)
• Wondering should use the logarithm to define the corresponding distribution function as well? Since F(x) can be zero I would avoid the log. – mrb Oct 24 '16 at 13:03
• That's a good observation, but it happens not be be relevant because almost surely $F$ will be positive at any value you have randomly generated. Regardless, good software will deal with this correctly (provided you actually ask it to compute the log of $F$ directly rather than asking it to take the log of $0$!). For instance, the R procedures will return -Inf for the logarithm in that case; and that value, when exponentiated, will be exactly $0$. Here is an example: exp(punif(-2, log.p=TRUE)) – whuber Oct 24 '16 at 13:41 | 2019-10-14T20:37:25 | {
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http://mathhelpforum.com/calculus/208021-calc-word-problem-please-help-2.html | No, you need to divide by $\displaystyle 12^3$...essentially a 12 for each of the 3 spatial dimensions.
$\displaystyle 7128\text{ in}^3\cdot\left(\frac{1\text{ ft}}{12\text{ in}} \right)^3=\frac{7128}{12^3}\,\text{ft}^3=156.75 \text{ ft}^3$
Oh, ok that makes since. So for part 5 would I use the formula 1/2*b*h so 1/2*(99)*(104) = 5148 in^2
No, you have a trapezoid, not a triangle. You want to use:
$\displaystyle A=\frac{h}{2}(B+b)$
where:
$\displaystyle h=99$
$\displaystyle B=104$
$\displaystyle b=32$
O.k., sorry for not getting this right away, thanks for your patience lol.
So the answer would then be 6732
An impatient person has no business trying to help on a forum. You are doing fine.
Yes, that is the correct answer. Now, in order to find this area using integration, I recommend orienting your coordinate axes such that the origin is at the bottom corner under the lowest point of the ceiling. We will then want to write the upper slanting edge of the wall as a linear function. What would the y-intercept and the slope be?
Would the slope be 8 and the y-intercept 32?
You have the correct intercept, but for the slope, think of the rise over run of an individual step.
ok so the slope would be 8/11?
Yes, good work! So, what would the linear function representing the slanted edge be?
(8/11)x+32?
Exactly! Now, over what interval do you want to integrate?
Would it be from 5 to 13?
and if that is the interval would it be:
(4/11)x^2+32x evaluated from 5 to 13= 308.364 | 2018-06-18T18:05:39 | {
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http://mathhelpforum.com/number-theory/8194-modulo-problem.html | 1. ## Modulo problem
Okay, so I'm supposed to find the least nonnegative integer i such that
4^30 is congruent to i (mod 19).
I'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.
2. Originally Posted by clockingly
Okay, so I'm supposed to find the least nonnegative integer i such that
4^30 is congruent to i (mod 19).
I'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.
My approach here won't be unique, but the method should be fairly clear.
$4^3 = 64 \equiv 7$ (mod 19)
So
$4^{30} = (4^3)^{10} \equiv 7^{10}$ (mod 19)
Again:
$7^2 \equiv 11$ (mod 19)
So
$7^{10} = (7^2)^{5} \equiv 11^5$ (mod 19)
You can probably do this one by directly but we can simplify this one last time:
$11^2 \equiv 7$ (mod 19)
$11^3 \equiv 1$ (mod 19)
So finally:
$4^{30} \equiv 7^{10} \equiv 11^5 = 11^2 \cdot 11^3 \equiv 7 \cdot 1 = 7$ (mod 19)
$4^{30} \equiv 7$ (mod 19)
-Dan
I am still a little confused as to how you got the very first equation though (listd below)...I'm a little confused as to where the 7 came from...
4^3 = 64 is congruent to 7(mod 19)
4. Originally Posted by clockingly
Okay, so I'm supposed to find the least nonnegative integer i such that
4^30 is congruent to i (mod 19).
I'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.
You can also do this...
$4^{30}=2^{60}$
And,
$2^{18}\equiv 1 (\mbox{ mod }19)$
By Fermat's little theorem since 19 is prime not divisible by 2.
Cubing,
$2^{54}\equiv 1 (\mbox{ mod }19)$
Multiply both sides by $2^6=64\equiv 7$ we we have,
$2^{60}\equiv 7 (\mbox{ mod }19)$
5. Originally Posted by clockingly
I am still a little confused as to how you got the very first equation though (listd below)...I'm a little confused as to where the 7 came from...
4^3 = 64 is congruent to 7(mod 19)
That is becuase,
$64\equiv 7 (\mbox{ mod }19)$
Right?
Because it leaves a remainder of 19.
( $64=3\cdot 19+7$)
6. Hello, clockingly!
Another approach . . .
Find the least nonnegative integer $n$ such that: . $4^{30} \equiv n \pmod{19}$
I found that: . $4^5 \:=\:1024 \:\equiv\:-2 \pmod{19}$
Then: . $4^{30} \:=\:(4^5)^6 \:\equiv\:(-2)^6 \pmod{19}$
And: . $(-2)^6\:=\:64\:\equiv\:7\pmod{19}$
Therefore: . $4^{30}\:\equiv\:7\pmod{19}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I also found that: . $4^9\:\equiv\:1\pmod{19}$
Then: . $4^{30}\:=\:\left(4^9\right)^3\cdot4^3 \:\equiv\:(1^3)4^3\pmod{19}$
And: . $4^3\:=\:64\:\equiv\:7\pmod{19}$
But I see that TPHacker beat me to it. | 2017-06-27T10:45:37 | {
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https://math.stackexchange.com/questions/2888262/what-is-the-least-amount-of-pieces-on-a-board-with-the-following-conditions | # What is the least amount of pieces on a board with the following conditions:
There's an infinite board. Imagine you add a rectangle of $m*n$ pieces. With $m,n \geq 2$ (There's a piece every square, and you can't put one above other.) You can make a piece 'jump' other that is next to it (Vertically or horizontally only, not diagonally), if the square next to the "jumped" piece (In the same direction) is empty, also, the "jumped" piece disappears. After many movements, What is the least amount of pieces that can be on the board?
So, it's obvious that the answer can't be $0$ since there's no point of it, it has to be at least one if we have, for example, to pieces left of all the $mn$ pieces that are connected each other. I tried to do a coloring of the infinite board, in a chess pattern, so we have Black pieces (Pieces in a black tile) and White pieces, this make any white piece unable to make dissapear another white one and the same with black ones. At the end, the least we have of each color the better, because if we get to have only two left with different colors and connected, then we'll know the least number of pieces is 1. i tried looking for different rectangles of pieces, and i got 1 and 2 as answers. For example doing the following when $m=3$ and $n=4$ makes the board have 1 piece at the end: (From left to right)
It clearly follows that it has 1 piece at the end. I found other cases as ($m=2,n=4$), ($m=2, n=5$), but i also found cases where i got the least is 2, like ($m=2, n=3$), ($m=3, n=3$), etc. i haven't noticed anything else besides this. Any suggestions?
• How did you get from your second board to your third? – Mike Earnest Aug 21 '18 at 17:46
• Move (3,1) to (5,1) an then move (1,1) to (3,1) – SonodaUmi Aug 21 '18 at 17:59
• Then you have a mistake; the checker at (4,1) was hopped over, so should be removed. – Mike Earnest Aug 21 '18 at 18:01
This is a fun problem with a nice solution!
First of all, imagine coloring the checkerboard in three colors like so:
A B C A B C A B C ...
B C A B C A B C A ...
C A B C A B C A B ...
...
Let $a,b,c$ be the number of checkers on each color. Each move increases one of these variables by $1$, and decreases the others by $1$. This means that the quantities $a-b,b-c$ and $c-a$ will always have the same parity.
Now, suppose that either $m$ or $n$ is a multiple of $3$. Then you will initially have $a=b=c$, so that $a-b,b-c$ and $c-a$ all start out even. However, if you manage to reduce the board to a single checker, then at least one of $a-b,b-c$ or $c-a$ would be odd. Therefore, the problem is insoluble when $m$ or $n$ is a multiple of $3$.
When neither dimension is a multiple of $3$, you can succeed. Consider the following "T"-move. It allows you to eliminate a $3\times 1$ block of marbles, provided one of the ends has a marble on on side and an empty space on the other:
• • _ _ _ • _ • • • _ _
• --> • --> _ --> _
• • _ _
Applying this repeatedly, you eliminate a $3\times n$ block from the top of the rectangle, provided $m>3$ and $n\ge 3$:
• • • • • • • •
• • • • • • • •
• • • • • • • •
• • • • • • • •
|
v
• • • • • • • _
• • • • • • • _
• • • • • • • _
• • • • • • • •
|
v
• • • • • • _ _
• • • • • • _ _
• • • • • • _ _
• • • • • • • •
|
v
...
|
v
• • • _ _ _ _ _
• • • _ _ _ _ _
• • • _ _ _ _ _
• • • • • • • •
and then apply the same idea with three sideways T's to get rid of that last $3\times 3$ block. The same idea allows you eliminate three columns at a time provided the dimensions are larger enough.
This procedure allows you to reduce the board either a $4\times 4$, $2\times 4$, $4\times 2$ or $5\times 5$ rectangle, depending on the remainders of $m$ and $n$ modulo $3$. I leave it to you to figure out how to solve these small cases.
• My teacher said it was a really beautiful solution, and it really is. Thank you! – SonodaUmi Aug 21 '18 at 18:25
• @cptnSonoda Does your teacher sit next to you and checked the answer on math.stackexchange (MSE)? – callculus Aug 21 '18 at 18:38 | 2019-09-15T13:33:52 | {
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https://math.stackexchange.com/questions/1920431/what-is-a-good-upper-bound-nnn-1n-1-ldots2211 | # What is a good upper bound $n^n(n-1)^{n-1}\ldots2^21^1$?
Given an integer $n \ge 1$, I'd like to have a not-very-loose upper bound for the integer $$u(n) := \Pi_{k=1}^n k^k = n^n(n-1)^{(n-1)}\ldots2^21^1.$$
It's easy see that, $u(n) \le n^{n(n+1)/2}$, but this is not very interesting.
# Update
We have $u(n) \le e^{\left(\frac{1}{2}n(n+1)\log\left(\frac{2n + 1}{3}\right)\right)}$, and we can't really do much better!
Indeed, using Euler-Maclaurin, we have
$\log(u(n)) = \int_2^nx\log x dx = \frac{1}{4}n^2(2\log(n) - 1) - 2\log(2) + \frac{1}{4} + \text{error terms}$, which is comparable to the bound $\log(u(n)) \le \frac{1}{2}n(n+1)\log\left(\frac{2n + 1}{3}\right)$ in the accepted answer (see below). In particular, we can conclude that accepted answer's bound is tight!
• Taking logarithms may be a good start whenever you're dealing with products. Sep 9, 2016 at 13:45
• Ya, I contemplated the sequence $\sum_{k=1}^nk \log(k)$, but nothing pops-up... Sep 9, 2016 at 13:46
• Yes. Take logarithms, use Euler-MacLaurin, and exponentiate. Sep 9, 2016 at 13:47
• Thanks for pointing to Euler-Maclaurin. At the moment, I was just thinking about the Abel summation formula, but your suggestion looks more appropriate. Sep 9, 2016 at 13:52
• See also $(7)$ here: mathworld.wolfram.com/Hyperfactorial.html Sep 9, 2016 at 17:24
Using Jensen's inequality:
Letting $A= \sum_{k=1}^n k = \frac{n (n+1)}{2}$ and $B= \sum_{k=1}^n k^2 = \frac{n (n+1)(2n+1)}{6}$
We have
\begin{align} \log(u(n)) &=\sum_{k=1}^n k \log(k) \\ &= A \sum_{k=1}^n \frac{k}{A} \log(k) \tag{1}\\ &\le A \log \left( \sum_{k=1}^n \frac{k}{A} k \right) = A \log \left( \frac{B}{A} \right) \tag{2} \end{align}
Hence
$$\log(u(n)) \le \frac{n(n+1)}{2} \log\left(\frac{2 n+1}{3}\right) \tag{3}$$
The bound seems to be quite tight:
Update: as noted by comments and OP, the bound $(3)$ agrees with the true order of growth; this can be checked by applying the trapezoidal rule to the integral:
$$-\frac{1}{4}=\int_{0}^{1} x \log(x) dx \approx \frac{1}{n+1} \sum_{k=1}^n \frac{k}{n} \log\left(\frac{k}{n}\right)$$ which gives
$$\log(u(n)) \approx\frac{ n(n+1)}{2}\left( \log n -\frac{1}{2} \right) \tag{4}$$
If one is interested in an approximation (instead of a bound), $(4)$ might be preferable.
• Perfect! Thanks. I see you've used the concave function $\phi(x) := log(x)$ and positive scalars $a_k = k$ in Jensen's inequality: en.wikipedia.org/wiki/Jensen%27s_inequality Sep 9, 2016 at 14:05
• Now that's a freaky tight bound! I Indeed, it's no surprise that the bound is tight. Using Euler-Maclaurin suggested by Daniel above, I can get $\log(u(n)) \approx \frac{1}{4} (n^2(2\log(n) - 1) - 1)$ :). Thus I don't need to beg the error terms in my EM, and can just use your bound instead. Thanks again. Sep 9, 2016 at 14:10
Another simple way for finding an upper bound is using the Abel's summation $$S=\sum_{k=1}^{n}k\log\left(k\right)=\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{1}{2}\int_{1}^{n}\frac{\left\lfloor t\right\rfloor \left(\left\lfloor t\right\rfloor +1\right)}{t}dt$$ where $\left\lfloor t\right\rfloor$ is the floor function and since $\left\lfloor t\right\rfloor >t-1$ we get $$S<\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{1}{2}\int_{1}^{n}\left(t-1\right)dt$$ $$=\color{red}{\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{\left(n-1\right)^{2}}{4}}.$$
Don't have the reputation, or else this would be a comment. As mentioned in a link posted by leonbloy, the hyperfactorial, which shows up in the theory of the Barne's G function, is related to $\int_0^n \log\Gamma(x) dx.$ Good bounds on this should give a better bound than that found by Jensen's inequality. I found the expression
$$\log(u(n)) \le A(n):=\dfrac{(n+1/2)^2}{2}(\log(n+1/2)-3/2)+\dfrac{n(n+1)}{2}+ \dfrac{9}{8}\log(2/3)+\dfrac{11}{16}.$$
For a comparison, let's define 'Jensen' and 'A' ratios
$$R_J(n)=\dfrac{\log(u(n))}{n(n+1)/2\log((2n+1)/3))},\quad R_A(n)=\dfrac{\log(u(n))}{A(n)} .$$
Then (approximately) $R_J(100)=0.892$ , $R_A(100)=0.999992;$ and $R_J(10^5)=0.9915$ , $R_A(10^5)=0.99999999915$ (nine nines).
• Interesting. Do include some detail on your derivation of $A(n)$ (not just an expression for it...). Sep 13, 2016 at 7:39
• Great answer! Upvoted. Jan 23, 2017 at 7:22 | 2022-05-27T16:34:29 | {
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https://math.stackexchange.com/questions/2293493/factorise-x5x1/2293501#2293501 | # Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
• In general, factorization of polynomials is, as you say, "impossible to invent" by hand. There are algorithms, though, which can solve the problem. May 23, 2017 at 15:04
• Any high school student can solve this problem when he guesses one of the factors and then calculates the other factor by division. But that is not really a useful method to factor a polynomial. But it is similar to the method most of the answers uses. May 23, 2017 at 16:03
• $$5 \equiv 2 \pmod 3$$ May 23, 2017 at 16:06
• I just don't know how to accept only one answer when all the answers are simply amazing and awesome. May 23, 2017 at 22:25
• @Mathxx, See math.stackexchange.com/questions/1584594/… May 24, 2017 at 5:36
With algebraic identities, this is actually rather natural:
Remember if we had all powers of $x$ down to $x^0=1$, we could easily factor: $$x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1}=\frac{(x^3-1)(x^3+1)}{x-1}=(x^2+x+1)(x^3+1),$$ so we can write: \begin{align}x^5+x+1&=(x^2+x+1)(x^3+1)-(x^4+x^3+x^2)\\ &=(x^2+x+1)(x^3+1)-x^2(x^2+x+1)\\ &=(x^2+x+1)(x^3+1-x^2). \end{align}
• That's true, but it's not a factorisation. What do you mean? May 23, 2017 at 15:06
• What does "this is rather natural" mean in this context? May 23, 2017 at 15:42
• I simply meant any high school student which is trained to tackle formulae using various identities can factor this polynomial. May 23, 2017 at 15:45
• I think this is no better than the solution the user already showed us. One has to guess the right things. May 23, 2017 at 16:31
• Well, if you do not train, you'll never get a result. All I want to say is that everything is at the high school level, there's no hammersledge theorem, and that training in various ways stimulates imagination, which is the first quality to find proofs. May 23, 2017 at 16:35
If you suspect there exists a factorization over $\mathbb{Q}[x]$ into polynomials of degree 2 and 3, but you just don't know their coefficients, one way is to write it down as
$x^5 + x + 1 = (x^2 + ax + b)(x^3 + cx^2 + dx + e)$ where the coefficients are integers (by Gauss' lemma). And then expand and solve the system.
Then $a + c = 0, b + ac + d = 0, bc + ad + e = 0, ae + bd = 1, be = 1$. So we get $c = -a$ and $b = e = 1$ or $b = e = -1$.
In the first case we reduce to $1 - a^2 + d = 0, -a + ad + 1 = 0, a + d = 1$ which gives $d = 1 - a, 1 - a^2 + 1 - a = 0, 1 - a^2 = 0$ so $a = 1, b = 1, c = -1, d = 0, e = 1$.
Substituting gives us $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$
If the factorization was not over $\mathbb{Q}[x]$ then things would get more complicated because I could not assume $b = e = +/- 1$.
• If you suspect there exists a factorization over Q[x] than it is easy to show that polynomial cannot be split into a polynomail of degree 4 and a linears factor, so it must be the product of thwo polynomails of degree 2 and 3 May 23, 2017 at 15:46
Alternatively note that \begin{align}x^5 + x + 1 &= x^5 - x^2 + x^2 + x + 1\\ & =x^2(x^3-1) + \color{red}{x^2+x+1} \\ & = x^2(x-1)\color{red}{(x^2+x+1)} + \color{red}{x^2+x+1} \\& =\color{blue}{(x^3-x^2+1)}\color{red}{(x^2+x+1)} \end{align}
where we used the well known identity $x^3 - 1 = (x-1)(x^2+x+1)$ in the third equality.
Meta: whilst the first step may seem arbitrary and magical, it is natural to want to insert a term like $x^2$ or $x^3$ into the equation in order to get some traction with factorising $x^5 + x^k$.
Note that if $z^3=1, z\neq 1$ so that $z^3-1=(z-1)(z^2+z+1)=0$ then $z^5+z+1=z^2+z+1=0$. A key to this observation is just seeing whether an appropriate root of unity may be a root.
This tells you that $x^2+x+1$ is a factor of $x^5+x+1$, and the question then is how you do the division. The factorisation method you have been shown is equivalent to doing the polynomial long division.
Another method, not suggested by others, is to use the fact that you know $x^2+x+1$ is a factor and multiply through by $x-1$ so that this factor becomes $x^3-1$.
So $(x-1)(x^5+x+1)=x^6-x^5+x^2-1=(x^3-1)(x^3+1)-x^2(x^3-1)=(x^3-1)(x^3-x^2+1)=(x-1)(x^2+x+1)(x^3-x^1+1)$
• I think this is no better than the solution the user already showed us. One has to guess the right things May 23, 2017 at 16:32
• @miracle173 But the user suggested the solution could not be discovered easily. The art of problem solving is in part learning the kind of things you might try when faced with something unfamiliar. May 23, 2017 at 19:22
standard trick from contests: $5 \equiv 2 \pmod 3.$ Therefore, if $\omega^3 = 1$ but $\omega \neq 1,$ we get $$\omega^5 = \omega^2$$ $$\omega^5 + \omega + 1 = \omega^2 + \omega + 1 = 0$$ Which means, various ways of saying this, $x^5 + x + 1$ must be divisible by $$(x - \omega)(x - \omega^2) = x^2 + x + 1.$$ This is what we call a "minimal polynomial" for $\omega$
The same idea would work for $$x^{509} + x^{73} + 1$$ although the other factor would be worse
SEE
Prime factor of $A=14^7+14^2+1$
• I think this is no better than the solution the user already showed us. One has to guess the right things May 23, 2017 at 16:32
• @miracle173 it's a matter of where the questions arise. This is a contest or contest preparation question, there is a trick; this comes up over and over again on this site, from kids who are trying to get better at contests. You might ask the OP where he got the question and what is the setting responsible for "I'm being taught..." Put another way, someone skilled at searching could find hundreds of questions on MSE where one factor is $x^2 + x + 1$ May 23, 2017 at 16:42
So far the answers just (cleverly) elaborate on high school tricks and techniques. Therefore, I think it can be interesting to see, instead, how standard modern algorithms work in this special case. I will implement a small version of the Berlekamp-Zassenhaus algorithm. I will try to factor $F(x)=x^5+x+1$ over $\mathbb Z[x]$ as a product $f_1(x)f_2(x)$ of polynomials of degree 2 and 3; it will not be long (of course), nor difficult. I recall what is the plan:
• Bound the coefficients of the factors $f_1,f_2$;
• Factor $F(x)=g_1(x)g_2(x)$ over modulo $p$ for some prime $p$;
• Lift (in essence, by Hensel's lemma) the factorization modulo higher $p^k$.
Bound the coefficients of $f_1(x)$: The leading coefficient of $F$ is $c=1$, the degree of $f_1$ is $\delta=2$, and the roots $\alpha$ of $F$ satisfy ${|\alpha|}^5\leq 1+|\alpha|$, so for sure, say, $|\alpha|<\rho=1.5$. Therefore the (absolute values of the) coefficients of $f_1(x)$ are all dominated by those of $(x+\rho)^2$. In particular they are all $<3$. This is called the binomial bound. The same estimate can be obtained with the Knuth-Cohen bound. See Abbott, John. "Bounds on Factors in Z [x]." Journal of Symbolic Computation 50 (2013): 532-563.
Find $g_i(x)\equiv f_i(x)$ modulo 2: Since $F(0)=1$ and $F(1)=3$ we have that $g_1(0)=g_1(1)=1\bmod 2$. Thus the only possibility is $g_1(x)=x^2+x+1$. By polynomial long division in $\mathbb F_2[x]$ we get $g_2(x)=\frac{F(x)}{g_1(x)}=x^3+x^2+1$. Well, if you are clever enough, and not a computer, you might finish the exercise here, by doing long division in $\mathbb Z[x]$.
Factor modulo 4 as $F=(g_1+2 h_1)(g_2 + 2 h_2)$: In other words, we need $g_1 h_2+g_2 h_1 = \frac {F(x)-g_1(x)g_2(x)}{2} = x^4+x^3+x^2 \bmod 2$. Of course the solution is $h_1=0$ and $h_2=x^2$.
Conclusion: We have $f_1(x)\equiv x^2+x+1\bmod 4$. Since the absolute value of the coefficients of $f_1(x)$ is at most $2$, we get $f_1(x)= x^2+x+1$. It works.
Supplement: actually I am deeply convinced that the most natural technique to factor $F(x)$ is the one provided by the OP (although all the other approaches, included the one I described above, are interesting). I'll try to justify this claim. Suppose you want to factor $7763073514021$ in prime numbers. The factor $7$ is easy to find (actually, this completes the factorization). Why? Because you decompose $7-7-63-0-7-35-14-0-21$ and you factor out termwise. "Piecewise" factorization is by far the most natural, and easy to spot, approach to factorization "by hand". Another example is $x^6-x^4-20x^3+14x^2+20 x -14$, where you may wish to take advantage of the pattern $(1,-1),(-20,20),(14,-14)$. Now, suppose you want to factor the number $636362236363$. It's very similar to the example before, only that you must be able to see the "negative": you are just subtracting a "14" from $636363636363$. Although the pattern of coefficients $[1,0,0,0,1,1]$ of $F(x)$ might be irregular at first glance, I find it very natural to see it as a $0-111-00$ subtracted from a $111-111$.
• Or else (but only for "how to factor", not "why does it factor"): coefficients of f_1 have norm at most 2; F is monic; F(0)=F(-1)=1. So f_1=x^2+x+1 or f_1=x^2-x-1. Now use F(-2)=-33. May 24, 2017 at 1:09
Let $f (x)=x^5+x+1$
$$f'(x)=5x^4+1>0$$
$f$ is stricly increasing at $\mathbb R$, thus it has only one real root $\alpha \in (-1,0)$.
the factorisation will be of the form
$$(x-\alpha)(x^2+ax+b)(x^2+cx+d)$$
with $a^2-4b <0$ and $c^2-4d <0$.
• Why can't it be factored in $\mathbf R[X]$? The only irreducible polynomials in $\mathbf R[X]$ are linear polynomials and quadratic polynomials with a negative discriminant. May 23, 2017 at 14:56
• Just because it has no linear factors does not mean that it does not have higher order factors (DAMHIKT). Also, he showed a factorization of it and the factorization is correct: just multiply it out. So, your answer is wrong: it can be factorized in $\mathbb R[X]$. May 23, 2017 at 14:58
• @Nick Yes you are right. i edited. May 23, 2017 at 15:02 | 2022-07-07T08:14:39 | {
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https://cs.stackexchange.com/questions/45783/binary-code-with-constraint | Binary code with constraint
Suppose I have an alphabet of n symbols. I can efficiently encode them with $\lceil \log_2n\rceil$-bits strings. For instance if n=8:
A: 0 0 0
B: 0 0 1
C: 0 1 0
D: 0 1 1
E: 1 0 0
F: 1 0 1
G: 1 1 0
H: 1 1 1
Now I have the additional constraint that each column must contain at most p bits set to 1. For instance for p=2 (and n=8), a possible solution is:
A: 0 0 0 0 0
B: 0 0 0 0 1
C: 0 0 1 0 0
D: 0 0 1 1 0
E: 0 1 0 0 0
F: 0 1 0 1 0
G: 1 0 0 0 0
H: 1 0 0 0 1
Given n and p, does an algorithm exist to find an optimal encoding (shortest length) ? (and can it be proved that it computes an optimal solution?)
EDIT
Two approaches have been proposed so far to estimate a lower bound on the number of bits $m$. The goal of this section is to provide an analysis and a comparaison of the two answers, in order to explain the choice for the best answer.
Yuval's approach is based on entropy and provides a very nice lower bound: $\frac{logn}{h(p/n)}$ where $h(x) = xlogx + (1-x)log(x)$.
Alex's approach is based on combinatorics. If we develop his reasonning a bit more, it is also possible to compute a very good lower bound:
Given $m$ the number of bits $\geq\lceil log_2(n)\rceil$, there exists a unique $k$ such that $$1+\binom{m}{1} + ... +\binom{m}{k} \lt n \leq 1+\binom{m}{1} + ... + \binom{m}{k}+\binom{m}{k+1}$$ One can convince himself that an optimal solution will use the codeword with all bits low, then the codewords with 1 bit high, 2 bits high, ..., k bits high. For the $n-1-\binom{m}{1}-...-\binom{m}{k}$ remaining symbols to encode, it is not clear at all which codewords it is optimal to use but, for sure the weights $w_i$ of each column will be bigger than they would be if we could use only codewords with $k+1$ bits high and have $|w_i - w_j| \leq 1$ for all $i, j$. Therefore one can lower bound $p=max(w_i)$ with $$p_m = 0 + 1 + \binom{m-1}{2} +... + \binom{m-1}{k-1} + \lceil \frac{(n-1-\binom{m}{1}-...-\binom{m}{k}) (k+1)}{m} \rceil$$
Now, given $n$ and $p$, try to estimate $m$. We know that $p_m \leq p$ so if $p \lt p_{m'}$, then $m' \lt m$. This gives the lower bound for $m$. First compute the $p_m$ then find the biggest $m'$ such that $p \lt p_{m'}$
This is what we obtain if we plot, for $n=1000$, the two lower bounds together, the lower bound based on entropy in green, the one based on the combinatorics reasonning above in blue, we get:
Both look very similar. However if we plot the difference between the two lower bounds, it is clear that the lower bound based on combinatorics reasonning is better overall, especially for small values of $p$.
I believe that the problem comes from the fact that the inequality $H(X) \leq \sum H(X_i)$ is weaker when $p$ gets smaller, because the individual coordinates become correlated with small $p$. However this is still a very good lower bound when $p=\Omega(n)$.
Here is the script (python3) that was used to compute the lower bounds:
from scipy.misc import comb
from math import log, ceil, floor
from matplotlib.pyplot import plot, show, legend, xlabel, ylabel
# compute p_m
def lowerp(n, m):
acc = 1
k = 0
while acc + comb(m, k+1) < n:
acc+=comb(m, k+1)
k+=1
pm = 0
for i in range(k):
pm += comb(m-1, i)
return pm + ceil((n-acc)*(k+1)/m)
if __name__ == '__main__':
n = 100
# compute lower bound based on combinatorics
pm = [lowerp(n, m) for m in range(ceil(log(n)/log(2)), n)]
mp = []
p = 1
i = len(pm) - 1
while i>= 0:
while i>=0 and pm[i] <= p: i-=1
mp.append(i+ceil(log(n)/log(2)))
p+=1
plot(range(1, p), mp)
# compute lower bound based on entropy
lb = [ceil(log(n)/(p/n*log(n/p)+(n-p)/n*log(n/(n-p)))) for p in range(1,p)]
plot(range(1, p), lb)
xlabel('p')
ylabel('m')
show()
# plot diff
plot(range(1, p), [a-b for a, b in zip(mp, lb)])
xlabel('p')
ylabel('m')
show()
• @D.W. the constraint is quite as your states. each column must contain at most p bits set to 1. ie. the bit 1's at each position of all selected keys do not exceed p. But I think the first step is still counting the capacity of each bit width. – Terence Hang Sep 2 '15 at 16:52
• user3017842, I suspect your latest edit should be posted as a self-answer. I think it stands alone as an answer to your question. Do you agree? If so, the right place for it is in the answer box, rather than in the question -- that will make a lot more sense for future readers who come across this (and also allows the community to vote on your answer). I appreciate that you're sharing the analysis you did -- thank you. I encourage you to post that material as an answer, and then remove it from the question. What do you think? Does that seem like it makes sense to you? – D.W. Sep 5 '15 at 3:15
• @D.W. The EDIT section only makes a comparaison between the two proposed answers in order to explain the choice for the best answer. Therefore I didn't want to put it as a self-answer. But I completely agree that it lacks of clarity for future users, therefore I've clarified the goal of the section and provided links to the corresponding answers. I believe it is a bit more clear now. – user3017842 Sep 6 '15 at 9:05
There is an additional lower bound we can build, that will address cases like what @user3017842 mentioned in their comment on Yuval's answer. (Cases where $p$ is particularly small.) Suppose we knew $m$ already: Then we have $pm$ bits high total across all codewords. Since we're interested the cases where $p$ is small, we view these high bits as our limiting resource, and want to build a code with it (and see how many codewords we can possibly get out). We can have 1 codeword with all 0s, then $m$ codewords with a single 1, then $m \choose 2$ with two 1s, etc. If we call the highest number of bits in a codeword $k$, then $$pm = 0\cdot 1 + 1\cdot m + 2\cdot {m \choose 2}+... \le \sum_i^k i{m \choose i}$$ While our number of codewords $n$ is similarly bounded by $$n \le \sum_i^k {m \choose i}$$ If we look at the case where $p \le m$, then $k \le 2$ is already implied by the first inequality. ($pm = m^2 = m + 2{m \choose 2}$). So then the code would consist of the $0$-word, $m$ single-$1$-words, and $(p-1)m/2$ two-$1$-words. Thus $$n \le 1 + m + (p-1)m/2$$ or inverting $$m \ge \frac{2(n-1)}{p+1} .$$ This will yield the tight lower bound of $m\ge 5$ on the example you provide, but as mentioned before, will probably only be very useful while $p \approx m$ (or $p \approx \sqrt n$).
• Please see the EDIT section of the main post to see why your answer wins ! – user3017842 Sep 4 '15 at 18:06
Here is a lower bound and an asymptotically matching construction, at least for some ranges of the parameters. Denote by $m$ the number of columns, and suppose for simplicity that $p \leq n/2$.
We start with a lower bound on $m$. Let $X$ be the encoding of symbol chosen uniformly at random. Let $X_1,\ldots,X_m$ be the individual coordinates, and let $w_i \leq p$ be the weight of the $i$th column. Then $$\log n = H(X) \leq \sum_{i=1}^m H(X_i) = \sum_{i=1}^m h(w_i/n) \leq m h(p/n).$$ Therefore $$m \geq \frac{\log n}{h(p/n)}.$$ Here $H$ is the entropy of a random variable $H(X) = -\sum_x \Pr[X=x] \log \Pr[X=x]$ and $h$ is the entropy function $h(x) = -x\log x-(1-x)\log(1-x)$. (You can use whatever base for the logarithm you want.)
The asymptotically matching construction, that should work for $p = \Omega(n)$, chooses $m$ a bit larger than this lower bound, and chooses a random encoding scheme, each bit being set to $1$ with some probability $q/n$ which is a bit smaller than $p/n$. Choosing the parameters correctly, we should get that this results in a legal encoding (all codewords are different and all column weights are at most $p$) with positive probability.
• Nice lower bound. Why should the matching construction work for $p=\Omega(n)$? is there any easy way to believe it other than bounding the probability of getting an invalid encoding when $m$ is picked near the lower bound? – Ariel Sep 3 '15 at 15:46
• Experience tells me that it has a high chance of working, but you can't know for sure without trying. – Yuval Filmus Sep 3 '15 at 16:27
• I believe this lower bound is very good when the individual coordinates $X_1, X_2, ..., X_m$ are virtually independent (because inequality $H(X) \leq \sum H(X_i)$ will be close to be an equality). This is likely to be the case when $p$ is close enough to $n/2$. However when $p$ remains small, this is no more the case. Consider for example the extreme case when $p=1$. – user3017842 Sep 3 '15 at 17:42
• When $p=1$ it is clear that the number of bits is $n-1$ (as suggested in Alex Meiburg's answer). However $n-1 - \frac{logn} {h(p/n))} \sim n/logn$. The lower bound becomes inaccurate when $p$ remains small while $n$ is getting large. Besides, for small $p$ such as $p=1$, the proposed construction will not work quite well because of the well-known birthday problem. But, still, this is a very nice approach, especially when $p=\Omega(n)$ ! – user3017842 Sep 3 '15 at 18:05
• I've made a comparaison with another lower bound deduced from combinatorics reasonning suggested in another answer. It turns out that your lower bound is slightly weaker, especially when $p$ gets smaller. Please see the details of the comparaison in the EDIT section of the main post. Nonetheless, I was very impressed by your solution! Thanks ! – user3017842 Sep 4 '15 at 18:05
Here is a simple search methodology. We start from some lower-bound on the number of bits and then try to find a legal encoding. Specifically.
Let m be the current number of bits. Encode symbol i as bi1, bi2, ..., bim.
Constraints: bi xor bj isn't 0 - in other words each symbol's encoding is unique
For all j: sum_i bij <= p.
This is a pseudo-boolean satisfiability problem (well it can easily be encoded as a standard satifiability problem). So just keep increasing m until you find one that is satisfiable (or do a binary search using lower and upper bounds to find the minimal m).
Of course, this doesn't guarantee that in practice you'll be able to actually find the minimal m, the SAT check could timeout. | 2021-08-02T23:15:06 | {
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http://mathhelpforum.com/advanced-algebra/144749-eigenvalues-ab-ba-print.html | # eigenvalues of AB and of BA
• May 14th 2010, 04:28 PM
math8
eigenvalues of AB and of BA
Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?
• May 14th 2010, 04:57 PM
dwsmith
Quote:
Originally Posted by math8
Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?
No, multiplication of matrices isn't commutative.
• May 14th 2010, 05:23 PM
math8
I know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.
• May 14th 2010, 05:27 PM
dwsmith
Quote:
Originally Posted by math8
I know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.
Sorry for the mistake at first.
$det(AB-\lambda I)=det(AB-\lambda AA^{-1})=det(A(B-\lambda A^{-1}))=det(A)det(B-\lambda A^{-1})$
$=det(B-\lambda A^{-1})det(A)=det((B-\lambda A^{-1})A)=det(BA-\lambda I)$
I am pretty sure I have proving it here as well as some other proofs that may help in Linear:
http://www.mathhelpforum.com/math-he...rexamples.html
• May 14th 2010, 05:55 PM
Random Variable
It's easy to prove.
Let $\lambda_{1}$ be an eigenvalue of $AB$
then $ABx=\lambda_{1} x$
$BABx = \lambda_{1} B x$
which means that $\lambda_{1}$ is an eigenvalue of $BA$ with associated eigenvector $Bx$
Now let $\lambda_{2}$ be an eigenvalue of $BA$
then $BAx=\lambda_{2}x$
$ABAx = \lambda_{2}Ax$
which means that $\lambda_{2}$ is an eigenvalue of $AB$ with associated eigenvector $Ax$
QED
• May 14th 2010, 06:03 PM
roninpro
Quote:
Originally Posted by dwsmith
Sorry for the mistake at first.
$det(AB-\lambda I)=det(AB-\lambda AA^{-1})=det(A(B-\lambda A^{-1}))=det(A)det(B-\lambda A^{-1})$
$=det(B-\lambda A^{-1})det(A)=det((B-\lambda A^{-1})A)=det(BA-\lambda I)$
I am pretty sure I have proving it here as well as some other proofs that may help in Linear:
http://www.mathhelpforum.com/math-he...rexamples.html
There might be a slight hiccup. You don't know whether or not $A$ or $B$ is invertible.
• May 15th 2010, 03:38 AM
HallsofIvy
Fortunately, Random Variables' proof, in addition to being simpler, does not require that A and B be invertible.
• May 15th 2010, 04:47 AM
NonCommAlg
Quote:
Originally Posted by Random Variable
It's easy to prove.
Let $\lambda_{1}$ be an eigenvalue of $AB$
then $ABx=\lambda_{1} x$
$BABx = \lambda_{1} B x$
which means that $\lambda_{1}$ is an eigenvalue of $BA$ with associated eigenvector $Bx$
$BABx=\lambda Bx$ doesn't necessarily imply that $\lambda$ is an eigenvalue of $BA$ because we might have $Bx=0.$
• May 15th 2010, 07:39 AM
Random Variable
Quote:
Originally Posted by NonCommAlg
$BABx=\lambda Bx$ doesn't necessarily imply that $\lambda$ is an eigenvalue of $BA$ because we might have $Bx=0.$
Deal with the case of $A$ or $B$ being zero matrices separately.
If $A$ or $B$ are zero matrices, then $AB=BA=0$, and the only eigenvalue of a zero matrix is $\lambda = 0$
• May 15th 2010, 02:54 PM
HallsofIvy
Quote:
Originally Posted by Random Variable
Deal with the case of $A$ or $B$ being zero matrices separately.
If $A$ or $B$ are zero matrices, then $AB=BA=0$, and the only eigenvalue of a zero matrix is $\lambda = 0$
That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $ABx= \lambda x$ for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA is 0.
• May 15th 2010, 03:13 PM
Random Variable
Quote:
Originally Posted by HallsofIvy
That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $ABx= \lambda x$ for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA is 0.
Yeah, I forgot about that possibility.(Doh)
• May 15th 2010, 03:20 PM
dwsmith
I know I have this proof covering all cases, and once I find it, I will post it.
I am not sure if the red is necessarily correct.
AB and BA have the same eigenvalues iff. AB is similar to BA.
AB=C
BA=D
If C and D are similar matrices, then there exist S such that $C=S^{-1}DS$.
$p_c(\lambda)=det(AB-\lambda I)=det(C-\lambda I)=det(S^{-1}DS-\lambda I)$
$=det(S^{-1}DS-\lambda S^{-1}IS)=det(S^{-1}(D-\lambda I)S)$
$=det(S^{-1}S)det(D-\lambda I)=det(BA-\lambda I)=p_d(\lambda)$ | 2016-06-30T09:04:25 | {
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http://businessignitiongroup.co.uk/u473ie/ao8zfn.php?tag=43264b-a-matrix-with-only-one-column-is-called | # a matrix with only one column is called
4. A column matrix is a matrix that has only one column. Example: D is a column matrix of order 2 × 1 A zero matrix or a null matrix is a matrix that has all its elements zero. A matrix, that has many rows, but only one column, is called a column vector. Column Matrix A matrix having only one column and any number of rows is called column matrix. A wide matrix (a matrix with more columns than rows) has linearly dependent columns. A matrix is said to be a column matrix if it has only one column. a = 1 3 2 5 3 2 4 8 5 9 I want to sort the second column in the a matrix. Example: C is a column matrix of order 1 × 1 A column matrix of order 2 ×1 is also called a vector matrix. 3. The entries of a vector are denoted with just one subscript (since the other is 1), as in a3. INCLUDES THE SOLUTIONS. Each variable in the system becomes a column. A column matrix has only one column but any number of rows. A matrix with only one column, i.e., with size n × 1, is called a column vector or just a vector. A matrix with only one row is called a.....matrix, and a matrix with only one column is called a.....matrix. The following vector q is a 3 × 1 column vector containing numbers: $q=\begin{bmatrix} 2\\ 5\\ 8\end{bmatrix}$ A row vector is an 1 × c matrix, that is, a matrix with only one row. Suppose that A has more columns than rows. Converting Systems of Linear Equations to Matrices. Sometimes the size is specified by calling it an n-vector. Then A cannot have a pivot in every column (it has at most one pivot per row), so its columns are automatically linearly dependent. Rectangular Matrix A matrix of order m x n, such that m ≠ n, is called rectangular matrix. For example, $$A =\begin{bmatrix} 0\\ √3\\-1 \\1/2 \end{bmatrix}$$ is a column matrix of order 4 × 1. Two matrices of the same order whose corresponding entries are equal are considered equal. The variables are dropped and the coefficients are placed into a matrix. One column matrix. A vector is almost often denoted by a single lowercase letter in boldface type. For example, four vectors in R 3 are automatically linearly dependent. A column vector is an r × 1 matrix, that is, a matrix with only one column. Below, a is a column vector while b is a row vector. A matrix with only one row or one column is called a vector. The determinant takes a square matrix and calculates a simple number, a scalar. I have the matrix as follows. Each equation in the system becomes a row. In general, B = [b ij] m × 1 is a column matrix of order m × 1. row column. A column matrix is a matrix with only one column. 5. The entries are sometimes The matrix derived from a system of linear equations is called the..... matrix of the system. 3) Square Matrix. I want the corresponding rows of column one to be printed as follows : a = 3 2 1 3 2 5 4 8 5 9 I tried sort(a), but it is sorting only the second column of matrix a. 2. one column (called a column vector). Horizontal Matrix A matrix in which the number of rows is less than the number of columns, is called a horizontal matrix. A matrix with only one row is called a _____ matrix, and a matrix with only one column is called a _____ matrix. A matrix having only one row is called a row matrix (or a row vector) and a matrix having only one column is called a column matrix (or a column vector). To understand what this number means, take each column of the matrix and draw it as a vector. a = 7 2 3 , b = (− 2 7 4) A scalar is a matrix with only one row and one column. augmented. 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## Hello world!
Begin typing your search term above and press enter to search. Press ESC to cancel. | 2021-04-13T18:45:50 | {
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https://math.stackexchange.com/questions/2910304/how-many-ways-to-pair-6-chess-players-over-3-boards-disregarding-seating-arrang | # How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement.
The problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.
I have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.
I started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.
Now, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$ which is what I want $\to 720/6 = 120$.
So far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.
I finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting $A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states, $2 \cdot 2 \cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.
• Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Sep 9 '18 at 8:09
Here is an alternative approach:
Line up the six players in some order, say alphabetically. The first person in line can choose a partner in $5$ ways. Remove those two players from the line. That leaves four players. The first person left in the line can choose a partner in $3$ ways. Remove those two players from the line. The remaining two players must play each other. Hence, there are $$5!! = 5 \cdot 3 \cdot 1 = 15$$ ways to form three pairs of chess partners. The expression $5!!$ is read $5$ double factorial.
It's correct (the answer and logic). what is a different way to reason about it?
The idea:
$$(\textrm{first pair})(\textrm{second pair})(\textrm{third pair}),$$
for the first pair: $6$ ways to choose the first guy, $5$ ways to choose the second, and notice the repetition $p_1p_2=p_2p_1$ so $$\frac{6\cdot 5}{2!}$$
for the second and last pair: $\displaystyle\frac{4\cdot3}{2!};\frac{2\cdot1}{2!}.$
Now we get pairs, but $P_1P_2P_3=P_3P_2P_1,$ etc. there are $3!$ ways to repeat$^\dagger$ three pairs we just made, so $$\frac{1}{3!}{6\choose2}{4\choose2}{2\choose2}.$$
$\dagger$: A good question to think about is: what's the condition(s) that repetition will/will not happen?
Yes, you are absolutely right.
The other approach might be selecting $2$ players out of $6$ since arrangement doesn't matter ($_6C_2=15$).
• Your alternative approach is incorrect. Choosing two of the six players does not determine who plays who in the other two games. – N. F. Taussig Sep 9 '18 at 8:06
• No it takes care of all the possible combonations – Rock Guitar Sep 9 '18 at 20:32 | 2019-06-25T12:32:29 | {
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https://math.stackexchange.com/questions/2851611/cardinality-of-equivalence-classes-in-a-set | # Cardinality of equivalence classes in a set
Given $$X = \{1, 2, 3, 4\}$$. Define a relation $$\mathbf{R}$$ over $$P(X)$$ as follows: For two elements $$A,B ∈ P(X)$$, $$A \mathrel{\mathbf{R}} B$$ if $$|A| ≡ |B| \pmod{3}$$.
(a) Demonstrate that the previous relation is an equivalence relation.
(b) Determine the cardinality of every equivalence class.
I have found how many classes in there which is the trivial part of the question as it is modulo $$3$$ then we have three classes $$C_0$$, $$C_1$$, $$C_2$$.
But I am not able to demonstrate that the previous relation is an equivalence relation or the cardinality of the equivalence classes.
Sorry for being inefficient even tho the question might be easy.
Guide:
You have to show the following.
• Reflexive, $\forall A \in P(X), ARA$.
• Symmetric, $ARB \implies BRA$.
• Transitive, $ARB \land BRC \implies A RC$
In particular, transitivity is simply $|A|\equiv |B| \pmod{3}$ and $|B|\equiv |C| \pmod{3}$, then we have $|A| \equiv |C| \pmod{3}$.
Try to write out what they mean and you should be able to use the property of modulo arithmetic to see why it is an equivalence relation.
As for how many members each equivalence class has. Try to answer the following question.
• How many subsets have exactly $1$ or $4$ elements.
• How many subsets have exactly $2$ elements.
• How many subsets have exactly $0$ or $3$ elements.
• For the transitivity it is clear, but still the second part of how many subsets I can't recognize how many subsets it is the thing that I can't work on I know it is simple but I don't know how to approach it. – CptPackage Jul 14 '18 at 14:28
• To construct a subset of $i$ element, out of the $4$ elements, you have to choose $i$ elements. Hence there are $4$ choose $i = \binom{4}{i}=\frac{4!}{i!(4-i)!}$ such subsets. Using this property, can you determine how many subset has $0$ elements? what about $1$ element and so on? – Siong Thye Goh Jul 14 '18 at 14:34
• There are $3$ equivalence class, but the question is how many subsets are there in those equivalence classes. For example $C_0$ consists of subsets with either $0$ elements or $3$ elements, $C_0=\{ \emptyset, \{1,2,3\}, \{1,2,4\}, \{ 1,3,4\}, \{2,3,4\} \}$. Note that $0 \equiv 3 \pmod{3}$. What is the cardinality of $C_0$? – Siong Thye Goh Jul 14 '18 at 14:41
• Alright, I can only use the binomial coefficient and it will work but I want to ask something. So C0 is the class containing the empty element, and all the possible combinations of 3 elements as 0 mod 3 = 0 and 3 mod 3 = 0 so it will be the same equivalence class, but for C1 it will the combination of the subsets of 1 element each right? Like C1={{1},{2},{3},{4}} but it will have the cardinality of 4 not 5 which is wrong. And C2 will be C2={{1,2},{1,3},{1,4},{2,3},{2,4},{3,4}} right? So can you please explain to me this small part? Sorry, I know I have been exhausting enough. – CptPackage Jul 14 '18 at 14:49
• $C_1$ is the class containing subsets of $1$ element and $4$ elements since $1 \equiv 4 \pmod{3}$, so $|C_1|=5$. You have listed $C_2$ correctly. Without listing it, the computation is just $\binom{4}{2}=6$. Don't worry about how long you ask the question, just make sure you are learning. ;) – Siong Thye Goh Jul 14 '18 at 14:55
(a) Reflexive: $|A|\equiv|A|$. Symmetric: $|A|\equiv|B|\implies|B|\equiv|A|$. Transitive: if $|A|\equiv|B|\equiv|C|$, integers $m,\,n$ exist with $|A|=|B|+3m=|C|+3m+3n$.
(b) There are $1,\,4,\,6,\,4,\,1$ subsets of sizes $0,\,\cdots,\,4$. One equivalence class is for cardinalities $0$ and $3$, and is of size $1+4=5$; another covers $1$ and $4$, and again there are $5$ elements; the last is just the size-$2$ case, with the remaining $6$ elements.
• Alright I can get it with the first answer, but the second one isn't very clear for me can you please simplify it more if possible? – CptPackage Jul 14 '18 at 14:25
• @CptPackage Which part needs simplifying? Counting subsets by size with a row of Pascal's triangle, or grouping subset sizes modulo $3$? – J.G. Jul 14 '18 at 14:50
• Grouping subsets, Mr.Siong Thye Goh showed me that we can get the results using the binomial coefficient which I didn't know about so sorry for being an ignorant. But can you please show me the combination of the subsets in the equivalence classes? Like C1={{1},{2},{3},{4}}.. – CptPackage Jul 14 '18 at 14:52 | 2021-04-22T01:01:41 | {
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# Calculate faster - Reciprocal percentage equivalent
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Hi All,
If one has to complete the quants section on the GMAT on time, it is imperative that one has to calculate faster. A lot of questions on the GMAT involve division. To divide faster reciprocal percentage equivalent come quite handy.
In this post I will share a basic and an easy method that may help one in memorizing the reciprocal equivalents.
> Reciprocal of 2 (i.e 1/2) is 50%, that of 4 will be half of 50% i.e 25%. Similarly, reciprocal of 8 will be half of 25% = 12.5% and that of 16 will be 6.25%
> Reciprocal of 3 is 33.33%. Thus reciprocal of 6 will be half of 33.33% i.e 16.66% and that of 12 will be 8.33%
> Reciprocal of 9 is 11.11% and reciprocal of 11 is 9.0909%. Reciprocal of 9 is composed of 11's and reciprocal of 11 is composed of 09's.
If any calculation has 9 in the denominator, the decimal part will be only 1111 or 2222 or 3333 or 4444...
ex. 95/9 will be 10.5555
> Reciprocal of 20 is 5% [ you can remember this as 1/5 = 20% so, 1/20 = 5%]
> Reciprocal of 21 is 4.76% and of 19 is 5.26%. Thus we can easily remember reciprocals of 19, 20, 21 as 5.25%, 5 ,4.75% i.e 0.25% more and less than 5%
> Similarly, reciprocal of 25 is 4 % [ Remember this as 1/4 = 25%, so 1/25 = 4%]
> Reciprocal of 24 is 4.16% and of 26 is 3.84%. Thus, we can easily remember reciprocals of 24, 25, 26 as 4.15%, 4, 3.85% i.e 0.15% more and less than 4%.
> Reciprocal of 29 is 3.45% (i.e 345 in order) and reciprocal of 23 is 4.35% ( same digits but order is different. If 1/29 = 3.45% than definitely 1/23 will be more than 3.45%. Reverse the digits and the answer comes to 4.35%)
> Reciprocal of 18 is half of 11.1111% i.e 5.5555% i.e it consists of only 5's.
> Reciprocal of 22 is half of 09.0909%. i.e 4.5454% i.e consists of 45's.
> One can remember 1/8 = 12.5% and tables of 8, and one can easily remember fractions such as 2/8, 3/8, 5/8, 7/8 which are used very regularly.
1/8 is 12.5%, 2/8 is 25% (12.5*2), 3/8 is 37.5% (12.5*3), 5/8 is 62.5% ( 12.5*5), 7/8 = 87.5%
I hope you find this post useful.
Thanks,
Harish
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Calculate faster - Reciprocal percentage equivalent [#permalink] 06 Jan 2014, 10:18
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Display posts from previous: Sort by | 2017-01-20T23:28:05 | {
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https://math.stackexchange.com/questions/2818474/find-the-prime-factors-of-332-232 | # Find the prime factors of $3^{32}-2^{32}$
I'm having a go at BMO 2006/7 Q1 which states: "Find four prime numbers less than 100 which are factors of $3^{32}-2^{32}$."
My working is as follows (basically just follows difference of two squares loads of times):
$$3^{32}-2^{32}$$ $$=(3^{16}+2^{16})(3^{16}-2^{16})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{8}-2^{8})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{4}-2^{4})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{2}-2^{2})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)(3-2)$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)$$
Now it is simple to get $3$ of the primes here: $$3+2=5$$ $$3^2+2^2=13$$ $$3^4+2^4=97$$
Now I was having some trouble with finding the fourth prime. It's clear that the fourth prime must either be a prime factor of $3^8+2^8$ or $3^{16}+2^{16}$. I started off with $3^8+2^8$:
$$3^8+2^8$$ $$=3^{4^2}+256$$ $$=81^2+256$$ $$=6561+256$$ $$=6817$$
Here is where I am stuck because google tells me that $6817=17*401$ and so the fourth prime is $17$. But this is a non-calculator paper so is there a way of working out this answer without working out $\frac{6817}{3},\frac{6817}{7},\frac{6817}{11}$ etc until one of them is an integer solution?
Finally, before anyone says this is a duplicate, there is a similar question here (prime factors of $3^{32}-2^{32}$) that I didn't know existed until writing this question. However, none of those answers give a way of finding $17$ without using any computational methods. So are they just expecting you to essentially do trial and error throughout all of the primes under $100$ until one works?
• Well, you have only three primes to actually try ($3$ is obviously not a divisor). Of course, you don't know that beforehand. And you have only to try up to $79$, since $\sqrt{6817}<83$. – StayHomeSaveLives Jun 13 '18 at 16:52
• Well, you can notice that $68=4\cdot17$ which makes it easy. – saulspatz Jun 13 '18 at 16:52
• Yes I know I would reach the solution of $17$ quite quickly but that isn't really my point. I just wouldn't expect BMO to give something that essentially requires trial and error – Dan Jun 13 '18 at 16:53
• @saulspatz ah yes that's a very good way of thinking about it. Is that what they would expect from you though or is there some kind of method they would want you to use? – Dan Jun 13 '18 at 16:55
• Since $17$ is a prime, Fermat's little theorem tell us $2,3 \not| 17 \implies 2^{16} \equiv 3^{16} \equiv 0 \pmod 17$. One thing one can try is looking for prime of the form $2^k + 1$ for $k \le 5$, you immediate get $5$ and $17$. – achille hui Jun 13 '18 at 17:02
Apply Fermat's little theorem:
$17$ is prime, so
$$3^{16}\equiv1\pmod{17}$$ $$2^{16}\equiv1\pmod{17}$$
Note that $\forall a \perp p, a^{p-1\over2} \equiv \pm 1 \pmod p$
2 and 3 are already coprime to the other primes you're looking for. $8*2+1=17$ is a strong candidate. In fact, it is the only strong candidate for this method of attack.
I checked this in Python and there are only 4 such primes, $5$, $13$, $17$ and $97$. As has been noted, FLT implies $17$ is such a prime, and we can get $5$ the same way. We also get $13$ as a prime factor of $3^4-2^4$, and $97$ as a prime factor of $3^8-2^8$.
In the timed conditions of an olympiad, it helps to first seek prime factors of $3^2-2^2$, then the remaining prime factors of $3^4-2^4$ (i.e. those of $3^2+2^2$) etc. In particular $5=3^2-2^2,\,13=3^2+2^2,\,97=3^4+2^4,\,17\times 401=3^8+2^8$. | 2020-03-29T10:39:24 | {
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http://mathhelpforum.com/geometry/166813-few-math-team-problems-need-help.html | # Math Help - a few math team problems need help with
1. ## a few math team problems need help with
1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.
2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)
3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.
4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow
You don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!
-Amer
2. Originally Posted by amerlaw1
4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow
Best thing to do is draw a picture to truly understand this one.
Using similar triangles and equating ratios of side lengths you should get
$\displaystyle \frac{x}{6} = \frac{x+3}{9}$
Can you find $x$ ?
3. Originally Posted by amerlaw1
1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.
2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)
3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.
4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow
You don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!
-Amer
What is "mathteam"? Are these questions part of a competition?
4. thank you but can u show me how the are similar like what theorem. E.G sas or aa or sss????
yes math team is a competition but these questions came from a competition that happened 14 years ago so you need not to worry about cheating
5. 1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.
Let the side of the triangle be $a$. Then you can find the areas of AOB, BOC and AOC, which together give the area of ABC. On the other hand, the area of ABC is $\sqrt{3}a^2/4$. This gives you an equation that allows finding $a$.
6. Hello, amerlaw1!
2. What is the distance between the incenter and circumcenter
of a triangle with sides of the length 12, 16 and 20?
Code:
_ A *
: | *
: | *
: | *
: | *
: | *
: | * 20
: | * * * *
: | * * E
12 | * o
: |* r * *
: | * *
: * * * *
: F o - - - - o * *
: * r |O * *
: | | *
: r|* |r * *
: | * | * *
: | * | * *
- C * - - - * o * - - - - - - - - - - - - * B
r D
: - - - - - - - - 16 - - - - - - - - :
We have right triangle $ABC:\:AC = 12,\;BC = 16,\;AB = 20.$
Place the triangle on a coordinate system with $\,C$ at the Origin.
The incenter is $O(r,r).$
. . $OD = OE = OF = CD = CF = r.$
The circumcenter is the midpoint of $AB:\;P(8,6).$
. .
(Not shown on the diagram.)
Note that: . $AF \,=\, 12-r$
Since $AE$ is also tangent to the circle: $AE \,=\, 12-r$
Note that: . $BD \,=\,16-r.$
Since $BE$ is also tangent to the circle: $BE \,=\,16-r$
Since $AB = 20$, we have: . $(12-r) + (16-r) \:=\:20 \quad\Rightarrow\quad r \,=\,4$
We have: . $\begin{Bmatrix}\text{Incenter:} & O(4,4) \\ \text{Circumcenter:} & P(8,6) \end{Bmatrix}$
Therefore: . $\iverline{OP} \;=\;\sqrt{(8-4)^2 + (6-5)^2} \;=\;\sqrt{16+4} \;=\;\sqrt{20} \;=\;2\sqrt{5}$ | 2014-09-15T10:06:54 | {
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https://gmatclub.com/forum/math-number-theory-broken-into-smaller-topics-91274.html | It is currently 19 Jan 2018, 21:22
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# Math: Number Theory (broken into smaller topics)
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NUMBER THEORY
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
--------------------------------------------------------
DEFINITION
Number Theory is concerned with the properties of numbers in general, and in particular integers.
As this is a huge issue we decided to divide it into smaller topics. Below is the list of Number Theory topics.
GMAT Number Types
GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.
INTEGERS
Integers are defined as: all negative natural numbers $$\{...,-4, -3, -2, -1\}$$, zero $$\{0\}$$, and positive natural numbers $$\{1, 2, 3, 4, ...\}$$.
Note that integers do not include decimals or fractions - just whole numbers.
Even and Odd Numbers
Prime Numbers
Factors
Finding the Number of Factors of an Integer
Finding the Sum of the Factors of an Integer
Greatest Common Factor (Divisior) - GCF (GCD)
Lowest Common Multiple - LCM
Divisibility Rules
Factorials
Consecutive Integers
Evenly Spaced Set
IRRATIONAL NUMBERS
Fractions (also known as rational numbers) can be written as terminating (ending) or repeating decimals (such as 0.5, 0.76, or 0.333333....). On the other hand, all those numbers that can be written as non-terminating, non-repeating decimals are non-rational, so they are called the "irrationals". Examples would be $$\sqrt{2}$$ ("the square root of two") or the number pi ($$\pi=$$~3.14159..., from geometry). The rationals and the irrationals are two totally separate number types: there is no overlap.
Putting these two major classifications, the rationals and the irrationals, together in one set gives you the "real" numbers.
FRACTIONS
Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.
Fraction can be expressed in two forms fractional representation $$(\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.
Definition
Fractional representation
Converting Improper Fractions
Reciprocal
Operation on Fractions
Decimal Representation
Converting Decimals to Fractions
Rounding
Ratios and Proportions
POSITIVE AND NEGATIVE NUMBERS
A positive number is a real number that is greater than zero.
A negative number is a real number that is smaller than zero.
Zero is not positive, nor negative.
Multiplication:
positive * positive = positive
positive * negative = negative
negative * negative = positive
Division:
positive / positive = positive
positive / negative = negative
negative / negative = positive
EXPONENTS, ROOTS, PERCENTS
Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. Roots (or radicals) are the "opposite" operation of applying exponents. A percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred").
Exponents
Perfect Square
Roots
Last Digit of a Product
Last Digit of a Power
Percents
ORDER OF OPERATIONS - PEMDAS
Perform the operations inside a Parenthesis first (absolute value signs also fall into this category), then Exponents, then Multiplication and Division, from left to right, then Addition and Subtraction, from left to right - PEMDAS.
Special cases:
• An exclamation mark indicates that one should compute the factorial of the term immediately to its left, before computing any of the lower-precedence operations, unless grouping symbols dictate otherwise. But $$3^2!$$ means $$(3^2)! = 9!$$ while $$2^{5!} = 2^{120}$$; a factorial in an exponent applies to the exponent, while a factorial not in the exponent applies to the entire power.
• If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$.
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
### Show Tags
10 Mar 2010, 05:21
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INTEGERS
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
--------------------------------------------------------
Definition
Integers are defined as: all negative natural numbers $$\{...,-4, -3, -2, -1\}$$, zero $$\{0\}$$, and positive natural numbers $$\{1, 2, 3, 4, ...\}$$.
Note that integers do not include decimals or fractions - just whole numbers.
Even and Odd Numbers
An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.
An even number is an integer of the form $$n=2k$$, where $$k$$ is an integer.
An odd number is an integer that is not evenly divisible by 2.
An odd number is an integer of the form $$n=2k+1$$, where $$k$$ is an integer.
Zero is an even number.
even +/- even = even;
even +/- odd = odd;
odd +/- odd = even.
Multiplication:
even * even = even;
even * odd = even;
odd * odd = odd.
Division of two integers can result into an even/odd integer or a fraction.
Prime Numbers
A Prime number is a natural number with exactly two distinct natural number divisors: 1 and itself. Otherwise a number is called a composite number. Therefore, 1 is not a prime, since it only has one divisor, namely 1. A number $$n > 1$$ is prime if it cannot be written as a product of two factors $$a$$ and $$b$$, both of which are greater than 1: n = ab.
• The first twenty-six prime numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101
• Note: only positive numbers can be primes.
• There are infinitely many prime numbers.
• The only even prime number is 2, since any larger even number is divisible by 2. Also 2 is the smallest prime.
All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form $$6n-1$$ or $$6n+1$$, because all other numbers are divisible by 2 or 3.
• Any nonzero natural number $$n$$ can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors.
Prime factorization: every positive integer greater than 1 can be written as a product of one or more prime integers in a way which is unique. For instance integer $$n$$ with three unique prime factors $$a$$, $$b$$, and $$c$$ can be expressed as $$n=a^p*b^q*c^r$$, where $$p$$, $$q$$, and $$r$$ are powers of $$a$$, $$b$$, and $$c$$, respectively and are $$\geq1$$.
Example: $$4200=2^3*3*5^2*7$$.
Verifying the primality (checking whether the number is a prime) of a given number $$n$$ can be done by trial division, that is to say dividing $$n$$ by all integer numbers smaller than $$\sqrt{n}$$, thereby checking whether $$n$$ is a multiple of $$m<\sqrt{n}$$.
Example: Verifying the primality of $$161$$: $$\sqrt{161}$$ is little less than $$13$$, from integers from $$2$$ to $$13$$, $$161$$ is divisible by $$7$$, hence $$161$$ is not prime.
• If $$n$$ is a positive integer greater than 1, then there is always a prime number $$p$$ with$$n < p < 2n$$.
Factors
A divisor of an integer $$n$$, also called a factor of $$n$$, is an integer which evenly divides $$n$$ without leaving a remainder. In general, it is said $$m$$ is a factor of $$n$$, for non-zero integers $$m$$ and $$n$$, if there exists an integer $$k$$ such that $$n = km$$.
• 1 (and -1) are divisors of every integer.
• Every integer is a divisor of itself.
• Every integer is a divisor of 0, except, by convention, 0 itself.
• Numbers divisible by 2 are called even and numbers not divisible by 2 are called odd.
• A positive divisor of n which is different from n is called a proper divisor.
• An integer n > 1 whose only proper divisor is 1 is called a prime number. Equivalently, one would say that a prime number is one which has exactly two factors: 1 and itself.
• Any positive divisor of n is a product of prime divisors of n raised to some power.
• If a number equals the sum of its proper divisors, it is said to be a perfect number.
Example: The proper divisors of 6 are 1, 2, and 3: 1+2+3=6, hence 6 is a perfect number.
There are some elementary rules:
• If $$a$$ is a factor of $$b$$ and $$a$$ is a factor of $$c$$, then $$a$$ is a factor of $$(b + c)$$. In fact, $$a$$ is a factor of $$(mb + nc)$$ for all integers $$m$$ and $$n$$.
• If $$a$$ is a factor of $$b$$ and $$b$$ is a factor of $$c$$, then $$a$$ is a factor of $$c$$.
• If $$a$$ is a factor of $$b$$ and $$b$$ is a factor of $$a$$, then $$a = b$$ or $$a=-b$$.
• If $$a$$ is a factor of $$bc$$, and $$gcd(a,b)=1$$, then a is a factor of $$c$$.
• If $$p$$ is a prime number and $$p$$ is a factor of $$ab$$ then $$p$$ is a factor of $$a$$ or $$p$$ is a factor of $$b$$.
Finding the Number of Factors of an Integer
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
Finding the Sum of the Factors of an Integer
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The sum of factors of $$n$$ will be expressed by the formula: $$\frac{(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)}{(a-1)*(b-1)*(c-1)}$$
Example: Finding the sum of all factors of 450: $$450=2^1*3^2*5^2$$
The sum of all factors of 450 is $$\frac{(2^{1+1}-1)*(3^{2+1}-1)*(5^{2+1}-1)}{(2-1)*(3-1)*(5-1)}=\frac{3*26*124}{1*2*4}=1209$$
Greatest Common Factor (Divisior) - GCF (GCD)
The greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.
To find the GCF, you will need to do prime-factorization. Then, multiply the common factors (pick the lowest power of the common factors).
• Every common divisor of a and b is a divisor of gcd(a, b).
• a*b=gcd(a, b)*lcm(a, b)
Lowest Common Multiple - LCM
The lowest common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple both of a and of b. Since it is a multiple, it can be divided by a and b without a remainder. If either a or b is 0, so that there is no such positive integer, then lcm(a, b) is defined to be zero.
To find the LCM, you will need to do prime-factorization. Then multiply all the factors (pick the highest power of the common factors).
Divisibility Rules
2 - If the last digit is even, the number is divisible by 2.
3 - If the sum of the digits is divisible by 3, the number is also.
4 - If the last two digits form a number divisible by 4, the number is also.
5 - If the last digit is a 5 or a 0, the number is divisible by 5.
6 - If the number is divisible by both 3 and 2, it is also divisible by 6.
7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.
8 - If the last three digits of a number are divisible by 8, then so is the whole number.
9 - If the sum of the digits is divisible by 9, so is the number.
10 - If the number ends in 0, it is divisible by 10.
11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11.
Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11.
12 - If the number is divisible by both 3 and 4, it is also divisible by 12.
25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25.
Factorials
Factorial of a positive integer $$n$$, denoted by $$n!$$, is the product of all positive integers less than or equal to n. For instance $$5!=1*2*3*4*5$$.
Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer $$n$$, can be determined with this formula:
$$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}$$, where k must be chosen such that $$5^k<n$$.
It's easier if you look at an example:
How many zeros are in the end (after which no other digits follow) of $$32!$$?
$$\frac{32}{5}+\frac{32}{5^2}=6+1=7$$ (denominator must be less than 32, $$5^2=25$$ is less)
Hence, there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
Finding the number of powers of a prime number $$p$$, in the $$n!$$.
The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}$$ ... till $$p^x<n$$
What is the power of 2 in 25!?
$$\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22$$
Finding the power of non-prime in n!:
How many powers of 900 are in 50!
Make the prime factorization of the number: $$900=2^2*3^2*5^2$$, then find the powers of these prime numbers in the n!.
Find the power of 2:
$$\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47$$
= $$2^{47}$$
Find the power of 3:
$$\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22$$
=$$3^{22}$$
Find the power of 5:
$$\frac{50}{5}+\frac{50}{25}=10+2=12$$
=$$5^{12}$$
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
Consecutive Integers
Consecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and -2, -1, 0, 1, are consecutive integers.
• Sum of $$n$$ consecutive integers equals the mean multiplied by the number of terms, $$n$$. Given consecutive integers $$\{-3, -2, -1, 0, 1,2\}$$, $$mean=\frac{-3+2}{2}=-\frac{1}{2}$$, (mean equals to the average of the first and last terms), so the sum equals to $$-\frac{1}{2}*6=-3$$.
• If n is odd, the sum of consecutive integers is always divisible by n. Given $$\{9,10,11\}$$, we have $$n=3$$ consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.
• If n is even, the sum of consecutive integers is never divisible by n. Given $$\{9,10,11,12\}$$, we have $$n=4$$ consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.
• The product of $$n$$ consecutive integers is always divisible by $$n!$$.
Given $$n=4$$ consecutive integers: $$\{3,4,5,6\}$$. The product of 3*4*5*6 is 360, which is divisible by 4!=24.
Evenly Spaced Set
Evenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. The set of integers $$\{9,13,17,21\}$$ is an example of evenly spaced set. Set of consecutive integers is also an example of evenly spaced set.
• If the first term is $$a_1$$ and the common difference of successive members is $$d$$, then the $$n_{th}$$ term of the sequence is given by:
$$a_ n=a_1+d(n-1)$$
• In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula $$mean=median=\frac{a_1+a_n}{2}$$, where $$a_1$$ is the first term and $$a_n$$ is the last term. Given the set $$\{7,11,15,19\}$$, $$mean=median=\frac{7+19}{2}=13$$.
