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http://math.stackexchange.com/questions/400403/solve-sqrtx4-sqrtx1-1-for-x | # Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$
Can someone give me some hints on how to start solving $\sqrt{x+4}-\sqrt{x+1}=1$ for x?
Like I tried to factor it expand it, or even multiplying both sides by its conjugate but nothing comes up right.
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$0$ of course, it's easy to guess ... square it, isolate the roots and then square it. – Santosh Linkha May 23 '13 at 17:10
This is pretty much a duplicate of math.stackexchange.com/questions/399199/… – Thomas May 23 '13 at 20:38
Start by squaring it to get
$$x+4-2\sqrt{(x+4)(x+1)}+x+1=1\;,$$
which simplifies to
$$\sqrt{(x+4)(x+1)}=x+2\;.$$
Now square again.
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Thanks for the help! – Regis May 23 '13 at 17:11
@Regis: You’re welcome! – Brian M. Scott May 23 '13 at 17:11
Note - when squaring like this, always check back to see that your solution works for the original equation, since squaring destroys the distinction between the positive and negative values of the square root, and you will generally get solutions corresponding to both. – Mark Bennet May 23 '13 at 17:32
HINT:
As $(x+4)-(x+1)=3 \ \ \ \ \$
$\implies (\sqrt{x+4}-\sqrt{x+1})(\sqrt{x+4}+\sqrt{x+1})=3$
$$\text{As }\sqrt{x+4}-\sqrt{x+1}=1\ \ \ \ \ (1)$$
$$\implies \sqrt{x+4}+\sqrt{x+1}=3\ \ \ \ \ (2)$$
Add/subtract $(1)$ and $(2),$ then square
Generalization :
$$\text{As }(ax+b)-(ax+c)=b-c$$
$$\text{If }\sqrt{ax+b}-\sqrt{ax+c}=d \ \ \ \ \ (1)$$
$$\text{As } (ax+b)-(ax+c)=(\sqrt{ax+b}-\sqrt{ax+c})(\sqrt{ax+b}+\sqrt{ax+c})$$
$$\implies \sqrt{ax+b}+\sqrt{ax+c}=\frac{b-c}d\ \ \ \ \ (2)$$
Add/subtract $(1)$ and $(2),$ then square
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(+1) neat approach – vadim123 May 23 '13 at 17:28
– lab bhattacharjee May 23 '13 at 17:58
very nice approach – srijan May 23 '13 at 18:32
Multiplying by the conjugate as you originally suggested does work here. If you multiply both sides by $\sqrt{x+4} + \sqrt{x+1}$ you get $$(x+4) - (x+1) = \sqrt{x+4} + \sqrt{x+1}$$ Which is the same as $$\sqrt{x+4} + \sqrt{x+1} = 3$$ Add this to the original equation and divide by $2$ to obtain $$\sqrt{x+4} = 2$$ Squaring you get $$x +4 = 4$$ Therefore $x = 0$ is the only solution.
Also note the similarity to lab bhattacharjee's method.
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Such equations, if slick tricks such as lab bhattacharjee's can't apply, are solved with a standard procedure:
\begin{align} &\sqrt{x+4}-\sqrt{x+1}=1\\[2ex] &\text{Rearrange}\\ &\sqrt{x+4}=1+\sqrt{x+1}\\[2ex] &\text{Square}\\ &x+4=1+2\sqrt{x+1}+(x+1)\\[2ex] &\text{Rearrange}\\ &2=2\sqrt{x+1}\\[2ex] &\text{Simplify}\\ &1=\sqrt{x+1}\\[2ex] &\text{Square}\\ &1=x+1\\[2ex] &x=0 \end{align}
We just need to ckeck that the solution makes the square roots existent, because at each "Square" stage we are dealing with non negative numbers. Of course the conditions are $$\begin{cases} x\ge-4\\ x\ge-1 \end{cases}$$ which boil down to $x\ge-1$, that's satisfied by our solution.
Some care has to be reserved in different situations, when there's no guarantee that at the "Square" staged we have non negative numbers.
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Note that lab's trick 'always' applies, in some sense; with an equation of form $\sqrt{x+a}-\sqrt{x}=b$ (obviously the given equation can be put into such a form by using $y=x+1$), then by factoring the left side of $x+a-x=a$ one gets $\sqrt{x+a}+\sqrt{x} = \frac{a}{b}$, and then adding/subtracting gives the answer. – Steven Stadnicki May 23 '13 at 17:47
@StevenStadnicki It doesn't work with $\sqrt{2x+a}-\sqrt{x}=b$, for instance; that's what I was referring to. I believe that a standard procedure should be taught, leaving slick tricks to the best students. We don't want to teach math as a collection of tricks, do we? – egreg May 23 '13 at 17:50
Let $a = \sqrt{x+4}$ and $b=\sqrt{x+1}$. So that $a^2 = x + 4$ and $b^2 = x + 1$.
From the given equation, $$\sqrt{x+4} - \sqrt{x+1}=1 \implies a - b = 1 \ \ \ \text{and} \ \ \ a^2-b^2=3.$$ So we have that, $$a^2-b^2 =(a-b)(a+b)=1 \cdot(a+b)=3 \implies a+b=3.$$ We see that $$(a+b)+(a-b) = 2a.$$ Also, $$(a+b)+(a-b)=3+1=4.$$ Therefore, $$2a=4 \implies a=2 \implies \sqrt{x+4}=2.$$ Solving for $x$, $$x+4=4 \implies x=0.$$
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Very interesting approach! – Gamma Function Jun 12 '13 at 5:30
I made an edit to add $\LaTeX$ and brief explanations. This is your answer and you should edit/modify it as you see fit. – Gamma Function Jun 12 '13 at 5:43
It's easy to guess that the answer is $x=0$. However, one should prove that this is the only root. Here is how.
1. Because $\sqrt {x+4} > \sqrt {x+1}$, this function is strictly increasing.
2. Function $y =1$ is constant.
3. It is easy to proove that monotoniously increasing function and a constant can cross at no more than 1 point (either cross at 1 point or do not cross at all).
Since we have a point of $x = 0$, then there are no other roots for this equasion.
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#4 is not true... $x$ and $x + {1 \over 2} \sin x$ are both monotone increasing but their graphs intersect repeatedly. – Zarrax May 23 '13 at 21:35
@Zarrax, Sorry for that. It is the best illustation of my mistake. I'll edit the answer. – Stoleg May 24 '13 at 13:57
You have two numbers $\sqrt{x+4}$ and $\sqrt{x+1}$. Consider their squares, which are (x+4) and (x+1). So, their squares differ by 3, with the larger square number equaling (x+4). Do we know of two square numbers which differ by 3? As you might recall, the square numbers belong to the sequence (1, 4, ...). So, the larger square number (x+4) equals 4, and the smaller square number (x+1) equals 1. Thus, x=0.
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I have to wonder what the people who use a bunch of algebraic manipulation to solve these sorts of problems think about solutions like these or if my answer would have gotten accepted had I managed to post it 20 hours earlier. – Doug Spoonwood May 24 '13 at 14:48
There's also the possibility that $x$ is not a whole number. There are many pairs of positive real numbers whose squares differ by $3$. – Zarrax May 24 '13 at 16:10
@Zarrax Your second statement "There are many pairs of positive real numbers whose squares differ by 3." is true. However, if there exists a possibility that x is not a whole number, then since x works to solve the equation, then there would exist a possibility that x=y and x=z where y does not equal z. In which case, we have a contradiction. So, unless your logic is inconsistent, x=0 here. – Doug Spoonwood May 24 '13 at 16:27
I have no idea what you're saying. At any rate, if you can cover the general real case I suggest adding it to your answer. – Zarrax May 24 '13 at 19:59 | 2015-09-04T20:59:54 | {
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https://stats.stackexchange.com/questions/130069/what-is-the-distribution-of-r2-in-linear-regression-under-the-null-hypothesis | # What is the distribution of $R^2$ in linear regression under the null hypothesis? Why is its mode not at zero when $k>3$?
What is the distribution of the coefficient of determination, or R squared, $R^2$, in linear univariate multiple regression under the null hypothesis $H_0:\beta=0$?
How does it depend on the number of predictors $k$ and number of samples $n>k$? Is there a closed-form expression for the mode of this distribution?
In particular, I have a feeling that for simple regression (with one predictor $x$) this distribution has mode at zero, but for multiple regression the mode is at a non-zero positive value. If this is indeed true, is there an intuitive explanation of this "phase transition"?
Update
As @Alecos showed below, the distribution indeed peaks at zero when $k=2$ and $k=3$ and not at zero when $k>3$. I feel that there should be a geometrical view on this phase transition. Consider geometrical view of OLS: $\mathbf y$ is a vector in $\mathbb R^n$, $\mathbf X$ defines a $k$-dimensional subspace there. OLS amounts to projecting $\mathbf y$ onto this subspace, and $R^2$ is squared cosine of the angle between $\mathbf y$ and its projection $\hat{\mathbf y}$.
Now, from @Alecos's answer it follows that if all vectors are random, then the probability distribution of this angle will peak at $90^\circ$ for $k=2$ and $k=3$, but will have a mode at some other value $<90^\circ$ for $k>3$. Why?!
Update 2: I am accepting @Alecos'es answer, but still have a feeling that I am missing some important insight here. If anybody ever suggests any other (geometrical or not) view on this phenomenon that would make it "obvious", I will be happy to offer a bounty.
• Are you willing to assume error normality? Dec 23, 2014 at 1:17
• Yes, I guess one has to assume it to make this question answerable (?). Dec 23, 2014 at 1:20
• Have you checked this davegiles.blogspot.jp/2013/05/good-old-r-squared.html ? Dec 23, 2014 at 2:35
• @Khashaa: in fact, I have to admit that I did find that blogspot page before posting my question here. Honestly, I still wanted to have a discussion of this phenomenon on our forum, so pretended I did not see that. Dec 23, 2014 at 9:46
• Strongly related CV question stats.stackexchange.com/questions/123651/… Dec 23, 2014 at 13:14
For the specific hypothesis (that all regressor coefficients are zero, not including the constant term, which is not examined in this test) and under normality, we know (see eg Maddala 2001, p. 155, but note that there, $k$ counts the regressors without the constant term, so the expression looks a bit different) that the statistic
$$F = \frac {n-k}{k-1}\frac {R^2}{1-R^2}$$ is distributed as a central $F(k-1, n-k)$ random variable.
Note that although we do not test the constant term, $k$ counts it also.
Moving things around,
$$(k-1)F - (k-1)FR^2 = (n-k)R^2 \Rightarrow (k-1)F = R^2\big[(n-k) + (k-1)F\big]$$
$$\Rightarrow R^2 = \frac {(k-1)F}{(n-k) + (k-1)F}$$
But the right hand side is distributed as a Beta distribution, specifically
$$R^2 \sim Beta\left (\frac {k-1}{2}, \frac {n-k}{2}\right)$$
$$\text{mode}R^2 = \frac {\frac {k-1}{2}-1}{\frac {k-1}{2}+ \frac {n-k}{2}-2} =\frac {k-3}{n-5}$$
FINITE & UNIQUE MODE
From the above relation we can infer that for the distribution to have a unique and finite mode we must have
$$k\geq 3, n >5$$
This is consistent with the general requirement for a Beta distribution, which is
$$\{\alpha >1 , \beta \geq 1\},\;\; \text {OR}\;\; \{\alpha \geq1 , \beta > 1\}$$
Note that if $\{\alpha =1 , \beta = 1\}$, we obtain the Uniform distribution, so all the density points are modes (finite but not unique). Which creates the question: Why, if $k=3, n=5$, $R^2$ is distributed as a $U(0,1)$?
IMPLICATIONS
Assume that you have $k=5$ regressors (including the constant), and $n=99$ observations. Pretty nice regression, no overfitting. Then
$$R^2\Big|_{\beta=0} \sim Beta\left (2, 47\right), \text{mode}R^2 = \frac 1{47} \approx 0.021$$
and density plot
Intuition please: this is the distribution of $R^2$ under the hypothesis that no regressor actually belongs to the regression. So a) the distribution is independent of the regressors, b) as the sample size increases its distribution is concentrated towards zero as the increased information swamps small-sample variability that may produce some "fit" but also c) as the number of irrelevant regressors increases for given sample size, the distribution concentrates towards $1$, and we have the "spurious fit" phenomenon.
But also, note how "easy" it is to reject the null hypothesis: in the particular example, for $R^2=0.13$ cumulative probability has already reached $0.99$, so an obtained $R^2>0.13$ will reject the null of "insignificant regression" at significance level $1$%.
To respond to the new issue regarding the mode of the $R^2$ distribution, I can offer the following line of thought (not geometrical), which links it to the "spurious fit" phenomenon: when we run least-squares on a data set, we essentially solve a system of $n$ linear equations with $k$ unknowns (the only difference from high-school math is that back then we called "known coefficients" what in linear regression we call "variables/regressors", "unknown x" what we now call "unknown coefficients", and "constant terms" what we know call "dependent variable"). As long as $k<n$ the system is over-identified and there is no exact solution, only approximate -and the difference emerges as "unexplained variance of the dependent variable", which is captured by $1-R^2$. If $k=n$ the system has one exact solution (assuming linear independence). In between, as we increase the number of $k$, we reduce the "degree of overidentification" of the system and we "move towards" the single exact solution. Under this view, it makes sense why $R^2$ increases spuriously with the addition of irrelevant regressions, and consequently, why its mode moves gradually towards $1$, as $k$ increases for given $n$.
• Its mathematical. For $k=2$ the first parameter of the beta distribution (the "$\alpha$" in standard notation) becomes smaller than unity. In that case the Beta distribution has no finite mode, play around with keisan.casio.com/exec/system/1180573226 to see how the shapes change. Dec 23, 2014 at 9:53
• @Alecos Excellent answer! (+1) Can I strongly suggest that you add to your answer the requirement for the mode to exist? This is usually stated as $\alpha>1$ and $\beta>1$ but more subtly, it's ok if equality holds in one of the two ... I think for our purposes this becomes $k \geq 3$ and $n \geq k + 2$ and at least one of these inequalities is strict. Dec 23, 2014 at 13:25
• @Khashaa Except if theory demands it, I never exclude the intercept from the regression -it is the average level of the dependent variable, regressors or no regressors (and this level is usually positive, so it would be a foolishly self-created misspecification to omit it). But I always exclude it from the F-test of the regression, since what I care about is not whether the dependent variable has a non-zero unconditional mean, but whether the regressors have any explanatory power as regards deviations from this mean. Dec 24, 2014 at 3:59
• +1! Are there results for the distribution of $R^2$ for nonzero $\beta_j$? Jan 6, 2016 at 11:44
I won't rederive the $\mathrm{Beta}(\frac{k-1}{2}, \, \frac{n-k}{2})$ distribution in @Alecos's excellent answer (it's a standard result, see here for another nice discussion) but I want to fill in more details about the consequences! Firstly, what does the null distribution of $R^2$ look like for a range of values of $n$ and $k$? The graph in @Alecos's answer is quite representative of what occurs in practical multiple regressions, but sometimes insight is gleaned more easily from smaller cases. I've included the mean, mode (where it exists) and standard deviation. The graph/table deserves a good eyeball: best viewed at full-size. I could have included less facets but the pattern would have been less clear; I have appended R code so that readers can experiment with different subsets of $n$ and $k$.
Values of shape parameters
The graph's colour scheme indicates whether each shape parameter is less than one (red), equal to one (blue), or more than one (green). The left-hand side shows the value of $\alpha$ while $\beta$ is on the right. Since $\alpha = \frac{k-1}{2}$, its value increases in arithmetic progression by a common difference of $\frac{1}{2}$ as we move right from column to column (add a regressor to our model) whereas, for fixed $n$, $\beta = \frac{n-k}{2}$ decreases by $\frac{1}{2}$. The total $\alpha + \beta = \frac{n-1}{2}$ is fixed for each row (for a given sample size). If instead we fix $k$ and move down the column (increase sample size by 1), then $\alpha$ stays constant and $\beta$ increases by $\frac{1}{2}$. In regression terms, $\alpha$ is half the number of regressors included in the model, and $\beta$ is half the residual degrees of freedom. To determine the shape of the distribution we are particularly interested in where $\alpha$ or $\beta$ equal one.
The algebra is straightforward for $\alpha$: we have $\frac{k-1}{2}=1$ so $k=3$. This is indeed the only column of the facet plot that's filled blue on the left. Similarly $\alpha < 1$ for $k<3$ (the $k=2$ column is red on the left) and $\alpha > 1$ for $k>3$ (from the $k=4$ column onwards, the left side is green).
For $\beta=1$ we have $\frac{n-k}{2}=1$ hence $k=n-2$. Note how these cases (marked with a blue right-hand side) cut a diagonal line across the facet plot. For $\beta > 1$ we obtain $k < n - 2$ (the graphs with a green left side lie to the left of the diagonal line). For $\beta < 1$ we need $k > n - 2$, which involves only the right-most cases on my graph: at $n=k$ we have $\beta=0$ and the distribution is degenerate, but $n=k-1$ where $\beta = \frac{1}{2}$ is plotted (right side in red).
Since the PDF is $f(x;\,\alpha,\,\beta) \propto x^{\alpha-1} (1-x)^{\beta-1}$, it is clear that if (and only if) $\alpha<1$ then $f(x) \to \infty$ as $x \to 0$. We can see this in the graph: when the left side is shaded red, observe the behaviour at 0. Similarly when $\beta<1$ then $f(x) \to \infty$ as $x \to 1$. Look where the right side is red!
Symmetries
One of the most eye-catching features of the graph is the level of symmetry, but when the Beta distribution is involved, this shouldn't be surprising!
The Beta distribution itself is symmetric if $\alpha = \beta$. For us this occurs if $n = 2k-1$ which correctly identifies the panels $(k=2, n=3)$, $(k=3, n=5)$, $(k=4, n=7)$ and $(k=5, n=9)$. The extent to which the distribution is symmetric across $R^2 = 0.5$ depends on how many regressor variables we include in the model for that sample size. If $k = \frac{n+1}{2}$ the distribution of $R^2$ is perfectly symmetric about 0.5; if we include fewer variables than that it becomes increasingly asymmetric and the bulk of the probability mass shifts closer to $R^2 = 0$; if we include more variables then it shifts closer to $R^2 = 1$. Remember that $k$ includes the intercept in its count, and that we are working under the null, so the regressor variables should have coefficient zero in the correctly specified model.
There is also an obviously symmetry between distributions for any given $n$, i.e. any row in the facet grid. For example, compare $(k=3, n=9)$ with $(k=7, n=9)$. What's causing this? Recall that the distribution of $\mathrm{Beta}(\alpha, \beta)$ is the mirror image of $\mathrm{Beta}(\beta, \alpha)$ across $x=0.5$. Now we had $\alpha_{k,n} = \frac{k-1}{2}$ and $\beta_{k,n} = \frac{n-k}{2}$. Consider $k'=n-k+1$ and we find:
$$\alpha_{k',n} = \frac{(n-k+1)-1}{2} = \frac{n-k}{2} = \beta_{k,n}$$ $$\beta_{k',n} = \frac{n-(n-k+1)}{2} = \frac{k-1}{2} = \alpha_{k,n}$$
So this explains the symmetry as we vary the number of regressors in the model for a fixed sample size. It also explains the distributions that are themselves symmetric as a special case: for them, $k' = k$ so they are obliged to be symmetric with themselves!
This tells us something we might not have guessed about multiple regression: for a given sample size $n$, and assuming no regressors have a genuine relationship with $Y$, the $R^2$ for a model using $k-1$ regressors plus an intercept has the same distribution as $1 - R^2$ does for a model with $k-1$ residual degrees of freedom remaining.
Special distributions
When $k=n$ we have $\beta=0$, which isn't a valid parameter. However, as $\beta \to 0$ the distribution becomes degenerate with a spike such that $\mathsf{P}(R^2 = 1)=1$. This is consistent with what we know about a model with as many parameters as data points - it achieves perfect fit. I haven't drawn the degenerate distribution on my graph but did include the mean, mode and standard deviation.
When $k=2$ and $n=3$ we obtain $\mathrm{Beta}(\frac{1}{2}, \, \frac{1}{2})$ which is the arcsine distribution. This is symmetric (since $\alpha = \beta$) and bimodal (0 and 1). Since this is the only case where both $\alpha < 1$ and $\beta < 1$ (marked red on both sides), it is our only distribution which goes to infinity at both ends of the support.
The $\mathrm{Beta}(1, \, 1)$ distribution is the only Beta distribution to be rectangular (uniform). All values of $R^2$ from 0 to 1 are equally likely. The only combination of $k$ and $n$ for which $\alpha = \beta =1$ occurs is $k=3$ and $n=5$ (marked blue on both sides).
The previous special cases are of limited applicability but the case $\alpha > 1$ and $\beta=1$ (green on left, blue on right) is important. Now $f(x;\,\alpha,\,\beta) \propto x^{\alpha-1} (1-x)^{\beta-1} = x^{\alpha-1}$ so we have a power-law distribution on [0, 1]. Of course it's unlikely we'd perform a regression with $k=n-2$ and $k>3$, which is when this situation occurs. But by the previous symmetry argument, or some trivial algebra on the PDF, when $k=3$ and $n > 5$, which is the frequent procedure of multiple regression with two regressors and an intercept on a non-trivial sample size, $R^2$ will follow a reflected power law distribution on [0, 1] under $H_0$. This corresponds to $\alpha=1$ and $\beta>1$ so is marked blue on left, green on right.
You may also have noticed the triangular distributions at $(k=5,n=7)$ and its reflection $(k=3,n=7)$. We can recognise from their $\alpha$ and $\beta$ that these are just special cases of the power-law and reflected power-law distributions where the power is $2-1=1$.
Mode
If $\alpha>1$ and $\beta>1$, all green in the plot, $f(x; \, \alpha, \, \beta)$ is concave with $f(0)=f(1)=0$, and the Beta distribution has a unique mode $\frac{\alpha-1}{\alpha+\beta-2}$. Putting these in terms of $k$ and $n$, the condition becomes $k>3$ and $n>k+2$ while the mode is $\frac{k-3}{n-5}$.
All other cases have been dealt with above. If we relax the inequality to allow $\beta=1$, then we include the (green-blue) power-law distributions with $k=n-2$ and $k>3$ (equivalently, $n>5$). These cases clearly have mode 1, which actually agrees with the previous formula since $\frac{(n-2)-3}{n-5}=1$. If instead we allowed $\alpha=1$ but still demanded $\beta>1$, we'd find the (blue-green) reflected power-law distributions with $k=3$ and $n>5$. Their mode is 0, which agrees with $\frac{3-3}{n-5}=0$. However, if we relaxed both inequalities simultaneously to allow $\alpha=\beta=1$, we'd find the (all blue) uniform distribution with $k=3$ and $n=5$, which does not have a unique mode. Moreover the previous formula can't be applied in this case, since it would return the indeterminate form $\frac{3-3}{5-5}=\frac{0}{0}$.
When $n=k$ we get a degenerate distribution with mode 1. When $\beta < 1$ (in regression terms, $n=k-1$ so there is only one residual degree of freedom) then $f(x) \to \infty$ as $x \to 1$, and when $\alpha < 1$ (in regression terms, $k=2$ so a simple linear model with intercept and one regressor) then $f(x) \to \infty$ as $x \to 0$. These would be unique modes except in the unusual case where $k=2$ and $n=3$ (fitting a simple linear model to three points) which is bimodal at 0 and 1.
Mean
The question asked about the mode, but the mean of $R^2$ under the null is also interesting - it has the remarkably simple form $\frac{k-1}{n-1}$. For a fixed sample size it increases in arithmetic progression as more regressors are added to the model, until the mean value is 1 when $k=n$. The mean of a Beta distribution is $\frac{\alpha}{\alpha+\beta}$ so such an arithmetic progression was inevitable from our earlier observation that, for fixed $n$, the sum $\alpha+\beta$ is constant but $\alpha$ increases by 0.5 for each regressor added to the model.
$$\frac{\alpha}{\alpha+\beta} = \frac{(k-1)/2}{(k-1)/2 + (n-k)/2} = \frac{k-1}{n-1}$$
Code for plots
require(grid)
require(dplyr)
nlist <- 3:9 #change here which n to plot
klist <- 2:8 #change here which k to plot
totaln <- length(nlist)
totalk <- length(klist)
df <- data.frame(
x = rep(seq(0, 1, length.out = 100), times = totaln * totalk),
k = rep(klist, times = totaln, each = 100),
n = rep(nlist, each = totalk * 100)
)
df <- mutate(df,
kname = paste("k =", k),
nname = paste("n =", n),
a = (k-1)/2,
b = (n-k)/2,
density = dbeta(x, (k-1)/2, (n-k)/2),
groupcol = ifelse(x < 0.5,
ifelse(a < 1, "below 1", ifelse(a ==1, "equals 1", "more than 1")),
ifelse(b < 1, "below 1", ifelse(b ==1, "equals 1", "more than 1")))
)
g <- ggplot(df, aes(x, density)) +
geom_line(size=0.8) + geom_area(aes(group=groupcol, fill=groupcol)) +
scale_fill_brewer(palette="Set1") +
facet_grid(nname ~ kname) +
ylab("probability density") + theme_bw() +
labs(x = expression(R^{2}), fill = expression(alpha~(left)~beta~(right))) +
theme(panel.margin = unit(0.6, "lines"),
legend.title=element_text(size=20),
legend.text=element_text(size=20),
legend.background = element_rect(colour = "black"),
legend.position = c(1, 1), legend.justification = c(1, 1))
df2 <- data.frame(
k = rep(klist, times = totaln),
n = rep(nlist, each = totalk),
x = 0.5,
ymean = 7.5,
ymode = 5,
ysd = 2.5
)
df2 <- mutate(df2,
kname = paste("k =", k),
nname = paste("n =", n),
a = (k-1)/2,
b = (n-k)/2,
meanR2 = ifelse(k > n, NaN, a/(a+b)),
modeR2 = ifelse((a>1 & b>=1) | (a>=1 & b>1), (a-1)/(a+b-2),
ifelse(a<1 & b>=1 & n>=k, 0, ifelse(a>=1 & b<1 & n>=k, 1, NaN))),
sdR2 = ifelse(k > n, NaN, sqrt(a*b/((a+b)^2 * (a+b+1)))),
meantext = ifelse(is.nan(meanR2), "", paste("Mean =", round(meanR2,3))),
modetext = ifelse(is.nan(modeR2), "", paste("Mode =", round(modeR2,3))),
sdtext = ifelse(is.nan(sdR2), "", paste("SD =", round(sdR2,3)))
)
g <- g + geom_text(data=df2, aes(x, ymean, label=meantext)) +
geom_text(data=df2, aes(x, ymode, label=modetext)) +
geom_text(data=df2, aes(x, ysd, label=sdtext))
print(g)
• Really illuminating visualization. +1 Dec 24, 2014 at 8:16
• Great addition, +1, thanks. I noticed that you call $0$ a mode when the distribution goes to $+\infty$ when $x\to 0$ (and nowhere else) -- something @Alecos above (in the comments) did not want to do. I agree with you: it is convenient. Dec 24, 2014 at 10:29
• @amoeba from the graphs we'd like to say "values around 0 are most likely" (or 1). But the answer of Alecos is also both self-consistent and consistent with many authorities (people differ on what to do about the 0 and 1 full stop, let alone whether they can count as a mode!). My approach to the mode differs from Alecos mostly because I use conditions on alpha and beta to determine where the formula is applicable, rather than taking my starting point as the formula and seeing which k and n give sensible answers. Dec 24, 2014 at 10:37
• (+1), this is a very meaty answer. By keeping $k$ too close to $n$ and both small, the question studies in detail, and so decisively, the case of really small samples with relatively too many and irrelevant regressors. Dec 24, 2014 at 12:39
• @amoeba You probably noticed that this answer furnishes an algebraic answer for why, for sufficiently large $n$, the mode of the distribution is 0 for $k=3$ but positive for $k>3$. Since $f(x) \propto x^{(k-3)/2}(1-x)^{(n-k-2)/2}$ then for $k=3$ we have $f(x) \propto (1-x)^{(n-5)/2}$ which will clearly have mode at 0 for $n>5$, whereas for $k=4$ we have $f(x) \propto x^{1/2}(1-x)^{(n-6)/2}$ whose maximum can be found by calculus to be the quoted mode formula. As $k$ increases, the power of $x$ rises by 0.5 each time. It's this $x^{\alpha-1}$ factor which makes $f(0)=0$ so kills the mode at 0 Dec 24, 2014 at 14:23 | 2022-05-18T00:59:00 | {
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https://math.stackexchange.com/questions/2121815/does-the-coin-flipping-game-terminate | # Does the coin flipping game terminate?
Imagine I were to toss a coin, so that
1. if it is tails, I will toss again.
2. if it is heads, I stop the game.
It is clear that if I would to play the game forever the probability of me tossing tails over and over again tends to $0$, because $\lim\limits_{n \rightarrow \infty}{\frac{1}{2^{n}}} = 0$ and therefore the game terminates. My question is if it would still be correct to say that the game terminates after a finite number of tosses, for the reason that after a finite number of tosses the probability still gets smaller than every $\epsilon>0$ (if i get it right). This is just confusing me for a while now and I would appreciate a detailed explanation.
• You cannot say "The game terminates in a finite number of moves" because you cannot define an upper bound on the number of moves the game must end in. Hence, the statement is incorrect. However, what you can say is that it is more likely that the game will end in a small number of moves rather than a large number, and the game is almost surely not going to go on infinitely. – астон вілла олоф мэллбэрг Jan 31 '17 at 2:42
• You can also say the expected number of flips to reach a game end is finite. – JMoravitz Jan 31 '17 at 2:42
• There is special jargon for this type of situation. We say that the game terminates "with probability 1". – MJD Jan 31 '17 at 12:18
You are correct that when the game terminates it does so in a finite number of steps. Now, is it possible that the game does not terminate? Sure, there is no reason to assume that you will always end up getting a Head.
However, the probability of the game not terminating is the probability of an infinite sequence of Tails, i.e.... zero.
Such is life with probability. There can be events that are possible (not impossible) but still have zero probability (for continuous random variables, many values are possible but usually all have zero probability).
Note that in practice you cannot perform this experiment: whatever resources you may have (millions of people tossing billions of coins for many many years) you are not guaranteed to ever get a Head. The above therefore relates to a thought experiment that is allowed to go on "forever".
How about the followings? The probability that the game will terminate after N tosses = 1 - probability that the game doesn't terminate after N tosses = $$1 - \frac{1}{2^{N}}$$
If you define $\epsilon$ be the upper bound of the probability, $$1 - \frac{1}{2^{N}} < \epsilon$$ $$2^{N} < \frac{1}{1 - \epsilon}$$ $$N < -\frac{\ln(1 - \epsilon)}{\ln2}$$
So if there are N tosses and $N < -\frac{\ln(1 - \epsilon)}{\ln2}$, the probability of the game terminating is less than $\epsilon$. As $N \rightarrow \infty, \epsilon \rightarrow 1.$ | 2019-12-16T11:08:15 | {
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https://canpi.cc/post/oj/poj3126/ | The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. — It is a matter of security to change such things every now and then, to keep the enemy in the dark. — But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know! — I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door. — No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime! — I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds. — Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened. — No unnecessary expenditure, please! I happen to know that the price of a digit is one pound. — Hmm, in that case I need a computer program to minimize the cost. You don’t know some very cheap software gurus, do you? — In fact, I do. You see, there is this programming contest going on… Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above. 1033 1733 3733 3739 3779 8779 8179 The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
## Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
## Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
## Sample
input:
1 2 3 4 3 1033 8179 1373 8017 1033 1033
output:
1 2 3 6 7 0
## Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 #include #include #include #include using namespace std; // based on the fact that: // if a number n, n!=(6x+1) && n!=(6x+5), // then n cannot be a prime // so we can only check (6x+1) and (6x+5) // while searching for factor, only consider these prime proposals // proof: // n can only \in {6x, 6x+2, 6x+3, 6x+4} // but they are all not prime bool is_prime(int n){ if(n==2 || n==3) return 1; if(n%6!=1 && n%6!=5) return 0; int n_sqrt=floor(sqrt((float)n)); for(int i=5;i<=n_sqrt;i+=6) if(n%(i)==0 | n%(i+2)==0) return 0; return 1; } // return power of 10 int pow_10[5]={1,10,100,1000,10000}; struct node{ int num; int depth; }; int BFS(int first, int last){ bool visit[9000]; memset(visit,0,sizeof(visit)); queue q; while(!q.empty()) q.pop(); node node0={first,0}; q.push(node0); while(!q.empty()){ node node1=q.front(); q.pop(); if(node1.num==last) return node1.depth; for(int i=0; i<4; ++i){ // which bit will be replaced int bit=node1.num%pow_10[4-i]/pow_10[3-i]; for(int j=0; j<10; j+=1){ // first bit cannot be 0 if(i==0 && j==0) continue; // itself if(bit==j) continue; // after replace int num_new=node1.num+((j-bit)*pow_10[3-i]); if(num_new==last) return node1.depth+1; if(visit[num_new-1000]==0 && is_prime(num_new)) q.push({num_new,node1.depth+1}); visit[num_new-1000]=1; } } } return -1; } int main(){ int N,first,last; char s_first[5],s_last[5]; while(scanf("%d",&N)!=EOF){ for(int i=0; i | 2023-03-25T04:35:21 | {
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http://math.stackexchange.com/questions/30253/one-to-one-correspondence | one-to-one correspondence
My textbook states the following proposition: Let $f:R \rightarrow S$ be a ring homomorphism and let $s$ be in the image of $f$. Then $\{r \in R \mid f(r) = s\}$ is in one-to-one correspondence with $\ker(f)$.
What does it mean to have one to one correspondence with $\ker(f)$? Does it mean the set $\{r \in R \mid f(r) = s\}$ and the set $\ker(f)$ have the same cardinality?
Thanks!
-
"One-to-one correspondence" is a poor (in my opinion, anyway) way of saying "bijection." – Jay Kopper Apr 1 '11 at 6:06
HINT $\rm\$ If $\rm\:f(r)\ =\ s\$ then $\rm\ f(r')\ =\ s\ \iff\ 0\ =\ f(r')-f(r)\ =\ f(r'-r)$
Hence $\rm\ \ f^{-1}(s)\ =\ r\ +\ ker\ f\ =\:$ particular + homogeneous solution, as in linear algebra.
-
@Bill cool, it's actually the proof! I see that you are saying each $r \in R$ corresponds to (r' + k) for each $k \in ker f$, which implies the same cardinality. but what about the cases where (r'+2k), (r'+3k), and so on? since f maps them to s as well, wouldn't that make the set R have more elements than ker f? I would appreciate it if you can point out what's wrong with this reasoning... – mathcat Apr 2 '11 at 2:19
@mat As above, $\rm\ z\to r+z\$ bijects $\rm\:f^{-1}(0)$ with $\rm\:f^{-1}(s)\:,\:$ i.e. the shifting the homogeneous solution space $\rm\:f^{-1}(0)$ by any particular solution $\rm\:r\:$ puts it in bijection with the nonhomogeneous solution space $\rm\:f^{-1}(s)\:.$ – Bill Dubuque Apr 2 '11 at 2:33
@Bill thanks! now I see the difference between our views. you are saying shifting the entire space by a particular solution to create a bijection, which is a nice clean way. I was thinking in terms of elements: each $r \in R$ is created by adding each $k\in$ ker f to a particular r'. after this, R is in bijection with ker f. adding 2k to r' doesn't create any new element in R, because f(2k)=f(2)*f(k)=f(2)*0=0, i.e. 2k was already in ker f, so 2k+r' is already in R. this is the fault with my previous reasoning. well, quite messy, but glad I got it. anyway, thanks a lot :) – mathcat Apr 2 '11 at 4:30
@mat Many times you say $\rm\: R\:$ but it should be $\rm\: f^{-1}(s)\:.\:$ $\rm\:R\:$ is partitioned into equivalence classes $\rm\:[r]\:$ where $\rm\ r'\in [r]\ \iff\ f(r') = f(r)\:.\:$ These equivalence classes are known as the kernel, or equalizer, or level-sets of $\rm\:f\:.$ – Bill Dubuque Apr 2 '11 at 4:49
@Bill well, the same thing applies to your space shifting as well. one (newbie like me) might ask, after shifting r1 to the homogeneous solutions to obtain a set of non-homogeneous solutions, why not then shift the homogeneous solutions by r2 to obtain even more solutions? well, let's say f(r2) = s, and r2 = r1 + r0. then f(r2) = f(r1)+f(r0). since f(r1) = s, then f(r0) = 0, which means $r0 \in$ ker f, which means r1+r0=r2 is already obtained. so shifting by r2 won't lead to any new solutions. sorry for the long comments, since they are not immediately clear to me xP – mathcat Apr 2 '11 at 4:50
It means that there is a map $g: \{r\in R | f(r) = s\} \rightarrow \operatorname{ker}f$ such that every element of $\operatorname{ker}f$ is the image of one and only one element of $\{r\in R | f(r) = s\}$. This implies that the cardinality is the same.
In this case, if $s'\in R$ is such that $f(s') = s$, then one such map is $g(r) = r - s'$.
@mathcat: No, what they are doing is fixing a single value for $s$ and defining the set in question as the set of elements mapped to $s$. – Alex Becker Apr 1 '11 at 5:48
think of a group homomorphism $f:G\to H$, then $G/\text{ker}f\cong\text{im}f$, ie everything in a coset gets mapped to the same element... – yoyo Apr 1 '11 at 15:30 | 2014-03-10T08:01:31 | {
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https://math.stackexchange.com/questions/2159076/can-someone-clarify-this-doubt-of-mine-regarding-calculation-of-average-for-a-p | # Can someone clarify this doubt of mine, regarding calculation of average for a probability distribution?
Suppose there is a question on probability which asks us to find the expected number of successful attempts or average number of successful attempts, with the probability of success given. On computation using the formula for expectation (average), the result is a mixed number, having integer and fractional part. Now, we know that the number of attempts has to be a whole number to be physically realizable. So, if the computation of mean using the formula yields a mixed number, do I round it off to the nearest integer or keep it as it is ? For instance, suppose the average number of successful attempts comes out to be 2.5. Do I mark the answer as 2.5 itself or mark it as 3 ?
No, you do not round the average.
For example, let's say I flip a coin and write down $0$ if I get heads and $1$ if I get tails. Now, clearly, the expected value of what I will get is $\frac12$, right?
What may be confusing you is that in every single experiment, I will get an integer value, i.e. $0$ or $1$. So how can I "expect" to get $\frac12$, when I clearly cannot?
The answer is that you don't expect to get $\frac12$ in any single throw of the coin. What you need to think about are repeats of this experiment. So, say I repeat the experiment $11$ times, then I might end up with a sequence like $$1,0,0,1,0,0,0,1,1,1,1$$
and the average of this sequence is $\frac{6}{11}$ which is not an integer, but it is pretty close to $\frac12$. And that, not the result of a single experiment, is what the average value tells me.
Remember, the mean (also called expected value) is not
"the value I predict will come up in my experiment",
it is more like
"the average of several values I will come up in my experiment".
• Thank you. That sure helped. Just for a clarification, could you refer to this question: Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of getting reservation in any attempt by a passenger, then the average number of attempts that passengers need to make to get a seat reserved is ? I found the expectation to be 2.5. So, I should mark 2.5 as the answer, right ? – LumosMaxima Feb 24 '17 at 7:50
• @SouvikChakraborty refer to what question? – 5xum Feb 24 '17 at 7:50
• The one in my last comment. I edited it. – LumosMaxima Feb 24 '17 at 7:52
• @SouvikChakraborty Yes, the answer is $2.5$ for that question. – 5xum Feb 24 '17 at 7:53
• @SouvikChakraborty Remember, if an answer is what you were looking for, it's usual to upvote and accept it. – 5xum Feb 24 '17 at 7:55 | 2019-10-18T18:17:16 | {
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https://math.stackexchange.com/questions/3703410/prove-or-disprove-that-the-ellipse-of-largest-area-centered-at-origin-inscribe/3703954 | # Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$.
I can show that $$x^2+y^2=\frac12(1+\log2)$$ is the equation of the circle of largest area inscribed in $$y=\pm e^{-x^2}$$:
The minimum distance $$r$$ (which will be the radius of the circle) between the origin and $$y=e^{-x^2}$$ can be found by finding the critical numbers of the derivative of the distance function. \begin{align} r&=\sqrt{x^2+(e^{-x^2})^2}\\ \frac{dr}{dx}&=\frac{2x-4xe^{-2x^2}}{2\sqrt{x^2+e^{-2x^2}}}\\ 0&=2x(1-2e^{-2x^2})\\ x&=0\quad\text{(obviously inadmissible, or)}\\ x&=\sqrt{\frac12\log2}\\ \\ r&=\sqrt{\frac12\log2+e^{-\log2}}\\ r^2&=\frac12\log2+\frac12\\ \implies\quad x^2+y^2&=\frac12\left(1+\log2\right) \end{align} as desired.