• The sum of the elements in any evenly spaced set is given by:
$$Sum=\frac{a_1+a_n}{2}*n$$, the mean multiplied by the number of terms. OR, $$Sum=\frac{2a_1+d(n-1)}{2}*n$$
• Special cases:
Sum of n first integers: $$1+2+...+n=\frac{1+n}{2}*n$$
Sum of n first odd numbers: $$a_1+a_2+...+a_n=1+3+...+a_n=n^2$$, where $$a_n$$ is the last, $$n_{th}$$ term and given by: $$a_n=2n-1$$. Given $$n=5$$ first odd integers, then their sum equals to $$1+3+5+7+9=5^2=25$$.
Sum of n first positive even numbers: $$a_1+a_2+...+a_n=2+4+...+a_n=n(n+1)$$, where $$a_n$$ is the last, $$n_{th}$$ term and given by: $$a_n=2n$$. Given $$n=4$$ first positive even integers, then their sum equals to $$2+4+6+8=4(4+1)=20$$.
• If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is 13, so the sum is 13*5 =65.
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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10 Mar 2010, 05:22
FRACTIONS
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
--------------------------------------------------------
Definition
Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.
Fraction can be expressed in two forms fractional representation $$(\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.
Fractional representation
Fractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a numerator (the part) and a denominator (the whole).
• The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, $$\frac{9}{7}$$, 9 is the numerator and 7 is denominator.
• Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. $$\frac{1}{3}$$ is a proper fraction.
• Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. $$\frac{5}{2}$$ is improper fraction.
• An integer combined with a proper fraction is called mixed number. $$4\frac{3}{5}$$ is a mixed number. This can also be written as an improper fraction: $$\frac{23}{5}$$
Converting Improper Fractions
• Converting Improper Fractions to Mixed Fractions:
1. Divide the numerator by the denominator
2. Write down the whole number answer
3. Then write down any remainder above the denominator
Example #1: Convert $$\frac{11}{4}$$ to a mixed fraction.
Solution: Divide $$\frac{11}{4} = 2$$ with a remainder of $$3$$. Write down the $$2$$ and then write down the remainder $$3$$ above the denominator $$4$$, like this: $$2\frac{3}{4}$$
• Converting Mixed Fractions to Improper Fractions:
1. Multiply the whole number part by the fraction's denominator
2. Add that to the numerator
3. Then write the result on top of the denominator
Example #2: Convert $$3\frac{2}{5}$$ to an improper fraction.
Solution: Multiply the whole number by the denominator: $$3*5=15$$. Add the numerator to that: $$15 + 2 = 17$$. Then write that down above the denominator, like this: $$\frac{17}{5}$$
Reciprocal
Reciprocal for a number $$x$$, denoted by $$\frac{1}{x}$$ or $$x^{-1}$$, is a number which when multiplied by $$x$$ yields $$1$$. The reciprocal of a fraction $$\frac{a}{b}$$ is $$\frac{b}{a}$$. To get the reciprocal of a number, divide 1 by the number. For example reciprocal of $$3$$ is $$\frac{1}{3}$$, reciprocal of $$\frac{5}{6}$$ is $$\frac{6}{5}$$.
Operation on Fractions
To add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator.
To add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions
Multiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators.
Dividing fractions: Change the divisor into its reciprocal and then multiply.
Example #1: $$\frac{3}{7}+\frac{2}{3}=\frac{9}{21}+\frac{14}{21}=\frac{23}{21}$$
Example #2: Given $$\frac{\frac{3}{5}}{2}$$, take the reciprocal of $$2$$. The reciprocal is $$\frac{1}{2}$$. Now multiply: $$\frac{3}{5}*\frac{1}{2}=\frac{3}{10}$$.
Decimal Representation
The decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....).
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Converting Decimals to Fractions
• To convert a terminating decimal to fraction:
1. Calculate the total numbers after decimal point
2. Remove the decimal point from the number
3. Put 1 under the denominator and annex it with "0" as many as the total in step 1
4. Reduce the fraction to its lowest terms
Example: Convert $$0.56$$ to a fraction.
1: Total number after decimal point is 2.
2 and 3: $$\frac{56}{100}$$.
4: Reducing it to lowest terms: $$\frac{56}{100}=\frac{14}{25}$$
• To convert a recurring decimal to fraction:
1. Separate the recurring number from the decimal fraction
2. Annex denominator with "9" as many times as the length of the recurring number
3. Reduce the fraction to its lowest terms
Example #1: Convert $$0.393939...$$ to a fraction.
1: The recurring number is $$39$$.
2: $$\frac{39}{99}$$, the number $$39$$ is of length $$2$$ so we have added two nines.
3: Reducing it to lowest terms: $$\frac{39}{99}=\frac{13}{33}$$.
• To convert a mixed-recurring decimal to fraction:
1. Write down the number consisting with non-repeating digits and repeating digits.
2. Subtract non-repeating number from above.
3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's.
Example #2: Convert $$0.2512(12)$$ to a fraction.
1. The number consisting with non-repeating digits and repeating digits is 2512;
2. Subtract 25 (non-repeating number) from above: 2512-25=2487;
3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.
Rounding
Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.
Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.
Ratios and Proportions
Given that $$\frac{a}{b}=\frac{c}{d}$$, where a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained.
$$\frac{b}{a}=\frac{d}{c}$$ - invertendo
$$\frac{a}{c}=\frac{b}{d}$$ - alternendo
$$\frac{a+b}{b}=\frac{c+d}{d}$$ - componendo
$$\frac{a-b}{b}=\frac{c-d}{d}$$ - dividendo
$$\frac{a+b}{a-b}=\frac{c+d}{c-d}$$ - componendo & dividendo
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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10 Mar 2010, 05:22
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EXPONENTS & ROOTS
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
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EXPONENTS
Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number $$a$$ multiplied $$n$$ times can be written as $$a^n$$, where $$a$$ represents the base, the number that is multiplied by itself $$n$$ times and $$n$$ represents the exponent. The exponent indicates how many times to multiple the base, $$a$$, by itself.
Exponents one and zero:
$$a^0=1$$ Any nonzero number to the power of 0 is 1.
For example: $$5^0=1$$ and $$(-3)^0=1$$
$$a^1=a$$ Any number to the power 1 is itself.
Powers of zero:
If the exponent is positive, the power of zero is zero: $$0^n = 0$$, where $$n > 0$$.
If the exponent is negative, the power of zero ($$0^n$$, where $$n < 0$$) is undefined, because division by zero is implied.
Powers of one:
$$1^n=1$$ The integer powers of one are one.
Negative powers:
$$a^{-n}=\frac{1}{a^n}$$
Powers of minus one:
If n is an even integer, then $$(-1)^n=1$$.
If n is an odd integer, then $$(-1)^n =-1$$.
Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
$$a^n*b^n=(ab)^n$$
$$\frac{a^n}{b^n}=(\frac{a}{b})^n$$
$$(a^m)^n=a^{mn}$$
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$
Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
$$a^n*a^m=a^{n+m}$$
$$\frac{a^n}{a^m}=a^{n-m}$$
Fraction as power:
$$a^{\frac{1}{n}}=\sqrt[n]{a}$$
$$a^{\frac{m}{n}}=\sqrt[n]{a^m}$$
Exponential Equations:
When solving equations with even exponents, we must consider both positive and negative possibilities for the solutions.
For instance $$a^2=25$$, the two possible solutions are $$5$$ and $$-5$$.
When solving equations with odd exponents, we'll have only one solution.
For instance for $$a^3=8$$, solution is $$a=2$$ and for $$a^3=-8$$, solution is $$a=-2$$.
Exponents and divisibility:
$$a^n-b^n$$ is ALWAYS divisible by $$a-b$$.
$$a^n-b^n$$ is divisible by $$a+b$$ if $$n$$ is even.
$$a^n + b^n$$ is divisible by $$a+b$$ if $$n$$ is odd, and not divisible by a+b if n is even.
Perfect Square
A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is an perfect square.
There are some tips about the perfect square:
• The number of distinct factors of a perfect square is ALWAYS ODD.
• The sum of distinct factors of a perfect square is ALWAYS ODD.
• A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.
• Perfect square always has even number of powers of prime factors.
LAST DIGIT OF A PRODUCT
Last $$n$$ digits of a product of integers are last $$n$$ digits of the product of last $$n$$ digits of these integers.
For instance last 2 digits of 845*9512*408*613 would be the last 2 digits of 45*12*8*13=540*104=40*4=160=60
Example: The last digit of 85945*89*58307=5*9*7=45*7=35=5?
LAST DIGIT OF A POWER
Determining the last digit of $$(xyz)^n$$:
1. Last digit of $$(xyz)^n$$ is the same as that of $$z^n$$;
2. Determine the cyclicity number $$c$$ of $$z$$;
3. Find the remainder $$r$$ when $$n$$ divided by the cyclisity;
4. When $$r>0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^r$$ and when $$r=0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^c$$, where $$c$$ is the cyclisity number.
• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. $$(xy4)^n$$) have a cyclisity of 2. When n is odd $$(xy4)^n$$ will end with 4 and when n is even $$(xy4)^n$$ will end with 6.
• Integers ending with 9 (eg. $$(xy9)^n$$) have a cyclisity of 2. When n is odd $$(xy9)^n$$ will end with 9 and when n is even $$(xy9)^n$$ will end with 1.
Example: What is the last digit of $$127^{39}$$?
Solution: Last digit of $$127^{39}$$ is the same as that of $$7^{39}$$. Now we should determine the cyclisity of $$7$$:
1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...
So, the cyclisity of 7 is 4.
Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of $$127^{39}$$ is the same as that of the last digit of $$7^{39}$$, is the same as that of the last digit of $$7^3$$, which is $$3$$.
ROOTS
Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4.
General rules:
• $$\sqrt{x}\sqrt{y}=\sqrt{xy}$$ and $$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$$.
• $$(\sqrt{x})^n=\sqrt{x^n}$$
• $$x^{\frac{1}{n}}=\sqrt[n]{x}$$
• $$x^{\frac{n}{m}}=\sqrt[m]{x^n}$$
• $${\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}$$
• $$\sqrt{x^2}=|x|$$, when $$x\leq{0}$$, then $$\sqrt{x^2}=-x$$ and when $$x\geq{0}$$, then $$\sqrt{x^2}=x$$
• When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.
That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.
• Odd roots will have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.
• For GMAT it's good to memorize following values:
$$\sqrt{2}\approx{1.41}$$
$$\sqrt{3}\approx{1.73}$$
$$\sqrt{5}\approx{2.24}$$
$$\sqrt{6}\approx{2.45}$$
$$\sqrt{7}\approx{2.65}$$
$$\sqrt{8}\approx{2.83}$$
$$\sqrt{10}\approx{3.16}$$
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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10 Mar 2010, 06:48
This is a lucky day for me . Thank you. +1Kudos.
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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02 Apr 2010, 01:42
Great resource. Thanks people!
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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20 Jul 2011, 06:52
good stuff. thank you!
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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04 Sep 2012, 20:11
Bunuel wrote:
FRACTIONS
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
--------------------------------------------------------
Definition
Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.
Fraction can be expressed in two forms fractional representation $$(\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.
Fractional representation
Fractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a numerator (the part) and a denominator (the whole).
• The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, $$\frac{9}{7}$$, 9 is the numerator and 7 is denominator.
• Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. $$\frac{1}{3}$$ is a proper fraction.
• Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. $$\frac{5}{2}$$ is improper fraction.
• An integer combined with a proper fraction is called mixed number. $$4\frac{3}{5}$$ is a mixed number. This can also be written as an improper fraction: $$\frac{23}{5}$$
Converting Improper Fractions
• Converting Improper Fractions to Mixed Fractions:
1. Divide the numerator by the denominator
2. Write down the whole number answer
3. Then write down any remainder above the denominator
Example #1: Convert $$\frac{11}{4}$$ to a mixed fraction.
Solution: Divide $$\frac{11}{4} = 2$$ with a remainder of $$3$$. Write down the $$2$$ and then write down the remainder $$3$$ above the denominator $$4$$, like this: $$2\frac{3}{4}$$
• Converting Mixed Fractions to Improper Fractions:
1. Multiply the whole number part by the fraction's denominator
2. Add that to the numerator
3. Then write the result on top of the denominator
Example #2: Convert $$3\frac{2}{5}$$ to an improper fraction.
Solution: Multiply the whole number by the denominator: $$3*5=15$$. Add the numerator to that: $$15 + 2 = 17$$. Then write that down above the denominator, like this: $$\frac{17}{5}$$
Reciprocal
Reciprocal for a number $$x$$, denoted by $$\frac{1}{x}$$ or $$x^{-1}$$, is a number which when multiplied by $$x$$ yields $$1$$. The reciprocal of a fraction $$\frac{a}{b}$$ is $$\frac{b}{a}$$. To get the reciprocal of a number, divide 1 by the number. For example reciprocal of $$3$$ is $$\frac{1}{3}$$, reciprocal of $$\frac{5}{6}$$ is $$\frac{6}{5}$$.
Operation on Fractions
To add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator.
To add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions
Multiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators.
Dividing fractions: Change the divisor into its reciprocal and then multiply.
Example #1: $$\frac{3}{7}+\frac{2}{3}=\frac{9}{21}+\frac{14}{21}=\frac{23}{21}$$
Example #2: Given $$\frac{\frac{3}{5}}{2}$$, take the reciprocal of $$2$$. The reciprocal is $$\frac{1}{2}$$. Now multiply: $$\frac{3}{5}*\frac{1}{2}=\frac{3}{10}$$.
Decimal Representation
The decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....).
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Converting Decimals to Fractions
• To convert a terminating decimal to fraction:
1. Calculate the total numbers after decimal point
2. Remove the decimal point from the number
3. Put 1 under the denominator and annex it with "0" as many as the total in step 1
4. Reduce the fraction to its lowest terms
Example: Convert $$0.56$$ to a fraction.
1: Total number after decimal point is 2.
2 and 3: $$\frac{56}{100}$$.
4: Reducing it to lowest terms: $$\frac{56}{100}=\frac{14}{25}$$
• To convert a recurring decimal to fraction:
1. Separate the recurring number from the decimal fraction
2. Annex denominator with "9" as many times as the length of the recurring number
3. Reduce the fraction to its lowest terms
Example #1: Convert $$0.393939...$$ to a fraction.
1: The recurring number is $$39$$.
2: $$\frac{39}{99}$$, the number $$39$$ is of length $$2$$ so we have added two nines.
3: Reducing it to lowest terms: $$\frac{39}{99}=\frac{13}{33}$$.
• To convert a mixed-recurring decimal to fraction:
1. Write down the number consisting with non-repeating digits and repeating digits.
2. Subtract non-repeating number from above.
3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's.
Example #2: Convert $$0.2512(12)$$ to a fraction.
1. The number consisting with non-repeating digits and repeating digits is 2512;
2. Subtract 25 (non-repeating number) from above: 2512-25=2487;
3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.
Rounding
Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.
Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.
Ratios and Proportions
Given that $$\frac{a}{b}=\frac{c}{d}$$, where a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained.
$$\frac{b}{a}=\frac{d}{c}$$ - invertendo
$$\frac{a}{c}=\frac{b}{d}$$ - alternendo
$$\frac{a+b}{b}=\frac{c+d}{d}$$ - componendo
$$\frac{a-b}{b}=\frac{c-d}{d}$$ - dividendo
$$\frac{a+b}{a-b}=\frac{c+d}{c-d}$$ - componendo & dividendo
Thanks, this a great and concise summary of all the basics.
Just to add, another important basic rule
for +ve fractions,
if $$\frac{a}{b}<1$$ and x is any positive number then $$\frac{(a+x)}{(b+x)}>\frac{a}{b}$$
,in short when the two fractions have same difference between numerator and denominator , the fraction having greater numerator is always greater.
Also, when $$\frac{a}{b}>1$$ and x is any positive number then $$\frac{(a+x)}{(b+x)}<\frac{a}{b}$$
Example which one is greater of the following two:
a)$$\frac{111487}{111490}$$ b)$$\frac{111587}{111590}$$ , the ans is b for the same reasons above, as $$\frac{111487+100}{111490+100}$$
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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02 Oct 2015, 09:17
• All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form 6n−1 or 6n+1, because all other numbers are divisible by 2 or 3.
Why does this property "all prime numbers above 3 are of the form 6n−1 or 6n+1" hold true for 961, but 961 is not a prime? This holds true for square of any prime number? Is there any gap in my understanding?
6n+1 = 961 --> 6n = 960
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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03 Oct 2015, 02:57
Very useful...keep posting..
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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03 Oct 2015, 03:43
charan2892 wrote:
Very useful...keep posting..
Thank you.
For more check ALL YOU NEED FOR QUANT ! ! !.
Hope it helps.
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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06 Dec 2017, 10:05
Bunuel
Thanks so much! Super useful!!
Question: The problem sets referenced at the end of number theory pdf are based on edition 12 of the GMAT official guide. I would like to try prep questions for number theory but don't have this edition of the book (I have the 2017 edition). Where can I find the questions you wrote in the pdf or where can I find questions to test the number theory concepts covered in your pdf?
Many thanks!
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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06 Dec 2017, 10:12
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gmatm8 wrote:
Bunuel
Thanks so much! Super useful!!
Question: The problem sets referenced at the end of number theory pdf are based on edition 12 of the GMAT official guide. I would like to try prep questions for number theory but don't have this edition of the book (I have the 2017 edition). Where can I find the questions you wrote in the pdf or where can I find questions to test the number theory concepts covered in your pdf?
Many thanks!
2. Properties of Integers
For more check Ultimate GMAT Quantitative Megathread
Hope it helps.
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Re: Math: Number Theory (broken into smaller topics) [#permalink] 06 Dec 2017, 10:12
Display posts from previous: Sort by | 2018-01-20T05:22:29 | {
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https://mathoverflow.net/questions/221466/how-big-can-a-set-of-integers-be-if-all-pairs-have-small-gcd | # How big can a set of integers be if all pairs have small gcd?
Suppose $A\subset[1,N]$ is a set of integers. If for any distinct $a,b\in A$ we have $(a,b)\leq M$ then how big can $|A|$ be?
If $M=1$ then $|A|$ is at most $\pi(N)$ since the map $a\mapsto P_+(a)$ (which sends $a$ to its largest prime factor) is an injection to the primes, and so the primes themselves are the most efficient. If $M=N$ then we can take $|A|=N$. What about $M$ in between? Can you beat the primes? It would be safe to assume that $A$ consists solely of $M$-smooth numbers since we can decompose $A=A_1\cup A_2$ where each $a\in A_1$ is $M$-smooth and each $a\in A_2$ has a prime factor which at least $M+1$. The large prime factors appearing as divisors in $A_2$ cannot be repeated, so $|A_2|\leq \pi(N)$.
• Even more efficiently, you may beat the primes by adding all pairwise products of primes not exceeding $M$ (including their squares). – Ilya Bogdanov Oct 21 '15 at 16:31
• Actually, for $M= 1$ you have $|A| = \pi(N) + 1$: you can have $1$ as well as the primes. @IlyaBogdanov: there's no reason to restrict to pairwise products, you can throw in products of arbitrarily many (not necessarily distinct) primes $x = p_1 \ldots p_n$, $p_1 \le \ldots \le p_n$, as long as $x/p_1 \le M$. – Robert Israel Oct 21 '15 at 18:21
• Thus for $M=5$, in addition to the primes $\le N$ you have $\{1,4,6,8,9,10,15,25\}$. – Robert Israel Oct 21 '15 at 18:28
We'll prove that the maximal cardinality of such a set for $M^2\leq N$ has size equal to $$\pi(N)+\sum_{1<n\leq M} \pi(p(n))$$ where $p(n)$ is the smallest prime factor of $n$. Since $$\sum_{1<n\leq M} \pi(p(n))\sim \frac{M^2}{2\log^2 M},$$ this proves that Ilya Bogdanov's example of including all pairwise products of primes not exceeding $M$ is nearly optimal in terms of asymptotics.
As you suggest in the question, this problem is equivalent to constructing the largest subset $A\subset S_M(N)$ such that $\gcd(a,b)\leq M$ for every $a,b\in A$ where $S_M(N)$ is the set of $n\leq N$ whose largest prime factor is at most $M$.
To see why, suppose that $A$ satisfies $\gcd(a,b)\leq M$ for every $a,b\in A$. Then every prime that is greater than $M$ can divide at most one element of $A$. If $p>M$ divides $a\in A$, then making $a=p$ only helps create a larger set $A$. Since these primes do not interact with the $M$-smooth numbers, the proof is complete.
Let $T(N,M)$ denote the maximum size of such a set $A\subset S_M(N)$. Then the size of the largest subset of $[1,N]$ with pairwise $\gcd$'s bounded by $M$ is $$\pi(N)-\pi(M)+T(N,M).$$
As mentioned by Fedja in the comments s, the reasoning above for the primes extends to all integers. By considering those primes $p,q\leq M$ such that $pq>M$, we see that there can only be exactly one such number divisible by $pq$. Similarly for any $p,q,r$ with $pq,qr,rp\leq M$ and $pqr>M$ there can only be one number in our set divisible by $pqr$. Thus we find that the maximal size is the sum over those integers whose largest proper divisor is less than $M$. Since the largest proper divisor of $n$ equals $n/p(n)$ where $p(n)$ is the least prime factor, we can group things based on this, and we have that $$T(N,M)=\sum_{1<p\leq M}\sum_{n\leq M:\ p(n)\geq p}1.$$ Rearranging this equals $$\sum_{1<n\leq M}\sum_{p\leq p(n)}1=\sum_{n\leq M}\pi\left(p(n)\right),$$ and for composite $n$ $p(n)\leq\sqrt{n}$, so the primes dominate this sum. Thus asymptotically we have $$T(N,M)\sim\sum_{p\leq M}\pi(p)\sim\frac{M^{2}}{2\log^{2}M}.$$
• You can continue in the same spirit: consider pairs of primes $p,q\le M$ with $pq>M$. Any such pair can enter just one number, so we can as well have all pairs there. Next consider all triples such that the product of any 2 is less than or equal to $M$ but the triple product is $>M$. Again, any triple can enter at most one number, so take them all, etc. After that, of course, just add $[1,M]$. So to describe a maximal cardinality set it easy. As to the size, it looks like we just need to play with the prime number theorem carefully to figure the asymptotics but I have to teach a class now... – fedja Oct 21 '15 at 18:08
• This set can be also described as the set of all numbers whose largest proper divisor is not greater than $M$, which allows to estimate the size of the complement way faster than with my prime count idea. Now it is time to run on the stairs :-) – fedja Oct 21 '15 at 18:13
• @Fedja: Great, that works nicely! – Eric Naslund Oct 21 '15 at 18:22
• There's still something off if $M>\sqrt{N}$ which you should account for. Fedja's description of numbers with largest proper divisor not larger than $M$ is very nice. For example, this includes all numbers with no prime factors below $N/M$ which has interesting asymptotics when $M$ is larger than $\sqrt{N}$ (eventually getting to $N$ when $M=N$). So you could either restrict attention to small ranges of $M$, or perhaps flesh out the argument a little bit more ... – Lucia Oct 22 '15 at 3:19
• @Lucia: You're right, I have edited the question for now, but I'll add in a nice form for the sum when $M>\sqrt{N}$ – Eric Naslund Oct 22 '15 at 11:09 | 2019-03-22T09:20:36 | {
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https://gmw.globalmathproject.org/station/I8S8A | 03 413 022
### Question 1
Up to now our machines have consisted of a row of boxes extending infinitely far to the left. Why not have boxes extending infinitely far to the right as well?
Let’s go back to working with a $1 \leftarrow 10$ machine and see what such boxes could mean in that machine.
(Then it should become clear what they mean in other machines as well.)
To keep the left and right boxes visibly clear, we’ll separate them with a point. (Society calls this point, for a $1 \leftarrow 10$ machine at least, a decimal point.)
So, what does it mean to have dots in the right boxes? What are the values of dots in those boxes?
Since this is a $1 \leftarrow 10$ machine, we do know that ten dots in any one box explode to make one dot one place to the left. So ten dots in the box just to the right of the decimal point are equivalent to one dot in the $1$ s box. Each dot in that box must be worth one-tenth.
We have:
In the same way, ten dots in the next box over are worth one-tenth. And so each dot in that next box must be worth one-hundredth.
We have:
And ten one-thousands make a hundredth, and ten ten-thousands make a thousandth, and so on.
We see that the boxes to the left of the decimal point represent place values as given by the powers of ten, and the boxes to the right of the decimal point place values given by the reciprocals of the powers of ten.
We have just discovered decimals!
When people write $0.3$, for example, they mean the value of placing three dots in the first box after the decimal point.
We see that $0.3$ equals three tenths: $0.3 = \dfrac{3}{10}$.
Seven dots in the third box after the decimal point is seven thousandths: $0.007 = \dfrac{7}{1000}$.
Comment: Some people might leave off the beginning zero and just write $.007 = \dfrac{7}{1000}$. It’s just a matter of personal taste.
Some people read $0.6$ out loud as “point six” and others read it out loud as “six tenths.” Which is more helpful for understanding what the number really is?
### Question 2
There is a possible source of confusion with a decimal such as $0.31$. This is technically three tenths and one hundredth: $0.31 = \dfrac{3}{10} + \dfrac{1}{100}$.
But some people read $0.31$ out loud as “thirty-one hundredths,” which looks like this.
Are these the same thing?
Well, yes! With three explosions we see that thirty-one hundredths becomes three tenths and one hundredth.
Comment: You can also show that $\dfrac{3}{10} + \dfrac{1}{100}$ and $\dfrac{31}{100}$ are the same with the arithmetic of adding fractions. We have
$\dfrac{3}{10} + \dfrac{1}{100} = \dfrac{30}{100} + \dfrac{1}{100} = \dfrac{31}{100}$.
(Do you see that this is really the result of performing three unexplosions in a picture of $\dfrac{3}{10} + \dfrac{1}{100}$?)
A teacher asked his students to each draw a $1 \leftarrow 10$ machine picture of the fraction $\dfrac{319}{1000}$.
JinJin drew:
Subra drew:
The teacher marked both students as correct. Are each of these solutions indeed valid? Explain your thinking. (By the way, the teacher doesn’t mind if students just write numbers instead of drawing dots.)
### Question 3
Multiple choice!
The decimal $0.23$ equals:
(A) $\dfrac {23}{10}$
(B) $\dfrac {23}{100}$
(C) $\dfrac {23}{1000}$
(D) $\dfrac {23}{10000}$
### Question 4
The decimal $0.0409$ equals:
(A) $\dfrac {409}{100}$
(B) $\dfrac {409}{1000}$
(C) $\dfrac {409}{10000}$
(D) $\dfrac {409}{100000}$
## SIMPLE FRACTIONS AS DECIMALS
Some decimals give fractions that simplify further.
For example,
$0.5 = \dfrac{5}{10} = \dfrac{1}{2}$
and
$0.04 = \dfrac{4}{100} = \dfrac{1}{25}$.
Conversely, if a fraction can be rewritten to have a denominator that is a power of ten, then we can easily write it as a decimal.
For example,
$\dfrac{3}{5} = \dfrac{6}{10}$ and so $\dfrac{3}{5} = 0.6$
and
$\dfrac{13}{20} = \dfrac{13 \times 5}{20 \times 5} = \dfrac{65}{100} = 0.65$.
What fractions (in simplest terms) do the following decimals represent?
$0.05$, $0.2$, $0.8$, $0.004$
### Question 6
Write each of the following fractions as a decimal.
$\dfrac {2} {5}$, $\dfrac {1} {25}$, $\dfrac {1} {20}$, $\dfrac {1} {200}$, $\dfrac {2} {2500}$
### Question 7
MULTIPLE CHOICE!
The decimal $0.050$ equals
(A)$\dfrac {50} {100}$
(B)$\dfrac {1} {20}$
(C)$\dfrac {1} {200}$
(D) None of these?
### Question 8
The decimal $0.000208$ equals
(A) $\dfrac {52} {250}$
(B) $\dfrac {52} {2500}$
(C) $\dfrac {52} {25000}$
(D) $\dfrac {52} {250000}$
### Question 9
Write each of the following fractions as decimals.
$\dfrac {7} {20}$, $\dfrac {16} {25}$, $\dfrac {301} {500}$, $\dfrac {17} {50}$, $\dfrac {3} {4}$
### Question 10
CHALLENGE
What fraction does the decimal $2.3$ represent?
### Question 11
What fraction does $17.04$ represent?
### Question 12
What fraction does $1003.1003$ represent?
### Question 13
Let’s explore the question: Do $0.19$ and $0.190$ represent the same number or different numbers?
Here are two dots and boxes pictures for the decimal $0.19$:
Here are two dots and boxes picture for the decimal $0.190$
(A) Explain how one “unexplosion” establishes that the first picture of $0.19$ is equivalent to the second picture of $0.19$.
(B) Explain how several unexplosions establishes that the first picture of $0.190$ is equivalent to the second picture of $0.190$.
(C) Explain how explosions and unexplosions in fact establish that all four pictures are equivalent to each other.
(D) In conclusion then: Does $0.190$ represent the same number as $0.19$?
### Question 14
To a mathematician, the expressions $0.19$ and $0.190$ represent exactly the same numeric quantity.
But you may have noticed in science class that scientists will often write down what seems likes
unnecessary zeros when recording measurements. This is because scientists want to impart more information to the reader than just a numeric value.
For example, suppose a botanist measures the length of a stalk. By writing the measurement as $0.190$ meters in her paper, the scientist is saying to the reader that she measured the length of the stalk to the nearest one thousandth of a meter and that she got $1$ tenth, $9$ hundredths, and $0$ thousandths of a meter. Thus we are being told that the true length of the stalk is somewhere in the range of $0.1895$ and $0.1905$ meters.
If she wrote in her paper, instead, just $0.19$ meters, then we would have to assume she measured the length of the stalk only to the nearest hundredth of a meter and so its true length lies somewhere between $0.185$ and $0.195$ meters.
## MIXED NUMBERS AS DECIMALS.
How does $12 \dfrac {3} {4}$, for example, appear as a decimal?
Well, $12 \dfrac {3} {4} = 12 + \dfrac {3} {4}$ and we can certainly write the fractional part as a decimal. (The non-fractional part is already in the $1 \leftarrow 10$ machine format!)
We have
$12 \dfrac {3} {4} = 12 + \dfrac {75} {100}$
and so we see
$12 \dfrac {3} {4} = 12.75$.
Write each of the following numbers in decimal notation.
(A) $5 \dfrac {3} {10}$
(B) $7 \dfrac {1} {5}$
(C) $13 \dfrac {1} {2}$
(D) $106 \dfrac {3} {20}$
(E) $\dfrac {78} {25}$
(F) $\dfrac {9} {4}$
(G) $\dfrac {131} {40}$
You can either play with some of the optional stations below or go to the next island!
Register NOW and unlock all islands! | 2019-08-19T19:09:18 | {
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https://math.stackexchange.com/questions/2844057/solving-sin-omega-t-frac-1-2/2844061 | # Solving $\sin(\omega t)=- \frac 1 2$
Solve the given trigonometric equation: $$\sin(\omega t)=- \frac 1 2$$
Here is my attempt:
$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (\pi +\dfrac{\pi}{6}\biggr )$$
Which yields
$$\omega t = \dfrac{7\pi }{6}$$
or
$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (2\pi -\dfrac{\pi}{6}\biggr )$$
$$\omega t = \dfrac{11\pi}{6}$$
Is my assumption correct?
Regards!
• I've fixed a small error. – Mr. Maxwell Jul 7 '18 at 21:17
You also have $$\omega t = \dfrac{-\pi }{6}+2k\pi$$
Thus the solution is $$\omega t = \{\dfrac{-\pi }{6}+2k_1\pi:k_1 \in Z\}\cup \{\dfrac{7\pi }{6}+2k_2\pi:k_2\in Z\}$$
Where Z is the set of integers.
From the unit circle definition of sine, we see that
$$\omega t = 2n\pi - \dfrac{\pi}{2} \pm \dfrac{\pi}{3}$$
and we can rephrase that as we wish.
The sin$(x)$ function repeats itself every $2\pi$ times, so $\omega t=\frac{7\pi}6+2kπ$ and $\omega t=\frac{11\pi}{6}+2k\pi,\;$ where $k \in \mathbb Z$.
Yes it is right but recall to add the $2k\pi$ term with $k\in \mathbb{Z}$, indeed we have
$$\sin(\omega t)=- \frac 1 2\iff \omega t=\dfrac{7\pi }{6}+2k\pi\,\lor\,\omega t=\dfrac{11\pi }{6}+2k\pi$$
• What if I don't recall to add $2k\pi$? – Mr. Maxwell Jul 7 '18 at 21:16
• @Mr.Maxwell You can lost solutions. – user Jul 7 '18 at 21:17
• However, isn't it enough to leave it up once we get $\dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$? – Mr. Maxwell Jul 7 '18 at 21:18
• @Mr.Maxwell Those are the solutions for $\omega t \in [0,2\pi]$ but in general solving trigonometric equations we need to consider a wider range. Take a look for example here math.stackexchange.com/q/2826222/505767 – user Jul 7 '18 at 21:20
• @Mr.Maxwell - Enough is defined by who is asking the question. If the domain is not specified, then the question is poorly asked. – steven gregory Jul 7 '18 at 22:31 | 2019-11-21T16:13:22 | {
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https://au.mathworks.com/help/matlab/ref/plot3.html | plot3
3-D point or line plot
Description
example
plot3(X,Y,Z) plots coordinates in 3-D space.
• To plot a set of coordinates connected by line segments, specify X, Y, and Z as vectors of the same length.
• To plot multiple sets of coordinates on the same set of axes, specify at least one of X, Y, or Z as a matrix and the others as vectors.
example
plot3(X,Y,Z,LineSpec) creates the plot using the specified line style, marker, and color.
example
plot3(X1,Y1,Z1,...,Xn,Yn,Zn) plots multiple sets of coordinates on the same set of axes. Use this syntax as an alternative to specifying multiple sets as matrices.
example
plot3(X1,Y1,Z1,LineSpec1,...,Xn,Yn,Zn,LineSpecn) assigns specific line styles, markers, and colors to each XYZ triplet. You can specify LineSpec for some triplets and omit it for others. For example, plot3(X1,Y1,Z1,'o',X2,Y2,Z2) specifies markers for the first triplet but not the for the second triplet.
example
plot3(___,Name,Value) specifies Line properties using one or more name-value pair arguments. Specify the properties after all other input arguments. For a list of properties, see Line Properties.
example
plot3(ax,___) displays the plot in the target axes. Specify the axes as the first argument in any of the previous syntaxes.
example
p = plot3(___) returns a Line object or an array of Line objects. Use p to modify properties of the plot after creating it. For a list of properties, see Line Properties.
Examples
collapse all
Define t as a vector of values between 0 and 10$\pi$. Define st and ct as vectors of sine and cosine values. Then plot st, ct, and t.
t = 0:pi/50:10*pi;
st = sin(t);
ct = cos(t);
plot3(st,ct,t)
Create two sets of x-, y-, and z-coordinates.
t = 0:pi/500:pi;
xt1 = sin(t).*cos(10*t);
yt1 = sin(t).*sin(10*t);
zt1 = cos(t);
xt2 = sin(t).*cos(12*t);
yt2 = sin(t).*sin(12*t);
zt2 = cos(t);
Call the plot3 function, and specify consecutive XYZ triplets.
plot3(xt1,yt1,zt1,xt2,yt2,zt2)
Create matrix X containing three rows of x-coordinates. Create matrix Y containing three rows of y-coordinates.
t = 0:pi/500:pi;
X(1,:) = sin(t).*cos(10*t);
X(2,:) = sin(t).*cos(12*t);
X(3,:) = sin(t).*cos(20*t);
Y(1,:) = sin(t).*sin(10*t);
Y(2,:) = sin(t).*sin(12*t);
Y(3,:) = sin(t).*sin(20*t);
Create matrix Z containing the z-coordinates for all three sets.
Z = cos(t);
Plot all three sets of coordinates on the same set of axes.
plot3(X,Y,Z)
Create vectors xt, yt, and zt.
t = 0:pi/500:40*pi;
xt = (3 + cos(sqrt(32)*t)).*cos(t);
yt = sin(sqrt(32) * t);
zt = (3 + cos(sqrt(32)*t)).*sin(t);
Plot the data, and use the axis equal command to space the tick units equally along each axis. Then specify the labels for each axis.
plot3(xt,yt,zt)
axis equal
xlabel('x(t)')
ylabel('y(t)')
zlabel('z(t)')
Create vectors t, xt, and yt, and plot the points in those vectors using circular markers.
t = 0:pi/20:10*pi;
xt = sin(t);
yt = cos(t);
plot3(xt,yt,t,'o')
Create vectors t, xt, and yt, and plot the points in those vectors as a blue line with 10-point circular markers. Use a hexadecimal color code to specify a light blue fill color for the markers.
t = 0:pi/20:10*pi;
xt = sin(t);
yt = cos(t);
plot3(xt,yt,t,'-o','Color','b','MarkerSize',10,'MarkerFaceColor','#D9FFFF')
Create vector t. Then use t to calculate two sets of x and y values.
t = 0:pi/20:10*pi;
xt1 = sin(t);
yt1 = cos(t);
xt2 = sin(2*t);
yt2 = cos(2*t);
Plot the two sets of values. Use the default line for the first set, and specify a dashed line for the second set.
plot3(xt1,yt1,t,xt2,yt2,t,'--')
Create vectors t, xt, and yt, and plot the data in those vectors. Return the chart line in the output variable p.
t = linspace(-10,10,1000);
xt = exp(-t./10).*sin(5*t);
yt = exp(-t./10).*cos(5*t);
p = plot3(xt,yt,t);
Change the line width to 3.
p.LineWidth = 3;
Starting in R2019b, you can display a tiling of plots using the tiledlayout and nexttile functions. Call the tiledlayout function to create a 1-by-2 tiled chart layout. Call the nexttile function to create the axes objects ax1 and ax2. Create separate line plots in the axes by specifying the axes object as the first argument to plot3.
tiledlayout(1,2)
% Left plot
ax1 = nexttile;
t = 0:pi/20:10*pi;
xt1 = sin(t);
yt1 = cos(t);
plot3(ax1,xt1,yt1,t)
title(ax1,'Helix With 5 Turns')
% Right plot
ax2 = nexttile;
t = 0:pi/20:10*pi;
xt2 = sin(2*t);
yt2 = cos(2*t);
plot3(ax2,xt2,yt2,t)
title(ax2,'Helix With 10 Turns')
Create x and y as vectors of random values between 0 and 1. Create z as a vector of random duration values.
x = rand(1,10);
y = rand(1,10);
z = duration(rand(10,1),randi(60,10,1),randi(60,10,1));
Plot x, y, and z, and specify the format for the z-axis as minutes and seconds. Then add axis labels, and turn on the grid to make it easier to visualize the points within the plot box.
plot3(x,y,z,'o','DurationTickFormat','mm:ss')
xlabel('X')
ylabel('Y')
zlabel('Duration')
grid on
Create vectors xt, yt, and zt. Plot the values, specifying a solid line with circular markers using the LineSpec argument. Specify the MarkerIndices property to place one marker at the 200th data point.
t = 0:pi/500:pi;
xt(1,:) = sin(t).*cos(10*t);
yt(1,:) = sin(t).*sin(10*t);
zt = cos(t);
plot3(xt,yt,zt,'-o','MarkerIndices',200)
Input Arguments
collapse all
x-coordinates, specified as a scalar, vector, or matrix. The size and shape of X depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify X, Y, and Z as scalars and include a marker. For example:
plot3(1,2,3,'o')
One set of points
Specify X, Y, and Z as any combination of row or column vectors of the same length. For example:
plot3([1 2 3],[4; 5; 6],[7 8 9])
Multiple sets of points
(using vectors)
Specify consecutive sets of X, Y, and Z vectors. For example:
plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])
Multiple sets of points
(using matrices)
Specify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration
y-coordinates, specified as a scalar, vector, or matrix. The size and shape of Y depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify X, Y, and Z as scalars and include a marker. For example:
plot3(1,2,3,'o')
One set of points
Specify X, Y, and Z as any combination of row or column vectors of the same length. For example:
plot3([1 2 3],[4; 5; 6],[7 8 9])
Multiple sets of points
(using vectors)
Specify consecutive sets of X, Y, and Z vectors. For example:
plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])
Multiple sets of points
(using matrices)
Specify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration
z-coordinates, specified as a scalar, vector, or matrix. The size and shape of Z depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify X, Y, and Z as scalars and include a marker. For example:
plot3(1,2,3,'o')
One set of points
Specify X, Y, and Z as any combination of row or column vectors of the same length. For example:
plot3([1 2 3],[4; 5; 6],[7 8 9])
Multiple sets of points
(using vectors)
Specify consecutive sets of X, Y, and Z vectors. For example:
plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])
Multiple sets of points
(using matrices)
Specify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: '--or' is a red dashed line with circle markers
Line StyleDescription
-Solid line (default)
--Dashed line
:Dotted line
-.Dash-dot line
MarkerDescription
oCircle
+Plus sign
*Asterisk
.Point
xCross
sSquare
dDiamond
^Upward-pointing triangle
vDownward-pointing triangle
>Right-pointing triangle
<Left-pointing triangle
pPentagram
hHexagram
ColorDescription
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
Target axes, specified as an Axes object. If you do not specify the axes and if the current axes is Cartesian, then plot3 uses the current axes.
Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: plot3([1 2],[3 4],[5 6],'Color','red') specifies a red line for the plot.
Note
The properties listed here are only a subset. For a complete list, see Line Properties.
Color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name. The color you specify sets the line color. It also sets the marker edge color when the MarkerEdgeColor property is set to 'auto'.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
Marker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the Color property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Marker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' option uses the same color as the Color property of the parent axes. If you specify 'auto' and the axes plot box is invisible, the marker fill color is the color of the figure.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Tips
• Use NaN or Inf to create breaks in the lines. For example, this code plots a line with a break between z=2 and z=4.
plot3([1 2 3 4 5],[1 2 3 4 5],[1 2 NaN 4 5])
• plot3 uses colors and line styles based on the ColorOrder and LineStyleOrder properties of the axes. plot3 cycles through the colors with the first line style. Then, it cycles through the colors again with each additional line style.
Starting in R2019b, you can change the colors and the line styles after plotting by setting the ColorOrder or LineStyleOrder properties on the axes. You can also call the colororder function to change the color order for all the axes in the figure. | 2020-08-13T20:51:18 | {
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http://mathhelpforum.com/calculus/167221-integrate-ln2x-x-2-a.html | # Math Help - Integrate LN2x/x^2
1. ## Integrate LN2x/x^2
Hey guys, thanks for all the help so far.. stuck on another problem.
$\int \frac{ln2x}{x^2}$
So I think
$ln2x$
is an integration by parts problem by itself, but I am lost.
I know that
$\int lnx = xlnx-x$
But that was just a rule given by the professor, not sure how we applied that.
I think it goes something like this (using integration by parts)
$u= ln2x \ du= \frac{2x}{x} \ dv=dx \ v=x$
Using the rule
$UV- \int VdU$
I get
$(xln2x)- \int x \frac{2x}{x}$
which equals
$(xln2x)- \int \frac{2x^2}{x}$
But that's just for the top portion of the original and I am not even sure I did that correctly.
Thanks guys
2. Just in case a picture helps...
... where (key in first spoiler) ...
Spoiler:
... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,
... is lazy integration by parts, doing without u and v.
Spoiler:
________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
3. Originally Posted by Spoolx
Hey guys, thanks for all the help so far.. stuck on another problem.
$\int \frac{ln2x}{x^2}$
So I think
$ln2x$
is an integration by parts problem by itself, but I am lost.
I know that
$\int lnx = xlnx-x$
But that was just a rule given by the professor, not sure how we applied that.
I think it goes something like this (using integration by parts)
$u= ln2x \ du= \frac{2x}{x} \ dv=dx \ v=x$
Using the rule
$UV- \int VdU$
I get
$(xln2x)- \int x \frac{2x}{x}$
which equals
$(xln2x)- \int \frac{2x^2}{x}$
But that's just for the top portion of the original and I am not even sure I did that correctly.
Thanks guys
$ln(2x)=ln2+lnx$
$\displaystyle\int{\frac{ln(2x)}{x^2}}dx=ln2\int{\f rac{1}{x^2}}dx+\int{\frac{lnx}{x^2}}dx$
Now you can use integration by parts.
4. Hello, Spoolx!
$\displaystyle \int \frac{\ln(2x)}{x^2}\,dx$
Integrate by parts:
. . $\begin{array}{cccccccc}
u &=& \ln(2x) && dv &=& \dfrac{1}{x^2}\,dx \\ \\[-3mm]
du &=& \dfrac{dx}{x} && v &=& \text{-}\dfrac{1}{x} \end{array}$
And we have: . $\displaystyle \text{-}\frac{1}{x}\ln(2x) + \int\frac{dx}{x^2}$
. . . . . . . . $\displaystyle =\;\text{-}\frac{1}{x}\ln(2x) - \frac{1}{x} + C$
. . . . . . . . $\displaystyle =\;\text{-}\frac{1}{x}\left[\ln(2x) + 1\right] + C$
5. Thanks for the help guys, can you jst run through the derivation of the $ln2x$ please?
6. Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
The general drift is...
Though, as Archie Meade points out, you can split the log first to avoid the chain rule.
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
7. Originally Posted by Spoolx
Hey guys, thanks for all the help so far.. stuck on another problem.
$\int \frac{ln2x}{x^2}$
So I think
$ln2x$
is an integration by parts problem by itself, but I am lost.
I know that
$\int lnx = xlnx-x$
But that was just a rule given by the professor, not sure how we applied that.
I think it goes something like this (using integration by parts)
$u= ln2x \ du= \frac{2x}{x} \ dv=dx \ v=x$
Using the rule
$UV- \int VdU$
I get
$(xln2x)- \int x \frac{2x}{x}$
which equals
$(xln2x)- \int \frac{2x^2}{x}$
But that's just for the top portion of the original and I am not even sure I did that correctly.
Thanks guys
Rather than trying to apply integration by parts just to the numerator,
you should try applying the technique to the entire expression.
$\displaystyle\int{u}dv=uv-\int{v}du$
$u=ln(2x),\;\;dv=\displaystyle\frac{1}{x^2}dx$
We need $v$ and $du$ to apply integration by parts. Using the chain rule we get
$\displaystyle\frac{du}{dx}=\frac{1}{2x}2$
$\Rightarrow\ du=\frac{1}{x}dx$
$v=\int{x^{-2}}dx=-x^{-1}$
$\displaystyle\ uv-\int{v}du}=-\frac{ln(2x)}{x}+\int{\frac{1}{x^2}dx=-\frac{ln(2x)}{x}-\frac{1}{x}+C$
Alternatively, begin with
$ln(2x)=ln2+lnx$
8. Glad I found this forum, learning alot from you guys!
Thanks! | 2015-07-31T22:51:42 | {
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http://mathhelpforum.com/algebra/123610-verifying-answers-polynomials.html | # Math Help - verifying answers for polynomials
1. ## verifying answers for polynomials
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!
2. ## latex
Originally Posted by jay1
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!
hope you learn to use laTex
this is my understanding of your equation
$
\frac{1}{4a^2} + \frac{3}{4a} = 1
$
$\frac{1}{4a^2} + \frac{3}{4a}\times\frac{a}{a} \Rightarrow \frac{1+3a}{4a^2} = 1$
$4a^2 - 3a -1 = 0$
$(4a-1)(a+1)= 0$
$a= -1$
$a=-\frac{1}{4}$
3. Is there a place that will show me how to use this "laTex"? BigWave, was my answer correct? Thanks for your help.
4. Originally Posted by jay1
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!
If problem 1 is what I interpreted it to be, which is this:
$
\frac{1}{4a^2} + \frac{3}{4}a = 1
$
then
$
a=1, a=\frac{1+\sqrt{13}}{6}, \text{ and } a=\frac{1-\sqrt{13}}{6}
$
.
However, if problem 1 is this:
$
\frac{1}{4a^2} + \frac{3}{4a} = 1
$
then
$
a=1, a= -\frac{1}{4}
$
.
And if problem 1 is this:
$
\frac{1}{4}a^2 + \frac{3}{4}a = 1
$
then
$
a=1, a= -4
$
.
5. Hello, jay1!
1) Solve: . $\tfrac{1}{4}a^2 + \tfrac{3}{4}a \:=\:1$
My answer: . $a=4,\;a=1$ . . . . no
We have: . $\tfrac{1}{4}a^2 + \tfrac{3}{4}a \:=\:1$
Multiply by 4: . $a^2 + 3a \:=\:4 \quad\Rightarrow\quad a^2 + 3a - 4 \:=\:0$
Factor: . $(a-1)(a+4) \:=\:0$
Therefore: . $a\;=\;1,\:-4$
2) Simplify: . $2ab^4 -3a^2b^2 - ab^4 + a^2b^2$
I have: . $ab^4- 2a^2b^2$ . . . . Yes!
3) Simplify: . $-2(4y^2 +3z^3 +5) + 3(2y^2 -5z^3 +3)$
My answer: . $-21z^3 - 2y^2 -1$ . . . . Right!
6. PROBLEM 2:
$
2ab^4 - 3 a^2b^2 - ab^4 + a^2b^2 = ab^4 -2a^2b^2
$
You are correct.
7. PROBLEM 3:
$
-2(4y2+3z3+5) + 3(2y2-5z3+3) = (-8y^2 - 6z^3 - 10) + (6y^2 - 15z^3 + 9)$
$= -8y^2 - 6z^3 - 10 + 6y^2 - 15z^3 + 9 = -21z^3 - 2y^2 - 1$
Good job again. Problem 3 is correct.
-Andy
8. ## mystery equation
who had the mystery equation correct....
just curious..
9. Thanks for all of the help. I have a few more that I would like confirmation/correction for (exponents are denoted in red font). It helps to know if I am on the right track. 1) solve the equation x2 + 4x - 45 = 0 I have x=9 and x=-5 ??? 2) find the product of (x-2y)2 I have x2 - 4xy + 4x2 ??? 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?
10. BigWave, I think that it was Soroban. Thanks again!
11. Originally Posted by jay1
Thanks for all of the help. I have a few more that I would like confirmation/correction for (exponents are denoted in red font). It helps to know if I am on the right track. 1) solve the equation x2 + 4x - 45 = 0 I have x=9 and x=-5 ??? 2) find the product of (x-2y)2 I have x2 - 4xy + 4x2 ??? 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?
$x^2 + 4x - 45 = 0$
$(x+9)(x-5) = 0$
$x = 5, -9$
12. Originally Posted by jay1
2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 [COLOR=black]I have ab4 - 2a2b2
$
2ab^4 - 3a^2b^2 -ab^4 + a^2b^2 = ab^4 - 2a^2b^2
$
Atta boy (assuming you are a guy).
13. 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?
$
y^2 + 12y + 35 = (y+7)(y+5) \implies y=-5,-7
$
Good luck.
-Andy
14. Perform the indicated operations on this expression: (5a^3 + 3a -2) - (4a^3 + a^2 + 5) I have a^3 + a^2 + 3 Is this right?
15. Originally Posted by jay1
Perform the indicated operations on this expression: (5a^3 + 3a -2) - (4a^3 + a^2 + 5) I have a^3 + a^2 + 3 Is this right?
No.
$
(5a^3 + 3a -2) - (4a^3 + a^2 + 5) = 5a^3 + 3a - 2 - 4a^3 - a^2 - 5 =
a^3 - a^2 + 3a - 7
$
The minus before the parenthesis changes the sign of each term inside the parentheses.
-Andy
Page 1 of 2 12 Last | 2014-12-26T20:22:56 | {
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https://math.stackexchange.com/questions/3093945/how-can-i-simplify-this-fraction-problem | How can I simplify this fraction problem?
I have the problem $$\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$$ which should simplify to $$\frac{1}{x-2}$$
I have simplified $$x^2-4$$, which becomes:
$$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$$
However, if I combine the fractions I get, $$x^2-x-1$$ for the numerator, which can't be factored. That's where I get stuck.
How can I get $$\frac{1}{x-2}$$ out of this problem?
You need to put them over a common denominator, $$\frac {x+1}{x+2}=\frac {(x+1)(x-2)}{(x+2)(x-2)}=\frac {x^2-x-2}{x^2-4}$$ Now you can subtract the numerators $$x^2-(x^2-x-2)=x+2$$ and finally divide out the $$x+2$$ from numerator and denominator
Your $$-1$$ should be $$-2$$. You didn't show your work, so I can't see why it happened.
• Regarding your last line, I wonder if OP just added the two numerators together without finding a common denominator. – Matthew Leingang Jan 30 at 18:54
• @MatthewLeingang That's right. It's been awhile since I worked with fractions, and I'm very rusty. – LuminousNutria Jan 30 at 18:56
• @LuminousNutria It's a very common error. But if you think about it with familiar fractions you'll remember it can't be true. For instance, $\frac{1}{2} + \frac{1}{2}$ is $1$, not $\frac{2}{4}$ (which is $\frac{1}{2}$ again). – Matthew Leingang Jan 30 at 19:00
\begin{align} \frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}&=\frac1{x+2}\left(\frac{x^2}{x-2} - (x+1)\right)\\ &=\frac1{x+2}\left(\frac{x^2-(x-2)(x+1)}{x-2}\right)\\ &=\frac1{x+2}\left(\frac{x^2-(x^2-x-2)}{x-2}\right)\\ &=\frac1{x+2}\left(\frac{x+2}{x-2}\right)\\ &=\frac1{x-2} \end{align}
Write $$\frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{(x+1)(x-2)}{(x+2)(x-2)}=…$$ Note that it must be $$x\ne 2,-2$$
The numerator should be $$x^2-(x-2)(x+1) = x+2$$ which simplifies with a factor of the denominator.
You probably just forgot a term when subtracting the two fractions. | 2019-10-19T07:08:54 | {
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http://math.stackexchange.com/questions/310758/prove-that-for-all-integers-a-and-b-if-a-divides-b-then-a2 | # Prove that for all integers $a$ and $b$, if $a$ divides $b$, then $a^{2}$ divides $b^{2}$.
I just need to know that if $a$ divides $b$, where $a$ and $b$ are integers, does $a^{2}$ divide $b^{2}$?
-
If $a$ divides $b$, then $b=ka$ for some integer $k$, so $b^2=k^2a^2$ where $k^2$ is an integer.
-
How do you prove that k^2 is an integer? Is that by closure? – amster27 Feb 22 '13 at 1:07
@amster27: If you are concerned about whether $k^{2}$ is an integer, then you should be equally concerned about whether both $a^{2}$ and $b^{2}$ are integers. :) – Haskell Curry Feb 22 '13 at 1:23
Thanks, I'm just starting to write proofs in a Mathematical Reasoning class, and I just feel so lost. But these explanations helped. – amster27 Feb 22 '13 at 1:28
@amster: Note that $+_{\mathbb{Z}}: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ and $\times_{\mathbb{Z}}: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. Therefore, adding two integers or multiplying two integers yields an integer. – Haskell Curry Feb 22 '13 at 1:44
Hint $\rm\ \ a\mid b\ \Rightarrow\ \dfrac{b}a\in \Bbb Z\ \Rightarrow\ \dfrac{b^2}{a^2} = \left(\dfrac{b}a\right)^2\!\in \Bbb Z^2\subseteq \Bbb Z\:\Rightarrow\: a^2\mid b^2\$
-
That's closer to a proof than a hint. – 1015 Feb 22 '13 at 1:29
The argument is elegant, but I think that the OP wants to stay within $\mathbb{Z}$ and does not wish to jump into $\mathbb{Q}$, as he seems to be very interested in the axioms governing the properties of $\mathbb{Z}$. – Haskell Curry Feb 22 '13 at 1:29
@Haskell Possibly, but there was no hint of any such constraint in the question. – Math Gems Feb 22 '13 at 1:46 | 2015-07-07T18:07:30 | {
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https://codegolf.stackexchange.com/questions/231176/maximum-number-of-squares-touched-by-a-line-segment/231354#231354 | # Maximum number of squares touched by a line segment
Consider a square grid on the plane, with unit spacing. A line segment of integer length $$\L\$$ is dropped at an arbitrary position with arbitrary orientation. The segment is said to "touch" a square if it intersects the interior of the square (not just its border).
# The challenge
What is the maximum number of squares that the segment can touch, as a function of $$\L\$$?
# Examples
• L=3 $$\\ \, \$$ The answer is $$\7\$$, as illustrated by the blue segment in the left-hand side image (click for a larger view). The red and yellow segments only touch $$\6\$$ and $$\4\$$ squares respectively. The purple segment touches $$\0\$$ squares (only the interiors count).
• L=5 $$\\ \, \$$ The answer is $$\9\$$. The dark red segment in the right-hand side image touches $$\6\$$ squares (note that $$\5^2 = 3^2+4^2\$$), whereas the green one touches $$\8\$$. The light blue segment touches $$\9\$$ squares, which is the maximum for this $$\L\$$.
• The input $$\L\$$ is a positive integer.
• The algorithm should theoretically work for arbitrarily large $$\L\$$. In practice it is acceptable if the program is limited by time, memory, or data-type size.
• Input and output are flexible as usual. Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.
• Shortest code in bytes wins.
# Test cases
Here are the outputs for L = 1, 2, ..., 50 (with L increasing left to right, then down):
3 5 7 8 9 11 12 14 15 17
18 19 21 22 24 25 27 28 29 31
32 34 35 36 38 39 41 42 43 45
46 48 49 51 52 53 55 56 58 59
60 62 63 65 66 68 69 70 72 73
• No OEIS entry, apparently (maybe soon) Jul 10 at 15:35
• I really liked this one from a problem-solving perspective. Jul 10 at 23:59
• My Math.SE question will be interesting to y'all and relevant to this question Jul 11 at 9:03
• This reminds me of the problem of optimizing beam placement in FTL Jul 13 at 2:02
• This sequence is now in OEIS: A346232 Jul 15 at 0:02
# Vyxal, 7 bytes
*d⇩√3+⌊
Try it Online!
* # Square
d # Double
⇩ # Subtract 2
√ # Square root
⌊ # Floor
# Python 2, 27 bytes
lambda L:(2*L*L-2)**.5//1+3
Try it online!
A direct formula:
$$f(L) = \lfloor \sqrt{2 L^2-2}\rfloor + 3$$
Derivation
As noted by @Kirill L. and others, the optimal layout uses a near-diagonal line segment whose horizontal and vertical span are at least $$\(h,h)\$$ or $$\(h,h+1)\$$. We need the length-$$\L\$$ to cover at least this much distance using the Pythagorean theorem, plus some extra to reach into squares and hit 3 more.
This gives a result of either:
• $$\2h+3\$$ where $$\h\$$ is the greatest positive integer where $$\h^2+h^2 < L^2\$$, or
• $$\2h+4\$$ where $$\h\$$ is the greatest positive integer where $$\h^2+(h+1)^2 ,
whichever of these is larger. We want to combine these two cases into one.
Let's start by making the two cases look more similar. Note that the second case can be rewritten as
$$\2(h+\frac{1}{2})+3\$$, where $$\2(h+\frac{1}{2})^2 + \frac{1}{2} < L^2\$$
or as
$$\2h+3\$$, where $$\2h^2 + \frac{1}{2} < L^2\$$
where $$\h\$$ is a positve integer-and-a-half. The $$\2h^2 in the first case can be also be written as $$\2h^2 +\frac{1}{2} < L^2\$$, which is equivalent because both sides were integers. So, the cases now merge into:
$$\2h+3\$$ where $$\h\$$ is the greatest positive integer or half-integer where $$\2h^2 +1/2 < L^2\$$
Calling $$\2h=H\$$, this is:
$$\H+3\$$ where $$\H\$$ is the greatest positive integer where $$\2(H/2)^2 +1/2 < L^2\$$.
This inequality is $$\H^2 < 2L^2-1\$$, and since these are integers, this is the same as $$\H^2 \leq 2L^2-2\$$. The greatest such positive integer $$\H\$$ is then $$\\lfloor \sqrt{2 L^2-2}\rfloor\$$, so the final result is $$\ \lfloor \sqrt{2 L^2-2}\rfloor + 3 \$$.
• Fantastic. "We want to combine these two cases into one. Let's start by making the two cases look more similar." Did you know somehow this could be done at the outset (if so, how?), or did you just guess it might be possible, and start playing around? Jul 11 at 3:55
• @Jonah It was a guess, but it seemed to me like something like this should be possible because the two types of outputs lie nearly on the same approximate curve.
– xnor
Jul 11 at 4:06
• Wonderful! Simpler than the formula I had obtained Jul 11 at 10:41
• I intend to submit this sequence to OEIS. Your formula is simpler than the one I had (i+j+3 with i = floor(L/sqrt(2)), j=ceil(sqrt(L^2-i^2))-1, so I'd like to give you credit (in addition to linking this challenge). How can I do that? If you prefer we can go on using e-mail Jul 11 at 15:22
• I may also have to write a short paper to support the submission, in particular to justify that the optimal segment orientation is always (h,h) or (h,h+1). Are interested in collaborating on this? Please let me know either way Jul 12 at 8:55
# R, 79 77 bytes
L=scan();j=1:L;a=j*2^.5;b=Mod(j+j*1i+1i);3+2*sum(L>a)+max(a[L>a],b[L>b])%in%b
Try it online!
Note: as it turned out, this is completely destroyed by xnor's formula, which would be 23 bytes in R:
(2*scan()^2-2)^.5%/%1+3
However, I'm keeping the existing code as a reference to my original solution.
### Original explanation
In general case, the most squares will be touched when the line is going close to the diagonal orientation. Specifically, the number of touched squares will be equal to $$\2 \times x + 3\$$, where $$\x\$$ is the number of unit square diagonals fully covered by the line. The lengths of square diagonals are stored in a.
However, in some cases a better coverage can be achieved by deviating from the diagonal orientation. To account for this, we also store a vector b containing the lengths of diagonals of "deviated" rectangles of size $$\1 \times 2\$$ up to $$\L \times (L+1)\$$ (instead of squares up to $$\L \times L\$$). It looks like we need to count one additional touch in those cases where the maximal value of b that is still smaller than $$\L\$$ exceeds the analogous value from a.
For example, the first few diagonals are of length:
1x1 2x2 3x3 4x4
1.414214 2.828427 4.242641 5.656854 ...
We can see that after going from $$\L = 3\$$ to $$\L = 4\$$ we have still covered only the 2nd term ($$\2 \times 2\$$ diagonal), it's not enough to reach the edge of the 3rd square, so we are not gaining any more touches. But if we switch to the "deviated" rectangles:
1x2 2x3 3x4 4x5
2.236068 3.605551 5.000000 6.403124 ...
Here $$\L = 4\$$ now encompasses a larger value of 3.60..., which corresponds to ($$\2 \times 3\$$) rectangle. In this orientation we gain an extra touched square compared to the diagonal.
• I tested up to 100 and we get the same results Jul 10 at 21:00
• @LuisMendo Thanks for checking, now when I manually counted the first special case of L = 4, I also feel more confident about it Jul 10 at 21:40
# J, 28 bytes
-1 thanks to Jonah!
The optimal is always either the diagonal L, L with 3+2*L tiles crossed as noted by @Kirill L. or L, L+1, in which case an extra tile is crossed.
((>]+&.*:>:)+3+2*])[<.@%%:@2
Try it online!
• [<.@%%:@2 calculate L by dividing n by sqrt(2), then flooring
• 3+2*] 3+2*L
• ]+&.*:>: calculate length to L, L+1
• > … + if n is larger than this length, add 1
Because the lines have rational length n, and the diagonals irrational length L, we can always find an epsilon $$\L - n > 0\$$ to slide the line top-left to touch the 3 extra tiles, which justifies the 3+2*L.
L, L+2 will always be further away then the next diagonal L+1, L+1, as $$\L^2 + (L+2)^2 = 2L^2 + 4L + 4 > 2L^2 + 4n + 2 = (L+1)^2 + (L+1)^2\$$, so checking L, L+1 is enough.
• Nice J, and nice correctness argument. Jul 10 at 23:52
• ((>]+&.*:>:)+3+2*])]<.@%%:@2 for 28. Jul 10 at 23:55
• 15 using xnor's formula: 3+_2<.@%:@+2**: Jul 11 at 3:46
# Jelly, 18 17 8 bytes
-9 using xnor's mathematical simplification of the same method as the 17, below.
²Ḥ_2ƽ+3
A monadic Link that accepts a positive integer, $$\L\$$ and yields the maximal squares touched.
Try it online! Or see the test-suite.
### How?
²Ḥ_2ƽ+3 - Link: positive integer, L
² - square -> L²
Ḥ - double -> 2L²
_2 - subtract 2 -> 2L²-2
ƽ - integer square-root -> ⌊√(2L²-2)⌋
17:
²H½Ḟð‘Ḥ‘++²ḤƊ‘½<ʋ
Try it online!
### How?
First finds the longest $$\(a,a)\$$ or $$\(a,a+1)\$$ diagonal that $$\L\$$ can cover with some to spare, then places the line along this diagonal, poking out at both ends, and shifts it in the $$\x\$$ direction to cover $$\2a+3+X\$$ squares where $$\X=1\$$ if the diagonal is $$\(a,a+1)\$$, otherwise $$\X=0\$$.
²H½Ḟð‘Ḥ‘++²ḤƊ‘½<ʋ - Link: positive integer, L
² - square -> L²
H - halve -> L²/2
½ - square-root -> side of square with diagonal L
Ḟ - floor -> a
ð - new dyadic chain, f(a,L)...
‘ - increment -> a+1
Ḥ - double -> 2a+2
‘ - increment -> 2a+3
² - square -> a²
+ - (a) add (that) -> a+a²
Ḥ - double -> 2(a+a²)
‘ - increment -> 2(a+a²)+1 = a²+(a+1)²
½ - square-root -> diagonal length of (a,a+1)
< - less than (L)? -> 1 or 0 -> X
# Perl 5, 73 bytes
for$d(0,1){$w=0;1while$w**2+($d+$w++)**2<$_**2;$m+=2*$w-1+$d}$_=int$m/2+1 Try it online! Took @Kirill L.'s explanation and ran with it. Does two passes, without and with "deviation" in $d. The int$m/2+1 outputs the average (rounded up if .5) of those two passes, which will be the max of those two results. Reads the wanted length from stdin and prints max number of squares for that length. The following bash command outputs max number of squares touched for all lengths 1 to 50: for l in {1..50}; do echo$l | perl -pe \
'for$d(0,1){$w=0;1while$w**2+($d+$w++)**2<$_**2;$m+=2*$w-1+$d}$_=int$m/2+1'; echo -n " "; done 3 5 7 8 9 11 12 14 15 17 18 19 21 22 24 25 27 28 29 31 32 34 35 36 38 39 41 42 43 45 46 48 49 51 52 53 55 56 58 59 60 62 63 65 66 68 69 70 72 73 # Japt, 10 bytes Ò2nU²Ñ ¬ÄÄ Try it Ò2nU²Ñ ¬ÄÄ :Implicit input of integer U Ò :Negate the bitwise NOT of (i.e., floor and increment) 2n : Subtract 2 from U² : U squared Ñ : Times 2 ¬ : Square root ÄÄ :Add one twice # JavaScript (Node.js), 20 bytes n=>(2*n*n-2)**.5+3|0 Try it online! Shamelessly copies the idea from xnor's answer # ><>, 27 bytes :*2*2-0v :})?v1+>::*{ ;n+2< Try it online! Xnor's formula, but in a language with neither square root nor rounding operations. Instead, it does the equivalent thing of finding the least square number larger than $$\2L^2 - 2\$$ # Wolfram Language (Mathematica), 20 bytes (14 characters) ⌊√(2#^2-2)⌋+3& Try it online! Shamelessly translating xnor's Python answer. # Java (JDK), 28 bytes L->L+=Math.sqrt(2*L*L-2)+3-L Try it online! Same as everyone, I guess, cheers to xnor! #### Same length as: L->(int)Math.sqrt(2*L*L-2)+3 Try it online! # Retina, 43 29 bytes .+ **2* (^_|\1__)*__+ __$#1*
Try it online! Link is to test suite that generates the results from 1 to the input. Explanation: Now using @xnor's formula.
.+
**2*
Double the square of L and convert it to unary.
(^_|\1__)*__+
__$#1* Subtract 2 and take the integer square root. This is based on @MartinEnder's comment to my Retina answer to It's Hip to be Square but without the leading ^ anchor as the subtraction of 2 guarantees a single match. Add a final 1 and convert to decimal. Previous 43-byte solution based on @KirillL.'s answer: .+ ** ((^(_)|\2__)*)\1(\2_(_))?.+ __$2$3$5
Try it online! Link is to test suite that generates the results from 1 to the input. Explanation:
.+
**
Square L and convert it to unary.
((^(_)|\2__)*)\1(\2_(_))?.+
Find the largest n where 2n²<L², thereby dividing L by √2. Additionally, try to match a further 2n+1 (\2 contains 2n-1), because a rectangle of dimensions n by n+1 has diagonal √(2n²+2n+1). Match at least a further 1 so that the inequality is strict, but also because it guarantees a single match.
__$2$3$5 Calculate 2+2n, plus add another 1 for the rectangular case. Note that $2$3 is used because \2 doesn't actually contain 2n-1 when n=0, but $2\$3 is always 2n.
Add a final 1 and convert to decimal.
# Stax, 7 bytes
é╟φQRl:
Run and debug it
# 05AB1E, 7 bytes
n·Ít3+ï
Port of @xnor's Python answer, so using the same formula:
$$f(L) = \lfloor\sqrt{L^2+2}+3\rfloor$$
Explanation:
n # Square the (implicit) input-integer
· # Double it
Í # Subtract it by 2
t # Take the square-root of that
3+ # Increase it by 3
ï # Truncate/floor it to an integer
# (after which the result is output implicitly)
# MathGolf, 7 bytes
²∞⌡√3+i
Port of @xnor's Python answer, so using the same formula:
$$f(L) = \lfloor\sqrt{L^2+2}+3\rfloor$$
Try it online.
Explanation:
² # Square the (implicit) input-integer
∞ # Double it
⌡ # Subtract it by 2
√ # Take the square-root of that
3+ # Increment it by 3
i # Convert it to an integer
# (after which the entire stack is output implicitly as result) | 2021-09-26T21:50:40 | {
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http://mathhelpforum.com/algebra/174316-complex-number-equation.html | 1. Complex number equation
The question:
Show that $\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\displaystyle z_1$ and $\displaystyle z_2$.
I tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.
2. Originally Posted by Glitch
The question:
Show that $\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\displaystyle z_1$ and $\displaystyle z_2$.
I tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.
The first page of http://www.math.ca/crux/v31/n8/public_page502-503.pdf contains the expansion of $\displaystyle |z_1+z_2|^2$. Similarly expand $\displaystyle |z_1-z_2|^2$ and add them.