The relationship between a circle and an ellipse is like that of a square and a rectangle: given a set perimeter, the more square the rectangle, the larger the area. But since this question is based on the curve $$y=e^{-x^2}$$ and not on any set perimeter, it does not seem like the same conclusion can be drawn. I would need something stronger (perhaps based on concavity?) to show whether $$x^2+y^2=\frac12(1+\log2)$$ is the largest ellipse or not.
• I think what you say holds good if the Bell curve has a single constant in its parametrixzation. But there are two constants in: $y= y_{max} e^{-k x^2}$ – Narasimham Jun 3 at 13:12
We can assume by symmetry and without loss of generality that the ellipse can be parametrized by $$(x,y) = (a \cos \theta, b \sin \theta), \quad a, b > 0, \quad \theta \in [0,2\pi).$$ We require tangency to the curve $$y = e^{-x^2}$$ as well as a single point of intersection in the first quadrant. That is to say, $$b \sin \theta = e^{-(a \cos \theta)^2}$$ has a unique solution for $$\theta \in (0, \pi/2)$$, and at this point, $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{b}{a} \cot \theta = -2x(\theta)e^{-x(\theta)^2} = -2a (\cos \theta )e^{-(a \cos \theta)^2}.$$ Consequently, $$-\frac{b}{a} \cot \theta = -2ab \cos \theta \sin \theta,$$ or $$\sin \theta = \frac{1}{a \sqrt{2}}.$$ Note if $$a < 1/\sqrt{2}$$, no such angle exists. The ellipse is "too narrow"--this is due to the fact that the point of tangency is at $$(x,y) = (0,1)$$. At the point of tangency, we also have $$\cos \theta = \sqrt{1 - (2a^2)^{-1}}$$ so that we now have $$\frac{b}{a \sqrt{2}} = e^{1/2 - a^2}$$ or $$b = a e^{1/2 - a^2} \sqrt{2}.$$ Finally, we seek to maximize the area of this family of ellipses parametrized by $$a$$. Since the area is proportional to $$ab$$, we need to maximize $$f(a) = a^2 e^{1/2-a^2}.$$ Computing the derivative with respect to $$a$$ and solving for critical values, we get $$0 = \frac{df}{da} = 2(a-1)a(a+1)e^{1/2-a^2},$$ hence $$a = 1$$ is the unique solution, with $$b = \sqrt{2/e}$$ and the ellipse has equation $$\frac{x^2}{1} + \frac{y^2}{2/e} = 1.$$ The area of this ellipse is simply $$\pi a b = \pi \sqrt{\frac{2}{e}}.$$ This is obviously not a circle.
For your enjoyment, I have included an animation of the family of ellipses $$\frac{x^2}{a^2} + \frac{y^2}{2a^2 e^{1-2a^2}} = 1,$$ for $$a \in [1/\sqrt{2},2]$$:
• Does the max area ellipse contact the Bell curve at its inflection point, by any chance? – Narasimham Jun 3 at 8:05
• @Narasimham Yes, it does. It is easy to show that for the maximal area, the point of tangency of the ellipse to the curve in the first quadrant is at $(2^{-1/2}, e^{-1/2})$. – heropup Jun 3 at 8:29
• Should n't the variable blue ellipse then be stationed after animation touching the above tangent point ? just a suggestion. – Narasimham Jun 3 at 9:23
• @Narasimham I tried but the looping of the animation doesn't seem to be working the way I want. – heropup Jun 3 at 9:25
• @heropup what software or program did you use for the animation? – user376343 Jun 3 at 11:02
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \implies \frac{dy}{dx}=-\frac{b^2}{a^2}\frac xy=-\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}}.$$
Hence we have the following system to find the tangent point: $$\begin{cases} e^{-x^2}=b\sqrt{1-\frac{x^2}{a^2}}\\ 2xe^{-x^2}=\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}} \end{cases}\implies 1-\frac{x^2}{a^2}=\frac{1}{2a^2}\implies x^2=a^2-\frac12.\tag1$$
Substituting this back into the equation of the tangent point one obtains equation to determine $$b$$: $$e^{\frac12-a^2}=\frac ba\sqrt{\frac12}\implies b=\sqrt{2e}\,ae^{-a^2}.$$ The area is respectively: $$A=\pi ab=\pi \sqrt{2e}\,a^2e^{-a^2}.\tag2$$ To find the extremum of the area we differentiate over $$a$$ to obtain: $$\frac1{\pi\sqrt{2e}}\frac{dA}{da}=2ae^{-a^2}-2a^3e^{-a^2}=2ae^{-a^2}(1-a^2),$$ meaning that the largest value of area is achieved at $$a=1$$: $$A_\text{max}=\pi\sqrt{\frac2e}.$$
Since $$a\ne b$$ the ellipse of the largest area is not a circle.
PS. In fact the system (1) has another solution: $$x=0, b=1$$. It can be however shown that the largest possible area of the inscribed ellipse in this case is $$\frac\pi{\sqrt2}$$ which corresponds to $$a=\frac1{\sqrt2}$$ in equation (2).
I think what you say holds good if the Bell curve has a single constant in its parametrization. But there are two constants in: $$y= y_{max} e^{- x^2/(2 \sigma^2)}$$
The touching curve is an ellipse, not a circle.
This is vague/intuitive at present,trying to prove that ellipse area is maximized when constrained on an envelope with symmetry of one axis with single inflection. We can proceed with intersections and differentiation etc., but for the present I proceed on this basis as given and may be shall ask a question separately about such an elegant general possibility.
Taking classical Bell curve as ( in present case $$\sigma = \frac{1}{\sqrt 2})$$
$$y= y_{max} e^{- x^2/(2 \sigma^2)}$$
Bell Curve by differentiation:
$$x_I= \sigma; y_I=y_m /\sqrt{e};\;y_I'= \dfrac{-xy}{\sigma^2} \rightarrow y_I'= -\dfrac{y_m}{\sigma \sqrt{e}}\tag1$$
Ellipse and derivatives: $$x_I^2/a^2+y_I^2/b^2=1; \;y_I'=-\dfrac{x_Ib^2}{y_Ia^2} \tag2$$
Eliminate $$y'$$ between (1),(2) simplifying and solving for $$(a^2,b^2)$$ we get $$(a,b)= (\sqrt 2 \sigma,y_m \sqrt {2/e} )\tag3$$ | 2020-09-29T03:50:30 | {
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https://math10blog.wordpress.com/tag/school/ | You are currently browsing the tag archive for the ‘school’ tag.
Wikipedia calls this a classic paradox of elementary probability theory.
Taken from Wikipedia.org:
Card version
Suppose you have three cards:
• black card that is black on both sides,
• white card that is white on both sides, and
• mixed card that is black on one side and white on the other.
You put all of the cards in a hat, pull one out at random, and place it on a table. The side facing up is black. What are the odds that the other side is also black?
The answer is that the other side is black with probability 2/3. However, common intuition suggests a probability of 1/2 because across all the cards, there are 3 white, 3 black. However, many people forget to eliminate the possibility of the “white card” in this situation (i.e. the card they flipped CANNOT be the “white card” because a black side was turned over).
In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 1/2; only 3 students correctly responded 2/3.[1]
Preliminaries
To solve the problem, either formally or informally, we must assign probabilities to the events of drawing each of the six faces of the three cards. These probabilities could conceivably be very different; perhaps the white card is larger than the black card, or the black side of the mixed card is heavier than the white side. The statement of the question does not explicitly address these concerns. The only constraints implied by the Kolmogorov axioms are that the probabilities are all non-negative, and they sum to 1.
The custom in problems when one literally pulls objects from a hat is to assume that all the drawing probabilities are equal. This forces the probability of drawing each side to be 1/6, and so the probability of drawing a given card is 1/3. In particular, the probability of drawing the double-white card is 1/3, and the probability of drawing a different card is 2/3.
In our question, however, you have already selected a card from the hat and it shows a black face. At first glance it appears that there is a 50/50 chance (ie. probability 1/2) that the other side of the card is black, since there are two cards it might be: the black and the mixed. However, this reasoning fails to exploit all of your information; you know not only that the card on the table has a black face, but also that one of its black faces is facing you.
Solutions:
Intuition
Intuition tells you that you are choosing a card at random. However, you are actually choosing a face at random. There are 6 faces, of which 3 faces are white and 3 faces are black. Two of the 3 black faces belong to the same card. The chance of choosing one of those 2 faces is 2/3. Therefore, the chance of flipping the card over and finding another black face is also 2/3. Another way of thinking about it is that the problem is not about the chance that the other side is black, it’s about the chance that you drew the all black card. If you drew a black face, then it’s twice as likely that that face belongs to the black card than the mixed card.
Labels
One solution method is to label the card faces, for example numbers 1 through 6.[2] Label the faces of the black card 1 and 2; label the faces of the mixed card 3 (black) and 4 (white); and label the faces of the white card 5 and 6. The observed black face could be 1, 2, or 3, all equally likely; if it is 1 or 2, the other side is black, and if it is 3, the other side is white. The probability that the other side is black is 2/3.
Bayes’ theorem
Given that the shown face is black, the other face is black if and only if the card is the black card. If the black card is drawn, a black face is shown with probability 1. The total probability of seeing a black face is 1/2; the total probability of drawing the black card is 1/3. By Bayes’ theorem,[3] the conditional probability of having drawn the black card, given that a black face is showing, is
$\frac{1\cdot1/3}{1/2}=2/3.$
Eliminating the white card
Although the incorrect solution reasons that the white card is eliminated, one can also use that information in a correct solution. Modifying the previous method, given that the white card is not drawn, the probability of seeing a black face is 3/4, and the probability of drawing the black card is 1/2. The conditional probability of having drawn the black card, given that a black face is showing, is
$\frac{1/2}{3/4}=2/3.$
Symmetry
The probability (without considering the individual colors) that the hidden color is the same as the displayed color is clearly 2/3, as this holds if and only if the chosen card is black or white, which chooses 2 of the 3 cards. Symmetry suggests that the probability is independent of the color chosen. (This can be formalized, but requires more advanced mathematics than yet discussed.)
Experiment
Using specially constructed cards, the choice can be tested a number of times. By constructing a fraction with the denominator being the number of times “B” is on top, and the numerator being the number of times both sides are “B”, the experimenter will probably find the ratio to be near 2/3.
Note the logical fact that the B/B card contributes significantly more (in fact twice) to the number of times “B” is on top. With the card B/W there is always a 50% chance W being on top, thus in 50% of the cases card B/W is drawn, card B/W virtually does not count. Conclusively, the cards B/B and B/W are not of equal chances, because in the 50% of the cases B/W is drawn, this card is simply “disqualified”. | 2018-01-16T09:25:38 | {
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https://math.stackexchange.com/questions/2120928/prove-that-the-square-root-of-a-positive-integer-is-either-an-integer-or-irratio | # Prove that the square root of a positive integer is either an integer or irrational
Is my proof that the square root of a positive integer is either an integer or an irrational number correct?
The proof goes like this:
Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt{n}$ must be rational. Since $\sqrt{n}$ is an integer, we can conclude that n is a square number, that is for some integer a. Therefore, if n is a square number, then $\sqrt{n}$ is rational.
Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.
We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\sqrt{n} = \frac{a}{b}$, where $a,b \in Z^+, b \neq 0$. We also suppose that $a \neq 0$, otherwise $\frac ab = 0$ , and n will be a square number, which is rational.
Hence $n = \frac {a^2}{b^2}$, so $nb^2 = a^2$.
Suppose $b=1$. Then $\sqrt n = a$ , which shows that n is a square number. So $b \neq 1$. Since $\sqrt n > 1$, then $a>b>1$.
By the unique factorization of integers theorem, every positive integer greater than $1$ can be expressed as the product of its primes. Therefore, we can write $a$ as a product of primes and for every prime number that exists in $a$, there will be an even number of primes in $a^2$. Similarly, we can express $b$ as a product of primes and for every prime number that exists in $b$, there will be an even number of primes in $b^2$.
However, we can also express $n$ as a product of primes. Since $n$ is not a square number, then there exist at least one prime number that has an odd number of primes. Therefore, there exists at least one prime in the product of $nb^2$ that has an odd number of primes. Since $nb^2=a^2$ , then this contradicts the fact that there is an even number of primes in $a^2$ since a number can neither be even and odd.
Therefore, this contradicts the fact that $\sqrt n$ is rational. Therefore, $\sqrt n$ must be irrational.
Is this sufficient? Or is there any parts I did not explain well?
• I think its correct and very well explained. – Shobhit Jan 30 '17 at 14:25
• It's a bit wordy, but logically you've got a solid proof, the even-ness of the powers of each prime is exactly what you're going for. – Adam Hughes Jan 30 '17 at 14:26
• The conclusion is not precise enough. Instead of "contradicts there is an even number of primes in $a^2$" we want to say that it contradicts the fact that the prime $p$ occurs to odd power in the unique factorization of $nb^2,\,$ but even power in $a^2,\,$ i.e. we are comparing the parity of the count of single prime, not the total of all primes – Number Jan 30 '17 at 14:28
• For example the argument shows that $\,\sqrt{3\cdot 5}\,$ is irrational because $\,3\,$ occurs to odd power in $\,3\cdot 5,\,$ but the total number of primes in $\,3\cdot 5\,$ is even. – Number Jan 30 '17 at 14:37
• On a side note, this result is known as Theaetetus' Theorem, and it's proven in Euclid's Elements here: aleph0.clarku.edu/~djoyce/elements/bookX/propX9.html "[S]quares which do not have to one another the ratio which a square number has to a square number also do not have their sides commensurable in length either." – Keshav Srinivasan Jan 30 '17 at 14:41
Your proof is very good and stated well. I think it can be made shorter and tighter with a little less exposition of the obvious. However, I would prefer students to err on the side of more rather than less so I can't chide you for being thorough. But if you want a critique:
"Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt{n}$ must be rational. Since $\sqrt{n}$ is an integer, we can conclude that n is a square number, that is for some integer a. Therefore, if n is a square number, then $\sqrt{n}$ is rational."
Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.
This can all be said more simply and to argue that if $\sqrt{n}$ is an integer we can conclude $\sqrt{n}$ is rational or that $n$ is therefore a perfect square, is a little heavy handed. Those are definitions and go without saying. However, it shows good insight and understanding to be aware one can assume things and all claims need justification so I can't really call this "wrong".
But it'd be enough to say. "If $n$ is a perfect square then $\sqrt{n}$ is a an integer and therefore rational, so it suffices to prove that if $n$ is not a perfect square, then $\sqrt{n}$ is irrational.
We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\sqrt{n}$=ab , where a,b∈Z+,b≠0. We also suppose that a≠0, otherwise ab=0, and n will be a square number, which is rational.
Terminologistically, to say "$n$ is a square number" is to mean $n$ is the square of an integer. If $n = (\frac ab)^2$ we don't usually refer to $n$ as a square (although it is "a square of a rational") We'd never call $13$ a square because $13 = (\sqrt{13})^2$.
Also you don't make the usual specification that $a$ and $b$ have no common factors. As it turns out you didn't need to but it is a standard.
Suppose b=1 . Then $\sqrt{n}$=a , which shows that n is a square number. So b≠1. Since $\sqrt{n}$>1, then a>b>1
This was redundant as $b=1 \implies$ $a/b$ is an integer and we are assuming that $n$ is not a perfect square.
.
By the unique factorization of integers theorem, every positive integer greater than 1 can be expressed as the product of its primes. Therefore, we can write a as a product of primes and for every prime number that exists in a, there will be an even number of primes in a2. Similarly, we can express b as a product of primes and for every prime number that exists in b, there will be an even number of primes in b2
Bill Dubuque in the comments noted what you meant to say was "each prime factor will be raised to any even power".
.
However, we can also express n as a product of primes. Since n is not a square number, then there exist at least one prime number that has an odd number of primes. Therefore, there exists at least one prime in the product of nb2 that has an odd number of primes. Since nb2=a2 , then this contradicts the fact that there is an even number of primes in a2 since a number can neither be even and odd.
Ditto:
Overall I think your proof is very good.
But I should point out there is a simpler one:
Assume $n = \frac {a^2}{b^2}$ where $a,b$ are positive integers with no common factors (other than 1). If $p$ is a prime factor of $b$ and $n$ is an integer, it follows that $p$ is a prime factor of $a^2$ and therefore of $a$. But that contradicts $a$ and $b$ having no common factors. So $b$ can not have any prime factors. But the only positive integer without prime factors is $1$ so $b = 1$ and $n= a^2$ so $\sqrt{n} = a$. So either for any integer either $n$ is a perfect square with an integer square root, or $n$ does not have a rational square root.
And a slight caveat: I'm assuming that your class or text is assuming that all real numbers have square roots (and therefore if there is not rational square root the square root must be irrational). It's worth pointing out, that it is a result of real analysis that speaking of a square root actually makes any sense and that we can claim every positive real number actually does have same square root value. But that's probably beyond the range of this exercise.
But if I want to be completely accurate, you (and I) have actually only proven that positive integer $n$ either has an integer square root or it has no rational square root at all. Which is the same thing as saying if positive integer $n$ has a square root, the root is either integer or irrational. But we have not actually proven that positive integer $n$ actually has any square root at all.
• Thank you for the critique! It definitely helped me a lot! – Icycarus Jan 31 '17 at 11:00
• Can you explain or state a case where we cannot find the square root of a positive integer $n$ (as you state in the last line)? – Astrobleme Apr 15 '17 at 12:56
• I didn't say there are positive integers without square roots. (There aren't.) I said we haven''t proven that the square root of any integer exists. And we haven't. To prove that we have to prove that $K = \{q \in \mathbb Q| q^2 < n\}$ is not empty, and bounded above and that if $z = \sup K$ then $z^2 - n$. All we have proven so far is that either there is an integer so that $m^2 = n$ or there is no rational $q$ so that $q^2 = n$. We have not proven that there is an irrational $z$ so that $z^2 = n$. – fleablood Apr 15 '17 at 15:54
• Proving that the square root of 17 isn't rational isn't the same thing as proving the square root of 17 is irrational, in the same way as proving the easter bunny is not aquatic is not the same thing as proving the easter bunny is land dwelling. – fleablood Apr 15 '17 at 15:58
• Square roots must exist because i) the real numbers are a continuum. (which the rationals are not) and because $z^2$ is continuous and increasing and unbounded on positive reals. So if $w^2 < n$ (and there is always such a w) and $y^2 > n$ (and there is always such a y) then $0 < w < y$ and there has to be a $z$ so that $w < z < y$ and $z^2 = n$. But we haven't proven that at all. Indeed, we haven't even discussed what irrational numbers are. Saying irrational numbers are numbers that aren't rational. Is like saying fairies are people that are not animals. – fleablood Apr 15 '17 at 16:08
We know that $\sqrt{4} = 2$ and $\sqrt{2} = 1.414...$ are rational and irrational respectively, so all we have to do is to show that if $n\in \mathbb{Z+}$ such that $\sqrt{n} = \dfrac{a}{b}$ where $a$ and $b$ are positive integers and the expression $\dfrac{a}{b}$ is in its simplest form then $\sqrt{n}$ is integral. Squaring both sides of the expression we get that $n = \dfrac{a^2}{b^2}$ since $a$ and $b$ have no common factors other than $1$ then $a^2 = n$ and $b^2 = 1$ therefore $b = 1$ hence $\sqrt{n}$ if rational it's an integer.
• How is that a critique of the OP's proof? – fleablood Jan 30 '17 at 18:51
• The problem lies in the first paragraph, the op argues that if $\sqrt{n}$ is an integer then it is rational which is very true but the op does not show that all the rational square roots to be integers in fact the op shows that the square roots are either rational or irrational which is obvious. Moreover the op in the second paragraph assumes that $n$ is not a square number which does not make sense as all integers have square roots but this roots are integral or irrational. – Tom Carter Jan 30 '17 at 20:27
• The op shows (correctly) that the square root of a non square is irrational. There is no need to show that rational roots are integers. Assuming n is not square makes perfect sense and the case for square n's has been covered. Your objections are not valid. Meanwhile you post is just ... weird. What do $\sqrt 4$ and $\sqrt 2$ have to do with anything. And your statement $n = a^2/b^2$ implies $a^2 = n$ and $b^2 =1$ is said without justification when the justification is the entire point. And you are merely pointing out an easier proof. Which is not a valid critique. – fleablood Jan 30 '17 at 22:57
• The purpose for showing that $\sqrt{4}$ and $\sqrt{2}$ are rational and irrational respectively was to show just that there are rational and irrational square roots so the only exercise left was to show that the rational roots are always integers. Secondly I already pointed out that in $\dfrac{a}{b}$ , $a$ and $b$have no common factors and therefore their squares will certainly have no common but we are equating this square $n$ which might as well be written as $\dfrac{n}{1}$ that's where my conclusion where $b = 1$ and $a = \sqrt{n}$ arises from. – Tom Carter Jan 31 '17 at 6:31
• I believe if you reread the op's post you especially the areas I pointed out you will understand where my problem arises but as you said before I was only simplifying the post and corrected the few areas which I thought needed to be repaired. – Tom Carter Jan 31 '17 at 6:34 | 2018-08-19T22:57:04 | {
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https://math.stackexchange.com/questions/3756351/finding-real-x-y-solutions-that-satisfies-a-system-of-equation/3756391 | Finding real $(x,y)$ solutions that satisfies a system of equation.
I was given:
$$x + y^2 = y^3 ...(i) \\ y + x^2 = x^3...(ii)$$
And was asked to find real $$(x,y)$$ solutions that satisfy the equation.
I substracted $$(i)$$ by $$(ii)$$:
$$x^3 - y^3 + y^2 - x^2 + x - y = 0$$
Then factored it out so I have:
$$(x-y)(x^2 + xy + y^2 - x - y + 1) = 0$$
Multiplying it by two, I get:
$$(x-y)(2x^2 + 2xy + 2y^2 - 2x - 2y + 2) = 0 \\ (x-y)((x^2 - 2x + 1) + (y^2 - 2y + 1) + (x^2 + 2xy + y^2)) = 0 \\ (x-y)((x-1)^2 + (y-1)^2 + (x+y)^2) = 0$$
I noticed that a solution exists only if $$x=y$$ because there are no real solutions for $$x$$ and $$y$$ that satisfies $$(x-1)^2 + (y-1)^2 + (x+y)^2 = 0$$.
Substituting $$x=y$$ into the first equation, I get: $$y(y^2-y-1)=0$$ where the roots are $$y= 0, \frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}$$. Hence, the real solutions of $$(x,y)$$ that satisfy are:
$$(x,y) = (0,0), (\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}), (\frac{1-\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2})$$.
What I would like to ask is: Is there a better way to solve the question? It's from a local university entrance test, where this kind of questions are aimed to be done in < 3 minutes. It took me a while to manipulate the algebraic stuffs above.
Someone in a local forum said something about symmetric systems which says that the solution does not exist for $$x \neq y$$. How do I know if the equation is a symmetric one? (Never heard of something before throughout high school here...) I would love to see a resource for this!
• Suppose $(x_1,y_1)$ is a solution, is $(y_1,x_1)$ a solution to your system? You'll see that it is, because one equation is just the flip of the other. This is what symmetry means Jul 14, 2020 at 8:46
• Do you have the mark scheme available? Jul 14, 2020 at 8:54
You can begin by noting that the $$2$$ functions are inverses of each other (and only involve odd non-zero exponents). Using the fact that inverse functions are reflections in the line $$y=x$$, we can now see that the intersection points must be along the line $$y=x$$. Substituting $$y$$ into $$x$$ or vice-versa, we obtain the equation you get and obtain the solutions you got. That would only take about 3 minutes. I hope that helps :) | 2022-05-24T00:31:52 | {
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https://math.stackexchange.com/questions/2328842/metric-space-consisting-two-elements | # Metric space consisting two elements.
Suppose that $S$ be a set consisting exactly $2$ elements. Suppose we define a function $\displaystyle d:S \times S \to [0,\infty)$ by $\displaystyle d(x,y)=\begin{cases}1 &\text{ , if }x\not=y\\0 &\text{ , if} x=y\end{cases}$
How I can show that $d$ defines a metric on $S$ ?
Problem is on triangular inequality..To prove triangular inequality we need at least three points. How I can show the triangular inequality ?
Same problem for a set consisting only $1$-element or empty set.
What's the idea behind these ?
• You can pick 3 elements among two, obviously at least two of them will be equal, but that shouldn't bother you. Notice when you prove the triangular inequality for bigger metric spaces, you do not require that the 3 points are different. – Smurf Jun 19 '17 at 18:54
The triangle inequality says: $$\text{For all } x, y, z, \;\; d(x,y) + d(y,z) \ge d(x,z).$$ It may seem like it to you (thinking of it as a triangle), but $x,y,z$ do not have to be distinct. Two of them can be the same point.
However, triangle inequality with two or one point(s) is trivial. For instance, suppose $x = y$. Then the triangle inequality says $$d(x,x) + d(x,z) \ge d(x,z)$$ which is true merely relying on the fact that $d(x,x) \ge 0$. The same thing happens if $y = z$. Finally if $x = z$, we get $$d(x,y) + d(y,x) \ge d(x,x),$$ which follows from the other property of a metric space, that $d(x,x) = 0$, and the facts $d(x,y) \ge 0$ and $d(y,x) \ge 0$.
The bottom line: The triangle inequality only says something "interesting" (not implied by the other properties of a metric space) when the three points of the triangle are distinct.
• A lawyer might write the Triangle Inequality as " For any point, hereinafter called x, and any point hereinafter called y, and notwithstanding that x and y are different names, the point called x and the point called y may or may not be the same point, and any point hereinafter called z..... (etc)" ...or something like that. – DanielWainfleet Jun 20 '17 at 5:18
You don't need three points. Here's the statment of the triangular inequality:
$$\forall x,y,z\in S,d(x,z)\le d(x,y)+d(y,z)$$
Remark: This metric is called the discrete metric, and it can be defined on any nonempty set. | 2020-01-21T02:40:09 | {
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https://math.stackexchange.com/questions/1304633/solution-to-y-y2-4 | # Solution to $y'=y^2-4$
I recognize this as a separable differential equation and receive the expression:
$\frac{dy}{y^2-4}=dx$
The issue comes about when evaluating the left hand side integral:
$\frac{dy}{y^2-4}$
I attempt to do this integral through partial fraction decomposition using the following logic:
$\frac{1}{(y+2)(y-2)} = \frac{A}{y+2}+\frac{B}{y-2}$
Therefore, $1=Ay-2A+By+2B$.
Since the coefficients must be the same on both sides of the equation it follows that:
$0=A+B$ and $1=-2A+2B$. Hence, $A=-B$, $B=\frac14$, $A=-\frac14$.
Thus the differential equation should be transformed into:
$-\frac{1}{4} \frac{dy}{y+2} + \frac14 \frac{dy}{y-2} = x+C$
Solving this should yield:
$-\frac14 \ln|y+2| + \frac14 \ln|y-2| = x+C$
which simplifies as:
$\ln(y-2)-\ln(y+2)=4(x+c)$
$\ln[(y-2)/(y+2)]=4(x+c)$
$(y-2)/(y+2)=\exp(4(x+c))$
$y-2=y*\exp(4(x+c)+2\exp(4(x+c))$
$y-y\exp(4(x+c))=2+2\exp(4(x+c))$
$y(1-\exp(4(x+c)))=2(1+\exp(4(x+c)))$
$y= 2(1+\exp(4(x+c)))/(1-\exp(4(x+c)))$
However, when done in Mathematica/Wolfram Alpha the result is given as (proof in the attached image)
$\frac14 \ln(2-y) -\frac14 \ln(2+y) = x + C$
$y= 2(1-\exp(4(x+c)))/(1+\exp(4(x+c)))$.
Can anyone figure out where I have made an error? The only thing I can think of is something with evaluating the absolute values of the natural logarithms.
• two constant solutions, $y=2$ and $y=-2.$ The other solutions split into three types, $y < -2,$ very similar $y > 2,$ rather different $-2 < y < 2$ – Will Jagy May 29 '15 at 22:03
• Ok but how does this help me in understanding why my answer is wrong? – Filip May 29 '15 at 22:09
• That is up to you. – Will Jagy May 29 '15 at 22:16
• If I am not mistaken, $y' = y^2 - 4$ is a Ricatti equation; you might check the wiki page on such; I have found it quite useful. As far as Wolfram Alpha goes, I have often found its solutions to be quite different than mine; I usually like mine better; at least I know where mine came from! – Robert Lewis May 29 '15 at 22:25
• I'm not sure you're answer is wrong. Did you try checking it the old-fashioned way, by hand? – Robert Lewis May 29 '15 at 22:26
Saturday morning: I do like robjohn's point about $\log |x|$ giving the student some incorrect expectations. What I wrote here is correct and careful, but the same conclusions come from dropping the absolute value signs and doing the calculations three times, $y < -2,$ $-2 < y < 2,$ $y > 2.$ At some point the vertical asymptotes will be revealed, how solutions with $y > 2$ reach a vertical asymptote and then jump below to $y < -2.$ That is how I usually do things, split into cases early.
Lost my train of thought. However, if I want $\int \frac{1}{x} dx,$ without absolute values, the answer would be $\log x$ for $x > 0,$ but $\log (-x)$ for $x < 0.$
$-\frac14 \log|y+2| + \frac14 \log|y-2| = x+C$
which simplifies as:
$\log|y-2|-\log|y+2|=4(x+c),$ or
$$\left| \frac{y-2}{y+2} \right| = k e^{4x}$$ with $k > 0$ for the nonconstant solutions. The item inside absolute value signs is a linear fractional or Mobius transformation, once we decide about the $\pm$ signs the inverse is given by the matrix $$\left( \begin{array}{rr} 2 & 2 \\ -1 & 1 \end{array} \right)$$ If $y > 2$ or $y < -2,$ those inequalities hold true forever, we have $$\frac{y-2}{y+2} = k e^{4x}$$ and we get robjohn's $$y = \frac{2k e^{4x} + 2}{-k e^{4x} + 1}.$$ There is a jump discontinuity: for some value of $x,$ we find that $-k e^{4x} + 1= 0,$ indeed $e^{4x} = 1/k,$ $4x = - \log k,$ x = $-(1/4) \log k.$ For $x < -(1/4) \log k,$ we have $y > 2.$ There is a vertical asymptote, then for $x > -(1/4) \log k,$ we have $y < -2.$
In all cases, the curves are asymptotic to the lines $y=2$ and $y=-2.$ Also, note that the ODE is autonomous. Given one of the solutions described, we get another solution by shifting $x$ by any constant we like. If preferred, we can drop the multiplier $k$ by emphasizing the horizontal shifts.
If, however, we begin with $-2 < y < 2,$ then those inequalites hold forever, we have $$\frac{y-2}{y+2} = - k e^{4x}$$ and we get the continuous $$y = \frac{-2k e^{4x} + 2}{k e^{4x} + 1}.$$
In the picture below, I took $k=1.$ To find all other solutions, replace $x$ by some $x - x_0.$ Put another way, we can take $e^{-4 x_0}= k;$ thus, we can account for all solutions either by varying $k,$ or by deleting $k$ and adjusting $x_0.$ Note also that the part with $y>2$ and the part with $y < -2$ are tied together, the same vertical asymptote.
\begin{align} x &=\int\frac{\,\mathrm{d}y}{y^2-4}\\ &=\frac14\int\left(\frac1{y-2}-\frac1{y+2}\right)\,\mathrm{d}y\\ &=\frac14\log\left(\frac{y-2}{y+2}\right)+C \end{align} Therefore, $$\frac{y-2}{y+2}=ke^{4x}$$ or, solving for $y$, $$y=2\,\frac{1+ke^{4x}}{1-ke^{4x}}$$ We get your form of the answer by letting $k\lt0$. It is hard to see this since you need to use a complex $c$ to get the equivalent answer. Check your answer in the original equation. You'll see that it works.
• but isn't your k guaranteed to be a positive value since it is e^C and e^x>0? – Filip May 29 '15 at 22:45
• @WillJagy: I dislike using $\log|y|$ as the antiderivative of $\frac1y$ since it suggests that $\int_{-1}^1\frac{\mathrm{d}y}y=0$, when, in fact, the integral diverges. Taking the Cauchy Principal Value of the integral does give $0$, but I prefer to use either an complex constant of integration for the case when $y\lt0$, or use $\log(y)+C$ and $\log(-y)+C$ depending on the domain, emphasizing that the domains should not be mixed. – robjohn May 29 '15 at 23:02
• @Filip: As I mentioned in my answer, if $C$ is complex, we can have $e^C\lt0$. – robjohn May 29 '15 at 23:05 | 2019-08-20T21:13:32 | {
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https://math.stackexchange.com/questions/3099028/binomial-distribution-probabilities-need-help-with-multiple-requirements-for-k | # Binomial Distribution Probabilities - Need help with multiple requirements for k
Could someone please check if the procedure for part a) is correct and I need a bit of help in regards to part b). I don't have the answer to refer to. Thank you for your time in advance!
An airport reports that on a stormy day, with wind speed between 60 and 70 kilometers per hour, 20% of the domestic flichts are cancelled, 30% are delayed and 50% are on time. Tomorrow 12 domestic flights are on schedule and the weather is expected to be stormy with average wind speed of 65 kilometer per hour.
a) Compute the probability that exactly 3 flights are cancelled.
So here we have Binomial Distribution with $$n=12$$ and $$p=0.2$$ Since we have only 2 possible outcomes as far as we are concerned for this case, either a flight is cancelled or a flight is not cancelled.
The formula for Binomial Distribution is as follows: $$P(k)=\binom{n}{k}p^k(1-p)^{n-k}$$ Hence we just plug in the values from above:
$$P(3)=\binom{12}{3}0.2^3(1-0.2)^{12-3}=0.23622$$
b) Determine the probability that less than 9 but at least 2 flights are punctual.
So for this probability space (again Binomial distribution), we have $$n=12$$ $$p=0.5$$. However, I am confused in regards to what to do here. Do I have to compute every single probability and then combine them? Aka P(2),P(3),P(4),P(5),P(6),(7),P(8)? That seems to be rather long and as if I am missing something here so any help would be much much appreciated! Thank you!
• Yes you are correct. We would usually use a calculator or check tables of cumulative values for part b) in this case. – Peter Foreman Feb 3 '19 at 20:13
• Oh, alright, thank you! – VRT Feb 3 '19 at 20:14
In part b) you could also calculate probability of the opposite event: probability that $$9$$ or more or less than $$2$$ flights are on time.
$$P_{complement} = P(0) + P(1) + P(9) + P(10) + P(11) + P(12)$$
Here you have to sum only $$6$$ probabilities and the probability of original event from question is:
$$P = 1 - P_{complement} = 1 - P(0) - P(1) - P(9) - P(10) - P(11) - P(12)$$
If this isn't an exam question you can also write a simple computer program that computes desired probabilities.
P.S. When I was learning probability and statistics first time I found writing computer programs that model problems from textbooks very insightful.
• Thank you so much! Unfortunately, I am preparing for an exam, hence, using a program is not an option :/ – VRT Feb 3 '19 at 20:35 | 2020-01-25T21:01:53 | {
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https://cs.stackexchange.com/questions/136602/onm-vs-on2m-time-complexity/136612 | O(n+m) vs O(n+2m) time complexity
Are $$O(n+m)$$ and $$O(n+2m)$$ the same?
If $$m>n$$, then both complexities are $$O(m)$$. Likewise, if $$n>m$$, then they are $$O(n)$$. Is this correct?
• Since $n+m \le n+2m \le 2(n+m)$, assuming $n$ and $m$ are non-negative, $O(n+m) = O(n+2m)$ under whatever reasonable definition of $O$ for two variables, as long as it is not the case that both $n$ and $m$ are bounded. Mar 14, 2021 at 4:20
I would like to try an intuitive explanation.
Assume your algorithm works in $$2m$$ time units on your current machine. If you run it on a machine twice as fast, then it takes $$m$$ times units. As we do not know (and do not care) on which machine(s) the algorithm will run, this makes the $$2$$ factor irrelevant. This is what the big-$$O$$ notation captures, in a sense.
(In addition, if we know that $$n>m$$ then we have $$3n > n+2m$$, and so anything below $$n+2m$$ also is below $$3n$$. As big-$$O$$ deals with upper bound, this means that $$O(n+2m)$$ complexity is, in this case, in $$O(3n)$$ complexity which, as explained above, is nothing but $$O(n)$$.)
Hope this helps.
Assuming $$n, m$$ independent variables let me use definition for big-$$O$$ for multiple variables, which I discussed in Big O of multiple variables:
$$O(f(n,m)),(n,m) \to \infty$$ is set of functions $$g$$, such, that $$\exists C_g>0$$ and $$\exists M_g>0$$ such that for $$\forall \left\Vert (n,m)\right\Vert_\infty > M_g$$ holds $$g(n,m) \leqslant C_g f(n,m)$$, where $$\left\Vert(m,n)\right\Vert_\infty = \max\{n,m \}$$ based on intention to infinity by squares. It's well known Chebyshev norm or the infinity norm and is equivalent to intention to infinity by rectangles.
As case $$O(n+m) \subset O(n+2m)$$ is obvious, let me go to reverse subset direction: assume we have $$f\in O(n+2m)$$, which means $$f(n,m)\leqslant B_f (n+2m)$$ in appropriate conditions for some constant $$B_f$$. Now we need such $$C_f$$ for which holds $$B_f (n+2m) \leqslant C_f(n+m)$$. It's easy to see, that any $$C_f \gt 2 B_f$$ satisfy this condition, so $$O(n+2m) \subset O(n+m)$$, which means, that we obtain equality.
At end let me say, that if $$n, m$$ are not independent variables, then it requires more context as well as if/when we decide to consider other types of intentions to infinity such as intention to infinity by circles, by triangles etc. | 2022-05-22T18:00:03 | {
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https://math.stackexchange.com/questions/1127137/polar-plots-and-square-roots | # Polar Plots and square roots
When I plot a polar plot of $r=\sin (3 \theta)$, and $r=\sqrt{\sin (3 \theta)}$ I get nearly identical graphs, both $3$ pedal rose type plots. In the case without the square root, it is easy to understand the plot. However, for the plot involving the square root of $\sin 3 \theta$, it is strange to me how the graph would handle thetas for which the $\sin$ of the $3 \theta$ is negative. It would seem that the negative values inside the square root should cause the 2 graphs to be dissimilar, yet it appears this is not the case.
Example: For $\theta = 65^\circ, r=\sin (3 \theta) = -0.259$; I would expect this it be an issue with the graph of $r= \sqrt{\sin (3 \theta)}$. Enclosed are images of the graphs of the 2 polar plots. URL for 2 graphs --> https://s3.amazonaws.com/grapher/exports/gtwzzdokst.png
• try plotting both for $0 \le \theta \le 2\pi/3$ – abel Jan 31 '15 at 1:25
These images might give you a better sense of things. Very often, the nature of a parametric plot is best understood as a path drawn as the parameter changes, instead of just a static set of points that satisfy a relation.
You'll notice that the $r = \sin 3\theta$ curve draws smoothly, actually plotting "backwards" through the pole for values of $\theta$ (for instance, $90^\circ$) when $r$ is negative.
On the other hand, the $r = \sqrt{\sin 3\theta}$ curve "pauses" as $\theta$ sweeps through values that make the radicand negative. (The red flashing is Mathematica expressing its frustration at being asked to plot imaginary values. Normally, I'd suppress that; however, here, it's actually directly relevant to the discussion, so I left it.)
• It's worth noting that the $r=\sqrt{\sin 3\theta}$ curve plots only non-negative $r$ values. This is because the square-root function itself never returns negative values. On the other hand, a related curve, $r^2 = \sin 3\theta$ would plot both a positive and a negative $r$ value for a given $\theta$, effectively creating a six-petal rose by drawing "positive" and "negative" three-petal roses simultaneously. – Blue Jan 31 '15 at 2:21
• Very, very useful. thank you very much!!! – user163862 Jan 31 '15 at 18:25
• very useful indeed, thank you – Thor Feb 19 at 1:57
Notice that if you plug in $r=|\sin\:(3θ)|$ instead of $r=\sin\:(3θ)$ you get a six-petal graph. The concept of a negative radius doesn't make much sense, so in reality the negative results of $r=\sin\:(3θ)$ are just not being drawn either.