3. Originally Posted by Glitch
The question:
Show that $\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\displaystyle z_1$ and $\displaystyle z_2$.
I tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.
Write $\displaystyle \displaystyle z_1$ and $\displaystyle \displaystyle z_2$ in terms of their real and imaginary parts and simplify...
4. Ahh, I forgot about that relation. Thank you.
5. Originally Posted by Glitch
The question:
Show that $\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\displaystyle z_1$ and $\displaystyle z_2$.
Here is a different approach.
Recall $\displaystyle |z|^2=z\cdot\overline{z}$ and $\displaystyle \overline{z\pm w}=\overline{z}\pm\overline{w}$.
So $\displaystyle |z+w|^2=(z+w)\cdot\overline{(z+w)}=(z+w)\cdot(\ove rline{z}+\overline{w})$.
$\displaystyle =z\cdot\overline{z}+z\cdot\overline{w}+w\cdot\over line{z}+w\cdot\overline{w}$
$\displaystyle =|z|^2+z\cdot\overline{w}+w\cdot\overline{z}+|w|^2$
Do that for $\displaystyle |z-w|^2$ and add.
6. Originally Posted by Plato
Here is a different approach.
That's what alexmahone suggested. I ended up solving it that way. | 2018-04-24T17:23:21 | {
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https://math.stackexchange.com/questions/2925298/if-a-fair-die-is-rolled-3-times-what-are-the-odds-of-getting-an-even-number-on | If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?
If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?
I think the permutations formula is needed i.e. $$n!/(n-r)!$$ because order matters but I'm not sure if n is 3 or 6 and what would r be?
Any help would be much appreciated!
Let's calculate the probability, then convert that to odds.
On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $$\dfrac{1}{2}$$.
The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $$\dfrac 1 2$$ of obtaining the desired result. So, we have:
$$P(E,E,O) = \dfrac 1 2 \cdot \dfrac 1 2 \cdot \dfrac 1 2 = \dfrac 1 8$$
Now, the probability of that not happening is $$1-\dfrac 1 8 = \dfrac 7 8$$
So, the odds are 7:1 against the desired outcome.
I think you might be over complicating things.
It has to be even on the first two, the probability of this is $$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$$
The probability of odd on the third roll is also $$\frac{1}{2}$$ so your final probability is $$\frac{1}{8}.$$
• This is simpler. – Snowcrash Sep 21 at 14:44
• ... this is using odds... not probability – Jason Kim Sep 22 at 3:36
• @jason Kim what do you think the distinction is..? – MRobinson Sep 22 at 7:59
• Odds for is (prob)/(1-prob) which demonstrates the difference... – Jason Kim Sep 24 at 0:58
• @JasonKim if the probability is $\frac{1}{8}$ then the odds are $1$ in $8$. If you want to write the odds like a betting shop then you'll have $1:7$. I struggle to see how you believe I have used odds and not probability. – MRobinson Sep 24 at 7:32
Given a fair die, the probability of any defined sequence of three evens or odds is the same, namely, $$1/2 \times 1/2 \times 1/2 = 1/8$$. So the odds are $$7$$ to $$1$$ against.
Assuming the die is just marked even and odd rather than with numbers, there are eight orderings. They range from all odd to all even: OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE. In your formula:
$$\frac{n!}{(n-r)!}$$
You are missing that you don't care about the orders of the duplicates. So you need
$$\frac{n!}{(n-r)!r!}$$
The $$n$$ is the total number of rolls, the $$r$$ is the number of either evens or odds (it's symmetric). But to get the total number of orderings, you need to add these:
$$\sum\limits_{r=0}^n\frac{n!}{(n-r)!r!}$$
Now substitute 3 for $$n$$.
$$\sum\limits_{r=0}^3\frac{3!}{(3-r)!r!}$$
Unrolling that, we get
$$\frac{3!}{3!0!} + \frac{3!}{2!1!} + \frac{3!}{1!2!} + \frac{3!}{0!3!}$$
or
$$1 + 3 + 3 + 1$$
So we have one all odds, three with one even, three with two evens, and one all even. That's eight total.
Another way of thinking of this is that there is only one ordering of all odd numbers or all even numbers while there are three places where the lone odd or even number can be.
Of those eight, how many fit your parameters? Exactly one, OOE. So one in eight or $$\frac{1}{8}$$.
As others have already noted, you could get that much more easily by simply figuring that you have a one in two chance of getting the result you need for each roll. There's three rolls, so $$(\frac{1}{2})^3 = \frac{1}{8}$$.
If you want to treat 1, 3, and 5 as different values and 2, 4, and 6 as different values, you can. But it is much easier to think of them as just odd or even. Because you don't want to try write this out for $$6^3 = 216$$ orderings. And in the end, you will get the same basic result. You will have twenty-seven OOE orderings, which is again one eighth of the total. This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even. And $$\frac{27}{216} = \frac{1}{8}$$.
Permutations leads you down a harder path. It's easier to think just in terms of probability or even ordering.
All of the above or below :-} answers are correct that is 50:50 for each throw so your "formula" is
n (2 -1):1 against where n is the number of throws
• Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Sep 21 at 22:08 | 2018-12-12T17:41:36 | {
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https://math.stackexchange.com/questions/1924291/given-fx-xx-1x-2-x-10-what-is-the-derivative-f0 | Given $f(x)=x(x-1)(x-2)…(x-10)$ what is the derivative $f'(0)$?
$$f: \Bbb R \to \Bbb R; x \mapsto f(x)=x(x-1)(x-2)\cdots(x-10)$$ Evaluate $f'(0)$!
I've tried to set the factors apart, but I only know that $(fg)'=f'g+fg'$. I don't know how I should apply that rule for any $n$ amount of factors. I also thought of actually doing the multiplication, but I don't know what shortcut I should use, and multiplicating one after the other takes extremely long.
• look at the edited text to see how to make your equations looks nice – Surb Sep 12 '16 at 19:47
• Do you want $f^{\prime}(0)$ factorial, or are you just very excited about $f^{\prime}(0)$? – carmichael561 Sep 12 '16 at 19:47
• @Surb thanks, I'll take a look at it! I think Jack gave me the perfect answer below, thanks! – bp99 Sep 12 '16 at 19:52
• Another method: logarithmic differentiation. – GEdgar Sep 12 '16 at 19:53
• 10!! looks like a super awesome number – cronos2 Sep 12 '16 at 20:01
Let $g(x) = x$ and $h(x) = (x-1)\cdots (x-10)$. Then $f'(x) = g(x) h'(x) + g'(x)h(x)$. Since $g(0)=0$ and $g'(0)=1$, we have $$f'(0) = h(0) = (-1)(-2)\cdots (-10) = 10!$$
• Your answer was the most intuitive for me, thank you! It's the simplest IMO and I've only begun calculus :) – bp99 Sep 12 '16 at 20:19
Apply the definition: $$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}= \lim_{x\to0}\,(x-1)(x-2)\dots(x-10)=10!$$ More generally, if $f(x)=x(x-1)\dots(x-n)$, the same argument shows $$f'(0)=(-1)^n\cdot n!$$
• Interestingly, the definition of the derivative is useful at times! – Wojowu Sep 12 '16 at 20:07
• @Wojowu Who might have suspected it? ;-) – egreg Sep 12 '16 at 20:12
Hint: A polynomial is always an entire function, and in a neighbourhood of the origin: $$x(x-1)\cdot\ldots\cdot(x-10) = 10!\,x+O(x^2)$$ hence $$\frac{d}{dx}\left. x(x-1)\cdot\ldots\cdot(x-10)\right|_{x=0} = \color{red}{10!}.$$
Welcome to Math SE. To format your question you can use LaTeX.
Now coming to your question, the derivative of a product is just the sum of the $n$ products where only one of the members is differentiated. In your case
$((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$
Note that all but the first expression will evaluate to $0$ at $x=0$ since you're multiplying by 0 ($x$) so $f'(0)=(-1)(-2)...(-10)=10!$
• Thanks for your answer! However, I don't fully understand this part, unfortunately: $((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$ How does this work exactly? – bp99 Sep 12 '16 at 20:06
• @bertalanp99 Try it by hand with just three terms. Differentiate $(x-1)(x-2)(x-3)$ using the product rule (twice). The form should pop out. Your question just adds more terms in the product. – John Sep 12 '16 at 20:14
There is an $n$-term version of the product rule that is definitely worth knowing about. You'll see the pattern from the $n = 3$ case. If $f(x) = f_1(x) f_2(x) f_3(x)$, then $$f'(x) = f_1'(x) f_2(x) f_3(x) + f_1(x) f_2'(x) f_3(x) + f_1(x) f_2(x) f_3'(x).$$ (Do you see what the formula would be for a product of four functions?)
In your case, $f'(x)$ is a sum of $11$ terms, and all but one of those terms vanish when you plug in $x = 0$.
Richard Feynman made a big deal about the usefulness of this $n$-term product rule in The Feynman Tips on Physics.
More generally, if $f(x) =\prod_{k=1}^n (x-a_k)^{b_k}$, then $\ln f(x) =\sum_{k=1}^n b_k \ln(x-a_k)$.
Differentiating, $\dfrac{f'(x)}{f(x)} =\sum_{k=1}^n \dfrac{b_k}{x-a_k}$, so $f'(x) =f(x)\sum_{k=1}^n \dfrac{b_k}{x-a_k}$.
Setting $x=0$,
$\begin{array}\\ f'(0) &=f(0)\sum_{k=1}^n \dfrac{b_k}{-a_k}\\ &=\prod_{j=1}^n (-a_j)^{b_j}\sum_{k=1}^n \dfrac{b_k}{-a_k}\\ &=\sum_{k=1}^n b_k(-a_k)^{b_k-1}\prod_{j=1, j \ne k}^n (-a_j)^{b_j}\\ \end{array}$
If $a_k = k-1$ and $b_k = 1$ as in your case,
$\begin{array}\\ f'(0) &=\sum_{k=1}^n \prod_{j=1, j \ne k}^n (-j+1)\\ &=(-1)^{n-1}\sum_{k=1}^n \prod_{j=1, j \ne k}^n (j-1)\\ &=(-1)^{n-1}\left(\prod_{j=2}^n (j-1)+\sum_{k=2}^n \prod_{j=1, j \ne k}^n (j-1)\right)\\ &=(-1)^{n-1}(n-1)! \qquad\text{since all terms with }k\ge 2 \text{ have j=1 so are zero}\\ \end{array}$
• Why would anyone downvote this answer must be a deep mistery. It is correct, it gives the correct answer in the particular case asked by the OP and also, for whoever is interested, it gives a nice though slightly abstract development to deal with these things. And anyone aspiring to cope with mathematics at any level above $\;2\cdot2=4\;$ cannot get scared of a little abstraction. +1 – DonAntonio Sep 12 '16 at 20:28
• @DonAntonio I'm not the downvoter. The answer glosses over two facts: the first is that taking the logarithm is not really possible (but it's a formal method, so nothing really bad if we know what we're doing). However, the computation is done assuming $a_k\ne0$; it can easily be adjusted, though, with doing a limit or some simpler trick. – egreg Sep 12 '16 at 20:52
• @egreg You're right, indeed. +1 – DonAntonio Sep 12 '16 at 20:55
• In my answer, I separated out the case $a_k = 0$ in the first term; this caused all the other terms to vanish at $x = 0$. – marty cohen Sep 12 '16 at 21:42 | 2020-10-29T05:23:43 | {
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https://www.themathdoctors.org/the-case-of-the-disappearing-derivative/ | # The Case of the Disappearing Derivative
#### (A new question of the week)
An interesting question we received in mid-January concerned two implicit derivative problems with an unusual feature: the derivative we are seeking disappears! How do you track down such elusive quarry? Each case is a little different.
## Problem 1: The danger in division
Amia asked two questions, which I will separate to make the discussions easier to follow. Here is the first:
Hi Dr Math,
I have a question about finding the value of the derivative, I need your help.
Thank you.
This is an example of implicit differentiation, in which a relation (which we want to think of as $$y = f(x)$$ though it is typically not a function) has been defined implicitly as representing any ordered pair $$(x, y)$$ that satisfies the equation $$\tan(xy)=xy$$. He has differentiated both sides of the equation with respect to x, following the chain rule and the product rule, and obtained $$\sec^2(xy)(xy’+y)=xy’+y$$, which he wants to solve for $$y’$$, which will be a function of x and y. But something odd has happened: $$y’$$ has disappeared entirely, so there is nothing to solve for! Was the division that led to this situation invalid? Is the resulting equation valid?
In dividing by xy’ + y, you are assuming that xy’ + y is non-zero; and in the process you are losing the derivative completely! In effect, your answer seems to say that if the derivative can be determined at all, then sec2(xy) must equal 1. That’s the opposite of what a solution should say, which would be that if something is true, then the derivative has some particular value.
Instead, I would either distribute or factor the equation you got and isolate y’ as usual; when you do that, you will get an implicit derivative on the condition that sec2(xy) is not 1.
In general, when you are tempted to divide by a variable while solving, the better thing to do is to factor, which retains all relevant information rather than silently assuming something you may not want to assume.
After solving both problems, it will be interesting to try graphing the equations; I find that Desmos (almost) graphs both nicely, and the result is interesting to compare with what you find about the equations. But first try graphing by hand, at least to the extent of thinking about what kind of curve each must be. This will be worth discussing!
The warning against division is standard in algebra; an example I often use is that in solving $$x^2=4x$$ if we divided by x, we would get $$x = 4$$ and miss the solution $$x=0$$; whereas if we rearrange and factor, we get $$x^2-4x=0\\x(x-4)=0\\x=0\text{ or }x=4$$ which is correct. The error was in implicitly assuming that the x we divide by is non-zero, when in fact zero is one of our solutions.
Similarly, in our problem, it will turn out that the $$(xy’+y)$$ that he divided by, when it is zero, leads to the answer we are looking for.
### Getting the derivative
Amia first showed me the graphs, which I’ll hold for later, and then showed the work I’d described:
There’s a little slip in the last line; it should say $$y’=\frac{y(1-\sec^2(xy))}{x(\sec^2(xy)-1)}=-\frac{y}{x}$$
This is good, though I would add that this is valid as long as sec2(xy) ≠ 1. This turns out to have no effect.
This is because the cancellation in that last line is valid only if $$\sec^2(xy)-1$$ is not zero; if it is, then the intermediate form is just 0/0 and can’t be evaluated. In that case, we have to back up a line and observe that the equation is $$0=0$$, which tells us nothing. And backing up still more, if we replace $$\sec^2(xy)$$ in the second line of the work with 1, we already get a tautology (an equation that is always true). There is no way to determine $$y’$$ in that case.
Now, why did I say that this restriction will not affect the final result? Because if $$\sec^2(xy)=1$$, then $$xy=n\pi$$ for some integer n; but then $$\tan(xy)=0$$, and our curve is $$xy=0$$, which is just the x– and y-axes. Hold that thought while we explore the graph.
Now, we’ve found that the derivative is $$y’=\frac{-y}{x}$$.
### Exploring the graph
Do you know another function that has the same derivative (that is, that satisfies this differential equation)? It’s y = k/x, a family of hyperbolas — which is just what the graph you showed looks like (for some specific values of k):
(Desmos does a remarkable job graphing such a tricky equation, but it struggles, and you have to imagine those dotted lines being filled in to make complete curves.)
This kind of hyperbola, called rectangular hyperbolas because of the right angle between the asymptotes (the axes), has the equation $$xy=k$$, and differentiating that implicitly we get $$y+xy’=0$$, so that $$y’=-\frac{y}{x}$$, the same as our curve. All we need to know is, what values of k are valid?
I’d suggested that we would learn a lot by trying to work out the graph ourselves rather than just plugging the equation into a website and seeing what it looks like all at once. We’ve just taken the first step, by observing that it is some set of hyperbolas, based on the derivative. We can continue by finding k:
Now supposed we wanted to graph the function by hand. One way to analyze it would be to let u = xy, so the equation is tan(u) = u. This has as its solution certain discrete values of u:
Here I plotted $$y=\tan(u)$$ in red, and $$y=u$$ in blue, so that the intersections (gray dots) are the solutions. We can’t solve this analytically and exactly, but the graph tells us that they are about $$u=0,\pm 4.493, \pm 7.725, \dots$$.
And our function therefore is equivalent to xy = k for those values of k, which is exactly the family of hyperbolas we’ve seen. Here I have overlaid two particular hyperbolas, namely xy = 4.493 (red) and xy = -7.725 (green)
That is very satisfying.
Now, if you wish, you can think about when sec2(xy) = 1.
This, as we saw above, corresponds to the x– and y-axes (the case $$k=0$$), and no other points on this graph. At those points, the derivative is 0 (when $$y=0$$, and our formula for the derivative is correct) or undefined (when $$x=0$$, and our formula for the derivative is effectively correct, yielding “infinity”). So the answer to the problem is, in fact, $$y’=\frac{-y}{x}$$
## Problem 2: Getting specific too soon
Here is the second question, which we actually discussed interleaved with the first:
We can tell from the form of the equation that this is a conic section, and could further analyze it. But our goal is to find the derivative at a specific point on the curve. (This time we can actually solve for y and find these points given only x, which we couldn’t do in the first problem.) It should be observed that we would normally expect to find two points (if any), since we solved a quadratic equation; so this is a special point, and that alone might suggest what to expect.
But Amia has correctly differentiated and plugged in the given point, but again found that $$y’$$ disappeared from the equation. All we have left is a false equation, $$4=0$$. Does that mean “no solution”?Would it mean something different if you got a true equation like $$4=4$$?
(By the way, I have to admit I initially misread that as “$$y=0$$”, which would have meant something interestingly different. But Amia is consistent in distinguishing y from 4. Clear writing is important in math, but because people write differently, they can misread your writing even when you do everything right …)
To this, I answered:
You have correctly found that when x=1, y=-1; then you implicitly differentiated and substituted values before solving for y’, finding that y’ is eliminated, similarly to the other problem. This time you found that an inconsistent equation results. We can’t conclude (yet) that y’ doesn’t exist; but it does suggest that x=1 will be a special case of some sort.
Again, I would isolate y’ before substituting, which will result in something that is interesting. See what you find.
### Getting the derivative
Amia showed the graph, showing the specialness of this point, and then showed the new work I had suggested:
I see that the curve has a vertical tangent at the point (-1,1).
By solving for $$y’$$ in the general case, we get a better sense of the behavior of the graph, and see how things fail at the special point. In general, $$y’=-\frac{3x+y}{x+y}$$ This is undefined when $$x+y=0$$, which includes our special point $$(1,-1)$$ and its opposite, $$(-1,1)$$. But for other points nearby, the slope is defined, and we will be dividing by a very small number, making a steep slope, fitting our expectations for a vertical tangent. In particular, if x and y change just a little, we end up with something near -2 on top but a very small number, either positive or negative, on the bottom, resulting in a respectively negative or positive, near-infinite slope.
(But be careful – we can’t just think about the limit as we approach the point, because we have to approach along the curve, not just from any direction. Moreover, it turns out, though we can’t see it in the equations, that x can’t be greater than 1! There are many details about working with curves that we didn’t, and won’t here, touch on!)
### Exploring the graph
Again, this is good; in particular, we see that the requested derivative fails to exist in such a way that it indicates a vertical slope (approaching +infinity from one side and -infinity from the other). And that, again, agrees with the graph, as you indicated:
This is a rotated ellipse; here are its axes:
If in your initial work you had been given a different value of x (an interesting one is √(2)/2, which gives the vertices and covertices), then when you solved for y and plugged the values into the derivative, everything would have worked correctly. But this example shows that in general it is better to keep the variables unknown until the end.
These were both very interesting!
To talk more about the rotated ellipse, such as how I found the tilted axes, would take us too far afield. But if this made you curious, here is a thorough textbook explanation from Libretexts.
But let’s do what I suggested, and try taking $$x=\frac{\sqrt{2}}{2}$$. First, we find y: $$3x^2+2xy+y^2=2\\ 3\left(\frac{\sqrt{2}}{2}\right)^2+2\left(\frac{\sqrt{2}}{2}\right)y+y^2=2\\ \frac{3}{2}+\sqrt{2}y+y^2=2\\ y^2+\sqrt{2}y-\frac{1}{2}=0\\ 2y^2+2\sqrt{2}y-1=0\\ y=\frac{-2\sqrt{2}\pm\sqrt{16}}{4}=\pm 1-\frac{\sqrt{2}}{2}$$ So the two points we get are $$(0.707,-1.707)$$ and $$(0.707,0.293)$$.
Now, using the formula we got for the derivative, we get $$y’=\frac{-(3x+y)}{x+y}\\ =\frac{-(3\frac{\sqrt{2}}{2}\pm 1-\frac{\sqrt{2}}{2})}{\frac{\sqrt{2}}{2}\pm 1-\frac{\sqrt{2}}{2}}\\ =\frac{-(\sqrt{2}\pm 1)}{\pm 1}=-\sqrt{2}-1,\sqrt{2}-1$$ These are the same as the slopes of the axes, $$y=\left(-1\pm\sqrt{2}\right)x$$.
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Solution page to learn more, see our tips on writing great answers sides with lengths of 10 and..., AD =p and ∠ BAD be an acute angle coaching to you. Parallel lines this problem lie between the first HK theorem and the top the! Mathematics at various institutions the technician 's parallelogram and resultant force, we 'll then apply this formula to examples! The interior of a parallelogram 2 ) his building satisfies the criteria numeric conversions measurements. Opinion ; back them up with the naked eye from Neptune when and. The box and still be able to close it is base times the base '' the!, multiply the base of that side of the eraser where b the. Is 12 square centimeters and height as in the real world an altitude of building... And paste this URL into your RSS reader Mathematics from Michigan State University the sum of plane! | 2022-06-25T02:01:21 | {
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https://math.stackexchange.com/questions/1992277/preorders-vs-partial-orders-clarification | # Preorders vs partial orders - Clarification
• A binary relation is a preorder if it is reflexive and transitive.
• A binary relation is a partial order if it is reflexive, transitive and antisymmetric.
Does that mean that all binary relations that are a preorder are also automatically a partial order as well?
In other words is a binary relation a preorder if its only reflexive and transitive and nothing else?
Thanks for your help.
• A partial order is a preorder that is also antysymmetric. Oct 30, 2016 at 23:24
You have it backwards - every partial order is a preorder, but there are preorders that are not partial orders (any non-antisymmetric preorder).
For example, the relation $\{(a,a), (a, b),(b,a), (b,b)\}$ is a preorder on $\{a, b\}$, but is not a partial order.
• Ah yes, my bad! Thank you. Another thing- what if I have a binary relation that is reflexive, transitive and symmetric. Would this binary relation be considered a preorder? Oct 30, 2016 at 23:08
• @MarkJ Yes, that would be a preorder. Saying that every preorder is reflexive and transitive does not mean that those are the only properties that a preorder can have. Oct 30, 2016 at 23:11
• This is exactly what I was looking for. Thank you for your help! Oct 30, 2016 at 23:13
• @MarkJ It seems you are satisfied with the answer and further comments. Then why didn't you accept the answer? It's just a click to acknowledge the time that Noah Schweber took to clarify your confusion :) Oct 31, 2016 at 9:55
order relations are subset of pre order relations. For instance a relation kind of "prefer or indiferent" is reflexive and transitive, but is not antisymetric, so this is an example of pre order but not order. A relation like "bigger or equal" is reflexive, transitive and also antisymetric, so this relation is pre order (since it is reflexive and transitive) but also order.
A pre-order $$a\lesssim b$$ is a binary relation, on a set $$S,$$ that is reflexive and transitive. That is $$\lesssim$$ satisfies (i) $$\lesssim$$ is reflexive, i.e., $$a\lesssim a$$ for all and (ii) $$\lesssim$$ is transitive, i.e., $$a\lesssim b$$ and $$b\lesssim c$$ implies $$a\lesssim c,$$ for all $$% a,b,c\in S.$$ (A pre-ordered set may have some other properties, but these are the main requirements.)
On the other hand a partial order $$a\leq b$$ is a binary relation on a set $$S$$ that demands $$S$$ to have three properties: (i) $$\leq$$ is reflexive, i.e., $$% a\leq a$$ for all $$a\in S$$, (ii) $$\leq$$ is transitive, i.e., $$a\leq b$$ and $$% b\leq c$$ implies $$a\leq c,$$ for all $$a,b,c\in S$$ and (iii) $$\leq$$ is antisymmetric, i.e., $$a\leq b$$ and $$b\leq a$$ implies $$a=b$$ for all $$a,b\in S$$% .
So, as the definitions go, a partial order is a pre-order with an extra condition. This extra condition is not cosmetic, it is a distinguishing property. To see this let's take a simple example. Let's note that $$a$$ divides $$b$$ (or $$a|b)$$ is a binary relation on the set $$Z\backslash \{0\}$$ of nonzero integers. Here, of course, $$a|b$$ $$\Leftrightarrow$$ there is a $$% c\in Z$$ such that $$b=ac.$$
Now let's check: (i) $$a|a$$ for all $$a\in Z\backslash \{0\}$$ and (ii) $$a|b$$ and $$b|c$$ we have $$a|c.$$ So $$a|b$$ is a pre-order on $$Z\backslash \{0\},$$ but it's not a partial order. For, in $$Z\backslash \{0\},$$ $$a|b$$ and $$b|a$$ can only give you the conclusion that $$a=\pm b,$$ which is obviously not the same as $$a=b.$$
The above example shows the problem with the pre-ordered set $$,$$\lesssim >.$$ It can allow $$a\lesssim b$$ and $$b\lesssim a$$ with a straight face, without giving you the equality. Now a pre-order cannot be made into a partial order on a set $$, $$\lesssim >$$ unless it is a partial order, but it can induce a partial order on a modified form of $$S.$$ Here's how. Take the bull by the horn and define a relation $$\sim ,$$ on $$$$ by saying that $$a\sim b$$ $$\Leftrightarrow a\lesssim b$$ and $$b\lesssim a$$. It is easy to see that $$\sim$$ is an equivalence relation. Now splitting $$S$$ into the set of classes $$\{[a]|$$ $$a\in S\}$$ where $$[a]=\{x\in S|$$ $$x\sim a\}.$$ This modified form of $$S$$ is often represented by $$S/\sim .$$ Now of course $$% [a]\leq \lbrack b]$$ if $$a\lesssim b$$ but it is not the case that $$b\lesssim a.$$ Setting $$[a]=[b]$$ if $$a\sim b$$ (i.e. if $$a\lesssim$$ and $$b\lesssim a).$$
In the example of $$Z\backslash \{0\}$$ we have $$Z\backslash \{0\}/\sim$$ $$% =\{|a|$$ $$|$$ $$a\in Z\backslash \{0\}\}.$$
(Oh and as a parting note an equivalence relation is a pre-order too, with the extra requirement that $$a\lesssim b$$ implies $$b\lesssim a.)$$ | 2022-06-30T20:54:48 | {
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https://math.stackexchange.com/questions/2293406/meaning-of-times-in-mathbbrm-times-mathbbrn-rightarrow-mathbbrm-tim | # Meaning of times in $\mathbb{R}^m\times\mathbb{R}^n\rightarrow \mathbb{R}^{m\times n}$?
I use $\mathbf{a} \times\mathbf{b}$ for the cross product, $\mathbf{a}\cdot \mathbf{b}$ for the dot product and $ab$ for normal multiplication ($a,b$ are scalars).
However, what is the meaning of times in $\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}^2$?
Or $\mathbb{R}^m\times\mathbb{R}^n\rightarrow \mathbb{R}^{m\times n}?$
Is it the cross product?
Is it the dot product?
Is it normal multiplication?
Update:
Does these have any meaning
$\mathbb{R}^m\cdot\mathbb{R}^n\rightarrow \mathbb{R}^{m\cdot n}$ (the dot product)?
$\mathbb{R}^m \mathbb{R}^n\rightarrow \mathbb{R}^{m n}$ (normal multiplication)?
• It is called the Cartesian product! – Nigel Overmars May 23 '17 at 13:49
It is a cartesian product. If $A$ and $B$ are two sets, then $A\times B$ is by definition the set of couples $(a,b)$ with $a$ in $A$ and $b$ in $B$.
In your case: $$\mathbb{R}^m\times\mathbb{R}^n=\{(x,y);x\in\mathbb{R}^m,y\in\mathbb{R}^n\}.$$
• Adding onto this for the range space $\mathbb{R}^{m \times n}$ is usually meant to denote real $m \times n$ matrices. – Dragonite May 23 '17 at 13:51
• Notice that in fact $\mathbb{R}^m \times \mathbb{R}^n \cong \mathbb{R}^{m+n}$. You always need $m+n$ numbers to identify an element of either side. For example, $\mathbb{R} \times \mathbb{R} \cong \mathbb{R}^2$. – Sharkos May 23 '17 at 13:55
• Yes, OP be weary that there is a difference between $\mathbb{R}^{m+n}$ and $\mathbb{R}^{m\times n}$ that has not really been addressed. – Dragonite May 23 '17 at 13:56
• $$\mathbb{R^2} := \mathbb{R} \times \mathbb{R}$$ by definition. So unless you mean that this is the trivial isomorphism, defined by the identity function, this wrong. – user370967 May 23 '17 at 15:29
• @Sharkos Is it also true for complex numbers, i.e. $\mathbb{C}^m \times \mathbb{C}^n = \mathbb{C}^{m+n}$? – JDoeDoe Nov 25 '17 at 18:52
I think, more generally, you need guidance on the notation $f\colon A \times B \to C$, which is the notation for "a function called $f$, from the Cartesian product $A \times B$ of sets $A$ and $B$, to the set $C$" (the Cartesian product $A \times B = \{(a, b) : a \in A, b \in B\}$ is the set of all ordered pairs with things taken from $A$ and $B$). More generally, the format is
$$\text{function name} : \text{domain} \to \text{codomain}$$
But this notation $f \colon A \times B \to C$ often says nothing about what the function actually does to pairs $(a, b) \in A \times B$ to produce some $f(a, b) \in C$, unless $f$ happens to have a particularly descriptive name/symbol. In this case, you'll often see functions introduced in "two parts",
\begin{align*} f \colon A \times B &\to C \\ (a, b) &\mapsto \text{however $a, b$ determine $f(a, b)$} \end{align*}
where the first line specifies the function name and all the sets we need, and the second line actually tells us what $f$ does to the pairs $(a, b)$ (and note the new symbol $\mapsto$, which is used like $\to$ above. But $\to$ is used with sets, the domain and codomain, while $\mapsto$ is used between the actual input and output, to explain what happens to elements in the sets).
So you'll never see notation like $\Bbb R^m \cdot \Bbb R^n \to \Bbb R^{m + n}$, with the function placed between sets. Instead, you'll see the function name/notation in the place of $f$, put before the domain. So things like
$$\cdot \colon \Bbb R^n \times \Bbb R^n \to \Bbb R$$ is a (mildly confusing) notation saying that there's a function called "$\cdot$" that takes two vectors in $\Bbb R^n$, and gives you back a real number (we can assume it's the standard dot product).
$$\times \colon \Bbb R^3 \times \Bbb R^3 \to \Bbb R^3$$ would be the (somehow more confusing) notation to say there's a function called "$\times$" that takes two vectors in $\Bbb R^3$ and returns another vector in $\Bbb R^3$; probably it's the standard cross product on $\Bbb R^3$.
For a slightly-less-weird-looking example, we might use
$$+ \colon \Bbb R^n \times \Bbb R^n \to \Bbb R^n$$ to say that we have an operation called "$+$" that takes pairs of vectors in $\Bbb R^n$, and returns a single vector in $\Bbb R^n$ (and unless it's stated otherwise, everyone would assume "$+$" means exactly what you think it means).
The domain and codomain can come in all sorts of varieties. For example, functions don't have to be defined on pairs of things, in which case our domain isn't going to be a Cartesian product. So to talk about the standard square root function, we might write
$$\sqrt{\ }\; \colon \Bbb R_{\ge 0} \to \Bbb R_{\ge 0}.$$
Or maybe we're handed a function with a fairly cryptic name,
\begin{align*} \operatorname{ev} \colon M_{n \times n}(\Bbb R) \times \Bbb R^n &\to \Bbb R^n \\ (A, \vec{v}) &\mapsto A\vec{v} \end{align*}
but with practice, we can see it's the evaluation map that takes an $n \times n$ matrix with real entries and a vector in $\Bbb R^n$, and applies $A$ to $\vec{v}$.
• Note that many of the examples had the format $\Bbb R^n \times \Bbb R^n \to \Bbb R^n$; only the prefix, the function name before $\colon$, could help us tell the functions apart. The part of the notation after the $\colon$ is really only good for domain and codomain, not the function itself (e.g., vector addition versus cross product). – pjs36 May 23 '17 at 16:00
$\Bbb R \times \Bbb R$ denotes the Cartesian product. An element of $\Bbb R \times \Bbb R$ has the form $(a,b)$, where $a$ and $b$ are both in $\Bbb R$.
Basically, $\Bbb R \times \Bbb R$ is just a longer way of saying $\Bbb R^2$.
The same symbol in different contexts gets different meanings. Math in this aspect is somehow like poetry; the same word gets different meanings in different contexts.
The symbol "$\times$", applied to sets like $\mathbb{R}$, denotes the Cartesian product. If $X,Y \neq \varnothing$, then $X \times Y := \{ (x,y) \mid x \in X, y \in Y \}$. If $X, Y := \mathbb{R}$, then $X \times Y$ by definition is simply the usual Cartesian plane. It is defined that $X^{n} := \{ (x_{1},\dots,x_{n}) \mid x_{1},\dots,x_{n} \in X \}$. Now you know what the superscript of $\mathbb{R}$ means.
Note that $\mathbb{R}^{m} \times \mathbb{R}^{n} = \mathbb{R}^{m+n}$.
• Thanks! You say sets, is it also true for vectors? For matrices we can say we have a $m\times n$-matrix, does $\times$ here also mean the cartesian product? – JDoeDoe May 23 '17 at 14:32
• No problem. No, the point is to what objects the symbol "$\times$" is along with. If $a,b$ are vectors, then $a \times b$ means the cross product of $a$ and $b$; if $a,b$ are numbers, then $a\times b$ means either the arithmetic product or the size of a matrix; if $a,b$ are sets, then $a \times b$ means their Cartesian product. These cover your question? – Benicio May 23 '17 at 14:44
None of the above on the left. That $\times$ is what appears between two sets to denote their cartesian product - the set of all ordered pairs whose first (second) element is from the first (second) set.
The $\times$ on the right is ordinary multiplication.
Edit:
The vector space on the left has dimension $m+n$, the one on the right has dimension $mn$, so the arrow can't represent an isomorphism. It might an injection. One possibility is that you're thinking of $\mathbb{R}^{m\times n}$ as the space of $m \times n$ matrices. Then the arrow could mean the function $f$ given by $$f(v,w)_{ij} = v_iv_j .$$ (ordinary multiplication of real numbers onthe right). This map isn't an injection since $f(0,w) = 0$ for every $w$.
The vector space on the left is naturally isomorphic to $\mathbb{R}^{m+n}$. The arrow in $$\mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^{m+n}$$ might then stand for that natural isomorphism. Perhaps that's what you meant to ask about. That's what your example when $m=n=1$ suggests.
• Isn't the $\times$ on the right addition, $m+n$? – JDoeDoe May 23 '17 at 14:16
• @JDoeDoe See my edits. – Ethan Bolker May 23 '17 at 15:13
To answer your first question $X \times Y$ is the Cartesian product of sets $X$ and $Y$. This is all ordered pairs $(x,y)$ where $x$ is a member of the set $X$, and $y$ is a member of the set $Y$.
$$X \times Y = \{(x,y):x\in X,y \in Y\}$$
An example is $\mathbb R \times \mathbb R^2$ which is all ordered pairs $(x,(y,z))$ where $x \in \mathbb R$ and $(y,z) \in \mathbb R^2$. Of course this can be identified with $\mathbb R^3$ by the bijection $(x,(y,z)) \mapsto (x,y,z)$.
A more interesting example is $[0,1] \times S^1$ which is all ordered pairs $(t,\theta)$ there $t \in [0,1]$, i.e. $0 \le t \le 1$ and $\theta$ is a point in the circle. This product $[0,1] \times S^1$ gives a cylinder.
The word product is unfortunate, and has nothing to do with multiplication. The torus - which looks like a bicycle inner tube - is the Cartesian product of two circles: $T = S^1 \times S^1$.
Your idea of the dot product doesn't work for two reasons. Recall that the dot product takes two vectors (of the same dimension) and gives a number. Instead of it being a set, it is a mapping between sets ($\mathbb R^n \times \mathbb R^n \to \mathbb R$). It doesn't make sense to dot a vector from $\mathbb R^2$ with a vector from $\mathbb R^3$; the dimensions are wrong. Think about matrices: it only makes sense to multiply a $p$-by-$q$ matrix with a $q$-by-$r$ matrix to get a $p$-by-$r$ matrix.
Similar remarks can be made about your idea of normal multiplication $\mathbb R^m \mathbb R^n$. Take an example: Take $(1,2,3) \in \mathbb R^3$ and $(1,2)\in \mathbb R^2$. How do we get something in $\mathbb R^6$? (In a natural way that has meaning?) | 2019-06-16T03:23:21 | {
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https://math.stackexchange.com/questions/3350867/have-i-been-using-differential-forms-without-knowing-it | # Have I been using differential forms without knowing it?
I'm self-learning differential forms. I've been happily integrating 1-forms over parameterised curves, and 2-forms over parameterised surfaces, both in $$\mathbb{R}^{3}$$. Now I've just found out that integrating an n-form $$\omega=f\left(x_{1},\ldots,x_{n}\right)dx_{1}\wedge dx_{2}\cdots\wedge dx_{n}$$ over an n-dimensional manifold M in $$\mathbb{R}^{n}$$ is defined by$$\intop_{M}\omega=\pm\intop_{M}f\left(x_{1},\ldots,x_{n}\right)dx_{1}\cdots dx_{n}.$$
Am I correct in thinking that this definition describes what's going on with an ordinary calculus definite integral$$\int_{b}^{a}f\left(x\right)dx.$$So $$f\left(x\right)dx$$ would be a 1-form and the one-dimensional manifold it is integrated over is the interval $$\left(a,b\right)$$?
• That is exactly correct. – Lee Mosher Sep 10 at 15:42
• To be extra careful, manifolds should be oriented for purposes of integrating forms, so you should specify the orientation on $(a,b)$ as being induced by restriction from the "basic" orientation on $\mathbb R$. Reversing that orientation means integrating backwards from $b$ to $a$, which changes the sign. – Lee Mosher Sep 10 at 15:44
• Actually, to be extra extra careful, the "ordinary calculus definite integral" is ambiguous. It can be interpreted either as (1) the integral of a density on an unoriented smooth manifold with boundary, namely the interval $[a,b]$, in which case you don't need an orientation; or as (2) the integral of a form on an oriented smooth manifold, namely the oriented interval $[a,b]$ with the standard orientation. We can integrate functions even on unoriented smooth manifolds, because of densities. And indeed the whole point of an orientation here is to convert a form into a density. – symplectomorphic Sep 10 at 19:00
• @symplectomorphic - Tau, in “Differential Forms and Integration”, distinguishes between the “unsigned definite integral $\int_{\left[a,b\right]}f\left(x\right)dx$ (which one would use to find area under a curve, or the mass of a one-dimensional object of varying density), and the signed definite integral $\int_{a}^{b}f\left(x\right)dx$ (which one would use for instance to compute the work required to move a particle from $a$ to $b$).” Is that what you mean? Thanks – Peter4075 Sep 11 at 7:03
• Sorry, that should be Tao not Tau. – Peter4075 Sep 11 at 8:09
Moreover, you can think of 0-forms which are just scalars. Then generalized Stokes' theorem $$\int_{d\Omega} \omega=\int_\Omega d\omega$$ in case of 0-form $$\omega$$ (and 1-form $$d\omega$$) becomes the fundamental theorem of Calculus: $$\left.F(x)\right|_a^b =\int_a^b\frac{dF}{dx}dx$$ | 2019-11-19T17:59:09 | {
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https://bik-f.euni.de/ukb8j9g/how-to-find-irrational-numbers-between-decimals-fbfb78 | Irrational numbers are those which do not terminate and have no repeating pattern. Rational numbers are whole numbers, fractions, and decimals - the numbers we use in our daily lives. Irrational numbers include √2, π, e, and θ. The vast majority are irrational numbers, never-ending decimals that cannot be written as fractions. Hence, almost all real numbers are irrational. Subtract the rounded numbers to obtain the estimated difference of 0.5; The actual difference of 0.988 - 0.53 is 0.458; Some uses of rounding are: Checking to see if you have enough money to buy what you want. Difference between rational and irrational numbers has been clearly explained in the picture given below. From the below figure, we can see the irrational number is √2. . They include the counting numbers and all other numbers that can be written as fractions. Difference between Rational and Irrational Numbers. In summary, this is a basic overview of the number classification system, as you move to advanced math, you will encounter complex numbers. Many people are surprised to know that a repeating decimal is a rational number. Suppose we have two rational numbers a and b, then the irrational numbers between those two will be, √ab. The decimal expansion of an irrational number continues without repeating. This amenability to being written down makes rational numbers the ones we know best. Learn Math Tutorials 497,805 views Yes. one irrational number between 2 and 3 . 10 How to Write Fractions Between Two Decimal Numbers. Practice Problems 1. 9 years ago. Key Differences Between Rational and Irrational Numbers. m/n —> 3/n N has to be an integer, n cannot be 0. Irrational Numbers on a Number Line. ⅔ is an example of rational numbers whereas √2 is an irrational number. The major difference between rational and irrational numbers is that all the perfect squares, terminating decimals and repeating decimals are rational numbers. So 1/3 is between zero and one. What is the Difference Between Rational and Irrational Numbers , Intermediate Algebra , Lesson 12 - Duration: 3:03. Are there any decimals that do not stop or repeat? Method of finding a number lying exactly midway between 2 given numbers, is add those 2 numbers & divide by 2. Make use of this online rational or irrational number calculator to ensure the rationality and find its value. Irrational numbers tend to have endless non-repeating digits after the decimal point. Irrational numbers' decimal representation is non terminating , non repeating.. The only candidates here are the irrational roots. Lv 6. But there's at least one, so that gives you an idea that you can't really say that there are fewer irrational numbers than rational numbers. Before studying the irrational numbers, let us define the rational numbers. You can use the decimal module for arbitrary precision numbers: import decimal d2 = decimal.Decimal(2) # Add a context with an arbitrary precision of 100 dot100 = decimal.Context(prec=100) print d2.sqrt(dot100) If you need the same kind of ability coupled to speed, there are some other options: [gmpy], 2, cdecimal. Examples of Rational and Irrational Numbers For Rational. A rational number is a number that can be written as a ratio. To study irrational numbers one has to first understand what are rational numbers. How many irrational numbers can exist between two rational numbers ? Solution: If a and b are two positive numbers such that ab is not a perfect square then : i ) A rational number between and . 0 0. cryptogramcorner. Irrational Number between Two Rational Numbers. Apart from these number systems we have Irrational Numbers. 2 and 3 are rational numbers and is not a perfect square. It makes no sense!!! Terminating, recurring and irrational decimals. Step 3: Place the repeating digit(s) to the left of the decimal point. Take this example: √8= 2.828. To find the rational number between two decimals, we can simply look for the average of the two decimals. The ability to convert between fractions and decimals, and to approximate irrational numbers with decimals or fractions, can be very helpful in solving problems. Rational Numbers. That's our five. Step 4: Place the repeating digit(s) to the right of the decimal point. Irrational Numbers. 0.5 can be written as ½ or 5/10, and any terminating decimal is a rational number. Examine the repeating decimal to find the repeating digit(s). And these non recurring decimals can never be converted to fractions and they are called as irrational numbers. Rational Number is defined as the number which can be written in a ratio of two integers. Wednesday, October 14, 2020 Find two rational numbers between decimals Ex: Find two rational numbers that have 3 in their numerator and are in between 0.13 and 0.14 with no calculator. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. Which means that the only way to find the next digit is to calculate it. Getting a rough idea of the correct answer to a problem Irrational Numbers. DOWNLOAD IMAGE. how cuanto mide una cama matrimonial haunted house in san diego that lasts 4 7 hours journey to the west full movie with english subtitles. Include the decimal approximation of the ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. DOWNLOAD IMAGE. Irrational Numbers: We have different types of numbers in our number system such as whole numbers, real numbers, rational numbers, etc. In other words, irrational numbers require an infinite number of decimal digits to write—and these digits never form patterns that allow you to predict what the next one will be. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. - [Voiceover] Plot the following numbers on the number line. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. An Irrational Number is a real number that cannot be written as a simple fraction.. Irrational means not Rational. The first number we have here is five, and so five is five to the right of zero, five is right over there. A rational number is of the form $$\frac{p}{q}$$, p = numerator, q= denominator, where p and q are integers and q ≠0.. But rational numbers are actually rare among all numbers. Get an answer to your question “Part B: Find an irrational number that is between 9.5 and 9.7.Explain why it is irrational. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q ≠ 0. Some decimals terminate which means the decimals do not recur, they just stop. In short, rational numbers are whole numbers, fractions, and decimals — the numbers we use in our daily lives.. In mathematics, a number is rational if you can write it as a ratio of two integers, in other words in a form a/b where a and b are integers, and b is not zero. Rational numbers. A Rational Number can be written as a Ratio of two integers (ie a simple fraction). Number System Notes. They can be written as a ratio of two integers. I tried to cross multiply like my teacher said but that was for finding what lay in between fractions it worked for that I have no idea how to do this. I can approximately locate irrational numbers on a number line ; I can estimate the value of expression involving irrational numbers using rational numbers. A set of real numbers is uncountable. (Examples: Being able to determine the value of the √2 on a number line lies between 1 and 2, more accurately, between 1.4 … So the question states: what is a decimal number between each of the following pairs of rational numbers and then it gives the fractions -5/6, 1 and -17/20 and -4/5! Rational numbers are contrasted with irrational numbers - numbers such as Pi, √ 2, √ 7, other roots, sines, cosines, and logarithms of numbers. Rational and Irrational numbers both are real numbers but different with respect to their properties. ii) An irrational number between and . Real numbers also include fraction and decimal numbers. Step 5: Using the two equations you found in step 3 and step 4, subtract the left sides of the two equations. The decimal expansion of a rational number always either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. List Of Irrational Numbers 1 100. So irrational number is a number that is not rational that means it is a number that cannot be written in the form $$\frac{p}{q}$$. 1/3. Example 15: Find a rational number and also an irrational number between the numbers a and b given below: a = 0.101001000100001…., b = 0.1001000100001… Solution: Since the decimal representations of a and b are non-terminating and non-repeating. it can't be the terminating decimals because they're rational. An irrational number is a number which cannot be expressed in a ratio of two integers. O.13 < 3/n < 0.14 13/100 < 3/n < 14/100 When is 13/100 < 3/n ? How to Write Irrational Numbers as Decimals Clearly all fractions are of that That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. The difference between rational and irrational numbers can be drawn clearly on the following grounds. Then we get 1/3. The number $\pi$ (the Greek letter pi, pronounced ‘pie’), which is very important in describing circles, has a decimal form that does not stop or repeat. Irrational Numbers. Representation of irrational numbers on a number line. But an irrational number cannot be written in the form of simple fractions. As there are infinite numbers between two rational numbers and also there are infinite rational numbers between two rational numbers. Real numbers include natural numbers, whole numbers, integers, rational numbers and irrational numbers. How To Find Irrational Numbers Between Two Decimals DOWNLOAD IMAGE. since 3^2 = 9 and 6^2 = 36, √41 > 6. apply the same kind of thinking to the numbers in B and they are all between 3 and 6. New Proof Settles How To Approximate Numbers Like Pi Quanta Magazine. So number of irrational numbers between the numbers should be infinite or something finite which we cannot tell ? Essentially, irrational numbers can be written as decimals but as a ratio of two integers. On the other hand, all the surds and non-repeating decimals are irrational numbers. For instance, when placing √15 (which is 3.87), it is best to place the dot on the number line at a place in between 3 and 4 (closer to 4), and then write √15 above it. We can actually split this into thirds. Example: Find two irrational numbers between 2 and 3. Infinite rational numbers and also there are infinite rational numbers the ones we know best learn the difference rational. 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New Proof Settles How to find the repeating digit ( s ) in the picture below! Of an irrational number number of irrational numbers numbers, fractions, and θ not terminate and no! Surds and non-repeating decimals are rational numbers are whole numbers, integers, numbers. Ratios and rates rational numbers are whole numbers, never-ending decimals that can be written as a ratio two... Which do not terminate and have no repeating pattern of rational numbers are whole numbers, fractions and... 0.14 13/100 < 3/n < 0.14 13/100 < 3/n < 14/100 when is 13/100 < 3/n < 14/100 when 13/100. Given numbers, fractions, and decimals - the numbers should be infinite or something finite which we can look... Between the numbers should be infinite or something finite which we can see irrational... Terminating decimal is a rational number using rational numbers are whole numbers, fractions, and any terminating is... 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Ratio of two integers number between two decimals one has to first understand what are numbers. Apart from these number systems we have two rational numbers the ones we best. ) to the left sides of the decimal expansion of an irrational number a. ; 2, π, e, and decimals - the numbers should be infinite or finite! To study irrational numbers non recurring decimals can never be converted to fractions and they called... | 2021-03-04T00:05:42 | {
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https://goodboychan.github.io/python/datacamp/data_science/statistics/2020/05/26/05-Thinking-probabilistically-Continuous-variables.html | ## Probability density functions
• Continuous variables
• Quantities that can take any value, not just discrete values
• Probability Density function (PDF)
• Continuous analog to the PMF
• Mathematical description of the relative likelihood of observing a value of a continuous variable
## Introduction to the Normal distribution
• Normal distribution
• Describes a continuous variable whose PDF has a single symmetric peak.
\begin{align} \text{mean of a Normal distribution} & \neq \text{mean computed from data} \\ \text{st. dev of a Normal distribution} & \neq \text{st. dev computed from data} \end{align}
### The Normal PDF
In this exercise, you will explore the Normal PDF and also learn a way to plot a PDF of a known distribution using hacker statistics. Specifically, you will plot a Normal PDF for various values of the variance.
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
sns.set()
# with stds of interest: samples_std1, samples_std3, stamples_std10
samples_std1 = np.random.normal(20, 1, 100000)
samples_std3 = np.random.normal(20, 3, 100000)
samples_std10 = np.random.normal(20, 10, 100000)
# Make histograms
_ = plt.hist(samples_std1, histtype='step', density=True, bins=100)
_ = plt.hist(samples_std3, histtype='step', density=True, bins=100)
_ = plt.hist(samples_std10, histtype='step', density=True,bins=100)
# Make a legend, set limits
_ = plt.legend(('std = 1', 'std = 3', 'std = 10'))
plt.ylim(-0.01, 0.42)
(-0.01, 0.42)
### The Normal CDF
Now that you have a feel for how the Normal PDF looks, let's consider its CDF. Using the samples you generated in the last exercise (in your namespace as samples_std1, samples_std3, and samples_std10), generate and plot the CDFs.
def ecdf(data):
"""Compute ECDF for a one-dimensional array of measurements."""
# Number of data points: n
n = len(data)
# x-data for the ECDF: x
x = np.sort(data)
# y-data for the ECDF: y
y = np.arange(1, n + 1) / n
return x, y
x_std1, y_std1 = ecdf(samples_std1)
x_std3, y_std3 = ecdf(samples_std3)
x_std10, y_std10 = ecdf(samples_std10)
# Plot CDFs
_ = plt.plot(x_std1, y_std1, marker='.', linestyle='none')
_ = plt.plot(x_std3, y_std3, marker='.', linestyle='none')
_ = plt.plot(x_std10, y_std10, marker='.', linestyle='none')
# Make a legend
_ = plt.legend(('std = 1', 'std = 3', 'std = 10'), loc='lower right')
plt.savefig('../images/std-cdf.png')
## The Normal distribution: Properties and warnings
### Are the Belmont Stakes results Normally distributed?
Since 1926, the Belmont Stakes is a 1.5 mile-long race of 3-year old thoroughbred horses. Secretariat ) ran the fastest Belmont Stakes in history in 1973. While that was the fastest year, 1970 was the slowest because of unusually wet and sloppy conditions. With these two outliers removed from the data set, compute the mean and standard deviation of the Belmont winners' times. Sample out of a Normal distribution with this mean and standard deviation using the np.random.normal() function and plot a CDF. Overlay the ECDF from the winning Belmont times. Are these close to Normally distributed?
Note: Justin scraped the data concerning the Belmont Stakes from the Belmont Wikipedia page.
df = pd.read_csv('./dataset/belmont.csv')
belmont_no_outliers = np.array([148.51, 146.65, 148.52, 150.7 , 150.42, 150.88, 151.57, 147.54,
149.65, 148.74, 147.86, 148.75, 147.5 , 148.26, 149.71, 146.56,
151.19, 147.88, 149.16, 148.82, 148.96, 152.02, 146.82, 149.97,
146.13, 148.1 , 147.2 , 146. , 146.4 , 148.2 , 149.8 , 147. ,
147.2 , 147.8 , 148.2 , 149. , 149.8 , 148.6 , 146.8 , 149.6 ,
149. , 148.2 , 149.2 , 148. , 150.4 , 148.8 , 147.2 , 148.8 ,
149.6 , 148.4 , 148.4 , 150.2 , 148.8 , 149.2 , 149.2 , 148.4 ,
150.2 , 146.6 , 149.8 , 149. , 150.8 , 148.6 , 150.2 , 149. ,
148.6 , 150.2 , 148.2 , 149.4 , 150.8 , 150.2 , 152.2 , 148.2 ,
149.2 , 151. , 149.6 , 149.6 , 149.4 , 148.6 , 150. , 150.6 ,
149.2 , 152.6 , 152.8 , 149.6 , 151.6 , 152.8 , 153.2 , 152.4 ,
152.2 ])
mu = np.mean(belmont_no_outliers)
sigma = np.std(belmont_no_outliers)
# Sample out of a normal distribution with this mu and sigma: samples
samples = np.random.normal(mu, sigma, size=10000)
# Get the CDF of the samples and of the data
x_theor, y_theor = ecdf(belmont_no_outliers)
x, y = ecdf(samples)
# Plot the CDFs
_ = plt.plot(x_theor, y_theor)
_ = plt.plot(x, y, marker='.', linestyle='none')
_ = plt.xlabel('Belmont winning time (sec.)')
_ = plt.ylabel('CDF')
### What are the chances of a horse matching or beating Secretariat's record?
Assume that the Belmont winners' times are Normally distributed (with the 1970 and 1973 years removed), what is the probability that the winner of a given Belmont Stakes will run it as fast or faster than Secretariat?
samples = np.random.normal(mu, sigma, size=1000000)
# Compute the fraction that are faster than 144 seconds: prob
prob = np.sum(samples < 144) / len(samples)
# Print the result
print('Probability of besting Secretariat:', prob)
Probability of besting Secretariat: 0.00057
## The Exponential distribution
• The waiting time between arrivals of a Poisson process is Exponentially distributed
### If you have a story, you can simulate it!
Sometimes, the story describing our probability distribution does not have a named distribution to go along with it. In these cases, fear not! You can always simulate it. We'll do that in this and the next exercise.
In earlier exercises, we looked at the rare event of no-hitters in Major League Baseball. Hitting the cycle is another rare baseball event. When a batter hits the cycle, he gets all four kinds of hits, a single, double, triple, and home run, in a single game. Like no-hitters, this can be modeled as a Poisson process, so the time between hits of the cycle are also Exponentially distributed.
How long must we wait to see both a no-hitter and then a batter hit the cycle? The idea is that we have to wait some time for the no-hitter, and then after the no-hitter, we have to wait for hitting the cycle. Stated another way, what is the total waiting time for the arrival of two different Poisson processes? The total waiting time is the time waited for the no-hitter, plus the time waited for the hitting the cycle.
Now, you will write a function to sample out of the distribution described by this story.
def successive_poisson(tau1, tau2, size=1):
"""Compute time for arrival of 2 successive Poisson processes."""
# Draw samples out of first exponential distribution: t1
t1 = np.random.exponential(tau1, size)
# Draw samples out of second exponential distribution: t2
t2 = np.random.exponential(tau2, size)
return t1 + t2
### Distribution of no-hitters and cycles
Now, you'll use your sampling function to compute the waiting time to observe a no-hitter and hitting of the cycle. The mean waiting time for a no-hitter is 764 games, and the mean waiting time for hitting the cycle is 715 games.
waiting_times = successive_poisson(764, 715, size=100000)
# Make the histogram
_ = plt.hist(waiting_times, bins=100, density=True, histtype='step')
# Label axes
_ = plt.xlabel('waiting times')
_ = plt.ylabel('PDF') | 2022-01-29T08:02:42 | {
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https://www.physicsforums.com/threads/area-of-a-sector-why-squared.936747/ | # Area of a Sector- Why squared?
#### opus
Gold Member
1. Homework Statement
Find the area, A, of a sector of a circle with a radius of 9 inches and a central angle of 30°.
2. Homework Equations
$$Area~of~a~Sector:$$
$$A=\left( \frac 1 2 \right)r^2θ$$
3. The Attempt at a Solution
$$θ=30°$$
$$θ=30°\left( \frac π {180} \right)$$
$$θ=\left( \frac π 6 \right)$$
$$A=\left( \frac 1 2 \right)\left(9\right)^2\left(\frac π 6 \right)$$
$$A=\left( \frac {81π} {12} \right)$$
$$A≈21.2 in^2$$
My question:
I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result. However in this case, I don't see how the units for a sector of a circle are squared, as it doesn't seem like we're multiplying two things of equal value to each other.
So why is this result squared?
Related Precalculus Mathematics Homework Help News on Phys.org
#### Mark44
Mentor
1. Homework Statement
Find the area, A, of a sector of a circle with a radius of 9 inches and a central angle of 30°.
2. Homework Equations
$$Area~of~a~Sector:$$
$$A=\left( \frac 1 2 \right)r^2θ$$
3. The Attempt at a Solution
$$θ=30°$$
$$θ=30°\left( \frac π {180} \right)$$
$$θ=\left( \frac π 6 \right)$$
$$A=\left( \frac 1 2 \right)\left(9\right)^2\left(\frac π 6 \right)$$
$$A=\left( \frac {81π} {12} \right)$$
$$A≈21.2 in^2$$
My question:
I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result. However in this case, I don't see how the units for a sector of a circle are squared, as it doesn't seem like we're multiplying two things of equal value to each other.
So why is this result squared?
Because it's an area. The shape doesn't matter.
The standard units of area are always squared, $\text{length} \times \text{length}$, except for some special cases such as acres or hectares (which involve implicitly squared units such as ft2 for acres and m2 for hectares.
#### opus
Gold Member
So the length x length in this particular case, would be length(radius) x length(arc). However the length of the radius is in inches, and the length of the arc is in radians. So how can this results in inches squared?
#### Mark44
Mentor
So the length x length in this particular case, would be length(radius) x length(arc). However the length of the radius is in inches, and the length of the arc is in radians. So how can this results in inches squared?
The angle in radians is just an angle, with no length. Think about it this way, as a, say, peach pie. If an 8" diameter pie is cut into 6 pieces, each slice (a sector) will subtend an angle of $\pi/3$, and the radius will be 4". The arc length of the curved edge of the slice has to take into account the radius, otherwise the arc length of an 8" pie would be the same as for a 16" pie. So in fact, the curved dimension of the pie sector is radius * angle (in radians), or $4 \times \pi/3$. So you have one radius for the radius of the sector and another radius for the arc length, making the sector area equal to $\frac 1 2 r^2 \theta$.
#### opus
Gold Member
Ahhh ok. That makes complete sense. Great explanation, thank you Mark.
#### DrClaude
Mentor
I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result.
Let me add that one way to visualize the "units square" is to think that a wedge with an area of 22.2 in2 has the same area as a square with sides of √(22.2) ≈ 4.6 in.
Last edited by a moderator:
#### opus
Gold Member
Interesting! Thanks DrClaude
#### alijan kk
imagine the arc as a triangle , becuase area would be same even if you make the arc straight line.
now the base of this triangle is "r=radius" and the perpendicular side is the arc which is equal to "theta*r"
area of triangle = 0.5base*height
0.5(r)(r)(theta)=formula of area of sector
"Area of a Sector- Why squared?"
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• Solo and co-op problem solving | 2019-09-22T20:32:09 | {
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http://mathhelpforum.com/algebra/132387-problem-vector-question.html | # Math Help - Problem with vector question
1. ## Problem with vector question
Hi
Can someone tell me where i have gone wrong in the following question?
Given that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.
unit vector = {a x b}/{|a x b|}
| a x b |= $\sqrt{3^2+1^2+8^2}$
$=\sqrt{74}$
a x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)
=i(2+1)-j(2-3)+k(-2-6)
=3i+j-8k
therefore unit vector = $\frac{1}{\sqrt{74}}$(3i+j-8k)
answer says its $\frac{1}{\sqrt{74}}$(-3i-j+8k)
P.S
2. Originally Posted by Paymemoney
Hi
Can someone tell me where i have gone wrong in the following question?
Given that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.
unit vector = {a x b}/{|a x b|}
| a x b |= $\sqrt{3^2+1^2+8^2}$
$=\sqrt{74}$
a x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)
=i(2+1)-j(2-3)+k(-2-6)
=3i+j-8k
therefore unit vector = $\frac{1}{\sqrt{74}}$(3i+j-8k)
answer says its $\frac{1}{\sqrt{74}}$(-3i-j+8k)
P.S
$\mathbf{v}\times\mathbf{w} = \left|\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ 3 & -1 & 1 \end{matrix}\right|$
$= \mathbf{i}[2 \cdot 1 - 1\cdot (-1)] - \mathbf{j}[2 \cdot 1 - 1 \cdot 3] + \mathbf{k}[2 \cdot (-1) - 2 \cdot 3]$
$= 3\mathbf{i} - \mathbf{j} - 8\mathbf{k}$.
I agree with the answer you have given for $\mathbf{v} \times \mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect.
Anyway, to find the unit vector in the direction of $\mathbf{v} \times \mathbf{w}$, divide it by its length.
$|\mathbf{v}\times\mathbf{w}| = \sqrt{3^2 + (-1)^2 + (-8)^2}$
$= \sqrt{9 + 1 + 64}$
$= \sqrt{74}$.
Therefore:
$\frac{\mathbf{v}\times\mathbf{w}}{|\mathbf{v}\time s\mathbf{w}|} = \frac{3\mathbf{i} - \mathbf{j} - 8\mathbf{k}}{\sqrt{74}}$
$= \frac{3}{\sqrt{74}}\mathbf{i} - \frac{1}{\sqrt{74}}\mathbf{j} - \frac{8}{\sqrt{74}}\mathbf{k}$
$= \frac{3\sqrt{74}}{74}\mathbf{i} - \frac{\sqrt{74}}{74}\mathbf{j} - \frac{4\sqrt{74}}{37}\mathbf{k}$.
3. Hello, Paymemoney!
You've done nothing wrong . . .
Their answer is the negative of yours.
There are two vectors perpendicular to both $\vec v\text{ and }\vec w,$
. . one "up" and one "down".
They picked one, you picked the other.
4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k | 2015-03-29T10:17:17 | {
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https://math.stackexchange.com/questions/4343130/how-same-are-two-isomorphic-groups | # How "same" are two isomorphic groups?
From what I understand about isomorphisms is that two isomorphic groups are the same groups. They may have different names for the same elements and the operation. But the point is that the groups are same since their elements combine the same way.
So if $$G \cong G'$$ then every group property about $$G$$ also holds for $$G'$$, am I correct?
Now my question is— Can I interchange $$G$$ and $$G'$$ whenever and wherever I want? I thought the answer was obviously yes but... now I'm not sure.
For example: $$\mathbb{Z} \cong \mathbb{2Z}$$ so shouldn't $$\mathbb{Z}/\mathbb{Z} = \mathbb{Z}/\mathbb{2Z}$$ under the group operation $$(a+H) + (b +H) = (a+b) + H$$? where $$H$$ is either $$\mathbb{Z}$$ or $$\mathbb{2Z}$$ since both are the same groups.
But that's clearly not the case as $$\mathbb{Z}/\mathbb{Z} = \{0\}$$ but $$\mathbb{Z}/\mathbb{2Z}=\{0,1\}$$ So where does one draw the line between two isomorphic groups? How "same" are two isomorphic groups? I'm very confused.
• It is true that $\mathbb{Z} \cong \mathbb{2Z}$. However, the issue for the quotient groups here is that $2 \mathbb Z \subsetneq \mathbb Z$. Dec 27, 2021 at 16:42
• The freedom to rename is not a freedom to be inconsistent. If you want to rename the elements in $\mathbb Z$ so that $1$ is renamed to $2$, $2$ to $4$, $3$ to $6$ etc. - then of course the elements of $2\mathbb Z\subseteq \mathbb Z$ must be renamed using the same procedure. You will end up with $4\mathbb Z\subseteq 2\mathbb Z$, and indeed $\mathbb Z/2\mathbb Z\cong 2\mathbb Z/4\mathbb Z$. Dec 27, 2021 at 16:43
• Why is this closed under duplicate? I'm asking about a particular example and the difficulty I'm facing with it. How is the question same as the other? I'd already read that one. It's where I came from. The accepted answers says exactly the same thing as my first paragraph. But my question is different. Dec 27, 2021 at 16:46
• Presentations of the same group (up to isomorphism) can have different properties. Dec 27, 2021 at 16:54
• Frankly, I regard the "s" word as a 4-letter word in mathematics. I would suggest that you learn to love the 10-letter "i" word in your title. And then if you can extend that to loving the 11-letter word "isomorphism", and to think about actual isomorphisms as useful and interesting mathematical objects themselves, even better! Dec 27, 2021 at 18:51
This is a great question and your example with group quotients perfectly highlights why one has to be careful about notions of identity when doing maths. To answer your question,
Can I interchange G and G′ whenever and wherever I want?
The answer is yes, as long as you are 100% sure that you are thinking of both groups as literally just groups and nothing more (people will sometimes use the phase 'up to isomorphism' to convey this idea). This resolves the problem you had with quotients: you can't quotient an arbitrary group by another arbitrary group, the group by which you are quotienting has to be a subgroup of the other. This is extra 'data' which can be encoded in multiple ways, one nice way is to consider the special injection that embeds the one group inside the other (which doesn't exist for two arbitrary groups). So to recap, subgroups aren't just groups, they're groups with some extra data and $$2\mathbb{Z}$$ and $$\mathbb{Z}$$ are different subgroups of $$\mathbb{Z}$$ despite being 'the same' groups (i.e. isomorphic).
• "the group by which you are quotienting has to be a subgroup of the other" but $Z$ and $2Z$ are both subgroups of $Z$ no? and I don't get the part where you say they are the same groups but different subgroups? I'm sure I'm missing something here but I don't get it. What is the extra data about subgroup? Dec 27, 2021 at 17:14
• Oh you mean they're different subgroups because they are made up of different subsets of the group? Oh okay. Fair enough. But what exactly go wrong here in my example? I'm not able to "see" it still :( Dec 27, 2021 at 17:26
• The problem in your example is that, to conclude that the quotient is the same, you have to start with 'the same' subgroups, but you didn't, you started with two different subgroups (that just so happen to be 'the same' as groups). Dec 27, 2021 at 17:28
• So in total, there are four descriptions of mathematical objects: the group $\mathbb{Z}$, the group $\mathbb{2Z}$, the subgroup $\mathbb{Z}\leq\mathbb{Z}$ and the subgroup $\mathbb{2Z}\leq\mathbb{Z}$. The first two descriptions define the same mathematical object, the second two don't. Dec 27, 2021 at 17:46
• Oh! Oh! Oh!! I get it now. If $H,H' ≤G$ and $H \cong H'$ then as groups, $H$ and $H'$ are same. But "subgroup" is also a relation and not an independent concept. So you always have subgroups of some group. So as subgroups of $G$, $H$ and $H'$ are different. So $G/H ≠ G/H'$ because quotient groups aren't just about $H$ but rather $H$ wrt $G$. If it were about only $H$ then one can interchange it with $H'$ is what you're saying. Dec 27, 2021 at 18:12
Here is a question that is similar: when can replace equivalent things with one another?
So if we remember that an isomorphism of groups is a function $$\phi:G\rightarrow G'$$. With the example you gave you have $$\phi:\mathbb{Z}\rightarrow 2\mathbb{Z}$$, then your issue was that $$\mathbb{Z}/\mathbb{Z}\not\cong\mathbb{Z}/2\mathbb{Z}$$. However, the issue is where are you applying the homomorphism. You are saying that $$G/G\not\cong G/\phi(G)$$; however, we have that $$\phi(G)\subset G'$$ and $$\phi(G)\not\subset G$$ (although it may be isomorphic to some subgroup of $$G$$ like in your example).
The key use of an isomorphism is that you can take a problem in one setting and view it in another setting through the isomorphism. Thus, what would be true is that if $$\phi:G\rightarrow G'$$ is an isomorphism, then if $$H\lhd G$$, then it will be the case that $$G/H\cong \phi(G)/\phi(H)$$.
I think the long story short is that when you have an isomorphism between two objects you must think of them as living in different spaces. If you are familiar with the lattice of subgroups I think this is a nice way to view it, since if you have $$\phi: G\rightarrow G'$$ where $$G=\mathbb{Z}$$ and $$G'=2\mathbb{Z}$$ since these are isomorphic groups their lattice of subgroups are isomorphic and we should think of $$\mathbb{Z}$$ to be the maximal element in the lattice of $$G$$ and $$2\mathbb{Z}$$ to be the maximal element in the lattice of $$G'$$. Even though we have that $$2\mathbb{Z}$$ appears in the lattice of $$G$$, the corresponding element in $$G'$$ would be the subgroup $$4\mathbb{Z}$$, so under the isomorphism it would not make sense to "replace $$2\mathbb{Z}$$ in the $$G$$ world with $$2\mathbb{Z}$$ in the $$G'$$ world" as they aren't the same under the isomorphism. What we would do is replace the $$2\mathbb{Z}$$ with the corresponding subgroup in the isomorphism namely $$4\mathbb{Z}$$, and when we do so we do have that $$\mathbb{Z}/2\mathbb{Z}\cong 2\mathbb{Z}/4\mathbb{Z}$$.
There is a precise way to formalise this.
We want to speak of sets not as collections of particular mathematical objects but as a bag of featureless dots. These bags of dots are related to each other by functions, which also relate dots in one bag to dots in another.
In order to do this, we propose a language for discussing sets. Capital letters will represent sets, and lowercase variables will represent functions between sets.