• The concept of negative radius does make sense in polar coordinates; you just look the the exact opposite way through the pole. (So, the point with polar coordinates $(-1,\theta)$ is identical to the point with polar coordinates $(1,\pi + \theta)$. This is just like complex numbers, and it's one of the things that makes polar graphs so cool. :) The "negative" portions of $r=\sin 3\theta$ are indeed being drawn, only "backwards" through the pole; as a result, they overlap the "positive" portions. (Continued.) – Blue Jan 31 '15 at 1:19
• (Part 2) However, the concept of an imaginary radius doesn't make sense in this context, so in graphing $r = \sqrt{\sin 3\theta}$, the grapher is skipping-over the angles at which the radicand is negative. This is not unlike graphing $y = \sqrt{x}$; a grapher simply draws nothing for $x < 0$. If you can find a grapher that allows you to see the graph being drawn as a path as $\theta$ increases, the dynamic here becomes more clear. – Blue Jan 31 '15 at 1:22 | 2019-10-24T01:48:46 | {
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https://www.physicsforums.com/threads/trouble-with-lorentz-transformations.393523/ | # Trouble with Lorentz transformations
1. Apr 8, 2010
### pc2-brazil
Good evening,
As an effort for trying to understand Lorentz transformations, I'm trying to use them to derive the "length contraction" result.
Consider two reference frames, O (non-primed) and O' (primed), moving with respect to each other with a velocity v. Consider them to be under http://en.wikipedia.org/wiki/Lorent...rmation_for_frames_in_standard_configuration".
Choosing the non-primed reference frame, the Lorentz transformations for position (x-axis) and time (t-axis) will be:
x = γ(x' + vt')
t = γ(t' + vx'/c²)
The inverse transformations will be:
x' = γ(x - vt)
t' = γ(t - vx/c²)
Where γ is the Lorentz factor and c is the speed of light in vacuum.
Now, I will try to derive the length contraction result.
Suppose I have a thin rod moving along with the primed reference frame. One end of the object is at x'1 x'2. Then, the length of the object in the primed reference frame is L' = x'2 - x'1.
To find the corresponding coordinates of the ends of the rod in the non-primed frame (x1 and x2), I will use the inverse transformation for position:
x1 = γ(x'1 + vt')
x2 = γ(x'2 + vt')
The length of the rod as measured by the non-primed frame will be:
x2 - x1 = γ(x'2 + vt') - γ(x'1 + vt')
x2 - x1 = γ(x'2 - x'1)
L = γL'
This is wrong. I should have obtained L' = γL, because the non-primed frame sees the rod shorter than the primed frame does. In other words, the length measured by the non-primed frame should be shorter than the proper length.
What am I thinking wrong?
Last edited by a moderator: Apr 25, 2017
2. Apr 8, 2010
### starthaus
You need to calculate x2 - x1 for the condition: t2 = t1. You are inadvertently calculating it for t'2 = t'1, this is why you get the error.
Last edited by a moderator: Apr 25, 2017
3. Apr 9, 2010
### pc2-brazil
But why shouldn't t'2 equal t'1?
I thought this was implied, because, for the primed referential, the times of measurement of the ends of the rod were the same.
4. Apr 9, 2010
### starthaus
Because of relativity of simultaneity. You are measuring in the unprimed frame, you need to mark the endpoints simultaneously in the unprimed frame. This means t2 equals t1 thus guaranteeing that t'2 is NOT equal to t'1.
Last edited: Apr 9, 2010
5. Apr 10, 2010
### pc2-brazil
Thank you, I understand it now.
So, in order to derive the length contraction result using the transformation with t' (x = γ(x' + vt')), I have to find the relation between t'1 and t'2, knowing that t1 = t2:
$$t_1=\gamma(t'_1+vx'_1/c^2)$$
$$t_2=\gamma(t'_2+vx'_2/c^2)$$
$$t_1=t_2$$
$$\gamma(t'_1+vx'_1/c^2)=\gamma(t'_2+vx'_2/c^2)$$
$$t'_1+vx'_1/c^2=t'_2+vx'_2/c^2$$
$$t'_2-t'_1=-\frac{v(x'_2-x'_1)}{c^2}$$
Now, I can make x2 - x1:
$$x_2-x_1=\gamma(x'_2+vt'_2-x'_1-vt'_1)$$
$$x_2-x_1=\gamma[x'_2-x'_1+v(t'_2-t'_1)]$$
Substituting the expression found for (t'2-t'1)
$$x_2-x_1=\gamma\left[x'_2-x'_1-\frac{v^2(x'_2-x'_1)}{c^2}\right]$$
$$x_2-x_1=(x'_2-x'_1)\gamma(1-\frac{v^2}{c^2})$$
$$x_2-x_1=(x'_2-x'_1)\gamma\times\gamma^{-2}$$
$$x_2-x_1=\frac{x'_2-x'_1}{\gamma}$$
which is the length contraction result.
This same result can be obtained in a faster way by using x'1 = γ(x1 - vt1) and x'2 = γ(x2 - vt2), and calculating x'2 - x'1 with t1 = t2.
6. Apr 10, 2010
### stevmg
The following two equations are the easiest way to remember Einstein's depiction of the Lorentz transformation (one-dimension). This\\ey appear on page 34 of his book "Relativity."
x' = gamma(x - vt)
t' = gamma(t - vx/c^2)
Thus x'_2 = gamma(x_2 - vt) and x'_1 = gamma(x_1 - vt)
We don't need the the equations for t_1, t'_1, t_2 and t'_2 from here on.
x'_2 - x'_1 = gamma(x_2 - x_1) or (x'_2 - x'_1)/gamma = x_2 - x_1
Remember, gamma = 1/[SQRT(1 - v^2/c^2)]. Gamma is always >=1, so 1/gamma is always <=1.
This makes sense. You give us x'_2 and x'_1 so x_2 - x_1 is always <= x'_2 - x'_1
L' = (x'_2 - x'_1) and L = (x_2 - x_1)
Thus L' = gamma*L which is the way it should be...
The same as starthaus above...
Steve G
7. Apr 10, 2010
### stevmg
H-E-L-P!
How do I partially quote a post?
How do I refer to a prior post on the same thread or different thread by hyperlink?
8. Apr 10, 2010
### pc2-brazil
Thank you for the clarifications in your previous post.
First, click the "Quote" button below the post you want to quote. Then, you enter a page similar to the "New reply" page in which the whole post you quoted appears written between two "QUOTE" tags. To partially quote, just erase the part of the quote that you don't want, keeping only the relevant part.
To refer to a prior post on the same thread, first you need to get its address. To do this, click on the small number that appears in the top-right side of the post. Then, copy the address in the address bar of your browser. For example, for the post marked with "#5" in this thread:
https://www.physicsforums.com/showpost.php?p=2665478&postcount=5
This hyperlink can be used to refer to this post.
9. Apr 10, 2010
### starthaus
No, with t'1 = t'2
10. Apr 10, 2010
### starthaus
No, that's clearly wrong. The correct answer is L'=L/gamma
No, you got a different result
11. Apr 11, 2010
### pc2-brazil
I'm measuring from the unprimed frame, so I need to mark the endpoints simultaneously in the unprimed frame. Thus, t2 equals t1.
With t1 = t2:
x'2 - x'1 = γ(x2 - vt2) - γ(x1 - vt1)
x'2 - x'1 = γ(x2 - x1)
L' = γL
This is the result I was expecting. The observer in the unprimed frame sees the rod shorter than its proper length (L').
12. Apr 12, 2010
### starthaus
When you write x'2 - x'1 it means that you are measuring in the primed frame.
One more time, L' = γL is wrong. If you did things correctly you should have gotten L' = L/γ
13. Apr 12, 2010
### stevmg
Then pc2-brazil had it right the first time...
L = $$\gamma$$L'
Starthaus, you are right... due to time dilation (with concomitant length contraction), L' will always be less than L. I got inverted. Damn, am I getting sloppy!
Steve G
14. Apr 12, 2010
### pc2-brazil
I still think that that is not necessarily true, since I'm analyzing the same situation as before, in which the unprimed frame measures the length of an object that is moving with the primed frame. Thus, t1 = t2.
So, I can choose to write x2 - x1 and, by Lorentz transformations, obtain an expression containing x'2 - x'1, or I can write x'2 - x'1 and obtain an expression containing x2 - x1. But, in both cases, I have to use t1 = t2 and obtain L' = γL, since the situation I want to analyze is the same, and not a different one where I'm measuring from the primed frame.
15. Apr 12, 2010
### starthaus
Yes, this is now fully correct. Post #14 is not.
16. Apr 12, 2010
### starthaus
Yes.
No.
17. Apr 12, 2010
### stevmg
DQ = Dumb question. Use the above
If I have a stick 1 meter long and it slides on a table frictionless.
If it moves at 0.6c, will it fall through a hole in the table 0.9 meter?
What is x_1, what is x_2, what is x'_1, what is x'_2 (therefore what is L and what is L'?)
My gut feeling is that L' = 1 m (moving at 0.6c). L calculates to 0.8 m by length contraction and it will fall through the 0.9 m hole. Now, using what you discussed above can you show us by use of the Lorentz transformations rather than just the length contraction formula.
In the text Special Relativity by AP Frenchf M.I.T., on page 97, he goes over this exact problem presented by pc2-brazil. He assumes the measurement of x'_1 and x'_2 are done simultaneously in the S' frame of reference and he derives L' = x'_2 - x'_1 = L/gamma = (x_2 - x_1)/gamma. Now it would appear that by length contraction that the moving frame (x'_2 - x'_1) should be greater than the "static" frame (x_2 - x_1) so that it would "contract" to x_2 - x_1 but that is the unit length in S' is less than the unit length in S and thus the length L in S is greater than the length L'. S is the "static" F.O.R. (the x_1 and x_2) and S' is the "moving" F.O.R. (the x'_1 and x'_2.)
I know, there is no such thing as a preferred or static F.O.R., but you know what I mean here.
18. Apr 12, 2010
### stevmg
You have to keep oriented as to who is what here. A rod which is in motion relative to a reference frame is shorter than it would be in its own frame. Thus, the 1 meter rod is actually 0.8 m in the reference frame while it is 1 m in its own (moviing) frame. It will fit through the 0.9 m gap in the table as described.
In this sense, the x'_2 - x'_1 is greater than the x_2 - x_1 which sort of contradicts what you guys have shown. What you have shown is that the moving frame measurement is smaller than that of the reference frame. It is true that the moving frame measurement transforms to something shorter in the reference frame even though it measures longer in its own frame.
Gotta be careful here and not blindly follow equations without keeping track of your orientation.
19. Apr 12, 2010
### starthaus
Last edited: Apr 12, 2010
20. Apr 13, 2010
### stevmg
Starthaus:
Here simple example which goes along with what you say:
Take the Earth and a star some 7.2 lt-yr away.
Assume there is a reference frame S' that moves at 0.6c to the right with respect to the "stationary" F.O.R. S. Assume there is a rocket ship that travels in S at 0.8c to the right for 9 hours.
As per convention let us assign x_1 = 0 and t_1 = 0 and x'_1 = 0. Now, we state that t_2 = 9. Thus, after 9 hours (t_2) in the S F.O.R. the coordinate of that rocket ship is x_2 = 9*0.8 = 7.2 lt-yr (which coincides with the position of that star.) Thus x_2 - x_1 = 7.2 - 0 = 7.2.
Because v (the velocity of S' is 0.6c, gamma = 1/SQRT[1 - 0.6^2] = 1/0.8 = 1.25
Using Lorentz for distance, x'_2 = gamma[(x_2 - vt_2)) = 0.8[7.2 - 9*0.6) = 2.25
Thus x'_2 - x'_1 = 2.25 - 0 = 2.25. This is less than the 7.2 lt-yr difference cited above for x_2 - x_1. I've set t_1 = t'_1 = 0 and t_1 = t'_1 (because we used the same instantaneous measurement of the x_2 - x_1 as you cited in one of your explanations.) I made no such assumption for t'_2.
Clearly, t'_2 = gamma[t_2 - vx_2/c^2] = 1.25 (9 - 0.6*7.2) = 5.85 years. This not = to t'_1 of 0.
5.85 - 0 = 5.85 or t'_2 - t'_1 which is not 9 or t_2 - t_1.
But, there is another question associated with this:
Assume a rod is of sufficient length that at 0.8c moving to the right, it measures 7.2 lt-yr. in the S F.O.R. This rod would have to be 12 lt-yr in length in the S' F.O.R. for it to do this.
It appears that this rod is longer in S' (12) or x'_2 - x'_1 than in S (7.2) or x_2 - x_1.
What gives?
H-E-L-P-!
Last edited: Apr 13, 2010 | 2018-12-18T13:39:07 | {
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https://www.physicsforums.com/threads/finding-an-interval-of-convergence.253266/ | # Finding an Interval of Convergence
1. Sep 2, 2008
### Battlemage!
1. The problem statement, all variables and given/known data
Find the interval of convergence of the infinite series:
(x-1)n / 2n
n = 1
2. Relevant equations
Using the ration test. It converges if the absolute value of the limit of f(n+1)/f(n) as n -> ∞ < 1.
Well, I hope that's how you write it. I'm sure you guys know how to use the ration test.
3. The attempt at a solution
I actually believe I have the answer. But I have no way of knowing if it is right.
Here is my answer:
The interval of convergence is -1 < x < 3.
I got this by using the ratio test, and then eliminating factors of (x-1)n and 2n (after expanding the exponents), leaving me with the limit of
lim (x-1)/2 as n -> ∞
Then, since there is no n, then that limit IS (x -1)/2 (unless I'm totally off)
And the absolute value of that has to be less than 1 for it to converge, so
-1 < (x-1)/2 < 1
-2 < x - 1 < 2
-1 < x < 3
Is this correct? Do you need me to post the part where I canceled out terms before I took the limit?
Thanks!
2. Sep 2, 2008
Looks good. The only thing the test doesn't tell you about is what happens at the endpoints ($$x = -1, x = 3$$). You have to try each one and determine what happens. (In general) You may find the series does not converge at either endpoint, it may converge at one but not the other, or it may converge at both.
Good work.
3. Sep 2, 2008
### Battlemage!
Thanks for the help, but I do have one more question. You see, it's been about 1.5 years since I've worked with series. This will be a very stupid question..
How exactly do you test if the two endpoints are included in the interval? I just don't know the method for doing that.
Thanks again!
4. Sep 2, 2008
Your most recent post begins:
Nonsense.
It continues
Good, common question. As an example, suppose I have the series
$$\sum_{n=1}^\infty \frac{x^n}{n}$$
Work similar to what you did in your example shows the radius of convergence to be $$1$$, so I know the series converges for $$-1 < x < 1$$. What about the endpoints here?
First, consider $$x = 1$$. Simply plug this value into the series and study the result. I get
$$\sum_{n=1}^\infty \frac{(1)^n}{n} = \sum_{n=1}^\infty \frac 1 n,$$
which I know diverges because it is the harmonic series. If I try $$= -1$$,
$$\sum_{n=1}^\infty \frac{(-1)^n}{n}$$
which is known to converge.
Putting all of this work together shows that the original series converges for $$-1 \le x < 1$$, or $$[-1, 1)$$ if you write it in interval notation.
I hope this helps. | 2017-01-22T14:16:07 | {
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https://www.physicsforums.com/threads/solving-an-equation-with-fractional-part-function.867739/ | Solving an equation with fractional part function
Tags:
1. Apr 19, 2016
1. The problem statement, all variables and given/known data
If $\{ x \}$ denotes the fractional part of x, then solve:
$\{ x \} + \{ -x \} = x^2 + x -6$
It's provided that there are going to be 4 roots of this equation. And two of them will be integers.
2. Relevant equations
$0 \lt \{ x \} \lt 1~~\text{if}~~x \not\in I$
$\{ x \} = 0~~\text{if}~~x \in I$
3. The attempt at a solution
CASE 1: When $x \in I$
⇒ $0 + 0 = x^2 +x -6$
⇒ $x = -3, 2$
I got two integer roots here.
CASE 2: When $x \not\in I$
$\text{Let, }~~x=i+f ,~~\text{where}~~i \in I~~\text{and}~~0<f<1$
⇒ $\{ i+f \} + \{ -(i+f) \} = (i+f)^2 + (i+f) - 6$
⇒ $f +f = (i^2 +f^2 +2if) + (i+f) -6$
⇒ $f^2 + (2i-1)f + (i^2 + i -6) = 0$
⇒ $f = \frac{(1-2i) \pm \sqrt{(2i-1)^2 -4(i^2 + i -6)}}{2}$
⇒ $f = \frac{(1-2i) \pm \sqrt{8i^2 -23}}{2}$
Now
$0<f<1$
⇒ $0<\frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} <1$
⇒ $\frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} >0 ~~ \text{and} ~~ \frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} <1$
⇒ $(1-2i) \pm \sqrt{8i^2 -23}>0 ~~ \text{and} ~~ (1-2i) \pm \sqrt{8i^2 -23}<2$
⇒ $\pm \sqrt{8i^2 -23}>2i-1 ~~ \text{and} ~~ \pm \sqrt{8i^2 -23}<2i+1$
Squaring,
⇒ $8i^2 -23>4i^2 -4i +1 ~~ \text{and} ~~ 8i^2 -23<4i^2 +4i +1$
⇒ $4i^2 +4i -24>0 ~~ \text{and} ~~ 4i^2 -4i -24<0$
⇒ $i^2 +i -6>0 ~~ \text{and} ~~ i^2 -i -6<0$
⇒ $(i+3)(i-2)>0 ~~ \text{and} ~~ (i-3)(i+2)<0$
⇒ $(i>2 \text{or} i<-3) ~~ \text{and} ~~ (i>-2 \text{and} i<3)$
⇒ $2<i<3$
but this gives no definite solution.
Thank you.
2. Apr 19, 2016
BvU
Hi,
Can you explain how $$\{ i+f \} + \{ -(i+f) \} = f + f \quad ?$$ For with e.g $x = 1.3$ I don't get $0.6$ !
3. Apr 19, 2016
Samy_A
You have to tell us how you define $\{ x \}$ for negative $x$.
4. Apr 19, 2016
Ray Vickson
Since, for example, we can write $-1.4$ as either $-1 - 0.4$ or as $- 2 + 0.6$, we could consider $\{-1.4\}$ as being either $-0.4$ or $+0.6$. Which one would you take?
5. Apr 19, 2016
BvU
Based on the first relevant equation I concluded $0.6$ ...
6. Apr 22, 2016
Sorry, I did a silly mistake there.
My teacher said, the fractional part function is defined as:
$\{ x \} = x - [x]$, where [x] is the greatest integer function.
@Ray Vickson , that makes $\{ -1.4 \} = +0.6$
So,
$\text{When}~~ x \not\in I \\ \{ x \} + \{ -x \} \\ \qquad = (x - [x] ) + (-x - [-x]) \\ \qquad= -[i+f] - [-(i+f)] \\ \qquad= -(i) - (-i-1)~~\text{or}~~ -(-i-1) - (i) \\ \qquad= 1\\ ⇒ 1 = x^2 +x -6 \\ ⇒x^2 +x -7 = 0$
and that gives the other two roots.
Thank you everyone! | 2018-03-21T19:32:16 | {
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https://math.stackexchange.com/questions/1499897/write-on-my-own-my-first-mathematical-induction-proof/1500024 | # Write on my own my first mathematical induction proof
I am trying to understand how to write mathematical induction proofs. This is my first attempt.
Prove that the sum of cubic positive integers is equal to the formula $$\frac{n^2 (n+1)^2}{4}.$$ I think this means that the sum of cubic positive integers is equal to an odd number. However, let's go on proving...
1) I start by proving the base case $n=1$ and I show that the formula holds.
2) I assume than any number $k$ other than $1$, which appartains at $N$, holds for the formula and I write the same formula but with $k$ which replaces $n$.
3) For mathematical induction, I assume that the formula holds also for $k+1$ = $n$ So, the left side of the equation should be:
$$\sum^{k+1}_{i=1} i^3 = 1^3 + 2^3 + 3^3 + ... + (k+1)^3$$
I am wondering about which one of these 2 forms (equivalents, I think) should have the right side :
this one, with $k+1$ in place of the $n$ of the original formula / or $k$ in the second version: $\frac{(k+1)^2[(k+1)+1]^2}{4}$ or this one: $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ ?
I think that, in order for the proof to be convincing, we should write an equivalent statement for the original form of the formula, namely $$\sum^{n}_{i=1} i^3= \frac{n^2(n+1)^2}{4}$$ and perhaps we do it by showing that after algebraic passages $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ is equal to $\frac{(k+1)^2[(k+1)+1]^2}{4}$ ?
Sorry for my soliloquy but it helps to understand and I would appreciate confirmation from you!
• "appartains at $N$" what on earth does that mean / have to do with anything? – MichaelChirico Oct 27 '15 at 12:04
• I am confused by your aside about odd numbers. 1 + 8 + 27 is not an odd number. – Eric Lippert Oct 27 '15 at 14:11
• As an aside, I've often found it helpful in induction to think about the inductive step first, and then the base case. In the inductive step, you show that if it works for n, then it works for n+1. Then you have the motivation for the base case: how do I know whether it works for, say, 17? Well, it would work for 17 if it works for 16, and it would work for 16 if it works for 15, and ..., all the way back to 1. Now just show it for 1 (or whatever the appropriate base case is), and you're all set. – Joshua Taylor Oct 27 '15 at 16:09
• @MichaelChirico I guess "appartains at $N$" is an attempt to translate appartient à $N$, which is French for belongs to $\mathbb{N}$. – yoann Oct 27 '15 at 20:55
• @yoann neat! $N$ instead of $\mathbb{N}$ threw me off the scent of that possibility. – MichaelChirico Oct 27 '15 at 21:51
Your inductive assumption is such that the formula marked $\color{red}{\mathrm{red}}$ (several lines below) holds for $i=k$: $$\sum^{i=k}_{i=1} i^3=\frac{k^2 (k+1)^2}{4}$$
You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^3=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$ To do this you cannot use: $$\sum^{i=n}_{i=1} i^3=\color{red}{\frac{n^2 (n+1)^2}{4}}$$ as this is what you are trying to prove.
So what you do instead is notice that: $$\sum^{i=k+1}_{i=1} i^3= \underbrace{\frac{k^2 (k+1)^2}{4}}_{\text{sum of k terms}} + \underbrace{(k+1)^3}_{\text{(k+1)th term}}$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{1}{4}k^2+(k+1)\right)$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{k^2+4k+4}{4}\right)$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{(k+2)^2}{4}\right)=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$
Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.
• @Alwayslearning Good to see you making a positive start with inductive techniques, well done. I have added something of my own here, I was a little late for the party, and I think the other users have already given you some good answers. – BLAZE Oct 27 '15 at 13:03
• Yes, I have understood it already but your answer is very precise and well written and targets exactly what was my doubt! However, I am just writing another question about another proof by induction;) – Always learning Oct 27 '15 at 13:13
• @Alwayslearning Thank you, that means a lot; basically your logic was totally correct when you asked "perhaps we do it by showing that after algebraic passages $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ is equal to $\frac{(k+1)^2[(k+1)+1]^2}{4}$?" You were right with that approach, you need to have more confidence in your ability. Well spotted :) – BLAZE Oct 27 '15 at 13:16
• To me this seems rather unfortunate to say that some formula holds for $i=k$, when the formula does not contain $i$ as a free variable, but as index of summation. – Martin Sleziak Nov 30 '15 at 11:41
• @Martin Sorry it took me so long to reply. By "unfortunate", I'm guessing you mean 'sub-optimal' or maybe even 'incorrect'. I think understand what you are saying about referring to summation index as if it was a variable in a formula. This does seem a little odd, but I simply could not think of any other way to word it more appropriately, and even more unfortunate is that this is the way I have seen induction techniques written in textbooks. Do you have any recommendations on an improved way of wording this? Thanks for pointing this out, I appreciate that. – BLAZE Dec 11 '15 at 7:47
It seems your issue is more conceptual than algebraic since you're stuck about which form of right hand side to use.
A proof by induction on a sum formula works by showing: (1) it holds in the base case, when the index is at its minimum; and (2) if it applies for the $n=k$ case, then it will also hold for the $n=k+1$ case.
With these two in hand we prove that the formula holds at any index. For example, we know it holds at $k=10$ because it holds at $k=1$ (via (1)) which implies it holds at $k=2$ (via (2)) which implies it holds at $k=3$ (via (2) again), and so on, repeatedly applying (2) until we reach 10.
We want to show that "the formula applies at $k$" implies that the formula holds at $k+1$, so our target is showing that $\sum_{i=1}^{k+1}i^3 = \frac{(k+1)^2((k+1)+1)^2}{4}$ and our ammunition is our assumption that $1^3+\ldots+k^3=\frac{k^2(k+1)^2}{4}$.
To connect the two, we notice that the left side of our target nests our assumption -- $\sum_{i=1}^{k+1}i^3 = (1^3+\ldots+k^3)+(k+1)^3$.
The rest is algebra.
• I think I have clarified all my doubts with this question! – Always learning Oct 27 '15 at 12:38
• @Michael Was it you who had the answer accepted first? In any case I felt bad as my answers in the past have been accepted then unaccepted so I know how it feels. So to counter this I have up-voted all answers in this thread (+1 for all) – BLAZE Oct 27 '15 at 14:09
• @BLAZE it wasn't, and thanks for the upvote, but the best answer should always get the tick. and yours is colorful! :p – MichaelChirico Oct 27 '15 at 14:58
You should use the second one: Suppose it holds for the first $k$ numbers. So their sum is equal to $\frac{k²(k+1)^2}{4}$. Then the first sum of the first $k+1$ is equal to $1^3 + 2^3 + 3^3 + ... + (k+1)^3=\frac{k²(k+1)^2}{4}+(k+1)^3=\frac{k²(k+1)^2}{4}+\frac{4(k+1)^3}{4}$ which is equal to $\frac{k²(k+1)^2+4(k+1)^3}{4}=\frac{(k+1)²(k²+4k+4)}{4}=\frac{(k+1)²(k+2)²}{4}$.
Which is precisely what you need.
• Thank you! But, perhaps, you could break your passages as the final ones go on the right column of the website where there are the related questions. Thus, I can't read all your passages. – Always learning Oct 27 '15 at 11:34
• Does this help? – Biouk Oct 27 '15 at 11:37
• yes, thanks. Now it is the other answer which goes on the right side of the website:) but it is written in big characters, so I can read it ;) Thank you both! I am looking at the algebraic passages in order to be sure that everything is clear to me. – Always learning Oct 27 '15 at 11:42
• After your final step, in order to make the sentence similar to the original one, I would write (k+1)^2 * [(k+1)+1]^2 / 4 thank you! – Always learning Oct 27 '15 at 12:08
You want to prove that $$\sum_{i=1}^{n}i^3 = \frac{n^2 (n+1)^2}{4}$$ using induction.
For $n=1$, $$\sum_{i=1}^{1}i^3 = 1^3=1=\frac{1^2(1+1)^2}4=1$$ So the formula does work for the base case $n=1$.
Now, assume the formula works for $n=k$ and show that this implies that the formula is correct for $n=k+1$ which will accomplish the prove by induction. Thus, \begin{align} \sum_{i=1}^{k+1}i^3 & = \left(\sum_{i=1}^{k}i^3\right) +(k+1)^3 \\ & = \frac{k^2 (k+1)^2}{4}+(k+1)^3 \\ & = \frac{k^2 (k+1)^2+4(k+1)^3}{4} \\ & = \frac{ (k+1)^2(k^2+4(k+1))}{4} \\ & = \frac{ (k+1)^2(k+2)^2}{4} \\ & = \frac{ (k+1)^2((k+1)+1)^2}{4} \end{align} and you are done.
• Thak you! But the final number is 1, not 2, I think! – Always learning Oct 27 '15 at 12:06 | 2019-10-23T14:03:07 | {
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https://math.stackexchange.com/questions/441034/all-distinct-subgroups-of-mathbbz-4-times-mathbbz-4-isomorphic-to-mat | All distinct subgroups of $\mathbb{Z}_4 \times \mathbb{Z}_4$ isomorphic to $\mathbb{Z}_4$
This question is from a past exam.
Find all distinct subgroups of $\mathbb{Z}_4 \times \mathbb{Z}_4$ isomorphic to $\mathbb{Z}_4$
Attempt/Thoughts?
Since $\mathbb{Z}_4$ is cyclic we are looking for cyclic subgroups of the given group. Then can I use the fundamental theorem of f.g.ab. groups to solve this problem?. I have a hard time imagining how the elements in $\mathbb{Z}_4\times \mathbb{Z}_4$ look like. Is there a standard way to proceed in this type of problem?.
Can somebody help?. I usually don't ask for detailed answers in here. But in this case I would really appreciate it. Thanks.
The elements of the product are all pairs $(a,b)$ with $a$ and $b$ in $\mathbb{Z}_4$. Go patiently over all the candidates for generator of the subgroup isomorphic to $\mathbb{Z}_4$.
Start with $(0,1)$. It generates a subgroup of order $4$, consisting of $(0,1)$, $(0,2)$, $(0,3)$, $(0,0)$.
Note that $(0,2)$ is no good, while $(0,3)$ generates a group isomorphic to $\mathbb{Z}_4$ that we have already.
Now continue with $(1,1)$: good. Now try $(1,2)$: good, we get in turn $(2,0)$, $(3,2)$, $(0,0)$.
Continue. Analysis will quickly speed up. Be careful not to list the same subgroup twice. And take advantage of symmetry: the analysis of the group generated by $(0,a)$ is automatic once you have dealt with the group generated by $(a,0)$.
Remark: There is nothing wrong with computing. One becomes intimately acquainted with the structures that way. After a while we find shortcuts.
• This is exactly what I needed. Get my hands dirty and do these computations. Once I see the computations, I can actually understand what nrpeterson and julien are talking about. So Thank you. – minibuffer Jul 11 '13 at 18:56
• Nicholas: Just so you know I passed my exam :). Your answers/comments helped me a great deal. This one especially. "There is nothing wrong with computation" really struck a chord with me. MSE discourages commenting just to thank someone, but I couldn't resist. Thank you for sharing your knowledge. – minibuffer Sep 23 '13 at 6:28
• You are welcome, and congratulations. The abstractions can be tremendously helpful, but one needs to remember that they are ultimately about very concrete objects. – André Nicolas Sep 23 '13 at 6:34
HINT:
You're making this problem much too hard!
Elements of $\mathbb{Z}_4\times\mathbb{Z}_4$ are of the form $(a,b)$, where $a,b\in \mathbb{Z}_4$... and addition is defined coordinate-wise.
The real question here is: if you're handed $(a,b)$, where $a,b\in\mathbb{Z}_4$, what is its order? Essentially, you need to find all elements of order 4, then partition them up in to cyclic subgroups (since no two distinct cyclic subgroups of the same order can intersect).
Now, it turns out that the order of $(a,b)$ is the least common multiple of the order of $a$ and the order of $b$. So, since elements of $\mathbb{Z}_4$ can only have order 1, 2, or 4, it must be the case that either $a$ or $b$ has order 4; and, the converse also holds.
See if you can finish it up from there. Find an element; generate its subgroup; find another element that you haven't seen so far; etc.
If you want to count them without computing any particular subgroup, you can do it as follows.
An element $(x,y)\in \mathbb{Z}_4\times \mathbb{Z}_4$ has order $4$ if and only if one of $x,y$ has order $4$. This yields $16-4=12$ order $4$ elements (counting those for which both $x,y$ have order $1$ or $2$ and subtracting them).
A cyclic group of order $4$ has two generators (each of order $4$ of course). And every order $4$ element is the generator a cyclic order $4$ subgroup.
Two cyclic subgroups of order $4$ are equal if and only if they share a generator.
So there is a natural partition of the set of order $4$ elements into pairs of generators of the same order $4$ cyclic subgroup.
So there are $\frac{12}{2}=6$ distinct cyclic order $4$ subgroups. You will find what they actually are below if you really need it.
If you want to be more explicit, note that the map (isomorphism $-Id$ of $\mathbb{Z}_4\times \mathbb{Z}_4$, actually) $(x,y)\longmapsto (4-x,4-y)$ induces an involution of the set of order $4$ elements onto itself which sends the generator of a given cyclic order $4$ group to the other generator. This will help you go over all these $6$ subgroups injectively.
Details: $(0,1)$ and $(0,3)$ will be the generators of the first cyclic subgroup of order $4$ in lexicographic order. Then $(1,0)$ and $(3,0)$ . Then $(1,1)$ and $(3,3)$. Then $(1,2)$ and $(3,2)$. Then $(1,3)$ and $(3,1)$. Then $(2,1)$ and $(2,3)$. Then...there is nothing left. | 2019-10-17T01:15:59 | {
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https://www.physicsforums.com/threads/show-that-a-is-identical.1042412/ | # Show that A is identical
• MHB
Gold Member
MHB
Hello! (Wave)
I want to prove that if a $m \times m$ matrix $A$ has rank $m$ and satisfies the condition $A^2=A$ then it will be identical.
$A^2=A \Rightarrow A^2-A=0 \Rightarrow A(A-I)=0$.
From this we get that either $A=0$ or $A=I$.
Since $A$ has rank $m$, it follows that it has $m$ non-zero rows, and so it cannot be $0$.
Thus $A=I$. Is everything right? Or could something be improved? (Thinking)
Homework Helper
MHB
Hello! (Wave)
I want to prove that if a $m \times m$ matrix $A$ has rank $m$ and satisfies the condition $A^2=A$ then it will be identical.
$A^2=A \Rightarrow A^2-A=0 \Rightarrow A(A-I)=0$.
From this we get that either $A=0$ or $A=I$.
Hey evinda!
I don't think we can draw that conclusion.
Consider $A=\begin{pmatrix}1&0\\1&0\end{pmatrix}$, what is $A(A-I)$? (Worried)
Gold Member
MHB
Hey evinda!
I don't think we can draw that conclusion.
Consider $A=\begin{pmatrix}1&0\\1&0\end{pmatrix}$, what is $A(A-I)$? (Worried)
It is $0$ although neither $A$ nor $A-I$ is $0$.
How else can we get to the conclusion that $A=I$ ? (Thinking)
Do we use the linear independence? (Thinking)
Homework Helper
MHB
It is $0$ although neither $A$ nor $A-I$ is $0$.
How else can we get to the conclusion that $A=I$ ?
Do we use the linear independence?
How about multiplying by $A^{-1}$?
$A$ is invertible isn't it? (Thinking)
Gold Member
MHB
How about multiplying by $A^{-1}$?
$A$ is invertible isn't it? (Thinking)
So is it as follows? (Thinking)
Since the rank of the $m\times m$ matrix $A$ is $m$ we have that at the echelon form the matrix has no zero-rows. This implies that at the diagonal all elements are different from zero and this implies that the determinant is different from zero, since the determinant is equal to the product of the diagonal elements. Since the determinant is not equal to zero, the matrix is invertible. So $A^{-1}$ exists.
Thus $A^2-A=0 \Rightarrow A(A-I)=0 \Rightarrow A^{-1}A(A-I)=0 \Rightarrow A-I=0 \Rightarrow A=I$.
Is it complete now? Could we improve something? (Thinking)
Homework Helper
MHB
So is it as follows?
Since the rank of the $m\times m$ matrix $A$ is $m$ we have that at the echelon form the matrix has no zero-rows. This implies that at the diagonal all elements are different from zero and this implies that the determinant is different from zero, since the determinant is equal to the product of the diagonal elements. Since the determinant is not equal to zero, the matrix is invertible. So $A^{-1}$ exists.
Thus $A^2-A=0 \Rightarrow A(A-I)=0 \Rightarrow A^{-1}A(A-I)=0 \Rightarrow A-I=0 \Rightarrow A=I$.
Is it complete now? Could we improve something?
All good. (Nod)
It may suffice to simply state that an $m\times m$ matrix of rank $m$ is invertible though.
It's a property of matrices:
If A is a square matrix (i.e., m = n), then A is invertible if and only if A has rank n (that is, A has full rank).
(Nerd)
Gold Member
MHB
All good. (Nod)
It may suffice to simply state that an $m\times m$ matrix of rank $m$ is invertible though.
It's a property of matrices:
If A is a square matrix (i.e., m = n), then A is invertible if and only if A has rank n (that is, A has full rank).
(Nerd)
I see... Thanks a lot! (Party) | 2023-03-21T05:17:58 | {
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https://math.stackexchange.com/questions/3727393/prove-x-sqrtx21 | # Prove $x<\sqrt{x^2+1}$.
Prove $$x<\sqrt{x^2+1}$$.
I am pretty sure this an easy question as the inequality seems obviously true, but I am not entirely convinced by my argument.
So I squared both sides (is this allowed?):
$$x^2, so $$0<1$$ so the inequality is obviously true.
However, I am unconvinced that this process is reversible due to the squaring, so could someone just explain whether this is correct?
• Consider the cases $x <0$ and $x \geq 0$. Jun 20, 2020 at 8:14
• You can start with $x^2 < x^2+1$, and square root both sides. Make sure to remember that $\sqrt{x^2}=|x|$ and $|x| \ge x$. Jun 20, 2020 at 8:14
• If $x<0$ your inequality is obvious, since the right hand side is $\ge 0$. If $x\ge 0$ your squaring is correct, the squared inequality is equivalent to the original one and your proof works. Jun 20, 2020 at 8:15
• Consider the 3 cases $x<0$,$x=0$,$x>0$ separately. Jun 20, 2020 at 8:15
• In fact the reverse implies that $|x|<\sqrt{x^2+1}$ from here it is easy to continue just consider two cases. Jun 20, 2020 at 8:16
You can not use squaring of the both sides directly because $$x$$ can be negative.
If so, you need to consider two cases: 1)$$x\geq0$$ and 2)$$x<0$$.
I think it's better to use a way without squaring:$$\sqrt{x^2+1}>\sqrt{x^2}=|x|\geq x.$$
By definition $$\sqrt {x^2+1}\ge0$$. If $$x <0$$, then $$x <0\le \sqrt {x^2+1} \implies x <\sqrt {x^2+1}$$. If $$x\ge0$$ we have: $$\underbrace {(\sqrt {x^2+1}+x)}_{\ge0}(\sqrt {x^2+1}-x)=1>0\implies x <\sqrt {x^2+1}.$$ | 2022-07-04T06:20:47 | {
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https://math.stackexchange.com/questions/3249127/how-does-left-sqrt23x-1-sqrt23x-2-right-become-left-frac3x-1 | # How does $\left|\sqrt{2+3x_1} - \sqrt{2+3x_2}\right|$ become $\left| \frac{3(x_1-x_2)}{\sqrt{2+3x_1} + \sqrt{2+3x_2}} \right|$?
Could someone explain to me this transformation? It is used frequently in my uni course, and I do not understand what's happening:
$$\left|\sqrt{2+3x_1} - \sqrt{2+3x_2}\right| = \left| \frac{3(x_1-x_2)}{\sqrt{2+3x_1} + \sqrt{2+3x_2}} \right|\qquad (x_1,x_2 \in [0,3))$$
Is there some obvious rule I don't see? Thanks in advance. (The transformation is used to prove Lipschitz continuity.)
• If $A_i = \sqrt{2+3x_i}$, then what is $(A_1-A_2)(A_1+A_2)$? – Don Thousand Jun 3 '19 at 2:58
Let $$A_i = \sqrt{2+3x_i}$$ for $$i=1,2$$ as suggested by Don Thousand in the comments. Then your original expression is $$|A_1 - A_2|$$.
Notice, then:
$$A_1 - A_2 = (A_1-A_2) \cdot \underbrace{\frac{A_1 + A_2}{A_1 + A_2}}_{=1} = \frac{A_1^2 - A_2^2}{A_1+A_2}$$
$$A_i^2 = 2+3x_i$$ for both $$i$$ indices and thus the difference in the numerator is $$2+3x_1 - 2 - 3x_2$$, or, more simply, $$3(x_1 - x_2)$$.
Then
$$A_1 - A_2 = \cdots = \frac{A_1^2 - A_2^2}{A_1+A_2} = \frac{3(x_1 - x_2)}{\sqrt{2+3x_1} + \sqrt{2+3x_2}}$$
Throw in the absolute values and you're done! | 2020-02-21T21:57:10 | {
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https://mathhelpboards.com/threads/a-nice-problem-2.472/ | # a nice problem #2
#### sbhatnagar
##### Active member
Find the limit.
$$\lim_{n \to \infty} \left\{ \left( 1+\frac{1^2}{n^2}\right)\left( 1+\frac{2^2}{n^2}\right)\left( 1+\frac{3^2}{n^2}\right) \cdots \left( 1+\frac{n^2}{n^2}\right)\right\}^{\dfrac{1}{n}}$$
#### Fernando Revilla
##### Well-known member
MHB Math Helper
If $L$ is the limit, then $\log L=\displaystyle\lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1}^n \log \left(1+(k/n)^2\right)=\displaystyle\int_0^1 \log (1+x^2)\;dx$ .
#### sbhatnagar
##### Active member
If $L$ is the limit, then $\log L=\displaystyle\lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1}^n \log \left(1+(k/n)^2\right)=\displaystyle\int_0^1 \log (1+x^2)\;dx$ .
Yeah that's right. The integral can be further simplified. $$\displaystyle \int_{0}^{1}\ln(1+x^2) dx= \cdots=\frac{\pi}{2}-2+\ln(2)$$.