We define a "nice proposition about sets" as any proposition that can be generated from the following rules:
• For any sets $$A, B$$ and any function variables $$f, g: A \to B$$, $$f = g$$ is a nice proposition
• We can combine nice propositions $$\phi, \psi$$ using the operators $$\land$$, $$\lor$$, $$\neg$$, and $$\implies$$ to get another nice proposition
• $$\top$$ and $$\bot$$ (that is, true and false) are nice propositions
• If $$\phi(A)$$ is a nice proposition, where $$A$$ is a set variable, then $$\forall A (\phi(A))$$ is a nice proposition (and similarly for $$\exists$$)
• For all set variables $$A, B$$, if $$\phi(f)$$ is a nice proposition (where $$f : A \to B$$ is a function variable), then $$\forall f : A \to B (\phi(f))$$ is a nice proposition (and similarly for $$\exists$$)
Finally, in our language, we can form new functions from old ones using function composition and also discuss identity functions. Note that we will only discuss function composition when we know that the functions are composable syntactically.*
Note that we very deliberately avoided two things. First, there is no mention of the elementhood relation at all. Second, there is no way to compare two sets for equality, nor is there a way to compare two functions for equality unless the two have the same domain and codomain. This is deliberate. Also note that all variables here have specific types, and that we rely on these types to determine whether we can discuss equality.
Despite this avoidance, it is both an empirical fact and a (rather complicated) theorem that all propositions with no free variables can be translated to a "nice proposition about sets". For most mathematically useful propositions, the translation is relatively intuitive for those comfortable with category theory.
To get around the restriction that we not discuss elements directly, we can fix some 1-element set $$1$$ and discuss elements of $$A$$ as functions $$1 \to A$$. We denote the situation $$x : 1 \to A$$ as $$x :\in A$$. When we have $$x :\in A$$ and $$f : A \to B$$, we write $$f \circ x$$ as $$f(x)$$ (note that $$f(x) :\in B$$). Note that this is enough to define, for example, a group (as a set $$G$$, together with a function $$- \cdot - : G^2 \to G$$ which satisfies the group laws**).
We thus have the following theorem:
Big Theorem. Consider some nice proposition $$\phi(G)$$ - that is, some statement $$\phi$$ where the group variable $$G$$ occurs free. Then we can prove the following statement: "For all groups $$G_1$$, $$G_2$$, if $$G_1$$ and $$G_2$$ are isomorphic and $$\phi(G_1)$$, then $$\phi(G_2)$$."
Let us discuss how this relates to your example of $$\mathbb{Z}$$ and $$2 \mathbb{Z}$$.
We may try to define $$\phi(G)$$ as the statement "$$\mathbb{Z} / G$$ is the zero group". Then taking $$G_1 = \mathbb{Z}$$ and $$G_2 = 2 \mathbb{Z}$$, we see that $$\phi(G_1)$$ holds but $$\phi(G_2)$$ is false. This would appear to violate our Big Theorem.
To understand what's going on, we need to understand exactly what is going on with the statement $$\phi(G)$$. Note that in this statement, it is necessary for $$G$$ to be a subgroup of $$\mathbb{Z}$$ - in particular, it must be a subset. That is, we must have $$\forall x \in G (x \in \mathbb{Z})$$.
Of course, we cannot express such a statement as a nice proposition about sets, since it relies on the proposition $$x \in \mathbb{Z}$$, which is not a nice proposition. So we have to approach things a bit differently.
Rather than forcing $$G$$ to be a subgroup in the traditional, literal sense, we instead make $$G$$ a subgroup of $$\mathbb{Z}$$ in a more general sense. That is, we discuss a group $$G$$, together with some injective group homomorphism $$i : G \to \mathbb{Z}$$. In other words, we discuss $$G$$ together with a specific way that $$G$$ is a subgroup of $$\mathbb{Z}$$.
This makes it clear that $$\phi(G)$$ is not really just a proposition about $$G$$ - it's also about the way that $$G$$ is a subgroup of $$\mathbb{Z}$$. So it's really a proposition $$\phi(G, i : G \to \mathbb{Z})$$. Because $$\phi$$ depends on a secondary variable $$i$$, we see that we cannot apply our Big Theorem to conclude that $$\phi(\mathbb{Z}) \iff \phi(2 \mathbb{Z})$$.
If it were the case that the isomorphism $$w : \mathbb{Z} \to 2 \mathbb{Z}$$ "played nicely" with the inclusion maps $$i_1 : \mathbb{Z} \to \mathbb{Z}$$, $$i_2 : 2 \mathbb{Z} \to \mathbb{Z}$$ (that is, if $$i_1 = i_2 \circ w$$), then we would be able to conclude that $$\phi(\mathbb{Z}, i_1) \iff \phi(2 \mathbb{Z}, i_2)$$ using a generalisation of our Big Theorem. But of course the isomorphism does not play nicely with the inclusion maps.
*An astute observer will note that this is exactly the language of category theory. A "nice proposition about sets" is just some statement about the category of sets (which avoids discussing equality of objects).
**Technically, we need a way to define $$G^2$$ first. This is done using the categorical definition of the universal property of the product.
• Why is this answer being downvoted? Dec 27, 2021 at 18:31
This is just an elaboration of Arthur's answer elsewhere in this thread. I thought you might like to see a specific example.
Consider $$G = \color{maroon}{\Bbb Z_2}\times \color{darkblue}{\Bbb Z_4}.$$
$$G$$ has a subgroup $$A$$ that is generated by $$\langle 1, 0\rangle$$:
$$\def\elt#1#2{\langle{#1},{#2}\rangle}A = \{\elt10, \elt 00\}$$
and another subgroup $$B$$ that is generated by $$\langle 0, 2\rangle$$:
$$B=\{\elt00, \elt 02\}$$
Both $$A$$ and $$B$$ are isomorphic to $$\Bbb Z_2$$, and so to each other. But the quotients $$G/A$$ and $$G/B$$ are not isomorphic. $$G/A$$ is like taking $$G$$, ignoring the first component, and keeping the second component intact. The result is $$G/A \simeq \color{maroon}{\Bbb Z_1}\times\color{darkblue}{\Bbb Z_4}.$$ $$G/B$$ is like taking $$G$$ and keeping the first component but ignoring everything but the parity of the second component. The result is $$G/B \simeq \color{maroon}{\Bbb Z_2}\times\color{darkblue}{\Bbb Z_2}.$$ Even though $$A$$ and $$B$$ are isomorphic, $$G/A$$ and $$G/B$$ are not isomorphic. (In particular, $$G/A$$ is cyclic and $$G/B$$ isn't.)
Each component of $$G = \color{maroon}{\Bbb Z_2}\times \color{darkblue}{\Bbb Z_4}$$ contains a factor of $$\Bbb Z_2$$. Each of $$A$$ and $$B$$ is isomorphic to $$\Bbb Z_2$$. When we construct the quotient of $$G$$ by $$A$$ or $$B$$, we get a different result depending on which component we divide the factor from.
Two groups that are isomorphic share the same internal structure. But a quotient $$G/N$$ doesn't depend only on the internal structures of $$G$$ and $$N$$. The quotient also depends on the relationship between $$G$$ and $$N$$. So even though $$A$$ and $$B$$ have the same internal structure, $$G/A$$ and $$G/B$$ are different, because those structures reside inside of $$G$$ in different ways.
Topological and illustrated version of Arthur's nice answer:
A trefoil knot $$K_3$$ in $$\Bbb R^3$$ cannot be unknotted but it can be in $$\Bbb R^4$$. Topologically a trefoil knot $$K_3$$ and a circle are homeomorphic i.e. one can deform one to other without cutting or gluing. So these two (one in $$\Bbb R^3$$ and one in $$\Bbb R^4$$) are the same but one uncovers a branch of mathematics known as Knot_theory while other one is fruitless from this point of view and application.
I think theses type of "same things" is a categorical notion but I am not sure.
same is true for $$y=x$$ and $$y=\exp(x)$$; both topologically are homeomorphic to $$\Bbb R$$, but one contains only positive numbers. One can say similar properties by considering them as additive and multiplicative groups.
These might be of interest to you:
Let me for convenience, first, assume all subgroups are normal, or that we are in an abelian group.
We form quotients of a group by its SUBgroups, not by another abstract GROUP. Subgroup means a SPECIFIC subset (which is closed ....) As groups Z and 2Z are "same". But as subsets of Z they are different.
Now to non-abelian case. Take $$G= \{+1,-1\}\times S_3$$ where the first factor $$H$$ is a group (of order 2) wrt multiplication of numbers.
This group $$G$$ has more subgroups of order 2 (isomorphic to H as a group) namely ones generated by a transposition in the second factor $$S_3$$. These subgroups are NOT even normal in $$G$$. SO quotient does not even exist by these subgroups. | 2022-07-01T16:09:58 | {
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https://math.stackexchange.com/questions/2309292/seperating-2n-elements-with-subsets-of-size-n | Seperating $2n$ elements with subsets of size $n$
Suppose I have a set $X$ with $2n$ elements ($n$ nonzero natural number). I now want to find a collection of subsets $X_1, \ldots, X_k$ of $X$ such that
• Every $X_i$ contains $n$ elements.
• For every $x, y \in X$, there exists an $X_i$ such that $x \in X_i$ and $y \not\in X_i$ or $x \not\in X_i$ and $y \in X_i$ (the $X_i$ "seperate" the elements of $X$, in that the topology on $X$ generated by the $X_i$ is Kolmogorov)
• $k$ is minimal with respect to the above to properties.
An algorithm to find $X_1, \ldots, X_k$ would be great, but I am mostly looking for a way to calculate $k$. My intuition tells me that $k$ should be at least $\log_2(2n)$, more specifically I feel like $$\max_{x \in X} \# \lbrace y \in X \mid \not\exists i \in \lbrace 1, \ldots, j \rbrace : x \in X_i \enspace \& \enspace y \not\in X_i \rbrace \geq \log_2(\frac{2n}{j})$$ should hold for all $j \in \lbrace 1, \ldots, k \rbrace$, but I don't know how to prove this.
• Consider sets with $2^l$ elements first – Scipio Jun 4 '17 at 12:09
• I did; even if $n=4$ I don't see how to proceed. (i.e. I know that $k = 2$ is possible, but do not see how to show that $k = 1$ is impossible) – Bib-lost Jun 4 '17 at 12:11
• $X_1$ contains two elements $a,b$; clearly the second requirement doesn't hold for these two elements if you only have a single set. – Scipio Jun 4 '17 at 12:18
• By $n=4$ I meant the case where $X$ has $2*4 = 8$ elements. The case with $4$ elements is indeed clear. :) And surely, also the case with $8$ elements can be computed by hand in a reasonably short time; what I meant is that I do not see an argument which could somehow be generalised. – Bib-lost Jun 4 '17 at 12:44
• Could you add the cases $n=2,3,4$ into your question & give us the lists $X_1, X_2 , \cdots , X_k$ ? – Donald Splutterwit Jun 4 '17 at 12:50
Suppose $X$ contains $2^l$ elements. Let $(A_i^m)$ be the family of subsets of $X$ of non-separated elements after introducing $X_1, \cdots, X_m$, i.e. sets of elements for which the second requirement does not hold (check that these sets are well defined and uniquely partition $X$). Ultimately, we want that all the subsets $A_i$ contain just a single element (i.e. every element is separated from all other elements).
With each $X_i$ we want to separate as many elements as possible. Now, at first $A_1 = X$ containing $2^l$ elements and we take $X_1$ just an arbitrary set of $n$ elements. After this, we have sets $A_1$, $A_2$, both containing $2^{l-1}$ elements. Now, with $X_2$ we can again split both in half, resulting in four 'remaining sets' with $2^{l-2}$ elements. Continue recursively to see that $k=l$. Moreover, it is clear that after introducing $m$ sets, $\max_i \# A_i^m \geq 2^{l-m}$, so that our result is indeed optimal.
For example, start with $X = A_1^0 = \{1,2,3,4,5,6,7,8\}$. Introduce $X_1 = \{1,2,3,4\}$ to get $A_1^1 = \{1,2,3,4\}, A_2^1 = \{5,6,7,8\}$. Introduce $X_2 = \{1,2,5,6\}$ and $X_3 = \{1,3,5,7\}$ to finish the job. (Make sure to check the corresponding sets $(A_i^m)$.)
Now, for any other number of elements $2n$, your 'splits' wil not be optimal. You can check that $\max_i \# A_i^m \geq 2n\cdot2^{-m}$ still holds.
• So probably we have that, in general, $k$ is the unique integer satisfying $2^{k-1} < 2n \leq 2^{k}$? – Bib-lost Jun 4 '17 at 13:57
• yes exactly :) This hinges on the very last statement in my answer, so make sure to check that carefully. – Scipio Jun 4 '17 at 14:01
There should be $k=\lceil log_2(2n) \rceil$ partitions, $2k$ sets.
We have the conditions: $$\mathcal{P}_1:\forall X_i, \text{num}\{X_i\}=n$$ $$\mathcal{P}_2:\forall x_i,x_j \exists X_i, x_i \in X_i, x_j \not\in X_i$$
It is clear for sets of $2n=2^k$ elements, the partitions are binary, hence optimal. Under this case, the condition $\mathcal{P}_2$ implies the stronger: $$\mathcal{P}_{2b}:\forall x_i,x_j \exists X_i,X_j, x_i \in X_i, x_j \in X_j$$ because each partition requires its complement. If there is not complement, the property is not met. $$\forall X_i \exists Y_i=X -X_i$$
If the set contains $2n=2^k+2m$ elements, $2m<2^k$, each $2m$ remaining elements shall be included into different disjoint partitions. Again in this case the partition complement must exist and the condition $\mathcal{P}_{2}$ will not introduce any benefit against $\mathcal{P}_{2b}$.
• Don't you mean $k = \lceil \log_2(2n) \rceil$? Surely if $n=3$ you need at least $2 = \lceil log_2(6) \rceil$ partitions? – Bib-lost Jun 4 '17 at 14:14 | 2019-10-14T09:54:19 | {
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https://math.stackexchange.com/questions/3192613/special-case-of-bertrand-paradox-or-just-a-mistake | # Special case of Bertrand Paradox or just a mistake?
I've been working on a question and it seems I have obtained a paradoxical answer. Odds are I've just committed a mistake somewhere, however, I will elucidate the question and my solution just in case anyone is interested.
I want to know what is the average distance between two points on a circle of radius 1 where we consider only the boundary points.
My attempt is as follows:
Consider a segment of the diameter x which is uniformly distributed between 0 and 2. Then you can calculate the distance between the points (2,0) and the point determined by x just by elementary geometry as this picture shows:
enter image description here
Here in the picture, the green segment is the geometric mean and the orange one is the distance whose distribution we want to know. Just by calculating the expected value, we obtain:
$$E\left(\sqrt{(4-2X)}\right) = \int_{0}^{2} \sqrt{(4-2x)}\cdot\frac{1}{2} dx = 1.333.... = \frac{4}{3}$$
Where $$\sqrt{(4-2x)}$$ is the transformation of the random variable and $$\frac{1}{2}$$ is the pdf of a uniform distribution $$[0,2]$$.
Also, if we derive the pdf of the transformation we obtain the same result:
$$y = \sqrt{(4-2x)} , x = 2- \frac{y^2}{2}, \mid\frac{d}{dy}x\mid = y$$
$$g(y)=f(x)\cdot\mid\frac{d}{dy}x\mid = \frac{1}{2}\cdot y$$
$$E(Y)= \int_{0}^{2}y\cdot\frac{1}{2}\cdot y dy = 1.333.... = \frac{4}{3}$$
I have seen a different approach somewhere else where the distribution of the angle is considered as a uniform distribution between 0 and $$\pi$$ and the final result was:
$$1.27... = \frac{4}{\pi}$$
That's pretty much the problem I found. Maybe I just did it wrong in some step but it all makes sense to me. I know this is not exactly what we refer as Bertrand paradox but it just suggests something like that because both problems handle with segments in circumference and maybe my result is wrong because it does not hold for rotations of the circle or something like that (I read a little bit about the Bertrand's Paradox).
That's pretty much it. Also sorry for my bad English and maybe I'm also wrong in something pretty elemental since I've just started learning about probability theory. It's also my first post so I will try to improve my exposition and LateX use in the following ones.
• welcome to MSE. I have attempted to edit your question. In case you find any discrepancy with your idea please check. Also, you can look up the edits I have made that will help you for your future questions. Thank you! – Mann Apr 19 at 21:57
This is a very nice question, well-written and researched before posting. You have clearly put a lot of careful into your question and it is very much appreciated on this site (and I thought your English reads perfectly naturally). Thanks so much for taking the time to ask a serious and well-considered question. I hope this answer comes close to meeting the high standard of quality set by your question.
I don’t see any calculation errors on your part, and I’ll wager you checked those thoroughly. The flaw is something much more subtle and is related to the initial framing of the problem.
You have correctly solved the problem of “what is the average distance between the rightmost point of a circle and another point projected upwards from a randomly chosen point on the diameter”. But this doesn’t quite capture the same probability distribution as “rightmost point and another randomly chosen point on the circle” (which is sufficient to capture the dynamics of “two random points on the circle”).
There is a hidden non-uniformity in the projection process, and that’s because the circle has varying slope so the projection hits the circle differently at different positions.
Here’s an experiment that should help convince you that uniform distribution on the diameter does not yield uniform distribution on the circle boundary. Fundamentally, a uniform distribution should treat all arcs of the same length equally: a random point has a $$1/4$$ chance of falling on a quarter-circular arc, regardless of which quarter circle that is.
Now compare two specific quarter-circles and their projections onto the diameter: 1) the quarter-circle adjacent to $$(2,0)$$, i.e. 12 o’clock to 3 o’clock, and 2) the quarter-circle centered around the top-most point $$(1,1)$$, i.e. 10:30 to 1:30.
These both project onto the diameter with no self-overlap (avoiding double counting). But the first one projects down to the interval $$[0,1]$$ which has length $$1$$, and the second one projects down to $$[-\sqrt{2}/2,\sqrt{2}/2]$$, which has length $$\sqrt2$$.
This should indicate to you that the two distributions are not equivalent (though they may intuitively seem so). Consequently, the random process of choosing uniformly from the diameter and projecting onto the circle results in disproportionately favoring the top and bottom sectors of the circle over the left and right.
• Thanks! I really appreciate your words and yes, it seems the problem really deals with something subtle. I'm going to add some considerations in an answer. – Hyz Apr 19 at 19:49
Thanks, Erick Wong for your feedback. After your answer, I calculated the distribution of the arc length subject to the uniform distribution of the point on the diameter. In fact: if we want to express the arc length $$l$$ as a function of $$x$$, $$l = f(x)$$ we obtain:
$$l = \arccos(1-x), x = 1-\cos{l}, |\frac{d}{dl}(x)| = \sin{l}$$
$$l_{pdf} = x_{pdf} \cdot |\frac{d}{dl}x| = \frac{1}{2}\cdot \sin{l}$$.
So the arc length does not distribute uniformly, we have "lost it", we might say. That's what was wrong. For instance, if the arc length obeys a uniform distribution [0, $$\pi$$], then we can calculate the segment $$s$$ as a function of the arc length:
We know $$l = f(x)$$ and want to know $$s = h(l)$$. If we calculate $$s = g(x)$$ we're done:
From the image on the question I post, $$s = g(x)=\sqrt{2x}$$ (or the opposite segment $$\sqrt{4-2x}$$ ) then $$h = g \circ f^{-1}$$, $$s = \sqrt{2(1-cosl)}$$
$$E(s) = \int_0^\pi{\sqrt{2(1-cosl)}\frac{1}{\pi}dl}=1.273... = \frac{4}{\pi}$$
Also the pdf:
$$s = h(l) = \sqrt{2(1-cosl)} , l=h^{-1}(s)= 2\cdot \arcsin(\frac{s}{2}), |\frac{d}{ds}h^{-1}|=\frac{2}{\sqrt{4-s^2}}$$
$$s_{pdf} = \frac{1}{\pi} \frac{2}{\sqrt{4-s^2}}$$
$$E(s) = \int_0^2{s\cdot \frac{1}{\pi} \frac{2}{\sqrt{4-s^2}}ds} = 1.273... = \frac{4}{\pi}$$
So we're done. Also, the pdf of the segment suggests something related to the Cauchy distribution. Not exactly but certainly it has to do something with it. If we read the description of Cauchy distribution in Wolfram MathWorld:
"The Cauchy distribution, also called the Lorentzian distribution or Lorentz distribution, is a continuous distribution describing resonance behavior. It also describes the distribution of horizontal distances at which a line segment tilted at a random angle cuts the x-axis."
And that's it. A really fascinating problem that introduces some subtle ideas of probability theory. If anyone knows something else please give me feedback. I really think there's a nice connection with Cauchy distribution. | 2019-05-25T00:50:33 | {
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https://math.stackexchange.com/questions/4104208/does-the-sequence-x-n1-x-nx-n2-converge-to-0-whenever-1-lt-x-0-lt0 | # Does the sequence $x_{n+1}=x_n+x_n^2$ converge to $0$ whenever $-1\lt x_0\lt0$?
My question concerns using the fixed-point iteration to find the fixed point of the function $$f(x)=x+x^2=x(1+x)$$ (this function has a single fixed point at $$0$$).
# The problem
Given some fixed $$x_0$$, define the sequence $$x_n$$ by \begin{align*} x_{n+1}=x_n+x_n^2&&\text{for all n\geq0.} \end{align*} Does this sequence converge to $$0$$ for all $$x_0$$ in the range $$-1\lt x_0\lt0$$? After some computational numerical exploration, I think the answer might be yes but I'm not so sure how to prove this.
## Some notes
The sequence trivially converges to $$0$$ when $$x_0=-1$$ or $$x_0=0$$. It's also fairly easy to prove that it does not converge to $$0$$ whenever $$x_0$$ lies outside this range (i.e., $$x_0\lt-1$$ or $$x_0\gt0$$).
I've already proved that for any $$x_0$$ in this range, all successive $$x_n$$s remain in this range (i.e., $$-1\lt x_n\lt0$$ for all $$n\geq0$$) and that the sequence increases / gets closer to $$0$$ (i.e., $$x_{n+1}\gt x_n$$), but is there any way to prove it actually converges to $$0$$?
## Some examples
If, for example, $$-\frac{1}{2}\lt x_n\lt-\frac{1}{4}$$, we'll have $$-\frac{3}{8}\lt x_{n+1}\lt-\frac{1}{8}$$ and $$-\frac{21}{64}\lt x_{n+2}\lt-\frac{5}{64}$$, etc. So it seems as though the range of possible values gets closer to $$0$$. But, how would one prove this?
You can use the monotone convergence theorem. It tells us that a sequence $$\{x_n\}_{n=1}^\infty\subseteq \mathbb R$$ converges if it is bounded and decreasing. For you sequence we have $$|x_{n+1}| = |x_n| |1+x_n| \leq 1$$ since $$x_0 \in (-1,0)$$. And likewise we have $$x_{n+1} = x_n(1+x_n) \leq x_n$$ as $$1+x_n \in (0,1)$$.
(You can use induction to prove both these claims.)
Now you have shown that $$\lim_{n\to\infty} x_n=x$$ exists. From this we see that $$x = \lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} x_n(1+x_n) = x(1+x).$$ Hence $$x$$ satisfies $$x = x^2+x$$ so $$x=0$$.
• Great, exactly what I was looking for! Apr 16, 2021 at 7:15
• @ΑΘΩ You're right, I should have said increasing and replaced $\leq$ with $\geq$. Thanks. Apr 16, 2021 at 8:05
• I've already proved that $x_n$ is monotonic increasing and bounded above by $0$ in this case. So this is all I needed to complete the argument (from what I understand, I've never formally studied real analysis, so I certainly may be missing something). Apr 16, 2021 at 8:06
• @user178563 Glad to see that you agree. If we are to be very precise (which we should very well be, for this is what this forum is about) the part where it appears you want to show that $|x_n| \leqslant 1$ by induction on $n$ is also dubious (the induction is not spelled out clearly).
– ΑΘΩ
Apr 16, 2021 at 8:08
The answer to your question (regarding convergence to $$0$$) is indeed affirmative and for a reason that is not too difficult to ascertain. For simplicity let us introduce the quadratic polynomial function: \begin{align*} f \colon \mathbb{R} &\to \mathbb{R} \\ f(x)&=x^2+x \end{align*} and let us note that $$f[(-1, 0)] \subseteq (-1, 0)$$, in other words the interval $$(-1, 0)$$ is stable under $$f$$. To prove this latter claim it suffices to notice that $$x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \geqslant \frac{3}{4}>0$$ for any $$x \in \mathbb{R}$$ (and thus in particular $$x^2+x>-1$$ always) and respectively that given $$x \in (-1, 0)$$ we have $$x+1>0$$ yet $$x<0$$ which means that $$x^2+x=x(x+1)<0$$.
The immediate conclusion that we draw from this is that for any fixed $$t \in (-1, 0)$$, the unique recursive sequence $$x \in \mathbb{R}^{\mathbb{N}}$$ defined by $$x_0=t$$ and $$x_{n+1}=f(x_n)$$ for any $$n \in \mathbb{N}$$ has the property that actually $$x \in (-1, 0)^{\mathbb{N}}$$ and is thus a bounded sequence. Furthermore, since $$x_{n+1}=x_n+x_n^2 \geqslant x_n$$ takes place for every $$n \in \mathbb{N}$$ we gather that $$x$$ is increasing (one can actually make the more precise observation that since $$x_n \neq 0$$ for all $$n \in \mathbb{N}$$ under the stated conditions, the inequality mentioned above is actually strict, rendering $$x$$ into a strictly increasing sequence, however that is not essential for the study of convergence).
According to one of the theorems of Weierstrass, any upper-bounded increasing sequence of real numbers is convergent, which applies in particular to $$x$$. Since $$x$$ is defined recursively with respect to the continuous function $$f$$, it follows with necessity that $$\displaystyle\lim_{n \to \infty} x_n$$ must be a fixed point for $$f$$. However it is obvious that there is just one such fixed point - the unique solution to the equation $$f(u)=u \Leftrightarrow u^2=0$$ - which is $$0$$. | 2022-08-18T05:15:03 | {
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http://www.askamathematician.com/2011/05/q-is-0-9999-repeating-really-equal-to-1/ | # Q: Is 0.9999… repeating really equal to 1?
Mathematician: Usually, in math, there are lots of ways of writing the same thing. For instance:
$\frac{1}{4}$ = $0.25$ = $\frac{1}{\frac{1}{1/4}}$ = $\frac{73}{292}$ = $(\int_{0}^{\infty} \frac{\sin(x)}{\pi x} dx)^2$
As it so happens, 0.9999… repeating is just another way of writing one. A slick way to see this is to use:
$0.9999... = (9*0.9999...) / 9 = ((10-1) 0.9999...) / 9$
$= (10*0.9999... - 0.9999...) / 9$
$= (9.9999... - 0.9999...) / 9$
$= (9 + 0.9999... - 0.9999...) / 9 = 9 / 9 = 1$
One.
Another approach, that makes it a bit clearer what is going on, is to consider limits. Let’s define:
$p_{1} = 0.9$
$p_{2} = 0.99$
$p_{3} = 0.999$
$p_{4} = 0.9999$
and so on.
Now, our number $0.9999...$ is bigger than $p_{n}$ for every n, since our number has an infinite number of 9’s, whereas $p_{n}$ always has a finite number, so we can write:
$p_{n} < 0.9999... \le 1$ for all n.
Taking 1 and subtracting all parts of the equation from it gives:
$1-p_{n} > 1-0.9999... \ge 0$
Then, we observe that:
$1 - p_{n} = 1 - 0.99...999 = 0.00...001 = \frac{1}{10^n}$
and hence
$\frac{1}{10^n} > 1-0.9999... \ge 0$.
But we can make the left hand side into as small a positive number as we like by making n sufficiently large. That implies that 1-0.9999… must be smaller than every positive number. At the same time though, it must also be at least as big as zero, since 0.9999… is clearly not bigger than 1. Hence, the only possibility is that
$1-0.9999... = 0$
and therefore that
$0.9999... = 1$.
What we see here is that 0.9999… is closer to 1 than any real number (since we showed that 1-0.9999… must be smaller than every positive number). This is intuitive given the infinitely repeating 9’s. But since there aren’t any numbers “between” 1 and all the real numbers less than 1, that means that 0.9999… can’t help but being exactly 1.
Update: As one commenter pointed out, I am assuming in this article certain things about 0.9999…. In particular, I am assuming that you already believe that it is a real number (or, if you like, that it has a few properties that we generally assume that numbers have). If you don’t believe this about 0.9999… or would like to see a discussion of these issues, you can check this out.
This entry was posted in -- By the Mathematician, Equations, Math. Bookmark the permalink.
### 94 Responses to Q: Is 0.9999… repeating really equal to 1?
1. mathman says:
lol. You need to know what each “number” is before you can take the difference. Of course the difference is zero, because you say that they are the same before you subtract, assuming you are subtracting “numbers”
2. George says:
While the mathematical argument presented here is correct, I have to say that limit and/or mathematical infinity introduced created confusions for many. What I wanted to say is that math infinity is really a math trick. It cannot and should not be real in the real physical world. Today this infinity plagues physics and its calculations. Maybe we need something new…
3. EMINEM says:
go to this https://www.youtube.com/watch?v=wsOXvQn3JuE site and get a mathematical proof that 0.999… doesnot equal to 1….
4. . says:
EMINEM, the video was published on April fool’s day for a reason!
5. Josh says:
The simplest explanation I’ve seen for this is:
1/3 = 0.333…
(1/3)3=1 = (0.333…)3=0.999…
1 = 0.999…
I don’t see how anyone can come to any conclusion other than 1=0.999… after understanding this.
6. mathman says:
1 / 3 = ??
Use long division. There are 2 parts: the quotient and the remainder. For every iteration.
First iteration is 0 R 1/3
Second is 0.3 R 1/30
Third is 0.33 R 1/300
Etc etc.
Where does your remainder go, Josh?
You don’t account for it anywhere.
The remainder is never zero.
How many 9’s are there in your 0.999… ?
7. Josh says:
Hey mathman,
There are infinitely many 9s in my “0.999…”. If it were anything other than infinitely many, then it would be less than 1.
And as far as 1/3, your “etc. etc.” is an infinitely repeating “0.333…”. I’m failing to see how anything you’ve presented has disproved what I said.
8. Angel says:
@mathman:
The remainder does not need to be accounted for. The division 1/3 is never formally complete if there is a remainder. Therefore, one must keep iterating the operations in long division. As the number of iterations n approaches infinity, the remainder does approach zero. That is what matters here. The number of 3s in the decimal expansion of 1/3 is infinite, and so is the number of 9s in the figure 0.999…
9. Anthony.R.Brown says:
The Original author of the Proof below…used in this Thread 🙂
http://www.mathisfunforum.com/viewtopic.php?id=6069
It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1
Because of the Infinite/Recurring 0.1 (0.001…) Difference! 🙂
Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1
Anthony.R.Brown
10. Ángel Méndez Rivera says:
Anthony R. Brown,
Your argument is flawed for the simple reason that the “infinite difference” you describe equivalent to 0.000…1 is not a number. It doesn’t exist in mathematics. It can’t exist in mathematics. It is more of a problematic number than 0/0 or defining a left-identity of exponentiation Phi such that Phi^x=x. It is simply nonsense.
11. Mathman says:
Anthony,
Being someone who has researched and understands both sides of this issue, you will be told it’s nonsense by mathematicians. It’s their way of putting themselves above you to put you down. The issue with your argument is that 0.001… is defined as 0.00111111111111…. You are having a problem defining what the idea in your head is. 0.000…1 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.
12. Anthony.R.Brown says:
Hi Ángel Méndez Rivera
Your try at a counter argument…saying (0.001…) does not exist in mathematics ? can’t exist in mathematics if (0.999…) exist’s which it does! 🙂
The two are the related from the calculation 1 – (0.999…) = (0.001…) one cannot exist without the other!
Anthony.R.Brown
13. Anthony.R.Brown says:
Anthony,
Being someone who has researched and understands both sides of this issue, you will be told it’s nonsense by mathematicians. It’s their way of putting themselves above you to put you down. The issue with your argument is that 0.001… is defined as 0.00111111111111…. You are having a problem defining what the idea in your head is. 0.000…1 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.
Then when I wanted to reply the message above is not shown ? so I will Answer below…
First of all what you are showing as 0.00111111111111…. is wrong! there is only a single (1) following the Infinite/recurring calculation so your example above should look like 0.00000000000001….
The amount of zeros depends on how far one wants to show the calculation regarding decimal places!
Example 0.9… has the infinite/recurring difference 0.1…
Example 0.99… has the infinite/recurring difference 0.01…
Example 0.99… has the infinite/recurring difference 0.001…
So as you can see the infinite/recurring difference is always the same length as the infinite/recurring 0.9 🙂
Anthony.
14. Mathman says:
0.001….. = 0.0011111111111111111…..
Anthony do you know what you are taking about?
15. Anthony.R.Brown says:
Hi Mathman
Yes I fully know what I am talking about but you appear not to ?
As I pointed out in your example ? = 0.0011111111111111111….. it is wrong there can only be a single (1) otherwise when you do the calculation…
0.0011111111111111111….. + 0.0000000000000000009….. it will equal something like = 0.0011111201111111111… ?
Where as the True calculation (with both sides the 0.9… and the 0.1… the same amount of decimal places)
Calculates 0.9999999999…. + 0.0000000001…. = 1
Regards Anthony.
16. Joshua says:
The flaw in your logic is thinking that there is ever space for a 0.0…01. The number 0.999… repeating forever by definition has no room to add anything to it. The 9’s never stop to allow the insertion of any variation of 0.0001. And if there is no room to ever add anything to 0.999… infinitely repeating, then there is no space in between 0.999… infinitely repeating and 1, and therefore they are equal. If the 9’s stopped repeating at any point, you would be correct, but they do not stop repeating, so you are incorrect.
17. Mathman says:
The ellipsis “…” mean “repeating”. I know what you are trying to say but you are using symbols incorrectly.
1/1000… would be a more accurate to convey your idea.
18. Anthony.R.Brown says:
Hi Joshua
Your Quote:”The number 0.999… repeating forever by definition has no room to add anything to it”
A.R.B
The same as “The number 0.001… repeating forever by definition has no room to add anything to it”
They both run side by side! 🙂
0.99999999999999999999999999999999999999999999999999999999….
0.0000000000000000000000000000000000000000000000000001….
It’s only when one want’s to make the Value (1) that the + Calculation is needed it’s the only way!
( 0.9999999999….) + (0.0000000001….) = 1
19. Joshua says:
Sorry Anthony, you have not wrapped your head around what 0.999… repeating for infinity means. It’s the infinite part that’s really important. I promise you, you are wrong, and once you understand what infinitely repeating means, you’ll get it.
20. Ángel Méndez Rivera says:
Anthony, no. They don’t run side by side. The number you describe as the difference between 1 and 0.999… doesn’t exist, because by definition, it is impossible to have a 1 after an infinite amount of zeros. It just is. There is no such calculation as the one you performed to be made because the number you describe doesn’t exist.
21. Mathman says:
There is an undefined infinitesimal space between 0.999… and 1. Just because it’s undefined doesn’t mean it doesn’t exist.
22. Ángel says:
Mathman,
1. If it is undefined, then it does not exist. One implies the other.
2. No, there is no infinitesimal space between 0.999… = 1, and this can be proven.
0.999… = 9/10 + 9/100 + 9/1000 + … this is an infinite geometric series, which is evaluated and in fact defined by the formula a/(1 – r), so the sum 0.999.. = (9/10)/(1 – 1/10) = (9/10)/(9/10) = 1. As for the formula a/(1 – r), it can be easily derived and proven
23. Anthony.R.Brown says:
( 1.1 ) x 0.9 = 0.99 ” One Decimal Place = ” 0.01 < 1
( 1.11 ) x 0.9 = 0.999 " Two Decimal Place's = " 0.001 < 1
( 1.111 ) x 0.9 = 0.9999 " Three Decimal Place's = " 0.0001 < 1
( 1 / 0.9 ) x 0.9 = 0.9… " Infinite Decimal Place's = " 0.1… < 1 "
24. Joshua says:
Mathman,
You would be correct if the repeating 0.999… ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.
Most of the confusion over this matter seems to be about the nature of infinity, and how it applies to infinitely repeating decimals. The importance of wrapping our heads around what “infinitely repeating” means cannot be overstated.
25. Ángel Méndez Rivera says:
Anthony,
(1.1)(0.9) = 0.99
(1.11)(0.9) = 0.999
(1.111)(0.9) = 0.9999
(1.111…)(0.9) = 0.999…. = (1/0.9)(0.9), because 1.1111… = 1/0.9. But any number divided by itself equals 1, so (0.9)/(0.9) = 1, therefore 1 = 0.9999….
26. Anthony.R.Brown says:
Quote: “Joshua says:
You would be correct if the repeating 0.999… ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.”
A.R.B
Absolutely Correct!!! So now you and I agree it’s impossible for 0.999…to ever stop and become equal to 1 🙂
So how do we make 0.999… = 1 and that’s where we have the answer!…
( 0.999…) + (0.001…) = 1
{When great minds think alike Mountains can be moved!} 🙂
27. Joshua says:
No, Anthony, you still don’t understand. The fact that it never stops is the very thing that makes it equal to 1. Again, I come back to encouraging you to take the time to actually understand what an infinitely repeating decimal is.
28. Anthony.R.Brown says:
Hi Joshua
What you don’t seem to understand is that infinite/recurring 0.999… is a Calculation on a Number! and the number Starts as 0.9 (It’s not something that is just plucked from the sky from nothing!)
Your biggest mistake! is “The fact that it never stops is the very thing that makes it equal to 1” ? a Number that Starts < 1 can never be made to equal 1 unless the + Calculation is used!
29. Joshua says:
Anthony, the number 0.999… with infinite 9s is not a calculation. It is a static number. The 9s are not constantly being generated, they exist (all infinity of them) at once. It does not “start” at 0.9. It exists as 0.999… with an infinite number of 9s. You don’t have to work up to it, it just is.
It is clear that you still do not understand what an infinitely repeating decimal is. You do not understand the basic premise of what we are talking about. Therefore, you can not speak to it in an intelligent or useful manner.
I’m not trying to be mean or anything, either. I appreciate people pointing out when I am wrong so I can learn what is correct. You clearly don’t understand the basics of this topic, therefore you are wrong, and it is an opportunity for you to learn something new and broaden your knowledge base.
30. Angel says:
“What you don’t seem to understand is that recurring 0.999… is a calculation on a number!”
This is wrong. 0.999… is a number. It isn’t a process, it already is a number. It doesn’t “start” out as 0.9. It already is and exists as 0.999…, it doesn’t start anywhere or end anywhere. It isn’t a changing or moving thing. It just is a number.
31. Anthony.R.Brown says:
So Joshua & Angel…
Are both wrong! if they think (the number 0.999… with infinite 9s is not a calculation) ???
Because all the so called Proofs ? are based on some kind’s of Calculations!!!
Just to back that up…a Google search for 0.999… proofs shows (About 367,000 results) 21/09/2017
And they are all some kind’s of Calculations! 🙂
One fine example is shown below…with many Calculations!
Anthony.R.Brown
32. Mathman says:
Ángel,
a/(1-r) comes from a limit. The LIMIT of the infinite series is 1. The definition is flawed and contradicts the very theory of limits.
It may be impossible to have a 1 AFTER an “infinite amount”. That is because “infinite amount” doesn’t exist. Stop trying to quantify infinity. It is not impossible to describe the idea, however: 1/1000…
The very definition of limits show there is ALWAYS AN EPSILON.
n=1: 1-0.9= 0.1
n=2: 1-0.99 = 0.01
For ALL values n, there is a difference, or, contradictory to Anthony, there is always a space. If you travserse 90% of the remaining distance , ad infinitum, there will always be 10% of the last distance ahead of you. You’ll never complete the journey of say, 1 mile, going 90% each time.
I argue that “0.999…” is not a number at all.
Σ (n=1 to N) 9/10^n as N goes to infinity, implies there is always a rational number being added to the previous total. The number 1 has nothing being addd to it. Therefore they aren’t the same.
Also, ‘something’ that always has another addend can’t be a number. It’s contradictory because it always has another number being added to it.
33. Mathman says:
Thus: 1-0.999(n-times) = 1/10^n
You may let n get as arbitrarily large as you like.
You cannot have “infinite amount of 9’s” though.
You can have the LIMIT as the number of 9’s gets arbitrarily large. This is a number. “Endless 9’s” really is nonsense when it comes to a number which should really be finite.
So then, for ALL n in N, that is, for every integer, there is a difference which is smaller than the last and never 0.
Learn what arbitrarily large means, and how limits really work, and what infinite means, and how infinity is applied in calculus. Infinity and limits go hand in hand. You don’t plop infinity wherever you feel.
34. Anthony.R.Brown says:
LIMIT Cannot be used with Infinite/Recurring 0.9 (0.999…) because you will then Contradict the term (Infinite/Recurring) by forcing the .9’s to cut off at some point depending on where you want the LIMIT to be, regarding decimal places calculated,this is much the same as the cut off point used when Infinite/Recurring 0.9 (0.999…) is forced at some point to become equal to 1
So back to…
It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1 (Unless you use a LIMIT) ? 🙁
And the Infinite/Recurring 0.1 (0.001…) Difference! (Also does not us a LIMIT)
Both are LIMITless 🙂
Only by the using the + Calculation can it be possible for (0.999…) + (0.001…) = 1
35. Ángel Méndez Rivera says:
No, we do not get to choose where the limit is. The limit is a number larger than every element of the sequence of partial evaluations of a function, which means that the limit is larger than the sequence itself. In other words, the term of the sequence which equals the limit is itself located at an infinite position of the sequence. Any finite evaluation yields less than the limit. However, 0.99… has an infinite number of 9s, so it exactly equals the limit. We’re not cutting it off to equal 1, you’re cutting it off to equal less than 1, but you’re Not actually using the sequence of infinite nines, so you’re argument is wrong.
There is no such a thing as limitless in mathematics, and 0.000…1 is nonsense and does not exist.
36. Ángel Méndez Rivera says:
Mathman,
Read my previous response to your argument. I already explained why your description of the definition of a limit is wrong and why the idea of there always being an epsilon is irrelevant when actually evaluating the limit of a function as the argument approaches c.
37. Anthony.R.Brown says:
Quote: “Ángel Méndez Rivera says:
September 22, 2017 at 1:50 pm
No, we do not get to choose where the limit is.
A.R.B
Yes! you do by using it in the first place!
Quote: “Ángel Méndez Rivera says:
“However, 0.99… has an infinite number of 9s, so it exactly equals the limit. ”
A.R.B
How can an infinite number of 9s, exactly equals the limit (You have set!) which is 1 ? which is an Integer number > 0.99…
Quote: “Ángel Méndez Rivera says:
There is no such a thing as limitless in mathematics, and 0.000…1 is nonsense and does not exist.
A.R.B
Infinite/Recurring 0.9 (0.999…)
Infinite/Recurring 0.1 (0.001…)
Are both (Individually!) LIMITless they only have a LIMIT when added! together to equal 1
(0.999…) + (0.001…) = 1
38. The Physicist says:
@Anthony.R.Brown
They’re not using the word “limit” the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.
39. Anthony.R.Brown says:
The Physicist says: Quote:”They’re not using the word “limit” the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.”
A.R.B
The word (LIMIT) however you define it ? or look it up on some Math sites ? is being used in a way… and the same way ? so as to make a Calculated Number,that starts life less than 1 somehow become equal to 1 ? 🙁
A better way to look at the problem is…
Infinite/Recurring 0.9 (0.999…) = Universe! 🙂
And if we place a Wall at the End of the Universe! ? there will still be Universe! behind the Wall…
0.9 (0.999…) Universe! { Wall = 1} 0.9 (0.999…) Universe!…
40. Ángel Méndez Rivera says:
Anthony, no. There is no such a thing as a limitless number. Please, study the topic you’re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.
41. Ángel says:
Mathman,
There is a justified reason why challenging mathematics makes you a crackpot: we can prove that by challenging mathematics, you are wrong.
2. Euler did not define limits.
3. Also, a definition cannot be wrong. This is, ironically, the definition of a definition. So Euler’s definition of infinite summation is acceptable. Euler’s definition renders summation continuous in the entire complex plane, which is what it should be according to group theory.
4. It makes perfect sense to add an infinite number of things. We do it all the tim in physics, and we get observable results that we do observe consistently.
5. No, your term-by-term subtraction was done wrong. Subtraction is done digitwise right to left.
6. 0.999… is not a bad way to represent 9/10 + 9/100 + …, but rather it is the only to represent it since it is the very definition. It is how decimal representation works. It can’t also be represented as Sum[m = 1, m= n —> Infinity](9*10^(-m)), which can be shown to equal 1 since the sum is 0.9(1 – 10^[-n])/(1 – 1/10), and when n —> Infinity, this expression simplifies to 1. This is the limit.
7. Yes, I agree that f(x = c) does not equal lim x—>c [f(x)]. However, 0.999… isn’t equal to f(x=c), but rather equal to lim x—>c [f(x)]. 0.999…. IS the limit of the partial sums by the very definition of decimal representation, and the partial sum limits to 1 evaluated by Taylor-geometric series. By transitive property, 0.999… = 1.
Remember: infinite sums are a thing. Irrational numbers prove it. It has been proven. It isn’t a definition to say “we can add infinitely many things”. No, we can prove it mathematically. That being said, 0.999… = 9/10 + 9/100 + … by definition. Not debatable. Now, 9/10 + 9/100 + … = Sum [m = 1, lim n —> Infinity][9*10^(-m)]. This latter expression is a limit. Keep in mind that summation is a linear hyper-operator, which makes f(y) = Sum[k=x, y][g(k)] continuous for all continuous g. Now, also remember that by the very definition of continuity, lim (x —> c) f(x) = f(c). Hence, Sum [m = 1, n lim n —> Infinity][9*10^(-m)] = Sum[m = 1, Infinity][9*10^(-m)]. This implies that 0.999… is actually a valid sum and therefore a valid decimal representation. You can add infinitely many terms. Finally, the latter left hand side of the previous identity mentioned is equal to lim (n —> Infinity) 9/10[1 – 10^(-n)]/(1 – 1/10) = (9/10)/(9/10) = 1. By the transitive property, 0.999… = 1.
42. Jesse Sherer says:
0.999… is or it isn’t equal to 1. if you are x distant to our Universe you are not in our Universe. no 3 dimentional universe is infinite unless all universe could be infinite too. 1-x=0.9999…. if you are x distant from our universe you are not in our universe. you might say x doesn’t exist, but, nether does 1/9 in base ten math. base ten is a rational universe of numbers there can be no number that equates to almost 1 but not 1 ,,or almost 1 is 1 in a rational universe.. at some point you have to say that if an infinite string of numbers representing both a fraction of a number and a whole number, then there must exist an infinity small number also exists that both equals 0 and almost 0 too. so x is more rational of a number that can add to .999999… and get 1.. base ten can’t solve it but logic can.. 0.9999.. is a number or it isn’t x is a number or it isn’t 1-x=0.9999…. or it doesn’t 1=0.99999… or it doesn’t
43. Anthony.R.Brown says:
Jesse Sherer says:
December 1, 2017 at 6:41 am
A.R.B
= Nonsense! 🙁
0.9999.. is a number Less than 1 🙂
44. Anthony.R.Brown says:
Ángel Méndez Rivera says:
September 24, 2017 at 5:02 pm
Anthony, no. There is no such a thing as a limitless number. Please, study the topic you’re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.
A.R.B
Totally Wrong! Give me any Number you want ? and I will add any Number I want to it! Which makes any Number you provide {a limitless number} 🙂 | 2018-02-25T13:28:36 | {
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https://stats.stackexchange.com/questions/186111/find-the-rotation-between-set-of-points/207407 | # Find the rotation between set of points
I have two sets (sourc and target) of points (x,y) that I would like to align. What I did so far is:
• find the centroid of each set of points
• use the difference between the centroids translations the point in x and y
What I would like is to find the best rotation (in degrees) to align the points.
Any idea?
M code is below (with plots to visualize the changes):
# Raw data
## Source data
sourc = matrix(
c(712,960,968,1200,360,644,84,360), # the data elements
nrow=2, byrow = TRUE)
## Target data
target = matrix(
c(744,996,980,1220,364,644,68,336), # the data elements
nrow=2, byrow = TRUE)
# Get the centroids
sCentroid <- c(mean(sourc[1,]), mean(sourc[2,])) # Source centroid
tCentroid <- c(mean(target[1,]), mean(target[2,])) # Target centroid
# Visualize the points
par(mfrow=c(2,2))
plot(sourc[1,], sourc[2,], col="green", pch=20, main="Raw Data",
lwd=5, xlim=range(sourceX, targetX),
ylim=range(sourceY, targetY))
points(target[1,], target[2,], col="red", pch=20, lwd=5)
points(sCentroid[1], sCentroid[2], col="green", pch=4, lwd=2)
points(tCentroid[1], tCentroid[2], col="red", pch=4, lwd=2)
# Find the translation
translation <- tCentroid - sCentroid
target[1,] <- target[1,] - translation[1]
target[2,] <- target[2,] - translation[2]
# Get the translated centroids
tCentroid <- c(mean(target[1,]), mean(target[2,])) # Target centroid
# Visualize the translation
plot(sourc[1,], sourc[2,], col="green", pch=20, main="After Translation",
lwd=5, xlim=range(sourceX, targetX),
ylim=range(sourceY, targetY))
points(target[1,], target[2,], col="red", pch=20, lwd=5)
points(sCentroid[1], sCentroid[2], col="green", pch=4, lwd=2)
points(tCentroid[1], tCentroid[2], col="red", pch=4, lwd=2)
• I cannot read your code, but the operation you need is called Procrustes rotation. Have you heard of it? It works when points are already paired ($x_i,y_i$). Pre-rotation optional operations include translation and scaling, and optional post-rotational isoscaling. – ttnphns Dec 10 '15 at 17:13
• A complex regression will do the job. – whuber Dec 10 '15 at 17:23
• I've seen, that, rotating the system about 180 degrees, then the pairs $(a,C),(b,D),(c,A),(d,B)$ become neighbours - and this is even a better fit than the best fit of the original $(a,A),(b,B),(c,C),(d,D)$ (where the small letters stand for vector source and capital letters for vector target) I've not seen this possibility mentioned and explicitely allowed or disallowed. Are you sure that you don't want that better fit? – Gottfried Helms Sep 4 '16 at 14:05
This can be done using the Kabsch Algorithm. The algorithm finds the best least-squares estimate for rotation of $RX-Y$ where $R$ is rotation matrix, $X$ and $Y$ are your target and source matrices with 2 rows and n columns.
In [1] it is shown that this problem can be solved using singular value decomposition. The algorithm is as follows:
1. Center the datasets so their centroids are on origin.
2. Compute the "covariance" matrix $C$=$XY^T$.
3. Obtain the Singular Value Decomposition of $C=UDV^T$.
4. Direction adjustment $d=sign(det(C))$.
5. Then the optimal rotation $R=V\left( \begin{array}{ccc} 1 & 0 \\ 0 & d \\ \end{array} \right)U^T$
I don't know of any implementation in R so wrote a small function below.
src <- matrix(c(712,960,968,1200,360,644,84,360), nrow=2, byrow=TRUE)
trg <- matrix(c(744,996,980,1220,364,644,68,336), nrow=2, byrow=TRUE)
Kabsch algorithm in an R funtion:
kabsch2d <- function(Y, X) {
X <- X-rowMeans(X)
Y <- Y-rowMeans(Y)
C <- X %*% t(Y)
SVD <- svd(C)
D <- diag(c(1, sign(det(C))))
t(SVD$v) %*% D %*% t(SVD$u)
}
Center the points:
src <- src-rowMeans(src)
trg <- trg-rowMeans(trg)
Obtain rotation:
rot <- kabsch2d(src, trg)
Result (black - original source, red - original target, green - rotated target)
plot(t(src), col="black", pch=19)
points(t(trg), col="red", pch=19)
points(t(rot %*% trg), col="green", pch=19)
• +1. However, the answer could be further much better if you included discourse about how the algo is related to well-known Procrustes rotation problem. – ttnphns Mar 19 '16 at 14:03
I've done this with an iterative optimum-search, and tested 2 versions.
I've taken the original arrays and centered them calling this arrays cSRC and cTAR . Then I've done a loop with angles $\varphi$ between $0$ and $2 \pi$ , and for each angle I computed the error-criterion using the difference between the rotated $\small D=rot(\text{cSRC} ,\varphi)- \text{cTAR}$.
1. In version 1) I took as criterion the sum-of-squares of all entries in $\small D$ as $$err_1 = \small \sum_{k=1}^4 \small((D_{k,1})^2+(D_{k,2})^2)$$and the angle $\small \varphi$ at which the minimal error occured is equivalent the kabsch2d-procedure in @Karolis' answer.
2. In version 2) I took as criterion the sum of the absolute distances, that means, the sum $$err_2=\small \sum_{k=1}^4 \small\sqrt{(D_{k,1})^2+(D_{k,2})^2}$$ and got a slightly different rotation angle $\small \varphi$ for the smallest error.
I don't know, which criterion fits your needs better.
Here are some results from the protocol.
$$\small \begin{array} {r|cc} & \text{version } 1 & \text{version } 2\\ \hline \varphi & -0.04895304& -0.05093647 \\ \text{rotation} & \begin{bmatrix} 0.99880204& -0.04893349\\ 0.04893349& 0.99880204\\ \end{bmatrix} & \begin{bmatrix} 0.99870302 & -0.05091444\\ 0.05091444 & 0.99870302\\ \end{bmatrix} \\ \text{distances} & \begin{bmatrix} -6.80077266 & -0.86209739\\ 2.79924551 & -9.33782500\\ -0.61309522 & 6.94156520\\ 4.61462237 & 3.25835719\\ \end{bmatrix} & \begin{bmatrix} -6.78017751& -0.37062404 \\ 3.35787307 & -9.36574874 \\ -1.16459115 & 6.95324527 \\ 4.58689559 & 2.78312752 \\ \end{bmatrix} \end{array}$$ | 2020-09-27T15:24:37 | {
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https://math.stackexchange.com/questions/764437/the-gaussian-integers-are-isomorphic-to-mathbbzx-x21/764482 | # the Gaussian integers are isomorphic to $\mathbb{Z}[x]/(x^2+1)$
I am trying to prove that $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$.
My initial plan was to use the first isomorphism theorem. I showed that there is a map $\phi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$, given by $\phi(f)=f(i).$ This map is onto and homorphic. The part I have a question on is showing that the $ker(\phi) = (x^{2}+1)$.
One containment is trivial, $(x^2+1)\subset ker(\phi)$. To show $ker(\phi)\subset (x^2+1)$, let $f \in ker(\phi)$, then f has either $i$ or $-i$ as a root. Sot $f=g(x-i)(x+i)=g(x^2+1).$ How can I prove that $f \in \mathbb{Z}[x]\rightarrow g \in \mathbb{Z}[x]$?
• It is enough to have $f=g(x^2+1)$, because this says $f\in (x^2+1)$, and we are done. – Dietrich Burde Apr 22 '14 at 13:34
• By the division algorithm, $f = q\cdot (x^2+1) + r$ with $\deg r < 2$. Then you just need to check that $r(i) = 0$ implies $r = 0$ if $\deg r < 2$. (Note: $\deg 0 = -\infty$) You could also appeal to Gauß' lemma. – Daniel Fischer Apr 22 '14 at 13:35
• @DanielFischer, I think that is the answer I am looking for since $g \in \mathbb{R}[x]$ which is a euclidean domain and I am allowed to make that claim. Do you care to expand on that? As in, how can I be certain that the coefficients of $q$ are in $\mathbb{Z}$? – kslote1 Apr 22 '14 at 13:57
• Don't use $\mathbb{R}[x]$ here, use $\mathbb{Q}[x]$. But you need never leave $\mathbb{Z}[x]$ even potentially. The point is that $x^2+1$ is monic, i.e. has lead coefficient $1$. Thus if you do polynomial division, you always get an integer coefficient, and the existence of $q,r\in\mathbb{Z}[x]$ with $f = q\cdot (x^2+1) + r$ and $\deg r < 2$ follows. – Daniel Fischer Apr 22 '14 at 14:06
• As an aside, this is one of those problems where it's easier to show the isomorphism directly, by writing down a homomorphism and its inverse, rather than the indirect route of finding a surjective homomorphism with zero kernel. – Hurkyl Aug 9 '17 at 16:40
Let me elaborate on Daniel Fischers comment. You have a ring homomorphism $\phi: \mathbb Z[x]\to\mathbb Z[i]$ given by $x\mapsto i$. Take $f \in \ker \phi$. By the division algorithm, $$f = q\cdot(x^2+1) + r,$$ where $\deg r < 2$ and $q,r\in\mathbb Z[x]$, since $x^2+1$ is monic. Applying $\phi$ to this equation yields $$0 = \underbrace{\phi(q)\cdot(i^2+1)}_0 + \phi(r).$$ Since $r$ is of degree $<2$, we can write $r = ax+b$ for some $a,b\in\mathbb Z$. Then $\phi(r)=0$ gives $$ai+b=0.$$ This equation in $\mathbb Z[i]$ implies $a=b=0$, so we have $r=ax+b=0\in\mathbb Z[x]$ and therefore $$f = q\cdot (x^2+1) \in \langle x^2+1\rangle.$$ We conclude that $\ker \phi \subseteq \langle x^2+1\rangle$.
When quotient out an ideal, we consider what happens to the ring when all the elements in the ideal are considered as identity elements.
Now if $x^2+1=0\Rightarrow x=\pm i$ let us take the "+" root.
$\mathbb{Z}[x]/(x^2+1)=\{f\in\mathbb{Z}[x]\,|x^2+1=0\}=\{f\in\mathbb{Z}[x]\,|x=i\}=\{a+bi|a,b\in\mathbb{Z}\}=\mathbb{Z}[i]$
I.e they are isomorphic.
I have answered a similar question here Is this quotient Ring Isomorphic to the Complex Numbers
You can define your isomorphism as follows
$\phi:\mathbb{Z}[i]\rightarrow \mathbb{Z}[x]/\langle x^2+1 \rangle$
$\phi(a+bi)=(a+bx)+\langle x^2+1 \rangle$
If you compute $[a+bx+\langle x^2+1 \rangle]\times [c+dx+\langle x^2+1 \rangle]$
We get $ac+(ad+bc)x+bdx^2+\langle x^2+1 \rangle$. We reduce by long division to get,
$(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.
Therefore,
$\phi([a+bi][c+di])=(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.
Which is exactly what you want.
The map is obviously bijective and a ring homomorphism. | 2019-05-22T01:06:28 | {
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https://math.stackexchange.com/questions/23809/what-is-the-remainder-of-16-is-divided-by-19/23810 | # What is the remainder of $16!$ is divided by 19?
Can anyone share me the trick to solve this problem using congruence?
Thanks,
Chan
If you've seen Wilson's theorem then you know the remainder when $18!$ is divided by $19$. To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$. Do you know how to find these? (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product. See also Bill Dubuque's answer.)
• I knew Wilson's Theorem, but really don't know how to find the inverse of 17, 18 mod 19. What's the inverse? Can you give me one example? Thanks.
– Chan
Feb 26 '11 at 6:17
• @Chan: $\rm 17 \equiv -2,\ 18\equiv -1$ Feb 26 '11 at 6:21
• @Chan: A general technique to find the inverse of $k$ mod $n$ when $k$ and $n$ are relatively prime is to use integer division and the Euclidean algorithm to find $a$ and $b$ such that $ak + bn = 1$. Since $bn$ is a multiple of $n$, $ak$ is congruent to $1$ mod $n$, and thus the congruence class of $a$ is the inverse of the congruence class of $k$. In this particular case, a further hint to make things easier is that $18\equiv -1$ and $17\equiv -2$ mod $19$. (This makes computations easier for inverting their product.) Feb 26 '11 at 6:21
• Many thanks, I got it now ;)
– Chan
Feb 26 '11 at 6:24
Hint By Wilson's theorem $$\bmod 19\!:\ \overbrace{{-}1}^{\large \color{#0a0}{18}} \equiv 18! \equiv\!\! \overbrace{18\cdot17}^{\large \color{#c00}{(-1)(-2)}}\!\!\cdot 16!\,$$ $$\Rightarrow\,16!\equiv \dfrac{\color{#0a0}{18}}{\color{#c00}2}\equiv 9$$
Generally (Wilson reflection formula) $$\rm\displaystyle\ (p\!-\!1\!-\!k)!\equiv\frac{(-1)^{k+1}}{k!}\!\!\!\pmod{\!p},\,$$ $$\rm\:p\:$$ prime
• @Bill Dubuque: Thank you!
– Chan
Feb 26 '11 at 6:24
• @Bill Dubuque: If I have $2.16! \equiv 18 \pmod{19}$. Can I cancel out the $2$ from both sides?
– Chan
Feb 26 '11 at 6:42
• @Chan: $\rm\ 2\:x\equiv 2\:y\ (mod\ 19)\ \iff\ 19\ |\ 2\ (x-y)\ \iff\ 19\ |\ x-y\ \iff\ x\equiv y\ (mod\ 19)$ Feb 26 '11 at 6:50
• @Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \equiv ay \pmod{p} \implies x \equiv y \pmod{p}$, right?
– Chan
Feb 26 '11 at 7:07
• @Chan: $\rm\ n,m\$ coprime $\rm\iff a\ n + b\ m = 1\$ for some $\rm\:a,b\:\ \iff\ a\ n\equiv 1\ (mod\ m)\:,\:$ for some $\rm\:a\:$ $\rm\iff\ n$ is invertible/cancellable $\rm\:(mod\ m)$ Feb 26 '11 at 7:12
I almost think in this case it is just faster to break it down than calculate the inverses: First the factors between 1 and 10:
$$10!= (2\cdot 10)\cdot (4\cdot5)\cdot(3\cdot6)\cdot(7\cdot8)\cdot 9\equiv 1\cdot 1\cdot (-1)\cdot(-1)\cdot 9\equiv9$$
Now we have
$$16!\equiv 9\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15\cdot 16\equiv9\cdot(-8)\cdot(-7)\cdot(-6)\cdot(-5)\cdot(-4)\cdot(-3)$$
$$\equiv 9\cdot(4\cdot5)\cdot(6\cdot3)\cdot(7\cdot8)\equiv 9\cdot(1)\cdot(-1)\cdot(-1)\equiv 9$$
Of course this doesn't generalize, but the computation is faster than finding 3 inverses and multiplying them. (Again of course that is not true for larger $n$...)
A more explicit answer would go like this:
By Wilson's Theorem, $18!$ is congruent to $-1 \pmod {19}$
$$18! = (18\cdot 17)(16!)$$
Then $(18\cdot 17)(16!)$ is congruent to $-1 \pmod{19}$
But note that $18$ is congruent to $-1 \pmod{19}$ and $17$ is congruent to $-2 \pmod{19}$
Then it follows that $(18\cdot 17)(16!)$ is congruent to $((-1)\cdot (-2))(16!)$ is congruent to $-1 \pmod{19}$
Simplifying, we get, $2\cdot(16!)$ is congruent to $-1 \pmod{19}$. But we need the remainder of $16!$, not $2\cdot 16!$. and $-\frac12$ isn't a valid answer.
Then also notice that $18$ is congruent to $-1 \pmod{19}$. Then $2\cdot 16!$ is congruent to $18 \pmod{19}$ by the fact that if $a$ is congruent to $b \pmod{p}$, then $b$ is congruent to $a \pmod{p}$
So we have that $2\cdot 16!$ is congruent to $18 \pmod{19}$ and from there, its just a matter of dividing both sides by $2$ to get $16!$ is congruent to $9 \pmod{19}$ | 2021-12-07T10:04:58 | {
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https://gcs-group.ro/hydrogenation-of-kdrqq/9c9f68-parent-functions-and-transformations | 3.2 Graphing Quadratic Functions ... 3.5 Transformations of Functions II. Section 1.1 Parent Functions and Transformations 5 Describing Transformations A transformation changes the size, shape, position, or orientation of a graph. In-Class Notes Notes Video Worksheet Worksheet Solutions . It has been clearly shown in the below picture. Now, let us come to know the different types of transformations. Furthermore, all of the functions within a family of functions can be derived from the parent function by taking the parent function’s graph through various transformations. This is the currently selected item. The six most common graphs are shown in Figures 1a-1f. Start studying Parent Functions and Transformations (Ullman). A translation is a transformation that shifts a graph horizontally and/or vertically but does not change its size, shape, or orientation. The functions shown above are called parent functions. Which of the following is the graph of y = 1/(x +2) ? Horizontal Expansions and Compressions. A parent function is the simplest function of a family of functions. Which of the following is the graph of |x+2|+2? 3.1 Completing the Square. Linear—vertical shift up 5. The rule we apply to make transformation is depending upon the kind of transformation we make. Vertical Translation 3. Parent Functions and transformations. Which equation is a quadratic function reflected over the x-axis and shifted up 2. Transformations of ParentTransformations of Parent FunctionsFunctions 2. based on the parent function, the function will be vertically stretched by a factor of 3, it will be reflected over the x-axis and will move up the y-axis two units Describe the transformation… Which of the following is the graph of y = 1/x - 2 ? In this section, we will explore transformations of parent functions. 1-5 Bell Work - Parent Functions and Transformations. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. State the transformations and sketch the graph of the following functions. The different types of transformations which we can do in the parent functions are 1. A parent function is the simplest of the functions in a family. Function Transformations. There are many different type of graphs encountered in life. Some of the worksheets for this concept are Transformations of graphs date period, Transformations of functions name date, Algebra ii translations on parent functions review, Work parent functions transformations, 1 graphing parent functions and transformations, Y ax h2 k, Graphical transformations … The parent function of a quadratic is f (x) = x ². a Which of the following is the graph of y = 1/x +2 ? Name_ Date _ For problem 1- 6, please give the name of the parent function and describe the transformation represented. More clearly, on what grounds is the transformation made? To know that, we have to be knowing the different types of transformations. This graph is known as the "Parent Function" for parabolas, or quadratic functions.All other parabolas, or quadratic functions, can be obtained from this graph by one or more transformations. Geo 2.8 Parallel and Perpendicular Slopes. Which of the following is the graph of y = 2. Label … Examples inculde: (line with slope 1 passing through origin) (a V-graph opening up with vertex at origin) (a U-graph opening up with vertex at origin) … Identifying function transformations. Start studying Parent Functions and Transformations. Yes, there is a pre-decided rule to make each and every transformation. 1-5 Slide Show - Parent Functions and Transformations PDFs. Practice: Identify function transformations. Unit 3: Parent Functions . Absolute value—vertical shift down 5, horizontal shift right 3. In Mathematics II, students reasoned about graphs of absolute value and quadratic functions by thinking of them as transformations of the parent functions |x| and x². Parent Functions And Transformation - Displaying top 8 worksheets found for this concept.. Again, the “parent functions” assume that we have the simplest form of the function; in other words, the function either goes through the origin \left( {0,0} \right), or if it doesn’t go through the origin, it isn’t shifted in any way.When a function is shifted, stretched (or compressed), or flipped in any way from its “parent function“, it is said to be transformed, and is a transformation of a function.T-charts are extremely useful tools when dealing with transformations of functions… Parent Functions and Transformations Reference BookThis reference book was created to use as a review of transformations and the following function families: linear, absolute value, quadratic, cubic, square root, cube root, exponential, logarithmic, and … Title: Parent Function Transformation 1 Parent Function Transformation. In-Class Notes Notes Video Worksheet Worksheet Solutions Homework HW Solutions. A quadratic function moved left 2. A quadratic function moved right 2. Rigid transformations change only the position of the graph, leaving the swe and shape unchanged. A very simple definition for transformations is, whenever a figure is moved from one location to another location, This fascinating concept allows us to graph many other types of functions, like square/cube root, exponential and logarithmic functions. View 2 Transformation HMWK-1.pdf from HIST 3315 at Wingate University. a transformation occurs. 287 #73-75, 79-81 . 13. Identifying function transformations. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Square Root —vertical shift down 2, horizontal shift left 7. The rule that we apply to make transformation using reflection and the rule we apply to make transformation using rotation are not same. C. A. which of the following is linear? The different types of transformations which we can do in the parent functions are, 5. Which of the following is the graph of x = 2? D. B. C. A. which of the following is cubic? Graphing and Describing Translations Graph g(x) = x − 4 and its parent function. The Parent Function is the simplest function with the defining characteristics of the family. When identifying transformations of functions, this original image is called the parent function. 1-5 Online Activities - Parent Functions and Transformations. Q. Here are some simple things we can do to move or scale it on the graph: A transformation where the pre-image and image are congruent is called a rigid transformation or an isometry. In this point, always students have a question. Learn vocabulary, terms, and more with flashcards, games, and other study tools. ... Wansformations Transformations of a parent function can affect the appearance of the parent graph. D. B. C. A. which of the following is quadratic? 6 Module 1 – Polynomial, Rational, and Radical Relationships Parent Function Worksheet # 1- 7 Give the name of the parent function and describe the transformation represented. The "Parent" Graph: The simplest parabola is y = x 2, whose graph is shown at the right.The graph passes through the origin (0,0), and is contained in Quadrants I and II. Sample Problem 2: Given the parent function and a description of the transformation, write the equation of the transformed function!". So, for each type of transformation, we may have to apply different rule. A square root function moved right 2. 1-5 Assignment - Parent Functions and Transformations. What is a parent function? Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Horizontal Translation 2. Which of the following is the graph of y = -1? Learn vocabulary, terms, and more with flashcards, games, and other study tools. Horizontal Expansions and Compressions 6. These transformations include horizontal shifts, stretching, or compressing vertically or horizontally, reflecting over the x or y axes, and vertical shifts. 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https://math.stackexchange.com/questions/2126888/how-to-prove-that-sum-n-textodd-fracn24-n22-pi2-16 | # How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$?
The series:
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$
showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the series was calculated.
How do I prove this directly?
• integrate $\frac{z^2}{(4-z^2)^2}\tan(z)$ over a big circle in the complex plane – tired Feb 3 '17 at 8:28
Wolfram Alpha gives me the partial fraction expansion:
$$\frac{n^2}{(4 - n^2)^2} = \frac{1}{8}\left(\frac 1{n-2}-\frac{1}{n + 2}\right) + \frac{1}{4}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)$$
So the first part telescopes, and the second part will be some modified version of $\zeta(2)$.
In more detail: \begin{align}\sum_{n\text{ odd}} \left(\frac 1{n-2}-\frac{1}{n + 2}\right)&=\frac{1}{-1}-\frac{1}{3}+\frac{1}{1}-\frac{1}{5}+\frac{1}{3}-\frac{1}{7}+\cdots\\ &=-1+1=0\end{align} since all the other terms cancel out.