Therefore $$\displaystyle L=2+e^{\dfrac{\pi}{2}-2}$$.
#### Sherlock
##### Member
The integral is just boring old integration by parts (the real key part was realizing that it's a Riemann sum). I'm sure that's why Dr Revilla didn't bother to provide us with the calculation. But we can make it interesting in such a way that the given problem is in the end done by writing it as sum then as an integral, then as a sum, and then finally as an integral. You don't believe me? Watch this:
\displaystyle \begin{aligned} & \begin{aligned} \log (\ell) & = \lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1} ^n \log \left(1+(k/n)^2\right) \\& = \int_0^1 \log (1+x^2)\;dx \\& = \int_0^1\sum_{k \ge 0}\frac{(-1)^kx^{2k+2}}{k+1}\;{dx} \\& = \sum_{k \ge 0}\int_{0}^{1}\frac{(-1)^kx^{2k+2}}{k+1}\;{dx} \\& = \sum_{k \ge 0}\bigg[\frac{(-1)^kx^{2k+3}}{(k+1)(2k+3)}\bigg]_0^1 \\& = \sum_{k \ge 0}\frac{(-1)^k}{(k+1)(2k+3)} \\& = \sum_{k \ge 0}\frac{(-1)^k(2k+3)-2(-1)^k(k+1)}{(k+1)(2k+3)} \\& = \sum_{k\ge 0}\bigg(\frac{(-1)^k}{k+1}-\frac{2(-1)^k}{2k+3}\bigg) \\& = \sum_{k\ge 0}\int_0^1\bigg((-1)^kx^{k}-2(-1)^kx^{2k+2}\bigg)\;{dx} \\& = \int_0^1\sum_{k\ge 0}\bigg( (-1)^kx^{k}-2(-1)^kx^{2k+2}\bigg)\;{dx} \\&= \int_0^1\bigg(\frac{1}{1+x}-\frac{2x^2}{1+x^2}\bigg)\;{dx} \\& = \int_0^1\bigg(\frac{1}{1+x}-\frac{2x^2+2-2}{1+x^2}\bigg)\;{dx} \\& = \int_0^1\bigg(\frac{1}{1+x}-2+\frac{2}{1+x^2}\bigg)\;{dx} \\& = \ln|1+x|-2+2\tan^{-1}{x}\bigg|_{0}^{1} \\& = \ln(2)-2+\frac{\pi}{2} \end{aligned} \\& \therefore ~ \ell = \exp\left(\ln{2}-2+\frac{\pi}{2}\right). \end{aligned}
#### Krizalid
##### Active member
I like the double integration approach:
\begin{aligned} \int_{0}^{1}{\ln (1+{{x}^{2}})\,dx}&=\int_{0}^{1}{\int_{0}^{{{x}^{2}}}{\frac{dt\,dx}{1+t}}} \\ & =\int_{0}^{1}{\int_{\sqrt{t}}^{1}{\frac{dx\,dt}{1+t}}} \\ & =\int_{0}^{1}{\frac{1-\sqrt{t}}{1+t}\,dt} \\ & =2\int_{0}^{1}{\frac{t-{{t}^{2}}}{1+{{t}^{2}}}\,dt} \\ & =-2+\frac{\pi }{2}+\ln 2. \end{aligned}
Now, integration by parts it's faster than any of these methods, perhaps you may consider it a boring way, but it's still valid and saves a lot of time. | 2020-09-29T16:48:41 | {
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https://math.stackexchange.com/questions/253696/can-a-limit-of-an-integral-be-moved-inside-the-integral | # Can a limit of an integral be moved inside the integral?
After coming across this question: How to verify this limit, I have the following question:
When taking the limit of an integral, is it valid to move the limit inside the integral, providing the limit does not affect the limits of integration?
For instance in the question, the OP is trying to determine that:
$$\lim_{n\to\infty}\int_{0}^{1}{\frac{dx}{(1+\frac{x}{n})^{n}}}=1-\rm{e}^{-1}$$
The answers to the question involve evaluating the integral and then taking the limit to prove the result; but I was wondering if it would be valid to move the integral inside the limit, that is:
$$\lim_{n\to\infty}{\int_{0}^{1}{\frac{dx}{(1+\frac{x}{n})^{n}}}}=\int_{0}^{1}{\lim_{n\to\infty}\frac{dx}{(1+\frac{x}{n})^{n}}}=\int_{0}^{1}{\frac{dx}{{\rm{e}}^x}}=\rm{e}^{0}-\rm{e}^{-1}=1-\rm{e}^{-1}$$
As required. So is this a valid technique, or is it just coincidental that this works?
Check :
\displaystyle\begin{align}\lim\limits_{n\to\infty}\int_0^1&\dfrac{dx}{\left(1+\dfrac xn\right)^n}=\lim\limits_{n\to\infty}\left[\dfrac{-n}{n-1}\left(1+\dfrac xn\right)^{1-n}\,\right]_0^1=\\ &=\lim\limits_{n\to\infty}\dfrac{-n}{n-1}\left[\left(1+\dfrac1n\right)^{1-n}-1\right]=\\ &=\lim\limits_{n\to\infty}\dfrac n{n-1}\left[1-\left(1+\dfrac1n\right)^{1-n}\,\right]=\\ &=\lim\limits_{n\to\infty}\dfrac n{n-1}\left[1-\dfrac{\left(1+\dfrac1n\right)}{\left(1+\dfrac1n\right)^n}\,\right]=\\ &=1\cdot\left[1-\dfrac1{\rm{e}}\right]=\\ &=1-\rm{e}^{-1} \end{align}
\displaystyle\begin{align}\int_0^1 \lim\limits_{n\to\infty}\dfrac{dx}{\left(1+\dfrac xn\right)^n}&=\int_0^1\dfrac{dx}{{\rm{e}}^x}=\bigg[\!-{\rm{e}}^{-x}\bigg]_0^1=\\ &=\rm{e}^0-\rm{e}^{-1}=1-\rm{e}^{-1} \end{align}
• If $\: f_n \longrightarrow f \:$ uniformly and the interval is finite, then you are allowed to do so. $\hspace{1.2 in}$
– user57159
Dec 8, 2012 at 11:31
Taking the limit inside the integral is not always allowed. There are several theorems that allow you to do so. The major ones being Lebesgue dominated convergence theorem and Monotone convergence theorem.
The uniform convergence mentioned in the comments is a special case of Dominated convergence theorem.
• If you know that the limit function is integrable, uniform convergence is enough. DCT and MCT go a step further and ensure that the limit is also integrable. Feb 4, 2019 at 21:53
• Can I get an explaination to why it is a special case of the Dominated convergence theorem?
– Algo
May 10 at 19:45
The limit can be moved inside the integral if the convergence of the integrand is uniform. In our case if $$f_n(x)=(1+\frac{x}{n})^{-n}$$, then $$\lim_{n\to +\infty}f_n(x)=e^{-x}=f(x)$$ We need to show that on $$[0,1]$$, $$\left\|f_n-f \right\|_{\infty}\to 0$$ But $$\left\|f_n-f \right\|_{\infty}=\sup_{x\in [0,1]}\left|f_n(x)-f(x)\right|= \sup_{x\in [0,1]}\left|(1+\frac{x}{n})^{-n}-e^{-x}\right|$$ We need to determine the maximum of $$g_n(x)=(1+\frac{x}{n})^{-n}-e^{-x}$$ on $$[0,1]$$ or at least show it converges to $$0$$.
$$g_n(0)=0$$, $$g_n(1)=(1+\frac1n)^{-n}-e^{-1}$$ and $$g_n^{\prime}(x_0)=0\Leftrightarrow (1+\frac{x_0}{n})^{-n-1}=e^{-x_0}$$ Whenever the latter is true, $$g_n(x_0)=\frac{x_0}{n}e^{-x_0}$$. Therefore, $$\left\|f_n-f \right\|_{\infty}=\max_{x\in [0,1]}\left|g_n(x)\right|=\max\left\{g_n(0),g_n(1),g_n(x_0)\right\}=\max\left\{0,(1+\frac1n)^{-n}-e^{-1},\frac{x_0}{n}e^{-x_0}\right\}\to 0$$ as $$n\to +\infty$$. Uniform convergence on $$[0,1]$$ is proven and the limit-integral interchange can be done
• Why the $g_n(x_0)$ is not equal to $e^{-x_0}\frac{x_0}{n}$ ? Sep 17, 2018 at 11:11
• Hi @Nameless! I'm obviously a newbie here. So pardon me to ask this. Why is it sufficient to show that $\| f_n-f\|_{\infty} \rightarrow 0$ to show that $f_n$ converges uniformly? :) Feb 28, 2019 at 2:56
• The expression $\|f_n - f\|_\infty \to 0$ is equivalent with $\forall \epsilon>0 :\exists N \in \mathbb N: \forall n > N: \sup_{x \in \mathbb R} |f_n(x) - f(x)| < \epsilon$. By the definition of supremum, we get that $|f_n(x) - f(x)| < \epsilon$ for all $x \in \mathbb R$. In other words, for all $\epsilon>0$, we find an $N$, independent of $x$ such that if $n>N$, $|f_n(x) - f(x) | < \epsilon$ for all $x \in \mathbb R$, which is exactly the definition of uniform convergence. Mar 13, 2019 at 14:13
There's a recent paper that exactly answers this question.
Kamihigashi, T. Interchanging a limit and an integral: necessary and sufficient conditions. J Inequal Appl 2020, 243 (2020).
https://doi.org/10.1186/s13660-020-02502-w
The main result gives a necessary and sufficient condition under which the limit can be moved inside the integral. The exact condition is somewhat complicated, but it's strictly weaker than uniform integrability.
• Interesting. Doesn't seem as easy to apply, but very interesting that it's both necessary and sufficient. And there's a nice example. If I were an analyst I'd be all over this thing. Apr 26, 2021 at 5:35 | 2022-05-29T00:26:30 | {
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https://math.stackexchange.com/questions/1976250/what-is-the-maximum-distance-of-k-points-in-an-n-dimensional-hypercube/3738931 | # What is the maximum distance of k points in an n-dimensional hypercube?
For this question, I'm thinking only about the euclidean distance:
Let $p_1 = (x_1^{(1)}, \dots, x_n^{(1)})$ and $p_2 = (x_1^{(2)}, \dots, x_n^{(2)})$ be $n$-dimensional points. The euclidean distance of $p_1$ and $p_2$ is $$d(p_1, p_2) = \sqrt{\sum_{i=1}^n {\left (x_i^{(1)} - x_i^{(2)} \right )}^2}$$
Lets say $\alpha(n, k)$ is the maximum distance for $k$ points in the unit-hypercube of $\mathbb{R}^n$:
$$\alpha(n, k) = \max( \left \{\min(d(p_i, p_j))| (p_1, \dots, p_k) \in [0, 1]^n, i, j \in \{1, \dots, k\} \right \})$$
## $n = 1$
• $\alpha(1, k = 2 = 2^n) = 1$
• $\alpha(1, k = 3)= 0.5$
• $\alpha(1, k) = \frac{1}{k-1}$
## $n = 2$
• $\alpha(2, k = 2) = \sqrt{2}$: The maximum distance is the diagonal and hence $\sqrt{1+1}$
• $\alpha(2, k = 3)=?$
• $\alpha(2, k = 4 = 2^n) = 1$: Putting each point at the corners of the square.
• $\alpha(2, k = 5)$: I guess like 4 but with one point in the center? (hence $\frac{\sqrt 2}{2}$?)
## n = 3
• $\alpha(3, k = 2) = \sqrt{3}$: The diagonal again and hence $\sqrt{1+1+1}$
• $\alpha(3, k = 2^n)$: The corners again and hence 1
## Arbitrary $n$
• $\alpha(n, k=2) = \sqrt{n}$
• $\alpha(n, 2^n) = 1$
What is $\alpha(n, k)$?
• if $k\leq n$ isn't it sufficent to consider the graph made up by the corners of the hypercube? – tired Oct 19 '16 at 20:33
• @tired: I'm not sure. This would mean $\alpha(2, 3) = 1$, but I'm relatively certain that you could place the points on the edges (not the corners) and get a bigger distance. – Martin Thoma Oct 19 '16 at 20:40
• @tired: No. This would mean $\alpha(2, 3) = 1$. But $p_1 = (0, 0.5)$, $p_2 = (0, 0.75)$, $p_3 = (1, 1)$ has a bigger distance than 1. – Martin Thoma Oct 19 '16 at 20:49
• Am I the only one not grasping how the (maximum) distance between $k>2$ points is defined here? – Jack D'Aurizio Oct 19 '16 at 20:51
• @JackD'Aurizio I've added a definition of the distance. I hope that helps. – Martin Thoma Oct 19 '16 at 21:01
(This should be a comment but I'll post it as an answer since I don't have enough reputation to comment.)
I'd like to answer the case with $n = 2$ and $k = 3$. The proof is really simple and can be found geometrically, if you assume two facts:
• the first point is on a vertex of the 2D hypercube, and the two others are on the opposite edges. It makes sense all 3 points should be on the edges to maximize distance.
• the resulting triangle is equilateral. This also makes sense, as in a non-equilateral triangle, one of the sides would have a smaller length and thus penalize the minimal distance between the points.
The triangle will look like this:
Equilateral triangle inside unit square
Solving for $x$ (using the fact than the triangle is equilateral), we find $x = 2-\sqrt{3} \approx 0.2679$, and finally:
$$a = \alpha(2, k=3) = \sqrt{6} - \sqrt{2} \approx 1.035276\dots$$
But solving for $\alpha(n, k)$ in the general case seems challenging and I did not find a solution on the internet, interesting problem!
Let's start with the $$2$$D case.
Consider to have $$m$$ circles of radius $$r$$.
Packing them inside a square $$C_r$$ of size $$(1+2r) \times (1+2r)$$, will be equvalent to ask that their centers be inside the unit square and at mutual distance $$r \le d$$.
Then your question is equivalent to:
what is the maximum $$r$$ such that it is still possible to pack $$m$$ circles in the square $$C_r$$ ?
Clearly the same holds for any dimension $$n$$ and of course, you can size down the square to become unitary and correspondingly the radius of the balls by $$\frac{1}{1+2r}$$, and ask
in $$n$$ dimension, what is the maximum $$\rho = r/(1+2r)$$ such that it is still possible to pack $$m$$ balls of radius $$\rho$$ in the unit cube ?
and that is a typical Packing problem, whose difficulty is well known.
Some results have been reached and published in specialized literature and sites, and some of them are proved.
For example, for $$n = 2$$ in this site it is stated that the minimum side of the square containing $$3$$ unitary circles is $$s = 2 + {{\sqrt 2 + \sqrt 6 } \over 2}$$ then the corresponding radius for your problem will be $${s \over 1} = {{1 + 2r} \over r}\quad \Rightarrow \quad r = {1 \over {s - 2}} = {2 \over {\sqrt 2 + \sqrt 6 }} = 0.5176 \ldots$$ | 2021-01-20T20:41:33 | {
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https://mathematica.stackexchange.com/questions/157464/confusion-on-using-lognormaldistribution-in-mathematica | # Confusion on using LognormalDistribution in Mathematica
I would like to fit a LogNormalDistribution[μ,σ] based on some data that I have. Roughly speaking, would I need to take the Log of these data's y-values before calculating the mean and standard deviation in order to determine $\mu$ and $\sigma$ above, or can I calculate $\mu$ and $\sigma$ directly from the data and supply these values to Mathematica's LogNormalDistribution-function?
Thanks.
Update: Added in @gwr 's suggestion about using SeedRandom so the one can obtain exactly the same answers as below.
It's not clear that your question has anything to do with Mathematica but rather should you use maximum likelihood or method of moments (which would be best asked at CrossValidated).
Consider a random sample from a log normal distribution:
SeedRandom[12345]
n = 100;
data = RandomVariate[LogNormalDistribution[1, 0.3], n];
One can use the FindDistributionParameters function or brute force it:
(* Maximum likelihood *)
FindDistributionParameters[data, LogNormalDistribution[μ, σ]]
{μ -> 0.940761, σ -> 0.291374}
(* Maximum likelihood the direct way *)
Mean[Log[data]]
(* 0.940761 *)
StandardDeviation[Log[data]] ((n - 1)/n)^0.5
(* 0.291374 *)
If you don't want to take the logs of the data, then consider the method of moments where you equate the sample moments to the population moments:
$$E(X)=e^{\mu+\sigma^2/2}$$ $$V(X)=e^{2\mu} e^{\sigma^2}(e^{\sigma^2}-1)$$
(* Method of moments *)
FindDistributionParameters[data, LogNormalDistribution[μ, σ], ParameterEstimator -> "MethodOfMoments"]
(* {μ -> 0.939427, σ -> 0.298547} *)
(* Method of moments the direct way *)
FindRoot[{Mean[data] == Exp[μ + σ^2/2],
Variance[data] (n - 1)/n == Exp[2 μ] Exp[σ^2] (Exp[σ^2] - 1)},
{{μ, Mean[Log[data]]}, {σ, StandardDeviation[data]}}]
(* {μ -> 0.939427, σ -> 0.298547} *)
As to which method (or some other method) is best, that would be a question for CrossValidated.
• Nice answer. I would suggest to use a defined SeedRandom when using RandomVariate so as to make your results replicable? – gwr Oct 11 '17 at 18:07
• @gwr Good suggestion! In fact, I thing I might open a question about that: Does the same SeedRandom selection get the same sequence of random numbers among versions of Mathematica and among different operating systems. – JimB Oct 11 '17 at 18:53
• I just tested RandomVariate[ NormalDistribution[ ] , 1000 ] given the same seed: Versions 10.4, 11.1.1, and 11.2 give the exact same values, which I confirmed by using Hash. – gwr Oct 12 '17 at 14:38
There are slight variations in the distribution obtained depending on the specific approach taken
Generating data
SeedRandom[0]
data = RandomVariate[LogNormalDistribution[3, 1.5], 1000];
FindDistribution will conclude that this data has a LogNormalDistribution
(dist1 = FindDistribution[data]) // InputForm
(* LogNormalDistribution[
2.875997585829534, 1.5560895409245938] *)
However, if you specify that it is a LogNormalDistribution
(dist2 = FindDistribution[data,
TargetFunctions -> {LogNormalDistribution}]) // InputForm
(* LogNormalDistribution[
2.904567956011853, 1.5056882462684804] *)
To force a fit to a LogNormalDistribution using FindDistributionParameters
(dist4 = LogNormalDistribution[m, s] /.
FindDistributionParameters[data,
LogNormalDistribution[m, s]]) // InputForm
(* LogNormalDistribution[
2.904567990925176, 1.505688384471444] *)
This is equivalent to
(dist5 = LogNormalDistribution[m, s] /.
FindDistributionParameters[Log[data],
NormalDistribution[m, s]]) // InputForm
(* LogNormalDistribution[
2.904567990925176, 1.505688384471444] *)
dist4 === dist5
(* True *)
Using the Mean and StandardDeviation of the Log of the data
(dist6 = LogNormalDistribution @@ (#[Log[data]] & /@ {Mean,
StandardDeviation})) // InputForm
(* LogNormalDistribution[
2.904567990925176, 1.5064417937677634] *)
Using TransformedDistribution on the Log of the data
(dist7 = TransformedDistribution[E^x,
x \[Distributed] FindDistribution[Log[data]]]) // InputForm
(* LogNormalDistribution[
2.904567990925176, 1.5056883844714437] *)
(dist8 = TransformedDistribution[E^x,
x \[Distributed]
FindDistribution[Log[data],
TargetFunctions -> {NormalDistribution}]]) // InputForm
(* LogNormalDistribution[
2.904567941143856, 1.505688383284438] *) | 2020-02-19T10:15:10 | {
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https://math.stackexchange.com/questions/1590780/find-all-solutions-to-the-functional-equation-fxy-fy-fracxyxy | Find all solutions to the functional equation $f(x+y)-f(y)=\frac{x}{y(x+y)}$
Find all solutions to the functional equation $f(x+y)-f(y)=\cfrac{x}{y(x+y)}$
I've tried the substitution technique but I didn't really get something useful.
For $y=1$ I have
$F(x+1)-F(1)=\cfrac{x}{x+1}$
A pattern I've found in this example is that if I iterate $n$ times the function $g(x)=\cfrac{x}{x+1}$ I have that $g^n(x)=\cfrac{x}{nx +1}$ ,which may be a clue about the general behaviour of the function ( ?) .
I am really kinda of clueless ,it seems like the problem is calling some slick way of solving it.
Can you guys give me a hint ?
• What can $x,y$ be? Anything as long as $y(x+y) \neq 0$? – wythagoras Dec 27 '15 at 17:37
• As the question is stated in my book,it doesnt make any restrictions . – Mr. Y Dec 27 '15 at 17:39
• Then we should assume that $x,y$ are real and $y\neq0$, $x+y \neq 0$. – wythagoras Dec 27 '15 at 17:41
Hint: Let $x=1-y$, hence $x+y=1$.
Then we get $$f(1)-f(y)=\frac{1-y}{y}$$
I will include the full solution in a spoiler, since you only asked for a hint.
This gives $f(0)=c$ and $f(1)=d$, then $f(y)=\frac{y-1}{y}+d$ for all other $y$.
This simplifies to $f(0)=c$ and $f(y)=\frac{y-1}{y}+d$ for all other $y$. We have to check whether every such function satisfies. Note that neither $y$ nor $x+y$ can ever be zero, so $f(0)$ can indeed be anything. Now we have to check the other values: Does
$$\left(\frac{x+y-1}{x+y} + d\right) - \left(\frac{y-1}{y} +d \right) = \frac{y}{y(x+y)} ?$$ $$\frac{y(x+y-1)-(y-1)(x+y)}{y(x+y)} = \frac{y}{y(x+y)} ?$$
It turns out that the functions do satisfy the functional equation.
• Give me some time to fully comprehend the meaning of it. – Mr. Y Dec 27 '15 at 17:45
• @Mr.Y Sorry I mistyped it. Now the equation is the correct one and you should now be able to solve it. There are a few tricky parts left. – wythagoras Dec 27 '15 at 17:48
• If I rearrange your hint I have that $f(1)-f(y)=\cfrac{1}{y}-1$ from which I see ,matching terms,that the requested function is $f(y)=-\cfrac{1}{y}$ since $y$ can be any value.Is this a bad way to extrapolate the solution ? – Mr. Y Dec 27 '15 at 18:08
• @wythagoras: Why are you entitled to assume that $x +y = 1$? – Geoff Robinson Dec 27 '15 at 18:14
• @GeoffRobinson We may assume that $x=1-y$ since $x,y$ can be anything but those satisfying $x+y=0$ and $y=0$. Rearranging $x=1-y$ gives $x+y=1$. – wythagoras Dec 27 '15 at 18:57
$f(x+y)-f(y) =\cfrac{x}{y(x+y)} =\frac1{y}-\frac1{x+y}$ so $f(x+y)+\frac1{x+y} =f(y)+\frac1{y}$.
Therefore, $f(x)+\frac1{x}$ is constant, so $f(x) =d-\frac1{x}$ for some $d$.
Substituting this, the $d$s cancel out, so any $d$ works, and the solution is $f(x) =d-\frac1{x}$ for any $d$.
• This is also really neat.Thank you marty cohen.I love when I get many beautiful answers on one question.I wish I could accept all of your answers. :) Thanks again ! – Mr. Y Dec 27 '15 at 21:13
Another approach: I assume that $f$ is a function of a single real variable.Write the defining equation as $\frac{f(y+x) - f(y)}{(y+x)- y} = \frac{1}{y(x+y)}$ (for $x,y,(x+y) \neq 0).$ Take the limit as $x \to 0$. On the one hand this limit is $\frac{1}{y^{2}}.$ On the other hand, the limit is, by definition, the derivative $f^{\prime}(y)$, (and we have proved that this exists for $y \neq 0$). An antiderivative of $\frac{1}{y^{2}}$ is $\frac{-1}{y}$. Hence $f(y) = \frac{-1}{y} + C$ for a constant $C$, as long as $y \neq 0$. (Strictly speaking this proves that any function $f$ which satisfies the equation has to be of the described form. It should be checked that such functions satisfy the equation- but they do).
• This is really a total different way to look at the problem,really neat !!Thank you Geoff Robinson! – Mr. Y Dec 27 '15 at 19:51
• The answer by Marty Cohen is more direct though! – Geoff Robinson Dec 28 '15 at 1:53 | 2019-08-24T15:23:45 | {
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http://math.stackexchange.com/questions/73250/how-to-evaluate-int-fracdx12-cos-x | # How to Evaluate $\int \! \frac{dx}{1+2\cos x}$ ? [duplicate]
Possible Duplicate:
How do you integrate $\int \frac{1}{a + \cos x} dx$?
I have come across this integral and I tried various methods of solving. The thing that gets in the way is the constant $2$ on the $\cos(x)$ term. I tried the conjugate (works without the 2$\cos x$), Weierstrass Substitution (not sure if I was applying it correctly), and others. Is there a way to solve this integral elegantly or some unknown (sneaky) trick when you come across families of similar integrals as this one?:
$$\int \! \frac{dx}{1+2\cos x}$$
-
## marked as duplicate by Willie WongJun 19 '12 at 12:50
Weierstrass substitution works for this integral, and it's not even that messy to work with.
Substitute $\tan \frac{x}{2} = t$, so that $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2dt}{1+t^2}$. Then the integral reduces to $$\int \frac{dx}{1+2 \cos x} = \int \frac{\frac{2}{1+t^2}}{1 + \frac{2(1-t^2)}{1+t^2}} dt = \int \frac{2}{1+t^2 + 2 - 2t^2} dt = \int \frac{2}{3-t^2} dt.$$ To evaluate the final integral, we can use the method of partial fractions: $$\frac{2}{3-t^2} = \frac{1}{\sqrt{3}} \left( \frac{1}{t + \sqrt{3}} - \frac{1}{t - \sqrt{3}} \right).$$
-
Thanks. Seems I did the same exact thing, I think I just messed up with the simplification after making the $\cos x$ and $\mathrm{d}x$ substitution. Good work. I will go back and make sure I get $\int \! \frac{2}{3-t^2}\,\mathrm{d}t$. Do you not get $-\int \! \frac{2}{t^2-3}$ after simplifying? Is there something I'm not seeing you pulled out? – night owl Oct 17 '11 at 5:14
@nightowl The expression $\dfrac{2}{3-t^2}$ is the same as: $\dfrac{-2}{t^2-3}$. – Srivatsan Oct 17 '11 at 5:15
@nightowl I have elaborated the relevant step a little. Hope that helps. – Srivatsan Oct 17 '11 at 5:27
$I=\int \frac{\mathrm{d}x}{1+2\cos x}$
$=\int\frac{\mathrm{d}x}{\sin^2(x/2)+\cos^2(x/2)+2(\cos^2(x/2)-\sin^2(x/2))}$
$=\int \frac{\mathrm{d}x}{3\cos^2(x/2)-\sin^2(x/2)}$
Multiply the Nr and the Dr of the integrand by $\sec^2 (x/2)$.
You will get:
$\int \frac{\sec^2(x/2)\mathrm{d}x}{3-\tan^2(x/2)}$
Substitution:
$z=\tan(x/2)$
$\mathrm{d}z=1/2 \sec^2(x/2) \mathrm{d}x$
Therefore,
Integral=$\int \frac {2\mathrm{d}z}{3-z^2}$
$=\int \frac{1}{\sqrt{3}}(\frac{1}{\sqrt{3}+z}+\frac{1}{\sqrt{3}-z})\mathrm{d}z$
-
Generalization:
Let's consider $\cos x = \frac{1-t^2}{1+t^2}; t = \tan\frac{x}{2}; dx=\frac{2}{1+t^2} dt.$ Then we get that our integral becomes:
$$J = \int \frac{2dt}{(a+b)+(a-b) t^2}$$
I. For the case $a>b$, consider $a+b=u^2$ and $a-b=v^2$, and obtain that: $$J = 2\int \frac{dt}{u^2+v^2 t^2}=\frac{2}{uv} \arctan\frac{vt}{u} +C.$$ Turning back to our notation we get: $$I=\frac{2}{\sqrt{a^2-b^2}} \arctan\left(\sqrt{\frac{a-b}{a+b}} \tan\frac{x}{2} \right) + C.$$
II. For the case $a<b$, consider $a+b=u^2$ and $a-b=-v^2$, and obtain that: $$J = 2\int \frac{dt}{u^2-v^2 t^2}=\frac{1}{uv}\ln\frac{u+vt}{u-vt} \ +C.$$
Turning back again to out initial notation and have that:
$$I=\frac{2}{\sqrt{b^2-a^2}} \ln\frac{b+a \cos x + \sqrt{b^2-a^2} \sin x}{a+b \cos x} + C.$$
Also, note that $x$ must be different from ${+}/{-}\arccos(-\frac{a}{b})+2k\pi$ if $|\frac{a}{b}|\leq1$.
Q.E.D.
-
It is a nice post and all, but is it necessary to post the same answer twice? :) – Willie Wong Jun 19 '12 at 12:51
@Willie Wong: i remember the first days i came here and it wasn't easy to find things i needed. For some new users, this would be of great benefit. Moreover, the problems are different, but the generalization works on both of them. – OFFSHARING Jun 19 '12 at 13:01 | 2015-11-29T21:26:48 | {
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http://mathhelpforum.com/calculus/16316-few-more-integral-questions.html | # Thread: Few more integral questions.
1. ## Few more integral questions.
First one, $\int x^4(x^5+2)^7dx$. Is that as straight forward as it seems? Expand that out $x^{39}+14x^{34}+84x^{29}+...$ and then integrate as normal? Or should I substitute, setting $u=x^5+2$ and $\frac{du}{5}=x^4dx$?
Second one, $\int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$. No idea where to start.
2. Originally Posted by cinder
First one, $\int x^4(x^5+2)^7dx$. Is that as straight forward as it seems? Expand that out $x^{39}+14x^{34}+84x^{29}+...$ and then integrate as normal? Or should I substitute, setting $u=x^5+2$ and $\frac{du}{5}=x^4dx$?
Yes substitute,
$u=x^5+2$ because its derivative appears almost outside.
Second one, $\int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$. No idea where to start.
Hint: First substitute $t=\sqrt{\theta}$.
Hint: There is a second substitute
3. Hello, cinder!
Looks like you're a bit shaky about substitution . . .
$\int x^4(x^5+2)^7\,dx$
Let $u = x^5+ 2\quad\Rightarrow\quad du = 5x^4\,dx\quad\Rightarrow\quad dx = \frac{du}{5x^4}$
Substitute: . $\int x^4\cdot u^7\left(\frac{du}{5x^4}\right) \;=\;\frac{1}{5}\int u^7\,du \;= \;\frac{1}{40}u^8 + C$
Back-substitute: . $\frac{1}{40}(x^5 + 2)^8 + C$
$\int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2\!(\ sqrt{\theta})} d\theta$
Let $\sqrt{\theta} = u\quad\Rightarrow\quad \theta = u^2\quad\Rightarrow\quad d\theta = 2u\,du$
Substitute: . $\int\frac{\cos u}{u\sin^2\!u}(2u\,du) \;=\;2\int\frac{\cos u}{\sin^2\!u}\,du \;=\;2\int\left(\frac{1}{\sin u}\!\cdot\!\frac{\cos u}{\sin u}\right)\,du$
. . $= \;2\int\csc u\cot u\,du \;=\;-2\csc u + C$
Back-substitute: . $-2\csc(\sqrt{\theta}) + C$
4. Originally Posted by Soroban
Hello, cinder!
Looks like you're a bit shaky about substitution . . .
Yes, I'm just not sure what I should be substituting.
5. Originally Posted by cinder
First one, $\int x^4(x^5+2)^7dx$. Is that as straight forward as it seems? Expand that out $x^{39}+14x^{34}+84x^{29}+...$ and then integrate as normal? Or should I substitute, setting $u=x^5+2$ and $\frac{du}{5}=x^4dx$?
Second one, $\int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$. No idea where to start.
Observe that $5 x^4$ is the derivative of $x^5+2$ so:
$\int x^4(x^5+2)^7dx=\frac{1}{5}\int \left[\frac{d}{dx}(x^5+2)\right](x^5+2)^7dx=\frac{1}{40}(x^5+2)^8$
For:
$\int \frac{\cos(\sqrt{\theta})}{\sqrt{\theta}\sin^2(\sq rt{\theta})} d\theta$,
ask yourself what is the derivative of:
$\frac{1}{\sin(\sqrt{\theta})}$
RonL
6. For the second integral you can make the substitution $\displaystyle \sin\sqrt{\theta}=u\Rightarrow \frac{\cos\sqrt{\theta}}{\sqrt{\theta}}d\theta=2du$.
Then $\displaystyle \int\frac{2}{u^2}du=-\frac{2}{u}+C$
So $\displaystyle \int\frac{\cos\sqrt{\theta}}{\sqrt{\theta}\sin ^2\sqrt{\theta}}d\theta=-\frac{2}{\sin\sqrt{\theta}}+C$ | 2017-05-27T08:40:27 | {
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https://math.stackexchange.com/questions/4022415/is-it-possible-to-split-the-natural-numbers-into-a-finite-number-of-sets-so-that | Is it possible to split the natural numbers into a finite number of sets so that no pair of numbers within a set adds up to a square?
My attempt:
$$\lbrace6,19,30\rbrace$$ is sufficient to show that two sets are impossible.
Using a computer program with a brute force method I found that separating the numbers $$1$$ through $$85$$ into three sets is possible as shown below:
$$\lbrace1,4,6,9,13,14,17,18,20,26,28,33,34,37,41,42,49,54,56,57,62,69,70,73,76,78,81,85\rbrace$$
$$\lbrace2,5,8,10,12,21,22,25,29,30,32,38,40,45,46,48,50,53,58,61,64,65,66,72,74,77,82,84\rbrace$$
$$\lbrace3,7,11,15,16,19,23,24,27,31,35,36,39,43,44,47,51,52,55,59,60,63,67,68,71,75,79,80,83\rbrace$$
but $$1$$ through $$86$$ is impossible.
Edit:
WhatsUp in the answer below provides the set of four numbers: $$\lbrace1058, 6338, 10823, 13826\rbrace$$ with an explanation of how he got them. This is a alternative non-brute force way of showing that separating the natural numbers into three sets is impossible. In a comment of this question a set of five numbers $$\lbrace 7442, 28658,148583,177458,763442\rbrace$$ is provided by the user Bob Kadylo. This shows that four sets are impossible.
Edit 2:
In a previous version of my post I made a proof showing that the number of sets needed for Natural numbers $$1$$ to $$N$$ so that no pair of numbers in the same set sums to a square is no more than $$\lfloor\sqrt{2N-1}\rfloor$$. I realized that I can do significantly better than this. In order to explain the method that has a smaller upper bound I have to transform the problem into graph theory. An equivalent formulation is to have $$N$$ vertices labeled from $$1$$ to $$N$$. A pair of points are connected iff the two points add up to a square. Then our goal is to color the vertices using the least number of colors so that no two vertices with the same color are connected. The first step in the greedy algorithm for coloring vertices is to make a list of colors with numbers. (ex. RED-1, BLUE-2, GREEN-3, YELLOW-4, etc.) If during the process more colors are required than are on the coloring list, add more colors to the list. The next step is to pick an uncolored vertex and use the lowest color number that isn't connected that the chosen vertex. Repeat the last step until all vertices are colored. The worst case scenario is to use one more color than the degree value of the vertex with the greatest degree (or tied with the greatest degree). If each vertex that is connected to the greatest degree vertex is a different color then the greatest degree vertex has to be a different color from all of those. The vertex with the greatest degree (or tied with the greatest) is $$3$$. It has degree $$\lfloor\sqrt{N+3}\rfloor-1$$. Therefore the Number of sets (or colors) required is no more than $$\lfloor\sqrt{N+3}\rfloor$$. We can do slightly better by using brooke's theorem which states that if a graph is simple, connected, not complete, and not an odd cycle, then the upper bound of the number of colors is equal to the degree of the greatest degree vertex. This means that the new upper bound is $$\lfloor\sqrt{N+3}\rfloor-1$$ sets. This is the significant improvment from $$\lfloor\sqrt{2N-1}\rfloor$$ I mentioned at the beginning.
End edits
For each natural number $$X$$ there are $$\lfloor\sqrt{2X-1}\rfloor-\lfloor\sqrt{X}\rfloor$$ numbers that are less than $$X$$ that when summed to $$X$$ results in a square. The expression: $$\lfloor\sqrt{2X-1}\rfloor-\lfloor\sqrt{X}\rfloor$$ increases as $$X$$ gets larger, because of this fact my guess is that separating the natural numbers into a finite number of sets so that no pair of numbers in a set doesn't sum to a square is impossible.
Without bruteforcing, I find the list $$\{1058, 6338, 10823, 13826\}$$ which shows that three sets is not enough.
My approach:
I start from the equations $$\begin{eqnarray} a + b &=& u^2\\ c + d &=& v^2\\ a + c &=& x^2\\ b + d &=& y^2\\ a + d &=& m^2\\ b + c &=& n^2. \end{eqnarray}$$ The matrix has rank $$4$$, which means that there are just $$6 - 4 = 2$$ linearly independent relations among $$u^2, \dots, n^2$$. They are: $$u^2 + v^2 = x^2 + y^2 = m^2 + n^2.$$
Also, if we assume $$a < b < c < d$$, then we have $$u^2 < x^2 < m^2, n^2 < y^2 < v^2$$.
In order to get positive solutions $$a, b, c, d$$, it suffices to have $$u^2 + x^2 > n^2$$.
Now I simply take the number $$N = 5 \times 13 \times 17 \times 29$$, which can be written as the sum of two squares in many ways. Somewhere in the middle, I take out these three: $$N = 86^2 + 157^2 = 109^2 + 142^2 = 122^2 + 131^2.$$ These are my candidates of $$u^2, \dots, n^2$$.
It only remains to solve $$a, b, c, d$$ back from the original equations, which gives the list in the very beginning.
Luckily, the solutions are integers. But even if we got non-integral solutions, we could always multiply all of them by some square to clear the denominators.
I also tried to extend this method to five numbers. It became a bit messy though, so I gave up midway.
• I did some exploring with your method and I found that a set of numbers with with an odd cardinality where every pair of numbers in the set adds to a square is more "messy" to deal with than with a set of an even cardinality. There are three different ways to sort four numbers into pairs. This is why you needed a number that was the sum of two squares in three different ways. If move to six numbers, there are 15 different ways of sorting these numbers into pairs. – quantus14 Feb 12 at 14:54
• ...So if we want six numbers where any two of them sums to square, then we need a number that is the sum of three squares in 15 different ways. If we continue this pattern to eight numbers, there are 105 ways of sorting 8 numbers into pairs. So we need a number that is the sum of four squares in 105 different ways. As we increase the size of our set with $|2N|$, the number of ways required that the big number is the sum of $N$ squares increases super exponentially – quantus14 Feb 12 at 14:55
If you want no solutions to $$a+b=c^2$$ with $$a,b,c$$ all in the same set, then $$16$$ sets suffice.
The sets are $$A_i := \{n \in \mathbb{N} : n \equiv i \pmod{5}\}$$ for $$i=1,3,4$$, $$B_i := \{n \in \mathbb{N} : n = m\cdot 5^{2^{2u}(2v+1)} \text{ with } m \equiv i \pmod{5}, u,v \ge 0\}$$ for $$i=1,2,3,4$$, $$C_i := \{n \in \mathbb{N} : n = m\cdot 5^{2^{2u+1}(2v+1)} \text{ with } m \equiv i \pmod{5}, u,v \ge 0\}$$ for $$i=1,2,3,4$$, $$D_i := \{n \in \mathbb{N} : n = m\cdot 5^u+2 \text{ with } m \equiv i \pmod{5}, u \ge 1\}$$ for $$i=1,2,3,4$$, and $$E := \{2\}.$$
Where on Earth did this come from? Page $$10$$ of this. | 2021-02-27T00:33:55 | {
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https://math.stackexchange.com/questions/4051050/let-a-subseteq-mathbbrk-and-let-a-x-in-mathbbrk-mid-x-in-partiala | # Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. How do I prove that $A'$ is closed?
I was asked this simple question a short while ago.
Let $$A\subseteq\mathbb{R}^k$$, and let $$A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$$. Prove that $$A'$$ is closed.
I think I was able to explain it, but I am not sure what I claim is true. (The following is not a formal solution because I am not sure the reasoning is correct) My explanation goes as follows:
Let $$x\in A'$$. Say $$x$$ is in the interior of $$A$$, then $$x\notin\partial(A\setminus\{x\})$$ because there exists a ball $$B(x;r)\subseteq A$$ by definition. By a smilar logic if x is in the exterior of $$A$$, I claim that $$x\notin\partial(A\setminus\{x\})$$. Thus $$x\in\partial A$$ (that is $$A'\subseteq\partial A$$). From here I am quite sure it is not too hard to explain that $$A'$$ is closed.
I doubt my reasoning, but I am not sure what have I missed. Care to shed some light over the matter?
• It's not true that $A'=\partial A$ in general, and not every subset of a closed set is closed, so there's definitely still an argument to be made. It seems actually that your observations might more directly prove that the complement of $A'$, namely $\big\{ x\in\Bbb R^k\colon x\notin\partial(A\setminus\{x\}) \big\}$, is open. – Greg Martin Mar 6 at 8:07
• Note that $A' = \{x \in \Bbb R^k \mid \forall r>0: B(x,r) \cap (A\setminus \{x\} \neq \emptyset\}$. – Henno Brandsma Mar 6 at 8:44
• I think I get it! By my claims I can show that $\mathbb{R}^k{\setminus}A'$ is open, which is very similar to what @HennoBrandsma showed in his proof and pointed in his comment. Thank you for your speedy replies everyone! – Gamow Drop Mar 6 at 9:23
• $\partial A$ can be $\Bbb R^k$ and not any subset of it is closed... – Henno Brandsma Mar 6 at 9:26
• @GamowDrop: are you thinking that any subset of a closed set must be closed? That's definitely not true. For a concrete example, suppose $A$ is the unit disk in $\Bbb R^2$, so that $\partial A$ is the unit circle. The subset of $\partial A$ consisting of all points strictly about the $x$-axis is not closed. Neither is the set of points in $\partial A$ that have both coordinates rational. – Greg Martin Mar 6 at 18:55
Let $$p \notin A'$$. This means that $$\exists r>0: B(p,r) \cap A \subseteq \{p\}$$. If in fact $$B(p,r) \cap A = \emptyset$$, for any $$x \in B(p,r)$$ we have a ball $$B(x,r') \subseteq B(p,r)$$ (open balls are open sets) and this witnesses also that $$B(x,r') \cap A= \emptyset$$ and so $$x \notin A'$$. So $$B(p,r) \subseteq \Bbb R^k \setminus A'$$. If on the other hand, $$B(p,r) \cap A= \{p\}$$, we also have $$B(p,r) \subseteq \Bbb R^k \setminus A'$$: if $$x \in B(p,r)$$ and $$x \neq p$$ (WLOG) then there is a ball $$B(x,r') \subseteq B(p,r)$$ so that $$p \notin B(x,r')$$. Then $$B(x,r') \cap A = \emptyset$$ and so $$x \notin A'$$, as required. So any $$p \in \Bbb R^k \setminus A'$$ is an interior point of it, so the complement of $$A'$$ is open, and $$A'$$ is closed.
Suppose that $$(x_n)_{n\in\Bbb N}$$ is a sequence of elements of $$A'$$ which converges to some $$x\in\Bbb R^k$$; I will prove that $$x\in A'$$. Take $$r>0$$ and consider the ball $$B(x;r)$$. It contains some $$x_n$$. Since $$x_n\in A'$$, $$x_n\in\partial(A\setminus\{x\})$$. So, every ball $$B(x_n;r')$$ contains elements of $$A$$. So, take $$r'$$ so small that $$B(x_n;r')\subset B(x;r)\setminus\{x\}$$. Then $$B(x_n;r')$$ contains elements of $$A\setminus\{x\}$$. This proves that $$x\in A'$$. Therefore, since $$A'$$ contains the limit of any convergent sequence of its elements, it is a closed set.
I believe you get a better insight if you forget about $$\mathbb{R}^k$$ and balls. Let's work in a (Hausdorff) topological space $$X$$, which your case is a specialization of.
What are the points $$x$$ such that $$x\in\partial(A\setminus\{x\})$$?
First of all they don't need to belong to $$A$$. For instance, with $$A=(0,1)\subseteq\mathbb{R}$$, $$0$$ belongs to the set $$A'$$.
Let's see: $$x\in\partial(A\setminus\{x\})$$ means that every neighborhood of $$x$$ intersects both $$A\setminus\{x\}$$ and its complement. However, $$x$$ surely belongs to the complement of $$A\setminus\{x\}$$, so the condition just reads $$\textit{every neighborhood of x intersects A\setminus\{x\}}$$ which is the standard definition of limit point.
Now it's easier, isn't it? You need to show that if $$y$$ has the property that each of its open neighborhoods intersects $$A'$$, then $$y\in A'$$.
Take such an element $$y$$ and an open neighborhood $$V$$ of $$y$$. Then there exists $$x\in A'$$ such that $$x\in V$$. Since the space $$X$$ is Hausdorff, there exists a neighborhood $$U$$ of $$x$$ such that $$U\subseteq V$$ and $$y\notin U$$.
Since $$U$$ is a neighborhood of $$x$$, there exists a point $$z\in A$$ (different from $$x$$, but it's irrelevant) such that $$z\in U$$. Now necessarily $$z\ne y$$, but $$z\in V$$ and we're done. | 2021-04-20T10:17:00 | {
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http://math.stackexchange.com/questions/201883/fibonacci-number-proof | # Fibonacci Number Proof
How can I prove this statement? Would I use induction?
"Given $n \geq 11$, show that $a_n > (3/2)^{n}$. $a_n$ is the $n$th Fibonacci number."
-
Since $\sqrt[n]{a_n} \to \phi=(1+\sqrt5)/2 \approx 1.618$, there is not much improvement you can make to $3/2$. The proof given by Old John proves that $a_n> t^n$ for all $t$ such that $t+1>t^2$. The hard part is the base case for the induction. The closer $t$ gets to $\phi$, the larger the base case. For instance, for $t=1.6$ you need $n\ge72$. – lhf Sep 25 '12 at 1:15
Yes, induction is the way to go. Assume the result is true for two consecutive integers $n$ and $n+1$ and then deduce that it must be true for $n+2$. The rest should be easy, after you find 2 consecutive values for which it is definitely true.
To explain a bit more:
Assume the result is true for $n$ and for $n+1$, i.e. assume we have $a_n > (3/2)^n$ and $a_{n+1} > (3/2)^{n+1}$.
Adding these two, we get $a_{n+2} = a_{n+1} + a_n > (3/2)^{n+1} + (3/2)^n = (3/2)^n(3/2 + 1) = (3/2)^n(5/2) > (3/2)^{n+2}$
at the last step we use the fact that $5/2 > 9/4 = (3/2)^2$
Now we know that if the result is true for $n$ and $n+1$, then it follows that it is true for $n+1$ and $n+2$.
-
So should I select n = 11 and n+1 = 12? My friend mentioned something about choosing n = 11 and n-1 = 10, since the definition of Fibonacci number is F(n+1) = F(n) + F(n-1). Would this also work, or is the first option easier? – user41419 Sep 24 '12 at 23:03
For the induction step you do not need to specify the values of $n$ and $n+1$ at all. For the "starting values" you just need to select the smallest value of $n$ for which $F(n)>(3/2)^n$ and $F(n+1)>(3/2)^{n+1}$ – Old John Sep 24 '12 at 23:06
@OldJohn I think you should say more about what type of induction you have in mind. I cannot infer anything about the specific proof that you have in mind from what little you have written in your two-sentence answer. – Bill Dubuque Sep 24 '12 at 23:16
So my base case would be n = 11 and n+1 = 12, and that's all I would need to test, correct? Then how would I go about the induction step given this information? (Sorry, I've never done a problem with strong induction before) – user41419 Sep 24 '12 at 23:17
@OldJohn It would be impossible for anyone but a mindreader to indubitably infer what proof was intended from those two sentences. In any case, I am glad to see that you did elaborate. But, alas, I'm puzzled by the tone of your comments. – Bill Dubuque Sep 25 '12 at 2:36
Hint $\$ The second order recurrence for $\rm\:f(n)\:$ yields one for $\rm\:f(n)-c^n,\:$ namely, more generally, $$\begin{eqnarray}\rm &&\rm f(n\!+\!2) &=&\rm\ a\ f(n\!+\!1)\ +\ b\ f(n)\\ \Rightarrow\ &&\rm f(n\!+\!2)-c^{n+2} &=&\rm\ a\,(f(n\!+\!1)-c^{n+1}\!)\ +\ b\,(f(n)-c^n)\ -\ c^n(\color{#C00}{c^2 - a\,c -b})\end{eqnarray}$$
So we can inductively infer positivity of the LHS from positivity of the $3\,$ summands on the RHS, which follows if $\rm\:a,b,c > 0\:$ and $\rm\:\color{#C00}{f(c)} < 0\:$ for the characteristic polynomial $\rm\:\color{#C00}{f(x)\, =\, x^2 - a\,x - b}.$
In your case $\rm\:a,b,c\, =\, 1,1,3/2\, >\, 0,\:$ and $\rm\:\color{#C00}{f(c)} = (3/2)^2\!-3/2-1 =\, \color{#C00}{-1/4} < 0,\:$ so it succeeds.
- | 2015-11-27T19:47:08 | {
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http://mathhelpforum.com/geometry/154183-triangles-20-non-straight-line-dots.html | # Thread: Triangles in 20 non-straight-line "dots"
1. ## Triangles in 20 non-straight-line "dots"
Twenty points/dots are given so that the three of them are never in a straight line. How many triangles can be formed with corners(as in vertex i think?) in the dots?
2. Calculate the number of ways to choose three points from twenty.
3. C(20,3) ? But that results in a very big number that seems unlikely to be correct
4. Originally Posted by grottvald
C(20,3) ? But that results in a very big number that seems unlikely to be correct
It is not a large number at all: $\displaystyle \binom{20}{3}=\frac{20!}{3!\cdot 17!}=1140$
5. Originally Posted by grottvald
C(20,3) ? But that results in a very big number that seems unlikely to be correct
You may be thinking of non-overlapping triangles in the picture with 20 dots.
That's a different situation.
Imagine the dots are placed apart from left to right, not on a straight line.
Pick the leftmost point.
To make a triangle, you can pick any 2 of the remaining 19 dots.
The number of ways to do this is $\binom{19}{2}=171$
Therefore, there are 171 triangles that can be drawn which include the leftmost point.
If you move on to the next point to the right and exclude the point previously chosen,
then you can draw another $\binom{18}{2}=153$ triangles.
These triangles do not include any of the previous 171,
since these 153 omit the leftmost point.
Notice that these triangles typically share sides of other triangles,
but they are made of 3 distinct points, hence the triangles are being counted only once.
hence, there are $\binom{19}{2}+\binom{18}{2}+\binom{17}{2}+........ .+\binom{2}{2}=\binom{20}{3}$ triangles.
6. Originally Posted by Archie Meade
You may be thinking of non-overlapping triangles in the picture with 20 dots.
That's a different situation.
Imagine the dots are placed apart from left to right, not on a straight line.
Pick the leftmost point.
To make a triangle, you can pick any 2 of the remaining 19 dots.
The number of ways to do this is $\binom{19}{2}=171$
Therefore, there are 171 triangles that can be drawn which include the leftmost point.
If you move on to the next point to the right and exclude the point previously chosen,
then you can draw another $\binom{18}{2}=153$ triangles.
These triangles do not include any of the previous 171,
since these 153 omit the leftmost point.
Notice that these triangles typically share sides of other triangles,
but they are made of 3 distinct points, hence the triangles are being counted only once.
hence, there are $\binom{19}{2}+\binom{18}{2}+\binom{17}{2}+........ .+\binom{2}{2}=\binom{20}{3}$ triangles.
Thank you so much! You explained it in an very easy and beautiful way. Thanks again! | 2016-10-22T03:38:58 | {
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Archit143 wrote:
A club collected exactly $599 from its members. If each member contributed at least$12, what is the greatest number of members the club could have?
(A) 43
(B) 44
(C) 49
(D) 50
(E) 51
Obviously club could not have 50 or more members, since $12*50=$600>$599. What about 49? If 48 members contributes$12 ($12*48=$576) and 1 member contributed the remaining $23, then the club would have is 48+1=49. Answer: C. OR: $$12x\leq{599}$$ --> $$x\leq{49\frac{11}{12}}$$ --> $$x=49$$ (since x must be an integer). Answer: C. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 12882 Followers: 561 Kudos [?]: 158 [0], given: 0 Re: A club collected exactly$599 from its members. If each [#permalink]
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Estimate, you don't need to divide here to get the answer:
12X<=600 --> X<=50, But we rounded up, so the real number must be <50 --> 49 (C)
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Re: A club collected exactly $599 from its members. If each [#permalink] ### Show Tags 21 Jun 2016, 09:34 Archit143 wrote: A club collected exactly$599 from its members. If each member contributed at least $12, what is the greatest number of members the club could have? (A) 43 (B) 44 (C) 49 (D) 50 (E) 51 Let's first divide= 599 by 12 (the minimum amount each member could contribute) and then use the remainder to finish the problem. 599/12 = 49 R 11 This means that: 49 people x$12 + 1 person x $11 =$599
We see that if 49 members each contribute $12, someone would have to contribute the extra$11. Note that, since each member contributed at least $12, the$11 could not have come from an additional member. Therefore, the extra $11 must have been contributed by one (or more) of the existing 49 members. Regardless of who contributed the extra$11, the maximum number of members the club could have is 49.
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Re: A club collected exactly $599 from its members. If each [#permalink] 21 Jun 2016, 09:34 Similar topics Replies Last post Similar Topics: 2 A local club has between 24 and 57 members. The members of the club ca 4 21 Oct 2016, 08:54 2 A charity collected$1,199 from donors during the last month. If each 4 29 Feb 2016, 09:34
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http://mathhelpforum.com/algebra/5264-arithmetic-progression-question.html | # Math Help - Arithmetic Progression question
1. ## Arithmetic Progression question
Hey guys, I'm having problems with this question:
The sum of the first 100 terms in an arithmetic progression with first term a and common difference d is T. The sum of the first 50 even-numbered terms, i.e. 2nd, 4th, 6th, ..., 100th, is 0.5T + 100. Find the value of a in terms of T. (Answer: a = T/100 - 198)
Thanks in advance if you could help me with the above question!
2. Originally Posted by margaritas
Hey guys, I'm having problems with this question:
The sum of the first 100 terms in an arithmetic progression with first term a and common difference d is T. The sum of the first 50 even-numbered terms, i.e. 2nd, 4th, 6th, ..., 100th, is 0.5T + 100. Find the value of a in terms of T. (Answer: a = T/100 - 198)
Thanks in advance if you could help me with the above question!
You have a sum,
$a+(a+d)+(a+2d)+...+(a+99d)$
This is what I like to do since I hate formulas because it is better to understand than to remember the facts. So I am not going to use the sum formula (in fact I do not even know it!) instead rewrite this like this.
$(\underbrace{a+a+...+a}_{100 \mbox{ times}})+d(1+2+...+99)$
You should know that the sum of the first "n" integers is:
$\frac{n(n+1)}{2}$
Thus,
$100a+d\left( \frac{100(99)}{2} \right)$
Thus,
$T=100a+4950d$ (1)
---
The second sum is,
$(a+d)+(a+3d)+...+(a+99d)$
I think it is easier to express it in "sigma notation".
$\sum_{k=1}^{50}a+(2k-1)d$
Thus,
$\sum_{k=1}^{50}a+2kd-d$
Use the fact, the sum of the first "n" integers,
In sigma notation it says,
$\sum_{k=1}^n k=\frac{n(n+1)}{2}$
Thus, in this case,
$50a+2\left( \frac{50(51)}{2} \right) d - 50 d$
Note the 50 in front on the "a" and "d" because you add them 50 times.
$50a+2500d=\frac{1}{2}T+100$ (2)
Now you have two equation (1) and (2) you should be able to solve for "a" in terms of "T".
3. Thanks ThePerfectHacker for your solution but I still dont quite get it.
EDIT: Got the answer, finally! Thanks once again!
RE-EDIT: No actually, what if I would to use formulas to solve this question? How would the solution be presented?
4. Originally Posted by margaritas
RE-EDIT: No actually, what if I would to use formulas to solve this question? How would the solution be presented?
The last two equations that you solve are identical.
5. Hello, margaritas!
I had to baby-talk my way through this one.
See if you like my approach . . .
The sum of the first 100 terms in an A.P. with first term $a$ and common difference $d$ is $T.$
The sum of the first 50 even-numbered terms is $\frac{1}{2}T + 100$
Find the value of $a$ in terms of $T.$ .(Answer: $a \:= \:\frac{T}{100} - 198$)
We're expected to know the formula for the sum of the first $n$terms of an A.P.
. . . $S_n\;=\;\frac{n}{2}\left[2a + d(n-1)\right]$
The sum of the first 100 terms is $T.$
. . $S_{100} \:=\:\frac{100}{2}\left[2a + 99d\right] \:=\:T$
. . which simplifies to: . $100a + 4950d\:=\:T$ [1]
The first fifty even terms begin with $a_2 = a + d$ and have a common difference of $2d.$
Hence: . $S_{\text{50 even}} \:=\:\frac{50}{2}\left[2(a+d) + 49(2d)\right] \:=\:\frac{1}{2}T + 100$
. . which simplifies to: . $100a + 5000d\:=\:T + 200$ [2]
Subtract [1] from [2]: . $50d = 200\quad\Rightarrow\quad d = 4$
Substitute into [1]: . $100a + 4950(4)\:=\:T\quad\Rightarrow\quad 100a = T - 19800$
Therefore: . $\boxed{a\:=\:\frac{T}{100} - 198}$
6. Yay thanks so much Soroban, now I know how to get the second equation | 2015-01-25T15:19:14 | {
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https://math.stackexchange.com/questions/2560497/conditional-probability-bayes-rule-question-with-urns | # Conditional Probability (Baye's Rule) question with urns
There are two urns A and B. Urn A contains 1 red ball and 2 white balls, whereas urn B contains 2 red balls and 1 white ball. Calculate the conditional probability that a randomly chosen ball belonged to urn A given that it is white.
I know, that the answer is 0.6666... But I can't figure the way how to apply the Baye's rule to the conditions.
• Have you tried anything? Do you have any thoughts on this problem? Dec 10 '17 at 20:09
• Are we to take that Urn A and Urn B may be chosen with equal probability, and one random ball extracted ? Dec 10 '17 at 20:12
P(From Urn A | it is white) $= \dfrac{P(A)\cdot P(W|A)}{P(A)\cdot P(W|A) + P(B)\cdot(P(W|B)} =\dfrac{\dfrac12 \cdot\dfrac23}{\dfrac12\cdot\dfrac23+\dfrac12\cdot\dfrac13}=\dfrac23$
$$P(A|W) = \frac{P(A \cap W)}{P(W)}$$ Assuming either urn is equally probable ..
$P(W) = \frac 12$ ( 3 out of 6 balls are white )
$P(A \cap W)=\frac12 \times \frac 23$ | 2022-01-24T17:15:57 | {
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http://math.stackexchange.com/questions/65666/finding-the-probability-that-red-ball-is-among-the-10-balls | # Finding the probability that red ball is among the $10$ balls
A box contains $20$ balls all of different colors including the red color. If we select $10$ balls randomly without replacement, what is the probability that the red ball will be among these $10$ balls?
What I think is that: If we let $X$ to be the number of balls we select until we get the red ball, then $X$ will be a random variable with range $1,2,3, \ldots ,20$, and the probability of getting the red ball will be $1/20$, so our probability will be $$\left(\frac{1}{20} \right)^{10} ,$$ is that right?! I'm not sure about the distribution of $X$?
-
Your argument is wrong. But how do we check that? Ok, instead of picking just 10 balls, suppose I pick all the 20 balls. Without any calculations, what do you think is the probability the red ball is among the 20 balls? Now, what does your argument tell you? – Srivatsan Sep 19 '11 at 2:10
If you select all the 20 balls then the probability is 1. So my argument is false! – Kelly Sep 19 '11 at 2:15
Not just that. As you pick more balls, the probability should clearly increase. Your answer $$\left(\frac{1}{20} \right)^{\text{number of balls picked}}$$ is decreasing as I pick more balls and so it must be wrong. (If you change the answer to $1$ minus that value, then it is still wrong, but it is slightly better.) – Srivatsan Sep 19 '11 at 2:19
Your calculation represents the probability of getting the red ball ten times in a row with replacement: this is a totally different problem. It should be easy to see that it is false by how small it is: the probability of getting the red in just the first pick is one out of twenty, so your chances of getting the red ball in the process of taking ten balls out should be even greater.
Here's how the reasoning with $X$ could have gone: the chance the $n$-th ball came up red is the probability the other $n-1$ before it came up nonred and then you got the red ball out of the remaining ones (the latter colored red in the calculations below). The probability the first pick was nonred is $19/20$, the probability the second pick was nonred is $18/20$, and so on, so we get the pattern
• $P(X=1)=\color{Red}{\frac{1}{20}}=1/20$
• $P(X=2)=\frac{19}{20}\times\color{Red}{\frac{1}{19}}=1/20$
• $P(X=3)=\frac{19}{20}\frac{18}{19}\times\color{Red}{\frac{1}{18}}=1/20$
• $P(X=4)=\frac{19}{20}\frac{18}{19}\frac{17}{18}\times\color{Red}{\frac{1}{17}}=1/20$
• $\qquad\qquad\cdots\cdots\cdots\cdots\cdots\cdots$
• $P(X=10)=\frac{19}{20}\frac{18}{19}\cdots\frac{11}{12}\times\color{Red}{\frac{1}{11}}=1/20$
Hence $P(X=1)+P(X=2)+\cdots+P(X=10)=\frac{1}{20}+\cdots\frac{1}{20}=10(\frac{1}{20})=1/2.$
Alternatively, we could have reasoned with symmetry as follows: every way of picking twenty balls out of twenty balls and keeping track of what order they come out will be like putting the balls in some order. The probability a specific ball will end up in a specific position is equal to the same probability for any other specific ball, say $p$. Since at least one ball must be in the position we have that the sum of each ball's probability being there equals one, or $20p=1$, hence $p=1/20$. Sum this over the first ten positions for the red ball and you get $10(1/20)=1/2$ as the probability the red ball is in the first ten positions. Since it makes no difference whether we actually take out the last ten balls or keep them in the box, this must be our desired value.
Two quicker ways could have been the following:
1. Symmetry: For every way of taking 10 balls out of 20 and leaving the other 10 in the box, there is exactly one way of taking those other 10 out of the box and leaving the original 10 in, hence by symmetry the probability of choosing the red ball equals the probability of not choosing the red ball. Thus the probability is $1/2$.
2. Counting: The number of ways of getting a red ball in a choice of 10 out of 20 is equal to the number of ways the other 9 non-red balls can be chosen out of the total 19 non-red balls: in other words, $19\choose9$. And the total number of ways of picking 10 balls out of 20 is $20\choose10$. So the probability of getting the red ball is $$\frac{{19\choose9}}{{20\choose10}}=\frac{19!}{9!10!} \frac{10!10!}{20!}=\frac{10}{20}=\frac{1}{2}.$$
Having these two techniques in your probability toolkit is important because problems that can be solved easily with them are ubiquitous in combinatorial situations. However it is also important that you don't split a probability problem into a large number of smaller cases (originally, we split ours up into 10 different cases) when doing so isn't actually necessary - this saves a lot of energy.
-
I don't think symmetry works in this case because I have a second part of the problem which generalize this to any box with n different objects. Thanks! – Kelly Sep 19 '11 at 2:30
@Kelly: Symmetry does in fact work "in this case" - is my description of it in this particular case not clear enough, should I elaborate on it further? The idea may or may not be applicable to your general problem, but it is certainly applicable to this specific one. – anon Sep 19 '11 at 2:36
Does X: the number of balls we select until we get the red ball, has a well-known distribution? – Kelly Sep 19 '11 at 2:59
@Kelly: I've edited my answer to be more comprehensive; it establishes that $X$ is uniformly distributed first using direct computation and then with symmetry. Please tell me if something is unclear. – anon Sep 19 '11 at 3:54
@Kelly: If "number of balls we select until we get the red" includes the successful selection, we are dealing with a very special case of the negative hypergeometric distribution. In that distribution, we have a group of size $N$, and $g$ of the members of the group are good. We draw without replacement until we get $k$ good. The numbers $N$, $g$, $k$ are fixed. The random variable $X$ is the number of draws. In your case, $N=20$ and $g=k=1$. I don't know whether the case $k=1$ has a special name. – André Nicolas Sep 19 '11 at 6:25 | 2014-12-18T06:05:18 | {
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https://dsp.stackexchange.com/tags/convolution/hot?filter=all | # Tag Info
## Hot answers tagged convolution
42
Convolution is correlation with the filter rotated 180 degrees. This makes no difference, if the filter is symmetric, like a Gaussian, or a Laplacian. But it makes a whole lot of difference, when the filter is not symmetric, like a derivative. The reason we need convolution is that it is associative, while correlation, in general, is not. To see why this ...
33
Adapted from an answer to a different question (as mentioned in a comment) in the hope that this question will not get thrown up repeatedly by Community Wiki as one of the Top Questions.... There is no "flipping" of the impulse response by a linear (time-invariant) system. The output of a linear time-invariant system is the sum of scaled and time-...
15
The Idea of Convolution My favorite exposition of the topic is in one of Brad Osgood's lectures on the Fourier Transform. The discussion of convolution begins around 36:00, but the whole lecture has additional context that's worth watching. The basic idea is that, when you define something like the Fourier Transform, rather than working directly with the ...
15
If you want the shifted output of the IFFT to be real, the phase twist/rotation in the frequency domain has to be conjugate symmetric, as well as the data. This can be accomplished by adding an appropriate offset to your complex exp()'s exponent, for the given phase slope, so that the phase of the upper (or negative) half, modulo 2 Pi, mirrors the lower ...
15
One of the definitive features of LTI systems is that they cannot generate any new frequencies which is not already present in their inputs. Please note that in this context a frequency refers to signals of the type $x(t)=e^{j\Omega_0 t}$ or $\cos(\Omega_0 t)$ which are of infinite duration, and are also referred to as eigenfunctions of LTI systems (...
13
One reason you see people designing FIR filters, rather than taking a direct approach (like both 1 and 2) is that the direct approach usually fails to take into account the periodicity in the frequency domain, and the fact that convolution implemented using an FFT is circular convolution. What does this mean? Suppose you have a signal $x = [ 1, 2, 3, 4]$ ...
12
There is probably a bit of a misconception here. In many application the signal is runnning all the time or is VERY long: modem, audio stream, video etc. In this case you can't really define the "length" of the signal. The relevant metric is here "number of operations per input sample" not the "total number of operations". If ...
11
Real-time low-latency partitioned convolution reverb with a long impulse response works by dividing the impulse response into unequally sized partitions. The shortest partitions (blocks) are at the beginning of the impulse response, and the partition length grows towards the end of the impulse response: Each partition length can be processed separately, ...
11
Turns out that convolution and correlation are closely related. For real signals (and finite energy signals): Convolution: $\qquad y[n] \triangleq h[n]*x[n] = \sum\limits_{m=-\infty}^{\infty} h[n-m] \, x[m]$ Correlation: $\qquad R_{yx}[n] \triangleq \sum\limits_{m=-\infty}^{\infty} y[n+m] \, x[m] = y[-n]*x[m]$ Now, in metric spaces, we like to use this ...
11
When talking about modeling, there are two things that usually get modeled: 1. the guitar amp, and 2. the speaker cabinet. Only the latter is modeled by an impulse response, which means that the cabinet is simply represented by an LTI system and implemented by convolution. This is of course an approximation but it works fairly well. You can find a lot of ...
10
One problem is dealing with infinite length transforms that wrap-around when using a finite length FFT. The Fourier transform of a finite length frequency response is an infinite length impulse response or filter kernel. Most people would like their filter to finish before they die or run out of computer memory, so need tricks to produce shorter FIR ...
10
Is there any trade-off in numerical precision or speed? Yes. For delays that are integer multiples of the sampling period method 1 is far superior: it's computationally efficient, it's bit-exact, it's easy to implement and it's almost fool proof. Method 2 is computationally expensive, you need to pick an FFT length (which is not trivial) it's subject to ...
9
Think of this... Imagine a drum which you are beating repeatedly to hear the music right? Your drum stick will land on the membrane for the first time and due to the impact it will vibrate. When you strike for the second time, the first impact's vibration has already decayed, to some extent. So whatever sound you will hear is the current beating and sum of ...
9
Yes, you are correct. Multiplication in time domain means convolution in frequency domain and vice versa. Multiplying your signals $x[n]$ and $y[n]$ will give an output: \begin{align} z[n]&=\{2\cdot 5, 4\cdot 1, 1\cdot 8\}\\ &= \{10, 4, 8\}\end{align} Remember that this output is in time domain. When you convolve $x[n]$ and $y[n]$, you will get $... 9 Since convolution describes the operation of a linear time-invariant (LTI) system, the question is if the effect of an LTI system can be compensated by another LTI system. In the discrete-time domain you can use the$\mathcal{Z}$-transform to analyze LTI systems. If a signal$x(n)$(with$\mathcal{Z}$-transform$X(z)$) is filtered by a system with impulse ... 9 Convolution in the time domain is equivalent to multiplication in the frequency domain. If you window two sinusoids in the time domain to get finite length waveforms, and the two sinusoids are exactly integer periodic in the window width, then the DFT will be impulses. If the frequencies are different, the impulses in the two DFT will be disjoint, and ... 9 You can make a simple algebraic argument, given the premise that you provided. If: $$Y(\omega) = X(\omega) H(\omega)$$ where$X(\omega)$is the spectrum of the input signal and$H(\omega$) is the frequency response of the system, then it's obvious that if there is some$\omega$in the input signal for which$X(\omega) = 0$, then$Y(\omega) = 0$as well; ... 9 Given a discrete time LTI system with impulse response$h[n]$, one can compute its response to any input$x[n]$by a convolution sum: $$y[n] = x[n] \star h[n] = \sum_{k=-\infty}^{\infty} {h[k]x[n-k]} \tag{1}$$ Without anything further stated, above definition is for the linear convolution (aperiodic convolution) between$h[n]$and$x[n]$, which are ... 9 The proof of associativity of discrete convolution relies on the assumption that multiple infinite sums can be evaluated in any order. This is not true if some of the involved sequences do not converge absolutely, which is the case for the given sequences$x_1[n]$and$x_2[n]$. Note that the convolution sum$x_1\star x_2$does not converge, i.e.,$x_3\star (...
8
As a student I was involved in the same problem as you are. Let me explain to you in the simplest words without any math. Convolution: It is used to convolute two function. May sound redundant but I´ll put an example: You want to convolute (in a non math term to "combine") a unit cell (which can contain anything you want: protein, image, etc) and a ...
8
What's not clear to me is what the fundamental difference (if any) is between simulating an analog filter and making a digital filter. Either way, these functions will produce "ba" transfer function outputs, but the b and a are totally different. For a 2nd-order filter, for instance, b = [b0, b1, b2] and a = [a0, a1, a2]. These are the coefficients of the ...
8
Whether the direct convolution or the FFT/IFFT method is faster depends on the length of the impulse response, $N_\mathrm{i}$ and the signal length $N_\mathrm{s}$. With the formulas taken from here I've created a small Matlab script that calculates the required number of real multiplications and additions for the direct convolution and FFT/IFFT method, ...
8
If you're an EE student, you will have encountered the term LTI System (or you certainly will soon enough!): A system that, no matter the absolute time, outputs, given the same input, the same output; if you scale the input by a factor, the output is scaled by the same factor. Linear, time-invariant, so to speak. LTI systems can be applied to time-domain ...
7
Flipping the impulse response is really just a matter of perspective. The LTI system doesn't care about perspective. In any case, here is a graphic showing a system that takes an input of color weighted impulses.
7
Linear convolution is the basic operation to calculate the output for any linear time invariant system given its input and its impulse response. Circular convolution is the same thing but considering that the support of the signal is periodic (as in a circle, hance the name). Most often it is considered because it is a mathematical consequence of the ...
7
Be careful, a median filter cannot be expressed as a convolution, and thus is not considered a kernel in this respect. This is because the median filter is based on order statistics of an image patch, and the resulting pixel at the output of a median filter is not a linear combination of other pixels within a patch. Otherwise, you are right, kernels are ...
7
Multiplication in your first sentence is term-by-term multiplication: $z[n] = x[n]y[n]$ for all $n$. Convolution, for discrete-time sequences, is equivalent to polynomial multiplication which is not the same as the term-by-term multiplication. Convolution also requires a lot more calculation: typically $N^2$ multiplications for sequences of length $N$ ...
7
I can tell you of at least three applications related to audio. Auto-correlation can be used over a changing block (a collection of) many audio samples to find the pitch. Very useful for musical and speech related applications. Cross-correlation is used all the time in hearing research as a model for what the left and ear and the right ear use to figure ...
7
Let me clarify some definitions first. There is no such thing as a "passive impulse response", there are only passive systems. If you like, you can call the impulse response of a passive LTI system "passive", but that's not common usage. From what I understand from your question, you probably mean by a "passive impulse response" an impulse response $h[n]$ ...
7
Normally, the variables $t$ and $f$ are used for continuous time and frequency. But from your question I understand that you're talking about discrete time and frequency, and their relation via the Discrete Fourier Transform (DFT). If you have discrete-time sequences $a[n]$ and $b[n]$ and their DFTs $A[k]$ and $B[k]$, then the DFT of the linear convolution \$...
Only top voted, non community-wiki answers of a minimum length are eligible | 2021-03-09T08:22:10 | {
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https://math.stackexchange.com/questions/296247/convergence-of-fixed-point-iteration-for-polynomial-equations/296275 | # Convergence of fixed point iteration for polynomial equations
I'm looking for the solution $x$ of
$$x^n+nx-n=0.$$
Thoughts: From graphing it for several $n$ it seems there is always a solution in the interval $[\tfrac{1}{2},1)$. For $n=1$, the solution is the fraction $\tfrac{1}{2}$ and for higher $n$, the solution shifts to the right.
I then saw that the equation reads $$F_n(x)=x$$ with
$$F:[0,1]\to[0,1],\ \ \ \ \ \ F_n(x):=1-\frac{x^n}{n}.$$
I think I got all conditions together for making the iteration of $F_n$'s the general soluton for the equation. I computed $$x_5=F_5(F_5(F_5(F_5(F_5(F_5(F_5(F_5(x_S)))))))),$$
with starting value $x_S=\tfrac{1}{2}$ and it seems to be the solution of the equation for $n=5$.
Related wikipedia links: Fixed point theorem, Banach fixed-point theorem, Fixed point iteration;
I wonder:
Have I found the soluton and can one evaluate the iteration to a closed form?
Might be that it involves polylogs. edit: At least Wolfram Alpha claims to know $$n(x)=W\left(\frac{-\log(x)}{x-1}\right)/\log(x),$$ even if $n(x)$ isn't too interesting.
In my case, how is the relation between the function $x(n)$ and (I think) the fixed point combinator for the iteration?
Does it matter what I choose for $x_S$ here? Can one relate it's value and the number for necessary iteration for a good agreement with the real value?
(Also, are there any results on which polynomial equations this technique works? The property $F:[0,1]\to[0,1]$ seemed accidental to me.)
• – Mhenni Benghorbal Feb 6 '13 at 13:07
• @Nick Kidman: Not a solution, but a remark: the fact that there is always a solution on $(1/2,1)$ for $n>1$ follows directly from the fact that $f_n(1/2)=1/2^n+n/2-n=1/2^n-n/2<0$ and $f_n(1)=1+n-n=1>0$. So, by the intermediate value theorem, there exists $c\in(1/2,1)$ such that $f_n(c)=0$. – Dennis Gulko Feb 6 '13 at 13:07
• It seems like you want to do something like Newton Raphson (or root approximation). Note that for $n=1$, you will not converge towards a solution if you use any starting value that is not $x_S = \frac {1}{2}$. – Calvin Lin Feb 6 '13 at 13:25
You can get a good approximation of the solution as $n \to \infty$ by supposing that $x$ can be written as an asymptotic series in powers of $1/n$, say
$$x \sim 1 + \sum_{k=1}^{\infty} \frac{a_k}{n^k},$$
then substituting this into the given equation and calculating the coefficients recursively. For example we can calculate $a_1$ and $a_2$ by writing
\begin{align*} 0 &\approx \left(1 + \frac{a_1}{n} + \frac{a_2}{n^2}\right)^n + n\left(1+\frac{a_1}{n} + \frac{a_2}{n^2}\right) - n \\ &= \left(1 + \frac{a_1}{n} + \frac{a_2}{n^2}\right)^n + a_1 + \frac{a_2}{n} \\ &= a_1 + e^{a_1} + \frac{2a_2 (1+e^{a_1}) - a_1^2 e^{a_1}}{2n} + O\left(\frac{1}{n^2}\right). \end{align*}
By sending $n \to \infty$ we get that $a_1 + e^{a_1} = 0$ and so
$$a_1 = -W(1),$$
where $W$ is the Lambert W function.
Then, setting the coefficient of $1/n$ to $0$ and substituting the above value of $a_1$ we find that
$$a_2 = \frac{W(1)^3}{2(1+W(1))}.$$
Thus we have
$$x \approx 1 - W(1) n^{-1} + \frac{W(1)^3}{2(1+W(1))} n^{-2}.$$
This approximation seems to be pretty good. Below is a plot which compares the numerical roots with the asymptotic formula for $1 \leq n \leq 10$.
This series might actually converge for large enough $n$ but I don't see how to prove it. If the implicit function theorem could be employed then you would be set.
Isn't this simply the result of the Banach fixed-point theorem? In you case one must prove that there exists $q$ such that: $$d(F_n(x),F_n(y))=\frac{|y^n-x^n|}{n} \le q |x-y|$$ Which is true for all $n>1, x>0$, with say, $q=0.9$, So that it doesn't matter what $x_S$ you choose.
Regarding a closed form - convergence of an iterative series says nothing about the existence of such a closed form solution.
• Yes, I obtained the solution that way, I haven't used that before though and so I wasn't sure, especially since I solved all $n$ equations simultaneously and I couldn't find an explicit reference regarding the value of starting values $x_S$. I added some references in the quation now too. / "Convergence of an iterative series says nothing about the existence of such a closed form solution." - well yes, that's why I'm asking for this specifc example :) Wolfram Alpha computes $n(x)$ explicitly in terms of polylogs but $n$ in terms of $x$ is some kind of useless and I don't know hot to invert it. – Nikolaj-K Feb 6 '13 at 13:51
• @NickKidman - if you knew that, what makes you think that your iterative series has anything to do with a closed form solution? the question should be re-titled "does a closed form solution for equation X exist". – nbubis Feb 6 '13 at 13:55
• I now added a remark why I think one might exist in the comment above. That's also why I wrote about polylogs in the question, I though it was long enough already. – Nikolaj-K Feb 6 '13 at 13:56 | 2019-10-18T16:51:55 | {
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https://math.stackexchange.com/questions/996543/if-a-bcd-show-that-c-1-da-1b/996548 | # If A = BCD show that $C^{-1}$ = $DA^{-1}B$
I came across this question in a past paper, If A = BCD show that $C^{-1}$ = $DA^{-1}B$. All these matrices are sqaure and have inverses.
I attempted a solution but I am not sure if-
1. The solution is correct and;
2. Is this the best solution
Here is my attempt: *
$A = BCD$
$A^{-1}A = I = A^{-1}BCD$ This implies that $D^{-1} = A^{-1}BC$ and now multiply both sides by $D$ and we have:
$DD^{-1} = I = DA^{-1}BC$ from which the result follows.
• Right-multiply both sides with the inverse of the rightmost matrix on the side where $C$ is until $C$ has moved (as $C^{-1}$) to the other side. Then left-multiply with the inverse of the leftmost matrix on that side until $C^{-1}$ stands alone to get the desired result.
– user139000
Oct 29 '14 at 11:16
• Thanks, but is my solution correct or no?
– dan
Oct 29 '14 at 11:18
• Yours works fine! Oct 29 '14 at 11:24
• Asking if your (correct) solution is "best" is mostly a matter of opinion. It is fairly terse, but some might object to a lack of "motivation". Certainly if there were a number of such formulas to verify, one might want to develop a more systematic approach. Oct 29 '14 at 12:49
• I agree my use of 'best' was slightly misleading, I was merely asking if my solution (the method) was appropriate :)
– dan
Oct 29 '14 at 17:58
$$A=BCD$$ $$B^{-1}A=CD$$ $$C^{-1}B^{-1}A=D$$ $$C^{-1}B^{-1}=DA^{-1}$$ $$C^{-1}=DA^{-1}B$$
Note that we do not consider the size of the matrices as they are all square, say if $A$ is $n \times n$ and B is not, the question itself would be not well-defined.
\begin{align} A&=BCD\\ B^{-1}A&=CD\\ C^{-1}B^{-1}A&=D\\ C^{-1}B^{-1}&=DA^{-1}\\ C^{-1}&=DA^{-1}B\\ \end{align} | 2022-01-26T13:33:29 | {
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https://math.stackexchange.com/questions/4260552/how-to-tell-if-a-space-is-second-countable | # How to tell if a space is second-countable
A topological space is called second-countable iff it has a countable basis.