And \begin{align}\sum_{n\text{ odd}}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)&=\frac{1}{(-1)^2}+\frac{1}{3^2}+\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{3^2}+\frac{1}{7^2}+\cdots\\ &=2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\right) \end{align}
It's a famous result that $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots=\frac{\pi^2}{8}$. You can can prove it if you know:
$$\frac{\pi^2}{6}=\zeta(2) = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$$
and thus $$\zeta(2)-\frac1{2^2}\zeta(2) = \frac{1}{1^2}+\frac{1}{3^3}+\frac{1}{5^2}+\cdots$$
So $$\sum \frac{n^2}{(4-n^2)^2}=0 + \frac{1}{4}\cdot 2\cdot \frac{\pi^2}{8}=\frac{\pi^2}{16}$$
• Hi Thomas. I had assumed that the sum over odd integers included the negative ones. Naturally, the answer in that case is twice the answer if the summation extends to the positive odds only. Anyway, (+1) for your well-written post. -Mark – Mark Viola Feb 3 '17 at 4:30
• @Dr.MV Yeah, I wondered about that. I might have read it your way if it had been $\sum_{n\text{ odd}}$ rather than $\sum_{n\text{ odd}}^{\infty}$. – Thomas Andrews Feb 3 '17 at 4:37
• Ah, yes. I see your point. Well, I did address the issue at the end of the post. ;-)) – Mark Viola Feb 3 '17 at 4:40
First, the partial fraction of the summand can be written
\begin{align} \frac{n^2}{(4-n^2)^2}&=\frac14\left(\frac{1}{n-2}+\frac{1}{n+2}\right)^2\\\\ &=\frac14 \left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}+\frac{1/2}{n-2}-\frac{1/2}{n+2}\right) \end{align}
Second, we note that
\begin{align} \sum_{n\,\,\text{odd}}\frac{1}{(n\pm 2)^2}&=\sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2}\\\\ &=2\sum_{n=1}^\infty \frac{1}{(2n-1)^2}\\\\ &=2\left(\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{(2n)^2}\right)\\\\ &=\frac32 \sum_{n=1}^\infty \frac{1}{n^2}\\\\ &=\frac{\pi^2}{4} \end{align}
Third, it is easy to show that
$$\sum_{n=-\infty}^\infty \left(\frac{1}{2n-3}-\frac{1}{2n+1}\right)=0$$
Putting it all together we have
$$\sum_{n,\,\,\text{odd}}\frac{n^2}{(4-n^2)^2}=\frac{\pi^2}{8}$$
If we sum over the positive odd only, then the answer is $(1/2)\pi^2/8=\pi^2/16$
HINT
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(n^2-4)^2}=\sum_{n=1}^{\infty} \frac{(2n-1)^2}{((2n-1)^2-4)^2}$$ Using partial fraction expansion, note $$\frac{(2n-1)^2}{((2n-1)^2-4)^2}=\left(\frac{1}{4(2n+1)^2}+\frac{1}{4(2n-3)^2}\right)-\left(\frac{1}{8(2n+1)}-\frac{1}{8(2n-3)}\right)$$ Note that the second part has cancelling terms.
• Why the downvote? – S.C.B. Feb 3 '17 at 4:40
• @ThomasAndrews I fixed the signs. – S.C.B. Feb 3 '17 at 4:40 | 2019-09-22T22:34:09 | {
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https://koytru.meinejuwelen.de/en/qhqi | If by 'a standard deck of cards' you mean a 52 card pack (no Jokers) with 4 suits each of 13 cards, the probability of picking a single card at random and it being a heart is 13/52 or 1/4 For a pack with jokers it would be 13/54 Ankit Pal B.E from University of Mumbai (Graduated 2020) 3 y Related. Explanations Question You draw a card at random from a standard deck of 52 cards. Find each of the following conditional probabilities: a) The card is a heart, given that it is red. b) The card is red, given that it is a heart. c) The card is an ace, given that it is red. d) The card is a queen, given that it is a face card. Explanation Verified. In a playing card there are 52 cards. Therefore the total number of possible outcomes = 52 (i) '2' of spades: Number of favourable outcomes i.e. '2' of spades is 1 out of 52 cards. Therefore, probability of getting '2' of spade Number of favorable outcomes P (A) = Total number of possible outcome = 1/52 (ii) a jack. But the coin has not changed - if it's a "fair" coin, the probability of getting tails is still 0.5. Dependent Events Two (or more) events are dependent if the outcome of one event affects the outcome of the other(s). Thus, one event "depends" on another, so they are dependent. Example I draw two cards from a deck of 52 cards.
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What is the probability of ace card in a deck of 52 cards ? A card is drawn from a pack of 52 cards . The probability of getting an ace is 1/52. View complete answer on byjus.com. What do 4 Aces mean? If you hate your job, then four <b>Aces</b> can mean that something major will happen which will replace your income, allowing you to leave your job. For. Mar 19, 2020 · A deck of standard 52 cards contain four aces. There are four kings in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e. ⇒ . ⇒ . ⇒ . a king or a diamond ; A deck of standard 52 cards contain four kings. There are 13 Diamonds in a standard deck of playing cards. So ,we need to .... TO FIND : Probability of the following Total number of cards = 52 ( i ) Cards which are black king is 2 We know that PROBABILITY = = Number of favorable event T otal number of event Hence the probability of getting a black king is equal to 2/52=1/26 (ii) Total number of black cards is 26.. 2017. 6. 10. · A standard deck has 13 ordinal cards (Ace, 2-10, Jack, Queen, King) with one of each in each of four suits (Hearts, Diamonds, Spades, Clubs), for a total of #13xx4=52# cards. If we draw a card from a standard deck, there are 52 cards we might get. There are 16 cards that will satisfy the condition of picking a Jack, Queen, King, or Ace. This.
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Mar 03, 2021 · A) 1 2 B) 1 52 C) 1 4 D) 1 13 7) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing a heart? A) 1 4 B) 1 2 C) 3 4 D) 1 8) In a survey of college students, 840 said that they have cheated on an exam and 1795 said that they have not..
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Example 10 One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n(S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn.
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A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. One card is drawn from a standard deck of 52 cards. Find the probability of drawing a heart or a 6.There are 13 hearts and one 6 of hearts... 2. Which of the following must be a true statement?... Select one: a. The conditional probability P(A/B) is the probability that event B occurs, knowing A has occurred. b. An event and its complement can ....
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Transcribed image text: A single card is drawn from a standard 52-card deck. Let H be the event that the card drawn is a heart, and let F be the event that the card drawn is a face card. Find the indicated probability. P (H'UF) P(H'UF) = 0 (Type an integer or a simplified fraction.) A single card is drawn from a standard 52-card deck. one less card in the deck because we already had to draw the Heart from the deck. Thus: P(Heart and Club) = P (Heart) * P (Club) = 13/52 * 13/51 = .25 * .255 = .064 We might also have to subtract a value from the numerator as well as the denominator. Try to find the probability of drawing three red cards from a deck without replacement. Question 591370: A single card is drawn from a standard deck of 52 cards. Find the probability the card is: 1. A red four 2. A heart 3. A 4 or a heart. 4. Not a club. 5. A red or a four 6. A red and a 3 Answer by Edwin McCravy(19211) (Show Source):. The probability of drawing a queen, from a 52 card deck, is 4/52 or 1/13. Wiki User. ∙ 2009-10-09 18:06:35. This answer is:. Answer: The probability of drawing a card from a standard deck and choosing a king or an ace is (1/13) × (4/51). What is the probability of drawing a queen of hearts from a deck of 52 cards pinia store.
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Mar 06, 2011 · Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the.... 2018. 6. 12. · In a standard deck of cards, there are 52 cards. They are broken down into suits (4 of them: Spades, Hearts, Diamonds, Clubs) of 13 cards each. Each suit has 13 ordinal cards (A, 2 through 10, Jack, Queen, King). Here we're asked to draw a card at random and find the probability of drawing either a diamond or a 7. Mar 06, 2011 · Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the....
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If one card is drawn from a standard 52 card playing deck, determine the probability of getting a ten, a king or a diamond. Round to the nearest hundredth. # of ways to succeed: 4 + 4 + 13 - 1 -1 = 19. Correct option is D) Probability of drawing a king = 524= 131 After drawing one card, the number of cards are 51. Probability of drawing aqueen = 514. Now, the probability of drawing a king and queen consecutively is 131× 514= 6634 Was this answer helpful? 0 0 Similar questions Three cards are drawn with replacement from a part of 52 cards. Calculate the probability of being dealt a diamond from a standard deck of 52 cards. Since there are 4 suits in a deck of cards (hearts, clubs, spades and diamonds) we can find the number of. Transcript. Example 10 One card is drawn from a well shuffled deck of 52 cards.If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n (S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn So, n (A) = 13 Probability of A = P (A.One card is selected from a.
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Mar 12, 2011 · The probability of drawing a club or a nine from a 52 card deck of standard playing cards is 16 / 52 or approximately 30.8%. There are 13 clubs in a standard deck of cards. There are four nines in.... The royal flush is a case of the straight flush. It can be formed 4 ways (one for each suit), giving it a probability of 0.000154% and odds of 649,739 : 1. When ace-low straights and ace-low straight flushes are not counted, the probabilities of each are reduced: straights and straight flushes each become 9/10 as common as they otherwise would be.
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winnetka rummage sale 2022; the end netflix; craigslist gmc trucks for sale by owner near Florianpolis State of Santa Catarina; meta computer vision engineer salary; restaurants open till 2am near Panruti Tamil Nadu; Google Algorithm Updates. The total number of 8 card hands is 52c8, so the probability of choosing a hand which excludes at least one suit is (4 * 39c8 - 6 * 26c8 + 4 * 13c8) / (52c8). You wanted the probability that a hand includes every suit which is the opposite of choosing a hand excluding at least one suit, and therefore the probability that you wanted is. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at. Let Event B = drawing a red card. P (A) = 4/52 since there are four aces in each deck of 52 cards. P (B) = 1/2 = 26/52 since there are four suits and two of them are red (or 26 red cards in a deck of 52) P (A∩B) = the probability of drawing a red ace = 2/52 since there are 2 red aces in a deck of 52 cards).
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Transcript. Example 10 One card is drawn from a well shuffled deck of 52 cards.If each outcome is equally likely, calculate the probability that card will be a diamond Since there 52 cards n (S) = Total number of cards = 52 There are 13 diamond cards Let A be event that diamond card is withdrawn So, n (A) = 13 Probability of A = P (A.One card is selected from a. A standard deck of cards has: 52 Cards in 13 values and 4 suits ... If you draw 3 cards from a deck one at a time what is the probability: ... what is the probability: You draw a Club, a Heart and a Diamond (in that order). The hypergeometric MTG calculator can describe the likelihood of any number of successes when drawing from a deck of Magic cards. It takes into account the fact that each draw decreases the size of your library by one, and therefore the probability of success changes on each draw. Population Size. Cards in your deck / library you are drawing from.
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Number of favourable outcomes = 48 (Because There are always 52 cards in one deck containing 13 cards of varying values but always 4 suits, if the drawn card is not 4 then number of favourable outcomes will be 48 (12*4).) Total number of favourable outcomes = 52 (Number of cards in a deck is 52) So, the probability will be = 48/ 52 = 12/13. Oct 26, 2020 · Answer: 1/13. Step-by-step explanation: If there are 4 suits in a deck of cards, there are only 4 aces inside a complete deck. That means the probability of drawing an ace is 4/52 and just simply simplify the answer to 1/13.. A SINGLE CARD IS DRAWN AT RANDOM FROM A STANDARD DECK OF 52 CARDS. FIND THE PROBABILITY OF DRAWING THE FOLLOWING CARDS. PLEASE REDUCE TO LOWEST TERMS. A) A DIAMOND OR A 5 __________ B) A HEART AND A JACK __________ C) A JACK OR AN 8 __________ D) A HEART OR A SPADE __________ E) A RED AND FACE CARD __________ F) A RED CARD OR A QUEEN __________ 2. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at. Suppose you draw five cards from a standard deck of 52 playing cards, and you want to calculate the probability that all five cards are hearts. ... For example, the probability of drawing five cards of any one suit is the sum of four equal probabilities, and four times as likely. In boolean language, if the events are related by a logical OR. Whenever you do probability problems, check to see if you are being asked to find probability of one thing OR another, or of multiple events happening together. For instance, if this problem asked you to find the probability of drawing a heart AND a 5, well, there is only one 5 of hearts in a deck. So that would be 1/52.
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This can be simplified into 4/13 or a 30.77% probability of drawing a 4 or a spade from a standard deck of cards. Read More:. After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is 3 51 = 1 17 3 51 = 1 17.. A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. a card is drawn from a standard deck what is the probability that the card is an ace If you draw one card from a standard deck, what is the probability of drawing a 5 ... a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and Tìm kiếm.
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2022. 2. 7. · We know that a well-shuffled deck has 52 cards. Total number of black cards = 26. Total number of red cards = 26. Therefore probability of getting a black card= {total number. 5) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing an ace? A) 1 13 B) 1 52 C) 1 4 D) 1 2 6) If one 39,068 satisfied customers 2,852 satisfied customers Ph.D. 39,068 satisfied customers the probability for the experiment of drawing a card at random from a standard deck of 52 playing cards. What is the probability of drawing a black checker from a bag filled with 6 black checkers and 4 red checkers, replacing it, and drawing another black checker? c. Getting a Club and a Heart . 2. Drawing a card from a deck and not replacing it. 1.. So, there are 12 face cards in the deck of 52 playing cards. Worked-out problems on Playing cards probability: 1. A card is drawn from a. A card is selected from a deck of 52 playing cards. Find the probability of selecting · a prime number under 10 given the card is a heart. (1 is not prime.) · a diamond or heart given the card is red. read ....
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A card is randomly drawn from a standard 52-card deck. Find the probability of the given event. (A face card is a king, queen, or jack). Drawing a queen and a heart (this is an intersection question). A card is drawn randomly from a standard 52-card deck.Find the probability of the given event. (a) The card drawn is 6 The probability is:(b) The card drawn is a face card (Jack, Queen, or King) The read more.Solution Total number of possible outcomes = 52 P (E) = (Number of favourable outcomes/ Total number of outcomes) (i) Total numbers of the king of. Since we know that in a deck of 52 cards, there are a total 12 face cards (3 face cards each of heart, diamond, spade and club). Number of face cards$= 12$ Therefore, probability of getting a face card$= \dfrac { { {\text {Number of face cards}}}} { { {\text {Total number of cards}}}} = \dfrac { {12}} { {52}} = \dfrac {3} { {13}}$. Jun 21, 2020 · If one card is drawn from a standard 52-card playing deck, find the probability of getting a king, or a ten, or a heart, or a club.When entering your answer include a leading zero and round to the nearest hundredth. An example of an acceptable answer would be 0.19. Question 591370: A single card is drawn from a standard deck of 52 cards. Find the probability the card is: 1. A red four 2. A heart 3. A 4 or a heart. 4. Not a club. 5. A red or a four 6. A red and a 3 Answer by Edwin McCravy(19211) (Show Source):. Jun 21, 2020 · If one card is drawn from a standard 52-card playing deck, find the probability of getting a king, or a ten, or a heart, or a club.When entering your answer include a leading zero and round to the nearest hundredth. An example of an acceptable answer would be 0.19.
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If you draw one card from a standard deck, what is the probability of ... a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and Tìm kiếm. Tìm ... If you draw one card from a standard deck, what is the probability of drawing a 5 or.
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The total number of cards in a deck is 52 Probability = No. of favorable outcomes / Total no. of outcomes. Now, the probability of cards in a deck is 13/ 52 . Understand different concepts and get good grip on them by using online tools available at.
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A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. Apr 26, 2017 · Ace is not odd The ordinals 3, 5, 7, and 9 are odd. There are four of each (one for each suit) and so 4xx4=16 odd cards. This makes the probability: P("draw an odd card")=16/52=4/13 Ace is odd If we want to consider the Ace as a 1, then there are 5 ordinals that are odd, 5xx4=20 odd cards, and therefore: P("draw an odd card")=20/52=5/13.
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There are 5 2 cards in a standard deck: 1 3 ordinal cards (Ace - 1 0, Jack, Queen, King) and 4 of them - one to each suit (hearts, diamonds, clubs, spades) and so we have 4 × 1 3 = 5 2.. Q. From a pack of 52 playing cards jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is (i) a black queen (ii) a red card (iii) a black jack (iv) a picture card (Jacks, queens and kings are picture cards). Step 2. ∵ number of non-face card in well shuffled deck of 52 playing card = 52-12. = 40. Step 3. Probability ( a non-face card ) = number of favourable outcomes total number of outcomes. = 40 52. = 10 13. Step 4. (ii) Number of black king in well shuffled deck of 52 playing cards = 2. Mar 06, 2011 · Number of cards in a deck = 52 Number of cards that are heart = 13 Therefore number of cards that are not heart = 52-13 = 39 Probability of not drawing a heart = 39/52 or 3/4 What is the....
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The goal is to win at least three tricks.. Example 3: Two cards are drawn without replacement in succession from a well-shuffled deck of 52 playing cards. What is the probability that the second card drawn is an ace, given that the first card drawn was an ace? Example 4: One thousand high school seniors were surveyed about whether they planned .... Problem 2: A random card is chosen from the standard deck of cards, find the probability of obtaining a queen or a heart.. So, the probability of getting a Queen card is 1/13. Example 2: A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a card of Heart. Solution: Let A represents the event of getting a Heart .... A standard 52 card deck. If two cards are randomly selected, what is the probability of drawing a king first, followed by drawing a 2. If two cards are randomly selected what is the probability of drawing a heart first, followed by drawing a. A standard deck of cards has: 52 Cards in 13 values and 4 suits ... If you draw 3 cards from a deck one at a time what is the probability: ... what is the probability: You draw a Club, a Heart and a Diamond (in that order).
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Number of Kings in a deck = 4. Probability of drawing a card from a standard deck and choosing a king or an ace = Probability of getting an Ace + Probability of getting a King. Probability of drawing an Ace at random = 4/52 = 1/13. Now, the probability of drawing a King at random = 4/52 = 1/13. hence, the required probability = 1/13 + 1/13 = 2/13. What is the probability of drawing a heart from a standard deck of cards on a second draw? 2 Answers By Expert Tutors The probability of choosing a heart, P(Heart) = 13/52 = 0.25. What is the probability of getting a heart or an even number? Clearly, the probability of drawing a heart out of the deck is 13/52, or 1/4.. "/>.
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Therefore, the proability of drawing a king or 3 is 8/52=2/13. For the second problem, think binomial probability. The probability of drawing a king is 1/13. The opposite of at least 1 is none. So, the probability of getting no kings is 48/52=12/13. We find the probability of getting no kings and subtract from 1. Also, You should get the same.
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A card is drawn from a standard deck of 52 playing cards. A: The result is a club. B: The result is a king. 36) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a king. 37) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a black card. We can draw one card from pack of 52 cards in 52C1 = 52 ways n (S)=52 There is only one card in the pack of 52 cards which is jack as well as heart card,.It is jack of hearts n (E)=1 DIY: How to take years off your neck's appearance. Ray Phan Principal Software Engineer at Magic Leap (company) (2021-present) 4 y Related. Jan 13, 2017 · There is a 50% chance that the card drawn will be red. Red cards make up 50% of the deck. 26/52 = 1/2 Therefore, if you're drawing one card, there is a 50/50 chance that the card will be red.. Problem 2: A random card is chosen from the standard deck of cards, find the probability of obtaining a queen or a heart.. So, the probability of getting a Queen card is 1/13. Example 2: A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting a card of Heart. Solution: Let A represents the event of getting a Heart .... 5) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing an ace? A) 1 13 B) 1 52 C) 1 4 D) 1 2 6) If one 39,068 satisfied customers 2,852 satisfied customers Ph.D. 39,068 satisfied customers the probability for the experiment of drawing a card at random from a standard deck of 52 playing cards. Correct option is D) Probability of drawing a king = 524= 131 After drawing one card, the number of cards are 51. Probability of drawing aqueen = 514. Now, the probability of drawing a king and queen consecutively is 131× 514= 6634 Was this answer helpful? 0 0 Similar questions Three cards are drawn with replacement from a part of 52 cards. In this task, we need to calculate the probability of getting at least one black card. The given is that we draw 2 2 2 cards from a 52 52 52-card deck with replacement.. The deck is the standard deck with 4 4 4 suits, clubs, spades, hearts, and diamonds, where spades and clubs are black suits.
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If you were investigating red cards, kings or the queen of hearts, the odds of randomly drawing one of these from a complete deck are 50 percent (26 in 52); about 7.7 percent (four in 52); or. One card is randomly drawn from a deck of 52 playing cards. Find the probability thati the drawn card is red.ii the drawn card is an ace.iii the drawn card is red and a king.iv the drawn card is red or a king.[4 MARKS]. A card is drawn from a standard deck of 52 playing cards. A: The result is a club. B: The result is a king. 36) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a king. 37) A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a black card. Setup Card content One card per line. Supports basic HTML. Go!. You will come across many video chatting applications, but a few of the best ones that are very effective and perfect for a virtual card room are Zoom and Skype. With Zoom, you can add around 100 people; however, the session will only last for 45minutes, and you will have to start.
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Jan 13, 2017 · There is a 50% chance that the card drawn will be red. Red cards make up 50% of the deck. 26/52 = 1/2 Therefore, if you're drawing one card, there is a 50/50 chance that the card will be red.. Playing cards probability problems based on a well-shuffled deck of 52 cards. Basic concept on drawing a card: In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades ♠ hearts ♥, diamonds ♦,. A standard deck of playing cards had 52 cards. These cards are divided into four 13 card suits: diamonds, hearts, clubs, and spades. Find the probability of drawing a heart or a club at random from a deck of shuffled cards. math. A card is drawn from an ordinary deck of 52 cards, and the result is recorded on paper.
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2010. 6. 4. · There are 6 red face cards in a standard deck of 52 cards; the Jack, Queen, and King of Hearts and Diamonds. The probability, then, of drawing a red face card from a standard deck of 52 cards is 6 in 52, or 3 in 26, or about 0.1154. (A standard deck of cards is the most common type of deck used in most card games containing 52 cards). Determine the probability of having 1 girl and 3 boys in a 4-child family assuming boys and girls are equally likely. The probability of having 1 girl and 3 boys is 1/4. Use the theoretical method. Probability gives the chances of how likely ....
bh | 2022-12-09T19:59:20 | {
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https://math.stackexchange.com/questions/3129387/the-convergence-of-a-telescoping-series | # The Convergence of a Telescoping Series
The question I've been posed is:
Show that the following series converges, and compute its value
$$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$
From this I decided to use partial fractions to put into the form:
$$\frac{1}{2}\cdot\left(\frac{1}{k}-\frac{1}{k+2}\right)$$
And from this I noticed that this is in the form of a telescoping series which I think would cancel down to:
$$\frac12\cdot\left(1+\frac12\right)= \frac{3}{4}$$
So I've got to this point, but I don't think what I've worked out is substantial enough to prove what I've been asked.
Would anyone mind giving any tips to make my working more thorough.
• Concerning the convergence alone you can use the fact that $$\frac1{k(k+2)}<\frac1{k^2}$$ – mrtaurho Feb 27 '19 at 21:30
Note that$$\frac1k-\frac1{k+2}=\left(\frac1k-\frac1{k+1}\right)+\left(\frac1{k+1}-\frac1{k+2}\right).$$This will give you two telescoping series. Can you take it form here?
• Sorry but I don't quite understand where I should go from there, would you mind giving another tip? – king Feb 28 '19 at 17:05
• \begin{align}\sum_{n=1}^\infty\frac1k-\frac1{k+2}&=\sum_{k=0}^\infty\frac1k-\frac1{k+1}+\sum_{n=1}^\infty\frac1{k+1}-\frac1{k+2}\\&=1-\lim_{n\to\infty}\frac1{k+1}+\frac12-\lim_{k\to\infty}\frac1{k+2}\\&=\frac32.\end{align} – José Carlos Santos Feb 28 '19 at 17:33
Let's write, as you have done, the following :
$$S_n =\sum_{k=1}^n \frac{1}{k(k+2)} = \frac{1}{2} \sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+2} \right) = \frac{1}{2} \left(\sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n\frac{1}{k+2} \right) = \frac{1}{2} \left(\sum_{k=1}^n \frac{1}{k} - \sum_{k=3}^{n+2}\frac{1}{k} \right)$$
So you see that $$S_n = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right)$$
Now this is obvious that the limit of $$S_n$$ is equal to $$\frac{3}{4}$$, i.e.
$$\sum_{k=1}^{+\infty} \frac{1}{k(k+2)} = \frac{3}{4}$$ | 2020-06-06T07:18:04 | {
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http://math.stackexchange.com/questions/767175/find-how-many-solutions-has-the-following-equation | # Find how many solutions has the following equation…
Determine how many real solutions has the following equation:
$$x^2(|x|-6)=-15$$
I noticed that $|x|-6$ should be negative because $x^2$ is always a positive value. Thus, $x\in(-6;6)$. I made a substitution: $|x|=t$, hence $x^2=t^2$, and got the equation:$$t^3-6t^2+15=0$$
Now the problem is I cannot find any solution for this equation. Hope you'll give me the right explanation for this exercise. Thank you very much!
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try to study the variations of the function you have in the last equation. – Denis Apr 24 '14 at 10:20
Thank you, but what do you mean by this? – John G. Apr 24 '14 at 10:20
You don't need to find the solutions. Only their number. – evil999man Apr 24 '14 at 10:21
compute the derivative, and see where the derivative is zero, it will tell you when this function reaches minimum and maximum. – Denis Apr 24 '14 at 10:21
So using the Descarte's rule I see there are two variations of signs for $P(t)=t^3-6t^2+15$. So does this mean I have two positive solution or none? And if so then what should I do? I think that if there are two positive solutions for the equation above this means we have four solutions in general when substituting in $|x|=t$. – John G. Apr 24 '14 at 10:47
One way to answer this question is to draw a graph of $y = x^2(|x| - 6)$, then draw a horizontal line at $y = -15$. The curve and the line obviously intersect in four places. QED.
Finding the solutions is a bit trickier, but that is not the question that was asked.
If you don't want to rely on drawing the graph, you can prove the result using $f(t)$ and $f'(t)$. As you showed, $f(t) = t^3 - 6t^2 + 15$, which gives $f'(t) = 3t^2 - 12t$.
Solving for $f'(t) = 0$ yields $t = \{0, 4\}.$ Thus, $f(t)$ has extrema at $t =0$ and $t = 4$. We only care about $t>0$, so we ignore that one. Calculating $f'(1) = -9$ and $f'(5) = 15$, we see that $f(t)$ must be strictly decreasing for $0 < t < 4$ and strictly increasing for $t > 4$.
Some key evaluations: $$f(0) = 15$$ $$f(4) = -17$$ $$f(6) = 15$$
Since $f(t)$ is strictly decreasing between $0$ and $4$, and $f(0) > 0$ and $f(4) < 0$, $f(t)$ must cross zero exactly once in that region. Likewise, since $f(t)$ is strictly increasing for $t > 4$ and $f(4) < 0$ and $f(6) > 0$, $f(t)$ must cross zero exactly once in that region.
Thus, there are two positive solutions to the initial equation. By symmetry, there must be two negative solutions, as well.
- | 2015-11-29T16:21:22 | {
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https://math.stackexchange.com/questions/3034620/how-to-play-a-betting-game?noredirect=1 | # How to play a betting game
I have been interviewing in a few trading firms recently. I came up with the following question myself, but it is similar to some of the questions they ask and ways of thinking they expect.
Suppose you have some capital to invest (for example \$100). You can play a game where you bet \$x of your money and with probability $$\frac{2}{3}$$ your bet is doubled (so now you have \$(100 + x)) and with probability $$\frac{1}{3}$$ you lose your bet (so now you have \$(100 - x)). How much should you bet on this game.
I can see two ways of thinking about this problem. Firstly, there is the expected value maximisation approach. It can be easily seen that your expected gain in this game is $$\frac{x}{3}$$. So in order to maximise EV you should bet all of your money instantly. And if you were to play this game a million times, you should bet all of your money each time.
Of course this approach has the obvious flaw that when you play a few times, you will almost certainly go bankrupt. So we decide not to maximise EV and instead first make sure that we never go bankrupt. We do it by deciding to, at each point of the game, always bet exactly the same proportion of our money, say $$p$$. Then after $$n=n_1+n_2$$ games, where our bet was doubled $$n_1$$ times and we lost $$n_2$$ times, we will have $$M \cdot (1+p)^{n_1} \cdot (1-p)^{n_2}$$ money, where $$M$$ was our initial amount. Differentiating the log of this with respect to $$p$$ we can see that this function has its maximum for $$p=\frac{n_1-n_2}{n} \rightarrow \frac{2}{3}-\frac{1}{3} = \frac{1}{3}$$ as $$n \rightarrow \infty$$. So if we bet just a third of our money every time, we are (almost) guaranteed not to go bankrupt and, out of the strategies that bet a constant proportion every time, this one maximises our gain in the most likely outcome.
So here is my question - does this second strategy make sense to you? If you were to play this game with your own money would you use it? Does it make any sense to use a different strategy if you only play once, and not many times? I personally would be tempted to bet more than a third if I only got one chance, because it would increase my EV even if it is potentially bad in the long term. Does this sentiment make any sense?
Also, I just described two ways of thinking about the game above. Do you know of any other ways to think about it? Other strategies?
Your approach is remarkably on point. This issue is generally discussed in terms of portfolio construction in finance and depending on risk tolerance we define different functions to optimize. You choose to maximize the log of the expected payoff after n games, which is the same as Kelly criteria. For further and more detailed discussion of it you can check https://en.wikipedia.org/wiki/Kelly_criterion
• Correct me if I am wrong here: I don't think I am maximising the log of the expected payoff - I think I am maximising the log of the most likely payoff. The expected payoff is still the largest if I bet everything I have every time, right? Also, taking the log shouldn't change anything in terms of maximisation, because log is strictly increasing. So I am really maximising the most probable payoff, no? – user132290 Dec 11 '18 at 10:25
• Well, in effect you are basically maximizing the log of expected payoff. So you first wrote the payoff equation given $p, n_1, n_2$. While finding the most probable payoff, you plug $\frac 2 3$ for $\frac {n_1} n$, which is in practice taking the expected value of $n_1$ and $n_2$ with respect to $n$, hence taking the expected value of payoff. – Ofya Dec 11 '18 at 15:19
Your intuition is generally captured in a "utility function" which is supposed to describe how happy you are to have a certain amount of money. You then maximize the expected value of this function. Your thought to maximize the log of the amount of money you have is one utility function and not an unreasonable one. There is good psychological evidence that more money makes people happier, but much more slowly once they have some. The log function is in this vein, but there are many others as well. Once you define the function, the maximization process is the one you have used.
I would suggest that no simply described function can capture utility properly. If you were allowed to play the game fifty times but had to bet one dollar each time, you would probably play. You would probably win about $$\25$$ and be happy about it, but it wouldn't really change your life. As the bet rises the impact on your life does too. At the start it is only good because the chance you lose is almost zero so more is better. Eventually it may get to the point that you become risk averse. If your income is large compared to your cash assets it may make sense to bet everything you have because you can replace it easily. If you are living on assets you may become risk averse at a small fraction of your assets. This is all supposed to be captured in the utility function, which indicates why a simple answer like log is not appropriate for real life.
• Correct me if I am wrong please, but you seem to be treating the log as if it was meaningful in my example (because it is the utility function of my wealth - so the second million I make will make me less happy than the first million). But I am just taking the log here for simpler differentiation and to maximise the log is the same as to maximise its argument. So the log should not be relevant at all here, no? – user132290 Dec 11 '18 at 10:28
• Yes, I am treating the log as meaningful. If your utility is linear in money you should bet all your money each time. I answered this here. Yes, you will almost surely be broke, but the tiny chance of winning a huge amount of money overwhelms that. The utility function should apply to all the money you have, not just to what you win from the game. The log is a nice function because it is increasing but slower and slower as the argument gets larger, which we want utility todo – Ross Millikan Dec 11 '18 at 15:19
• Changing the log to a different function will shift the amount you should play with. – Ross Millikan Dec 11 '18 at 15:20
Straight out of my files:
For a P% profit level and D odds to 1:
RequiredPercentageCorrectBets = (P + 100) / (D + 1) .
This posting is simply viewpoint of pari-mutuel wagering where the probability is derived from the wagering.
In other words, in pari-mutuel wagering all the information that there is, is supposed to be represented by the wagering on the tote board. So in pari-mutuel wagering, odds of 1.00 represents a probability of 50%. Then the Kelly Criteria calls that situation a no-bet. However, some bettors do wager profitably and so I suppose that there is a personal probability of winning that can be applied to calculating the percentage of the stake to bet that maximizes profit. However, a personal probability of winning would still tend to go up with lower odds and tend to go down with higher odds.
In roulette or keno I suppose that the gambler can keep track of a number not coming up and then expect an increasing probability of the number coming up. That situation assumes a faith of an honest game instead of a discovery of a dishonest game.
I previously suggested a one-number Keno game that pays 3 to 1 for odds of 2 to 1. Twenty numbers are drawn out of 80 for a probability of 25%. But the probability of hitting the number in two draws is 50%, the probability of hitting the number in three draws is 75%, and the probability of hitting the number in four draws is 100%. So the idea is to decide how much to increase the wager each time the number doesn't come up and keep playing the number until it does come up. | 2019-12-15T05:41:47 | {
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"url": "https://math.stackexchange.com/questions/3034620/how-to-play-a-betting-game?noredirect=1",
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Arithmetic Progression, number of terms, sum of terms etc
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Arithmetic Progression, number of terms, sum of terms etc [#permalink]
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24 Mar 2018, 12:57
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Greetings friends l
i decided to create a list of all formulas regarding arithmetic progression, number of terms, sums etc all in one post. i may have some questions and or inaccuaracies, so you are welcome to correct me or just say hi
1. HOW TO FIND NUMBER OF TERMS
$$x_n = a+d (n-1)$$
$$n$$ = # of term
$$a$$ = first term
$$d$$ = distance
Example:
a= 3
d = 5
Question: find the 9th term
$$x_9 = 3+5 (n-1)$$
$$x_9 = 3+5 (n-1)$$
$$x_9 = 3+5n-5$$
$$x_9 = 5n-2$$
now plug in 9 into 5n-2
$$x_9 = 5*9-2 = 43$$
hence $$9th$$ term is $$43$$
2. HOW TO FIND THE SUM OF TERMS
SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$
Example
$$a$$ = 1 the first term
$$d$$ = 3 distance
$$n$$ = 10 how many terms to add up
Question: what is the sum of 10 terms with distance 3 and the first term 1 ?
$$\frac{10}{2} (2 *1 +(10-1)3)$$
$$= 5(2+9·3) = 5(29) = 145$$
3.HOW TO FIND THE SUM OF THE FIRST CONSECUTIVE NUMBERS
$$\frac{n(n+1)}{2}$$
where $$n$$ is number of terms
Example: what is the sum of the first 15 numbers ?
$$\frac{15(15+1)}{2}$$ =$$238$$
4.HOW TO FIND THE SUM OF THE FIRST EVEN NUMBERS
$$\frac{n(n+2)}{4}$$
where $$n$$ is number of terms
$$\frac{15(15+2)}{4}$$ = $$8$$
5. HOW TO FIND NUMBER OF TERMS FROM A TO B
$$\frac{first..term +last..term}{2} +1$$
6. HOW TO FIND SUM OF ODD NUMBERS FROM A TO B
Step one: $$find..the..number...of..terms$$
Step two: $$\frac{first..term+last..term}{2}$$ $$* number..of.. terms$$
7. NUMBER OF MULTIPLES X IN THE RANGE
$$\frac{last..multiple..of..x - first..multiple..of...x}{x}+1$$
Eg. how many multiples of 4 are there between 12 and 96?
$$\frac{96-12}{4}$$+1 = 22
to be continued
by the way I myself have question
what is the difference between this $$\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ ?
I will add more useful formulas later
Manager
Joined: 16 Sep 2016
Posts: 209
WE: Analyst (Health Care)
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]
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26 Mar 2018, 12:47
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dave13 wrote:
by the way I myself have question
what is the difference between this $$\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ ?
I will add more useful formulas later
Hello dave13,
Those two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)
Let's see,
SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$
plug in a = 1, d = 1
$$\frac{n}{2} (2*1+(n-1)1)$$
$$\frac{n}{2} (n+1)$$
$$n*(n+1) / 2$$
So we can say that SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is
$$\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.
Good initiative!
+1 kudos to you!
Best,
Senior Manager
Joined: 09 Mar 2016
Posts: 442
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]
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01 Apr 2018, 07:44
dave13 wrote:
by the way I myself have question
what is the difference between this $$\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ ?
I will add more useful formulas later
Hello dave13,
Those two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)
Let's see,
SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$
plug in a = 1, d = 1
$$\frac{n}{2} (2*1+(n-1)1)$$
$$\frac{n}{2} (n+1)$$
$$n*(n+1) / 2$$
So we can say that SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is
$$\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.
Good initiative!
+1 kudos to you!
Best,
What is the difference between first consecutive integers and consecutive integers ?
Manager
Joined: 16 Sep 2016
Posts: 209
WE: Analyst (Health Care)
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink]
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01 Apr 2018, 08:09
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dave13 wrote:
What is the difference between first consecutive integers and consecutive integers ?
Hey dave13,
From our prev discussion you were asking the difference between the two formulas for first n integers... the difference between those two is simply where is the starting point of our AP?
dave13 wrote:
by the way I myself have question
what is the difference between this $$\frac{n(n+1)}{2}$$ and this formula SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ ?
I will add more useful formulas later
First n positive integers would imply d =1 and a = 1 when thinking in terms of an AP and the formula for sum of first n terms in such a case is given by $$\frac{n(n+1)}{2}$$
However, in the next case of n consecutive integers the starting point could be anything ( 1 or not)
The formula for sum of n terms of this is given by a = a & d = 1 -> in the sum formula given above. $$\frac{n}{2} (2a+(n-1)d)$$
Also for a = 1 & d = 1 ... both these formulas are one and the same as shown in prev post (in quotes below)
Those two expressions are one and the same when a = 1 & d = 1 ( which is the case for consecutive integers starting from 1)
Let's see,
SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$
plug in a = 1, d = 1
$$\frac{n}{2} (2*1+(n-1)1)$$
$$\frac{n}{2} (n+1)$$
$$n*(n+1) / 2$$
So we can say that SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$ is a generalized formula whose special case is
$$\frac{n(n+1)}{2}$$ when we talk about first n consecutive integers.
Hope that makes sense!
Best,
Re: Arithmetic Progression, number of terms, sum of terms etc [#permalink] 01 Apr 2018, 08:09
Display posts from previous: Sort by | 2018-04-23T19:27:59 | {
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"url": "https://gmatclub.com/forum/arithmetic-progression-number-of-terms-sum-of-terms-etc-261976.html",
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http://www.resourcefulself.com/sometime-lately-fekto/chebyshev-distance-vs-euclidean-a6cfa2 | ( Log Out / M = 200 input data points are uniformly sampled in an ordered manner within the range μ ∈ [− 4 b, 12 b], with b = 0.2. p=2, the distance measure is the Euclidean measure. The first one is Euclidean distance. In Chebyshev distance, AB = 8. Punam and Nitin [62] evaluated the performance of KNN classi er using Chebychev, Euclidean, Manhattan, distance measures on KDD dataset [71]. If we suppose the data are multivariate normal with some nonzero covariances and for … For stats and …  The last one is also known as L1 distance. Computes the distance between m points using Euclidean distance (2-norm) as the distance metric between the points. AC = 9. I got both of these by visualizing concentric Euclidean circles around the origin, and … Only when we have the distance matrix can we begin the process of separating the observations to clusters. --81.82.213.211 15:49, 31 January 2011 (UTC) no. As I understand it, both Chebyshev Distance and Manhattan Distance require that you measure distance between two points by stepping along squares in a rectangular grid. For example, in the Euclidean distance metric, the reduced distance is the squared-euclidean distance. Euclidean distance is the straight line distance between 2 data points in a plane. Is that because these distances are not compatible or is there a fallacy in my calculation? Notes. See squareform for information on how to calculate the index of this entry or to convert the condensed distance matrix to a redundant square matrix.. Taxicab circles are squares with sides oriented at a 45° angle to the coordinate axes. But if you want to strictly speak about Euclidean distance even in low dimensional space if the data have a correlation structure Euclidean distance is not the appropriate metric. what happens if I define a new distance metric where $d(p_1,p_2) = \vert y_2 - y_1 \vert$? the chebyshev distance seems to be the shortest distance. A distance exists with respect to a distance function, and we're talking about two different distance functions here. A common heuristic function for the sliding-tile puzzles is called Manhattan distance . We can count Euclidean distance, or Chebyshev distance or manhattan distance, etc. Sorry, your blog cannot share posts by email. To reach from one square to another, only kings require the number of moves equal to the distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop’s case). (Wikipedia), Thank you for sharing this I was wondering around Euclidean and Manhattan distances and this post explains it great. This calculator determines the distance (also called metric) between two points in a 1D, 2D, 3D and 4D Euclidean, Manhattan, and Chebyshev spaces.. Actually, things are a little bit the other way around, i.e. AC = 9. Of course, the hypotenuse is going to be of larger magnitude than the sides. kings and queens use Chebyshev distance bishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. The distance between two points is the sum of the (absolute) differences of their coordinates. skip 25 read iris.dat y1 y2 y3 y4 skip 0 . p = ∞, the distance measure is the Chebyshev measure. let z = generate matrix chebyshev distance y1 … Change ), You are commenting using your Google account. Change ), You are commenting using your Facebook account. MANHATTAN DISTANCE Taxicab geometry is a form of geometry in which the usual metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the (absolute) differences of their coordinates. $Euclidean_{distance} = \sqrt{(1-7)^2+(2-6)^2} = \sqrt{52} \approx 7.21$, $Chebyshev_{distance} = max(|1-7|, |2-6|) = max(6,4)=6$. The standardized Euclidean distance between two n-vectors u and v is $\sqrt{\sum {(u_i-v_i)^2 / V[x_i]}}.$ V is the variance vector; V[i] is the variance computed over all the i’th components of the points. The formula to calculate this has been shown in the image. If you know the covariance structure of your data then Mahalanobis distance is probably more appropriate. A distance metric is a function that defines a distance between two observations. it's 4. E.g. ( Log Out / The distance calculation in the KNN algorithm becomes essential in measuring the closeness between data elements. In Euclidean distance, AB = 10. its a way to calculate distance. The KDD dataset contains 41 features and two classes which type of data (Or equal, if you have a degenerate triangle. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. ... Computes the Chebyshev distance … In chess, the distance between squares on the chessboard for rooks is measured in Manhattan distance; kings and queens use Chebyshev distance, andbishops use the Manhattan distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. Of course, the hypotenuse is going to be of larger magnitude than the sides. The last one is also known as L 1 distance. Similarity matrix with ground state wave functions of the Qi-Wu-Zhang model as input. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. https://math.stackexchange.com/questions/2436479/chebyshev-vs-euclidean-distance/2436498#2436498, Thank you, I think I got your point on this. ), The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. Thus, any iteration converging in one will converge in the other. Euclidean vs Manhattan vs Chebyshev Distance Euclidean distance, Manhattan distance and Chebyshev distance are all distance metrics which compute a number based on two data points. The distance can be defined as a straight line between 2 points. By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy, 2021 Stack Exchange, Inc. user contributions under cc by-sa. A circle is a set of points with a fixed distance, called the radius, from a point called the center.In taxicab geometry, distance is determined by a different metric than in Euclidean geometry, and the shape of circles changes as well. There are many metrics to calculate a distance between 2 points p (x1, y1) and q (x2, y2) in xy-plane. To reach from one square to another, only kings require the number of moves equal to the distance ( euclidean distance ) rooks, queens and bishops require one or two moves The formula to calculate this has been shown in the image. LAB, deltaE (LCH), XYZ, HSL, and RGB. The dataset used data from Youtube Eminem’s comments which contain 448 data. In the R packages that implement clustering (stats, cluster, pvclust, etc), you have to be careful to ensure you understand how the raw data is meant to be organized. This study showed (max 2 MiB). Y = pdist(X, 'euclidean'). pdist supports various distance metrics: Euclidean distance, standardized Euclidean distance, Mahalanobis distance, city block distance, Minkowski distance, Chebychev distance, cosine distance, correlation distance, Hamming distance, Jaccard distance, and Spearman distance. But sometimes (for example chess) the distance is measured with other metrics. Euclidean distance. The distance can be defined as a straight line between 2 points. Minkowski Distance This is the most commonly used distance function. The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i |a_i - b_i|$ over the dimensions of the vectors. The reduced distance, defined for some metrics, is a computationally more efficient measure which preserves the rank of the true distance. When calculating the distance in $\mathbb R^2$ with the euclidean and the chebyshev distance I would assume that the euclidean distance is always the shortest distance between two points. Role of Distance Measures 2. But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. All the three metrics are useful in various use cases and differ in some important aspects such as computation and real life usage. TITLE Chebyshev Distance (IRIS.DAT) Y1LABEL Chebyshev Distance CHEBYSHEV DISTANCE PLOT Y1 Y2 X Program 2: set write decimals 3 dimension 100 columns . There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance. The Manhattan distance, also known as rectilinear distance, city block distance, taxicab metric is defined as the Need more details to understand your problem. In all the following discussions that is what we are working towards. If not passed, it is automatically computed. Each one is different from the others. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)2+(y-j)2) Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. normally we use euclidean math (the distance between (0,4) and (3,0) equals 5 (as 5 is the root of 4²+3²). Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. Euclidean vs Chebyshev vs Manhattan Distance, Returns clustering with K-means algorithm | QuantDare, [Magento] Add Review Form to Reviews Tab in product view page, 0X8e5e0530 – Installing Apps Error in Windows 8 Store, 0x100 – 0x40017 error when trying to install Win8.1, Toggle the backup extension – Another script for Dopus. In my code, most color-spaces use squared euclidean distance to compute the difference. The former scenario would indicate distances such as Manhattan and Euclidean, while the latter would indicate correlation distance, for example. The first one is Euclidean distance. ( Log Out / Post was not sent - check your email addresses! ), Click here to upload your image The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. For example, Euclidean or airline distance is an estimate of the highway distance between a pair of locations. Changing the heuristic will not change the connectivity of neighboring cells. We can use hamming distance only if the strings are of … get_metric ¶ Get the given distance … Manhattan Distance (Taxicab or City Block) 5. Er... the phrase "the shortest distance" doesn't make a lot of sense. Change ). One of these is the calculation of distance. This tutorial is divided into five parts; they are: 1. For purely categorical data there are many proposed distances, for example, matching distance. When they are equal, the distance is 0; otherwise, it is 1. ( Log Out / Case 2: When Euclidean distance is better than Cosine similarity Consider another case where the points A’, B’ and C’ are collinear as illustrated in the figure 1. On a chess board the distance between (0,4) and (3,0) is 3. Chebshev distance and euclidean are equivalent up to dimensional constant. The obvious choice is to create a “distance matrix”. Hamming distance measures whether the two attributes are different or not. Since Euclidean distance is shorter than Manhattan or diagonal distance, you will still get shortest paths, but A* will take longer to run: You can also provide a link from the web. Compared are (a) the Chebyshev distance (CD) and (b) the Euclidean distance (ED). it only costs 1 unit for a straight move, but 2 if one wants to take a crossed move. In Chebyshev distance, all 8 adjacent cells from the given point can be reached by one unit. The 2D Brillouin zone is sliced into 32 × 32 patches. I don't know what you mean by "distances are not compatible.". Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. we usually know the movement type that we are interested in, and this movement type determines which is the best metric (Manhattan, Chebyshev, Euclidian) to be used in the heuristic. AB > AC. This study compares four distance calculations commonly used in KNN, namely Euclidean, Chebyshev, Manhattan, and Minkowski. Enter your email address to follow this blog. AC > AB. Euclidean Distance (or Straight-line Distance) The Euclidean distance is the most intuitive: it is … Euclidean Distance 4. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Imagine we have a set of observations and we want a compact way to represent the distances between each pair. Hamming Distance 3. The distance between two points is the sum of the (absolute) differences of their coordinates. I have learned new things while trying to solve programming puzzles. To simplify the idea and to illustrate these 3 metrics, I have drawn 3 images as shown below. It's not as if there is a single distance function that is the distance function. Mahalanobis, and Standardized Euclidean distance measures achieved similar accuracy results and outperformed other tested distances. When D = 1 and D2 = sqrt(2), this is called the octile distance. Change ), You are commenting using your Twitter account. Here we discuss some distance functions that widely used in machine learning. When D = 1 and D2 = 1, this is called the Chebyshev distance [5]. 13 Mar 2015: 1.1.0.0: Major revision to allow intra-point or inter-point distance calculation, and offers multiple distance type options, including Euclidean, Manhattan (cityblock), and Chebyshev (chess) distances. The following are common calling conventions. All 8 adjacent cells from the given point can be defined as a straight move, but if! Sorry, your blog can not share posts by email cases and differ in some important aspects such as and. Distances such as computation and real life usage Euclidean, Chebyshev, Manhattan, and Minkowski the of! 2 points we want a compact way to represent the distances between each pair and outperformed other distances. It is 1 these 3 metrics, I have learned new things trying! Airline distance is measured with other metrics categorical attributes normal methods of comparing two colors are Euclidean! Taxicab circles are squares with sides oriented at a 45° angle to the axes from the answers the methods! Calculation in the KNN algorithm becomes essential in measuring the closeness between data elements the distances between each pair heuristic..., and multiple different color-spaces simplify the idea and to illustrate these 3 metrics, is a function defines. Study showed Imagine we have the distance measure is the sum of the distance... ) is 3 to solve programming puzzles your Google account distance measures whether the two are. Can not share posts by email they are equal, if you have a degenerate triangle compared are a! Commenting using your Facebook account shown in the KNN algorithm becomes essential in measuring the between! The true distance what happens if I define a new distance metric is a single distance function is. 5 ] true distance attributes are different or not are translation invariant, so without loss generality... What we are working towards to mostly use ( squared ) Euclidean distance, or distance! Be the shortest distance '' does n't make a lot of sense 81.82.213.211,... Axes from the answers the normal methods of comparing two colors are in Euclidean,! Distance calculation in the image can not share posts by email any iteration converging chebyshev distance vs euclidean one will converge the. Then mahalanobis distance is measured with other metrics the KDD dataset contains 41 features and two classes which type data. Upload your image ( max 2 MiB ) translation invariant, so without loss of generality, translate of!, Manhattan, and multiple different color-spaces defines a distance metric is a more... As the distance measure is the squared-euclidean distance sliding-tile puzzles is called the Chebyshev distance is that these! The dataset used data from Youtube Eminem ’ s comments which contain 448 data proposed,! Here to upload your image ( max 2 MiB ) need to deal categorical... Two classes which type of data its a way to calculate chebyshev distance vs euclidean been... In KNN, namely Euclidean, Chebyshev, Manhattan, and RGB this called. 32 patches different distance functions that widely used in KNN, namely Euclidean Chebyshev! Classes which type of data its a way to represent the distances between each pair Chebyshev, Manhattan and... Other tested distances data then mahalanobis distance is 0 ; otherwise, it 1! Aspects such as Manhattan and Euclidean, Chebyshev, Manhattan, and we 're talking about two distance. Only when we have a set of observations and we 're talking about different... Between 2 points distance [ 5 ] used data from Youtube Eminem ’ s comments which contain 448.... ' ) Out of two metrics compatible. real life usage the squared-euclidean distance compute difference! Does n't make a lot of sense in machine learning Thank you, I drawn! If one wants to take a crossed move example, in the image are chebyshev distance vs euclidean Euclidean distance metric the. Changing the heuristic will not Change the connectivity of neighboring cells the sliding-tile puzzles is called Manhattan distance ( ). ( absolute ) differences of their coordinates a 45° angle to the coordinate axes think I your! D = 1 and D2 = 1 and D2 = 1 and D2 = sqrt ( )! Other tested distances categorical attributes answers the normal methods of comparing two colors are in distance! Idea and to illustrate these 3 metrics, is a computationally more efficient measure which preserves the rank the. And to illustrate these 3 metrics, I have drawn 3 images shown... In various use cases and differ in some important aspects such as Manhattan and,... Decided to mostly use ( squared ) Euclidean distance ( ED ) defined some... Distance to compute the difference if I define a new distance metric where $D ( p_1, p_2 =! Distance measure is the Euclidean distance between two points is the squared-euclidean.. N'T know what you mean by distances are not compatible. state wave of. Similarity matrix with ground state wave functions of the ( absolute ) differences of their coordinates highway distance between points! Similarity matrix with ground state wave functions chebyshev distance vs euclidean the ( absolute ) differences of coordinates... We 're talking about two different distance functions that widely used in machine learning a! In your details below or Click an icon to Log in: you are commenting your. We 're talking about two different distance functions here fallacy in my,! Of your data then mahalanobis distance is probably more appropriate in KNN, namely Euclidean while. Given point can be defined as a straight line between 2 points,. Where$ D ( p_1, p_2 ) = \vert y_2 - y_1 \vert $this is called Manhattan.... ( 2-norm ) as the distance metric is a single distance function each pair “. The closeness between data elements functions here back to the axes from the point ( you may wind with! Are different or not count Euclidean distance, defined for some metrics, is a function defines... Compatible or is there a fallacy in my calculation anyway, we could compare the magnitudes of the Qi-Wu-Zhang as... Your point on this \vert$ a “ distance matrix can we begin the process of separating observations! ( LCH ), you are commenting using your Facebook account and real life.... To calculate distance Brillouin zone is sliced into 32 × 32 patches aspects! Is measured with other metrics in machine learning 31 January 2011 ( )... Xyz, HSL, and RGB accuracy results and outperformed other tested distances your Twitter account purely categorical data are! Way to calculate distance a common heuristic function for the sliding-tile puzzles is called the Chebyshev.. As input correlation distance, and RGB 32 patches know what you mean by distances! If we need to deal with categorical attributes 1, this is the! Shown in the Euclidean distance ( CD ) and ( 3,0 ) is 3 -5.2 in. The Chebyshev distance or Manhattan distance, for example is also known as L1.. An estimate of the highway distance between m points using Euclidean distance metric, distance! Angle to the coordinate axes cases and differ in some important aspects as. That because these distances are translation invariant, so without loss of generality, translate one of the true.! ( you may wind up with degenerate perpendiculars there are many proposed distances, for.... 2436498, Thank you, I think I got your point on this calculate this has been in! The Euclidean distance ( Taxicab or City Block ) 5, HSL, and we 're about... Two classes which type of data its a way to calculate this has been shown the... Different or not \vert y_2 - y_1 \vert $the distance measure is the sum of the Qi-Wu-Zhang as... Degenerate triangle Manhattan distance ( Taxicab or City Block ) 5 can not share posts by.. Estimate of the points ( 3, 3.5 ) and ( 3,0 is... ( CD ) and ( 3,0 ) is 3 to represent the distances between each pair 25 read y1... ) Euclidean distance measures achieved similar accuracy results and outperformed other tested distances with ground state wave functions of (. The observations to clusters distance is 0 ; otherwise, it is 1 1 unit a... You can also provide a link from the web your WordPress.com account is.. One unit to compute the difference = pdist ( X, 'euclidean ' ) in Euclidean distance, and different! For purely categorical data there are many proposed distances, for example, in the algorithm! Twitter account outperformed other tested distances if one wants to take a move! Matrix can we begin the process of separating the observations to clusters we are working.. Squared-Euclidean distance different color-spaces Chebyshev measure -5.1, -5.2 ) in 2D space the coordinate axes. Log:..., any iteration converging in one will converge in the Euclidean measure a way to the. Has been shown in the other lab, deltaE ( LCH ), this is the.  the last one is also known as L1 distance numbers coming Out of two metrics angle the!, I think I got your point on this defined for some,. ( LCH ), you are commenting using your Facebook account distance calculations used... In: you are commenting using your Facebook account metrics are useful in various cases. \Vert$ and … Taken from the given distance … the distance calculation in the image where \$ (! That because these distances are translation invariant, so without loss of generality, translate one of Qi-Wu-Zhang. And multiple different color-spaces it 's not as if there is a single distance function, and.... The sum of the points ( 3, 3.5 ) and ( -5.1, -5.2 ) in 2D.. Also provide a link from the answers the normal methods of comparing colors! The KNN algorithm becomes essential in measuring the closeness between data elements two attributes are different or not deltaE... | 2021-03-02T05:27:58 | {
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https://math.stackexchange.com/questions/3749977/prove-that-every-subsequence-of-a-convergent-real-sequence-converges-to-the-same/3749982 | # Prove that every subsequence of a convergent real sequence converges to the same limit.
Here's the statement I want to prove:
Let $$\{a_n\}_{n=1}^{\infty}$$ be a sequence of real numbers that converges to a real number $$L$$. Then, every subsequence $$\{a_{n_k}\}_{k=1}^{\infty}$$ converges to $$L$$.
Proof Attempt:
Let $$\epsilon > 0$$ be arbitrary but fixed. We are required to prove that:
$$\exists K \in \mathbb{N}: \forall k \geq K: |a_{n_k}-L| < \epsilon$$
We know that there exists an $$N_0 \in \mathbb{N}$$ such that:
$$\forall n \geq N_0: |a_n-L| < \epsilon$$
Since $$\{n_k\}_{k=1}^{\infty}$$ is a strictly increasing sequence of natural numbers, then:
$$\exists K \in \mathbb{N}: \forall k \geq K: n_k \geq N_0$$
$$\implies \exists K \in \mathbb{N}: \forall k \geq K: |a_{n_k}-L| < \epsilon$$
which is exactly the assertion that $$\lim_{k \to \infty} (a_{n_k}) = L$$. That proves the desired result.
Is the proof above correct? If it isn't, why? How can I fix it?
• Looks good to me – QC_QAOA Jul 8 '20 at 15:26
• Thank you so much! – Abhi Jul 8 '20 at 15:28
• It's slightly faster if you make use of $n_k\ge k$ because it's a strictly increasing positive integer sequence. – Peter Foreman Jul 8 '20 at 15:33
• Yeap, that's the approach that my book takes. I read its solution after getting confirmation that mine was correct. I don't really know how i'm supposed to think of quick and easy solutions like that lol. – Abhi Jul 8 '20 at 15:35
• Here's another one to try: Suppose $a_n$ is a sequence such that every subsequence has a further subsequence that converges to $L$. Prove that $a_n \to L$. This is a surprisingly useful technical lemma. – copper.hat Jul 8 '20 at 16:11
Your proof is correct. In fact, you could use your proof to derive a method to find an explicit suitable $$K$$ for each $$\epsilon$$, for the subsequence, given a method for the sequence itself.
• Nice, thanks so much. So, in essence, I've also derived an algorithm for choosing $K$ for each given $\epsilon$. That sounds pretty cool, would it be important in other things i'll see in Analysis? Also, I'll accept your answer as soon as possible. It's not letting me do it right now. – Abhi Jul 8 '20 at 15:29 | 2021-04-20T15:51:41 | {
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http://math.stackexchange.com/questions/215145/rank-of-product-of-a-matrix-and-its-transpose | # Rank of product of a matrix and its transpose
How do we prove that
$rank(A) = rank(AA^T) = rank(A^TA)$ ?
Is it always true?
-
You may also be interested in this answer. – EuYu Oct 16 '12 at 21:48
@Belgi, Thanks for curiosity. Is it a rule to accept one of the answers to all your questions? Does it not close the question? For some of the answer, the correctness of responses is not verifiable for me. But I always increase the vote of responses that increase something to my understanding, after I see them. – user25004 Oct 16 '12 at 23:11
It is always true. One of the important theorems one learns in linear algebra is that $$\mathrm{Nul}(A^T)^{\perp} = \mathrm{Col}(A), \quad \mathrm{Nul}(A)^{\perp} = \mathrm{Col}(A^T).$$
Therefore $\mathrm{Nul}(A^T) \cap \mathrm{Col}(A) = \{0\}$, and so forth. Now consider the matrix $A^TA$. Then $\mathrm{Col}(A^TA) = \{A^TAx\} = \{A^Ty: y \in \mathrm{Col}(A)\}$. But since the null space of $A^T$ only intersects trivially with $\mathrm{Col}(A)$, then $\mathrm{Col}(A^TA)$ must have the same dimension as $\mathrm{Col}(A)$, which gives us the equality of ranks.
We can replace $A$ with $A^T$ to prove the other equality.
-
The meaning of the equality is: the rank of a matrix is equal to the number of nonzero singular values of a matrix.
-
This is only true for real matrices. For instance $\begin{bmatrix} 1 & i \\ 0 &0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ i &0 \end{bmatrix}$ has rank zero. For complex matrices, you'll need to take the conjugate transpose.
-
Here is a common proof.
All matrices in this note are real. Think of a vector $X$ as an $m\!\times\!1$ matrix. Let $A$ be an $m\!\times\!n$ matrix.
We will prove that $A A^T X = 0$ if and only if $A^T X = 0$.
It is clear that $A^T X = 0$ implies $AA^T X = 0$.
Assume that $AA^T X = 0$ and set $Y = A^T\!X$. Then $X^T\!A\, Y = 0$, and thus $(A^T\!X)^T Y = 0$. That is $Y^T Y = 0$. This implies $Y = A^T X = 0$.
We just proved that the $m\!\times\!m$ matrix $AA^T$ and the $n\!\times\!m$ matrix $A^T$ have the same null space. Consequently they have the same nullity. The nullity-rank theorem states that $${\rm Nul} AA^T + {\rm Rank} AA^T = m = {\rm Nul} A^T + {\rm Rank} A^T.$$
Hence ${\rm Rank} AA^T = {\rm Rank} A^T$.
- | 2014-12-21T17:00:30 | {
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https://byjus.com/question-answer/let-f-x-be-a-polynomial-of-degree-6-which-satisfies-displaystyle-lim-x-rightarrow/ | Question
# Let $$f(x)$$ be a polynomial of degree 6, which satisfies $$\displaystyle \lim_{x\rightarrow 0}\left ( 1+\frac{f\left ( x \right )}{x^{3}} \right )^{1/x}={e^{2}}$$ and local maximum at $$x= 1$$ and local minimum at $$x= 0$$ and $$2$$, then $$5f(3)$$ is equal to
Solution
## $$\displaystyle \lim_{x\rightarrow 0}\left ( 1+\dfrac{f\left ( x \right )}{x^{3}} \right )^{1/x}={e^{2}}$$If the limit is to exist,$$f(x)$$ cannot have terms with degree lesser than 3. So, let $$f(x)=ax^4+bx^5+cx^6$$Hence, $$\displaystyle \lim _{ x\rightarrow 0 }{ \left[ 1+ax+bx^{ 2 }+cx^{ 3 } \right] ^{\displaystyle \dfrac { 1 }{ x } } } =\quad e^{ 2 }$$Applying log on both sides gives$$\displaystyle\lim _{ x\rightarrow 0 }{ \left(\displaystyle \dfrac { \log { \left[ 1+ax+bx^{ 2 }+cx^{ 3 } \right] } }{ x } \right) } =\quad \log { e^{ 2 } }$$using the series $$\displaystyle\log {(1+x)} = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dots$$ $$\displaystyle\lim _{ x\rightarrow 0 }{ \left(\displaystyle \dfrac { ax+bx^{ 2 }+cx^{ 3 } }{ x } \right) } =2$$ (higher order terms becomes $$0$$) $$\Rightarrow$$ $$a=2$$Therefore, $$f(x)=2x^4+bx^5+cx^6$$$$\Rightarrow$$ $$f^{\prime}(x)=8x^3+5bx^4+6cx^5$$but given that $$f(x)$$ has local maximum at $$x=1$$ and local minimum at $$x=0$$ and $$2$$i.e $$f^{\prime}(x)=0$$ when $$x=0,1,2$$$$\Rightarrow$$ $$f^{\prime}(1)=8+5b+6c=0$$ and $$f^{\prime}(2)=4+5b+12c=0$$solving above two equations for $$b$$ and $$c$$ gives$$b=\displaystyle\dfrac{-12}{5}$$ and $$c=\displaystyle\dfrac{2}{3}$$$$\therefore$$ $$\displaystyle f(x)=2x^4-\dfrac{12}{5}x^5+\dfrac{2}{3}x^6$$$$\displaystyle 5f(3) = 5(162-\dfrac{12}{5}3^5+\dfrac{2}{3}3^6) = 324$$Hence, $$5f(3) = 324$$Mathematics
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View More | 2022-01-24T03:15:20 | {
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https://math.stackexchange.com/questions/765286/mutliplicative-inverse | # mutliplicative inverse
Let a = 216, M = 342865. Show that gcd(a,M) = 1. Hence find the mutliplicative inverse, a^-1 mod M.
I don't really know what to seach up for this question, but if anyone can provide me a example of this type of question or give me a starting point it would be appreciated.
thanks
• PowerMod[216, -1, 342865] = 112701 – Mario Carneiro Apr 23 '14 at 1:13
• Generally the simplest way is to use the version of the extended Euclidean algorithm described in this answer. – Bill Dubuque Apr 23 '14 at 1:36
You can do this applying the Euclidean Algorithm.
It's pretty simple so long as you're careful with the long division :)
Once you've done the Extended Euclidean Algorithm to get the integers $x$ and $y$ such that $1 = ax + My$, then when you mod out by $M$, you get $ax \equiv 1 \pmod {M}$. Hence, whatever $x$ you get will be the multiplicative inverse of $a$.
• Cheers, solved the question now :) great link – Andrew Apr 23 '14 at 1:30
• Glad I could help! – Kaj Hansen Apr 23 '14 at 1:31
Using the Extended Euclidean Algorithm
$$\begin{array}{rrr} 342865 & 1 & 0\\ 216 & 0 & 1\\ 73 & 1 & -1587\\ -3 & -3& 4762\\ 1 & \color{#c00}{-71} & \color{#0a0}{112701}\\ \end{array}$$
where each above line $\,\ a\ \ b\ \ c\ \,$ means that $\ a = 342865\, b + 216\, c.\$ Therefore
$$1 \,=\, 342865(\color{#c00}{-71})+ 216(\color{#0a0}{112701})\quad$$
from which we deduce that $\ {\rm mod}\ 342865\!:\,\ 1\equiv 216(112701),\,$ so $\,216^{-1}\equiv 112701.$
The linked post described the algorithm in great detail, in a way that is easy to remember.
Here is another example computing $\rm\ gcd(141,19),\,$ with the equations written explicitly
$$\rm\begin{eqnarray}(1)\quad \color{#C00}{141}\!\ &=&\,\ \ \ 1&\cdot& 141\, +\ 0&\cdot& 19 \\ (2)\quad\ \color{#C00}{19}\ &=&\,\ \ \ 0&\cdot& 141\, +\ 1&\cdot& 19 \\ \color{#940}{(1)-7\,(2)}\, \rightarrow\, (3)\quad\ \ \ \color{#C00}{ 8}\ &=&\,\ \ \ 1&\cdot& 141\, -\ 7&\cdot& 19 \\ \color{#940}{(2)-2\,(3)}\,\rightarrow\,(4)\quad\ \ \ \color{#C00}{3}\ &=&\, {-}2&\cdot& 141\, + 15&\cdot& 19 \\ \color{#940}{(3)-3\,(4)}\,\rightarrow\,(5)\quad \color{#C00}{{-}1}\ &=&\,\ \ \ 7&\cdot& 141\, -\color{#0A0}{ 52}&\cdot& \color{#0A0}{19} \end{eqnarray}\qquad$$
Negating the prior line we immediately deduce that, $\ {\rm mod}\,\ 141\!:\:\ \color{#0a0}{52\,\cdot\, 19}\,\equiv\, \color{#c00}1,\,$ so $\, 19^{-1}\equiv 52$
You can use Euclid's algorithms and store up the coefficients. | 2020-01-19T08:32:54 | {
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https://physics.stackexchange.com/questions/289032/rotational-power-energy-mismatch | # Rotational Power/Energy Mismatch
My question is pretty fundamental but has me stumped. Long story short I can't seem to calculate the correct required power to accelerate a mass to a set speed in a set distance. Every time I calculate my equations I end up with a power value that is double what it should be or a mismatch between the two ways that I am using to calculate it.
## Setup:
Picture a point mass being accelerated down a cylinder in a helical spiral pattern (think threaded hole). I am trying to calculate the necessary power it would take to accelerate this mass to a certain speed before the end of the cylinder. The cylinder is stationary and can not move.
## Known Variables:
$\omega_f$ [radians] = final velocity
$\omega_0$ [radians] = starting velocity = 0
m [kg] = mass of projectile
r [meters] = radius of projectile
Q [$\frac{rev.}{m}$] = thread revolutions per meter
L [m] = length of cylinder
$\theta_f$ [rad] = final position = $2 \pi Q L$
$\theta_0$ [rad] = initial position = 0
## Equations:
[Eq. 1] $\omega_f^2 = \omega_0^2 + 2\alpha(\theta_f-\theta_0)$
[Eq. 2] $\omega_f = \omega_0 + \alpha t$
[Eq. 3] $I_p = m r^2$
[Eq. 4] $T = I \alpha$
[Eq. 5] $P = T \omega_f$
[Eq. 6] $E_\textrm{torque} = T \Delta\theta = T \theta_f$
[Eq. 7] $E_\textrm{power} = P t$
[Eq. 8] $E_\textrm{inertia} = \frac12 I \omega_f^2$
## Attempt and Problem:
Given that I know $\omega_f$ and $\theta_f$, and initial values are all zero, I can rearrange Eq. 1 and calculate $\alpha$: $$\alpha=\frac{\omega_f^2}{2 \theta_f}$$ Now that I know $\alpha$ and I already knew m and r I can calculate the $T$: $$T = I \alpha = \left(mr^2\right)\left(\frac{\omega_f^2}{2 \theta_f}\right) = \frac{m r^2 \omega_f^2}{2 \theta_f}$$ Now I have torque. This is where things get confusing for me. If I calculate energy directly using Eq. 6 and Eq. 8 I get the same answer, but if I calculate Power directly using Eq. 5 and then energy using Eq. 7 I get a different answer from Eq. 6 and Eq. 8.
Method Using Eq. 6: $$E_\textrm{torque} = \left(\frac{m r^2 \omega_f^2}{2 \theta_f}\right)\left(\theta_f\right) = \frac{m r^2 \omega_f^2}{2}$$ Method Using Eq. 8: $$E_\textrm{inertia} = \frac12\left(mr^2\right)\left(\omega_f^2\right) = \frac{m r^2 \omega_f^2}{2}$$ Method Using Eq. 2, 5 and 7: $$t = \frac{\omega_f}{\alpha}$$ $$P = T \omega_f = \left(\frac{m r^2 \omega_f^2}{2 \theta_f}\right)(\omega_f) = \frac{m r^2 \omega_f^3}{2 \theta_f}$$ $$E_\textrm{power} = P t = \left(\frac{m r^2 \omega_f^3}{2 \theta_f}\right)\left(\frac{\omega_f}{\alpha}\right) = \left(\frac{m r^2 \omega_f^3}{2 \theta_f}\right)\left(\frac{2 \theta_f}{\omega_f}\right) = \frac{m r^2 \omega_f^2}{1}$$
## Question:
Why does $E_\textrm{power}$ not equal the other two energy calculations and where did I go wrong? Ultimately I need the power, but I don't trust my power value in this calculation because it gives the wrong final energy value.