How to prove or at least make an assumption about whether a space does, or does NOT have countable basis? Which properties of a space can imply that it is, or isn´t second countable? Is second-countability intuitively something like "small cardinality" or "weaker countability"?
Of particular interest are $$l^2$$ Hilbert spaces - are they second countable? I assume not, but am not sure.
Thank you!
• A metric space is second countable iff it is separable. A pre-hilbertian space is separable iff it admits a countable Hilbert basis. If X is a separable space then $L^2(X, \mathcal{B}(X),\mu)$ is also separable. Sep 26 at 8:04
• @Eevee, isn't that basis of Kronecker deltas you gave a Hilbert basis in the sense of linear algebra, which has little to do with the topological "basis" (I prefer "base")? Sep 26 at 12:29
• Oh, you're right, my bad. Had some wires crossed last night; sorry about that. Sep 26 at 18:27
Any metric space that has a countable dense subset has a countable base, this is a classic fact. I show this and more here; it could have been more succinctly stated as $$w(X)=d(X)=l(X)=nw(X)$$, etc. for metrisable $$X$$, e.g.
So $$\ell^2$$ and in fact all $$\ell^p$$ for $$p\neq \infty$$ are separable (so have countable weight) while $$\ell^\infty$$ has not, because it has a closed discrete subset of size continuum. The reasoning is: if $$X$$ has a countable base, this holds for all subspaces too, but a large discrete subspace does not have a countable base:
Suppose $$X$$ is discrete, and let $$\mathcal{B}$$ be a base for $$X$$. It's clear that for each $$x \in X$$, $$\{x\}$$ is in $$\mathcal B$$ because that's the only way to write to open singleton as a union of base elements. So $$|\mathcal B| \ge |X|$$ and so if $$X$$ is uncountable, $$X$$ does not have a countable base.
The minimal size of a base (weight) or dense set (density) etc. are just a few of the many ways to "measure" the "size" of a space. In analysis many spaces are indeed of countable base; but e.g. weak topologies on Banach spaces often are not.
• So is the $l^2$ second-countable, or not? You are saying only it is separable. EDIT: Separable = has countable dense subset = has countable base = is second-countable.... Sorry, I get it. Sep 26 at 11:16
• @TerezaTizkova both because it’s metric. Sep 26 at 11:44
Whether a space is second-countable is a purely topological property, because it's defined in terms of its open sets (whether it has a countable basis, that is, whether its open sets are generated by a countable basis). So to find out whether a given space is second-countable, you'll have to look at it's topology. Intuitively, second-countable spaces are 'smaller' and more 'well-behaved' than ones that aren't.
For example, $$\mathbb{R}^n$$ is second countable since it has many countable bases for its topology - for example the open balls with rational radii. Subspaces of second countable spaces are second-countable, so all subsets of $$\mathbb{R}^n$$ with the subspace topology are second-countable too.
For non second-countable spaces, you can consider the uncountable product of $$\mathbb{R}$$ (with either the box or product topology) or the Long line.
Also see Henno's answer on $$\ell^p$$ spaces for second-countable and non second-countable Hilbert spaces.
If a space is second-countable, the usual way to prove this will be to exhibit a specific countable basis.
If a space is not second-countable, proving this can be more intricate, and what methods are promising will vary a lot with the kind of space you are looking at. Any second-countable space is separable, and disproving separability can be easier. However, there are plenty of separable non-second-countable spaces. | 2021-10-25T17:34:58 | {
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http://desertdingo.com/l7ocfce/viewtopic.php?1b9ddc=transpose-of-matrix-product | » JavaScript » PHP Now this is pretty interesting, because how did we define these two? By using this website, you agree to our Cookie Policy. tcrossprod () function in R Language is used to return the cross-product of the transpose of the specified matrix. Properties of transpose » LinkedIn » C++ » Embedded Systems 'k_��� j�v�}sS�, P��~p:(Ȭ�ٸC'�+*1^���rQ�%�G������=NJZm���w"~��6GܠH��v�x�B���L �+��t�A$E�I̮F�Ͳ. » C More: » C++ Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. In linear algebra, an mxn matrix A is multiplied with its transpose AT then the resultant matrix is symmetric. » DBMS Transpose of the Matrix. Repeat this step for the remaining rows, so the second row of the original matrix becomes the second column of its transpose, and so on. A = [ 7 5 3 4 0 5 ] B = [ 1 1 1 − 1 3 2 ] {\displaystyle A={\begin{bmatrix}7&&5&&3\\4&&0&&5\end{bmatrix}}\qquad B={\begin{bmatrix}1&&1&&1\\-1&&3&&2\end{bmatrix}}} Here is an example of matrix addition 1. Let, A is a matrix of size m × n and At is the transpose of matrix A, Matrix addition and subtraction are done entry-wise, which means that each entry in A+B is the sum of the corresponding entries in A and B. )'s unnecessarily.They also return symmetricMatrix classedmatrices when easily detectable, e.g., in crossprod(m), the oneargument case. The transpose of the product of two matrices is equivalent to the product of their transposes in reversed order: (AB) T = B T A T . In the case of the matrix, transpose meaning changes the index of the elements. (If A is m × n, then x ∈ Rn, y ∈ Rm, the left dot product is in Rm and the right dot product is in Rn .) » Articles » CS Organizations share. The same is true for the product of multiple matrices: (ABC) T = C T B T A T. Example 1: Find the transpose of the matrix and verify that (A T) T = A. The transpose of a matrix is calculated by changing the rows as columns and columns as rows. In this program, the user is asked to enter the number of rows r and columns c. Their values should be less than 10 in this program. In this tutorial, we are going to check and verify this property. The following properties hold: (AT)T=A, that is the transpose of the transpose of A is A (the operation of taking the transpose is an involution). We can transpose the whole thing as I indicated before. : n 1 matrices). The transpose of a matrix is obtained by interchanging the rows and columns of the given matrix. Free matrix transpose calculator - calculate matrix transpose step-by-step This website uses cookies to ensure you get the best experience. » Node.js If I say roughly then the process of taking transpose of a matrix is something equivalent to changing the rows of the matrix into columns and columns of the matrix into rows. %���� Eigenvalues of a triangular matrix. Transpose Matrix. Are you a blogger? » Certificates x��[Ys#�~ׯ���P�%��Pʕ��QN*q�XUy���K���4�I����� �M/h����h���ht�5 ����0+&�W�_���NNW�r�x>l�\�4j�����r/~Z-q��W��_/��J"��^/�̸!�٨�����]1� ����EJw��Jhӷ.��p"(��אUl:/+Y=0�sU�>T"�+�1���=���*Sf1A$�ͫ���KJ�Ԅ����IU�(�j��u�q_��P!����Ȉe��!>$�2Dp{������p��D�l������91RNf���*,&3)�RLf�ˉ*�����w�����77�w��>W�#xR�P�#vO�@-A�=L�hn����~^^��$�!��4���>fnjQ�dD��\���i]2n��ĵy4�ֺd&�i}�K�$V��g��$���7��� �٣����Dq�73"m)߽��$�M���#�K�`A��f��D�@���8m \&� This is the definition of a transpose. » Internship Home » So we now get that C transpose is equal to D. Or you could say that C is equal to D transpose. A + B = [ 7 + 1 5 + 1 3 + 1 4 − 1 0 + 3 5 … Just like before, we would get the transpose of the product. I know this statement seems stupid, but keep reading. Now I can say, "lets transpose the product of the vectors":$\mathbf{A}^T=(\mathbf{v}*\mathbf{v}^T)^T$But as you can distribute the transpose over the multiplication, you can say: Transposes of sums and inverses. Linear Algebra 11w: Introduction to the Transpose of a Matrix - Duration: 7:40. » Java » O.S. Transpose of the product of two matrices is equal to the product of their transposes taken in the reverse order, The transpose of the matrix products can be extended to several matrices The inverse of a transpose matrix is equal to the transpose of its inverse, » Web programming/HTML The transpose of a matrix is a new matrix that is obtained by exchanging the rows and columns. » DBMS Two matrices can only be added or subtracted if they have the same size. Transpose of a vector. Now note that (AB)x ⋅ y = A(Bx) ⋅ y = Bx ⋅ A⊤y = x ⋅ B⊤(A⊤y) = x ⋅ (B⊤A⊤)y. » SQL The same is true for the product of multiple matrices: (ABC) T = C T B T A T . » Java Submitted by Anuj Singh, on June 06, 2020. » Facebook 3 0 obj << » Networks To transpose a matrix, start by turning the first row of the matrix into the first column of its transpose. » Privacy policy, STUDENT'S SECTION For example, if A(3,2) is 1+2i and B = A. » Python » CSS » Machine learning » SEO When we take transpose, only the diagonal elements don’t change place. » C# Linear Algebra 11y: The Transpose of a Triple Product - Duration: 2:33. » C Article Summary X. And we notice also this is 2 x 4 times a 4 x 7 gives you a 2 x 7 matrix. » C++ STL Ad: » C++ Interview que. » Content Writers of the Month, SUBSCRIBE /Filter /FlateDecode Thus, the matrix B is known as the Transpose of the matrix A. » Feedback The solver that is used depends upon the structure of A.If A is upper or lower triangular (or diagonal), no factorization of A is required and the system is solved with either forward or backward substitution. » Data Structure For example, element at position a12 (row 1 and column 2) will now be shifted to position a21 (row 2 and column 1), a13 to a31, a21 to a12and so on. Thus Transpose of a Matrix is defined as “A Matrix which is formed by turning all the rows of a given matrix into columns and vice-versa.” If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. returns the nonconjugate transpose of A, that is, interchanges the row and column index for each element.If A contains complex elements, then A.' i.e., (AT) ij = A ji ∀ i,j. The basic matrix product, %*% is implemented for all ourMatrix and also forsparseVector classes, fully analogously to R'sbase matrixand vector objects. We said that our matrix C is equal to the matrix product A and B. Now, we will understand the transpose matrix by considering two matrices P … B = A.' Numpy.dot () is the dot product of matrix M1 and M2. Enter rows and columns of matrix: 2 3 Enter elements of matrix: Enter element a11: 1 Enter element a12: 2 Enter element a13: 9 Enter element a21: 0 Enter element a22: 4 Enter element a23: 7 Entered Matrix: 1 2 9 0 4 7 Transpose of Matrix: 1 0 2 4 9 7 » About us A transpose will be denoted by original matrix with “T” in superscript, like Aᵀ. So: The columns of AT are the rows of A. » C & ans. $$A, B) Matrix division using a polyalgorithm. » Cloud Computing Thus, (AB)⊤ = B⊤A⊤. (A+B)T=AT+BT, the transpose of a sum is the sum of transposes. This is one of the most common ways to generate a symmetric matrix. The transpose of matrix A is represented by \(A'$$ or $$A^T$$. Then, the user is asked to enter the elements of the matrix (of order r*c). The meaning of transpose is to exchange places of two or more things. » Kotlin » CS Basics Transpose of a matrix can be found by changing all the rows into columns or vice versa. © https://www.includehelp.com some rights reserved. Numpy.dot () handles the 2D arrays and perform matrix multiplications. Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. » DOS Solved programs: The transpose of a matrix is an important phenomenon in the matrix theory. Languages: Example: If A= 1 2 3 4 5 6 , then AT = 2 4 1 4 2 5 3 6 3 5: Convention: From now on, vectors v 2Rn will be regarded as \columns" (i.e. » C#.Net & ans. » Ajax does not affect the sign of the imaginary parts. » Puzzles For permissions beyond … tcrossprod() takes the cross-product of the transpose of a matrix.tcrossprod(x) is f… The diagonal elements of a triangular matrix are equal to its eigenvalues. For input matrices A and B, the result X is such that A*X == B when A is square. Transpose & Dot Product Def: The transpose of an m nmatrix Ais the n mmatrix AT whose columns are the rows of A. » Contact us ... Now, I'm going to define the transpose of this matrix as a with this superscript t. And this is going to be my definition, it is essentially the matrix A with all the rows and the columns swapped. The transpose () function from Numpy can be … stream 1. Hence, the transpose of matrix for the above matrix is : (Image to be added soon) Properties of Transpose of Matrices. The program must accept two positive integers M and N as the input. This is one of the most common ways to generate a symmetric matrix. Linear Algebra using Python, Linear Algebra using Python | Product of a Matrix and its Transpose Property: Here, we are going to learn about the product of a matrix and its transpose property and its implementation in Python. MathTheBeautiful 3,285 views. The following statement generalizes transpose of a matrix: If $$A$$ = $$[a_{ij}]_{m×n}$$, then $$A'$$ = $$[a_{ij}]_{n×m}$$. So now, the transpose of matrix$\mathbf{A}$will still be a square matrix,$\mathbf{A}^T\$. » Java » Subscribe through email.
Machine Learning For Absolute Beginners 2nd Edition, Reflexive Relation In Discrete Mathematics, Westland Rose Food Enriched With Horse Manure, Social Work Articles 2020, Baiyai Dinner Plain, How To Make Burley For Bream, 4x2 Grow Tent Kit, Saffron Price Per Ounce 2020, Mc32j7035ct Convection Microwave Oven, Caulerpa Taxifolia Kingdom, Hsc Biology 2018, Rooster And Snake Meaning, | 2021-01-23T12:25:06 | {
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https://mathematica.stackexchange.com/questions/128184/how-to-determine-specific-values-of-a-function-on-each-streamline/128198 | # How to determine specific values of a function on each streamline
I was thinking if there is a way to obtain the values of a function from a streamline plot. This is what I mean:
Consider a typical streamline plot from the documentation: StreamPlot[{-1 - x^2 + y, 1 + x - y^2}, {x, -3, 3}, {y, -3, 3}, StreamPoints -> {{{{1, 0}, Red}, {{-1, -1}, Green}, Automatic}}] with a typical output
My Question: is there a way one could dynamically display the value of the functions that were plotted for the green streamline for instance, as you hover your cursor along the line (green streamline)? Thank you.
• You mean, display each point that went into drawing a particular stream line? – J. M.'s torpor Oct 8 '16 at 4:47
• I understand you mean something analoguous to: right click on the graph and select "get coordinates". – Dr. Wolfgang Hintze Oct 8 '16 at 12:51
• Thanks for your comment @Dr.WolfgangHintze. What I meant is what bbgodfrey actually showed. To display the corresponding value of the functions at a given coordinate. – D. Andrew Oct 8 '16 at 14:44
• In V12.2, due to a change in StreamPlot, one has to add the option StreamColorFunction -> None to reproduce the figure in the question. – Michael E2 Jan 11 at 2:36
Since ToolTip does not appear to work here, try
Dynamic[{loc = MousePosition["Graphics", {0, 0}],
{-1 - x^2 + y, 1 + x - y^2} /. {x -> First@loc, y -> Last@loc}}]
which gives a result that looks like
{{x, y}, {xstream, ystream}}
where x and y are the positions on the plot (the coordinates provided by Get Coordinate Tool as suggested by Dr. Wolfgang Hintze in a comment above), and xstream and ystream are the corresponding values of the functions there. For instance, when the mouse is at the tip of the Red streamline, the coordinates and values are approximately
((0.0308851, 1.0987}, {0.00331156, 0.101788}}
The output can be dressed up a bit, if desired, by
Dynamic[StringForm["loc = , val = ",
loc = MousePosition["Graphics", {0, 0}],
val = {-1 - x^2 + y, 1 + x - y^2} /. {x -> First@loc, y -> Last@loc}]]
• Thanks, @bbgodfrey. This is what I actually meant.This concept could be useful to gain some insight beforehand of what the solutions of some complex functions should be even before attempting to solve them. – D. Andrew Oct 8 '16 at 14:52
• @bbgodfrey excellent solution (+1). – Dr. Wolfgang Hintze Oct 9 '16 at 7:37 | 2021-08-02T06:49:37 | {
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https://kokecacao.me/page/Course/F20/21-127/Lecture_015.md | # Lecture 015
Let $m\in \mathbb{N}$, define a relation on $\mathbb{Z}$ by $(\forall a, b \in \mathbb{Z})$: $$a \equiv b (\mod m) \iff m|a-b$$
read as "a is congruent to b modulo m" (a and b have the same meaning under modulo m)
Claim: this is an equivalence relation Proof: show reflexive, symmetric, transitive
• R: a-a=0=m*0
• S: yes
• T: yes
## Equivalence Class
Definition: let R be an equivalence relation on a set S. For every $x \in S$, we define "equivalence class of x under R", denoted $[x]_R$ by
• all elements in two equivalence are related iff the equivalence classes are equal (need to prove)
[x]_R = \{y \in S | (x, y) \in R\}
($[x]_R$ is a collection of all elements that are equivalent to x under the specific notion of equivalence being discussed.)
## Congruent Modulo of All Equivalence Classes
Definition: $S/R = \{[x]_R | x \in S\}$
• $\mathbb{Z}/m\mathbb{Z}$ is for congruent modulo m
• There are only m distinct equivalence classes in $\mathbb{Z} /m \mathbb{Z} = \{[0]_m, [1]_m, ... [m-1]_m\}$
• each of equivalence classes is an infinite set
• if we union them together, $=\mathbb{Z}$
• if we intersect them together, $=\emptyset$
• so $\mathbb{Z}/m\mathbb{Z}$ is a partition on $\mathbb{Z}$ (like $\mathbb{Z}/m \mathbb{Z} = \{[0]_3, [1]_3, [2]_3\}$)
Consider: define ~ on $\mathbb{N}^2$. by $(a, b) ~ (c, d) \iff a+d = b+c$ (same thing as a-b = c-d, but we can't use minus because we haven't define negative number)
So:
• $[(0, 0)]_\sim = {(0, 0), (1, 1), (2, 2) ...}$
• $[(0, 1)]_\sim = {(0, 1), (1, 2), (2, 3) ...}$
• $[(1, 0)]_\sim = {(1, 0), (2, 1), (3, 2) ...}$
we then associate each with a negative integer
• -2 = $[(0, 2)]_\sim = [(1, 3)]_\sim$
• -1 = $[(0, 1)]_\sim$
• 0 = $[(0, 0)]_\sim$
• 1 = $[(1, 0)]_\sim$
• 2 = $[(2, 0)]_\sim$
so we can also define + and * of negative integer
• $[(a, b)]_\sim + [(c, d)]_\sim = [(a+c, b+d)]_\sim$
• $[(a, b)]_\sim \times [(c, d)]_\sim = [(ac+bd, bc+ad)]_\sim$
We check if this + and * is well defined
Table of Content | 2023-03-22T19:43:41 | {
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https://math.stackexchange.com/questions/2966925/probability-of-a-4th-die-roll-being-higher-than-one-of-the-first-3-rolls | # Probability of a 4th die roll being higher than one of the first 3 rolls?
If I roll 3 dice with n sides, and then roll a 4th die of the same size, what are the odds of it being higher than at least one of the previous rolls?
I'm thinking it should be something like the odds of rolling a 1 in the first 3 dice times the odds of rolling a 2 or higher on the last die, giving me this equation.
$$(1 - ((n-1)/n)^3)*((n-1)/n)$$
Is this equation correct? I feel like I'm missing something.
• I guess I misread the title at first glance, thinking you wanted the fourth (and last) roll to be higher than all the first three rolls. But as I reread everything, it seems more likely you ask only that the last roll exceeds at least one of the first three rolls. In that case you want the distribution of the smallest of three rolls, from which you can derive the chance that the fourth roll is above that. – hardmath Oct 23 '18 at 1:25
• @hardmath exactly, I'm unsure of how to turn that into an equation though. – william porter Oct 23 '18 at 1:29
There are $$n^3$$ sequences for the first three rolls, of which
• 1 sequence has minimum roll $$n$$
• $$2^3-1^3=7$$ sequences have minimum roll $$n-1$$. The expression as a difference of cubes can be seen by drawing a table of the minimum function
• $$3^3-2^3=19$$ sequences have minimum roll $$n-2$$, etc.
Thus the probability that the fourth roll is less than or equal to the minimum of the first three rolls is $$\frac1{n^3}\sum_{k=1}^n((n-k+1)^3-(n-k)^3)\frac kn=\frac1{n^4}\sum_{k=1}^nk^3=\frac1{n^4}\left(\frac{n(n+1)}2\right)^2=\frac{(n+1)^2}{4n^2}$$ where the simplification of the sum comes first from the difference of cubes telescoping and then the formula for the sum of consecutive cubes. Therefore the probability that the fourth roll is higher than at least one of the previous three rolls is the complement of this, or $$1-\frac{(n+1)^2}{4n^2}=\frac{(n-1)^2}{4n^2}$$.
What is the probability that the 4th roll is not the smallest?
One of the 4 dice must be the smallest. It is just as likely to be the 4th die rolled as one of the previous 3.
But there is a possibility that there is a tie.
As n gets to be very large, I would expect our probability to approach $$\frac 14$$
There are $$n^4$$ possibilities for the 4 dice to come up.
if you roll a $$k$$ on your last die there are $$(n-k)$$ ways to roll better than that, and $$(n-k)^3$$ to roll 3 dice better than that.
$$\sum_\limits{k=1}^n (n-k)^3 = \sum_\limits{k=1}^{n-1} k^3 = \frac {n^2(n-1)^2}{4}$$
$$\frac {(n-1)^2}{4n^2}$$ | 2019-08-21T17:48:23 | {
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https://math.stackexchange.com/questions/2658434/two-definitions-of-saddle-point | # Two definitions of Saddle point
Let $f\colon[a,b]\to\mathbb{R}$ be a differentiable function and $x\in[a,b]$ with $f'(x)=0$. Is there a counterexample with respect to the equivalence
$x$ is not a local extremum of $f$
$\Leftrightarrow$
There is an $\epsilon>0$ such that $f'$ is monotonically increasing on $[x-\epsilon;x]$ and monotonically decreasing on $[x;x+\epsilon]$ or
there is an $\epsilon>0$ such that $f'$ is monotonically decreasing on $[x-\epsilon;x]$ and monotonically increasing on $[x;x+\epsilon]$.
In other words: Are the two definitions of a saddle point
$x$ is a saddle point of a differentiable function $f$ iff $x$ is a stationary point which is not an local extremum.
and
$x$ is a saddle point of a differentiable function $f$ iff $x$ is a stationary inflection point
really equivalent?
The two definitions are not equivalent. Look at the graph of the differentiable function $$x\mapsto x^2 \sin\frac1x \qquad (x\neq 0); \qquad 0\mapsto 0$$
$0$ is a stationary point, but neither a local extremum nor a stationary inflection point.
• For completeness, note how $f'(x)$ is non-monotonic on both of the intervals $[-\epsilon, 0]$ and $[0, \epsilon]$ for any $\epsilon>0$, making it a direct counterexample for the proposed equivalence. Feb 20, 2018 at 11:20 | 2022-07-02T11:03:05 | {
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http://koyk.namastegroup.it/energy-calculator-physics.html | # Energy Calculator Physics
Efficiency is the ratio of the amount of useful energy produced (energy output- Eout) to the amount of energy used (energy input- Ein), expressed as a percentage. KE = 1/2*m*v^2. 5mv^2 and also compare E to the energy at the bottom of the projectile, E3 = 0. The principle of hydro electricity generation is quite simple. 2) Gittewitt, Paul. Multiple Choice with ONE correct answer 1. 7 Work and Kinetic Energy. The unit of energy is the joule. a ball of mass 100g falls from a height of 5m and rebounds to a height of 3m. Potential Energy vs. Not Sure About the Answer? Get an answer to your question "Calculate the kinetic energy of a 4. 9b10196 Cite This Page :. Work Equations and Formulas Calculator Science - Physics. Build Your Understanding - This is how to measure the specific heat capacity of a metal. The energy of motion can be calculated using mass and velocity. Kinetic energy Energy associated with the state of motion of an object. OP and Valued/Notable Contributors can close this post by using /lock command. A 100 kg lock is started with a speed of 2. This ultra calculator is special by allowing you to choose among a great variety of units. Yuri Gershtein named a Senior LHC Physics Center Distinguished Researcher Yuri Gershtein, Kristjan Haule, and Saurabh Jha Elected Fellows of the APS Jed Pixley wins SCES-2019 Neville Mott Prize. Physics; Calculating Energy? Calculate the work done in lifting a 500-N barbell 2. A 100 kg lock is started with a speed of 2. Here, you'll learn about how it's one of the most useful concepts in physics. Work Energy transferred "to" or "from" an object by means of a force acting on the object. The coefficient of kinetic friction between the block and the belt is 0. ICSE Solutions Selina ICSE Solutions. In physics , the kinetic energy of an object is the energy that it possesses due to its motion. (See also the KE of a Car). KINETIC ENERGY CALCULATOR. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second. Assume that the earth is a uniform sphere and that its path around the sun is circular. Energy and Power Conversion Calculator. This chemical energy in the cyclist is then converted to kinetic energy on the bike pedal due to the cyclist applying a downward force upon the bike pedal. Potential energy is the energy that exists by virtue of the relative positions (configurations) of the objects within a physical system. Yes you use the specific heat capacity of water to calculate, the formula for specific heat is Q = cmθ, Q - is the energy require to raise 1 Celsius of 1 kg o substance c - is the specific heat o a substance m - is the mass of the substance θ - is. Energy Transfer. 5 J Heat required to raise the temperature of ice from -10 °C to 0 °C = 522. A team led by the Department of Energy's Oak Ridge National Laboratory used a simple process to implant atoms precisely into the top layers of ultra-thin crystals, yielding two-sided structures. The formula for the energy of motion is KE =. 525 nF which is correct. Walter Lampl, a research scientist associated with the High Energy Physics (HEP) group in the Department of Physics at the University of Arizona, has received an "Outstanding Achievement Award" from the ATLAS collaboration at the Large Hadron Collider (LHC. 0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 10. In the International System of Units, the unit of power is the watt, equal to one joule per second. Baseball Trajectory Calculator--new version (updated, August 20, 2016) Click on the link to download an Excel spreadsheet that can be used to calculate baseball trajectories. Written by teachers for teachers and students, The Physics Classroom provides a wealth of resources that meets the varied needs of both students and teachers. Search for: 7. Quantity of heat This calculator can find missing values in the relationship between heat and temperature: heat added or removed, specific heat, mass, initial temperature and final temperature person_outline Timur schedule 2017-07-09 04:45:21. (force)× (distance) = (pressure)× (volume) The surface energy of the gas bubble is due to the difference between the bubble filled with gas and the bubble filled with liquid. ) It does this by the use of a counterweight that falls. In older works, power is sometimes called activity. Bullet Kinetic Energy Calculator. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. We are confronted every day by the notions of Energy and Power: Cars and motors are sold by Horsepower, lightbulbs by Watts, natural gas by Therms, electricity by Kilowatt-Hours, and air conditioners by Tons or BTUs per hour. 9 × 3000 = 27000. 098931 u * 931. Savings are for wattages shown. Energy against Gravity Calculator getcalc. Here are some practice questions that you can try. Your actual savings will vary based on the wattages you purchase. the bungee cord has more potential energy when it is stretched out than when it is slack. Calculate the kinetic energy of a body of mass 0. Your answer should be stated in joules, or J. The reactor physics does not need this fine division of neutron energies. A photon is characterized by either a wavelength, denoted by λ or equivalently an energy, denoted by E. Conservation of Energy 7 2. Physics Calculators Mechanics Statics and Dynamics -- Statics (stress analysis, beam analysis, resultant of vectors), Dynamics (motion on curved paths, projectile motion), and Fluid Dynamics (ideal flow, shock tube flows, airfoil flows, compressible aerodynamics, thermodynamics of air, boundary layer analysis, heat transfer). (Yes, even Passive House projects. If you heat a balloon (carefully), the molecules of air in the balloon gain energy and strike the inner walls of the balloon with greater force. 2 Chapter4 ChemicalEnergy amoleofoxygenatomswouldhaveamassof16grams. Here are some practice questions that you can try. This is hard to calculate directly, but for an ideal gas, for example, there is no potential energy, so the internal energy is simply U = 3nRT/2. h = m, then the velocity just before impact is v = m/s. Horsepower is a unit of measurement of power developed by engineer James Watt in the late 18 th century. Calculate the speed (in m/s) and the mass (in kg) of the particle. Gravitational potential energy is the energy stored in an object due to its location within some gravitational field, most commonly the gravitational field of the Earth. Work is the transfer of energy. Physics, 26. Specific energy of selected substances when combined with oxygen* * Explosives undergo decomposition and do not release energy by combining with oxygen. When one energy form is changed into another it is called a. The energy E in kilowatt-hours (kWh) per day is equal to the power P in watts (W) times number of usage hours per day t divided by 1000 watts per kilowatt: E (kWh/day) = P (W) × t (h/day) / 1000 (W/kW). KE = 1/2 mv 2 is the formula for the translational kinetic energy (KE) of the object. It relates to mass, force and motion. 00 V? (See Figure 2. making this calculator quite versatile. A boy weighing 50 kg climbs up a vertical height of 100 m. A 70 kg jogger is running up the stairs of the building and makes it up 48m vertically in 1 minute. The same difference as distance and velocity. Where: PE g or PE = gravitational potential energy. The first semester of an undergraduate physics course invariably spends a lot of time on two big ideas: The momentum principle and the work energy principle. Please spread the word about this completely free resource by linking to us. 626x10-34 J s). We provide step by step Solutions for ICSE Physics Class 10 Solutions Pdf. At its peak, a severe storm may have a total power near to 10 15 Watts: about 3,000 times the total electrical power generated in the world. Only input whole numbers, do not use a comma or point. Chapter 1 Energy and Power, and the Physics of Explosions! Energy’is’the’ability’to’do’work. Potential energy is stored energy. The Kinetic energy calculator allows you to calculate kinetic energy by entering mass and velocity in varying units. The energy goes from the. Our aim is to help students learn subjects like physics, maths and science for students in school , college and those preparing for. Concepts like how to calculate energy released from the solar fusion of deuterium or energy released from the fusion of deuterium and proton have also been covered in the material. Learning Objectives. 6 N / 60sec = 550 watt. , green plants convert solar energy to chemical energy (commonly of oxygen) by the process of photosynthesis. Assuming that half the energy is used to heat the ground and propel the debris away, the other half is used to start the ground moving. Kinetic Energy Calculator. Now if you remember the frequency f, the wavelength λ and the speed c of all electromagnetic radiations are related by the equation. Physics 03-01 Work and the Work-Energy Theorem Name: _____ Created by Richard Wright – Andrews Academy To be used with OpenStax College Physics Homework 1. If the velocity is non-uniform all you can say is what the average speed is. Physics Calculators The well-known American author, Bill Bryson, once said: “Physics is really nothing more than a search for ultimate simplicity, but so far all we have is a kind of elegant messiness. Calculate Energy Density Collection of important formulas of math and physics, with calculators: energy density. You need to make sure the units of work and energy match. LE PHYSICS 111 CONSERVATION OF MECHANICAL ENERGY Calculate the potential energy, kinetic energy, mechanical energy, velocity, and height of the ball at the various locations. In physics, mechanical work is the amount of energy transferred by a force. JEE Main Physics Work, Energy And Power Previous Year Questions with Solutions. Formulary Math and Physics Calculators. Having gained this energy during its acceleration , the body maintains this kinetic energy unless its speed changes. Both deal with forces acting on an. Physics Calculators Mechanics Statics and Dynamics -- Statics (stress analysis, beam analysis, resultant of vectors), Dynamics (motion on curved paths, projectile motion), and Fluid Dynamics (ideal flow, shock tube flows, airfoil flows, compressible aerodynamics, thermodynamics of air, boundary layer analysis, heat transfer). Use our free online unit converters to easily convert between different units of measurement. 38 eV on a wavelength of 520 nm. 282J of work is done against friction as the mass reaches the bottom of the plane. Nuclear physics involves the study of many nuclear reactions, where one atom or particle turns into another, or where. Nevertheless, this would demonstrate the concept of conversion of zero-point energy in principle if the Casimir effect attribution to zero-point energy is correct (which is debatable). The action of stretching a spring or lifting a mass is performed by an external force that works against the force field of the potential. There is not one way to calculate speed from potential energy and mass. Work and energy both use the standard unit of Joules, but the calculator above is unit less to allow you to input any unit. 00 V? (See Figure 2. At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertical shaft through a distance of 132 m. Horsepower is a unit of measurement of power developed by engineer James Watt in the late 18 th century. This silver. Metabolic energy = metabolic power (input) x time = 18. Students explore the physics exploited by engineers in designing today's roller coasters, including potential and kinetic energy, friction and gravity. Along the way, we'll talk about work, kinetic energy, potential energy, conservation of energy, and mechanical advantage. Your answer should be stated in joules, or J. Three things that affect the extent of heat transfer to an object are mass, type of substance and the amount of heat applied. If the velocity is non-uniform all you can say is what the average speed is. Calculate the resolution (FWHM) of the low-energy collimator described in Example 16-1, at source depths b = 0 and b = 10 cm, assuming it has a septal thickness of 0. Energy information Administration), but you can input whatever rate you pay per KWH found on your energy bill. First, they learn that all true roller coasters are completely driven by the force of gravity and that the conversion between potential and kinetic energy is essential to all roller coasters. Physics › Mechanics ›. Spring physics calculator solving for potential energy given spring force constant and spring stretch length. In a collision, a large force acts between two objects for a short time. The Formula to calculate mechanical energy is given by:. This page contains a JavaScript calculator of Hawking radiation and other parameters of a Schwarzschild black hole. Since kinetic energy was the first form identified, he attached a modifier to the form of energy he discovered. 40 minutes , and then calculate the amount of energy that an 11. The term is a little ambiguous because it can refer to the amount of energy you can get from breaking atomic bonds, as in the energy you get out of propane when you burn it, or the amount of energy stored as an ionic potential between molecules, such as in a neuron. E = energy measured in joules (J) m = mass measured in kilograms (kg) c = the speed of light 3 x 10 8 ms-1. Remember that no energy transforming device is 100% efficient:. To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke's Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by: $\text{PE}_{\text{el}}=\frac{1}{2}kx^2\\$. Calculate the kinetic energy of the moving football? 7. Home » Nuclear-physics » Compton-scattering Science Calculators. a) what is the joggers power output while running up the stairs? This I worked out as Power = work/time = 32961. This is hard to calculate directly, but for an ideal gas, for example, there is no potential energy, so the internal energy is simply U = 3nRT/2. Step 3: Finally, the kinetic energy of the object will be displayed in the output field. For this purpose, some useful derived value features have been introduced. If wavelength is measured in centimeters, then Planck's constant is 6. Energy is energy. George Stephans. U = Spring Potential Energy, k = Spring Force Constant, x = Spring Stretch Length. Efficiency can be explained as the amount of work done by an object to the total energy spent. In order to work a problem using Conservation of Energy, you need to know either that there are no significant forces taking energy out of the system or the size of those forces. For example, pressure is the intensity of force as it is force/area. Examples of energy include light energy, heat energy, mechanical energy, electrical energy, sound energy and many more. Efficiency is the ratio of the amount of useful energy produced (energy output- Eout) to the amount of energy used (energy input- Ein), expressed as a percentage. 1099 as of January 2011 (found at U. The next part of the question is asking me to calculate the maximum possible speed? I have used the equation Ek (Kinetic Energy) = 1/2 X mass X V (Speed) but I cannot seem to get the. 00 C of charge passes through a pocket calculator's solar cells in 4. 0 V battery and deliver a current of 0. (g=10ms-2). Solve Physics Problems Online Free with BYJU's Physics Calculator. This form of energy has. If a first object is the agent that gives energy to a second object, then the first object does work on the second object. Enter a wattage value for your own product for the most accurate estimate. A hammer:. Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. In simple words, it is the addition of kinetic and potential energy. When a football is kicked it gains kinetic energy. also calculate mean energy at 0 k (zinc electron effective mass me=0. The term work was first introduced by the French mathematician GaspardGustave Coriolis in 1826. E = energy measured in joules (J) m = mass measured in kilograms (kg) c = the speed of light 3 x 10 8 ms-1. Online calculators and converters have been developed to make calculations easy, these calculators are great tools for mathematical, algebraic, numbers, engineering, physics problems. Conservation of Mechanical Energy problems relate speed of an object at different positions. Conversion II: Steam and gas power cycles, the physics of power plants (PDF - 2. (a) Calculate the change in gravitational potential energy of the ski jumper between points X and Y. 0 scientific calculator , medium energy photon undergo Thomson or Compton scattering, and high energy photons pair produce. Energy, in physics, the capacity for doing work. Efficiency The ratio of energy which was transferred to a useful form compared to the total energy initially supplied is called the efficiency of the device. 6: carbon (to CO) 22. Plug in the values, and solve for the potential energy. The texts Katz and Giancoli use E for Total Energy, U for Potential Energy and K for Kinetic Energy. Dark Energy, Dark Matter In the early 1990s, one thing was fairly certain about the expansion of the universe. 5 × m × v 2 where KE is kinetic energy in joules, m is mass in kilograms and v is velocity in meters per second. The action of stretching a spring or lifting a mass is performed by an external force that works against the force field of the potential. Circuit waterworks provides the necessary pressure of water supplied to the turbine blades, which drives a generator, producing electricity. Disintegration energy is the energy released during radioactive decay. 66 ev) -find the lowest energy of an electron confined to move in a three dimensional potential box of lenght 0. 9 x 10^5 J/mol). The term is a little ambiguous because it can refer to the amount of energy you can get from breaking atomic bonds, as in the energy you get out of propane when you burn it, or the amount of energy stored as an ionic potential between molecules, such as in a neuron. The term work was first introduced by the French mathematician GaspardGustave Coriolis in 1826. One of the most well-known applications of half-life is carbon-14 dating. It might have enough energy density to stop its expansion and recollapse, it might have so little energy density that it would never stop expanding, but gravity was certain to slow the expansion as time went on. Course Material Related to This Topic: Read lecture notes, pages 1-2. University Physics Volume 1. The energy stored between the plates of a charged capacitor is electrical potential energy. Is this correct? What is another way to calculate loss of energy?. The more usual formula is given for an ideal gas. Every building has them. h = m, then the velocity just before impact is v = m/s. Physics of Wind Turbines. (g=10ms-2). In the calculation of Q value we require mass of reactants and products. Divide top and bottom by the radius. To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke's Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by: $\text{PE}_{\text{el}}=\frac{1}{2}kx^2\\$. I show you the equation and triangle that you need and then give you a worked example followed by three. Physics College +5 pts. The amount of energy possessed by a body is equal to the amount of work it can do when its energy is released. Planck is small, and the energy of one photon from a green light is only 2. At time 2, the gravitational potential energy equals the kinetic energy. Generate Kinetic Energy and Momentum. We have a basketball and a superball. 145 foot pounds?. OP and Valued/Notable Contributors can close this post by using /lock command. The underlying physics behind Technique 2 is discussed here; click here for the spreadsheet template described there. Physics, 26. Efficiency can be explained as the amount of work done by an object to the total energy spent. In older works, power is sometimes called activity. Calculate the kinetic energy of a 4. Classical physics refers to the physics of routine life and everyday phenomena of nature, that we can observe with our unaided senses involuntarily. 40 kW toaster oven, Wtoaster , which is used for 5. The Digital Dutch Unit Converter - Online conversion of area, currency, density, energy, force, length, mass, power, pressure, speed, temperature, volume and bytes. Home » Nuclear-physics » Compton-scattering Science Calculators. Spring physics calculator solving for potential energy given spring force constant and spring stretch length. 4 calculate the fermi energy in zinc. 9 × 3000 = 27000. Here are some practice questions that you can try. Compare the energy consumption of two commonly used items in the household. h = m, then the velocity just before impact is v = m/s. ) for inputs as well as output (J, kJ, MJ, Cal, kCal, eV, keV, C, kC, MC). This physics video tutorial provides a basic introduction into power, work, and energy. 01L Physics I: Classical Mechanics, Fall 2005 Dr. Therefore, the kinetic energy is going to be 27000 joules. Physics Calculators, Also tutorials, formulas and answers on many physics topics. By using this website, you agree to our Cookie Policy. Plot the square of the experimental velocity, vexp;ave2, as a function of mh Mc+mh. Free online physics calculators, mechanics, energy, calculators. Learn more about energy in this article. If the velocity is non-uniform all you can say is what the average speed is. Kinetic energy is the energy of an object in motion. Physics Calculators The well-known American author, Bill Bryson, once said: “Physics is really nothing more than a search for ultimate simplicity, but so far all we have is a kind of elegant messiness. Introduction to Power, Work and Energy - Force, Velocity & Kinetic Energy, Physics Practice Problems This physics video tutorial provides a basic introduction into power, work, and energy. A 100 kg lock is started with a speed of 2. The energy, E, of a photon is related to its frequency, ν, by Planck's constant (h = 6. These are very simple problems that can be solved without the use of a calculator. 0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 10. Calculate the kinetic energy again when the speed is doubled. Energy Range: Energy Range 0. 01L Physics I: Classical Mechanics, Fall 2005 Dr. At time 2, the gravitational potential energy equals the kinetic energy. The Kinetic energy calculator allows you to calculate kinetic energy by entering mass and velocity in varying units. Step 2: Now, apply the E= mc 2 equation to calculate nuclear binding energy. Where "g" is the standard acceleration of gravity which equals 9. Supports multiple metrics like meters per second (m/s), km per hour, miles per hour, yards and feet per second. Kinetic Energy: Calculate: Clear Form: Southern Maryland Waterfowl Season - 2014/2015. This chemical energy in the cyclist is then converted to kinetic energy on the bike pedal due to the cyclist applying a downward force upon the bike pedal. Full disclaimer. In fact the total amount of energy stays the same: Energy can't be created or destroyed. Enter any two of the values of mass, velocity and kinetic energy. The gas would then be doing work and transferring energy to the container. Here are my results:. Where KE is kinetic energy; m is mass ; and v is velocity; For guns, the mass of the bullet is typically measured in grains. 0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 10. This potential energy calculator enables you to calculate the stored energy of an elevated object. 9b10196 Cite This Page :. Although its original purpose was to compare the output of steam engines with the power of horses (hence its name), it has since been adopted as a unit of measurement for all sorts of engines used to power things such as vehicles, lawn mowers, boats, chainsaw, and airplanes. The principle of hydro electricity generation is quite simple. Statics and Dynamics-- Statics (stress analysis, beam analysis, resultant of vectors), Dynamics (motion on curved paths, projectile motion), and Fluid Dynamics (ideal flow, shock tube flows, airfoil flows, compressible aerodynamics, thermodynamics of air, boundary layer analysis, heat transfer). Do not use calculations for anything where loss of life, money, property, etc could result from inaccurate. (b) Calculate the energy liberated per mole of deutrium remembering that the gas is diatomic and compare with the ehat of combustion of hydrogen (about 2. How do I calculate the energy used to create one action potential along an axon with length 5cm diameter of 30 um with membrane diameter of. Power is a scalar quantity. 40 minutes , and then calculate the amount of energy that an 11. The Industrial Refrigeration Consortium is a collaborative effort between the University of Wisconsin Madison and industry. 00kg object traveling at 16. Practice problems for physics students on potential energy and kinetic energy. Learn what thermal energy is and how to calculate it. Compare the energy consumption of two commonly used items in the household. Next Page. The 9 that you see means that the kinetic energy is multiplied by 9. Physics College +5 pts. 8: coal, anthracite > 32. (force)× (distance) = (pressure)× (volume) The surface energy of the gas bubble is due to the difference between the bubble filled with gas and the bubble filled with liquid. v = velocity of a body. 00kg object traveling at 16. 09 J/g·°C specific heat of water = 4. Formula to calculate efficiency: Here is an simple example problem to calculate efficiency:. Òthe dragÓ v(t)2 = 3. A only kinetic energy B both kinetic and potential energy C mostly sound and heat energy: 9. µ k = Part II: Predicting D The distance, D pre, the friction block slides after the launch cart’s (plunger bar) trigger is released can be determined from the principle of conservation of energy. To calculate the amount of heat released, you use the specific heat formula. Although its original purpose was to compare the output of steam engines with the power of horses (hence its name), it has since been adopted as a unit of measurement for all sorts of engines used to power things such as vehicles, lawn mowers, boats, chainsaw, and airplanes. The 9 that you see means that the kinetic energy is multiplied by 9. There is an inverse relationship between the energy of a photon and the wavelength of the light given by the equation. This page explains how!. The short-term environmental benefits of the COVID-19 crisis, including declines in carbon emissions and local air pollution, have been documented since the early days of the crisis. "Energy can neither be created nor be destroyed, but can be converted from one form to another. The Physics of Current and Energy in a pocket calculator? A pocket calculator may work from a 3. ICSE Solutions for Class 10 Physics - Force, Work, Power and Energy ICSE SolutionsSelina ICSE Solutions APlusTopper. Gather your home's historical kWh usage month by month for the past 12 months. 098931) * (3*10 8 ms -1) 2 Or could simply calculate nuclear binding energy directly by converting it into MeV by, = 0. Efficiency The ratio of energy which was transferred to a useful form compared to the total energy initially supplied is called the efficiency of the device. Since the body had this kinetic energy when it was moving, that energy had to go somewhere when it stopped, probably to heat and sound energy. Next Page. Assume that the calculator operates for a time of one hour. Learn more Accept. In physics , the kinetic energy of an object is the energy that it possesses due to its motion. This calculator is generic in that it assumes averages for various appliances. µ k = Part II: Predicting D The distance, D pre, the friction block slides after the launch cart’s (plunger bar) trigger is released can be determined from the principle of conservation of energy. Energy Units and Conversions by Dennis Silverman U. The term is most commonly used in relation to atoms undergoing radioactive decay, but can be used to describe other types of decay, whether exponential or not. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. (force)× (distance) = (pressure)× (volume) The surface energy of the gas bubble is due to the difference between the bubble filled with gas and the bubble filled with liquid. If we know the power in watts of an appliance and how many seconds it is used we can calculate the number of joules of electrical energy which have been converted to sortie other. In physics, you can convert kinetic energy into potential energy and back again using conservation of energy. Welcome to CALCULATOR EDGE, an online FREE Engineering Calculators for Engineers and Students worldwide, Our website features more than few hundred calculators for solving complex equations and formulas in field of Electrical, Mechanical, Chemical, Electronics, Civil, Metallurgy, Oil & Gas, Optical, Plastics, Ceramics, Physics, Maths and many more to come, if you have any questions or. For (b), we calculate the classical kinetic energy (which would be close to the relativistic value if v were less than a few percent of c ) and see that it is not the same. Current (I) is measured in amps (A), using an ammeter. Efficiency is the ratio of the amount of useful energy produced (energy output- Eout) to the amount of energy used (energy input- Ein), expressed as a percentage. Letus consider the motion of anincompressible fluid of density. 8: coal, anthracite > 32. Binding energy = (1. Unlike the old version , this one can only be used for batted balls, not for pitched balls. 4 cm Thus, for b = 0 Rc = 0. This is called Conservation of Energy: energy just gets transformed and the total stays constant. We have also taken a look at how to calculate energy and mass balances with COMSOL Multiphysics in order to check the accuracy of simulation results. KE = 1/2*m*v^2. kilowatt-hour to watt-hour (kW·h—W·h) measurement units conversion. Energy which an object processes due to its motion is called Kinetic Energy and to explain further it is work done to accelerate a body of given mass from rest to its current or desired velocity, 'KE' represent Kinetic Energy. You can use that to prove that a mass of 1 u is equivalent to an energy of 931. The graduate program in the School of Physics provides the background and training needed to conduct and complete high quality, world-recognized research. 5 times mass times velocity squared. 9 x 10^5 J/mol). What are the units for specific heat capacity? 2. What is Meant by Kinetic Energy? In Physics, the kinetic energy is defined as the energy possessed by the object due to its motion. Using this value and our given mass, we can calculate the velocity from our original kinetic energy equation. 525 nF which is correct. Low Energy Implantation into Transition-Metal Dichalcogenide Monolayers to Form Janus Structures. Supports multiple measurement units (mv, V, kV, MV, GV, mf, F, etc. Objective. Solve for the unknown variable. In physics, you can convert kinetic energy into potential energy and back again using conservation of energy. First, they learn that all true roller coasters are completely driven by the force of gravity and that the conversion between potential and kinetic energy is essential to all roller coasters. The term work was first coined in the 1830s by the French mathematician Gaspard-Gustave Coriolis. Since kinetic energy was the first form identified, he attached a modifier to the form of energy he discovered. Low Energy Implantation into Transition-Metal Dichalcogenide Monolayers to Form Janus Structures. 2 Chapter4 ChemicalEnergy amoleofoxygenatomswouldhaveamassof16grams. Moment of Inertia & Rotational Energy Physics Lab IX. At time 2, the gravitational potential energy equals the kinetic energy. to calculate the kinetic friction coefficient μ k. If you want to know how to calculate energy, or even understand the Planck's equation, keep reading. Having gained this energy during its acceleration , the body maintains this kinetic energy unless its speed changes. Energy is the capacity to do work. The Physics of Current and Energy in a pocket calculator? A pocket calculator may work from a 3. Potential and Kinetic Energy Energy. 1 J = 1 N × 1 m. The units for energy are Joules. 17\times 10^7 m^3$of dirt gets moved. A Trebuchet is a siege engine that transfers gravitational energy into kinetic energy. If a first object is the agent that gives energy to a second object, then the first object does work on the second object. The specific heat capacity of a substance is the amount of energy required to raise the temperature of one kilogram of the substance by one degree Celsius. The maximum energy stored in the inductor is LI2/2 with I = I MAX. 660540 x 10-27 kg (per a. This page contains a JavaScript calculator of Hawking radiation and other parameters of a Schwarzschild black hole. According to the work-energy theorem if an external force acts upon an object, causing its. Work is a special name given to the (scalar) quantity = ∫ → ⋅ →. The strip of solar cells just above the keys of this calculator convert light to electricity to supply its energy needs. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second. Mass × Gravitational acceleration × Height is the rule to calculate: A gravitational potential energy B kinetic energy C heat energy: 11. For JEE Main other Engineering Entrance Exam Preparation, JEE Main Physics Work, Energy And Power Previous Year Questions with Solutions is given below. Bond dissociation energy (BDE) is a measure of the bond strength in a chemical bond. In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. This equation says that the energy of a particle of light (E), called a photon, is proportional to its frequency (), by a constant factor (h). 00 keV Energy Range 2. A boy weighing 50 kg climbs up a vertical height of 100 m. Physics 03-01 Work and the Work-Energy Theorem Name: _____ Created by Richard Wright – Andrews Academy To be used with OpenStax College Physics Homework 1. What is the formula used to calculate kinetic energy? b. 66 ev) -find the lowest energy of an electron confined to move in a three dimensional potential box of lenght 0. This is called Conservation of Energy: energy just gets transformed and the total stays constant. force in (b) force. We cover most electrical devices and home appliances, our online calculators can be edited to fit any home appliance and accurately calculate power costs. Electricity bill calculation Energy consumption calculation. Only input whole numbers, do not use a comma or point. Potential energy is one of several types of energy that an object can possess. 2020 10:57, john9727 Calculate the energy used by a 1. Play this game to review Physics. The questions on this page test your ability to use the formula:. In fact the total amount of energy stays the same: Energy can't be created or destroyed. Posted on March 20, 2019 Author Loveth Idoko Categories Physics Tags acceleration due to gravity , calculator encyclopedia , height , mass , nickzom calculator , physics. Kinetic energy is the energy of an object in motion. What is the potential energy of the barbell when it is lifted to this height?. Energy which an object processes due to its motion is called Kinetic Energy and to explain further it is work done to accelerate a body of given mass from rest to its current or desired velocity, 'KE' represent Kinetic Energy. For a body moving at a uniform velocity you can calculate the speed by dividing the distance traveled by the amount of time it took, for example one mile in 1/2 hour would give you 2 miles per hour. Elastic potential energy is the energy stored in compressed or stretched objects, elastic potential energy is used in calculations of mechanical equilibrium. ACS Nano , 2020; 14 (4): 3896 DOI: 10. A watt is one joule…. A boy weighing 50 kg climbs up a vertical height of 100 m. In physics , energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat , the object. This equation is important! For example, an object of mass 500kg has a velocity of 12 m/s, what is its kinetic energy? KE = 0·5 x 500 x 12 2 = 36,000J. 6256 X 10-34 m 2 kg/sec. Electricity bill calculation Energy consumption calculation. Solve Eqn-1 & Eqn-2 simultaneously for μ k and write the result below in terms of variables. If you want to know how to calculate energy, or even understand the Planck's equation, keep reading. After you input the necessary data such as the bullet weight, bullet velocity, powder charge weight, and the firearm weight it will output the recoil impulse, recoil velocity, and the recoil energy of the firearm. Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Use the work-energy theorem and the conservation of energy to calculate the speed of an object after falling through a given vertical height. That's not how physics is done. Tweets by @HuntOnly. Our appliance and electronic energy use calculator allows you to estimate your annual energy use and cost to operate specific products. The SI unit of energy is the joule , which is the energy transferred to an object by the work of moving it a. Where KE is kinetic energy; m is mass ; and v is velocity; For guns, the mass of the bullet is typically measured in grains. Tes Global Ltd is registered in England (Company No 02017289) with its registered office at 26 Red Lion Square London WC1R 4HQ. Physics of Wind Turbines. By using this website, you agree to our Cookie Policy. The energy E in kilowatt-hours (kWh) per day is equal to the power P in watts (W) times number of usage hours per day t divided by 1000 watts per kilowatt: E (kWh/day) = P (W) × t (h/day) / 1000 (W/kW). University Physics Volume 1. 13X103 kg /m3 and. b) If the joggers work efficiency is just 3% at what rate were they using metabolic energy? This I am having trouble with as I cannot find any formula's in. This page explains how!. Think of kinetic energy (KE) as the hammer and your arrow as the nail. The Intensity, Impedance and Pressure Amplitude of a Wave. 0 V battery and deliver a current of 0. Individuals at the UCMP and the Berkeley Department of Integrative Biology are leading experts in this field, which applies the laws of physics to organisms in an effort to understand how organisms function, and to perhaps answer questions such as : "How do organisms. The Physics of Current and Energy in a pocket calculator? A pocket calculator may work from a 3. Formula to calculate efficiency: Here is an simple example problem to calculate efficiency:. The kinetic energy of an object is directly proportional to its mass. Using the density of the crust of$2700 kg/m^3$, then about$2. In physics, you can convert kinetic energy into potential energy and back again using conservation of energy. Three things that affect the extent of heat transfer to an object are mass, type of substance and the amount of heat applied. The Digital Dutch Unit Converter - Online conversion of area, currency, density, energy, force, length, mass, power, pressure, speed, temperature, volume and bytes. edu is a platform for academics to share research papers. A team led by the Department of Energy's Oak Ridge National Laboratory used a simple process to implant atoms precisely into the top layers of ultra-thin crystals, yielding two-sided structures. When comparing incandescent or halogen bulbs to ENERGY STAR qualified bulbs or fixtures, compare the light. Household Energy Use: Simple Machines: Index. The standards-based unit in the International System of Units (SI) is the joule (J). Enjoy interesting facts about different forms of energy. Yuri Gershtein named a Senior LHC Physics Center Distinguished Researcher Yuri Gershtein, Kristjan Haule, and Saurabh Jha Elected Fellows of the APS Jed Pixley wins SCES-2019 Neville Mott Prize. There is an inverse relationship between the energy of a photon and the wavelength of the light given by the equation. Introduction to Power, Work and Energy - Force, Velocity & Kinetic Energy, Physics Practice Problems This physics video tutorial provides a basic introduction into power, work, and energy. Calculate how much useful energy is produced in one hour. In the International System of Units, the unit of power is the watt, equal to one joule per second. 5) wheremstands for mass andhstands for height. Calculating Planetary Energy Balance & Temperature. 00 C of charge passes through a pocket calculator's solar cells in 4. org is the ultimate resource for unit conversion. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second. How it is different from potential energy. kWh calculator. Yes you use the specific heat capacity of water to calculate, the formula for specific heat is Q = cmθ, Q - is the energy require to raise 1 Celsius of 1 kg o substance c - is the specific heat o a substance m - is the mass of the substance θ - is. zip: 1k: 09-11-03: Velocity after Falling enter initial height while an object is at rest, and the final height, calculate the final velocity of. The cost per KWH is found at the top, the default is the national U. Of course, once we have transferred some charge, an electric field is set up between the plates which opposes any further. By the end of this section, you will be able to: Calculate the kinetic energy of a particle given its mass and its velocity or momentum; Evaluate the kinetic energy of a body, relative to different frames of reference. It clearly lays out the course content and laboratory requirement and describes the exam and the AP Program in general. 1021/acsnano. Use the calculator below to calculate the kinetic energy of a bullet. In order to determine the kinetic energy, you must know the velocity of the bullet and the weight of the bullet in grains. If wavelength is measured in centimeters, then Planck's constant is 6. 40 minutes , and then calculate the amount of energy that an 11. 6: ethanol: 29. Physics problems: thermodynamics ; Problem 5. In older works, power is sometimes called activity. The wattage values provided are samples only; actual wattage of products varies depending on product age and features. Calculate the resolution (FWHM) of the low-energy collimator described in Example 16-1, at source depths b = 0 and b = 10 cm, assuming it has a septal thickness of 0. Trajectory Calculator This is a link to a page describing the latest version of my Trajectory Calculator, which is now fully 3-dimensional and utilizes drag and lift coefficients that have been optimized using Statcast data. average cost per KWH \$0. Important: Jump-Start Your Practice Order the Official SAT Subject Test Study Guide in Physics and get two full-length practice tests, detailed answer explanations, tips, and more. Kinetic energy is one half the objects mass times its linear velocity squared, but when an object like a water wheel is traveling in circular motion the rotational Kinetic energy has to be considered. Practice questions A bowling ball is lifted to a height …. Efficiency (%) = (useful energy out ÷ total energy in) x 100. Standard unit is the joule. If a 40 watt lamp is turned on for one hour, how many. Potential energy also includes other forms. 13g 3/2 s H b 1/2 t ! drag total velocity = paddle drop in +slope of- Òthe curlÓ breaker height duration of ride energy dissipation Future Research ÐAre the best surf spots in areas of narrow continental margin?. The units of power are watts, the units of energy are joules. Think of kinetic energy (KE) as the hammer and your arrow as the nail. First, we calculate the relativistic factor $$\gamma$$, and then use it to determine the relativistic kinetic energy. For more information, see the help screen. Walter Lampl, a research scientist associated with the High Energy Physics (HEP) group in the Department of Physics at the University of Arizona, has received an "Outstanding Achievement Award" from the ATLAS collaboration at the Large Hadron Collider (LHC. DARK ENERGY A major discovery in astrophysics in the late 1990s was the finding from type Ia supernovae redshift-luminosity observations that the expansion of. Coronavirus (COVID-19) Update: Bulbs. The units for energy are Joules. Nuclear physics involves the study of many nuclear reactions, where one atom or particle turns into another, or where. • When an object has the LEAST potential energy, it has the MOST kinetic energy. W=Fd Work is force times distance, measured in Joules. Energy Units and Conversions by Dennis Silverman U. 8 m//s^(2)). Thus, as shown in the Manuscript, “Trebuchet Mechanics,” the greatest range possible is Rm = 2*(m1/m2) h, where h is the distance the counterweight of. Einstein's famous equation relates energy and mass: E = mc2. Calculate values of different energies such as Potential, Kinetic energy etc. BMI Calculator » Triangle Calculators » Length and Distance Conversions » SD SE Mean Median Variance » Blood Type Child Parental Calculator » Unicode, UTF8, Hexidecimal » RGB, Hex, HTML Color Conversion » G-Force RPM Calculator » Chemical Molecular Weight Calculator » Mole, Moles to Grams Calculator » R Plot PCH Symbols » Dilution. com is committed to health and safety. If you heat a balloon (carefully), the molecules of air in the balloon gain energy and strike the inner walls of the balloon with greater force. This electric potential energy calculator calculates the electric potential energy of an object based on the object's charge, q, the electric field, E, of the object, and the distance, d, between the charged object we are measuring the electric potential energy of against another charge to which we are comparing it, according to the formula shown above. 09 J/g·°C)x(10 °C) q = 522. Click on the equation for more information. The coefficient of kinetic friction between the block and the incline is µk. Efficiency can be explained as the amount of work done by an object to the total energy spent. How high will they bounce if I drop them on the floor? (almost as high as they were dropped from). So it lost 30 joules. 4 calculate the fermi energy in zinc. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Your answer should be stated in joules, or J. In order to determine the kinetic energy, you must know the velocity of the bullet and the weight of the bullet in grains. Why Kinetic Energy Matters. org is the ultimate resource for unit conversion. Calculator that calculate the photon energy using Plancks constant. Solve for the unknown variable. The electron is ejected from its orbital position and the x-ray photon loses energy because of the interaction but continues to travel through the material along. To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke's Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by: $\text{PE}_{\text{el}}=\frac{1}{2}kx^2\\$. In order to work a problem using Conservation of Energy, you need to know either that there are no significant forces taking energy out of the system or the size of those forces. If wavelength is measured in centimeters, then Planck's constant is 6. g = acceleration due to gravity. I calculated the capacitance to be 12. com provides ICSE Solutions for Class 10 Physics Chapter 1 Force, Work, Power and Energy for ICSE Board Examinations. 5mv^2 and also compare E to the energy at the bottom of the projectile, E3 = 0. Electric power calculator calculation general basic electrical formulas mathematical voltage electrical equation formula for power calculating energy work power watts calculator equation power law current charge resistance converter ohm's law and power law power formulae formulas understandimg general electrical pie chart two different equations to calculate power electricas ohms law audio. After you input the necessary data such as the bullet weight, bullet velocity, powder charge weight, and the firearm weight it will output the recoil impulse, recoil velocity, and the recoil energy of the firearm. Compare the energy consumption of two commonly used items in the household. Capacitor charge and energy formula and equations with calculation examples. (g=10ms-2). You can think of a trebuchet as a see saw! Yes, a see saw is really all that a trebuchet is. kilowatt-hour to watt-hour (kW·h—W·h) measurement units conversion. Year 7 lesson on energy in food Investigating the energy content in food practical instructions not included as it was not my own resource. Solve Physics Problems Online Free with BYJU's Physics Calculator. We enter all the details of the building, set the design conditions, and get the heating and cooling. P E g = m g h. This physics video tutorial provides a basic introduction into power, work, and energy. Use our energy savings calculator to estimate energy costs savings over the life of a bulb when switching to a more energy efficient option such as LED lighting. Potential energy is stored energy. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. 7 Work and Kinetic Energy. Solving For Work Kinetic Energy Formulas Calculator Force Equations Physics Calculator Physics Equations Formulas Calculators Potential Energy Formulas Calculator Gravity Equations Calculator Friction Equations Calculator Projectile Motion Calculator Newton Second Law of Motion. 66 ev) -find the lowest energy of an electron confined to move in a three dimensional potential box of lenght 0. The chemical energy of a (chemical) substance can be converted to other forms of energy by a chemical reaction. The full name of this effect is gravitational potential energy because it relates to the energy which is stored by an object as a result of its vertical position or height. ATOMIC PHYSICS — 10% (such as properties of electrons, Bohr model, energy quantization, atomic structure, atomic spectra, selection rules, black-body radiation, x-rays, atoms in electric and magnetic fields) SPECIAL RELATIVITY — 6%. Here are some practice questions that you can try. Output in Joules (J, kJ, MJ), Watt-hours (Wh, kWh), calories (Cal, kCal) and foot-pounds (ft-lbs). Engineering Physics Resources. 8 m//s^(2)). Can you calculate the energy needed to increase the temperature of 100kg of iron by 40°C? Extension. In physics , energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat , the object. Learn more Accept. KINETIC ENERGY CALCULATOR. 6x10^6 js / 18x10^3 js. If wavelength is measured in centimeters, then Planck's constant is 6. KE = 1/2*m*v^2. Physics Calculator is available here for free use. Energy can be in many forms! Here we look at Potential Energy (PE) and Kinetic Energy (KE). A particle in the lower energy state absorbs a photon and ends up in the upper energy state. This chemical energy in the cyclist is then converted to kinetic energy on the bike pedal due to the cyclist applying a downward force upon the bike pedal. ATOMIC PHYSICS — 10% (such as properties of electrons, Bohr model, energy quantization, atomic structure, atomic spectra, selection rules, black-body radiation, x-rays, atoms in electric and magnetic fields) SPECIAL RELATIVITY — 6%. To +W From -W - Constant force: F x ma x d v v v v axd ax 2 2 2 0 2 2 0 2 Work done by the force = Energy. Exothermic reactions release energy by transferring heat to their surroundings. In older works, power is sometimes called activity. the bungee cord has more potential energy when it is stretched out than when it is slack. , green plants convert solar energy to chemical energy (commonly of oxygen) by the process of photosynthesis. Practice questions A bowling ball is lifted to a height […]. kinetic energy = mass * velocity² / 2 E = m * v² / 2. The maximum energy stored in the inductor is LI2/2 with I = I MAX. Efficiency can be explained as the amount of work done by an object to the total energy spent. You can think of a trebuchet as a see saw! Yes, a see saw is really all that a trebuchet is. Example of Few questions where you can use this Mechanical Energy Formula calculate the Mechanical energy of the object have mass 10 kg and velocity 3m/s and height above the ground is 10 m calculate the Kinetic Energy,Potential energy and Mechanical energy of the object have mass 1 kg and velocity 2m/s and height above the ground is 50 m. Kinetic energy = Joules Kinetic Energy Calculator is a free online tool that displays the kinetic energy of the object. Contribute to Ghostlydestinypolice/Physics_Energy development by creating an account on GitHub. We cover most electrical devices and home appliances, our online calculators can be edited to fit any home appliance and accurately calculate power costs. 2 Kinetic Energy. Current (I) is measured in amps (A), using an ammeter. The units of power are watts, the units of energy are joules. The kinetic energy just before impact is equal to its gravitational potential energy at the height from which it was dropped: K. 94u2rotkh0qgw ts2269o179 taplsbcnrzp02 tqj6cos4hnt038 862fohcw5vw 0kib5khqvn7x7u imx0rwjjl87j 7v6n0lvam9qps6 1ep3ku4gpb4 7jfyw1bfipc qndzw85umfgmg 3hdpzbzs6my yeu4magtr0k3kw0 e9wp756pxxgsa akrv0bgii4hb 5o2tlid2tvyo1 wuokdswbla3 5l1wnz0qxp z51o2y1d0z5 ff022i7pygcok f0hrxwv11v20 vubj5vlrkybs scwb930zomjiv9 1charfjnbcnye1 omblwc9sydmb8n2 b3c248fb9a0rz2 | 2020-12-04T16:48:27 | {
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https://math.stackexchange.com/questions/2823510/why-cant-i-get-an-accurate-polynomial-equation-for-these-data-points | # Why can't I get an accurate polynomial equation for these data points?
I have 30 data points that I have digitised from the red dashed line in the graph below. My goal is to find an approximate equation to represent the line.
I have tried to get a 6 degree polynomial trendline through Excel for these points, but if I then plot the trendline equation in Excel or Wolfram, the numbers are very clearly incorrect. I have tried several other websites to get a more accurate polynomial expression but so far I have not been able to get an equation that looks anything like the original graph and I am wondering what I could be doing wrong.
The below points are what I have been pasting into various regression analysis calculator websites (for example Desmos). I have then retyped the 6 degree polynomial from these websites into Excel and fed a couple of x coordinates into them. I have tried 8 different sites now and I can't replicate the line in the graph. Can anyone recommend a different approach or provide some advice?
14510.28 500.01
23609.49 443.79
36769.24 426.06
63088.75 407.6
85648.33 393.39
130767.49 377.42
198446.22 361.99
294324.44 347.57
432245.5 333.84
647381.85 320.45
891025.3 310.05
1113844.52 303.03
1359458.84 296.97
1614059.8 291.76
1891078.16 287.06
2166769.49 283.11
2454404.12 279.59
2730758.96 276.68
3024738.48 273.68
3301051.85 271.28
3583046.59 269.03
3875657.6 266.85
4158315.85 264.97
4442467.02 263.13
4733585.11 261.35
5015579.85 259.88
5314535.74 258.56
5590849.11 257.29
5889805 255.94
6036275.7 255.91
• why degree 6? have you tried higher degrees? – Glougloubarbaki Jun 18 '18 at 9:56
• The function doesn't look like a polynomial would be the first choice to approximate it. Have you tried fitting a sum of inverse powers? Or perhaps just subtract out the singularity at the origin and approximate the rest with a polynomial. Do you have any information (other than the data in the graph) about the type of the singularity at the origin? – joriki Jun 18 '18 at 11:05
• Unfortunately the graph is all I have. But I did end up taking out a couple of the plot values from the start and end of the line and have gotten the equation 1289.3859x^(-0.1039) which fits it pretty well. It seems some of the initial values were throwing it of. – david_10001 Jun 18 '18 at 13:35
Looking at the data, I thought that something as simple as $$y=a+\frac b {x^c}$$ could be reasonable (this is very similar to what you wrote in a comment).
A nonlinear regression leads to $R^2=0.999822$ which is quite good and the parameters are $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 101.314 & 26.3362 & \{47.1788,155.449\} \\ b & 1605.95 & 157.906 & \{1281.37,1930.53\} \\ c & 0.14960 & 0.01667 & \{0.11535,0.18386\} \\ \end{array}$$
If, as you did, we remove the $a$ term, we get $R^2=0.999775$ (almost the same) and the parameters are $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ b & 1306.83 & 24.8952 & \{1255.65,1358.00\} \\ c & 0.10489 & 0.00145 & \{0.10192,0.10786\} \\ \end{array}$$ very close to what you wrote in your comment.
In order to better represent the left part of the plot, we could consider $$y=\frac b {(x-d)^c}$$ which would lead to $R^2=0.999914$ and $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ b & 1166.87 & 21.3244 & \{1122.96,1210.79\} \\ c & 0.09707 & 0.00132 & \{0.09435,0.09978\} \\ d & 7658.20 & 743.875 & \{6126.16,9190.24\} \\ \end{array}$$
1. Polynomial regression is based on Taylor's expansion at the origin, hence as further your data goes from the $0$, the less accurate the model.
2. Try to fit an exponential model instead of the polynomial, i.e., $$\mathbb{E}[y_i|x]=\beta_0 \exp\{-(\mathbf{x}^T \beta)\},$$
then $$\ln\mathbb{E}[y_i|x] = \ln \beta_0-\mathbf{x}^T\beta,$$ now estimate the parameters and then exponentiation the predicted values. These values will still be biased, and if the bias severe you can use the OLS' values as initial values to the non-linear regression.
• It looks more like an inverse power than an exponential to me. – joriki Jun 18 '18 at 12:34
• Maybe you are right. He models time to failure, hence the first intuitive go-to model is commonly an exponential. I haven't bothered to fit the model to his data, so I cannot say anything concrete – V. Vancak Jun 18 '18 at 12:44
• But not time to failure as a random variable (which might be expected to be exponentially distributed), but the relationship between stress amplitude and (presumably average) time to failure (where I wouldn't know what to expect). – joriki Jun 18 '18 at 13:13 | 2021-03-05T13:44:12 | {
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https://math.stackexchange.com/questions/3529794/why-is-sum-p-in-s-n-2cp-equal-to-n1/3529843 | # Why is $\sum_{p \in S_n} 2^{c(p)}$ equal to $(n+1)!$?
It is obvious that $$\sum_{p \in S_n} 1=n!$$ because it is just counting how many permutations there are of $$n$$ symbols.
But I have also observed that $$\sum_{p \in S_n} 2^{c(p)}=(n+1)!$$, where $$c(p)$$ is the number of cycles of $$p$$.
What is the combinatorial interpretation of this identity?
An example. In $$S_3$$ we have one permutation with 3 cycles, three permutations with 2 cycles and two with 1 cycle. Then $$1\times 2^3+3\times 2^2+2\times 2^1=24=4!$$
• What do you mean by the number of cycles of $p$? – Rob Arthan Jan 31 '20 at 21:02
• @RobArthan When a permutation acts on the set $\{1,...,n\}$ it will have invariant subsets where its action is transitive. These are the cycles. – thedude Jan 31 '20 at 21:03
• Take any permutation in $S_n$. Try to create a permutation in $S_{n+1}$. You need to add $n+1$. You want a scheme of adding $n+1$ so that it is a distinct permutation. For each cycle, you either include $n+1$ or you don't (possibly merging cycles). My guess would be that there is a scheme where this produces a distinct element of $S_{n+1}$. – InterstellarProbe Jan 31 '20 at 21:06
• This answer generalises your result. – Milten Jan 31 '20 at 21:10
• This also has a probabilistic interpretation - if $\sigma$ is a random permutation in $S_n$, then $E(2^{c(\sigma)}) = n+1$ – Michael Lugo Jan 31 '20 at 21:52
The sum in question counts the number of pairs $$(f, p)$$ where $$f$$ is a function from $$\{1, 2, \dots, n\}$$ to $$\{1, 2\}$$, and $$p \in S_n$$ such that $$f \circ p = f$$. I don't know if there is an easy way to see that this is equal to $$(n + 1)!$$, but here is an approach:
First let's verify that the sum does actually count what I say that it does. Suppose that we have already chosen the permutation $$p$$. A function $$f$$ satisfies the conditions above if and only if for each cycle of $$p$$, we have that $$f$$ maps every element of that cycle to the same element of $$\{1, 2\}$$. We can thus determine $$f$$ by choosing the image of each cycle, and there are two options for the image of each cycle, giving us $$2^{c(p)}$$ functions in total.
Let's now instead determine the sum by counting the pairs $$(f, p)$$ such that there are $$k$$ elements of $$\{1, 2, \dots, n\}$$ that are mapped to $$1$$ under $$f$$. There are $$\binom{n}{k}$$ ways to choose which $$k$$ elements these are. There are then $$k!$$ ways to choose how $$p$$ permutes these $$k$$ elements, and $$(n - k)!$$ ways to choose how $$p$$ permutes the remaining elements. This gives us $$\binom{n}{k} k! (n - k)! = n!$$ ways to choose the pair $$(f, p)$$. Noting that $$k$$ can take any value from $$0$$ to $$n$$, this gives us $$(n + 1) n! = (n + 1)!$$ pairs in total.
• I hadn't heard of Polya's Enumeration Theorem before, but now that I have, I think that this may be essentially the same solution as @Milten's. – Dylan Jan 31 '20 at 22:03
• For your last paragraph, suppose that $f^{-1}(1)=\{x_1,x_2,\ldots,x_k\}$ and $f^{-1}(2)=\{y_1,y_2,\ldots,y_l\}$ with $x_1<x_2<\ldots<x_k$ and $y_1<y_2<\ldots<y_l$. There is a $1$-$1$ correspondence between the pairs $(f,p)$ and the permutations of $(0,1,2,\ldots,n)$ by taking $$(f,p)\mapsto \big(p(x_1),p(x_2),\ldots,p(x_k),0,p(y_1),p(y_2),\ldots,p(y_l)\big).$$ This is an alternative way to show that there are $(n+1)!$ pairs $(f,p)$. – Batominovski Jan 31 '20 at 22:36
• If I am not mistaken, your proof also shows that $$\sum_{p\in S_n} r^{c(p)}=\frac{(n+r-1)!}{(r-1)!}$$ for all non-negative integer $r$. Hence, it follows that $$\sum_{p\in S_n} x^{c(p)} =x(x+1)(x+2)\cdots (x+n-1)$$ as a polynomial identity. – Batominovski Jan 31 '20 at 22:47
• @Dylan Indeed it seems we had pretty much the same idea. I think you basically proved Pólya's theorem in your answer. I would say yours is probably cleaner as a combinatorial proof though :) – Milten Feb 1 '20 at 9:16
It can be seen as an application of Pólya's enumeration theorem.
Let us take $$n$$ beads and colour each of them black or white. We consider two colourings equivalent, if there are the same number of black and white beads. That is, if some permutation $$\sigma\in S_n$$ takes one of the colourings to the other. It is clear that there are then $$n+1$$ distinct colourings (since there can be $$0,1,\ldots, n$$ black beads). Pólya's theorem then says that: $$n+1 = \frac{1}{|S_n|}\sum_{\sigma\in S_n}2^{c(\sigma)} = \frac{1}{n!}\sum_{\sigma\in S_n}2^{c(\sigma)}$$ The $$2$$ in the formula is the number of colours.
I guess this is somewhere between a combinatorial interpretation and just a proof... Depends on your attitude towards Pólya's theorem I guess.
EDIT: Let me just include the immediate generalisation. If we have $$m$$ colours, then there are $$\binom{n+m-1}{n}$$ different colourings (by stars and bars). So: $$\sum_{\sigma\in S_n}m^{c(\sigma)} = n!\binom{n+m-1}{n} = \frac{(n+m-1)!}{(m-1)!} = m(m+1)\cdots(n+m-1)$$ This only works for $$m\in\mathbb N$$ of course. But as WE Tutorial School points out in a comment, this shows that the left and right hand sides are equal as polynomials in $$m$$, so the identity is proved for any value of $$m$$ (even complex and whatnot).
The combinatorial class of permutations with the number of cycles marked is
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}( \mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z}) + \mathcal{U}\times \textsc{CYC}_{=2}(\mathcal{Z}) + \mathcal{U}\times \textsc{CYC}_{=3}(\mathcal{Z}) + \cdots).$$
This gives the EGF
$$G(z, u) = \exp\left(uz+u\frac{z^2}{2} + u\frac{z^3}{3}+\cdots\right) \\ = \exp\left(u\log\frac{1}{1-z}\right).$$
A permutation on $$n$$ elements and having $$k$$ cycles is represented by $$u^k \frac{z^n}{n!}.$$ For the sum we set $$u=2$$ and obtain
$$H(z) = \exp\left(2\log\frac{1}{1-z}\right) = \frac{1}{(1-z)^2}.$$
We then get for the answer
$$n! [z^n] H(z) = n! [z^n] \frac{1}{(1-z)^2} = n! {n+1\choose 1} = (n+1)!.$$
This is the claim.
Addendum. With two being replaced by $$m$$ we obtain
$$n! [z^n] H(z) = n! [z^n] \frac{1}{(1-z)^m} = n! \times {n+m-1\choose m-1} = n! \times {n+m-1\choose n}.$$
• I apologise for using comments in this unintended way, but: I just wanted to tell you that this answer inspired me to begin reading Analytic Combinatorics by Flajolet & Sedgewick. I am now 200 pages deep and I love it. I had never heard of these methods before, so thank you :) – Milten Sep 3 '20 at 9:05 | 2021-01-25T19:31:27 | {
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https://mathoverflow.net/questions/350979/permuting-n-points-in-a-2-manifold | # Permuting $n$ points in a $2$-manifold
Given $$n$$ points on a connected $$2$$-manifold $$M$$, I'd like to consider the homotopy classes of paths that "permute" these points.
Edit (Clarifying what I mean by this):
Given a set of $$n$$ distinct points $$T=\{x_{1},\ldots,x_{n}\}\subset M$$, to each point we assign a continuous simple curve $$\gamma_{i}:[0,1]\to M$$ such that $$\gamma_{i}(0)=x_{i}, \gamma_{i}(1) \in T$$ and $$\gamma_{i}(s)\neq\gamma_{j}(s)$$ for all $$s\in[0,1]$$ (where $$i\neq j$$). I'd like to consider the homotopy classes of all such possible curves.
1. It seems obvious that these homotopy classes should constitute the elements of a group. Is that right? If so, what's the name of this group? I'm inclined to simply call this the motion group $$\text{Mot}_{n}(M)$$ of the $$n$$ points on $$M$$. Does this coincide with the mapping class group of $$n$$ points in $$M$$? Also, do I need any more restrictions on $$M$$? If so, why?
• E.g. considering $$3$$-space for a moment, it is obvious that $$\text{Mot}_{n}(\mathbb{R}^{3})\cong S_{n}$$ (where $$S_{n}$$ is the permutation group).
• It is also obvious that $$\text{Mot}_{n}(\mathbb{D}_{2})\cong \text{Mot}_{n}(\mathbb{R}^{2})\cong B_{n}$$, where $$\mathbb{D}_{2}$$ is the $$2$$-disk with boundary and $$B_{n}$$ is the braid group.
1. Consider a presentation of $$\text{Mot}_{n}(M)$$ with relations $$R$$.
(i) Is it true that $$\text{Mot}_{n}(M)\cong B_{n}(M)$$, where $$B_{n}(M)$$ is the surface braid group for $$M$$?
(ii) Under what conditions will it be true that the generator relations $$G$$ of $$B_{n}$$ will be a subset of $$R$$?
For instance, I'm sure that $$\text{Mot}_{n}(S^{2})\cong B_{n}(S^{2})$$ : in which case, we do have $$G\subset R$$ (in fact, $$B_{n}(S^{2})$$ is a quotient of $$B_{n}$$).
Relevant Resources:
A survey of surface braid groups and the lower algebraic K-theory of their group rings
I think my notion of $$\text{Mot}_{n}(M)$$ coincides with the Definition in Section 2.2 of the above paper. If so, then the answer to 2(i) is yes (according to the paper). Building on this, I believe Theorems 12 and 13 of Bellingeri (for $$m=0$$) may provide a partial answer to 2(ii).