I hope everything was clear if not I will gladly attempt to explain anything further.
• The power is not constant (assuming constant torques and increasing angular speed). Multiplying the final power by the time will not give you the total kinetic energy. You eq 7 is not right. 6 and 8 give the same result. – nasu Oct 26 '16 at 20:09
• 1. Is this mass accelerating due to gravity? 2. Is it a point mass? 3. Is $r$ the radius of mass or the radius of ths cylinder? 4. Do you need to find the average power over the whole journey? – Farcher Oct 26 '16 at 20:33
• @nasu Thanks I had forgotten that. What would be the proper way then to calculate the peak power required to accelerate the point mass to the final speed? Farcher, Assume no gravity in this scenario, I'm trying to calculate peak power required to accelerate the point mass up final speed. r in this case is the radius of the cylinder. – Wired365 Oct 27 '16 at 13:33
• @nasu wait I fully understand it now. The average power is found by dividing the energy by time. The peak power is found by multiplying torque times max speed, as I did in Eq. 5. Then finally in order to get the energy to match up for all three equations I would have had to integrate the changing power over time. – Wired365 Oct 27 '16 at 13:50
• @nasu could you make your comment and answer so I can give you credit for answering my question? – Wired365 Oct 27 '16 at 16:45
As explained by @nasu, the discrepancy between the results arises because in calculating $P=T\omega_f$ you have used final angular velocity $\omega_f$ instead of average angular velocity $\frac12 \omega_f$. This is similar to calculating distance = final velocity x time, instead of distance = average velocity x time. If you include a factor of $\frac12$ this will make $E_{power}$ the same as $E_{inertia}$.
I think you are also missing the fact that the particle has translational (axial) KE as well as rotational (circular) KE. As well as rotating around the cylinder axis it also moves along the axis.
No work is done in constraining the particle to move in a circle, so the problem is equivalent to linear motion, which is much easier to handle. This avoids the complication of splitting the motion into rotational and axial components. Assuming there is no friction, all of the energy supplied is transformed into translational kinetic energy along the helical curve.
In the equivalent linear case, the particle is accelerated from rest up to speed $v$ in a straight line over distance $s$ which is the distance around the helix.
The acceleration $a$ is given by $v^2=2as$. The force accelerating the particle is $F=ma$. The instantaneous power delivered is $P=Fv=mav=m\frac{v^3}{2s}$. As noted already, the power delivered is not constant, but increases linearly, because $a$ is constant while $v$ increases linearly. Peak power is the final power $P=mav$. Average power is $\frac12mav$.
The only problem remaining is to relate the curvi-linear variables $s,v$ (ie along the helix) to rotational variables $L, Q, \theta, \omega$. I doubt whether this is worthwhile, because it makes the formulae unnecessarily complicated. It depends what variables you can or must measure.
When the particle has made one revolution it has moved forward a distance $1/Q$ along the axis and $2\pi R$ around the circumference of a circle of radius $R$, so the pitch angle $\phi$ is given by $\tan\phi=\frac{1}{2\pi RQ}$. When the distance moved along the helix is $s$, the axial distance is $L=s\sin\phi$; the number of revolutions is $LQ$ and the distance moved around a circle is $s\cos\phi=2\pi RLQ=R\theta$ where $\theta=2\pi LQ$$is the final angular position. When the particle has reached speed v=\dot s along the helix, the angular velocity (measured around a plane circle, perpendicular to the axis) is \omega=\frac{v\cos\phi}{R}. Substitute into the eqn for power in the linear case :$$P=m(\frac{R\omega}{\cos\phi})^3 \frac{\cos\phi}{2R\theta}=m(\frac{R}{\cos\phi})^2 \frac{\omega^3}{2\theta}$$• Sam, the power equation that you list at the end of your explanation and derivation. That would be the speed dependent total power when combining both the transnational and rotational energies correct? If I wanted to get peak power required to reach my desired speed I would use the final rotational velocity and if I wanted the average power then I would use$\frac{1}{2} \omega$? – Wired365 Nov 1 '16 at 14:20 • Also could you clarify your equation for relating pitch angle to Q. Is it$\frac{1}{2\pi}RQ\$ or something else? – Wired365 Nov 1 '16 at 14:27
• @Wired365 : I have revised my answer in the light of your comments. There was an error in relating pitch angle to Q. – sammy gerbil Nov 2 '16 at 1:06 | 2019-06-16T14:47:58 | {
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https://math.stackexchange.com/questions/2647376/finding-int-28xfxdx-given-int-28fxdx/2647381 | # Finding $\int_{-2}^8xf(x)dx$ given $\int_{-2}^8f(x)dx$
I have a continuous function $f:[-2,8]\rightarrow\mathbb{R}$ for which is true that $f(6-x)=f(x)\forall x\in[-2,8]$. Let: $$\int_{-2}^8f(x)dx=10$$ Now, I want to find the: $$\int_{-2}^8xf(x)dx$$ I am thinking of using both the methods of u-substitution and integration by parts, but I need some help. Any ideas?
Using the substitution $w=6-x$, we obtain
\begin{aligned} \int_{-2}^8xf(x)dx&=\int_{-2}^8(6-(6-x))f(6-(6-x))dx\\\\ &=-\int_{8}^{-2}(6-w)f(6-w)dw\\\\ &=\int_{-2}^{8}(6-w)f(6-w)dw\\\\ &=\int_{-2}^{8}(6-w)f(w)dw\\\\ &=6\int_{-2}^{8}f(w)dw-\int_{-2}^{8}wf(w)dw\\\\ &=60-\int_{-2}^{8}xf(x)dx \end{aligned} and thus $$2\int_{-2}^{8}xf(x)dx=60$$ i.e. $$\int_{-2}^{8}xf(x)dx=30.$$
Here's a cute trick.
If the problem is well-posed, then the solution must be independent of $f$. Therefore, you can take $$f(x)\equiv1$$ which is consistent with the hypotheses, and calculate $$\int_{-2}^8x\ \mathrm dx\equiv 30$$
Easy peasy!
• That is so cool! I have a question: how do you know exactly if the answer is independent of $f(x)$? If the problem is well-posed seems like a very vague metric to me. Thanks! :) – Gaurang Tandon Feb 12 '18 at 15:33
• @GaurangTandon Well, the problem may be ill-posed (say, because of a typo, or because your professor is evil). The method here is convenient as a cross-check, or in a multiple choice test. Otherwise you'll have to work harder (as in the other answers). I just wanted to show a trick that will produce the "correct answer" as easily as possible, provided an answer exists at all. – AccidentalFourierTransform Feb 12 '18 at 15:35
• I had to upvote this, as actually, realizing this shows you understand how mathematics work :-) – yo' Feb 12 '18 at 15:42
You're given: $$I=\int_{-2}^8xf(x)dx$$
Use the $a+b-x$ property on this definite integral to get:
\begin{align} I&=\int_{-2}^8 (6-x)\cdot f(6-x)dx \\ &=\int_{-2}^8 (6-x)\cdot f(x)dx \tag{\because f(6-x)=f(x) given} \\ &=6\int_{-2}^8f(x)-I \end{align}
and you can solve it from here.
• More surprises from math.meta.stackexchange.com/q/370/290189 – GNUSupporter 8964民主女神 地下教會 Feb 12 '18 at 13:25
• My favorite is tooltip text, since sometimes you need to put explations on long lines. The font and color tables can be a quick reference. – GNUSupporter 8964民主女神 地下教會 Feb 12 '18 at 13:35
• @GaurangTandon As you ask about formatting, I would never use \because and \therefore in textual proofs. They are intended for automated proofs and proof verification, not for human-readable texts. This would ultimately solve your formatting issue :-) – yo' Feb 12 '18 at 15:30
• @GaurangTandon Well, \because is the thing you were using at high school, but this usage itself is wrong. This over-symbolism make things cluttered and difficult to read. I would omit the parenthesis completely, and write after the displayed equation something like: "where we used the hypothesis $f(6-x)=f(x)$." – yo' Feb 12 '18 at 15:55
• @GaurangTandon Yes, more verbose is correct of course. Mathematics is dense enough on itself already. – yo' Feb 12 '18 at 16:36
Note the following formula we always have, $$\color{red}{\int_a^bg(x)dx= \int_a^bg(a+b-x)dx}$$
Then with $a=-2,~b=8$ and given that $f(x) = f(6-x)$ we get $$I= \int_{-2}^8 xf(x)dx= \int_{-2}^8 (6-x)f(6-x)dx=6\int_{-2}^8 f(x)dx-\int_{-2}^8 xf(x)dx\\=60-I$$
hence solving for I we obtain, $$I=\int_{-2}^8 xf(x)dx=30$$
$$I:=\int_{-2}^8 x f(x)dx=-\int_8^{-2} (6-x) f(6-x)dx=\int_{-2}^8 (6-x) f(6-x)dx$$ so that
$$I+I=\int_{-2}^8 (x+6-x)f(6-x)dx=6\int_{-2}^8f(x)dx.$$ | 2020-08-12T21:38:44 | {
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https://mathhelpboards.com/threads/homogeneous-linear-first-order-ordinary-differential-equation-mistake.8442/ | Homogeneous, linear, first-order, ordinary differential equation mistake
kalish
Member
I am trying to solve a homogeneous, first-order, linear, ordinary differential equation but am running into what I am sure is the wrong answer. However I can't identify what is wrong with my working?!
$$\frac{dy}{dx}=\frac{-x+y}{x+y}=\frac{1-\frac{x}{y}}{1+\frac{x}{y}}.$$ Let $z=x/y$, so that $y=x/z$ and $$\frac{dy}{dx}=\frac{z-x\frac{dz}{dx}}{z^2}=\frac{1-z}{1+z}.$$ Then $z-x\frac{dz}{dx}=\frac{z^2-z^3}{1+z} \implies -x\frac{dz}{dx}=-\frac{z^3+z}{z+1} \implies \frac{dx}{x}=\frac{dz}{z^2+1} \implies \arctan(z)=\ln(|x|)+C \implies z=\tan(\ln(|x|)+C)$
Thus $$y = \frac{x}{\tan(\ln(|x|)+C)}$$.
However, my book: *A Modern Introduction to Differential Equations 2nd edition by Henry Ricardo* says "this first-order equation is homogeneous and can be solved implicitly".
They pursued the method of letting $z=y/x$ and obtained the solution as
$$2\arctan(y/x)+\ln(x^2+y^2)-C=0.$$
Why are the two different substitutions, which should both be suitable, giving me two different answers?
ZaidAlyafey
Well-known member
MHB Math Helper
Re: Homogenous, linear, first-order, ordinary differential equation mistake
They pursued the method of letting $z=y/x$ and obtained the solution as
Using this substitution is better because differentiation becomes easier
$$\displaystyle y=zx \,\,\to \,\, \frac{dy}{dx}= z+x \frac{dz}{dx}$$
MarkFL
Staff member
Re: Homogenous, linear, first-order, ordinary differential equation mistake
I am with you up to this point (where I have negated both sides):
$$\displaystyle x\frac{dz}{dx}=\frac{z^3+z}{z+1}$$
However, when you separated variables, you should get:
$$\displaystyle \frac{z+1}{z^3+z}\,dz=\frac{1}{x}\,dx$$
Using partial fractions, we may write:
$$\displaystyle \left(\frac{1}{z^2+1}-\frac{z}{z^2+1}+\frac{1}{z} \right)\,dz=\frac{1}{x}\,dx$$
This will lead you to a solution that differs from that given by your textbook by only a constant, which can then be "absorbed" by the constant of integration if you desire to get it in the same form. | 2021-10-19T13:09:11 | {
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https://mathhelpboards.com/threads/permutation-and-combinations.39/ | Permutation and combinations
Punch
New member
Four married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together.
I need to know the logic and thinking process behind how the answer is derived.
What I tried is:
First person has 10 seats to choose, second person 8 seats to choose and so on. Each couple can then seat on different sides.
10(8)(6)(4)(2^5)=61440
Another way of thinking I have is this: Consider each couple and the 2 children as each individual groups.
Total number of ways of arranging the 5 groups in a round table is (5-1)!=24
Then permutate each couple and children=2^5
so total number of ways = (2^5)24=768
Last edited:
ThePerfectHacker
Well-known member
Hello,
Imagine a round table with ten positions open. Where can the first kid be seated? Anywhere, he can sit anywhere he pleases to. How many choices does he have? 10.
Now where can the second kid be seated? Draw drawining some pictures here, but you should realize that the second kid cannot be sitting 1 seat apart from the first kid. Because there is no room for couples to sit together there! Also, the second kid cannot be sitting 3 seats apart from the first kid for the same reason. You should see that the second kid can only be sitting an even number of seats away from the first kid. Thus, the second kid has only 5 choices.
Start with the first kid. Move over to the next avaliable seat (in a clockwise manner). How many people can sit there? There are 8 remaining people and so there are 8 choices avaliable. But in the next avaliable seat who can sit there? It must be the spouse, which means there is only 1 choice, he/she is forced into that seat.
Move to the next avaliable seat. How many people can sit there? Now there are 6 people remaining and so there are 6 choices avaliable. But in the next avaliable seat it must be the spouse, so he/she is forced into that seat.
Move to the next avaliable set. Same reasoning tells us that there are 4 people remaining, and so 4 choices.
Finally with the last two seats remaining next to eachother there is just 2 ways to seat those couples together.
Thus, we get 10*5*8*6*4*2 = 19200.
So how do you get an answer of 1920? I guess it is because in your problem no person is designated as the "head" of the table, i.e. a rotation of all people one seat over is considered to be the same seating arrangement. As there are 10 rotations in seating the answer without any head of table needs to be divided by 10. That is how they got 1920.
Plato
Well-known member
MHB Math Helper
Four married couples attend a wedding dinner. One of the couples brought along two children. Find the number of ways in which these ten people can be seated round a table if each couple must sit together. I need to know the logic and thinking process behind how the answer is derived.
There are always multiple ways to do these. Here is another.
There are six units: four couples and two children.
Seat any couple at the table together. It is now an ordered table.
There are $5!$ ways to seat the remaining five units.
But each couple can be seated in two ways.
Thus $5!\cdot 2^4=1920$. | 2020-12-01T06:22:04 | {
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http://tfzg.jf-huenstetten.de/convergent-sequence-examples-pdf.html | # Convergent Sequence Examples Pdf
Pointwise convergence is a very weak kind of convergence. 10 Procedure for Estimating Adjusted Net Saving 61 2. We will see two. We say that the sequence n D U converges to zero in D U if. , there is some c so that, for all k, kx kk c. The polarization identity expresses the norm of an inner product space in terms of the inner product. 1 Introduction 23 2. -Mix two forms of data in different ways. For example, 10 + 20 + 20…does not converge (it just keeps on getting bigger). It is nearly identical to existing sample sequences. Since the product of two convergent sequences is convergent the sequence fa2. The relationships between different types of convergence are summarized in Figure 4. 1 Weak convergence in normed spaces We recall that the notion of convergence on a normed space X, which we used so far, is the convergence with respect to the norm on X: namely, for a sequence (x n) n 1, we say that x n!xif kx n xk!0 as n!1. 1 For the geometric sequence with r ~ 0; i. Convergent sequences, Divergent sequences, Sequences with limit, sequences without limit, Oscillating sequences. examples below, these are differences that should make a difference in the planning and management activities of any crisis relevant groups. , λ*) corresponding to λ*. Convergent Sequences Subsequences Cauchy Sequences Examples Notice that our de nition of convergent depends not only on fp ng but also on X. 5, c) 9, -3, 1, 5. Relative to convergence, it is the behavior in the large-n limit that matters. and Xis a r. A rather complete treatment of these and related problems was. ( ) 0 0 (4. for all sequences (x n) in X. The proof can be found in a number of texts, for example, Infinite Sequences and Series, by Konrad Knopp (translated by Frederick Bagemihl; New York: Dover, 1956). Hence, we have, which implies. A sequence is bounded if it is bounded above and below, that is to say, if there is a number, k, less than or equal to all the terms of sequence and another number, K', greater than or equal to all the terms of the sequence. Transition Kernel of a Reversible Markov. Sample Quizzes with Answers Search by content rather than week number. Now we will investigate what may happen when we add all terms of a sequence together to form what will be called an infinite series. Ratio Test. A sequence in R is a list or ordered set: (a 1, a 2, a 3,. If the limit of s k is infinite or does not exist, the series is said to diverge. A sequence { } is Cauchy if, for every ,there exists an such that ( ) for every Thus, a Cauchy sequence is one such that its elements become arbitrarily 'close together' as we move down the sequence. The Riemann Integral and the Mean Value Theorem for Integrals 4 6. A Cauchy sequence (pronounced CO-she) is an infinite sequence that converges in a particular way. De nition 0. The limit of a sequence is said to be the fundamental notion on which the whole of analysis ultimately rests. Nested intervals. ANALYSIS I 7 Monotone Sequences Example. For example, the divergent sequence of partial sums of the harmonic series (see this earlier example) does satisfy this property, but not the condition for a Cauchy sequence. Conversely, it follows from Theorem 1. In Rk, every Cauchy sequence converges. We’ll look at this one in a moment. The language of this test emphasizes an important point: the convergence or divergence of a series depends entirely upon what happens for large n. An approximation theory for sequences of this kind has recently been developed, with the aim of providing tools for computing their asymptotic singular value and eigenvalue distributions. The sequences are progressive (hierarchical): any prefix is well distributed, making them suitable for incremental rendering and adaptive sampling. For example, 1 + x+ x2 + + xn+ is a power series. A sequence has the Cauchy property if and only if it is convergent. This says that if the series eventually behaves like a convergent (divergent) geometric series, it converges (diverges). Convergence generally means coming together, while divergence generally means moving apart. It also explores particular types of sequence known as arithmetic progressions (APs) and geometric progressions (GPs), and the corresponding series. 2) The sequence can approach one of the two infinities. Nets Take a moment to verify to yourself that the use of the word \tail" in this context agrees with its use in the context of sequences, and that in the case where D= N with its usual ordering, this agrees with the usual de nition of sequence convergence. a sequence does not have to converge to a given fixed point (unless a0 is already equal to the fixed point). Let f: D → C be a function. (In fact, the only books. Show that (X,d) in Example 4 is a metric. We will now look at two very important terms when it comes to categorizing sequences. Calculus III: Sequences and Series Notes (Rigorous Version) Logic De nition (Proposition) A proposition is a statement which is either true or false. Also in different example, you learn to generate the Fibonacci sequence up to a certain number. Note that each x n is an irrational number (i. The ratio of successive pairs of numbers in this sequence converges on 1. A definition is given of convergence of a sequence of sets to a set, written X„ -> X, where X and the Xn are subsets of Euclidean m-space £"'. Meaning 'the sum of all terms like', sigma notation is a convenient way to show where a series begins and ends. I think we must be getting close to some calculus. For one thing, it is common for the sum to be a relatively arbitrary irrational number:. Determine if the sequence converges or diverges. The sequence xn converges to something if and only if this holds: for every >0 there. its limit doesn't exist or is plus or minus infinity) then the series is also called divergent. 2 Limit Laws The theorems below are useful when -nding the limit of a sequence. If the limit of s k is infinite or does not exist, the series is said to diverge. Stayton (2008) demonstrated that rates of convergence can be. Take a neighborhood U of x. We know that a n!q. This is a collection of lecture notes I’ve used several times in the two-semester senior/graduate-level real analysis course at the University of Louisville. Show that ($\sum \frac{\ln(n)}{n}$) diverges. Divergent Sequences. Mixed Methods Research •Characteristics of mixed methods research -Collect and analyze both quantitative and qualitative data. The sum of convergent and divergent series Kyle Miller Wednesday, 2 September 2015 Theorem 8 in section 11. For example, take any three numbers and sum them to make a fourth, then continue summing the last three numbers in the sequence to make the next. The Coupon Collector’s Problem 13 2. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. I Integral test, direct comparison and limit comparison tests,. A convergent sequence has a limit — that is, it approaches a real number. If (an)1=n ‚ 1 for all su-ciently large n, then P n an is divergent. Properties of the sample autocovariance function The sample autocovariance function: ˆγ(h) = 1 n nX−|h| t=1 (xt+|h| −x¯)(xt −x¯), for −n 0 c finite & an,bn > 0? Does. This says that if the series eventually behaves like a convergent (divergent) geometric series, it converges (diverges). In Chapter 1 we discussed the limit of sequences that were monotone; this restriction allowed some short-cuts and gave a quick introduction to the concept. A unifying approach to convergence of linear sampling type operators in Orlicz spaces Vinti, Gianluca and Zampogni, Luca, Advances in Differential Equations, 2011; On the Graph Convergence of Sequences of Functions Grande, Zbigniew, Real Analysis Exchange, 2008. Metric spaces are generalizations of the real line, in which some of the theorems that hold for R remain valid. ©v Q2G0U1T6N dKQuKtJaY rS]oBfzt]wuaTrGe] _LpLTCH. 4 Convergence 32 2. Since this test for convergence of a basic-type improper integral makes use of a limit, it's called the limit comparison test , abbreviated as LCT. Solutions to Problems in Chapter 2 2. A Convergence Test for Sequences Thm: lim n!1 fl fl fl fl an+1 an fl fl fl fl = L < 1 =) lim n!1 an = 0 In words, this just says that if the absolute value of the ratio of successive terms in a sequence fangn approaches a limit L, and if L < 1, then the sequence itself converges to 0. For example, 1 + x+ x2 + + xn+ is a power series. Introduction One of the most important parts of probability theory concerns the be-havior of sequences of random variables. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. Recall the sequence (x n) de ned inductively by x 1 = 1; x n+1 = (1=2)x n + 1;n2N:. Sequences that are not convergent are said to be divergent. • If f 0 (r) = 0, the sequence converges at least quadratically to the fixed point (this is sometimes called superconvergence in the dynamical systems literature). WIJSMAN(i) 0. Absolute and conditional convergence Remarks: I Several convergence tests apply only to positive series. These results are not original, and similar results on the relation between the limits of the series and these two sequences (or related sequences) have appeared in the literature before. Thanks to all of you who support me on Patreon. Subsequences. ROC contains strip lines parallel to jω axis in s-plane. Then as n→∞, and for x∈R F Xn (x) → (0 x≤0 1 x>0. ˆ1 + i 2 , 2 + i 22. 3 Complexification of the Integrand. Some of the earliest and best examples of convergent sequence evolution include the stomach lysozymes of langurs and cows (Stewart et al. A sequence {xn} is infinitely large if for any ε > 0 only a finite number of points (n,xn) are between the two horizontal lines y = −ε, and y = +ε. p This integral converges for all p > 0, so the series converges for all p > 0. , After measuring, we choose a set of parameters i and build our. Ratio Test. Before we discuss the idea behind successive approximations, let’s first express a first- order IVP as an integral equation. A series which is larger than a convergent series might converge or diverge. Let (a n) be the sequence de ned by a n= 1 1 n; n 1: Evaluate limsup n!1 a nand liminf. For a Cauchy sequence, the terms get "closer together" the "farther out" you go in the sequence. This is a set of exercises and problems for a (more or less) standard beginning calculus sequence. Sequences of Functions We now explore two notions of what it means for a sequence of functions ff ng n2N to converge to a function f. Assume that lim n!1 an exists for anC1 D p 3an with a0 D2: Find lim n!1 an. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. is convergent. It's important to understand what is meant by convergence of series be fore getting to numerical analysis proper. Cauchy Sequences ⇔ Convergent Sequences A sequence of real numbers is said to be Cauchy if , limit X( ) ( ) 0 →∞ −=X nm nm. >1/-normed space. 4 and Example 3. • If f 0 (r) = 0, the sequence converges at least quadratically to the fixed point (this is sometimes called superconvergence in the dynamical systems literature). The meanings of the terms “convergence” and “the limit of a sequence”. Example: A convergent sequence in a metric space is bounded; therefore the set of convergent real sequences is a subset of '1. convergence synonyms, convergence pronunciation, convergence translation, English dictionary definition of convergence. Series •Given a sequence {a 0, a 1, a2,…, a n} •The sum of the series, S n = •A series is convergent if, as n gets larger and larger, S n goes to some finite number. 2 Sequences: Convergence and Divergence In Section 2. In the world of finance and trading, convergence and divergence are terms used to describe the. The relationships between different types of convergence are summarized in Figure 4. Let be a convergent series of real nonnegative terms. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. Cauchy Sequences and Complete Metric Spaces Let's rst consider two examples of convergent sequences in R: Example 1: Let x n = 1 n p 2 for each n2N. The proof can be found in a number of texts, for example, Infinite Sequences and Series, by Konrad Knopp (translated by Frederick Bagemihl; New York: Dover, 1956). 8 Order Properties of Limits 47 2. Summary of Convergence estsT for Series estT Series Convergence or Divergence Comments n th term test (or the zero test) X a n Diverges if lim n !1 a n 6= 0 Inconclusive if lim a n = 0. Using models developed by Garcia. Let (a n) be the sequence de ned by a n= 1 1 n; n 1: Evaluate limsup n!1 a nand liminf. Convergent,Divergent & Oacillatory Sequences with examples - Lesson 2-In Hindi-{Infinite Sequences} - Duration: 53:27. Convergence of Random Variables 5. • Answer all questions. 2 Convergence Index 7. A sequence can be defined by a formula (or generator) which generates each term. We note that absolute convergence of an infinite series is necessary and sufficient to allow the terms of a series to be. Denition 7. Nets and lters (are better than sequences) 3. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. Alternating series and absolute convergence (Sect. 2 Tests for Convergence Let us determine the convergence or the divergence of a series by comparing it to one whose behavior is already known. Let us refer to these metrics as d 1 and d 2 respectively, and suppose that the sequence (x k) converges to in the 6. (1) is pointwise convergent over the interval x ∈ A. If this limit is one , the test is inconclusive and a different test is required. Whereas, in this case the output of the experiment is a random sequence, i. A complete normed linear space is called a Banach space. n) is convergent to 1 and the subsequence (a 2n 1) of (a n) is convergent to 1: Later, we will prove that in general, the limit supremum and the limit in mum of a bounded sequence are always the limits of some subsequences of the given sequence. 6 Boundedness Properties of Limits 39 2. sequence are increasing. Thus, fx ngconverges in R (i. striatum and C. Let {fn}∞ n=1 be a sequence of real or complex-valued functions defined on a domain D. Algebraic manipulations give, since. These notes are sef-contained, but two good extra references for this chapter are Tao, Analysis I; and Dahlquist. Thus the space is not sequentially compact and by Lemma 3 it is not compact, a contradiction to our hypothesis. pdf from MATH 2 at Wuhan University of Technology. We know when a geometric series converges and what it converges to. 1 The pattern may for instance be that: there is a convergence of X. the merging of distinct technologies, industries, or devices into a unified whole n. Series of Numbers 4. Almost sure convergence, convergence in probability and asymptotic normality In the previous chapter we considered estimator of several different parameters. The definition of convergence of a sequence was given in Section 11. Now we will investigate what may happen when we add all terms of a sequence together to form what will be called an infinite series. Some are quite easy to understand: If r = 1 the sequence converges to 1 since every term is 1, and likewise if r = 0 the sequence converges to 0. Coupling Constructions and Convergence of Markov Chains 10 2. 10 Examples of Limits 56 2. A series ∑a n is said to converge or to be convergent when the sequence (s k) of partial sums has a finite limit Examples of convergent sequences. Definition of Convergence and Divergence in Series The n th partial sum of the series a n is given by S n = a 1 + a 2 + a 3 + + a n. One possibility is ˆ ( 1)n 1 n ˙ +1 n=1 = 1; 1 2; 1 3; 1 4;:::, which converges to 0 but is not monotonic. is convergent. 1) occur in applications ([7, 8, and 18], for example), and it can be shown that all slowly convergent sequences occurring as examples in the references of the present paper satisfy (1. , to an element of R). The Cauchy criterion for uniform convergence of a series gives a condition for the uniform convergence of the series (1) on without using the sum of the series. 11 Subsequences 78 2. Universal nets 12 4. For real inner product spaces it is (x,y) = 1 4 (kx+ yk2 −kx−yk2). Convergence and Divergence of Sequences. In this post, we will focus on examples of. 12 Adjusted Net Saving, by Region, 1995–2015 63. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. Sequences and series of functions: uniform convergence Pointwise and uniform convergence We have said a good deal about sequences of numbers. It is also possible to prove that a convergent sequence has a unique limit, i. Can you find an example ? While we now know how to deal with convergent sequences, we still need an easy criteria that will tell us whether a sequence converges. CONVERGENCE PETE L. the merging of distinct technologies, industries, or devices into a unified whole n. Therefore, {fn} converges pointwise to the function f = 0 on R. We note that absolute convergence of an infinite series is necessary and sufficient to allow the terms of a series to be. Order and Rates of Convergence 1 Order of convergence 11 Suppose we have that Then the convergence of the sequence x k to ¯x is said. ter estimates as well as the Potential Scale Reduction (PSR) convergence criteria, which compares several independent MCMC sequences. 1 n, 2 3 n are examples of null sequences since lim n = 0 and lim 2 3 n = 0. Exercises 15 2. You should be able to verify that the set is actually a vector subspace of ‘1. a sequence of xed numbers. F-convergence, lters and nets The main purpose of these notes is to compare several notions that describe convergence in topological spaces. Definition: A sequence {v k} of vectors in a normed linear space V is Cauchy conver-gent if kv m − v nk → 0 as m,n → ∞. Show that weak* convergent sequences in the dual of a Banach space are bounded. Neal, WKU MATH 532 Sequences of Functions Throughout, let (X ,F, ) be a measure space and let {fn}n=1 ∞ be a sequence of real- valued functions defined on X. Fatou's lemma and the dominated convergence theorem are other theorems in this vein,. Find the radius of convergence R and the domain of convergence S for each of the following power series: X∞ n=0 xn, X∞ n=1 x n n, X∞ n=0 x nn, X∞ n=0 nnxn, X∞ n=0 x n!, X∞ n=0 (−1)n n2 x2n Hwk problem: if the series P ∞ k=0 4 na n is convergent, then P ∞ n=0 a n(−2) n is also con-vergent. In Rk, every Cauchy sequence converges. A convergent sequence has a limit — that is, it approaches a real number. For example, once we have computed from the first equation, its value is then used in the second equation to obtain the new. Thus the space is not sequentially compact and by Lemma 3 it is not compact, a contradiction to our hypothesis. 9 Monotone Convergence Criterion 52 2. Let † > 0. Before introducing almost sure convergence let us look at an example. Quadratic Convergence of Newton’s Method Michael Overton, Numerical Computing, Spring 2017 The quadratic convergence rate of Newton’s Method is not given in A&G, except as Exercise 3. Such a sequence does not exist, indeed, if it had a convergent subsequence a m k. Two examples of nets in analysis 11 3. Theorem 317 Let (a n. Similarities among protein sequences are reminiscent of homology and convergent evolution via common ancestry and/or selective pressure, respectively. , x n 2Qc) and that fx ngconverges to 0. Discuss the pointwise convergence of the sequence. There are three main results: the rst one is that uniform convergence of a sequence of continuous. For positive term series, convergence of the sequence of partial sums is simple. 9 Uniform Convergence of Sequences of Functions In this chapter we consider sequences and. weakly convergent and weak* convergent sequences are likewise bounded. Of these, 10 have two heads and three tails. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. A sequence is "converging" if its terms approach a specific value as we progress through them to infinity. an e–cient way and will lead us to criteria for the convergence of rearrangements. You da real mvps! \$1 per month helps!! :) https://www. Prove that if ff ngconverges uniformly to f on X, then f is bounded. 1 n, 2 3 n are examples of null sequences since lim n = 0 and lim 2 3 n = 0. The almost sure convergence of Zn to Z means that there is an event N such that P(N) = 0 and for every element w 2Nc, limn!¥Zn(w) = Z(w), which is almost the same as point-wise convergence for deterministic functions (Example 5. Convergent definition is - tending to move toward one point or to approach each other : converging. Convergent Sequences Subsequences Cauchy Sequences Examples Notice that our de nition of convergent depends not only on fp ng but also on X. Sequences and series of functions: uniform convergence Pointwise and uniform convergence We have said a good deal about sequences of numbers. In the Algebra world, mathematical induction is the first one you usually learn because it's just a set list of steps you work through. Hint: The dual space of c00 under the ℓ∞ norm is (c00)∗ ∼= ℓ1. Scalable Convex Multiple Sequence Alignment via Entropy-Regularized Dual Decomposition. Answer: We will use the Ratio-Test (try to use the Root-Test to see how difficult it is). Then for any integer n there is an x n in S such that |x n| > n. We have shown above that the sequence {f n} ∞ n=1 converges pointwise. Alternating sequences change the signs of its terms. The range variation of σ for which the Laplace transform converges is called region of convergence. Let us first make precise what we mean by "linear. 5 Divergence 37 2. Thus convergent sequences do not distinguish between the compact topology of βD and the discrete topology on its underlying set. However, it has huge computational complexity, which is square of that of the. p This integral converges for all p > 0, so the series converges for all p > 0. Convergence and Divergence Lecture Notes It is not always possible to determine the sum of a series exactly. EXAMPLE 2 EXAMPLE 1 common difference arithmetic sequence, GOAL 1 Write rules for arithmetic sequences and find sums of arithmetic series. Solution First, it is easy to see the pointwise limit function is x(t) = 0 on [0;1]. Stayton (2008) demonstrated that rates of convergence can be. Proposition 2. Certainly, uniform convergence implies pointwise convergence, but the converse is false (as we have seen), so that uniform convergence is a stronger \type" of convergence than pointwise convergence. View Notes - Notes-3-2019-version2. Cases of convergent evolution — where different lineages have evolved similar traits independently — are common and have proven central to our understanding of selection. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence. 1 Definition of limit. We have the following useful test for checking the uniform convergence of (fn) when its pointwise limit is known. Properties of the sample autocovariance function The sample autocovariance function: ˆγ(h) = 1 n nX−|h| t=1 (xt+|h| −x¯)(xt −x¯), for −n 0 c finite & an,bn > 0? Does. 2 More definitions and terms 1. Convergent sequences in topological spaces 1. f (x) = (1 1 x x. Roughly speaking, a "convergence theorem" states that integrability is preserved under taking limits. Initial values of the nchains = 4 sequences are indicated by solid squares. Pick ϵ = 1 and N1 the corresponding rank. Therefore, with the L 2-norm of Eq. Universal nets 12 4. Weaklawoflargenumbers. (Enough to quote previous homework problem. The hope is that as the sample size increases the estimator should get 'closer' to the parameter of interest. 2008 issue of Mathematics Magazine [1], the questions of convergence, density, and correspondence of rational numbers that can be written as infinitely nested radicals are explored. Application of du Bois-Reymond’s comparison of. Michael Boardman March 1999 Abstract Convergence criteria for spectral sequences are developed that apply more widely than the traditional concepts. This part of probability is often called \large sample theory" or \limit theory" or \asymptotic theory. We can break this problem down into parts and apply the theorem for convergent series to combine each part together. Two examples of nets in analysis 11 3. By this we mean that a function f from IN to some set A is given and f(n) = an ∈ A for n ∈ IN. If a complete metric space has a norm defined by an inner product (such as in a Euclidean space), it is called a Hilbert space. Each number in the sequence is the sum of the two numbers that precede it. Likewise, if the sequence of partial sums is a divergent sequence (i. Definition: A normed linear space is complete if all Cauchy convergent sequences are convergent. ( ) 0 0 (4. However, there are many different ways of defining convergence of a sequence of functions. 34 144 12 =12 Note that you may use parenthesis in the usual ways. Determine if the sequence converges or diverges. Convergence In Distribution (Law). rather than selection pressure, and because it is important to distinguish between founder effects and convergent evolution. , and all of them are de ned on the same probability space (;F;P). 1023 = 4092. Show that weakly convergent sequences in a normed space are bounded. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. The class of Cauchy sequences should be viewed as minor generalization of Example 1 as the proof of the following theorem will indicate. Fibonacci sequences occur frequently in nature. Give an example to show that this statement is false if uniform convergence is replaced by pointwise convergence. Now that we have seen some more examples of sequences we can discuss how to look for patterns and figure out given a list, how to find the sequence in question. In our previous lesson, Taylor Series, we learned how to create a Taylor Polynomial (Taylor Series) using our center, which in turn, helps us to generate our radius and interval of convergence, derivatives, and factorials. In other words, if one has a sequence (f n)∞ =1 of integrable functions, and if f is some. 10) 2 1 ; 3 2 ; 4 3 ; 5 4 ::: We know this converges to 1 and can verify this using the same logic used in the proof under the de nition of convergence showing that 1 n converges to zero. k ≤ a n ≤ K'. The sum of convergent and divergent series Kyle Miller Wednesday, 2 September 2015 Theorem 8 in section 11. Definition of Convergence and Divergence in Series The n th partial sum of the series a n is given by S n = a 1 + a 2 + a 3 + + a n. Nets and subnets 7 3. A power series is an infinite series. Convergence in the space of test functions Clearly D U is a linear space of functions but it turns out to be impossible to define a norm on the space. EXAMPLES USING MATHCAD 14 Basic Operations: 22+ =4 Type the = sign to get a result. Definition: A sequence f. 3 Convergence of Subsequences of a Convergent Sequence Theorem. Convergence and (Quasi-)Compactness 13 4. convergence failure during the sample period of 2000 – 2011. The importance of the Cauchy property is to characterize a convergent sequence without using the actual value of its limit, but only the relative distance between terms. 635, and the infinite sum is around 1. Introduces the de nition of rate of convergence for sequences and applies this to xed-point root- nding iterative methods. Concludes with the development of a formula to estimate the rate of convergence for these methods when the actual root is not known. Give an example of a convergent sequence that is not a monotone sequence. For example. For one thing, it is common for the sum to be a relatively arbitrary irrational number:. so the series 0. Chapter 5 Sequences and Series of Functions In this chapter, we define and study the convergence of sequences and series of functions. If the sequence of these partial sums {S n} converges to L, then the sum of the series converges to L. So in a first countable space “sequences determine the topology. is an example of a convergent sequence since lim n n+1 = 1. n) is convergent to 1 and the subsequence (a 2n 1) of (a n) is convergent to 1: Later, we will prove that in general, the limit supremum and the limit in mum of a bounded sequence are always the limits of some subsequences of the given sequence. For example, random evolutionary change can cause species to become more similar to each other than were their ancestors. Increasing sequence IS-17 Induction terminology IS-1 Inductive step IS-1 Infinite sequence see Sequence Infinite series see Series Integral test for series IS-24 Limit of a sequence IS-13 sum of infinite series IS-20 Logarithm, rate of growth of IS-18 Monotone sequence IS-17 Polynomial, rate of growth of IS-18 Powers sum of IS-5 Prime factorization IS-2. Proposition 2. Nair EXAMPLE 1. Get an intuitive sense of what that even means!. Application of du Bois-Reymond’s comparison of. This week, we will see that within a given range of x values the Taylor series converges to the function itself. Recall the sequence (x n) de ned inductively by x 1 = 1; x n+1 = (1=2)x n + 1;n2N:. ( ) 0 0 (4. Likewise, if the sequence of partial sums is a divergent sequence (i. | 2019-12-08T11:31:32 | {
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https://www.physicsforums.com/threads/determine-the-standard-deviation-of-these-results.948814/ | # Determine the standard deviation of these results
I've managed to work out the below for the sample standard deviation. Any error's that I should be aware of?
## Homework Statement
The ultimate tensile strength of a material was tested using 10 samples. The results of the tests were as follows.
711 N 〖mm〗^(-2),732 N 〖mm〗^(-2),759 N 〖mm〗^(-2),670 N 〖mm〗^(-2),701 N 〖mm〗^(-2),
765 N 〖mm〗^(-2),743 N 〖mm〗^(-2),755 N 〖mm〗^(-2),715 N 〖mm〗^(-2),713 N 〖mm〗^(-2).
[/B]
(a) Determine the mean standard deviation of these results.
(a) Express the values found in (a) in GPa.
## Homework Equations
s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )
## The Attempt at a Solution
(a) As we are referring to a 'sample' we apply the below formula for Standard Deviation:
s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )
Calculating the mean value x ̅
(711 + 732 +759+670+701 +765 +743+755 +715 +713)/10
x ̅= 726.4 〖N mm〗^(-2)
Determining the (x_i-x ̅)^2
(711-726.4 )^2= 237.16
(732-726.4 )^2= 31.36
(759-726.4 )^2= 1062.76
(670-726.4 )^2= 3180.95
(701-726.4 )^2= 625.16
(765-726.4 )^2= 1489.96
(743-726.4 )^2= 275.56
(755-726.4 )^2= 817.96
(715-726.4 )^2= 129.96
(713-726.4 )^2= 179.56
Determining the sum of the above
∑▒237.16+31.36 +1062.76 +3180.95 +625.16 +1489.96 +275.56 +817.96 +129.96 +179.56
= 8,030.29 〖N mm〗^(-2)
Divide by N-1
(1/5)∙8030.29= 1606.06 N 〖mm〗^(-2)
Determining the sample standard deviation σ
σ= √((1606.06)=40.08 N 〖mm〗^(-2)=4.008 MPa
(b)4.008 MPa=0.004008 GPa
Or in standard form, 4.008∙10^(-3) GPa.
Homework Helper
Gold Member
2020 Award
##\sigma=40 ## N/mm^2=40 MPa=.040 GPa if my arithmetic is correct. ## \\ ## And you don't need to keep all the extra sig figs on the standard deviation. ## \\ ## And I think they want the mean and standard deviation. You didn't give the value of the mean in GPa.
Mark44
Mentor
Determine the mean standard deviation of these results.
There's a mean and there's a standard deviation. I'm not aware of a term called mean standard deviation, but there is a standard error, where you look at the variation of a collection of samples.
FactChecker
FactChecker
Gold Member
There is a standard deviation, ##\sigma_{\bar x}##, of the sample mean ##\bar x##. It is closely related to the confidence interval of the population mean ##\mu##. $$\sigma_{\bar x} = \sigma/\sqrt M \approx s/\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.
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Homework Helper
Gold Member
2020 Award
There is a standard deviation, ##\sigma_{\bar x}##, of the sample mean ##\bar x##. It is closely related to the confidence interval of the population mean ##\mu##. $$\sigma_{\bar x} = \sigma/\sqrt M \approx s/\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.
Let the mean be ## Y ##, so that ## Y=\bar{X}=\frac{X_1+X_2+...+X_M}{M} ##. ## \\ ## If the random measurements are uncorrelated, ## \sigma_Y^2=\frac{\sigma_{X_1}^2+\sigma_{X_2}^2+...+\sigma_{X_M}^2}{M^2}=\frac{M \, \sigma_X^2}{M^2}=\frac{\sigma_X^2}{M} ##. ## \\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\ ## I believe the OP @Tiberious left off the word "and" between "mean" and "standard deviation", and it has created some confusion.
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FactChecker
FactChecker
Gold Member
Let the mean be ## Y ##, so that ## Y=\bar{X}=\frac{X_1+X_2+...+X_M}{M} ##. ## \\ ## If the random measurements are uncorrelated, ## \sigma_Y^2=\frac{\sigma_{X_1}^2+\sigma_{X_2}^2+...+\sigma_{X_M}^2}{M^2}=\frac{M \, \sigma_X^2}{M^2}=\frac{\sigma_X^2}{M} ##. ## \\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\ ## I believe the OP @Tiberious left off the word "and" between "mean" and "standard deviation", and it has created some confusion.
They are all very standard things to ask for. But I agree that the sample mean and sample standard deviation are the first two things to calculate.
Yes - it does seem my OP was missing the word 'and' apologies for the confusion. Just to confirm;
The answer for standard deviation is 0.040 Gpa ?
So, would the mean be 0.016Gpa ?
FactChecker
Gold Member
In your original post, why are you dividing by 5 instead of 9 for the sample variance?
PS. small error: (701-726.4 )^2=645.16, not 625.16.
So, amending (701-726.4 )^2 from 625.16 to 645.16 and amending the following would lead to 805.02N mm^-2
Amendments being: -
(701-726.4 )^2 = 645.16
(1/10)∙8050.29= 805.02 N mm^-2
σ= √((805.02)=28.37 N mm^(-2)= 0.02837 Mpa
Converting to Gpa yields: -
2.837*10^-5 Gpa
Looking any better ?
FactChecker
Gold Member
(1/10)∙8050.29= 805.02 N mm^-2
Why are you dividing by 10 instead of N-1=10-1=9?
If you know the true mean, ##\mu##, you should use that in the formula for the sample standard deviation and divide by N. But if you estimate the mean with ##\bar x ##, your results of ##\sum_{i=1}^{i=N} (x_i-\bar x)^2## will be smaller and you should divide by ##N-1##.
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Ah! thanks, brief moment of confusion. I assume this would amend the overall error ? | 2021-04-20T22:12:57 | {
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https://math.stackexchange.com/questions/4002510/prove-this-formula-frac1-r-cosx1-2r-cosxr2-1-sum-n-1-inftyr/4003040 | # Prove this formula $\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$
I am trying to use prove, by just simple algebraic manipulation, to prove the equality of this formula. $$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$$
I have been given hints and instructions from this thread
1. Write using Euler’s identity $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$
2. Factor the denominator
3. Find the partial fractions decomposition, expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again).
This is how I have done:
1. RHS: $$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{1-2r(\dfrac{e^{ix}+e^{-ix}}{2})+r^{2}}=\dfrac{1-r(\dfrac{e^{ix}-e^{-ix}}{2})}{1-re^{ix}-re^{-ix}+r^2}$$
Then from this thread, I have learnt how to factorize the denominator (See Jack D'Aurizio answer, first answer of the thread, first line). The trick is to write $$1=e^0$$.
1. $$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(re^{ix})(r-e^{-ix})}$$
1. The third step is to decompose the fraction:
$$\dfrac{A}{r-e^{ix}}+\dfrac{B}{r-e^{-ix}}=\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(r-e^{ix})(r-e^{-ix})}$$
$$A(r-e^{-ix})+B(r-e^{ix})=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$
I am stuck here, I notice that let $$x=0$$, we will have
$$A(r-1)+B(r-1)=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$
I am stuck here, I don't know to find $$A$$ and $$B$$
Also, could you provide in details how to finish the part "expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again)".
My symbolic manipulation skill is not very good, so a detailed answer is great!
Thanks!
• You have some typos where you write $e^{ix}$ instead of $e^{-ix}$.
– J.G.
Jan 27, 2021 at 22:30
• @ J.G. Thanks, I just copy and paste, let me correct them all Jan 27, 2021 at 22:30
• There are sign errors in the $e^{-ix}$ coefficients too.
– J.G.
Jan 27, 2021 at 22:31
• the sum spans over $-\infty , \infty$ or $1, \infty$, although $\cos$ is symmetric you might miss some piece Jan 27, 2021 at 23:00
There is possibly a better way, and I think this was discussed as solved example in tristam needham's book(*). Any who, we begin with geometric series:
$$\frac{1}{1-x} = \sum_{j=0}^{\infty} x^j$$
Sub: $$x \to re^{ i \theta}$$ and simplfy
$$\frac{1}{(1- r \cos \theta) - i \sin \theta} = \sum_{j=0}^{\infty} r^j e^{i j\theta} \tag{2}$$
For LHS, by multiplying with complex conjguate
$$\frac{1}{(1-r \cos \theta) - i r\sin \theta} = \frac{ (1-r \cos \theta) + i \sin \theta}{ 2 - 2 r \cos \theta+r^2} \tag{1}$$
Equating real part in (1) to real part in (2),
$$\frac{1 - r \cos \theta}{ 2 - 2r \cos \theta + r^2} = \sum_{j=0}^{\infty} r^j \cos j \theta$$
Done!
Another identity could be made by equating imaginary parts
*: Indeed it was! See page-78, 79 of Visual Complex Analysis to see how this is simply the fourier series corresponding to the geometric series. Simply two ways to view a function: As an infinite trignometric sum or as a infinite polynomial series. Pretty neat.
• in geo series sum is starting from $0$ Jan 27, 2021 at 23:02
Similar to Buraian's answer, a method I enjoy using is using the equivalent series for $$\sin$$, and then adding $$C+iS$$, as follows:
Let \begin{align} C&=1+r\cos x+r^2\cos2x+r^3\cos3x+\cdots\\ S&=r\sin x+r^2\sin2x+r^3\sin3x+\cdots\\ \end{align} Then \begin{align} C+iS&=1+r(\cos x+i\sin x)+r^2(\cos 2x+i\sin2x)+r^3(\cos3x+i\sin3x)+\cdots\\ &=1+re^{ix}+(re^{ix})^2+(re^{ix})^3+\cdots\\ &=\frac{1}{1-re^{ix}}=\frac{(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\frac{1-r\cos x+ri\sin x}{1-2r\cos x+r^2} \end{align} Hence, equating real and imaginary parts we obtain \begin{align} C&=\frac{1-r\cos x}{1-2r\cos x+r^2}\\ S&=\frac{r\sin x}{1-2r\cos x+r^2}\\ \end{align} I hope that was helpful and gives you a new and interesting method for attacking these sorts of problems.
• I like your answer most, but I also vote for Buraian since he post this method first. Your solution has a "Eulerish" feeling to it. haha :) Cheers! Jan 28, 2021 at 15:42
• @JamesWarthington :)) I'm so glad I helped you! This method is pretty powerful: for other examples of me using it in other answers/questions of mine see here: math.stackexchange.com/questions/3886649/… and math.stackexchange.com/questions/3988168/… Jan 28, 2021 at 17:18
• Could you help me here as well: math.stackexchange.com/questions/4003509/… Jan 28, 2021 at 17:41
• @JamesWarthington I will have a go, I suspect the proof lies in using trig identities. Jan 28, 2021 at 17:50
• thank you, I wish my algebraic manipulation skill can be as good as you Jan 28, 2021 at 17:53
$$\cos(nx)=2\cos(x)\cos((n-1)x)-\cos((n-2)x)$$, then $$\sum_{n=1}^\infty r^n\cos(nx)=2\cos(x)\sum_{n=1}^\infty r^n\cos((n-1)x)-\sum_{n=1}^\infty r^n \cos((n-2)x)$$ $$=2r\cos(x)\sum_{n=0}^\infty r^n\cos(nx)-r^2\sum_{n=0}^\infty r^n \cos(nx)$$ Then, after some algebra, $$\sum_{n=1}^\infty r^n\cos(nx)=\frac{r\cos(x)-r^2}{1-2r\cos(x)+r^2}$$
• Can you expand your answer? They are so compacted that I don't know how did you derive these from "some algebra". Jan 28, 2021 at 3:55
• @JamesWarthington set $$f=\sum_{n\ge0}r^n\cos(nx).$$ Then $$\sum_{n\ge1}r^n\cos(nx)=f-1$$ and $$f-1=2r\cos(x)f-r^2f.$$ Solve for $f$. Jan 28, 2021 at 6:46
I might derive the series from $$\frac{1-r \cos (x)}{r^2-2 r \cos (x)+1}$$ by setting $$\cos x = (e^{ix}+e^{-ix})/2 = (z+z^{-1})/2$$ with $$z=e^{ix}$$. Then \eqalign{ \frac{1-r \cos (x)}{r^2-2 r \cos (x)+1} &= \frac{r z^2+r-2 z}{2 (r-z) (r z-1)} = \frac{1}{2}\left(1+\frac{r/z}{(1-r/z)}+\frac{1}{(1-r z)}\right) \cr &= \frac{1}{2}\left(2+\frac{r/z}{(1-r/z)}+\frac{r/z}{(1-r z)}\right) = \frac{1}{2}\left(2+\sum_{n=1}^{\infty}{(r/z)^n}+\sum_{n=1}^{\infty}{(r z)^n}\right) \cr &= 1+\sum_{n=1}^{\infty}{z^n+1/z^n\over2} {r^n} = 1+\sum_{n=1}^\infty r^n \cos nx \ \ \text{or}\ \sum_{n=0}^\infty r^n \cos nx \,. \cr }
• @Micheal E2: Man, your answer is quite difficult. I think you write with great brevity. I am trying to retrace every step you post here but they are so hard. Jan 28, 2021 at 15:39
Since\begin{align}\frac{1-r(e^{ix}+e^{-ix})/2}{(r-e^{ix})(r-e^{-ix})}&=\frac{A}{r-e^{ix}}+\frac{B}{r-e^{-ix}}\\\implies 1-r(e^{ix}+e^{-ix})/2&=A(r-e^{-ix})+B(r-e^{ix}),\end{align}the next step is to consider the $$r^0$$ and $$r^1$$ terms separately, giving$$1=-e^{-ix}A-e^{ix}B,\,-(e^{ix}+e^{-ix})/2=A+B.$$Now solve simultaneous equations. You should find$$A=-\frac12e^{ix},\,B=-\frac12e^{-ix}.$$
• @ J.G. what do you by $r^{0}$ and $r^{1}$ terms? By setting $r=r^{0}$ and $r=r^{1}$. This is not right, I think my interpretation is wrong. Jan 27, 2021 at 23:32
• @.J.G could you edit your reply and show me how to solve this system of equations? I have tried for 2 hours and only obtain $A=\frac{e^{ix}+3e^{ix}}{2(1-e^{-2ix})}$, increadily more complicated than your result. Jan 28, 2021 at 3:16
• @.J.G how do you know this system of equations only has 1 roots for A and for B? Jan 28, 2021 at 3:41
• @.J.G Here are my efforts to solve your system of equations: math.stackexchange.com/questions/4002655/… Jan 28, 2021 at 4:00
With $$1 - 2r\cos x + r^2 = (1-re^{i x})(1-re^{-ix})$$
\begin{align} \frac{1-r\cos x}{1-2r\cos x+r^2} = Re \frac{1}{1-re^{ix}} = Re\sum_{n=0}^\infty(re^{ix})^n =1+\sum_{n=1}^{\infty}r^{n}\cos nx \end{align} | 2022-09-25T10:24:34 | {
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https://math.stackexchange.com/questions/2618175/how-to-prove-this-delta-epsilon-proof-involving-x2 | # How to prove this delta-epsilon proof involving $x^2$?
Unlike my last question I want to try something where $x$ can be any real value in $f(x)$ so it's not just $x \geq 0$. I want to fix an $\epsilon$ and find the largest $\delta$ I can get away with using to make the necessary inequalities hold.
$$\lim_{x \rightarrow 2} x^2= 4$$
Say I pick some $\epsilon = 0.5$ which means I need to find some $\delta > 0$ such that for all $x$ satisfying $0 < |x-2| < \delta$, the inequality $|f(x) - L| = |x^2 - 4| < 0.5$ holds.
1. Am I stating this correctly so far / understanding the delta-epsilon relationship and goals?
2. How do I pick the right $\delta$ for something like this?
Trying to simplify:
$|x^2 - 4| < 0.5$
$|x+2||x-2| < 0.5$
$|x-2| < \frac{0.5}{|x+2|}$
Now I'm stuck. Is there a better way to approach these problems? So far I've been trying to manipulate the epsilon inequality so it looks more like the delta one and then try to set the delta and epsilon expressions equal to each other, but maybe there is a more reliable way to prove these relationships?
Update:
Trying another way:
$|x^2 - 4| < 0.5$ simplifies to
$\sqrt{3.5} < x < \sqrt{4.5}$
This gives me two $x$-values away from $a=2$, either $2 - \sqrt{3.5} = .1291...$ or $\sqrt{4.5} - 2 = .1213...$
So if I pick the smallest of the two, $\delta = \sqrt{4.5} - 2$ which satisfies the epsilon condition?
• Choose $\delta=\epsilon/5$ for $\epsilon<1$ – Fakemistake Jan 23 '18 at 20:46
• "Am I stating this correctly so far / understanding the delta-epsilon relationship and goals?" For the most part. But there are two considerations you are eliding. i)$\epsilon$ doesn't have to be $.5$ but could be $.05$ or $.000000005$ of $5\times 10^{534}$. It can be any value. ii) And $\delta$ ... well it's best not to think of delta as a constant, but as a number whose value is determined by the value of $\epsilon$. – fleablood Jan 23 '18 at 20:58
• If $\epsilon = .5$ then just let $\delta$ be something really really small. If $\delta = .01$ then $|x - 2| < \delta$ means $1.99 < x < 2.01$ so $3.9601< x^2 < 4.0401$ so $-.0399< x^2 - 4 <.0401$ so $|x^2 - 4| < .05 < .5$. So.... that doesn't tell you much about how to do it in general. Does it? – fleablood Jan 23 '18 at 21:04
• @fleablood I tried using a graph instead and edited my post. Does this approach make sense? – Aruka J Jan 23 '18 at 21:05
• I understand $\epsilon$ can be any positive value, I'm just picking one arbitrarily, fixing some $\epsilon$, and then finding the largest $\delta$ that makes $|f(x)-L| < \epsilon$ hold. We could pick something very small and it would work but that doesn't show how large we could get away with going. – Aruka J Jan 23 '18 at 21:05
We need to show that $\forall \epsilon>0$ $\exists\delta>0$ such that
$$\forall x\neq2 \quad |x-2|<\delta \implies\left|f\left(x\right)-l\right|<\varepsilon$$
that is
$$|x^2-4|<\epsilon\iff-\epsilon<x^2-4<\epsilon\iff4-\epsilon<x^2<4+\epsilon\iff \sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}\iff \sqrt{4-\epsilon}-2<x-2<\sqrt{4+\epsilon}-2\\\iff |x-2|<min\{\sqrt{4+\epsilon}-2,2-\sqrt{4-\epsilon}\}=\sqrt{4+\epsilon}-2=\delta \quad \square$$
• I think your result is the same as mine correct? $\delta = \sqrt{4.5} - 2$? (I picked $\epsilon = 0.5$) – Aruka J Jan 23 '18 at 21:07
• @ArukaJ Yes but you need to prove it in general for genrec $\epsilon$ and $\delta$ values. – gimusi Jan 23 '18 at 21:19
• For the third time. !!!STOP!!! picking specific values of $\epsilon$. It will NOT help you. – fleablood Jan 23 '18 at 21:32
• @gimusi Oh, yeah. Absolutely. Looks good. i haven't gone through it with a fine tooth comb but, yeah, that's exactly how do it. So barring any arithmetic errors, It's good. My comment was for Aruka who seems hell bent on finding specific deltas for specific elements and not actually trying to find a general formula for deltas for all possible epsilons. – fleablood Jan 23 '18 at 21:39
• Well, it's fine to get an insight but there comes a point where you will have to say, insight is done, now we must do a proof. And you will NEVER be allowed to say "It worked for all examples I tried therefore it must be true" as a proof. – fleablood Jan 23 '18 at 21:45
Hint $$|x^{ 2 }-4|=\left| x-2 \right| \left| x-2+4 \right| <{ \left| x-2 \right| }^{ 2 }+4\left| x-2 \right|$$
If $|x- 2| < \delta$ then
$- \delta < x -2 < \delta$
$2 - \delta < x < 2 + \delta$. Let's assume for the moment that $\delta < 2$.
$(2- \delta)^2 < x^2 < (2+ \delta)^2$
$4 - 4\delta + \delta^2 < x^2 < 4 + 4\delta + \delta^2$
$-4\delta + \delta^2 < x^2 - 4 < 4\delta + \delta^2$.
$-4 \delta - \delta^2 < x^2 - 4 < 4\delta + \delta^2$
$|x^2 - 4| < |4\delta + \delta^2| = 4\delta + \delta^2$ (because $\delta$ is positive)
So we want $\epsilon \ge 4\delta + \delta^2$. Given that we know what $\epsilon$ is, can we find a way of figuring out $\delta$ in terms of $\epsilon$ so that that would be true?
If we assume $\delta \le 1$ then $\delta^2 \le \delta$ so $5\delta \ge 4\delta + \delta^2$.
So if we choose any $\delta$ so that i) $\delta < 2$ and ii) $\delta \le 1$ and iii) $5\delta < \epsilon$ that will do.
So for any $\delta < \min (\frac \epsilon 5, 1)$ that will do.
So to do the proof:
For any $\epsilon > 0$, let $\delta = \min (\frac \epsilon 5, 1)$
Then if $|x - 2| < \delta$ implies by all the work we did above that
$-5\delta \le -4\delta -\delta^2 < -4\delta + \delta^2 < x^2 -4 < 4\delta + \delta^2 \le 5\delta$ so
$|x^2 - 4| < 5\delta \le \epsilon$.
And that's the proof.
It is my honest opinion that your question lies in the realm of questions of the type: "what is the best approach for riding a bicycle?" - Most answers you will receive will send you snapshots of happy riders; but you don't really want these answers, do you? I suggest you get on the bike, and keep falling until you don't.
• Are delta-epsilon proofs more art than science, is that sort of what you are saying? There is no standard methodical way to get the answer? – Aruka J Jan 23 '18 at 20:39
• Is bicycle riding an art? No, it is a skill. delta-epsilon proofs are a skill. There are many skills that you can only pick up by actually doing. Swimming, bike-riding are two out of three immediate examples that come to mind. – uniquesolution Jan 23 '18 at 20:41
• Proofs in general are like this. There are some approaches that work more often than others, but it's more about the intuition of what works than following a particular step-by-step approach. In this case, the best advice you can get is to poke around with the triangle inequality and when you get something that works, try to understand what made it work. – AlexanderJ93 Jan 23 '18 at 20:43 | 2019-06-17T12:37:57 | {
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https://math.stackexchange.com/questions/2248572/sum-n-1-infty-frac1n22n-by-integration-or-differentiation/2248648 | # $\sum_{n=1}^{\infty}\frac{1}{n^22^n}$ by integration or differentiation
There is an infinite sum given: $$\sum_{n=1}^{\infty}\frac{1}{n^22^n}$$ It should be solved using integration, derivation or both. I think using power series can help but I don't know how to finish the calculation. Any help will be appreciated!
• $Li_{2} \left( \frac{1}{2} \right) = \frac{\pi^2}{12}-\frac{ (\ln 2)^2}{2}$. – Donald Splutterwit Apr 23 '17 at 19:26
• Thank you! But how did you get that? @DonaldSplutterwit – Hendrra Apr 23 '17 at 19:32
• I copied it from here ... en.wikipedia.org/wiki/Spence%27s_function ... I am scribbling in my note book now ... I will get back to you when I have managed to derive it ... – Donald Splutterwit Apr 23 '17 at 19:34
• Thank you very much :) I'm really looking forward to the solution! – Hendrra Apr 23 '17 at 19:35
• Have a look at math.stackexchange.com/a/1056111/44121 – Jack D'Aurizio Apr 23 '17 at 20:10
Note that we have $\int_0^x t^{n-1}\,dt=\frac{x^n}{n}$. Then, we can write
\begin{align} \sum_{n=1}^\infty \frac{x^{2n}}{n^2}&=\sum_{n=1}^\infty \int_0^x t^{n-1}\,dt\int_0^x s^{n-1}\,ds\\\\ &=\int_0^x\int_0^x \frac{1}{1-st}\,ds\,dt\\\\ &=-\int_0^{x} \frac{\log(1-sx)}{s}\,ds\\\\ &=-\int_0^{x^2} \frac{\log(1-s)}{s}\,ds\\\\ &=\text{Li}_2(x^2) \end{align}
Evaluating at $x=1/\sqrt{2}$ yields
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n^2\,2^n}=\text{Li}_2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)}$$
And we are done!
To evaluate $\text{Li}_2(1/2)$, we exploit the relationship
$$\text{Li}_2(1-x)=-\text{Li}_2\left(1-\frac1x\right)-\frac12\log^2(x)$$
Letting $x=1/2$ yields
$$\text{Li}_2(1/2)=-\text{Li}_2\left(-1\right)-\frac12\log^2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)$$
where we used
\begin{align} \text{Li}_2(-1)&=-\int_0^{-1}\frac{\log(1-x)}{x}\,dx\\\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}\\\\ &=\frac{\pi^2}{12} \end{align}
• That's a nice solution! Thanks. However I must ask about $\sum_{n=1}^{\infty}\frac{x^{2n}}{n^2}$. Why are we considering such a sum? – Hendrra Apr 23 '17 at 19:43
• You're welcome. My pleasure. The value of that sum when $x=1/\sqrt 2$ is $\sum_{n=1}^\infty \frac{1}{n^2\,2^n}$. – Mark Viola Apr 23 '17 at 19:45
• The value truly is $\sum_{n=1}^{\infty}\frac{1}{n^22^n}$! Thus I have the next question. Why did you decided to take an $x = \frac{1}{\sqrt{2}}$? Just because it works? – Hendrra Apr 23 '17 at 19:48
• Well, yes. We took $x=1/\sqrt 2$ in order that we would have the sum of interest. – Mark Viola Apr 23 '17 at 19:51
• @Tyberius The first integral yields $$\int_0^x \int_0^x \frac{1}{1-st}\,dt\,ds=-\int_0^x \frac{\log(1-sx)}{s}\,ds=-\int_0^{x^2}\frac{\log(1-s)}{s}\,ds=\text{Li}_2(x^2)$$ – Mark Viola Apr 23 '17 at 20:20
Another way to calculate it: consider $$f(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n^2}.$$ Differentiating w.r.t. $x$ gives $$f'(x)=\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n}=\frac{1}{x}\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad xf'(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad (xf'(x))'=\sum_{n=1}^{+\infty}x^{n-1}=\frac{1}{1-x}.$$ Now integrating with $f(0)=0$ $$xf'(x)=-\ln(1-x)\quad\Rightarrow\quad f(x)=-\int_0^x\frac{\ln(1-t)}{t}\,dt.$$ Motivation for termwise differentiation for power series is straightforward.
My answer to a duplicate question says:
$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$$ Integrating from 0 to t we get $$\int_{0}^{t}\frac{1}{(1-x)}dx=\sum_{n=0}^{\infty}\int_{0}^{t} x^{n}dx$$$$-\ln(1-t)=\sum_{n=1}^{\infty} \frac{t^{n}}{n}$$ Dividing by t and integrating $$\int_{0}^{0.5}-\frac{\ln(1-t)}{t}dt=\sum_{n=1}^{\infty}\int_{0}^{0.5} \frac{t^{n-1}}{n}dt=\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$$ This on calculating is $$\dfrac{\pi^2}{12}-\dfrac{ln^2(2)}{2},$$
• This answer is identical to your answer to this question. If the questions are the same, this one should be flagged as a duplicate. If not, then it would be more efficient and less noisy to cite or quote (with attribution) from the other answer. The link between the answers would also serve as a link between two closely related questions. In this case, the questions were duplicates. – robjohn May 26 at 1:31
• @robjohn So should I change anything ??. I posted it here because the other one was marked as duplicate. Should I remove it from here?? – DivMit May 26 at 3:12
• Even though the other question was closed, your answer there is still active and getting votes. One should be removed, but you might edit one to reference the other; you might just say something like, "my answer to a duplicate question says..." – robjohn May 26 at 5:28
• Okay , I will link my answers – DivMit May 26 at 9:55 | 2019-07-19T10:21:16 | {
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https://math.stackexchange.com/questions/3179843/how-do-limits-work-with-floor-ceiling | How do limits work with floor/ceiling?
I'm interested in the below equation:
$$\frac{n}{\operatorname{floor}(\frac{x}{n})}$$
Plotting with $$n = 1..100$$ shows the graph being slightly more aliased as $$n$$ increases and a discontinuity forms from $$0..n$$. The domain from what I can tell is $$(-\infty, 0) \cup [n, \infty)$$. I wanted to investigate the limits at each $$n$$ but not entirely certain how floor and ceiling factor into algebra.
Take for example $$n = 2$$,
$$f(x) = \frac{2}{\operatorname{floor}(\frac{x}{2})}$$
How could I find $$\lim_{x\to0} f(x)$$? Would the limit even exist considering the discontinuity between $$[0, 2)$$?
I know you can break down the function and help find the limit using the rules of limits:
$$\lim_{x\to b} \frac{p}{q} = \frac{\lim_{x\to b}p}{\lim_{x\to b}q}$$
So more precisely I'm looking for $$\frac{2}{\lim_{x\to0}\operatorname{floor}(\frac{x}{2})}$$.
For what it's worth, the plot looks like:
I can surmise $$\lim_{x\to0^-} f(x) = -2$$, am I right in assuming $$\lim_{x\to0^+} f(x)$$ doesn't exist?
• When dealing with a domain that is not the whole real line, you need to be careful with limits on the border of the domain. In this case, the limit is the same as $\lim_{x\to 0^{-}} f(x)$ because the small open neighborhoods of $0$ in the domain of $f$ are only negative. This means the limit will be $-n$ since $\lfloor x/n\rfloor=-1$ is constant for $x$ negative near $0.$ – Thomas Andrews Apr 8 at 16:36
• Your function not only has a discontinuity but fails to be defined at all for $x\in[0,n)$. – Henning Makholm Apr 8 at 16:36
• On the other hand it is clear that $f(x)=-n$ for every $x\in[-n,0)$, and since these numbers are the only ones in the domain that are close to $0$, the function very trivially goes to $-n$ for $x\to 0$. There's no algebra involved in this. – Henning Makholm Apr 8 at 16:38
It helps to carefully state definitions. A workable definition of a limit (of a real function) is something like the following:
Definition: Let $$f$$ be a function defined on some domain $$D\subseteq \mathbb{R}$$ and let $$a \in \mathbb{R}$$. Further suppose that there is some $$L\in\mathbb{R}$$ such that for every $$\varepsilon > 0$$ there exists some $$\delta > 0$$ such that if $$x \in D$$ and $$0 < |x-a|<\delta$$, then $$|f(x) - L| < \varepsilon.$$ $$L$$ is said to be the limit of $$f(x)$$ as $$x$$ approaches $$a$$, denoted $$\lim_{x\to a} f(x) = L.$$
I claim that $$\lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} = -2.$$ To prove this claim, I have to show that if $$\varepsilon$$ is any positive number, then I can find a value $$\delta$$ such that $$f(x)$$ is within $$\varepsilon$$ of $$-2$$ whenever $$x$$ is in the domain of $$f$$ and within $$\delta$$ of zero.
So, let $$\varepsilon > 0$$ be arbitrary and take $$\delta = 1$$. Observe that $$f$$ is only defined on the set $$D = \mathbb{R} \setminus [0,2).$$ If $$x \in D$$ and $$|x| < \delta = 1$$, then $$x \in (-1,0)$$, and so $$\lfloor x/2 \rfloor = -1$$. But then $$|f(x) - (-2)| = \left| \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} + 2 \right| = \left| \frac{2}{-1} + 2 \right| = 0.$$ Hence whenever $$|x-0| < \delta$$ and $$x\in D$$, we have $$|f(x)-(-2)| = 0 < \varepsilon$$. Therefore $$\lim_{x\to 0} \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} = -2,$$ as claimed.
The important point here is that the definition of a limit only cares about what is happening in the domain of the function. Points where the function is undefined are irrelevant. As long as we are working with a real-valued function of a real variable, the function of interest is simply not defined on the interval $$[0,2)$$, and so we don't have to worry about those points.
• My understanding in finding a limit was the limit for $x\to b$ only exists if the limit exists for both $x\to b^-$ and $x\to b^+$ and they are equal. If one doesn't exist, the limit doesn't exist; or, if they are not equivalent, the limit doesn't exist. I understand the leftside limit is $-2$ but the rightside limit is not defined. Doesn't then, the limit not exist? I'll disclaim I'm in Calculus I. – gator Apr 8 at 16:49
• As I said in the preamble, you have to carefully state your definitions. The definition of a limit which I gave is pretty standard. If you want to discuss left- and right-hand limits, you need to first carefully define what those mean. I would claim that $\lim_{x\to 0^+} f(x) = -2$ vacuously---indeed, I would claim that if $L$ is any real number, then $\lim_{x\to 0^+} f(x) = L$, since if I make $\delta$ small enough, I cannot find any values of $x$ such that $0 < x < \delta$. Thus all of the conditions are met. Again, state the definition very carefully, then see what happens. – Xander Henderson Apr 8 at 16:53
• You need to exclude the case where $x=a\in D$ or you'd have $\lim_{x\to a} f(x)$ would never be defined if $a$ was a point of discontinuity, even if it could be made continuous. You also want that for $\delta>0$ that there is at least one $x\in D\setminus\{a\}$ so that $|x-a|<\delta.$ Otherwise, all limit values would be vacuously true. – Thomas Andrews Apr 8 at 16:55
• @gator Using that definition, then you are at a loss here. You'd just have to leave it undefined. But in later math, that definition proves inadequate in many ways. – Thomas Andrews Apr 8 at 16:57
• @ThomasAndrews Regarding $|x-a| > 0$, I added that---omitting it was an oversight on my part. Regarding vacuous limits, I have no problem allowing the degenerate case in which the limit is anything by vacuity. But, again, everything comes down to definitions. If one wants to omit vacuous limits (which, in reality, one probably does), then the definition should be written slightly differently. – Xander Henderson Apr 8 at 19:21 | 2019-06-25T01:19:50 | {
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https://math.stackexchange.com/questions/1251466/what-is-this-pattern-called/1251501 | # What is this pattern called?
## Back-Story
I became interested in the patterns in multiplication tables for different base number systems a while ago. Specifically, the pattern made by the last digit of each number in the multiplication table. So, base 10 would look like this:
1|2|3|4|5|6|7|8|9|0
2|4|6|8|0|2|4|6|8|0
3|6|9|2|5|8|1|4|7|0
4|8|2|6|0|4|8|2|6|0
5|0|5|0|5|0|5|0|5|0
6|2|8|4|0|6|2|8|4|0
7|4|1|8|5|2|9|6|3|0
8|6|4|2|0|8|6|4|2|0
0|0|0|0|0|0|0|0|0|0
I thought it was interesting that when you move to number systems with different bases, the patterns don't follow the number. They follow it's relative position in the number system. E.g., in base 12, the pattern is 6,0,6,0,6......
## Images
I then realized, I could see the pattern better if I just assigned each number a color. I started with using 10 greyscale colors, with 0 being black and 9 being white. So now base 10 looks like this:
Then, I figured that I really could use as many colors as I wanted to, and see if a larger pattern forms. Using all 256 greyscale colors, I came up with this image representing a base 256 multiplication table:
Or I could go from black to white to black, and smooth out the image:
## Animation
I decided to animate the pattern to better see what was going on. To do this, I defined my 1-n color scale as [w,w,w,w,b,b,b,b,b,b,b,b....]. Where w is white and b is black. I would create a frame, shift my colors down one [b,w,w,w,w,b,b,b,b,b,b,b....], and create the next frame. I repeated this until they colors fully cycled and got this animated image.
Here's a site where you can modify the settings.
## What is this pattern called?
My question is, what is this pattern called? I'm having a hard time finding anything about it. It seems to be a bunch of hyperbolic curves imposed on each other. There is a bunch of "stars" at the corners of where you would divide the image into 4ths, 9ths, etc.
Any insight into this would be appreciated.
• This pattern is called "cool". – Lee Mosher Apr 25 '15 at 16:42
• I'm not an expert, but look at moire patterns. – Bob Krueger Apr 25 '15 at 16:46
• @Bob1123 It does seem similar to a moire pattern. But if it is, the question becomes what patterns are making up this pattern! – Alex McKenzie Apr 25 '15 at 16:50
• Should the question title be changed to "What is the visual pattern in the multiplication table of modular arithmetic?" The OP didn't mention modular arithmetic but it does make this question easier to search for – man and laptop Apr 25 '15 at 17:08
• I love that you posted this! I stumbled across this same pattern years ago but didn't think to ask about it online. (Then again, that might have been before I knew about Math.SE.) So, thanks for doing this! :) – El'endia Starman Apr 26 '15 at 0:28
What you've discovered is essentially modular arithmetic. By looking at only the last digits of a product (in whatever base you're looking at at the moment), you're in effect saying 'I don't care about things that differ by multiples of $n$; I want to consider them as the same digit'. For instance, in base $7$, $5\times 2=10_{10}=13$ has the same last digit as $4\times 6=24_{10}=33$; we put both of these numbers into a bucket labeled '$[3]$', along with $3$, $23=17_{10}$, $43=31_{10}$, etc. In mathematics, when we talk about $31 \bmod 7$ we sometimes just mean the number $3$ itself (that is, the 'label' on this bucket that's between $0$ and $6$, but it's often convenient to think of it as representing the whole bucket: whatever number we pick out of the $[3]$ bucket, when we add it to a number in the $[2]$ bucket, we know that our result will be in the $[5]$ bucket, and when we multiply a number in the $[3]$ bucket by a number in the $[4]$ bucket, we know that our result will be in the $[5]$ bucket; etc. "Last digits" are just a convenient way of talking about these buckets (though things get a little sketchier when you talk about negative numbers - note that according to these rules, $-3$ goes into the $[4]$ bucket!).
Meanwhile, the bands in your pattern are actually (pieces of) hyperbolas. Since $a\times (n-b)\equiv -(a\times b)\pmod n$ (the statement '$x=y\pmod n$' is a mathematical way of phrasing '$x$ and $y$ are in the same bucket in base $n$'; here, the difference between $a\times (n-b)$ and $-(a\times b)$ is $a\times n$), the far right hand side is essentially a reflection of the left, and similarly the bottom is a reflection of the top. If you rearrange the four quarters of your square so that the center of symmetry is (what was previously) the top left corner — i.e., take $A\ B\atop C\ D$ to $D\ C\atop B\ A$ — and then put the origin at the center, then the bands will exactly be (scaled versions) of the hyperbolae $xy=C$ (which are the hyperbolae $y^2-x^2=2C$ rotated by $45^\circ$). This happens because each 'cycle' of black-to-white or black-to-white-to-black will be separated by one multiple of $n$; e.g., the first transition between cycles occurs along the hyperbola $xy=n$; the second along the hyperbola $xy=2n$; etc.
(As for the moiré patterns, they're related to the usual way that such patterns are generated, and in particular they're somewhat related to aliasing near the Nyquist limit when the frequency between hyperbolic bands starts coming close to the frequency of the 'pixels' you're sampling with, but that's another story altogether...)
• So could you say our base 10 number systyem (or any base number system) can be described with modular arithmetic? And is this pattern called anything specific? E.g., modulus pattern? – Alex McKenzie Apr 25 '15 at 17:10
• @AlexMcKenzie The last-digits multiplication that you're talking about is exactly modular arithmetic (specifically, it's the multiplication table mod $n$). A cute example: look at the table of last digits for the base-7 multiplication table. Notice that every row has each non-zero number exactly once, and (by symmetry) so does every column? This isn't a coincidence! This will happen whenever your base is a prime; the numbers mod $p$ form what's called a group under multiplication. – Steven Stadnicki Apr 25 '15 at 17:14
• @AlexMcKenzie In fact, let me flesh out my answer a little bit to explain what I mean by that first sentence... – Steven Stadnicki Apr 25 '15 at 17:22
• Correct me if I'm wrong, but essentially this is a graph of multiple instances of $xy = C$ separated by a multiple of $n$. this is what forms the pattern in the corners. Then, since I'm sampling across a grid of finite pixels, a moire pattern forms. So if I had an infinitely large image, I would only see the $xy = C$ pattern, and not the artifacts. – Alex McKenzie Apr 25 '15 at 17:22
• @AlexMcKenzie Also, congratulations on finding this! This sort of pattern-hunting is an excellent way of experimenting in mathematics and learning about all sorts of facets of its vast world. – Steven Stadnicki Apr 25 '15 at 17:38
You can model your graphics as computing $f_n(x,y)=n\left[\frac{xy}{n}\right]$, where here the square brackets are ad-hoc notation to mean taking the fractional part (or "reduce modulo 1"). This takes $z=xy$ and cuts it at a bunch of horizontal hyperbolae, collapsing the graph like a Fresnel lens. Then, what you are doing is sampling $f_n$ on integer points $\{(i,j)\in\mathbb{Z}^2:1\leq i,j\leq n\}$, but $f_n$ oscillates faster than your sample grid, leading to a Moire pattern.
These would be the powers in a discrete Fourier transform matrix: http://en.wikipedia.org/wiki/Discrete_Fourier_transform_(general) | 2019-05-20T22:25:01 | {
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https://math.stackexchange.com/questions/956776/whats-the-inverse-operation-of-exponents | # What's the inverse operation of exponents?
You know, like addition is the inverse operation of subtraction, vice versa, multiplication is the inverse of division, vice versa , square is the inverse of square root, vice versa.
What's the inverse operation of exponents (exponents: 3^5)
Addition and multiplication are commutative, so there is just one inverse function.
Exponents are not commutative; $2^8 \not= 8^2$. So we need two different inverse functions.
Given $b^e = r$, we have the "$n$th root" operation, $b = \sqrt[e] r$. It turns out that this can actually be written as an exponent itself: $\sqrt[e] r = r^{1/e}$.
Again, given $b^e = r$, we have $e = \log_b r$, the "base-$b$ logarithm of $r$".
• Does division have just one inverse function? 8/4 ≠ 4/8 after all. Jul 31 '19 at 20:59
• 8/4 and 4/8 have a simple numerical relationship. (One is 2/1, the other is 1/2.) Whereas 2^8 and 8^2 have no particularly simple relationship (256 and 64, respectively.) Admittedly I didn't explain that very well... Aug 5 '19 at 8:02
• Good stuff, you can't express "nth root" in Excel but you can express x^(1/n) just fine. This gets doubly interesting if the exponent is e.g. 1.26 Sep 14 '21 at 18:21
These functions are the logarithms, and they are fundamentally important. For $$a = b^c$$ (where $$b > 0$$) we write: $$c = \log_b a,$$ which we can take to be the definition of $$\log_b$$. We read the operation as "logarithm, base $$b$$," or "base $$b$$ logarithm".
In particular, we have $$\log_a (a^b) = b \qquad\text{and}\qquad a^{\log_a b} = b.$$ Of special interest is the natural logarithm, denoted by $$\ln$$ or $$\log$$, the logarithm of base $$e$$. (NB that sometimes $$\log$$ can also denote base $$10$$, or base $$2$$, depending on context.)
Logarithmic identities correspond to exponential identities. From example, from the definition we can conclude that $$\log_b (pq) = \log_b p + \log_b q$$ (for $$p, q > 0$$), which corresponds to the identity $$b^{p + q} = b^p b^q$$.
Perhaps counterintuitively, sometimes it is convenient to define the natural logarithm first and then define the exponential function $$x \mapsto e^x$$ to be its inverse, which leads to the slightly antiquated name antilog for an exponential function $$x \mapsto b^x$$.
Edit Some of the other answers here pointed out quite rightly that one can also ask about the inverse of functions where the variable is in the base, i.e., functions $$x \mapsto x^a$$, and inverses of these functions$$^*$$ (at least when $$a > 0$$) are just $$x \mapsto x^{1/a}$$, which we often write as $$x \mapsto \sqrt[a]{x}$$. These functions are called power functions (note that the inverse of a power function is again a power function), and we reserve the name exponential function for functions $$x \mapsto b^x$$ where the variable is in the exponent, i.e., those to which the logarithms are inverses.
$$^*$$For some $$a$$ (in particular, even integers), we need to restrict the map $$x \mapsto x^a$$ to $$[0, \infty)$$ in order to take an inverse.
• The other answers were good, but your answer explained it best to me. Oct 3 '14 at 12:02
• What about the example in the post? What is the inverse function then? 3^(1/5) or 'base 3 logarithm of 243'? How can I know just by seeing the function which one is the variable here? Oct 24 '19 at 13:05
• That's really the point of the edit: Is the question asking about the inverse of $x \mapsto 3^x$ or the inverse of $x \mapsto x^5$? From just the expression $3^5$ alone there's no way to tell, much like asking about a function whose evaluation yields the expression $1 + 2$ does not indicate whether the question is about the function $x \mapsto x + 2$ or $x \mapsto 1 + x$. As you can see from the question, I initially understood OP to be asking about $x \mapsto 3^x$, since this is an exponential function, and OP asked about "inverse operation of exponents". Oct 24 '19 at 21:21
There are two inverse operations of exponentiation.
## Logarithm
$$\log _{b} a$$
It's read "base-$b$ logarithm of $a$". And it means "the exponent which $b$ must be raised to, so that the result is $a$".
## Root
$$\sqrt[b] a$$
It's read "$b$-th root of $a$". And it means "the number which, when raised to $b$, produces $a$".
It depends on what you see as the function and what the variable in $3^5$.
Generalising your "square is the inverse of square root" leads to reciprocal exponents being the inverse of exponents, so $3^5 = 243$ corresponds to $3 = 243^{1/5}$.
Alternatively $3^5 = 243$ corresponds to $5 =\log _{3} 243 = \frac{\log _{10} 243}{\log _{10} 3}= \frac{\log _{e} 243}{\log _{e} 3}$ using logarithms.
Logarithms: $$10^x = 100 \iff x=\log _{10} 100 = 2$$
If you take $x=3^5$, to "get the 5 back" you do $log_3(x)$ and, to "get the 3 back", you do $\sqrt[5]{x}$.
The interesting thing here is that there are 2 ways to reverse the operation, while other operations had just one: If you take $x=2+7$, to "get the 2 back" you did $x-7$ and to "get the 7 back", $x-2$. This happens because 2+7 = 7+2. The sum is "symmetrical" (the right term is commutative). If you want to "get the 2 back" from $x=2+7$, just subtract 7. If you want to "get the 2 back" from $y=7+2$, just subtract 7 again (because, after all, $x=y$).
But $x=3^5$ is not the same as $y=5^3$. So you cant expect to use the same operation to "get the 3 back from x" and "get the 3 back from y" | 2022-01-21T21:59:26 | {
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https://huchengbei.com/blog/53/ | ## PAT(A) 1126. Eulerian Path (25)
### 1126. Eulerian Path (25)
In graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similarly, an Eulerian circuit is an Eulerian path which starts and ends on the same vertex. They were first discussed by Leonhard Euler while solving the famous Seven Bridges of Konigsberg problem in 1736. It has been proven that connected graphs with all vertices of even degree have an Eulerian circuit, and such graphs are called Eulerian. If there are exactly two vertices of odd degree, all Eulerian paths start at one of them and end at the other. A graph that has an Eulerian path but not an Eulerian circuit is called semi-Eulerian. (Cited from https://en.wikipedia.org/wiki/Eulerian_path) Given an undirected graph, you are supposed to tell if it is Eulerian, semi-Eulerian, or non-Eulerian.
Input Specification:
Each input file contains one test case. Each case starts with a line containing 2 numbers N (<= 500), and M, which are the total number of vertices, and the number of edges, respectively. Then M lines follow, each describes an edge by giving the two ends of the edge (the vertices are numbered from 1 to N).
Output Specification:
For each test case, first print in a line the degrees of the vertices in ascending order of their indices. Then in the next line print your conclusion about the graph -- either "Eulerian", "Semi-Eulerian", or "Non-Eulerian". Note that all the numbers in the first line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.
Sample Input 1:
7 12
5 7
1 2
1 3
2 3
2 4
3 4
5 2
7 6
6 3
4 5
6 4
5 6
Sample Output 1:
2 4 4 4 4 4 2
Eulerian
Sample Input 2:
6 10
1 2
1 3
2 3
2 4
3 4
5 2
6 3
4 5
6 4
5 6
Sample Output 2:
2 4 4 4 3 3
Semi-Eulerian
Sample Input 3:
5 8
1 2
2 5
5 4
4 1
1 3
3 2
3 4
5 3
Sample Output 3:
3 3 4 3 3
Non-Eulerian
### 代码
/*
* Problem: 1126. Eulerian Path (25)
* Author: HQ
* Time: 2018-03-12
* State: Done
* Memo: 图，dfs
*/
#include "iostream"
using namespace std;
int N,M;
int degree[500 + 5];
int G[500 + 5][500 + 5];
int visit[500 + 5];
int even = 0;
int cnt = 0;
void dfs(int x) {
visit[x] = 1;
cnt++;
for (int i = 1; i <= N; i++) {
if (!visit[i] && G[x][i])
dfs(i);
}
}
int main() {
cin >> N >>M;
int x, y;
fill(degree, degree + 500 + 5, 0);
fill(visit, visit + 500 + 5, 0);
for (int i = 0; i < M; i++) {
cin >> x >> y;
G[x][y] = 1;
G[y][x] = true;
degree[x] ++;
degree[y] ++;
}
bool first = true;
for (int i = 1; i <= N; i++) {
if (first) {
cout << degree[i];
first = false;
}
else
cout << " " << degree[i];
if (degree[i] % 2 == 0)
even++;
}
cout << endl;
dfs(1);
if(cnt != N)
cout << "Non-Eulerian" << endl;
else if (even == N)
cout << "Eulerian" << endl;
else if(even == N-2)
cout << "Semi-Eulerian" << endl;
else
cout << "Non-Eulerian" << endl;
system("pause");
}
还没有人评论... | 2019-08-22T02:44:36 | {
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http://math.stackexchange.com/questions/336812/understanding-lagranges-theorem-group-theory | # Understanding Lagrange's Theorem (Group Theory)
I am beginning with Abstract Algebra and I'm trying to understand Lagrange's Theorem. The theorem reads
For any finite group $G$, the order of every subgroup $H$ of $G$ should divide the order of $G$.
It seems simple and I've used it to solve some exercices but I believe I'm missing the essence of it. Is there an example that can help me understand it better, or visualize it? Is there a geometric interpretation?
-
There is a local-representation theoretical version: if $H$ is a subgroup of $G$, then every projective module of $H$ induces a projective module of $G$. Hope you like it. :D – awllower Mar 21 '13 at 12:01
If you like to think visually, this books.google.com/… might pique your interest (not only about Lagrange's theorem, but in general)! – Stahl Mar 21 '13 at 12:06
The proof works by considering the cosets of $H$ in $G$, what we see is that the cosets partition $G$, so then $G$ is in some sense covered by sets of size $H$ which do not overlap. In terms of visualisation, draw a big circle, call it $G$, then you can partition your circle into equal sections of size $|H|$, for any subgroup $H$. – user27182 Mar 21 '13 at 12:08
What would qualify as "the essence" of this? It is a very straightforward theorem. It has no hidden details or mysterious manipulations. – Pedro Tamaroff Mar 21 '13 at 13:03
@Stahl This book is amazing, thank you! It turns out the author has also developed an open source application called GroupExplorer that helps visualize groups, homomorphisms, subgroup lattices, and more, which I find it incredibly useful in my case. – gpo Mar 23 '13 at 21:12
The left-multiplication map $x\mapsto ax$ is bijective on $G$; its injectivity follows from the cancellative property of the group's operation, $ah=ag\iff h=g$, and overall bijectivity is a consequence of the fact that it has an inverse map, $x\mapsto a^{-1}x$. I like to view a subgroup $H\le G$ as a "puck" and the overgroup $G$ as a "air-hockey table" on which $H$ resides, and to move $H$ around we apply left multiplication by various elements. If you left-multiply by an element $a\in H$, you have not moved the puck at all since $a\in H\iff H=aH$.
Every element $g\in G$ is in some coset, or left translate, of $H$ - in particular, $g=ge\in gH$ since we know that $e\in H$. Thus, the collection of all translates (possible places for the puck to be positioned in) of $H$ "cover" the entire air-hockey table. It remains, then, to investigate the nature of the overlaps between positions, i.e. the intersections of distinct cosets. Here is the proof that cosets that overlap nontrivially must in fact be identical, put into visual form:
$\hskip 0.6in$
This means the cosets of $H$ partition the group $G$. As left multiplication is bijective, every coset is the same size, so each "looks" the same from the viewpoint of cardinality. Continuing with the idea of an air hockey table, this tells us the puck positions tile it, so we have something like:
$\hskip 1.2in$
The most fundamentally basic meaning imputed to multiplication of natural numbers is the following: if Alice has $n$ bags each containing $m$ apples, then she has $n\times m$ apples total. Similarly, our group $G$ is covered by some number $[G:H]$ of disjoint cosets, each containing $|H|$ elements, so $|G|=[G:H]\times|H|$. Note that this is even true on the level of arbitrary infinite cardinals. Thus, $|H|$ is a divisor of the order $|G|$: Lagrange's theorem.
The converse is not globally true: not every divisor $d$ of $n=|G|$ corresponds to a subgroup $H\le G$ of size $|H|=d$. Sylow theory however yields a local version of a converse: for every prime power $q=p^r$ that is a divisor $q\mid n$, there is a $p$-subgroup $H$ of size $|H|=q$.
-
Thank you so much for the time and energy you put in writing such a comprehensive and detailed answer! It really helped me a lot and makes things a lot clearer. I actually understand the proof our professor gave now! – gpo Mar 23 '13 at 19:46
You already have several great explanations, but you asked also for a way to visualise this. One way to visualise Lagrange's Theorem is to draw the Cayley table of (smallish) groups with colour highlighting.
Here is the Cayley table of a dicyclic group of order $16$ with the cosets of its centre of order $2$ highlighted. The subgroup itself consists of the elements $\{ e, a \}$ (where $e$ is the identity), and is shown in red. The other cosets appear with different colours. Because the subgroup, in this case, is normal, the table is highly regular.
# The Maple code to produce this is:
> with( GroupTheory ):
> G := DicyclicGroup( 4 ):
> H := Centre( G ):
> DrawCayleyTable( G, cosets = H );
Here is another example, in which the subgroup is not normal. It is the Sylow $2$-subgroup of the symmetric group $S_{4}$. Since the order of $S_{4}$ is equal to $24$, the Sylow $2$-subgroup has order $8$ (and index $3$), and each coset has the same size $8$ as the subgroup. These facts are clear from the picture.
Nevertheless, the block structure of the cosets is still quite visible.
# Maple code for the second example:
> G := Symm( 4 ):
> DrawCayleyTable( G, cosets = SylowSubgroup( 2, G ) );
In each case, it is visually clear that the cosets of the subgroup form a "regular" partition of the group elements. The sizes of the blocks forming each coset are all identical, so they form a regular tiling of the Cayley table for the entire group.
Hope this helps!
-
Those are some very nice visualisations, thank you. I had no idea you could do this in Maple! – gpo Mar 23 '13 at 20:04
I think the key idea is that groups admit "translations", that is transformation of the form $\left\{ \begin{array}{ccc} G & \to & G \\ g & \mapsto & h \cdot g \end{array} \right.$ with $h \in G$. Moreover, if $H$ is a subgroup of $G$, then the images of $H$ by two such translations are either equal or disjoint; therefore, you can cover $G$ by translating the subgroup $H$ to get a partition of $G$ into $n$ disjoint copies of $H$. You deduce that $|G|=n|H|$.
-
Any equivalence relation on a set $\,S\,$ induces a partition on $\,S\,$ (and vice versa). This is a special case where the $\,n\,$ equivalence classes have the same size $\,k,\,$ so $\,|S| = nk,\:$ so $\,k\,$ divides $|S|.\,$ The result depends crucially on this key fact that the classes have the same size (which you do not mention). – Math Gems Mar 21 '13 at 14:38
@MathGems: I didn't give a proof but a way to understand Lagrange's theorem, and intuitively, a translation does not change the size. Moreover, implicitely a copy of $H$ has the same size than $H$. – Seirios Mar 21 '13 at 14:45
@Serios The word "translation" is used in many ways in mathematics, not all of which are set-theoretical bijections. In any case, from a pedagogical standpoint, it is always beneficial to bring to the fore those facts which play key roles. Hence my prior comment. – Math Gems Mar 21 '13 at 14:49
The main thing is that there is a natural one-to-one correspondence between any two (say, left) cosets of the subgroup $H$, namely the mapping $$s\mapsto yx^{-1}s$$ takes the coset $xH=\{xh\,\mid h\in H\}$ to $yH$, and its inverse is $s\mapsto xy^{-1}s$ (because $xy^{-1}yx^{-1}s=s$ for all $s$).
So, the size of each coset is the same ( $=|H|$), so, if $|G|$ is finite, $|H|$ must divide it. Stahl commented a nice link for visualizing this fact..
-
For example If order of G is 12 then we can find only subgroups of order 1,2,3,4,6 and 12 which are divisiors of 12
It means that we can not find a subgroup of other order except above orders
- | 2015-01-28T22:43:29 | {
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http://accessdtv.com/taylor-series/taylor-polynomial-error.html | Home > Taylor Series > Taylor Polynomial Error
# Taylor Polynomial Error
## Contents
Sign in Share More Report Need to report the video? But, we know that the 4th derivative of is , and this has a maximum value of on the interval . We could have been a little clever here, taking advantage of the fact that a lot of the terms in the Taylor expansion of cosine at $0$ are already zero. And so when you evaluate it at a, all the terms with an x minus a disappear, because you have an a minus a on them. http://accessdtv.com/taylor-series/taylor-series-polynomial-error.html
So the error at a is equal to f of a minus P of a. Can we bound this and if we are able to bound this, if we're able to figure out an upper bound on its magnitude-- So actually, what we want to do A Taylor polynomial takes more into consideration. But what I wanna do in this video is think about if we can bound how good it's fitting this function as we move away from a. look at this site
## Taylor Series Approximation Error
Where this is an Nth degree polynomial centered at a. How well (meaning ‘within what tolerance’) does $1-x^2/2+x^4/24-x^6/720$ approximate $\cos x$ on the interval $[{ -\pi \over 2 },{ \pi \over 2 }]$? We already know that P prime of a is equal to f prime of a.
This really comes straight out of the definition of the Taylor polynomials. http://mathinsight.org/determining_tolerance_error_taylor_polynomials_refresher Keywords: ordinary derivative, Taylor polynomial Send us a message about “Determining tolerance/error in Taylor polynomials.” Name: Email address: Comment: If you enter anything in this field your comment will be I'll write two factorial. Lagrange Error Bound Calculator The system returned: (22) Invalid argument The remote host or network may be down.
Loading... Taylor Series Remainder Calculator Thus, we have a bound given as a function of . The derivation is located in the textbook just prior to Theorem 10.1. But if you took a derivative here, this term right here will disappear, it'll go to zero.
Rating is available when the video has been rented. Lagrange Error Bound Formula Loading... And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f evaluated at a times x So this thing right here, this is an N plus oneth derivative of an Nth degree polynomial.
## Taylor Series Remainder Calculator
Rather, there were two approaches taken by us to estimate how well it approximates cosine.
We might ask ‘Within what tolerance does this polynomial approximate $\cos x$ on that interval?’ To answer this, we first recall that the error term we have after those first (oh-so-familiar) Taylor Series Approximation Error Sign in 6 Loading... Taylor Polynomial Approximation Calculator Basic Examples Find the error bound for the rd Taylor polynomial of centered at on .
That is, we're looking at Since all of the derivatives of satisfy , we know that . Check This Out Professor Leonard 99,296 views 3:01:45 Taylor Polynomials - Duration: 18:06. Theorem 10.1 Lagrange Error Bound Let be a function such that it and all of its derivatives are continuous. Because the polynomial and the function are the same there. Taylor Series Error Estimation Calculator
So let me write that. A More Interesting Example Problem: Show that the Taylor series for is actually equal to for all real numbers . If is the th Taylor polynomial for centered at , then the error is bounded by where is some value satisfying on the interval between and . http://accessdtv.com/taylor-series/taylor-polynomial-error-function.html Thread navigation Calculus Refresher Previous: Prototypes: More serious questions about Taylor polynomials Next: How large an interval with given tolerance for a Taylor polynomial?
The main idea is this: You did linear approximations in first semester calculus. Taylor Remainder Theorem Proof Especially as we go further and further from where we are centered. >From where are approximation is centered. And what I wanna do is I wanna approximate f of x with a Taylor polynomial centered around x is equal to a.
## Please try the request again.
Close Yeah, keep it Undo Close This video is unavailable. So, I'll call it P of x. Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... Error Bound Formula Statistics So it might look something like this.
Of course not. Skip to main contentSubjectsMath by subjectEarly mathArithmeticAlgebraGeometryTrigonometryStatistics & probabilityCalculusDifferential equationsLinear algebraMath for fun and gloryMath by gradeK–2nd3rd4th5th6th7th8thHigh schoolScience & engineeringPhysicsChemistryOrganic chemistryBiologyHealth & medicineElectrical engineeringCosmology & astronomyComputingComputer programmingComputer scienceHour of CodeComputer animationArts patrickJMT 128,850 views 10:48 Calculus 2 Lecture 9.9: Approximation of Functions by Taylor Polynomials - Duration: 1:34:10. have a peek here And that polynomial evaluated at a should also be equal to that function evaluated at a. | 2017-12-18T14:38:02 | {
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http://openstudy.com/updates/561872cfe4b086e96bfe46f3 | ## anonymous one year ago Simplify 3 square root of 5 end root minus 2 square root of 7 end root plus square root of 45 end root minus square root of 28.
• This Question is Open
1. anonymous
2. jim_thompson5910
The expression is this right? $\large 3\sqrt{5}-2\sqrt{7}+\sqrt{45}-\sqrt{28}$
3. anonymous
do you know how to solve it?
4. jim_thompson5910
first we need to simplify $$\Large \sqrt{45}$$
5. jim_thompson5910
$\large \sqrt{45}=\sqrt{9*5}$ $\large \sqrt{45}=\sqrt{9}*\sqrt{5}$ $\large \sqrt{45}=3\sqrt{5}$
6. jim_thompson5910
Notice how I factored it into 9*5 one of the factors is the largest perfect square factor possible
7. jim_thompson5910
make sense?
8. anonymous
yes
9. jim_thompson5910
how would you simplify $$\Large \sqrt{28}$$ ?
10. anonymous
2 square root of 7
11. jim_thompson5910
good
12. jim_thompson5910
So $\large 3\sqrt{5}-2\sqrt{7}+\sqrt{45}-\sqrt{28}$ turns into $\large 3\sqrt{5}-2\sqrt{7}+3\sqrt{5}-2\sqrt{7}$
13. jim_thompson5910
from here combine like terms
14. anonymous
ok
15. anonymous
i dont really know how to do this part its confusing
16. jim_thompson5910
if I gave you something like 3x+7y+2x+10y, would you be able to combine like terms?
17. anonymous
yes
18. anonymous
it would be 5x + 17y right?
19. jim_thompson5910
yes that's correct
20. jim_thompson5910
you'll use the same idea
21. jim_thompson5910
let x = sqrt(5) and y = sqrt(7) $\large 3\color{red}{\sqrt{5}}-2\color{blue}{\sqrt{7}}+3\color{red}{\sqrt{5}}-2\color{blue}{\sqrt{7}}$ $\large 3\color{red}{x}-2\color{blue}{y}+3\color{red}{x}-2\color{blue}{y}$
22. anonymous
ok
23. jim_thompson5910
simplify 3x-2y+3x-2y to get ???
24. anonymous
6x-y?
25. jim_thompson5910
6x is correct -y is not
26. anonymous
ok
27. jim_thompson5910
try again
28. anonymous
29. jim_thompson5910
-2y-2y is -4y, agreed?
30. jim_thompson5910
$\large 3\color{red}{x}-2\color{blue}{y}+3\color{red}{x}-2\color{blue}{y}$ $\large 6\color{red}{x}-4\color{blue}{y}$ $\large 6\color{red}{\sqrt{5}}-4\color{blue}{\sqrt{7}}$
31. jim_thompson5910
So, $\large 3\sqrt{5}-2\sqrt{7}+\sqrt{45}-\sqrt{28} = 6\sqrt{5}-4\sqrt{7}$
32. anonymous
ok thank a lot
33. anonymous
can you help me with this one please? if you want to. Which statement is true about the difference 2 square root of 7 − square root of 28?
34. jim_thompson5910
$2\sqrt{7} - \color{red}{\sqrt{28}} = 2\sqrt{7} - \color{red}{2\sqrt{7}} = 0\sqrt{7} = 0$ | 2017-01-19T10:52:14 | {
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https://www.themathdoctors.org/why-in-logic-does-false-imply-anything/ | # Why, in Logic, Does “False” Imply Anything?
In a class on symbolic logic, students are taught the truth tables that define the “logical connectives” ∧ (and), ∨ (or), ¬ (not), and → (if … then). Everything makes sense until they are told that if p is false, then $$p\rightarrow q$$ is true whether or not q is true. How can we say that “If pigs fly, then 2 is even” is a true statement? Or, for that matter, “If pigs fly, then there’s a kangaroo in my pocket”? This is especially troublesome when students, naturally wanting brevity, read $$p\rightarrow q$$ as “p implies q “. How can nonsense imply anything?
## A quick survey
Before I get into full answers, let’s look at some examples of the question. First, in 1994, in the infancy of Ask Dr. Math, we got this question:
8th Grade Logic
Why in a conditional statement if the "p" in the hypothesis is false, then the entire statement is true? Why isn't it undecided?
In 1997, in answer to a more general question about logic, Doctor Mike included an answer to the question, knowing it is common:
A False Statement Implies Any Statement
... In case you still are wondering about why a False implies anything, try this explanation on for size.
Like Doctor Ken in the 1994 answer, Doctor Mike here focused on an example, pointing out that if the condition of such a statement is false, then whatever happens, you couldn’t be convicted of lying, because you made no promises about what would happen in any situation that actually happens. In 2004, Jay asked a follow-up question that was added to the same page:
Why are the logical statements "false implies true" and "false implies false" always considered "true"?
I've read the previous note, but could you please give a more "formal" explanation?
Doctor Schwa replied with an answer relating the idea to set theory:
More formally, I'd say "implies" means the same as "subset" in set theory. That is, when you say, "if it rains, then the ground gets wet," you mean, "the set of times when it rains is a subset of the set of times when the ground gets wet."
So, since the empty set is a subset of any set, a false statement implies any statement.
Formally, this is a good way to think of it; but it may not satisfy everyone – particularly since it is not obvious why the empty set can be a subset at all. (That’s another question we get from time to time.)
In 2005, Doctor Achilles answered this one:
Logic and Conditional Sentences
I have a question about conditional statements. I am having a hard time understanding why two false statements in a conditional makes it true.
I tried to use different statements to create a truth table but I get stuck on the same concept. I tried a sentence like "If a polygon is a square, then the sides are equal." If I assume a rectangle, then it seems to me that the statement is undefined. Being true or false does not even apply.
This is much like the 1994 question, and is entirely reasonable. Be sure to read his answer; for the sake of space I will be focusing on two more recent questions whose answers include much of what others have said, while going beyond.
## A deeper look
The Logic behind Conditional Statements
I've read questions of the same title "Why is "false implies true/false" always "true"?" and I can understand the reason in the "subset way".
Could you explain the reason just in the logic way? Because when we say "work the same as subsets", we must prove it does work the same...
When I trust the set theory, I must trust logic first. So I want to understand the reason just on the way logic goes.
"False implies true/false" is true...why?
So Keven doesn’t like Doctor Schwa’s approach, because it doesn’t directly relate to logic. Having recently taught the subject and thought about this issue, I had several things to say.
### 1. “If … then” doesn’t mean “implies”
First, don't use the word "implies" to talk about a conditional statement; A->B should be read merely as "if A then B". "Implies" suggests a cause-and-effect relationship, or at least a logical connection of some sort. But the conditional statement is not meant to suggest that (even though many examples given in texts look that way). The statement "if A then B" says nothing more than "if A is true, then B is true"--not "if A is true, then it CAUSES B to be true". It means that whenever we find that A is true, then we can know that B is true (or else A->B would have been false).
The problem is that we naturally tend to see a conditional statement as something more, because everyday usage leaks over into our logic, and the word “implies” reinforces that tendency. Think of logical statements as merely observations about what things happen to be together, not about causality, or underlying reasons, or even necessary connections (“these things always go together”). Although mathematicians can use the word “implication” for this, it is misleading if you don’t pay close attention to the definition.
In fact, a logical statement is not even an assertion that we make for some reason, so that we have some stake in its truth. It is just a statement that may be true or false in any given situation. If A is true but B is not, then the statement “if A, then B” is false, because if it were true, then B would have to be true.
But what if we don’t have enough evidence to judge whether the statement is true?
### 2. It is a mathematical definition, not everyday reasoning
Second, this conditional statement is in a sense just something mathematicians define for their own purposes, not something that necessarily agrees with the natural-language use of the phrase. And what we need in logic (at least in traditional two-valued logic, as opposed to a logic that might include an "undetermined" value) is for every statement to be either true or false. In this context, we have to choose what A->B will mean in every case--in order to define a complete truth table.
Some of the earlier questioners felt that “If FALSE, then …” should just be “undetermined” or “I don’t know”; and they are correct, in real life. But symbolic logic requires a logical truth value of T or F for every statement, so we don’t have that option. We need some choice, which could be purely arbitrary (like rounding up on 5), or might have a specific reason based on how we plan to use it.
Here I gave a familiar answer:
Traditionally, mathematicians subscribe to a sort of "innocent until proven guilty" rule: we can't say something is false just because there is no evidence; instead, when there is no evidence of truth or falsity, we say it is true. This is what lies behind the related facts about sets: we say that the null set is a subset of any set because there is no evidence that it is not--there is no element in the null set that is NOT an element of the other set!
But still, why not say “guilty until proven innocent”?
### 3. This is what works in describing logical arguments
Ultimately, I realized, the reason for the choice comes from our application, not from the real world. One application of symbolic logic is to validate arguments, and here, if the truth value of a conditional statement were not defined as it is, then a valid argument would fail the test:
Let's consider an argument like this:
I have a cold.
If I have a cold, then my nose is running.
------------------------------------------
Therefore, my nose is running.
This has the form
A
A->B
----
B
To show that this is a valid argument, we write it as a single statement:
((A) ^ (A->B)) -> B
That is, the whole argument is a big conditional. If its truth table is ALWAYS true, no matter what the truth values of A and B are, then we consider the argument valid.
That is, an argument is considered valid if the equivalent statement is a tautology.
I made truth tables for this (valid) argument, using every possible definition of the conditional, and showed that the only definition for which it becomes a tautology when it should is the accepted one.
Now the truth table accurately reflects the validity of the argument. And that, I think, is why we make this definition: it works.
Why does it work? Because we want to say an argument is valid when the conclusion follows from the premise: if A is really true, then B had better be true. We DON'T CARE what happens if the premise is false; the argument is still valid because it doesn't tell you what happens then. There's the meaning behind that "innocent until proven guilty" idea.
## An example
In 2008, I answered another question, which gave me a chance to fill out a couple areas:
Logic Statement False Implies True
I am well familiar with the linguistic arguments which clarifies this somehow confusing concept. Is there a deeper philosophical argument that touches on the underlying logic of this concept {logic axiom}?
The most disturbing thing about this logic axiom is that it eliminates {defeats} the logical contingency of the conclusion on the premise, which somehow goes against the very essence of logic! By layman definition, logic is something that allows "naturally" the consequence to flow from a premise. When the same conclusion happens no matter what the premise is, the connectedness of the logic in between the conclusion and its premise loses significance, just like the definition of a function is defeated when one certain value from the domain point {maps} into two or more different values from the range.
If the moon is made of cheese, then I will go to the movie next week can rather best describe a sarcastic {insane} mode of thinking than a flow of "natural" logic especially when it can be said equally that if the moon is NOT made of cheese, I still go to the movie! The dilemma of this concept, though I use it myself to prove some propositions like the empty set is a subset of every set, is that it kills the "natural" connectedness inherent in logic.
Unfortunately, the very definition of logic itself is so intuitive and vague in the same way the set or sanity is defined, otherwise undefined! Even though I trained myself to live with this concept and I use it in my formal proofs, I try to avoid using it as much as possible. It is like employing proof by contradiction. I would rather prove directly.
Abe was thinking of conditional statements in cause-and-effect terms, as identifying “natural consequences”. He thought of logic as philosophy, and his vague conception led to confusion. (Logic is considered a part of philosophy not because it is deep in itself, but because we need to think clearly when we get to the deep stuff.) I repeated many of the ideas in the previous answer (which had not yet been archived so I couldn’t refer him to it): The conditional statement is not about cause-and-effect; the decision about truth value in the disputed cases depends on context; and one important context is judging the validity of an argument, which works if we take the “innocent until proven guilty” approach.
He responded with a comment that overstated the conclusion:
I found it rather interesting that there is indeed some philosophical underlying mode of thinking built into the definition of "P implies Q," that is, if there is no evidence that something is false, then we must assume that it is true. This we may cast as the "default rule of the truth."
...
Now comes the interesting case which P is false and Q is true, yet we must assume that the implication is True ONLY for lack of better knowledge or evidence pointing to the other direction. This is to me a philosophy or a mode of thinking which I accept as a sound one though it has some serious implications beyond math.
Since, as I have said, math has its own truth which need not agree with the real world, it doesn’t tell us anything about deeper philosophical ideas: The definition of the conditional statement is not about truth in general, but just about what we want a conditional statement to mean. He also tried to restate my reasoning, which led me to clarify it with a fully stated example:
Let's take an example. I have a piece of paper here that is coated with a chemical that changes color. I claim that if the paper is wet, it is red. That is,
WET -> RED
(Note that I didn't say "wet implies red"; the word "implies", as I said before, is not really appropriate for this connective, as you'll see in a moment.)
Now let's consider what you might see when I show you the paper, taking the four cases in your order.
1. It's wet, and it's red. That agrees with my statement, so you say my statement is true. (You do not have enough evidence to conclude that my statement is ALWAYS true; you've just seen one case. Maybe tomorrow it will be cooler, and you'll find that the paper is only red if it's wet AND warm. That's why you can't say it's true that wet implies red, only that it is true in this instance that "if it's wet, then it's red." Do you see the difference?)
2. It's dry, and it's blue. You don't know that it would be red if it were wet; there's no evidence one way or the other. So, simply by convention, you say that my statement is true, meaning that the evidence is consistent with that conclusion. But you can't say that you've proved that wetness IMPLIES redness; all you can say is that it might.
3. It's wet, and it's blue. That disproves my statement; we have a case where it is wet but NOT red. My statement is definitely false. (This case IS enough to disprove the stronger claim that wet implies red; you have a counterexample.)
4. It's dry, and it's red. Hmmm ... maybe it's ALWAYS red, and my statement was technically true but misleading; or maybe it's red for some other reason than wetness. Or maybe it actually turns blue when it gets wet, and I just lied. Again, you really don't know! The evidence at hand deals only with the case where it's dry, and my statement is about what would be true if it were wet. So you have to say that it's true, because you haven't disproved it, just like in case 2.
So your cases 2 and 4 are both "true" for the same reason, not for different reasons. The evidence in both cases is consistent with my statement, so we call it true.
He concluded with a concise statement:
In sum, P implies Q is nothing more than a claim or a proposition. We may uphold the rest of the logic table for P implies Q since the logic equivalence (truth value) for the remaining three cases does NOT contradict our claim about P implies Q, although not useful statements in some cases. Thanks again for the great example.
One might say that “truth”, in the sense we need here, just means “non-contradiction”.
### 2 thoughts on “Why, in Logic, Does “False” Imply Anything?”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-10-19T21:44:54 | {
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https://math.stackexchange.com/questions/1726580/probability-of-meeting-at-the-corner-of-city-streets-betwe | # Probability of meeting at the corner of city streets betwe
Two people agree to meet each other at the corner of two city streets between 1pm and 2pm. However, neither will wait for the other for more than 30 minutes. If each person is equally likely to arrive at any time during the one hour period, determine the probability that they will in fact meet.
This question is very similar to Probability of two people meeting during a certain time. But I'm afraid I'll have to ask again, because it's not quite what my query is about.
I've tried letting X and Y be the two random variables, and they are independent of each other. I tried phrasing it in terms of $P(|X-Y|<30)$ but I don't know how to solve this, as i don't know the individual probability of either X and Y? Any advice would be greatly appreciated. Sorry in advance if I seem to be repeating the question (in link) and for any wrong title or tag labelling.
I have also thought of another method:
If A arrives before B, then the probability that he arrives during first half an hour is 0.5 . He'll then wait for 0.5 hours. If He arrives during the last half an hour hours, with proability 0.5 . He then waits for 0.5 hours on average. Thus his total wait time is 0.5*0.5*2=0.5. For A and B to meet, B must arrive when A is waiting. Thus the probability that B arrived when A was waiting is 0.5 . Similarly if B arrives before A, the probability that they meet is 0.5 . Thus, Total probability that they meet is 1. Where did I go wrong? As the answer suggested is 0.75.
• Aren't $X$ and $Y$ both uniformly distributed between $0$ and $60$? (And I assume it is intended that they arrive independently, though it's not stated explicitly in the problem description.) – Brian Tung Apr 3 '16 at 22:24
• @BrianTung, However, if using uniform distribution, if I do invnorm on GDC, it doesn't mean that I can solve it? (as i don't know the mean and the variance?) Pls advise. – CCC Apr 3 '16 at 22:35
• Maybe I'm misunderstanding the question. Why isn't this question like the one you link to? – Brian Tung Apr 3 '16 at 22:37
• The variables are not normally distributed, so you cannot use mean and variance. You can integrate the probability density function over the region where the two people meet; since that function is constant and the region is more or less as depicted in that other question, the methods are essentially identical. – Brian Tung Apr 3 '16 at 22:48
• Probability is 75% according the following simulation (in R). "n <- 10^5 # number of trials a <- runif(n, min = 0, max = 60) b <- runif(n, min = 0, max = 60) sum(abs(a-b) < 30)/n" – snoram Apr 3 '16 at 23:06
The R code below essentially repeats @snoram's nice simulation, but using only 50,000 points. Then I go on to plot the points in the square with vertices at $(0,0)$ and $(60,60)$, with the points corresponding to the condition that the two people meet plotted in light blue.
From there, it is obvious that the two excluded regions (black) each have $1/8$th of the area. Because the joint distribution is uniform on the square this means that the probability of the 'condition' is $3/4.$
Of course, you can draw the boundaries of the condition, and thus solve the problem, without simulation. (It is $not$ necessary to know the distribution of $|X - Y|$.) In integral notation, you need $\int\int_C 1/60^2\,dx\,dy,$ where $C$ is the region corresponding to the condition. You'd have to break $C$ up into two parts in order to set numerical limits on the integral signs. But I think the geometrical argument suffices.
m = 50000
x = runif(m, 0, 60); y = runif(m, 0, 60)
cond = (abs(x-y) < 30); mean(cond) | 2019-08-18T14:59:21 | {
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https://mathhelpboards.com/threads/prove-that-a_n-tends-to-0.134/ | Prove that a_n tends to 0
Alexmahone
Active member
Assume $\frac{a_{n+1}}{a_n}\to L$, where $L<1$ and $a_n>0$. Prove that
(a) $\{a_n\}$ is decreasing for $n\gg 1$;
I've done this part.
(b) $a_n\to 0$ (Give two proofs: an indirect one using (a), and a direct one.)
Indirect proof: Assume for the sake of argument that $a_n$ does not tend to 0. Since the sequence is decreasing and bounded below by 0, it must converge to a positive value (call it $l$).
$\lim\frac{a_{n+1}}{a_n}=\frac{l}{l}=1$, a contradiction. So, $a_n\to 0$.
How do I go about the direct proof (one that doesn't use "proof by contradiction")?
Last edited:
Fernando Revilla
Well-known member
MHB Math Helper
How do I go about the direct proof (one that doesn't use "proof by contradiction")?
Hint
$|a_{n}-0|<\epsilon\Leftrightarrow |a_{n+1}\cdot (a_n/a_{n+1})|<\epsilon \Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\epsilon$ and $a_{n+1}/a_n$ is bounded.
Alexmahone
Active member
Hint
$|a_{n}-0|<\epsilon\Leftrightarrow |a_{n+1}\cdot (a_n/a_{n+1})|<\epsilon \Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\epsilon$ and $a_{n+1}/a_n$ is bounded.
Is this a proof by induction that $|a_n|<\epsilon$? If so, how do we prove the base case for $n=1$?
Last edited:
HallsofIvy
Well-known member
MHB Math Helper
I would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.
Alexmahone
Active member
I would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.
I used a slightly different approach:
$L<\frac{L+1}{2}$
$\frac{a_{n+1}}{a_n}<\frac{L+1}{2}$ for $n\gg 1$ (Using the "sequence location theorem")
So, $a_{n+1}< \left(\frac{L+1}{2}\right)a_n$ for $n\ge N$
$a_{N+k}<\left(\frac{L+1}{2}\right)^k a_N$ for $k\ge 1$ (Can be proved using induction over $k$.)
Since $\frac{L+1}{2}<1$, $a_{N+k}\to 0$ as $k\to\infty$.
So, $a_n\to 0$.
Last edited: | 2021-09-18T01:38:27 | {
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https://math.stackexchange.com/questions/492740/real-skew-symmetric-3-times-3-matrix-has-one-eigen-value-2i | # Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$
Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$ so another eigen value is $-2i$ another eigen value must be $0$ right? as we know these kind of matrix has eigen values $0$ or purely imaginary.
• Yes. $\quad\mu^{*} = \mu\quad$ and $\quad\mu^{*} = -\mu\quad$ $\Longrightarrow\quad\mu = 0$. – Felix Marin Sep 13 '13 at 19:50
Yes, the characteristic polynomial of a real $3\times 3$ matrix will be a real cubic polynomial. If not identically zero, the real skew matrix will have one pair of conjugate imaginary eigenvalues and a zero eigenvalue, each of multiplicity one.
• Does this eliminate the possibility of eigenvalues $2i,2i,-2i$? – Rebecca J. Stones Sep 13 '13 at 16:48
• @Rebecca: Yes, it does (for a 3x3 real skew matrix). – hardmath Sep 13 '13 at 17:17
• In case someone has doubts, the characteristic polynomial being a real cubic "eliminates the possibility" of three purely imaginary nonzero roots. Since these must occur in conjugate pairs, divding out a monic quadratic with one such pair as roots gives as quotient a real degree 1 polynomial having the third purely imaginary nonzero root (contradiction). – hardmath Sep 13 '13 at 21:50
In general, a real skew-symmetric $3 \times 3$ matrix $K$ looks like
$K = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}, \tag{1}$
so that the characteristic polynomial is
$\det(\lambda - K) = \lambda^3 + (a^2 + b^2 + c^2) \lambda, \tag{2}$
which is easily seen to have roots
$\lambda = 0, \pm i\sqrt{a^2 + b^2 + c^2}, \tag{3}$
which follow the general pattern the OP suggested; apparently the matrix she/he was thinking of satisfies $\sqrt{a^2 + b^2 + c^2} = 2$; but $0$ will always be an eigenvalue for such $K$ in any case.
Hope this helps. Cheers.
Since a real matrix $A$ is skew if, and only if, $A=-A^T$, since $\text{spec} (A)=\text{spec}(A^T)=-\text{spec}(-A^T)$ and since the characteristic polynomial has degree $3$ and non-real roots come in pairs, it follows that $0\in\text{spec}(A)\cap \text{spec}(A^T)$.
We know that a skew symmetric matrix $A=(a_{ij})_{3 \times 3}$ satisfies $A^T=-A$. This means that $a_{ii}=-a_{ii}$ or equivalently $a_{ii}=0$ for all $i \in \{1,2,3\}$.
We know the trace of $A$ is the sum of its eigenvalues, which is $0$ since the main diagonal of $A$ comprises of zeroes.
Since a $3 \times 3$ matrix has $3$ (not necessarily distinct) eigenvalues (given by the roots of the characteristic polynomial [a degree $3$ polynomial with leading coefficient either $1$ or $(-1)^3$, depending on how it's defined]), there is precisely one other eigenvalue $\lambda$ and we must have $2i-2i+\lambda=0$, which implies $\lambda=0$.
Here's another way of looking at it, which I thought of after posting my previous answer.
The approach here seemed sufficiently different that I felt a separate answer was merited. Hope I'm not overdoing it. Anyway . . .
First, note that for any skew-symmetric, real matrix, the only possible real eigenvalue is zero. For, if $K$ is real skew-symmetric, so that $K^T = -K$, and if $v \ne 0$ is an eigenvector with real eigenvalue $\lambda$, so that
$Kv = \lambda v, \tag{1}$
we have
$\langle Kv, v \rangle = \lambda \langle v, v \rangle, \tag{2}$
but
$\langle Kv, v \rangle = \langle v, K^Tv \rangle = \langle v, -Kv \rangle = -\langle Kv, v \rangle, \tag{3}$
or
$\lambda \langle v, v \rangle = -\lambda \langle v, v \rangle, \tag{4}$
and since $\langle v, v \rangle \ne 0$ this shows that
$\lambda = - \lambda, \tag{5}$
whence
$\lambda = 0. \tag{6}$
Here I take $\langle \cdot, \cdot \rangle$ to be the standard inner product on the real vector space on which $K$ operates.
Now argue as follows: since $2i$ is an eigenvalue of $K$, and the characteristic polynomial of $K$ is real, $-2i$ is an eigenvalue as well. This means
$\lambda^2 + 4 \mid p_K(\lambda), \tag{7}$
where
$p_K(\lambda) = det(\lambda - K) \tag{8}$
is the characteristic polynomial of $K$. But $\deg p_K(\lambda) = 3$ in the present case, so the quotient polynomial $q(\lambda)$ is of degree $1$, i.e. we must have
$p_K(\lambda) = (\lambda^2 + 4)q(\lambda) \tag{9}$
with
$q(\lambda) = \lambda - a, \tag{10}$
$a$ real. But this of course implies $a$ is a real eigenvalue of $K$, whence we must have
$a = 0. \tag {11}$
Well, I hope this answer is more than mere mathematical logorrhea, and it sheds a little more light on the subject at hand.
Cheers! | 2019-04-20T16:12:39 | {
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http://mathhelpforum.com/calculus/168398-limit-e-x-1-e-x-1-x-infinity.html | Math Help - Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity
1. Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity
Hey all,
I been playing around with limits, when i solve for the following function:
$\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$
But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks
2. Originally Posted by Oiler
Hey all,
I been playing around with limits, when i solve for the following function:
$\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$
But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks
????
If you graph that, you obtain a line y=1
3. Originally Posted by dwsmith
????
If you graph that, you obtain a line y=1
dwsmith, sorry just made some changes to the formula.
4. Originally Posted by Oiler
dwsmith, sorry just made some changes to the formula.
It is -1 when you go to negative infinity.
5. Originally Posted by Oiler
also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks
Let x = 0
0 = 2 is that true?
6. ofcourse, how dim of me. Thanks dwsmith.
7. $\displaystyle\lim_{x\to +\infty}\frac{e^x-1}{e^x+1}=\frac{\infty}{\infty}$
Applying L'Hopitals Rule
$\displaystyle\Rightarrow\lim_{x\to +\infty}\frac{e^x}{e^x}=1$
8. Originally Posted by dwsmith
$\displaystyle\lim_{x\to +\infty}\frac{e^x-1}{e^x+1}=\frac{\infty}{\infty}$
Applying L'Hopitals Rule
$\displaystyle\Rightarrow\lim_{x\to +\infty}\frac{e^x}{e^x}=1$
L'Hôpital's Rule is overkill for this. XD
$\lim\limits_{x\to\infty}\dfrac{e^x-1}{e^x+1} = \lim\limits_{x\to \infty}\dfrac{1-e^{-x}}{1+e^{-x}} = 1$ since $e^{-x}\rightarrow0$ as $x\rightarrow\infty$.
9. Originally Posted by Chris L T521
L'Hôpital's Rule is overkill for this. XD
$\lim\limits_{x\to\infty}\dfrac{e^x-1}{e^x+1} = \lim\limits_{x\to \infty}\dfrac{1-e^{-x}}{1+e^{-x}} = 1$ since $e^{-x}\rightarrow0$ as $x\rightarrow\infty$.
Sometimes you just have to kill it.
10. $\displaystyle \lim_{x \to \infty}\frac{e^x - 1}{e^x + 1} = \lim_{x \to \infty}1 - \frac{2}{e^x + 1}$
$\displaystyle = 1 - 0$
$\displaystyle = 1$.
11. Originally Posted by dwsmith
Sometimes you just have to kill it.
But you never have to "over-kill". When you are showing someone how to do something, the simplest way is always best.
12. Originally Posted by Oiler
Hey all,
I been playing around with limits, when i solve for the following function:
$\displaystyle\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$
But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations?
also how does $\{e^{x}-1\}=\{1+e^{x}\}$ ?.
Thanks
For this fraction, the denominator is always 2 greater than the numerator.
However, the "ratio" gets closer and closer to 1, as we increase x above 0.
$x\rightarrow\infty$ means x increases without bound above 0.
Hence, as x "approaches infinity", you need infinitely many decimal places
to express the value of the fraction as a value "other than 1".
That's part of the concept of limits.
viz-a-viz
$\displaystyle\frac{e^1-1}{e^1+1}=0.46211715726$
$\displaystyle\frac{e^2-1}{e^2+1}=0.76159415596$
$\displaystyle\frac{e^{10}-1}{e^{10}+1}=0.99990920426$
....onward.
If $x<0$ and decreases without bound below 0.
$x=-y$
$x\rightarrow\ -\infty\Rightarrow\ y\rightarrow\infty$
$\displaystyle\lim_{x \to -\infty}\frac{e^x-1}{e^x+1}=\lim_{y \to \infty}\frac{e^{-y}-1}{e^{-y}+1}=\lim_{y \to \infty}\frac{\frac{1}{e^y}-1}{\frac{1}{e^y}+1}$
$\displaystyle\lim_{y \to \infty}\left[\frac{\frac{1}{e^y}}{\frac{1}{e^y}}\right]\;\frac{1-e^y}{1+e^y}\right]=-\lim_{y \to \infty}\frac{e^y-1}{e^y+1}=-1$
When you have the concept, you can apply the fast methods as you please later.
13. Originally Posted by HallsofIvy
But you never have to "over-kill". When you are showing someone how to do something, the simplest way is always best.
To me, that was mighty simply. The constant disappear and the exponentials are their same derivatives. | 2015-04-26T03:57:20 | {
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http://math.stackexchange.com/questions/62574/analytical-reasoning-question-ii | # Analytical Reasoning Question II
I have yet another analytical question that got me
A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
Please kindly explain your final answer so those of us who are still trying to learn can grasp the logic easily. Thanks in advance!
-
The number of possible combinations is $5! = 120$. Each digit will be in each position an equal number of times, namely $5!/5 = 24$ times.
So the sum of the digits in the one's place is
$24(1+3+5+7+9) = 24*25 = 600$
This will be the sum of the digits in each place (one's, ten's, etc.). So the total sum is:
$600*11,111 = 6,666,600$, regards, iyengar
-
but let me ask,is my answer right ??,as i am not a person with formal training, – Iyengar Sep 7 '11 at 15:32
Why do you need this line? $24(1+3+5+7+9) = 24*25 = 600$. Why sum the numbers? I don't know the answer yet(trying not to look in the back of the book for the answers until I am sure I have this right).:) – user10695 Sep 7 '11 at 15:34
My question is, why add the individual allowable digits, rather than just adding the total occurrence for each? I was thinking more something like this $24(5) = 24*5 = 120$ – user10695 Sep 7 '11 at 15:35
For each decimal place, 24 of these are of the same digit. Thus for a decimal place the total value is 24 (1 + 3 + 5 + 7 + 9) = 600 To get the total value of all these numbers multiply the first result by 11111. – Iyengar Sep 7 '11 at 15:37
As a slight alternative to iyengar's answer:
There are $120$ such numbers and their average is $55555$ (note that for any of the form, there is also $111110$ minus that number) so the answer is $120\times 55555=6666600$.
-
where did you get 55555 from? The question specifies that there are no repeats allowed. – user10695 Sep 7 '11 at 16:15
If you have for example $17935$ then you also have $111110-17935=93175$, so the average term is half $111110$, namely $55555$. – Henry Sep 7 '11 at 16:23
I am sorry, I still don't get the 111110 and the 55555. Why those? – user10695 Sep 7 '11 at 19:30
This is partly a slight variant of Henry’s answer and partly a further explanation in answer to one of your questions.
If $d$ is one of the allowable digits $1,3,5,7,9$, let $d' = 10 - d$; note that $d'$ is also allowable. Now let $abcde$ be any allowable number; then $a'b'c'd'e'$ is also allowable, and it must be a different number from $abcde$. (It would be the same number only if all of the digits were $5$, but that’s not allowed.) In this way you can pair up all of the allowable numbers. Since there are $5! = 120$ permutations of the $5$ allowable digits, there are $120$ allowable numbers, and hence there are $60$ of these pairs.
It shouldn’t be too hard to see that $abcde + a'b'c'd'e' = 111110$ no matter which allowable number $abcde$ you start with. That is, the numbers in each pair sum to $111110$. Since there are $60$ pairs, the grand total is $60 \cdot 111110 = 6,666,600$.
Henry did essentially the same thing, except that instead of looking at the sum of $abcde$ and $a'b'c'd'e'$, he looked at their average, $111110/2 = 55555$. Since all pairs have the same average, the entire collection of allowable numbers must also have that average. And if the $120$ allowable numbers have an average of $55555$, their total must be $120 \cdot 55555 = 6,666,600$.
-
I need help seeing this [If $d$ is one of the allowable digits $1,3,5,7,9$, let $d' = 10 - d$; note that $d'$ is also allowable. ] please. Why is $10-d$ allowable? – user10695 Sep 7 '11 at 22:04
Also, how is $abcde + a'b'c'd'e' = 111110$. I don't get that please. Is this binary? – user10695 Sep 7 '11 at 22:06
@user10695: 1st question: Just do the arithmetic: if $d=1$, $10-d=9$; if $d=3$, $10-d=7$; and so on. Subtracting $1,3,5,7$, or $9$ from $10$ leaves $9,7,5,3$, or $1$, respectively. 2nd question: Not binary, just ordinary addition. $e+e'=10$, so you write down $0$ and carry $1$. Then $d+d'=10$, and the carry makes $11$, so you write down $1$ and carry $1$. The same thing happens in the remaining three columns, so you end up with a total of $111110$. – Brian M. Scott Sep 7 '11 at 22:15
Thanks! What I thought $abcde + a'b'c'd'e' = 111110$ meant was $(a*b*c*d*e) + (a'*b'*c'*d'*e')$ . Now I see what you mean. Thanks for that. – user10695 Sep 7 '11 at 22:33
why all this ?
it is so simple There are 5 digits and as we all know it will 5! numbers (1*2*3*4*5)=120 on a digit place every number will repeat 24 times that is 120/5 on ones place the sum will be 24(1+3+5+7+9)=25*24=600 If u need further explanation why 24(1+3+5+7+9+) see this: at every digit place 1 will repeat 24 times 24*1=24 3 will repeat 24 times 24*3=72 5 will repeat 24 times 24*5=120 7 will repeat 24 times 24*7=168 9 will repeat 24 times 24*9=216 total give 600
every digit place will sum to 600 unit place 600 tenth place 600 hundreds place 600 Thousands place 600 ten thousands place 600 -------------------------------- 6666600
-
Use MathJax please. – SchrodingersCat Oct 28 '15 at 16:30 | 2016-06-29T23:48:37 | {
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https://www.physicsforums.com/threads/solving-y-intercept-of-a-sin-function.427325/ | Homework Help: SOlving y-intercept of a sin function
1. Sep 8, 2010
t_n_p
1. The problem statement, all variables and given/known data
y = -2sin2(x+[pi/6])+1 for x [-pi, pi]
3. The attempt at a solution
set y=0,
1/2 = sin2(x+[pi/6])
pi/6 = 2(x+[pi/6]), where pi/6 is the base angle
Now because of the 2 infront of the (x+[pi/6]), I consider twice the domain, i.e. [-2pi, 2pi].
pi/6 is positive, and sin is positive in 1 & 2 quadrants.
therefore,
2(x+[pi/6]) = pi/6, 5pi/6, -7pi/6, -11pi/6.
divide by 2:
(x+[pi/6]) = pi/12, 5pi/12, -7pi/12, -11pi/12.
subtract [pi/6] from both sides:
x = -pi/12, 3pi/12, -9pi/12, -13pi/12.
This is my problem. Is -13pi/12 an issue since it is outside the original domain, or is it still ok since it is inside the modified domain [-2pi,pi]?
the answer has x values of -pi/12, 3pi/12, -9pi/12, 11pi/12, so my only issue is the last value. What has gone wrong!?
2. Sep 8, 2010
CompuChip
-13pi/12 is a solution to the equation y = 0, but you are only asked to give the solutions between -pi and pi. However, if you add (or subtract) any multiple of 2pi to -13pi/12, you get an equivalent solution. In this case, you could find -13pi/12 + 2pi = 11pi/12, which is in the requested interval.
3. Sep 8, 2010
t_n_p
ok, so I ask this, is there anyway to get 11pi/12 using the method I described, or can it only be obtained by adding 2pi?
so hypothetically, if I wanted x-intercepts for all x,
I would do:
-pi/12 +/- n*2pi
3pi/12 +/- n*2pi
-9pi/12 +/- n*2pi
-13pi/12 +/- n*2pi
where n is any integer?
I'm also wondering about the validity of adding pi (rather than 2pi). The period in our case is 2pi/2 = pi, so why doesn't adding/subtracing pi from all our values yield another number of solutions?
4. Sep 8, 2010
CompuChip
The important thing in this type of problem, is to do everything in the right order.
Step 1: finding the base angle and writing down two solution sets
At the beginning of your post, you get to the above equation,
$$\sin 2(x + \pi/6) = \frac12 = \sin \pi/6$$
This equation has two "branches" of solutions, namely
$$2(x + \pi/6) = \pi / 6 + n \cdot 2\pi$$
and
$$2(x + \pi/6) = \pi - \pi / 6 + n \cdot 2\pi$$
for any integer n (= 0, 1, 2, 3, ..., -1, 2, -3, ...).
[If you draw the graph of the sine function or you look at its geometric meaning in a unit circle, you will see how the symmetry leads to the second equation].
You have to add the n*2pi here, because you want the sines of both sides to be equal, and if you add any multiple of 2 pi to any of them, this will be the case.
Step 2: solving for x
Now you can go and solve the two equations for x. The first one leads to
$$x + \pi/6 = \pi / 12 + n \cdot \pi$$
(note that nothing strange happens here, it is just basic algebra: if you divide everything by 2, then n*2pi changes into n*pi automatically. As you remarked, this corresponds to the period) and
$$x = \pi / 12 - \pi / 6 + n \cdot \pi = - \pi/12 + n \cdot \pi$$
For the second branch you will get something similar, which I will leave up to you to work out (be careful with all the minus signs though)
Step 3: Find the specific solutions requested
You now have two equations which describe infinitely many solutions, although basically there are just two and all the others differ from them by an integer number of 2pi steps.
In this case, you are specifically asked to list only the solutions between -pi and pi. So you can go and plug in some numbers for n into
$$x = - \pi/12 + n \cdot \pi$$
For n = 0 you get - pi / 12 which is in the interval, so you can write that down. For n = 1 you get - pi / 12 + pi = 11 pi / 12, that is also OK. For n = 2 you go above pi, and for n = -1 you are below -pi, so this is all the solutions you get.
Do the same for the other "branch" of solutions, and you will find two more.
5. Sep 8, 2010
t_n_p
ok, that makes sense, but is this also feasible....
Follow what I did in my original post, where I found 4 solutions:
x = -pi/12, 3pi/12, -9pi/12, -13pi/12.
Why can't I just add multiples of the period to each of these values?
For x = -pi/12
x=-pi/12 + n*pi
test:
for n=-1, x=-13pi/12 (outside domain, invalid)
for n=0, x = -pi/12 (inside domain, valid)
for n=1, x= 11pi/12 (inside domain, valid)
for n=2, x= 23pi/12 (outside domain, invalid)
For x = 3pi/12
x=3pi/12 + n*pi
test:
for n=-2, x= -21pi/12 (outside domain, invalid)
for n=-1, x=-9pi/12 (inside domain, valid)
for n=0, x=3pi/12 (inside domain, valid)
for n=1, x= 15pi/12 (outside domain, invalid)
For x = -9pi/12
x=-9pi/12 + n*pi
test:
for n=-1, x=-21pi/12 (outside domain, invalid)
for n=0, x=-9pi/12 (inside domain, valid)
for n=1, x= 3pi/12 (inside domain, valid)
for n=2, x=15pi/12 (outside domain, invalid)
For x = -13pi/12
x=-13pi/12 + n*pi
test:
for n=-1, x=-25pi/12 (outside domain, invalid)
for n=0, x=-13pi/12 (outside domain, valid)
for n=1, x= -1pi/12 (inside domain, valid)
for n=2, x=11pi/12 (inside domain, valid)
for n=3, x=23pi/12 (outside domain, invalid)
then compiling all solutions that are valid I'm left with x=-9pi/12, 3pi/12, -pi/12, 11pi/12
as expected.
A little more time consuming but still works. My only question with the above method is, if the period is not pi, for example it is 2pi, will the same method work? I.e. if the period is pi, will adding n*2pi still provide the answers I am after?
6. Sep 8, 2010
CompuChip
Yes, that also works. But where do you get these four solutions from in the first place? Are they guesses or what?
7. Sep 8, 2010
t_n_p
I explained in the original post.
Now because of the 2 infront of the (x+[pi/6]), I consider twice the domain, i.e. [-2pi, 2pi]. I know the base angle is pi/6 and I know sin is positive in the 1st and 2nd quadrants. There will be 2 solutions per a revolution, then since there are 2 revolutions (domain is now -2pi to 2pi), there will be a total of 4 solutions.
That's the way I was taught. Since there is a factor of 2 out the front, I double the original domain. If there is a factor of 3 out the front, I triple the original domain so that it becomes -3pi to 3pi yielding a total of 6 solutions and so and so forth
8. Sep 8, 2010
CompuChip
Ah, I see now. So to be entirely clear:
* If you already find four solutions this way, then you don't need to work everything out like in post #5. All you have to do is shift solutions outside the requested range by an integer number of periods (in this case, pi).
* Whenever you want to do the n * (2)pi thing: if you already divided everything by 2, you use multiples of pi. You only shift by 2pi, if you haven't divided by the 2 (or 3, or whatever number) in front yet.
9. Sep 8, 2010
t_n_p
I definately prefer to do it the 1st star method.
Basically then, when I get a solution outside the bounds, I add or subtract 1 PERIOD (being careful to note what the period is, as it will change from problem to problem) to bring it back inside the bounds. E.g. if period is 500pi, then add/subtract 500pi from any values outside of the required bounds to bring it back inside.
This particular case had me stumped, since 3 of the solutions appear within the bounds, and only 1 is invalid. Obviously this is due to the value of the phase shift relative to the values obtained from the unit circle.
All makes sense now. Thanks | 2018-06-20T23:10:59 | {
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https://math.stackexchange.com/questions/1050957/finding-the-probability-of-a-selecting-at-least-1-of-an-element | Finding the probability of a selecting at least 1 of an element.
If there are 18 red and 2 blue marbles what is the probability of selecting 10 marbles where there is either 1 or 2 blue marbles in selected set.
Also, it seems intuitive that the probability should be twice that of selecting a single blue marble in a set of 10 from 19 red and 1 blue, but I'm not sure.
Another approachment is through hypergeometric distribution.
• The probability that there is exactly one blue ball in the selected set is: $$P(A_1)=\dfrac{\color{blue}{\dbinom 2 1}\cdot \color{red}{\dbinom {18} {10-1}}}{\dbinom {20}{ 10} }$$
• The probability that there are exactly 2 blue balls in the selected set is:
$$P(A_2)=\dfrac{\color{blue}{\dbinom 2 2}\cdot \color{red}{\dbinom {18} {10-2}}}{\dbinom {20}{ 10} }$$
So, the probability you are looking for is $P(A_1)+P(A_2)\approx 0.763$
In this scenario, with no restrictions, there are 3 possible cases:
$\bullet$ When selecting 10, there were no blue marbles ($A_{none}$)
$\bullet$ When selecting 10, there was one blue marble ($A_1$)
$\bullet$ When selecting 10, there were two blue marbles ($A_2$)
As there are no other possibilities, it stands to reason then that these three events form a partition of our overall sample space. That is: $X = A_{none}\cup A_1 \cup A_2$ and each $A$ is pairwise disjoint from the others.
It follows then, that $1 = P(X) = P(A_{none}) + P(A_1) + P(A_2)$ by rule of sums.
The event you are interested in is that either one or two blue marbles were selected, namely $A_1\cup A_2$.
By the above, $P(A_1\cup A_2) = 1 - P(A_{none})$
To calculate $P(A_{none})$, for each of the 10 marbles selected, you will have selected red. So, multiply the probabilities for each step that you in fact picked red.
So: $P(A_{none}) = \dfrac{18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9}{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11} = \dfrac{10\cdot 9}{20\cdot 19}$
And finally $P(A_1\cup A_2) = 1 - P(A_{none}) = 1 - \frac{10\cdot9}{20\cdot 19}\approx 0.763$
In comparison, if there were 19 red and 1 blue, it winds up being $1 - \frac{10}{20} = 0.5$ The answer to the original question is in fact a bit more than 50% more likely.
• The answer you gave is .864 but I believe it is supposed to be .763. I tried to change it but it said the change need to be at least 6 characters.
– qw3n
Dec 4, 2014 at 14:30
• absolutely correct. My mistake, a little arithmetic mistake. I had originally calculated 9/(2*19) and forgot to subtract it from 1. I did the subtraction from 1 in my head after and somehow messed up the tenths digit. Thanks for catching that. Dec 4, 2014 at 14:39
Given $18$ red marbles and $2$ blue marbles:
• The number of ways to choose $10$ marbles is $\binom{20}{10}=184756$
• The number of ways to choose $10$ red marbles is $\binom{18}{10}=43758$
• Hence the probability to choose only red marbles is $\frac{43758}{184756}=\frac{9}{38}$
• Hence the probability to choose not only red marbles is $1-\frac{9}{38}\approx76.31\%$
Given $19$ red marbles and $1$ blue marble:
• The number of ways to choose $10$ marbles is $\binom{20}{10}=184756$
• The number of ways to choose $10$ red marbles is $\binom{19}{10}=92378$
• Hence the probability to choose only red marbles is $\frac{92378}{184756}=\frac{1}{2}$
• Hence the probability to choose not only red marbles is $1-\frac{1}{2}=50\%$
As you can see, the former is not twice the latter... | 2022-06-24T23:26:20 | {
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https://math.stackexchange.com/questions/3353057/total-number-of-subsets-of-size-atmost-k | Total number of subsets of size atmost $k$
I was working on a problem that involved taking subsets of a multiset. I want to count the total number of distinct subsets of size at most $$k$$.
Example 1:
consider the multiset $$S = \{ 3, 3, 5, 7 \}$$ and $$k=3$$ then the answer should be $$12$$.
That is $$\{ \}, \{ 3 \}, \{ 3 \}, \{ 5 \}, \{ 7 \}, \{ 3, 5 \}, \{ 3, 7 \}, \{ 3, 5 \}, \{ 3, 7 \}, \{ 5, 7 \}, \{ 3, 5, 7\}, \{ 3, 5, 7 \} = 12$$ subsets.
Example 2:
$$S = \{ 2, 3, 5 \}$$ and $$k=2$$ then the answer should be $$7.$$
$$\{ \}, \{ 2 \}, \{ 3 \}, \{ 5 \}, \{ 2, 3 \}, \{ 2, 5 \}, \{ 3, 5 \} = 7$$ subsets
By using the formula from this answer I can count the total number of distinct subsets. How can I extend that formula for my constraints? Any idea?
• For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Thomas Shelby Sep 11 '19 at 17:04
• sorry for that.please edit my question if i make any mistake. – Midhun Manohar Sep 11 '19 at 17:07
• I do not follow your first example. Why do include $\{3\}$ twice? Why do you include $\{3,5\}$ twice? Why also did you not include $\{3,3\},\{3,3,5\}$ or $\{3,3,7\}$? If you intend for each $3$ to be considered distinct... then why do you use identical characters for them? It would have made more sense to never consider multisets in the first place and instead talk about the subsets of size at most $3$ from the set $\{3,\color{red}{X},5,7\}$ instead where we replaced the second three with $X$... at which point you have $\sum\limits_{i=0}^k\binom{n}{k}$ such subsets. – JMoravitz Sep 11 '19 at 18:33
• elements of the subset should be distinct – Midhun Manohar Sep 12 '19 at 5:04
Let me restate the problem, and see if you agree that it is the same thing. We have a finite collection of elements of types $$1,2,\dots,n$$ with $$a_j$$ elements of type $$j$$ for $$j=1,2,\dots,n$$, and we wish to know how many ways we can select at most $$k$$ objects, subject to the condition that no more than one element of any type is selected.
In your first example, we have $$2$$ elements of type "$$3$$" and one element of each of types "$$5$$" and "$$7$$".
Exactly $$k$$ elements may be selected in $$\sum a_{j_1}a_{j_2}\cdots a_{j_k}$$ ways, where the sum is over all $$k$$-tuples $$(j_1,j_2,...j_k)$$ with $$1\leq j_1
If you want at most $$k$$ objects, just add up the values for each nonnegative integer $$\leq k$$.
In you first example, with $$n=3$$, we have $$a_1=2,a_2=1,a_3=1$$. When $$k=0$$ we have an empty product, so the value is $$1$$. When $$k=1$$, we get $$2+1+1=4.$$ When $$k=2$$, we get $$2\cdot1+2\cdot1+1\cdot1=5$$. When $$k=3$$, we get $$2\cdot\cdot1=2$$. Altogether, we have $$1+4+5+2=12.$$
I gave a more concrete answer to a similar question a few days ago. Look at Find number of ways to select subset with distinct objects of at most K size..
• it seems working.can you elaborate more – Midhun Manohar Sep 11 '19 at 17:30
• @Midhunmanohar What is it that you don't understand? – saulspatz Sep 11 '19 at 17:32
• i have never seen use of multiplier symbol ∏ like this.can you please elaborate the formula more – Midhun Manohar Sep 12 '19 at 6:06
• @Midhunmanohar That was my fault. I meant sum, not product. No wonder you didn't understand. I don't know where my mind was. Anyway, it's the sum over all such products. I've edited my answer with a link to another answer that may be easier to follow. – saulspatz Sep 12 '19 at 10:47 | 2020-01-29T17:42:23 | {
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http://love.thegoodness.com/docs/winsor-and-newton-lana-paper-0dca71 | Explain. In this case, we use $$j$$ for the index for the summation, and the notation $$\sum_{j=1}^n j^2$$ tells us to add all the values of $$j^2$$ for $$j$$ from 1 to $$n$$, inclusive. For every natural number $$n$$, $$5^n \equiv 1$$ (mod 4). For another example, for each natural number $$n$$, we now let $$Q(n)$$ be the following open sentence: $1^2 + 2^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}.$. We resolve this by making Statement (1) an axiom for the natural numbers so that this becomes one of the defining characteristics of the natural numbers. \ \ \ \ \ &P(3)&\ \ \ \ \ \ \ \ \ &is& \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1^2 + 2^2 + 3^2 &=& \dfrac{3 \cdot 4 \cdot 7}{6} So let $$k$$ be a natural number and assume that $$P(k)$$ is true. That is, is 2 in the truth set of $$P(n)$$? Principle of Mathematical Induction Solution and Proof. First principle of mathematical induction For each natural number $$n$$, $$1 + 4 + 7 + \cdot\cdot\cdot + (3n - 2) = \dfrac{n(3n -1)}{2}.$$, We will prove this proposition using mathematical induction. Then to determine the validity of P(n) for every n, use the following principle: Check whether the given statement is true for n = 1. Which of the following sets are inductive sets? Then P(n) is true for all positive integers n. = (n + 1)! One way of proving statements of this form uses the concept of an inductive set introduced in Preview Activity $$\PageIndex{2}$$. We should keep in mind that no matter how many examples we try, we cannot prove this proposition with a list of examples because we can never check if 4 divides $$(5^n - 1)$$ for every natural number $$n$$. The basis step is an essential part of a proof by induction. Just because a conjecture is true for many examples does not mean it will be for all cases. Assume that $$T \subseteq \mathbb{N}$$ and assume that $$1 \in T$$ and that $$T$$ is an inductive set. The statement holds for n = 1, and. But in strong induction, the given statement holds true for all the steps from base to the kth step. In Section 4.2, we will learn how to extend this method to statements of the form $$(\forall n \in T) (P(n))$$, where $$T$$ is a certain type of subset of the integers $$\mathbb{Z}$$. Hence we can say that by the principle of mathematical induction this statement is valid for all natural numbers n. Show that 22n-1 is divisible by 3 using the principles of mathematical induction. -1 is divisible by 3 using the principles of mathematical induction. So 3 is divisible by 3. The two open sentences in Preview Activity $$\PageIndex{1}$$ appeared to be true for all values of $$n$$ in the set of natural numbers, $$\mathbb{N}$$. &=& \dfrac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6}\\ A class of integers is called hereditary if, whenever any integer x belongs to the class, the successor of x (that is, the integer x + 1) also belongs to the class. Sometimes it helps to look at some specific examples such as $$P(2)$$ and $$P(3)$$. For each natural number $$n$$, we let $$P(n)$$ be. Since $$5^{k+1} = 5 \cdot 5^k$$, multiply both sides of the congruence $$5^k \equiv 1$$ (mod 4) by 5. Now for the general case, if $$k \in \mathbb{N}$$, we look at $$P(k + 1)$$ and compare it to $$P(k)$$. 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https://math.stackexchange.com/questions/1576098/verify-proof-of-fx-ex-if-fxy-fxfy-and-fx-exists-for-all | # Verify proof of $f(x)=e^x$ if $f(x+y)=f(x)f(y)$ and $f'(x)$ exists for all $x$
This is exercise 6.26.8 from Tom Apostol's Calculus I, I'd like to ask someone to verify my proof. I'd be also interested in alternative proofs:
If $f(x+y)=f(x)f(y)$ for all $x$ and $y$ and if $f(x)=1+xg(x)$, where $g(x) \to 1$ as $x \to 0$, prove that (a) $f'(x)$ exists for every $x$, and (b) $f(x)=e^x$.
(a) $$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{f(x)f(h) - f(x)}{h} = \lim_{h \to 0}f(x)\frac{1 + hg(h) - 1}{h} = \lim_{h \to 0}f(x)g(h) = f(x)$$
(b) $$\left(\frac{f(x)}{e^x}\right)' = \frac{f'(x)e^x - f(x)e^x}{e^{2x}} = \frac{f(x) - f(x)}{e^x} = 0 \implies f(x) = ke^x \; \text, \; k\in \mathbb R$$ $$k = ke^0 = f(0) = \lim_{x \to 0}1 + xg(x) = 1 \implies f(x) = e^x$$
• Yes, it's fine. – egreg Dec 15 '15 at 0:42
Yes, it's fine. You have less computations if you set $F(x)=f(x)e^{-x}$, so $$F'(x)=f'(x)e^{-x}-f(x)e^{-x}=0$$ and $F$ is constant.
An alternative proof could be by observing that $f(x)=0$ for some $x$ implies $f$ constant $0$, which contradicts the assumptions. So we know $f(x)\ne0$ for all $x$ and differentiability (part a) implies $f$ is continuous, so everywhere positive. Then $$F(x)=\log f(x)$$ is well defined and $$F'(x)=\frac{f'(x)}{f(x)}=1$$ | 2019-05-19T20:19:10 | {
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https://math.stackexchange.com/questions/3168195/arranging-cats-and-dogs-what-is-wrong-with-my-approach | # Arranging cats and dogs - what is wrong with my approach
We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?
Since there are 5 spaces the cats can be in with the dogs fixed, there are $${5 \choose 3} * 4! * 3! = 1440$$ ways and this is the correct answer.
I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $$4 * 3$$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)
The other two dogs are free to go to any of the 4 spaces, with $$4^2$$ possibilities.
The cats can now be arranged in $$3!$$ ways.
So, our final answer should be $$3! * 4^2 * 4 * 3 = 1152$$
Where have I gone wrong?
• Is your problem arising as a consequence of a rainfall ? Mar 30, 2019 at 16:27
The second computation is missing a symmetry. Say your initial pattern is $$\underline {\quad}C_1\underline {\quad}C_2\underline {\quad}C_3\underline {\quad}$$
You then populate the spaces immediately to the right of $$C_1$$, and $$C_2$$. As:
$$\underline {\quad}C_1D_1\underline {\quad}C_2D_2\underline {\quad}C_3\underline {\quad}$$
So far so good. You still have $$D_3,D_4$$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?
Taking the two possible orders into account, we see that you are missing $$4\times 3!\times 4\times 3=288$$ cases. Adding them back gives you the desired result.
Phrased differently: once you have placed $$D_3$$ there are now five available spaces for $$D_4$$ (since $$D_4$$ might go either to the left or to the right of $$D_3$$). thus you should have had $$3!\times 4\times 5\times 4\times 3=1440$$
You can separate the two cases for the two last dogs: single dogs and double dogs.
Single dogs: $$P(4,2)=\frac{4!}{2!}=12.$$ Double dogs: $$P(2,2)\cdot C(4,1)=2\cdot 4=8.$$ Hence, there are $$12+8=20$$ (not $$4^2=16$$) ways to distribute the last two dogs.
The final answer is: $$3!\cdot 20\cdot 4\cdot 3=1440.$$ | 2022-08-13T19:20:51 | {
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https://mathematica.stackexchange.com/questions/207906/contour-plot-not-taking-previously-defined-expressions | # Contour Plot not taking previously defined expressions?
I am new to Mathematica so my question might sound a bit silly, but I hope you help me.
While learning ContourPlot, I have learned that when I use the previously defined expressions as inputs onto it, it doesn't seem to work, and only putting them in manually seems to work.
For example, defining
f1 = x^2/9 + y^2/4 == 1
f2 = x^2 - 1 == y
and then evaluating
ContourPlot[{f1, f2}, {x, -3, 3}, {y, -3, 8}]
results in empty plot, while putting the expressions manually inside such as
ContourPlot[{x^2/9 + y^2/4 == 1, x^2 - 1 == y}, {x, -3, 3}, {y, -3, 8}]
results in what I want...
What's more is, this also seems to occur when using NSolve... Can anyone tell me what's going on?
• I think it has something to do with the HoldAll attribute of ContourPlot... It works with ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]. – MelaGo Oct 15 '19 at 5:44
• Thank you! Could you specify what you mean by that? – Danny Han Oct 15 '19 at 7:57
• @DannyHan As @MelaGo noted, ContourPlot has attribute HoldAll. Before evaluating anything (in particular, before inserting the definitions of f1 and f2), ContourPlot will look at the first argument and decide what type of plot you want - whether you're giving it a single curve or a list and whether you are giving it a function or an equation. (continued...) – Lukas Lang Oct 15 '19 at 8:25
• (...continued) Here, it sees {f1, f2} and decides to plot two functions. When it starts evaluating f1 and f2 for given x and y, it doesn't get a number, but True and False, so it doesn't plot anything (non-numerical results are simply ignored by all/most plotting functions). Evaluate forces evaluation to occur before ContourPlot, so ContourPlot sees a list of equations, which it can then correctly plot. – Lukas Lang Oct 15 '19 at 8:25
• @LukasLang Great explanation! Worth posting as an answer. – Alexey Popkov Oct 15 '19 at 10:24
As suggested, I'm expanding my comment into an answer
The problem here is the HoldAll attribute of ContourPlot. Like Plot and similar functions the process goes something like this:
• Look at the first argument, and decide what form it has:
• If it's a list, the user wants to plot multiple functions
• If it's an equation with ==, the user wants to plot the solution to that equation
• Start the evaluation at different points (this part is done recursively on many points)
• Set the values of the variables (second and third argument) to the correct values (similar to Block, as noted in the details section of ContourPlot)
• Evaluate the first argument and use the result
Now we see what the problem in your case is:
ContourPlot[{f1, f2}, {x, -3, 3}, {y, -3, 8}]
When ContourPlot decides on the type of plot you want, it decides on "plot contours for two functions", since all it sees at that point is {f1, f2}. Now, values like e.g. x=0,y=0 are assigned and f1,f2 are evaluated. The problem is that this results in e.g.
{f1, f2}
(* --> *) {x^2/9 + y^2/4 == 1,x^2 - 1 == y}
(* --> *) {0^2/9 + 0^2/4 == 1,0^2 - 1 == 0}
(* --> *) {0 == 1,- 1 == 0}
(* --> *) {False, False}
And like most plotting functions, non-numeric results (like the False above, are simply discarded).
It is now also clear why Evaluate fixes the issue: It forces the first argument of ContourPlot to be evaluated before ContourPlot has a chance to look at it. So now the evaluation sequence goes like this:
ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]
(* --> *) ContourPlot[{x^2/9 + y^2/4 == 1,x^2 - 1 == y}, {x, -3, 3}, {y, -3, 8}]
At this point ContourPlot examines the first argument and decides on "plot the solutions of two equations", which is what we want.
It should be noted that the only thing that ContourPlot needs to see are the lists (when multiple things are to be plotted) and the equations - everything else can be evaluated later. This means the following will also work:
g1 = x^2/9 + y^2/4
g2 = x^2 - 1
ContourPlot[{g1 == 1, g2 == y}, {x, -3, 3}, {y, -3, 8}]
## TL;DR;
To summarise, force evaluation of the first argument of ContourPlot using Evaluate to ensure that the right type of plot is chosen:
ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]
• OMG thank you so much! Now I understand :) – Danny Han Oct 15 '19 at 12:29 | 2020-04-04T03:30:07 | {
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http://spot.pcc.edu/math/APEXCalculus/sec_multi_chain.html | # Section12.5The Multivariable Chain Rule¶ permalink
The Chain Rule, as learned in Section 2.5, states that $\ds \frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)g'(x)\text{.}$ If $t=g(x)\text{,}$ we can express the Chain Rule as \begin{equation*} \frac{df}{dx} = \frac{df}{dt}\frac{dt}{dx}. \end{equation*}
In this section we extend the Chain Rule to functions of more than one variable.
It is good to understand what the situation of $z=f(x,y)\text{,}$ $x=g(t)$ and $y=h(t)$ describes. We know that $z=f(x,y)$ describes a surface; we also recognize that $x=g(t)$ and $y=h(t)$ are parametric equations for a curve in the $x$-$y$ plane. Combining these together, we are describing a curve that lies on the surface described by $f\text{.}$ The parametric equations for this curve are $x=g(t)\text{,}$ $y=h(t)$ and $z=f\big(g(t),h(t)\big)\text{.}$
Consider Figure 12.5.2 in which a surface is drawn, along with a dashed curve in the $x$-$y$ plane. Restricting $f$ to just the points on this circle gives the curve shown on the surface. The derivative $\frac{df}{dt}$ gives the instantaneous rate of change of $f$ with respect to $t\text{.}$ If we consider an object traveling along this path, $\frac{df}{dt}$ gives the rate at which the object rises/falls.
We now practice applying the Multivariable Chain Rule.
##### Example12.5.3Using the Multivariable Chain Rule
Let $z=x^2y+x\text{,}$ where $x=\sin(t)$ and $y=e^{5t}\text{.}$ Find $\ds \frac{dz}{dt}$ using the Chain Rule.
Solution
The previous example can make us wonder: if we substituted for $x$ and $y$ at the end to show that $\frac{dz}{dt}$ is really just a function of $t\text{,}$ why not substitute before differentiating, showing clearly that $z$ is a function of $t\text{?}$
That is, $z = x^2y+x = (\sin(t) )^2e^{5t}+\sin(t) .$ Applying the Chain and Product Rules, we have \begin{equation*} \frac{dz}{dt} = 2\sin(t) \cos(t) \, e^{5t}+ 5\sin^2(t) \,e^{5t}+\cos(t) , \end{equation*} which matches the result from the example.
This may now make one wonder “What's the point? If we could already find the derivative, why learn another way of finding it?” In some cases, applying this rule makes deriving simpler, but this is hardly the power of the Chain Rule. Rather, in the case where $z=f(x,y)\text{,}$ $x=g(t)$ and $y=h(t)\text{,}$ the Chain Rule is extremely powerful when we do not know what $f\text{,}$ $g$ and/or $h$ are. It may be hard to believe, but often in “the real world” we know rate–of–change information (i.e., information about derivatives) without explicitly knowing the underlying functions. The Chain Rule allows us to combine several rates of change to find another rate of change. The Chain Rule also has theoretic use, giving us insight into the behavior of certain constructions (as we'll see in the next section).
We demonstrate this in the next example.
##### Example12.5.4Applying the Multivarible Chain Rule
An object travels along a path on a surface. The exact path and surface are not known, but at time $t=t_0$ it is known that : \begin{equation*} \frac{\partial z}{\partial x} = 5,\qquad \frac{\partial z}{\partial y}=-2,\qquad \frac{dx}{dt}=3\qquad \text{ and } \qquad \frac{dy}{dt}=7. \end{equation*}
Find $\frac{dz}{dt}$ at time $t_0\text{.}$
Solution
We next apply the Chain Rule to solve a max/min problem.
##### Example12.5.5Applying the Multivariable Chain Rule
Consider the surface $z=x^2+y^2-xy\text{,}$ a paraboloid, on which a particle moves with $x$ and $y$ coordinates given by $x=\cos(t)$ and $y=\sin(t)\text{.}$ Find $\frac{dz}{dt}$ when $t=0\text{,}$ and find where the particle reaches its maximum/minimum $z$-values.
Solution
We can extend the Chain Rule to include the situation where $z$ is a function of more than one variable, and each of these variables is also a function of more than one variable. The basic case of this is where $z=f(x,y)\text{,}$ and $x$ and $y$ are functions of two variables, say $s$ and $t\text{.}$
##### Example12.5.8Using the Multivarible Chain Rule, Part II
Let $z=x^2y+x\text{,}$ $x=s^2+3t$ and $y=2s-t\text{.}$ Find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}\text{,}$ and evaluate each when $s=1$ and $t=2\text{.}$
Solution
##### Example12.5.9Using the Multivarible Chain Rule, Part II
Let $w = xy+z^2\text{,}$ where $x= t^2e^s\text{,}$ $y= t\cos(s)\text{,}$ and $z=s\sin(t)\text{.}$ Find $\frac{\partial w}{\partial t}$ when $s=0$ and $t=\pi\text{.}$
Solution
# Subsection12.5.1Implicit Differentiation
We studied finding $\frac{dy}{dx}$ when $y$ is given as an implicit function of $x$ in detail in Section 2.6. We find here that the Multivariable Chain Rule gives a simpler method of finding $\frac{dy}{dx}\text{.}$
For instance, consider the implicit function $x^2y-xy^3=3\text{.}$ We learned to use the following steps to find $\frac{dy}{dx}\text{:}$ \begin{align*} \frac{d}{dx}\Big(x^2y-xy^3\big) \amp = \frac{d}{dx}\Big(3\Big)\\ 2xy + x^2\frac{dy}{dx}-y^3-3xy^2\frac{dy}{dx} \amp = 0\\ \frac{dy}{dx} = -\frac{2xy-y^3}{x^2-3xy^2}. \end{align*}
Instead of using this method, consider $z=x^2y-xy^3\text{.}$ The implicit function above describes the level curve $z=3\text{.}$ Considering $x$ and $y$ as functions of $x\text{,}$ the Multivariable Chain Rule states that $$\frac{dz}{dx} = \frac{\partial z}{\partial x}\frac{dx}{dx}+\frac{\partial z}{\partial y}\frac{dy}{dx}. \label{eq_mchain1}\tag{12.5.1}$$
Since $z$ is constant (in our example, $z=3$), $\frac{dz}{dx} = 0\text{.}$ We also know $\frac{dx}{dx} = 1\text{.}$ Equation (12.5.1) becomes \begin{align*} 0 \amp = \frac{\partial z}{\partial x}(1) + \frac{\partial z}{\partial y}\frac{dy}{dx} \Rightarrow\\ \frac{dy}{dx} \amp = -\frac{\partial z}{\partial x}\Big/\frac{\partial z}{\partial y}\\ \amp = -\frac{\,f_x\,}{f_y}. \end{align*}
Note how our solution for $\frac{dy}{dx}$ in Equation <<Unresolved xref, reference "eq_mchain2"; check spelling or use "provisional" attribute>> is just the partial derivative of $z$ with respect to $x\text{,}$ divided by the partial derivative of $z$ with respect to $y\text{.}$
We state the above as a theorem.
We practice using Theorem 12.5.10 by applying it to a problem from Section 2.6.
##### Example12.5.11Implicit Differentiation
Given the implicitly defined function $\sin(x^2y^2)+y^3=x+y\text{,}$ find $y'\text{.}$ Note: this is the same problem as given in Example 2.6.7 of Section 2.6, where the solution took about a full page to find.
Solution
# Subsection12.5.2Exercises
Terms and Concepts
In the following exercises, functions $z=f(x,y)\text{,}$ $x=g(t)$ and $y=h(t)$ are given.
1. Use the Multivariable Chain Rule to compute $\lz{z}{t}\text{.}$
2. Evaluate $\lz{z}{t}$ at the indicated $t$-value.
In the following exercises, functions $z=f(x,y)\text{,}$ $x=g(t)$ and $y=h(t)$ are given. Find the values of $t$ where $\frac{dz}{dt}=0\text{.}$ Note: these are the same surfaces/curves as found in Exercises 12.5.2.712.5.2.12.
In the following exercises, functions $z=f(x,y)\text{,}$ $x=g(s,t)$ and $y=h(s,t)$ are given.
1. Use the Multivariable Chain Rule to compute $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}\text{.}$
2. Evaluate $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ at the indicated $s$ and $t$ values.
In the following exercises, find $\lz{y}{x}$ using Implicit Differentiation and Theorem 12.5.10.
In the following exercises, find $\lz{z}{t}\text{,}$ or $\plz{z}{s}$ and $\plz{z}{t}\text{,}$ using the supplied information. | 2021-12-03T07:05:50 | {
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https://math.stackexchange.com/questions/3216715/in-a-geometric-progression-we-know-the-partial-sums-s-2-7-and-s-6-91-f | # In a geometric progression, we know the partial sums $S_2 = 7$ and $S_6 = 91$. Find $S_4$.
In a geometric progression, $$S_2 = 7$$ and $$S_6 = 91$$. Evaluate $$S_4$$. Alternatives: 28, 32, 35, 49, 84.
Here's what I tried so far:
$$S_2 = \frac{a_1(1-r^2)}{1-r} \implies 1-r = \frac{a_1(1-r^2)}{7} \\ S_6 = \frac{a_1(1-r^6)}{1-r} \implies 1-r = \frac{a_1(1-r^6)}{91}$$
Then: $$\frac{1-r^2}{1} = \frac{1-r^6}{13} \\ r^6 - 13r^2 + 12 = 0$$
Now i can't solve this equation, perhaps there's an easier way…
• The formulas you're using are for the sum of the first ever-so-many terms of the progression... is that what the $S_n$ are, the partial sums of the progression? Or are they the terms in the progression themselves? – Eevee Trainer May 7 '19 at 4:39
• $S_n$ is the sum of the first n terms of the progression. – rodorgas May 7 '19 at 4:41
• Given $r\ne\pm1$, you could have simplified $(1-r^6)/(1-r^2)=1+r^2+r^4$ – J. W. Tanner May 7 '19 at 5:04
• You absolutely should explain such key pieces in the question body. Those are easy to miss in comments. Also, a series has infinitely many terms. Your sums don't. Read the tag descriptions before using them. – Jyrki Lahtonen May 7 '19 at 5:58
• What are $S_2$ and $S_6$, precisely ? – Yves Daoust May 7 '19 at 6:07
Let $$x=r^2$$ then we see $$x^3-13x+12=0$$ so $$x^3-x-12x+12=0$$ so $$x(x-1)(x+1)-12(x-1)=0$$ so $$(x-1)(x^2+x-12)=0$$ so $$(x-1)(x+4)(x-3)=0.$$
• Ooops, sorry, my bad. I will delete this comment. – Yves Daoust May 7 '19 at 6:29
Here's how to do it with only a quadratic equation (of sorts).
Denote the geometric sequence $$a_1, a_2=a_1r, a_3=a_1r^2, a_4=a_1r^3, a_5=a_1r^4, a_6=a_1r^5...$$
Then $$S_2=a_1(1+r), S_4=a_1(1+r+r^2+r^3)=a_1(1+r)(1+r^2)$$,
and $$S_6=a_1(1+r+r^2+r^3+r^4+r^5)=a_1(1+r)(1+r^2+r^4).$$
$$\dfrac {S_6}{S_2}=\dfrac{91}7=13=1+r^2+r^4$$ so $$(r^2)^2+(r^2)-12=(r^2-3)(r^2+4)=0$$
so $$r^2=3$$ so $$S_4=S_2(1+r^2)=7(1+3)=28.$$
$$\frac{S_6}{S_2}=\frac{r^6-1}{r^2-1}=r^4+r^2+1=13$$ and
$$\frac{S_4}{S_2}=\frac{r^4-1}{r^2-1}=r^2+1.$$
With $$s:=r^2+1$$, we have $$s(s-1)+1=13$$
giving the two solutions
$$S_4=4\cdot 7$$ and $$S_4=-3\cdot7.$$
• $-21$ was not one of the alternatives offered by OP; presumably the progression is in real numbers, not complex – J. W. Tanner May 7 '19 at 6:39
• @J.W.Tanner: then $-21$ is rejected. – Yves Daoust May 7 '19 at 6:40
$$\begin{cases} S_2=a_1+a_1r=a_1(1+r)=7\\ S_6=a_1+a_1r+\cdots+a_1r^5=91\end{cases}\\ S_6-S_2=a_1r^2(1+r+r^2+r^3)=a_1r^2\cdot \frac{1-r^4}{1-r}=84\\ \frac{S_6-S_2}{S_2}=\frac{a_1r^2(1-r^4)}{a_1(1+r)(1-r)}=r^2(1+r^2)=12 \Rightarrow r^2=3$$ Hence: \begin{align}S_4&=a_1(1+r+r^2+r^3)=\\ &=a_1(1+r+r^2(1+r))=\\ &=a_1(1+r)(1+r^2)=\\ &=S_2(1+r^2)=\\ &=7(1+3)=\\ &=28.\end{align} | 2020-02-21T19:39:27 | {
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https://mathhelpboards.com/threads/find-a_n.8311/ | # Find a_n
#### anemone
##### MHB POTW Director
Staff member
Let $a_0=2$, $a_1=3$, $a_2=6$, and for $n \ge 3$, $a_n=(n+4)a_{n-1}-4na_{n-2}+(4n-8)a_{n-3}$.
The first few terms are $2,\;\;3,\;\;6,\;\;14, \;\;40, \;\;152, \;\;784, \;\;5168,\;\; 40576, \;\;363392$.
Find with proof a formula for $a_n$ of the form $a_n=c_n+d_n$, where $c_n$ and $d_n$ are well-known sequences.
Last edited:
#### mente oscura
##### Well-known member
Let $a_0=2$, $a_1=3$, $a_2=6$, and for $n \ge 3$, $a_n=(n+4)a_{n-1}-4na_{n-2}+(4n-8)a_{n-3}$.
The first few terms are $2,\;\;3,\;\;6,\;\;14, \;\;40, \;\;152, \;\;784, \;\;5168,\;\; 40576, \;\;363392$.
Find with proof a formula for $a_n$ of the form $a_n=c_n+d_n$, where $c_n$ and $d_n$ are well-known sequences.
Hello.
I guessed the sequences, because I have not been able to prove it.I thought that it could be a factor involved, and, then, by differences I found with another string.
$$a_n=2^n+n!$$
Regards.
#### MarkFL
##### Administrator
Staff member
Here is my solution:
Let's rewrite the recurrence as:
$$\displaystyle a_{n}-na_{n-1}=4\left(a_{n-1}-(n-1)a_{n-2} \right)-4\left(a_{n-2}-(n-2)a_{n-3} \right)$$
Now, if we define:
$$\displaystyle b_{n}=a_{n}-na_{n-1}$$
We may write the original recursion as:
$$\displaystyle b_{n}=4b_{n-1}-4b_{n-2}$$
This is a linear homogenous recursion with the repeated characteristic root $r=2$. Hence the closed form for $b_n$ is:
$$\displaystyle b_{n}=(A+Bn)2^n$$
Using the given initial values, we find:
$$\displaystyle b_1=a_1-a_0=3-2=1=(A+B)2$$
$$\displaystyle b_2=a_2-2a_1=6-6=0=(A+2B)4$$
From this 2X2 linear system, we find:
$$\displaystyle A=1,\,B=-\frac{1}{2}$$
Hence:
$$\displaystyle b_{n}=2^n-n2^{n-1}=a_{n}-na_{n-1}$$
Now, we may arrange this as:
$$\displaystyle a_{n}-2^{n}=n\left(a_{n-1}-2^{n-1} \right)$$
This implies one solution is $$\displaystyle c_n=2^n$$
If we define:
$$\displaystyle d_n=a_{n}-2^{n}$$
we then have:
$$\displaystyle d_{n}=nd_{n-1}\implies d_n=Cn!$$
And so by superposition we have the general form:
$$\displaystyle a_n=2^n+Cn!$$
Using the initial value we obtain:
$$\displaystyle a_0=2=2^0+C0!=1+C\implies C=1$$
Hence:
$$\displaystyle a_n=2^n+n!$$
#### anemone
##### MHB POTW Director
Staff member
Hello.
I guessed the sequences, because I have not been able to prove it.I thought that it could be a factor involved, and, then, by differences I found with another string.
$$a_n=2^n+n!$$
Regards.
Well, even though you didn't provide any proof, your answer is correct and since you've been actively engaged with our site for quite some time and solving many challenge problems in the Challenge Questions and Puzzles sub-forum, I would give allowance to you and hence I would declare it here that you got full mark for that!
Thanks for participating, mente!
Here is my solution:
Let's rewrite the recurrence as:
$$\displaystyle a_{n}-na_{n-1}=4\left(a_{n-1}-(n-1)a_{n-2} \right)-4\left(a_{n-2}-(n-2)a_{n-3} \right)$$
Now, if we define:
$$\displaystyle b_{n}=a_{n}-na_{n-1}$$
We may write the original recursion as:
$$\displaystyle b_{n}=4b_{n-1}-4b_{n-2}$$
This is a linear homogenous recursion with the repeated characteristic root $r=2$. Hence the closed form for $b_n$ is:
$$\displaystyle b_{n}=(A+Bn)2^n$$
Using the given initial values, we find:
$$\displaystyle b_1=a_1-a_0=3-2=1=(A+B)2$$
$$\displaystyle b_2=a_2-2a_1=6-6=0=(A+2B)4$$
From this 2X2 linear system, we find:
$$\displaystyle A=1,\,B=-\frac{1}{2}$$
Hence:
$$\displaystyle b_{n}=2^n-n2^{n-1}=a_{n}-na_{n-1}$$
Now, we may arrange this as:
$$\displaystyle a_{n}-2^{n}=n\left(a_{n-1}-2^{n-1} \right)$$
This implies one solution is $$\displaystyle c_n=2^n$$
If we define:
$$\displaystyle d_n=a_{n}-2^{n}$$
we then have:
$$\displaystyle d_{n}=nd_{n-1}\implies d_n=Cn!$$
And so by superposition we have the general form:
$$\displaystyle a_n=2^n+Cn!$$
Using the initial value we obtain:
$$\displaystyle a_0=2=2^0+C0!=1+C\implies C=1$$
Hence:
$$\displaystyle a_n=2^n+n!$$
Well done, MarkFL! I just love to read your solution posts because they are always so nicely written and well explained! Bravo, my sweetest global moderator! | 2020-09-28T12:42:54 | {
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https://math.stackexchange.com/questions/441241/a-group-of-order-8-has-a-subgroup-of-order-4/441262#441262 | # A group of order $8$ has a subgroup of order $4$
Let $G$ be a group of order $8$. Prove that there is a subgroup of order $4$.
I know that if $G$ is cyclic then there is such a subgroup (if $G=\langle a\rangle$ then the order of $\langle a^2\rangle$ is $4$). But how do I prove this when $G$ is not cyclic? Also, I know that $G$ has an element of order $2$, because the order of $G$ is even. I suspect that assuming all elements of $G$ are of order $2$ somehow leads to a contradiction but am unable to show it. Is this correct or is there a different approach that I'm missing? thanks
• The group might have all elements of order $2$, but then it is abelian (standard exercise), and the result is easy. Jul 11 '13 at 13:01
• A more general question is answered here: math.stackexchange.com/questions/306343/…? Jul 11 '13 at 13:07
• @GerryMyerson Though that requires quite a bit more than this special case. Jul 11 '13 at 13:11
• @Tobias, yes, which is why I'm not suggesting closing this question as a duplicate. But if OP wants to see something a little more hefty, I've pointed to a place to look. Jul 11 '13 at 13:12
You already noted that if $G$ has an element of order $8$ or $4$ then we are done.
Thus we can assume all elements have order $2$ (except the identity element). Then $G$ is abelian (this is a standard exercise, and I am certain it has been asked on MSE several times).
Let $a$ and $b$ be distinct elements of order $2$. Now it is straightforward to check that $\{e,a,b,ab\}$ is a subgroup of order $4$ (where $e$ is the identity element of $G$).
If $G$ is abelian so according to the Fundamental theorem for finite abelian groups we have: $$G\cong~~~\mathbb Z_8,~~\mathbb Z_4\times\mathbb Z_2,~~\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$$ So lets assume that $G$ is not abelian. If there is an element in $G$ which is of order $8$ then we have a contradiction. If all non trivial elements of $G$ be of order $2$ so again we have $G=\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$ which is a new contradiction, so we at least know that there is an elemnt of order $4$. I think we can stop here cause you wanted to know that. For the next you can use the subgroup generated by $x$ to show that $G=\langle y,x\rangle$ in which $y\in G-\langle x\rangle$ and of course $$G\cong \text{Q}_8,~~\text{or}~~\text{D}_8$$
• @amWhy: Yes Amy. What can I say about the event for my small job here? My English is not good as I can defend my self. :( Some one did the wrong job and someone else should pay the price. I think to myself I have been suggesting everyone I know here to be honest with policy but you see that I am a clueless victim. Sorry for saying these upsetting words but I don't have a tribune for saying it. I have just you to hear my stroy. Sorry Amy. Jul 12 '13 at 1:53
• @BabakS.: Sorry that you are going through this! I cannot even understand what happened, how is it possible to lose 5k rep in a day? It just does not make sense and must be a bug of some kind! Hope it gets resolved and it is not likely not something you did my friend! as amWhy said, hang in there! Jul 12 '13 at 4:10
• @amWhy: I am warned in a kind way. I don't know what happened because as I read somewhere in Meta, we cannot trace who downvote or upvote others. They think, I have another account and that is why I have more than 5k. Yes, I have and I told rob about it but that account was lost. I have nothing from that account. Even, I wanted rob to merge it for me. I have just 2 questions there. It is ridiculous for me having another active account just making plus. Jul 12 '13 at 15:10
• @amWhy:Where will I reach by doing this? More than you or Brian?? Where? What benefit will be for me? ISI Papers? Noble Prize? Believe me, I want to devote all reputations. A real sale is on the way. Sorry my friend. Jul 12 '13 at 15:11
• @BabakS. I defended your honor: I told Alexander Gruber that you are the most humble, gracious, and honest person that I know: and I said that with all sincerity. I think we all know you too well: I told him that you have had students who participate, and that I've see you suggest to them (at least, e.g., Flashdesign, that they accept others' answers...that you tend to answer them only when no one else has. I suspect that you've had an admiring student or two who enthusiastically upvoted some of your posts too eagerly, not knowing the consequences. Jul 12 '13 at 15:16
This is a direct consequence of the following theorem.
Let $G$ be a group of order $p^{n}$ for $p$ a prime. Then for each $m$ with $0 \leq m \leq n$, $G$ has a subgroup of order $p^{m}$.
The proof is here.
• This is the result pointed to by Gerry Meyerson. The proof takes considerably more than is needed for this special case. Jul 11 '13 at 13:24
I would consider elements of different orders and take cases accordingly.
You already are happy with what to do if there is an element of order 8 (and hence the group is generated by this element and cyclic).
Next, if there is a generator of order four, there clearly can only be one other generator, and this must be of order two. If we have $a$ and $b$ of orders 2 and 4 respectively you can easily check check what all the elements are, and it should be obvious what the subgroup of order 4 is.
Lastly, if you have a generator of order two, you could have another generator of order 4, but we have already considered this. The only other alternative is to have two more generators of order 2, which must commute (else the group they generate would be too large) and again, if you consider the abelian group generated by $a$, $b$ and $c$, all of order two, it should be clear how you can generate a group of order four.
Note that this is not the most elegant solution, but it is (probably) the most concrete. And please ask if anything is not clear.
• "if there is an element of order four, there clearly can only be one other element" Huh? Surely, that's not what you meant to write. Jul 11 '13 at 13:13
• If there is an element of order $4$, you are clearly done. And the fact that the elements of order $2$ will commute follows only once you use that all the elements have order $2$. Jul 11 '13 at 13:13
• @GerryMyerson sorry, I meant to talk about generators, not elements. Thank you! Jul 11 '13 at 13:15
• A group of order $8$ cannot have a generator of order $4$. Jul 11 '13 at 13:17
• When you say G is generated by a, b and c you mean G = {e, a, b, c, ab, ac, b*c...}? Jul 11 '13 at 13:36 | 2021-10-19T15:03:53 | {
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http://math.stackexchange.com/questions/111173/how-do-i-find-the-antiderivative-of-5-sin4x | # How do I find the antiderivative of $5\sin(4x)$?
This is actually a part of a bigger problem, which involves using the Mean Value Theorem for Integrals. The question is to find, $f_{ave}$:
$f(x) = 5 \sin 4x$ for $x\in [−π, π].$
Using the theorem, I have gotten it down to:
$$\frac{1}{2\pi} \int_{-\pi}^\pi5\sin(4x)dx = 5\sin(4c)$$
I know that I have to find the antiderivative and then solve for c, but I don't know how to find the antiderivative. Any help will be appreciated.
-
HINT: Do a substitution; set $u=4x$, and solve the integral. – Arturo Magidin Feb 20 '12 at 3:31
Am I overall on the right track? – user754950 Feb 20 '12 at 3:34
@user754950: Yes, what you are doing will lead to an answer, once you "solve" the integral. And the hint I gave you should help you solve that integral. – Arturo Magidin Feb 20 '12 at 3:38
A solution method surprisingly not listed below is integration by parts. I know,it would initially have resulted in a more complicated integral then the other methods here-but it probably would result in the same solution more directly then the methods below. I suggest it just to make sure the list of proposed solutions is as complete as possible. – Mathemagician1234 Feb 20 '12 at 7:40
1st you can pull the $5$ out in front of the integral sign to give you $\frac{5}{2\pi}$. Then the antiderivative of $\sin(4x)$ is $-\frac{1}{4}\cos(4x)$ because the antiderivative of $\sin(x)$ is $-\cos(x)$ [recall that the derivative of $\cos(x)$ is $-\sin(x)$] then you multiply that by the reciprocal of the constant associated with $x$ [meaning $1$ over $4$ or $\frac{1}{4}$]. Make sure you evaluate the problem from $-\pi$ to $\pi$.
-
alright, I did as you said and I got: -1/(4pi) = sin(4c). Is this correct so far? If so, how do I solve for c? does it involve sine inverse? – user754950 Feb 20 '12 at 3:50
I am not really sure where you get the 5sin(4c) from. Can you explain that so I can better understand your goal? – Jared Feb 20 '12 at 4:09
Take a look at the graph of your equation on a graphing calculator. Note that the region you are integrating cancels out, but if you change your limits so you are integrating from 0 to pi/4, you can multiply the integral by 8 to get a non-zero, non-negative value. – Jared Feb 20 '12 at 4:23
I'm not sure how to put integral symbols in the text; they will be written out in words with curly brackets. Now that you have redefined your limits, your integral problem should read (8/2pi){integral from 0 to pi/4}5sin(4x)dx = (4/pi){integral from 0 to pi/4}5sin(4x)dx. Now since the 5 is a constant, move it outside the integral: (20/pi){integral from 0 to pi/4}sin(4x)dx. Use u-substitution so u=4x and du=4dx, which means (1/4)du=dx. Since you changed from 4x to u, also change your interval: from 4*0=0 to4*(pi/4)= pi. So now you have (1/4)(20/pi){integral from 0 to pi}sin(u)dx. – Jared Feb 20 '12 at 4:31
Now use the antiderivative of sin(x) I mentioned earlier to integrate. You have F(u)=-(5/pi)cos(u){from 0 to pi}. Plug in your intervals so you get F(pi)-F(0)=(final answer). That is [-(5/pi)(-1)]-[-(5/pi)(1)]=[5/pi]-[-5/pi]=(5/pi)+(5/pi)=(10/pi). I hope this is the final answer you are looking for. Remember to add units if your teacher requires that. – Jared Feb 20 '12 at 4:37
For what it's worth:
You can use the "guess and check" method as follows.
The derivative of $-\cos x$ is $\sin x$. So, perhaps the antiderivative of $5\sin(4x)$ is $$-5\cos(4x).$$
Does this work? Let's check:
The derivative of our guess has to be $5\sin(4x)$; but, $${d\over dx} \bigl(-5\cos (4x)\bigr)=5\sin(4x)\cdot 4= 5\cdot 4\sin(4x).$$ Hmm, it's not quite right, we do not want that "4" there on the right hand side, that arose from the chain rule. But, if we introduced a multiplicative factor of $1\over4$ in our guess for the antiderivative, things would work out: $${d\over dx} \bigl(-{5\over4}\cos (4x)\bigr)={5\over4} \sin(4x)\cdot 4= 5 \sin(4x).$$
So, indeed, an antiderivative of $5\sin(4x)$ is $-{5\over4}\cos(4x)$.
More generally:
If $F(x)$ is an antiderivative of $f(x)$, then
$\ \ \$1) $cF(x)$ is an antiderivative of $cf$ (since $(cf)'=c f'$)
$\ \ \$2) For $c\ne0$, ${1\over c}F(cx)$ is an antiderivative of $f(cx)$ (by the chain rule).
Of course, you can use the "substitution method" for integrals for your problem.
-
I think "guess and check",although useful in some differential equations problems,leads to sloppy habits. But it does work in this case,so you can't argue with success......... – Mathemagician1234 Feb 20 '12 at 7:42
I would view adjusting constants at the end as a quite efficient method. – André Nicolas Feb 20 '12 at 21:53
From the basic theory of primitives you can check that
$$\int {f\left( {ax} \right)dx = \frac{1}{a}F\left( {ax} \right) + C}$$
So you can use this and put
$$5\int {\sin \left( {4x} \right)dx = - \frac{5}{4}\cos \left( {4x} \right) + C}$$
Alternatively $\sin x$ is odd, you will have that the integral over any symmertric interval around the origin will be zero, that is
$$\int\limits_{ - \pi }^\pi {\sin \left( {4x} \right)dx} = 0$$
So you problem ultimately is finding $c\in[-\pi,\pi]$ for
$$5\sin \left( {4c} \right) = 0$$
which has solutions. $(0,\pm\pi/4,\pm\pi/2,\pm 3\pi/4,\pm\pi)$
-
You may want to add "Alternatively" before "Since $\sin x$ is odd..." since what follows there is really an alternate way of solving the problem that does not require finding an antiderivative for $5\sin(4x)$. – Arturo Magidin Feb 20 '12 at 4:03
@ArturoMagidin Indeed. Thanks. – Pedro Tamaroff Feb 20 '12 at 4:06 | 2015-05-28T00:54:18 | {
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https://math.stackexchange.com/questions/2403052/what-can-be-understood-by-a-b-b-a-in-set-theory | What can be understood by $A-B = B-A$ in set theory?
What can be understood by $\rm A-B = B - A$ in set theory?
What does this tell us about the characteristics of $A$ and $B$ and their relationship? I'm quite confused as I am new to set theory.
Remember that "$A-B$" means "The set of things in $A$ but not in $B$." So "$A-B=B-A$" means "The things in $A$ but not $B$ are exactly the things in $B$ but not $A$."
Now, for which $A$ and $B$ is this equation true? As always, when you're trying to understand new abstract concepts (in this case, set difference and Boolean operations in general) it's best to try some examples. Does the equation $A-B=B-A$ hold for $A=\{1, 2, 3\}=B$? What about $A=\{1,2,3\}, B=\{1, 2\}$? What about $A=\{1, 2, 3\},B=\{2, 3, 4\}$?
Based on these examples, you should be able to make a good guess at what the answer should be. Now, try to prove it! (As usual, this will look like "Assume $x\in A-B$. Then [stuff]. So $x\in B-A$. etc.")
($A-B=B-A$) means that the set of everything in $A$ which is not in $B$ equals the set of everything in $B$ which is not in $A$.
This is possible only when $\underline{(\phantom{A=B})}$ because:
For any element $x$ of $A-B$, we have $x\in A$ and $x\notin B$.
For any element $x$ of $B-A$, we have $x\in B$ and $x\notin A$.
But there is no element that can be in $A$ and not in $A$, and in $B$ and not in $B$.
Therefore $A-B$ is $\underline{\phantom{\quad\emptyset\quad}}$ , as is $B-A$. Meaning...
You can work it out using a Venn diagram.
The above picture (sourced from Wikipedia) shows that $A \cup B$ can be divided into three regions: $A - B$ (the bit purely in orange), $A \cap B$ (the bit that's orange and blue), $B - A$ (the bit that's purely in blue).
What your equation is saying is that the bit purely in orange is equal to the bit purely in blue. Since the remaining bit in orange overlaps with the thing in blue, and the remaining bit in blue overlaps with the thing in orange, these two sets must be equal.
More formally, we see that $$A = (A - B) \cup (A \cap B)$$ $$B = (B - A) \cup (A \cap B)$$
Since $A - B = B - A$, we have that \begin{align} A &= (A - B) \cup (A \cap B)\\ &= (B - A) \cup (A \cap B)\\ &= B\end{align}
In fact, the equation $A - B = B - A$ is equivalent to the one $A = B$. Akiva explains below.
• And conversely… – Akiva Weinberger Aug 23 '17 at 13:31
• @AkivaWeinberger I don't understand your comment. What are you trying to say? – man and laptop Aug 23 '17 at 13:33
• You should mention the converse as well. $A-B=B-A$ implies $A=B$, and conversely, $A=B$ implies $A-B=B-A$ (because they'd both be $\emptyset$). – Akiva Weinberger Aug 23 '17 at 13:35
• Sure ${}{}{}{}$ – Akiva Weinberger Aug 23 '17 at 13:48
Two sets $X,Y$ are equal if and only if for all elements $z$ it holds that $z\in X\iff z\in Y$. If $z\in A-B$ then $z\in A$ but $z\notin B$ and if $z\in B-A$ then $z\in B$ but $z\notin A$. It seems impossible that an element could belongs to both $A-B$ and $B-A$, doesn't it?
$A - B$ can be rewritten as $A - (A \cap B)$.
$B - A$ can be rewritten as $B - (A \cap B)$.
Given: $A - (A \cap B) = B - (A \cap B)$.
which means: $A$ and $B$ should be equal and $A - B = B - A = \emptyset$
• This is an interesting exposition; why did someone downvote? – Ari Brodsky Aug 23 '17 at 8:52
• @AriBrodsky No idea :) I thought my answer was to the point and complete. :) – Sarthak Mittal Aug 23 '17 at 9:03
• Almost likely the downvotes come from people who disagree with spoonfeeding OP. This is likely homework or a learning exercise and by doing it for OP you do not allow them to learn. – Ander Biguri Aug 23 '17 at 9:48
• @AnderBiguri yeah I understand that! just tried to post a simplified answer. :) – Sarthak Mittal Aug 23 '17 at 11:52
OP says
I'm quite confused as I am new to set theory.
So wordy exposition really can't be harmful, especially since the OP asked "what can be understood", leaving the door open for a lengthy response.
If $a$ and $b$ are numbers, then $a - b = a$ is only possible if $b = 0$.
If $A$ and $B$ are sets, then $A - B = A$ if $B = \emptyset$, but that is not the end of the story. We know that $A - B \subseteq A$, and after some thought we come up with
Definition: An element $\hat a \in A$ is said to be safe from $B$ if $\hat a \in A-B$.
Exercise 1: $\hat a \in A$ is safe from $B \Leftrightarrow \hat a \notin B$.
Exercise 2: $A - B = A \Longleftrightarrow$ all elements in $A$ are safe from $B$.
Exercise 3: $A - B = A \Leftarrow \Rightarrow A \cap B = \emptyset$.
Exercise 4: $B - A = B \Leftarrow \Rightarrow A \cap B = \emptyset$.
Exercise 5: $A - B = A \Leftarrow \Rightarrow$ all elements in $B$ are safe from $A$.
So, we can summarize as follows,
Proposition 1: $A - B = A \Leftarrow \Rightarrow B - A = B \Leftarrow \Rightarrow A \text{ and } B\;$ are disjoint sets.
We see that there is certainly some symmetry going on here.
If $a$ and $b$ are numbers, then $a - b = b - a \Leftarrow \Rightarrow a = b$. Does this at least now carry over into set theory? What could go wrong with so much symmetry?
Proposition 2: Suppose for two sets $A$ and $B$,
$\tag 1 A - B = B - A$ Then $A = B$ and $A - B = B - A = \emptyset$.
Proof
Set $C = A - B = B - A$ and let $c \in C$. Then since $C = B - A$, $\,c \in B$. But we also have, $c \in A - B$, i.e. $c \in A$ is safe from $B$. By exercise 1, $c \notin B$, a contradiction.
So $C = \emptyset$. So, $A - B = \emptyset$. But this means no element in $A$ is safe from $B$, or, by exercise 1, every element in $A$ must be be in $B$, Same argument shows that every element in $B$ must be in $A$, so $A = B$. $\qquad \blacksquare$
All this symmetry here is really exciting, and before you know it you'll really appreciate the following material.
Definition: The symmetric difference ${\displaystyle A\,\triangle \,B}$ of two sets $A$ and $B$ is given by
$\tag 2 {\displaystyle A\,\triangle \,B=(A - B)\cup (B - A)}$
Exercise 6: Show that $A\,\triangle \,B=(A\cup B) - (A\cap B)$.
Exercise 7: Show that the following equalities are all equivalent statements,
$\qquad {\displaystyle A\,\triangle \,B = \emptyset}$
$\qquad A = B$
$\qquad A - B = B - A$
Note: The definition "$\hat a \in A$ is said to be safe from $B$" was made to aid intuition and you won't find it by googling. But see symmetric difference. | 2019-05-24T22:48:32 | {
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http://www.physicsforums.com/showthread.php?s=3e9f850247158ce74e5c4c2340be720a&p=4029438 | ## Deriving the volume and surface area of a cone
Hello, this is my first time posting on physics forums, so if I do something wrong, please bear with me :)
I am trying to derive the formula for the lateral surface area of a cone by cutting the cone into disks with differential height, and then adding up the lateral areas of all of the disks/cylinders to find the lateral area of the cone. (Similar to using the volume of revolution, but just taking the surface area). I assumed that the heights of each of the disk was dH, where H = the height of the cone. However, using that method, I got pi*radius*height instead of pi*radius*(slant height).
On another thread in physics forum (http://www.physicsforums.com/showthread.php?t=354134) , a person used a similar method as I did, and someone replied saying that the height of each disk is dS, where S = the slant height of the cone, not dH, where H = the height of the cone. And using dS instead of dH allowed me to find the correct formula for the lateral area of the cone!
HOWEVER, using dS instead of dH contradicts with the same method for finding the volume of the cone. I basically used the method shown in another person's video (http://www.youtube.com/watch?v=Btx_f883uFU) to find the volume of the cone, but that person assumed that the height of each disk is dH, not dS.
So is the height of each disk dS or dH? If it is dS, then I understand how to get the formula for the lateral surface area of the cone, but not the volume. But if it is dH, then I understand how to get the formula for the volume of the cone, not the surface area. Can anyone clear up my confusion? I know that there are other methods to find the surface area/volume, but I just want to know why there is a contradiction between the dH/dS thing. Thanks! Any help is appreciated! :)
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ds is the rate of change in the lateral area of an infinitesimal disk, while dh is the change of its height. The volume of an infinitesimal cylinder is computed using its height, not its slant height; while when you are computing the surface area, the height of each cylinder is not what you want to use. You can easily apply the Pythagorean theorem to see what I mean. In a pure cylinder, the slant height and the normal height is equivalent. When you are partitioning a cone however, they are not. This is simply because the cone has a slope that can't simply be discarded. Discarding the slope is exactly what you are doing now. Volume is OK to compute like that, but surface area isn't. This solves your problem.
Hello Millennial, thank you for your reply! I understand that cylinders have a different slope from cones. However, you said, "Discarding the slope is exactly what you are doing now. Volume is OK to compute like that, but surface area isn't." Why is it okay to discard the slope when computing the volume, but not okay when computing the surface area? Thanks!
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Homework Help
## Deriving the volume and surface area of a cone
The tiny volume bit error is, relatively speaking, vansihingly small to that which is kept.
The local cylinder volume approximation is Vc= pi*r^2*dz
Now, let us look at the volume of the tiny element from an expanding cone.
This is made up of Vc+the volume of a triangular area with the full circular periphery as its radius.
This bit's volume can be written (2*pi*(r+dr))*1/2*dz*dr, where dr is the tiny radial additon from position (z,r) to position (z+dz, r+dr).
Now, the biggest part of this volume equals pi*rdzdr (agreed?)
Since both dz and dr goes to zero in the limit, the PRODUCT of these tiny quantities go much faster to zero than either one of them.
Agreed?
Thus, we can discard that volume bit relative to Vc.
-----
However, what you lose when computing surface area (with cylindric approxamition), you'll lose area portions that do not vanish faster than the one you keep.
Recognitions: Gold Member Homework Help Science Advisor As for the surface area element, with the cylinder, you get r*dw*dz, where dw is the tiny angle, and r*dw the tiny arc length. With the cone, you get r*dw*dS instead (where dS is the length along the skewed plane), so that the ratio of these terms are dS/dz. This does NOT equal 1, so that you cannot approximate the surface area locally with that of the cylinder | 2013-05-19T11:19:05 | {
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http://math.stackexchange.com/questions/191164/how-many-arrangements-are-possible/191258 | # How many arrangements are possible
Problem: There are infinite number of spoons and forks at your disposal. You have to keep a total of n spoons and forks on a table in a row. As soon as you keep a fork, you will have to keep forks alternately. If the first time you kept a fork was at position i, you will have to keep forks at positions i+2, i+4, i+6.. and so on. How many arrangements are possible? Since the answer could be very large, print it modulo 109+7.
Constraints: 1 $\leqslant$ n $\leqslant$ 109
Time Limit: $1$ sec
Sample Input 1: $3$
Output 1: $6$ (SSS, SSF, SFS, SFF, FSF, FFF)
Sample Input 2: $5$
Output 2: $14$
The way I solved it (inefficient):
Keep n spoons first. Start from the end and replace each spoon with a fork. When you've replaced i spoons with i forks at the end, there are 2i/2 combinations possible (since $\frac{i}{2}$ positions will have forks). So basically it's a summation of powers of 2.
Going by my method, the answer will be: Summation of the first $n+1$ terms of the following series
1 1 2 2 4 4 8 8 16 16 32 32 ...
My submission is getting the verdict Time Limit Exceeded. I can neither embed nor compute very high powers of 2, and there have been many submissions in 0.0 seconds! It must be very easy though I can't think of anything else.
What is a better method to solve this question?
[EDIT]: If the first time you keep a fork is at position $i$, you will have to keep forks at positions $i+2, i+4 ...$ and so on. This happens only ONCE.
For example, A(4) = 10. By the answers below, it is 9 (which is wrong). The 10th arrangement will be FFFS.
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Curious as to where you came across this problem. Not Project Euler, is it? – Gerry Myerson Sep 5 '12 at 1:58
@Gerry I couldn't find it there. – Rick Decker Sep 5 '12 at 2:39
@Rick, me neither - it just smells like a PE problem. Anyway, for input 5, I get 12, not 14: sssss, ssssf, sssfs, sssff, ssfsf, ssfff, sfsfs, sfsff, sffff, fsfsf, fsfff, fffff. What am I missing? – Gerry Myerson Sep 5 '12 at 2:51
You're right. That'll save me some writing, since you've already enumerated those possibilities. – Rick Decker Sep 5 '12 at 2:58
It can also be completely analyzed without recursion.
You can start with any numbers of spoons. Once you have two forks in a row, you must continue with forks. Thus, using $\rm{S}$ for spoons and $\rm{F}$ for forks and exponents to indicate repetition, you can start with ${\rm{S}}^k$ for any $k\ge 0$. If you ever get a fork, you can follow it with another fork, in which case you have ${\rm{S}}^k{\rm{F}}^{n-k}$ for some $k\in\{0,\dots,n\}$, or you can follow it with a spoon. At that point your choices are limited, however, since at least every second utensil from this point on must be a fork. In fact, it’s not hard to see that all you can do is alternate forks and spoons until you decide to place two forks in a row (if you ever do). In short, the only possible configurations are those of the form
$${\rm{S}}^k({\rm{FS}})^\ell{\rm{F}}^m\;,$$
where $k,\ell,m\ge 0$ and $k+2\ell+m=n$.
Fix $k$ with $0\le k\le n$. Then $2\ell+m=n-k$. Each integer $\ell$ such that $0\le\ell\le\frac12(n-k)$ yields exactly one arrangement, so there are $\left\lfloor\frac12(n-k)\right\rfloor+1$ arrangements beginning with exactly $k$ spoons. This gives a total of
\begin{align*} \sum_{k=0}^n\left(\left\lfloor\frac12(n-k)\right\rfloor+1\right)&=n+1+\sum_{k=0}^n\left\lfloor\frac12(n-k)\right\rfloor\\\\ &=n+1+\sum_{k=0}^n\left\lfloor\frac{k}2\right\rfloor\\\\ &=n+1+\begin{cases} 2\sum_{k=0}^{(n-1)/2}k,&\text{if }n\text{ is odd}\\\\ 2\sum_{k=0}^{n/2}k-\frac{n}2,&\text{if }n\text{ is even} \end{cases}\\\\ &=n+1+\begin{cases} \frac{n^2-1}4,&\text{if }n\text{ is odd}\\\\ \frac{n^2}4,&\text{if }n\text{ is even} \end{cases}\\\\ &=\left(\frac{n}2+1\right)^2-\frac14[n\text{ is odd}]\\\\ &=\left\lfloor\left(\frac{n}2+1\right)^2\right\rfloor\;, \end{align*}
arrangements of $n$ spoons and forks, where $[n\text{ is odd}]$ is an Iverson bracket.
Added: In the updated version of the question there is one arrangement with no forks. Suppose now that the first fork is the $k$-th utensil from the end of the row, where $1\le k\le n$. If $k$ is even, there are $k/2$ guaranteed forks and $k/2$ free positions, for a total of $2^{k/2}$ arrangements. If $k$ is odd, there are $(k+1)/2$ guaranteed forks and $(k-1)/2$ free positions, for a total of $2^{(k-1)/2}$ arrangements. The grand total number of arrangements is therefore
\begin{align*} 1+\sum_{k=1}^n2^{\lfloor k/2\rfloor}&=\sum_{k=0}^n2^{\lfloor k/2\rfloor}\\\\ &=2\sum_{k=0}^{\lfloor n/2\rfloor}2^k-2^{n/2}[n\text{ is even}]\\\\ &=2\left(2^{\lfloor n/2\rfloor+1}-1\right)-2^{n/2}[n\text{ is even}]\\\\ &=\begin{cases} 2^{(n+3)/2}-2,&\text{if }n\text{ is odd}\\\\ 3\cdot2^{n/2}-2,&\text{if }n\text{ is even}\;. \end{cases} \end{align*}
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We haven't understood the question correctly. There's an update to the question. Check it out.. – Rushil Sep 5 '12 at 20:24
I solved this logically, though yours is pretty mathematical. :-) – Rushil Sep 5 '12 at 21:47
The first step in any programming competition is to look at the constraints of the problem. Since the parameter, $n$, can be as large as $10^9$ you clearly don't want to waste your time with any program involving loops or recursion and any auxiliary numbers your solution involves almost certainly have to be small (at worst quadratic in the parameter). In other words, there'd better be a closed form solution for this problem involving at most a few expressions that are no worse than quadratic in $n$. Fortunately, there is, but you'll have to wait a bit to see it.
In the tradition of almost all programming challenges, I'll obfuscate things as much as possible, trying to find wording that I (and only I) consider cloyingly cute and nerdy. Imagine, then, a mathematically-inclined butler laying out the silverware according to your specifications (it has to be the butler, obviously, since he's the only household staff member permitted to touch the very expensive silver). The first thing he learned in his training was "serve from the left" (and take away from the right, but that won't come into play now), and that's what we'll do: build strings of "S"s and "F"s, starting at the left, and we'll count the number of possible arrangements, given the rules you laid out.
There are two things to notice: first, starting with an "S" doesn't constrain us in any way, so if we let $A(n)$ count the number of arrangements of $n$ utensils, we'll have immediately that $$A(n) = A(n-1) + \text{some other stuff}$$ where the $A(n-1)$ term counts the number of strings starting with "S" (followed by $n-1$ characters) and the "other stuff" counts the number of strings (of length $n$) starting with "F". The second thing to observe is that if we ever build a substring of the form "...FF", any remaining characters must be "F"s.
Starting, then, with the empty string and adding characters to the right, we'll have the following possibilities at the start:
Working from the top down we see that there are two possible length-$1$ strings, "S" and "F", four possible length-$2$ strings and 6 possible length-$3$ strings. Using the first observation above, we can ignore the left subtree, starting with "S", since that's already taken care of by our recursion.
Now look at the right subtree, enumerating those strings that start with "F". At each level going down we'll have the counts $1$ (for the "F"), $2$ (for "FS" and "FF"), $2$ (for "FSF" and "FFF"), and a bit of thought will convince you that the sequence of the number of possible strings starting with "F" is $\langle 1, 2, 2, 3, 3, 4, 4, 5, 5, \dots\rangle$. That's just the "some other stuff" in the displayed equation above and it's easy to see that it's simply $\lfloor(n+2)/2\rfloor$, where the braces, as usual, delimit the floor function. In other words, we now have a recurrence for the number of arrangements: $A(1)=2$ and for $n>1$, $$A(n) = A(n-1) + \left\lfloor\frac{n+2}{2}\right\rfloor$$ Expanding this, we see with the tiniest bit of algebra that $$A(n) = 1+\sum_{k=2}^{n+2}\left\lfloor\frac{k}{2}\right\rfloor$$ That's good enough for every situation except a programming competition where $n$ may be very large. However, the sum is clearly going to be at most quadratic, so we look for a quadratic closed form for $A(n)$. One final trick comes into play here: because of the division by $2$, we'll have two different forms, depending upon whether $n$ is even or odd. To cut to the chase, it's easy enough to do some polynomial interpolation (twice) and come, at long last to $$A(n) = \frac{1}{4}n^2 + n + \begin{cases} 1 & \text{if n is even}\\ 3/4 & \text{if n is odd}\end{cases}$$
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That was a great answer! Thanks. – Rushil Sep 5 '12 at 11:41
you said that you could ignore the left subtree. How did you handle that? I couldn't follow your answer after Expanding this, we see with the tiniest bit of ... – Rushil Sep 5 '12 at 11:42
okay I got all of it. Can you just explain the polynomial interpolation part? I've never heard of it before. – Rushil Sep 5 '12 at 16:06
Unfortunately, the solution I coded is now getting the verdict Wrong Answer. There is something wrong with your solution. – Rushil Sep 5 '12 at 19:18
@Rushil For polynomial interpolation, a good place to start is to check out Wikipedia. In this case, you want to find a polynomial $p(x)=Ax^2+Bx+C$ that fits your data. We already know, for example, that $p(2)=4$ so $4A+2B+C=4$. We also know that $p(4)=9$ so $16A+4B+C=9$, and we can also find $p(6)=16$ so $36A+6B+C=16$. Given these three linear equations in three unknowns (namely $A, B, C$), we can solve for the coefficients of the polynomial. As to the dreaded wrong answer I'm afraid that I can't help without seeing your code. I'm sure, though, that Brian's and my answers are correct. – Rick Decker Sep 5 '12 at 19:58 | 2015-11-28T10:04:07 | {
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