• Questions about the definition. Let $T\subset M$ be a finite subset of cardinality $n$.This gives a basepoint for the configuration space of $n$ unordered points in $M$. Call this configuration space $C_n(M)$. Then it seems to me what you are describing is $\pi_0$ of the loop space $\Omega C_n(M)$. Is that right? If so, then your group is nothing more than $\pi_1(C_n(M))$. Right? – John Klein Jan 23 at 2:15
• Without further care in formulating definitions, I do not know what it might mean for a path to permute points. I presume by a "path" you mean a continuous function $f : [0,1] \to M$, in which case "permuting points" is not something a path ordinarily does. – Lee Mosher Jan 23 at 3:48
• @Lee: The OP probably meant a path in the configuration space of $n$-tuples of points. Of course he needs to make ir clear. – Mark Sapir Jan 23 at 10:04
• I've clarified what I meant in the post now. @JohnKlein I don't have a clear understanding of why $\pi_{1}(C_{n}(M))$ doesn't restrict to the trivial permutation? – Meths Jan 23 at 13:39
• Isn't your description just an equivalence class of path $\gamma: [0,1] \to C_n(M)$ such that $\gamma(0) = T = \gamma(1)$, where $C_n(M)$ is the space of subsets of cardinality $n = |T|$? – John Klein Jan 23 at 14:03
The comments seem to have answered questions 1 and 2i, to show that the group $$\operatorname{Mot}_n(M)$$ is indeed the surface braid group $$B_n(M)$$.
To answer 2ii, consider a disk $$D \subset M$$ such that $$T \subset D$$. Then the inclusion map $$D \hookrightarrow M$$ induces a homomorphism $$B_n \to B_n(M)$$, so the relations in $$B_n$$ always hold in $$B_n(M)$$.
• Bump... quick question: for your (induced homomorphism) argument to be true, doesn't this also mean there has to be a continuous map from the (unordered) configuration space of $D$ to that of $D\subset M$? That doesn't seem immediately obvious to me. Sorry for the posthumous enquiry! – Meths Feb 22 at 11:45
• It's not immediately obvious, but It does induce a continuous map between unordered configuration spaces. One way to see this is to first check it on the ordered configuration spaces, and then make sure the induced inclusion respects the quotient maps. However, it's clearer if you use the viewpoint you presented in your original question, of looking at elements of the surface braid group as a collection of paths on $D$ or $M$. Then a collection of paths in $D$ is clearly a collection of paths in $M$ once you include $D$ into $M$. – Ty Ghaswala Feb 24 at 3:56 | 2020-03-31T14:08:58 | {
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http://openstudy.com/updates/55d1f70ae4b0c5fe98066c0c | ## anonymous one year ago Half-Life reaction help?
1. anonymous
The purpose of this hands-on lab is to model the concept of half-life using a sample to represent radioactive atoms. Materials 200 M&M® candies, pennies, or other small candy/item with two distinct sides shoe box or other small box with a lid Procedure Place 200 candies in the shoe box, lettered sides up. The candies will stand for atoms of a hypothetical radioactive element. Cover the box and shake it vigorously for three seconds. This is one time interval. Remove the lid and take out any candies (atoms) that have that are now showing lettered sides down. These candies represent the atoms that decayed during the time interval. Count and record in a data table the number of decayed atoms and the number of remaining, not decayed, atoms. Continue repeating steps two and three until all atoms have decayed or you have reached 30 seconds on the data table. Repeat the entire experiment (steps 1–4) a second time and record all data. Data and Observations Create and complete a data table, like the one below, for each trial. Time (seconds) | Radioactive atoms remaining (not decayed) | Atoms decayed 0 | 200 | 0 3 | 107 | 93 6 | 51 | 56 9 | 29 | 27 12 | 18 | 11 15 | 7 | 11 18 | 6 | 1 21 | 4 | 2 24 | 2 | 2 27 | 2 | 0 30 | 2 | 0 Calculations Time (seconds) | Radioactive atoms remaining (not decayed) | Atoms decayed 0 | 200 | 0 3 | 113 | 87 6 | 46 | 41 9 | 31 | 15 12 | 16 | 15 15 | 8 | 8 18 | 5 | 3 21 | 3 | 2 24 | 1 | 2 27 | 1 | 0 30 | 0 | 1
2. anonymous
Determine the average number of atoms remaining (not decayed) at each three-second time interval by adding the results from the two trials and dividing by two. 3 = 163.5 6 = 48.5 9 = 30 12 = 17 15 = 7 18 = 5.5 21 = 2.5 24 = 1.5 27 = 1.5 30 = 1 Create a table that compares time to the average number of atoms remaining at each time interval.
3. anonymous
After how many time intervals (shakes) did one-half of your atoms (candies) decay? Trial 1: It took 8 Trial 2: 6 shakes What is the half-life of your substance? <-------------------(This I need help with)
4. anonymous
The answer is B
5. cuanchi
@alante in the first shake you went from 200 to 107-113 remaining atoms, this is almost 50% of the original amount . This will be the 1/2 life of your atom, is the time that takes the initial amount decrease in 1/2 of the original value.
6. anonymous
@Woodward would you mind helping me with this? Cauanchi is offline. I'm not sure if almost half would cut it, especially with it not being something like 101, 107 seems to high to cut it as "almost"
7. anonymous
too*
8. anonymous
I think that's part of the fun of the experiment. It's not exact, but that's how it was and we must accept the deviation of reality from the theory. So we didn't have exactly half, but it is not too far away. $\frac{107}{200} *100\% = 53.5\%$ So you expected 50% to be left, so instead you had 3.5% more. I don't think that's that bad but I guess this is kind of subjective. It's certainly much closer than 60% which would make me start to wonder. I don't know if that really answers your question or not...?
9. anonymous
See, that's the confusing part. This part of the lesson didn't explain anything along the lines of "close enough" so, I expected it to want something exactly 50%. There's more to follow, but I still need an understanding on what the half life is
10. anonymous
Not to mention there's also the other trial I needed to do where 113 is halflife being 56% of the 200
11. anonymous
would you mind if I were to just ask for the next questions? I'll eventually figure it out
12. anonymous
I'll just post these xD If the half-life model decayed perfectly, how many atoms would be remaining (not decayed) after 12 seconds? If you increased the initial amount of atoms (candies) to 300, would the overall shape of the graph be altered? Explain your answer. Go back to your data table and for each three-second interval divide the number of candies decayed by the number previously remaining and multiply by 100. Show your work. The above percentage calculation will help you compare the decay modeled in this experiment to the half-life decay of a radioactive element. Did this activity perfectly model the concept of half-life? If not, was it close? Compare how well this activity modeled the half-life of a radioactive element. Did the activity model half-life better over the first 12 seconds (four decays) or during the last 12 seconds of the experiment? If you see any difference in the effectiveness of this half-life model over time, what do you think is the reason for it?
13. anonymous
Sure you can try, I'm a bit distracted at the moment so no guarantees though! Half-life is an amount of time. So for instance 10 seconds or 5,730 years are possible values for a half-life. So now that we know that it's a length of time, what's supposed to happen during this time? Well during this time, half of what you started with will decay! So let's say we have 50 pounds of some substance that has a half-life of 10 seconds. Then that means 10 seconds later there's only about 25 pounds left! If we wait another 10 seconds (20 seconds in all) we will only have about 12.5 pounds of it left! It disappears quite rapidly, and in fact that's because this is a form of exponential decay.
14. anonymous
I understand, I appreciate any help man. Alright well 107 is at 3 seconds, but a "close enough" half is with 51 at 6 seconds. then we have 29 at 9 seconds. But what does this prove to a half life?
15. anonymous
I'm still really confused ._.
16. abb0t
that the whole decay process is not a constant rate half-life is just a description whence the amount is halved (or close) of the original number of species from when the decay process started at a given time.
17. anonymous
but it ask what the half life is to the substance? There's 2 trials and I'm still confused on what exactly I do to find the half life of that substance. Do I use both data? Separate data for 2 substance?
18. abb0t
it asked you to use both data then divide by two
19. anonymous
But that was for the first question. I also don't have substances for question 3 to use nor any weight. I know we can use the 2 trials but it just isn't making since to me what to do with it. Is it as simple as putting 107 and 113 by 2? Or 3 for 3 seconds?
20. anonymous
I got kind of lazy with the rest of the questions, sorry about that. I'm still not 100% sure of what to do on number 3 though.
21. cuanchi | 2017-01-19T15:30:12 | {
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http://mathhelpforum.com/calculus/130818-differentiation-need-help-latex.html | # Thread: Differentiation and Need HELP with LATEX!
1. ## Differentiation and Need HELP with LATEX!
g(x) = {-x, x<=0
3x^2,x>0
(a) Evaluate the limit of {g(x+delta x) - g(x)}/{delta x) for x<=0 and x>0 as delta x tends to 0
(b) evaluate the limit of {g(delta x) - g(0)}/{delta x) as delta x tends to 0 from the right and as delta x tends to 0 from the left
(c) sketch the graph of g'(x)
Is g(x) continuous at x=0?
I need help with (b) and (c) only. Is there any user-friendly programmes to type the equations? I find the syntax of latex very difficult to learn. Btw, how do you type the above equations (using latex)?
Thanks!
2. B is simply asking you to compute the separate limits for the left and right side of the equation corresponding the respective piece of your piece-wise function. You are meant to see whether the limit from the right is the same as the limit from the left.
3. Originally Posted by ANDS!
B is simply asking you to compute the separate limits for the left and right side of the equation corresponding the respective piece of your piece-wise function. You are meant to see whether the limit from the right is the same as the limit from the left.
Does this mean I must find the limit for $-x$, $x\leq 0$ as
delta x approaches 0 from the right and from the left as well as the limit for $3x^2$, $x>0$ as delta x approaches 0 from the right and from the left?
Can anyone show me the working??? Thank you.
4. Originally Posted by cyt91
Does this mean I must find the limit for $-x$, $x\leq 0$ as
delta x approaches 0 from the right and from the left as well as the limit for $3x^2$, $x>0$ as delta x approaches 0 from the right and from the left?
Can anyone show me the working??? Thank you.
For (a), you want to take x< 0 (not $\le 0$) and can assume that delta x is small enough that x+ delta x is also negative- then you can use "-x" as the formula for both g(x) and g(x+ delta x). If g(x)= -x, what is g(x+ delta x)?
Then take x> 0 and assume that delta x is small enough that x+ delta x is also positive- then you can use " $3x^2$ as the formula for both g(x) and g(x+ delta x). If $g(x)= 3x^2$, what is g(x+ deltax)?
For (b), where x=0, taking the limit "from the left" means using g(0+ delta x)= -delta x and taking the limit "from the right" means using $g(0+ delta x)= 3(delta x)^2$.
5. Originally Posted by HallsofIvy
For (a), you want to take x< 0 (not $\le 0$) and can assume that delta x is small enough that x+ delta x is also negative- then you can use "-x" as the formula for both g(x) and g(x+ delta x). If g(x)= -x, what is g(x+ delta x)?
Then take x> 0 and assume that delta x is small enough that x+ delta x is also positive- then you can use " $3x^2$ as the formula for both g(x) and g(x+ delta x). If $g(x)= 3x^2$, what is g(x+ deltax)?
For (b), where x=0, taking the limit "from the left" means using g(0+ delta x)= -delta x and taking the limit "from the right" means using $g(0+ delta x)= 3(delta x)^2$.
For (a) :
x<0,
g(x+delta x)= -(x+delta x)
Is this correct?
Then, the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
{-(x+delta x)-[-x]}/{delta x}=-1
Correct?
x>0,
g(x+delta x)=3(x+delta x)^2
Is this correct?
Then the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
{3(x+delta x)^2-3x^2}/[delta x]=6x
Correct?
For (b):
when delta x approaches 0 from the left, the limit of
{g(0+delta x)-g(0)}/{delta x} = {-(0+delta x)-[-0]}/[delta x]
=-1
and when delta x approaches 0 from the right, the limit of
{g(0+delta x)-g(0)}/{delta x}
= {3(0+delta x)^2-3(delta x)^2}/[delta x]
=0
Are my workings correct?
How about the graph? Is it g'(x)=-1 for x<=0 and g'(x)=6x for x>0?
Hence, is g(x) not continuous at x=0 since g'(x) for x<=0 and g'(x) for x>0 are not equal?
Thank you!
6. Originally Posted by cyt91
For (a) :
x<0,
g(x+delta x)= -(x+delta x)
Is this correct?
Then, the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
{-(x+delta x)-[-x]}/{delta x}=-1
Correct?
Yes, if x< 0 and $|\delta x|< |x|$, which we can assume since we are taking the limit as $\delta x$ goes to 0, $x+ \delta x$ is also < 0 so $f(x+ \delta x)- f(x)= -(x+\delta x)- x= -\delta x$.
x>0,
g(x+delta x)=3(x+delta x)^2
Is this correct?
Yes, if x> 0 and $\delta x> x$, which we can assume since we are taking the limit as $\delta x$ goes to 0, $x+ \delta x$ is also > 0 so $f(x+ \delta x)- f(x)= 3(x+ \delta x)^2- 3x^2$.
[quote]Then the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
{3(x+delta x)^2-3x^2}/[delta x]=6x
Correct?[/tex]
Yes. 3(x+ \delta x)^2= 3(x^2+ 2x\delta x+ \delta x^2)= 3x^2- 6x\delta x+ \delta x^2[/tex]. $3(x+ \delta x)^2- 3x^2= 6x\delta+ \delta x^2$ so $(3(x+ \delta x)^2- 3x^2)/\delta x= 6x+ \delta x$ and the limit of that, as $\delta x$ goes to 0 is 6x.
Very good!
For (b):
when delta x approaches 0 from the left, the limit of
{g(0+delta x)-g(0)}/{delta x} = {-(0+delta x)-[-0]}/[delta x]
=-1
and when delta x approaches 0 from the right, the limit of
{g(0+delta x)-g(0)}/{delta x}
= {3(0+delta x)^2-3(delta x)^2}/[delta x]
=0
Both correct.
Are my workings correct?
How about the graph? Is it g'(x)=-1 for x<=0 and g'(x)=6x for x>0?
Hence, is g(x) not continuous at x=0 since g'(x) for x<=0 and g'(x) for x>0 are not equal?
Thank you!
No, "continuous at x= a" mean $\lim_{x\to a} g(x)= g(a)$. It has nothing to do with g' because most continuous functions are not differentiable! And, in fact, to have a derivative, a function must be continuous.
Here, the point is that $\lim_{x\to 0^-} g(x)= \lim_{x\to 0} -x= 0$ and $\lim_{x\to 0^+} g(x)= \lim_{x\to 0} 3x^2= 0$.
So $\lim_{x\to 0} g(x)= 0= g(0)$ and g is continuous at x= 0. | 2016-12-06T12:17:36 | {
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https://math.stackexchange.com/questions/2777325/integrating-a-function-using-u-substitution | # Integrating a function using u-substitution
I'm trying to figure out how to integrate this function. I tried several tricks from my toolkit, but I can't seem to figure it out.
$$\int\ \frac {e^{2x}-6e^x}{e^x+2}\ dx$$
So let's say that I factor out some terms and split the equation to:
$\displaystyle\int\ \frac {e^{2x}}{e^x+2}\ dx\;$ and $\;\displaystyle -6\int\ \frac {e^x}{e^x+2}\ dx$
Now I cannot see how the derivative of $e^{2x}$ is $e^x+2$ or vice versa. When looking at the other side, it seems more reasonable to set $u$ to $e^x$. So let's do that:
$\displaystyle -6\int\ \frac {e^x}{e^x+2}\ dx$ = $\displaystyle -6\int\ \frac {u}{u+2}\ dx$
Okay, so as you can see, I didn't get anywhere here. I'm really new to $u$-substitution and partial integration, I probably missed some crucial step.
$$\int\ \frac {e^{2x}-6e^x}{e^x+2}\ dx$$
Let $\displaystyle e^x + 2 = u$.$^{\color{blue}{(1)}}$
So $\displaystyle e^x = u-2$, and so $\displaystyle e^{2x} = (u-2)^2$.
$\color{blue}{(1)}$ Now $\displaystyle e^x \ dx = du \iff \ dx = \frac{du}{e^x}$, but recall from above, that $\displaystyle e^x = u-2$.
So in fact, $\displaystyle \color{blue}{dx = \frac{du}{u-2}}$.
That gives us the integral
\begin{align} \int \frac{(u-2)^2 - 6(u-2)}{u\cdot \color{blue}{(u-2)}}\ \color{blue}{du} & = \int \frac{(u-2) - 6}{u}\,du \\ & = \int \frac{u-8}{u}\,du \\ &= \int \left(1- \frac 8u\right)\ du = \end{align}
Can you take it from here?
• Thanks! I get it now :). – user472288 May 11 '18 at 23:43
• Your welcome! ${}$ – Namaste May 11 '18 at 23:46
The trick is to see that $e^{2x}$ can be easily changed and factored. We start with the integral. $$\int \frac {e^{2x}-6e^x}{e^x+2}~ {\rm d}x$$ We can use the fact that $a^{mn} = (a^m)^n$ to change this to $$\int \frac {(e^x)^2-6e^x}{e^x+2}~ {\rm d}x = \int \frac {e^x(e^x - 6)}{e^x+2}~ {\rm d}x$$ We can then use the substitution $u = e^x +2$ and ${\rm d}u = e^x{\rm d}x$ and solve from there. $$\int \frac {u - 8}{u}~ {\rm d}u = \int \left(1 - \frac{8}{u}\right){\rm d}u = u - 8\ln|u| + C$$ We then substitute the equivalent of $u$. $$e^x + 2 - 8\ln|e^x + 2| + C$$ We can also now say that the $2$ can become part of the constant $C$ and remove the absolute value, since $e^x + 2 > 0$ for $x\in\mathbb{R}$ giving us a final answer of $$e^x - 8\ln\left(e^x + 2\right) + C$$
$$\int{\frac{e^{2x}-6e^x}{e^x+2}dx}=\int{\frac{e^{2x}}{e^x+2}dx}-6\int{\frac{e^x}{e^x+2}dx}$$ Substitution: $u=e^x+2$, then $\frac{du}{dx}=e^x=u-2\to dx=\frac{1}{u-2}du$
Sub into first half of integral for:
$$\int{\frac{(u-2)^2}{u(u-2)}du}=\int{\frac{u-2}{u}du}=\int{1-\frac{2}{u}du}=u-2\ln|u|+C$$
And second half of integral will get:
$$\int{\frac{u-2}{u(u-2)}du}\to\int{\frac{1}{u}du}=\ln|u|+C$$
Thus overall you get: $$u-2\ln|u|-6\ln|u| +C=u-8\ln|u|+C$$
Hint: $$\frac{e^{2x}-6e^x}{e^x+2}= \frac{e^{2x}}{e^x+2}-\frac{6e^x}{e^x+2}$$
Further hint: Let $u=e^x+2$ then $dx=\frac{du}{e^x}$
So when you substitute this back in the first one, it becomes $\int\frac{u-2}{u}du$. Double-check this carefully.
• Thanks Pure, that does make sense but what about the left side? $u$ at $e^x+2$ would result in a derivative that is not $e^{2x}$ – user472288 May 11 '18 at 23:13
• It does not result in that derivative, however it does not need to, you still get a cancellation – pureundergrad May 11 '18 at 23:16
• Your problem is that you didn't replace $dx$ with $du/e^x$, once you do that you are nearly done – pureundergrad May 11 '18 at 23:19
• pureundergrad $$u = e^x+2 \iff e^x = u-2 \iff e^{2x} = (u-2)^2$$ Also, dx is expressed in terms of both x and du. That might be confusing. OP: But you do need to use $$du = e^x dx \iff dx = \frac {du}{e^x} = \frac{du}{u-2}$$ So when substituting, where you see $dx$ in the integral, replace it with $\frac {du}{u-2}$ – Namaste May 11 '18 at 23:25
It looks to me like you are almost done. You have reduced the problem to integrating $-6\int \frac{e^x}{e^x+ 2}dx$. Now let $u= e^x+ 2$. Then $du= e^xdx$ so that integral becomes $-6\int \frac{1}{u}du$. | 2019-10-24T01:52:25 | {
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http://mathhelpforum.com/statistics/224045-very-very-hard-probability-problem.html | # Math Help - A VERY VERY HARD probability problem.!!!
1. ## A VERY VERY HARD probability problem.!!!
Inside a box, there are 16 marbles in which 7 are red and the rest are blue. 16 people that consists of 11 males and 5 females lined up to randomly pick one marble at a time. The females occupied the 3rd, 4th, 8th,9th, and 14th spot and the males occupied the rest.
Assuming that there is no replacement, what is the probability that all female would have red marbles?
Please HELP!!! Thank you <3.
2. ## Re: A VERY VERY HARD probability problem.!!!
Originally Posted by Kaloda
Inside a box, there are 16 marbles in which 7 are red and the rest are blue. 16 people that consists of 11 males and 5 females lined up to randomly pick one marble at a time. The females occupied the 3rd, 4th, 8th,9th, and 14th spot and the males occupied the rest. Assuming that there is no replacement, what is the probability that all female would have red marbles?
If you had a string $BBBBBBBBBRRRRRRR$ how many ways are there to arrange that string?
Of those rearrangements, how many have an $R$ in the 3rd, 4th, 8th,9th, and 14th spots?
3. ## Re: A VERY VERY HARD probability problem.!!!
The only thing 'hard' about this problem is recognizing that most of the information given is irrelevant. The "a-priori" probability that a given person has a red marble is independent of their place in line. It is sufficient to imagine that the females are first in line. The probability that the first female in line gets a red marble is 7/16, the probability that the second female gets a red marble is (7-1)/(16- 1)= 6/15, the probability the third female gets a red marble is (6- 1)/(15-1)= 5/14, fourth 4/13, fifth 3/12. The probability all 5 females get red marbles is
$\frac{7}{16}\frac{6}{15}\frac{5}{14}\frac{4}{13}\f rac{4}{13}\frac{3}{12}$.
4. ## Re: A VERY VERY HARD probability problem.!!!
Originally Posted by HallsofIvy
The only thing 'hard' about this problem is recognizing that most of the information given is irrelevant. The "a-priori" probability that a given person has a red marble is independent of their place in line. It is sufficient to imagine that the females are first in line. The probability that the first female in line gets a red marble is 7/16, the probability that the second female gets a red marble is (7-1)/(16- 1)= 6/15, the probability the third female gets a red marble is (6- 1)/(15-1)= 5/14, fourth 4/13, fifth 3/12. The probability all 5 females get red marbles is
$\frac{7}{16}\frac{6}{15}\frac{5}{14}\frac{4}{13}\f rac{4}{13}\frac{3}{12}$.
Thanks for the reply but I think your solution is flawed because you completely ignored the fact that there are also male candidates. I mean, what if the the first two males which are ahead of the first female picked the red ones? Won't the probability of "the first female in line" that she will pick a red marble be reduced to 5/14 instead of 7/14 (not 7/16 as you already stated).
5. ## Re: A VERY VERY HARD probability problem.!!!
Originally Posted by Kaloda
Thanks for the reply but I think your solution is flawed because you completely ignored the fact that there are also male candidates. I mean, what if the the first two males which are ahead of the first female picked the red ones? Won't the probability of "the first female in line" that she will pick a red marble be reduced to 5/14 instead of 7/14 (not 7/16 as you already stated).
No, actually both solutions are correct and equivalent.
Here are the calculations both.
6. ## Re: A VERY VERY HARD probability problem.!!!
Originally Posted by Plato
No, actually both solutions are correct and equivalent.
Here are the calculations both.
Okay. Got it. | 2015-01-25T11:14:30 | {
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https://math.stackexchange.com/questions/2030202/s-not-varnothing-implies-varnothing-subset-s | # $S \not= \varnothing \implies \varnothing \subset S$?
If $S$ is a nonempty set, is the following statement correct: $\varnothing \subset S$? It's confusing me becuase $\varnothing$ does not contain any elements so I'm struggling with the logic behind this statement.
Yes, it's correct. To say "$X \subseteq Y$" means that everything which is an element of $X$ is also an element of $Y$. Now imagine how you would demonstrate this wrong: you would supply an element of $X$ that is not an element of $Y$.
So, to show that $\emptyset \subseteq S$ is false, we would have to supply an element of $\emptyset$ that is not an element of $S$. Since we can't supply an element of $\emptyset$ at all, no such counterexample can be found; so the statement is true.
Now, the symbol $\subset$ denotes proper subset - $X \subset Y$ if and only if $X \subseteq Y$ and $X \neq Y$. So $\emptyset \subset S$ if and only if $\emptyset \subseteq S$ (which is always true) and $\emptyset \neq S$ (which is true if $S$ is nonempty).
• So $A\subset B$ should be interpreted as "there are no $x\in A$ such that $x \notin B$ instead of "all $x \in A$ also satisfy $x \in B$" since there might be no $x \in A$? – David Nov 25 '16 at 14:32
• @David By convention, the two statements are considered equivalent; but the first might be more intuitive. – Reese Nov 25 '16 at 14:35
• Indeed it is, thanks a lot. – David Nov 25 '16 at 14:35
The statement is vacuously true: suppose the proposition is false: therefore $\emptyset \not\subset S$. Therefore, there must exist $s\in \emptyset$ such that $s \not\in S$. But $\emptyset$ does not contain any element, meaning $s$ cannot exist. It must be true because stating otherwise is impossible.
• Does one always have to consider a statement to be either true or false? Can't it be undefined under certain circumstances? – David Nov 25 '16 at 14:41
$\varnothing\subset S$ for any set $S$, since you cannot find a counter-example, even if $S=\varnothing$: an element in $\varnothing$ which would not be in $S$.
Note: The correct mathematical notation is $\varnothing$ (code \varnothing), not $\emptyset$.
• -1 for saying that $\varnothing$ is correct, but $\emptyset$ isn't. – Magdiragdag Nov 25 '16 at 14:23
• From context, it seems that OP is distinguishing between $\subset$ and $\subseteq$; so $\subset$ means strict subset. – Reese Nov 25 '16 at 14:24
• You do as you please, but it's more of a computer science notation. I stick to Bourbaki's notation. – Bernard Nov 25 '16 at 14:24
• @Reese: Well, strict is denoted $\subsetneq$ or $\varsubsetneq$. – Bernard Nov 25 '16 at 14:26
• Why do you say $\emptyset$ is not correct? It is written \emptyset. The meaning is clear. +1 because the answer is correct. – mfl Nov 25 '16 at 14:27
It is true because it is not false that every object in $\emptyset$ belongs to $S$.
Let me elaborate.
$x \in \emptyset$ but $\emptyset$ is empty so whatever you want to say about any object inside $\emptyset$ will always be true since none exist, you might as well say that $x \in \emptyset$. | 2019-07-22T07:46:26 | {
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http://bootmath.com/how-to-prove-the-squeeze-theorem-for-sequences.html | # How to prove the Squeeze Theorem for sequences
The formulation I’m looking at goes: If $\lbrace x_n\rbrace$, $\lbrace y_n\rbrace$ and $\lbrace z_n \rbrace$ are sequences such that $x_n \le y_n \le z_n$ for all $n \in \mathbb N$, and $x_n \to l$ and $z_n \to l$ for some $l \in \mathbb R$, then $y_n \to l$ also.
So we have to use the definition of convergence to a limit for a sequence:
$$\forall \varepsilon > 0, \space \exists N_\varepsilon \in \mathbb N, \space \forall n \ge N_\varepsilon, \space |a_n – l| < \varepsilon$$
I’ve been trying to say something like: $|y_n – l| < |x_n – l| + |z_n – l| \le \frac\varepsilon 2 + \frac\varepsilon 2 = \varepsilon$ for every $\varepsilon > 0$, but I’m not sure how to get there or if there may be a better way to prove the theorem. Any help would be greatly appreciated.
#### Solutions Collecting From Web of "How to prove the Squeeze Theorem for sequences"
Write
$$|y_n-l|\le |y_n-x_n|+|x_n-l|\le( z_n-x_n)+|x_n-l|$$
Can you take it from here?
Let $\varepsilon > 0$. Since $x_n \to l$, there exists $N_1 = N_1(\varepsilon)$ such that $|x_n – l| < \varepsilon$ for all $n \ge N_1$. Since $z_n \to l$, there exists $N_2 = N_2(\varepsilon)$ such that $|z_n – l| < \varepsilon$ for all $n \ge N_2$. Set $N = \max\{N_1,N_2\}$. If $n \ge N$, then $$y_n – l \le z_n – l < \varepsilon$$ and $$y_n – l \ge x_n – l > -\varepsilon$$ Hence $|y_n – l| < \varepsilon$ for all $n \ge N$. Since $\varepsilon$ was arbitrary, $y_n \to l$.
Ok, so this is what I have now:
Let $x_n \le y_n \le z_n \space\space \forall n \in \mathbb N$ and $\lim_{n \to \infty} x_n = \lim_{n \to \infty} z_n = l$ for some $l \in \mathbb R$.
Then we have:
\begin{align} &\forall \varepsilon_1 > 0,\, \exists N_{\varepsilon_1} \in \mathbb N,\, \forall n \ge N_{\varepsilon_1}\space |x_n – l| < \varepsilon_1 \\ &\forall \varepsilon_2 > 0,\, \exists N_{\varepsilon_2} \in \mathbb N,\, \forall n \ge N_{\varepsilon_2}\space |z_n -l| < \varepsilon_2 \end{align}
Let $N = \max\lbrace N_{\varepsilon_1},N_{\varepsilon_2} \rbrace$ and $\varepsilon = \min\lbrace \varepsilon_1, \varepsilon_2 \rbrace$, so that we have
$$\forall \varepsilon > 0,\, \exists N \in \mathbb N,\, \forall n \ge N\space |x_n – l| < \varepsilon \space \text{and} \space |z_n -l| < \varepsilon$$
Let $\frac\varepsilon3 > 0$. Then $\exists N \in \mathbb N,\, \forall n \ge N\space |x_n -l| < \frac\varepsilon3 \text{ and } |z_n -l| < \frac\varepsilon3$ so that
$$|z_n – x_n| = |z_n – l + l – x_n| \le |z_n – l| + |x_n – l| < \frac\varepsilon3 + \frac\varepsilon3 = \frac{2\varepsilon}3$$
By assumption,
\begin{align} x_n \le&\, y_n \le z_n \\ 0 \le&\, y_n – x_n \le z_n – x_n \\ &|y_n – x_n| \le |z_n – x_n| \\ |y_n – l| = |y_n – x_n + x_n – l| \le& |y_n – x_n| + |x_n – l| \le |z_n – x_n| + |x_n – l| < \frac{2\varepsilon}3 + \frac\varepsilon3 = \varepsilon \end{align}
And so $\lbrace y_n \rbrace$ also converges to $l$, thus, Q.E.D.
I think that that’s satisfactory. If there are any ambiguities or mistakes, please let me know. Thanks to everybody for your advice and contributions.
How about
$$|y_n-l|\leq\max\{|x_n-l|,|z_n-l|\}<\varepsilon$$
where last inequality of the above holds for $n\geq N$ if $N$ is big enough?
An alternative proof is the following:
Proof: Since $x_n \leq y_n \leq z_n$ then $0\leq y_n-x_n\leq z_n-x_n$, thus $|y_n-x_n|\leq z_n-x_n$.
Combining the above with the fact that $\lim(z_n-x_n)=\lim z_n-\lim x_n=l-l=0 \$, we get: $$\lim(y_n-x_n)=0$$
Now we can write the terms of $(y_n)$ as the sum of the terms of two converging sequences: $y_n=(y_n-x_n)+x_n$, so we have:
$$\lim y_n=\lim\big((y_n-x_n)+x_n \big)=\lim(y_n-x_n)+\lim x_n=0+l=l$$
As $n$ grows, you can get the distance from $x_n$ to $l$ to be less than any $\epsilon > 0$, and the distance from $z_n$ to $l$ to be less than $\epsilon$ as well. Just take the max $N$ of the two indices for $x_n$ and $z_n$ that guarantee this, and you will get that both $x_n$ and $z_n$ are within $\epsilon$ of $l$ for $n \geq N$. What does that say about the distance between $y_n$ and $l$, for $n \geq N$? | 2018-06-22T12:53:48 | {
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http://www.itl.nist.gov/div898/software/dataplot/refman2/auxillar/empqua.htm | Dataplot Vol 2 Vol 1
# EMPIRICAL QUANTILE FUNCTION
Name:
EMPIRICAL QUANTILE FUNCTION (LET)
Type:
Let Subcommand
Purpose:
Compute the empirical quantile function.
Description:
The quantile function is the inverse of the cumulative distribution function, F,
$$Q(u) = F^{-1}(u) \hspace{0.2in} 0 < u < 1$$
Given a set of ordered data, x1x2 ... ≤ xn, an empirical estimate of the quantile function can be obtained from the following piecewise linear function
$$\hat{Q}(u) = (nu - j + \frac{1}{2}) x_{(j+1)} + (j + \frac{1}{2} - nu) x_{j}$$
$$\frac{2j - 1}{2n} \le u \le \frac{2j + 1}{2n}$$
This will be computed for a specified number of equi-spaced points between the lower and upper limits. Dataplot will use the number of points in the sample if this is greater than 1,000. Otherwise 1,000 points will be used.
Syntax:
LET <y> <u> = EMPIRICAL QUANTILE FUNCTION <x>
<SUBSET/EXCEPT/FOR qualification>
where <x> is the response variable;
<y> is a variable containing the empirical quantile function;
<u> is a variable containing the values where the empirical quantile function is computed;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
Examples:
LET Y U = EMPIRICAL QUANTILE FUNCTION X
LET Y U = EMPIRICAL QUANTILE FUNCTION X SUBSET X > 0
Default:
None
Synonyms:
None
Related Commands:
EMPIRICAL QUANTILE PLOT = Generate an empirical quantile plot. EMPIRICAL CDF PLOT = Generates an empiricial CDF plot. KAPLAN MEIER PLOT = Generates a Kaplan Meier plot. PROBABILITY PLOT = Generates a probability plot. INFORMATIVE QUANTILE FUNCTION = Compute the informative quantile function.
References:
"MIL-HDBK-17-1F Volume 1: Guidelines for Characterization of Structural Materials", Depeartment of Defense, pp. 8-36, 8-37, 2002.
Parzen (1983), "Informative Quantile Functions and Identification of Probability Distribution Types", Technical Report No. A-26, Texas A&M University.
Applications:
Distributional Analysis
Implementation Date:
2017/02
Program:
. Step 1: Define some default plot control features
.
title offset 2
title case asis
case asis
label case asis
line color blue red
multiplot scale factor 2
multiplot corner coordinates 5 5 95 95
.
. Step 2: Create 50, 100, 200, and 1000 normal random numbers and
. compute the empirical quantile funciton
.
let nv = data 50 100 200 1000
let p = sequence 0.01 0.01 .99
let y2 = norppf(p)
.
. Step 3: Loop through the four cases and compute and plot the
. empirical quantile funciton with overlaid NORPPF
.
multiplot 2 2
loop for k = 1 1 4
let n = nv(k)
let x = norm rand numb for i = 1 1 n
let y u = empirical quantile function x
title N: ^n
plot y u and
plot y2 p
end of loop
end of multiplot
.
justification center
move 50 97
text Empirical Quantile Functions (blue) Overlaid with ...
NORPPF (red) for Normal Random Numbers
move 50 5
text u
direction vertical
move 5 50
text Q(u)
NIST is an agency of the U.S. Commerce Department.
Date created: 07/20/2017
Last updated: 07/20/2017 | 2017-10-24T02:27:51 | {
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https://nl.mathworks.com/help/symbolic/use-symbolic-matrix-variables.html | # Create Symbolic Matrix Variables
Since R2021a
Symbolic matrix variables represent matrices, vectors, and scalars in compact matrix notation. When mathematical formulas involve matrices and vectors, writing them using symbolic matrix variables is more concise and clear than writing them componentwise. When you do this, you can take vector-based expressions and equations from textbooks, enter them in Symbolic Math Toolbox™, perform mathematical operations on them, and derive further equations from them.
Derived equations involving symbolic matrix variables are displayed in typeset as they would be in textbooks. For example, create three symbolic matrix variables $\text{A}$, $\text{x}$, and $\text{y}$ by using syms. Find the differential of the expression ${\text{y}}^{T}\text{A}\text{x}$ with respect to the vector $\text{x}$.
syms A [3 4] matrix
syms x [4 1] matrix
syms y [3 1] matrix
eq = y.'*A*x
eq = ${y}^{\mathrm{T}} A x$
D = diff(eq,x)
D = ${y}^{\mathrm{T}} A$
### Comparison Between Matrix of Symbolic Scalar Variables and Symbolic Matrix Variables
Symbolic matrix variables are an alternative to symbolic scalar variables. The two options are of different types and displayed differently.
For example, create two 3-by-4 matrices of symbolic scalar variables by using syms. For brevity, matrices of symbolic scalar variables are sometimes called symbolic matrices. These matrices are displayed by listing their components.
syms A B [2 3]
A
A =
$\left(\begin{array}{ccc}{A}_{1,1}& {A}_{1,2}& {A}_{1,3}\\ {A}_{2,1}& {A}_{2,2}& {A}_{2,3}\end{array}\right)$
B
B =
$\left(\begin{array}{ccc}{B}_{1,1}& {B}_{1,2}& {B}_{1,3}\\ {B}_{2,1}& {B}_{2,2}& {B}_{2,3}\end{array}\right)$
A matrix of symbolic scalar variables is of type sym.
class(A)
ans =
'sym'
Applying symbolic math operations to these matrices can result in a complex solution expressed in terms of the matrix components. For example, multiply the matrices A and B'.
C = A*B'
C =
$\left(\begin{array}{cc}{A}_{1,1} \stackrel{‾}{{B}_{1,1}}+{A}_{1,2} \stackrel{‾}{{B}_{1,2}}+{A}_{1,3} \stackrel{‾}{{B}_{1,3}}& {A}_{1,1} \stackrel{‾}{{B}_{2,1}}+{A}_{1,2} \stackrel{‾}{{B}_{2,2}}+{A}_{1,3} \stackrel{‾}{{B}_{2,3}}\\ {A}_{2,1} \stackrel{‾}{{B}_{1,1}}+{A}_{2,2} \stackrel{‾}{{B}_{1,2}}+{A}_{2,3} \stackrel{‾}{{B}_{1,3}}& {A}_{2,1} \stackrel{‾}{{B}_{2,1}}+{A}_{2,2} \stackrel{‾}{{B}_{2,2}}+{A}_{2,3} \stackrel{‾}{{B}_{2,3}}\end{array}\right)$
To create symbolic matrix variables of the same size, use the syms command followed by the variable names, their size, and the matrix keyword. Symbolic matrix variables are displayed in bold to distinguish them from symbolic scalar variables.
syms A B [2 3] matrix
A
A = $A$
B
B = $B$
Symbolic matrix variables are of type symmatrix.
class(A)
ans =
'symmatrix'
Applying symbolic math operations to symbolic matrix variables results in a concise display. For example, multiply A and B'.
C = A*B'
C = $A {\left(\stackrel{‾}{B}\right)}^{\mathrm{T}}$
### Mathematical Operations with Symbolic Matrix Variables
Symbolic matrix variables are recognized as noncommutative objects. They support common math operations, and you can use these operations to build symbolic matrix variable expressions.
syms A B [2 2] matrix
A*B - B*A
ans = $A B-B A$
For example, check the commutation relation for multiplication between two symbolic matrix variables.
isequal(A*B,B*A)
ans = logical
0
Check the commutation relation for addition.
isequal(A+B,B+A)
ans = logical
1
If an operation has any arguments of type symmatrix, the result is automatically converted to type symmatrix. For example, multiply a matrix A that is represented by symbolic matrix variable and a scalar c that is represented by symbolic scalar variable. The result is of type symmatrix.
syms A [2 2] matrix
syms c
class(A)
ans =
'symmatrix'
class(c)
ans =
'sym'
M = c*A
M = $c A$
class(M)
ans =
'symmatrix'
Multiply three matrices that are represented by symbolic matrix variables. The result X is a symmatrix object.
syms V [2 1] matrix
X = V.'*A*V
X = ${V}^{\mathrm{T}} A V$
class(X)
ans =
'symmatrix'
You can pass symmatrix objects as arguments to math functions. For example, perform a mathematical operation to X by taking the differential of X with respect to V.
diff(X,V)
ans = ${V}^{\mathrm{T}} {A}^{\mathrm{T}}+{V}^{\mathrm{T}} A$
### Create Symbolic Matrix Variable from Array of Symbolic Scalar Variables
You can convert an array of symbolic scalar variables to a single symbolic matrix variable using the symmatrix function. Symbolic matrix variables that are converted in this way are displayed elementwise.
syms A [3 4]
class(A)
ans =
'sym'
B = symmatrix(A)
B =
class(B)
ans =
'symmatrix'
### Convert Symbolic Matrix Variable into Array of Symbolic Scalar Variables
You can create symbolic matrix variables, derive equations, and then convert the result to arrays of symbolic scalar variables using the symmatrix2sym function.
For example, find the matrix product of two symbolic matrix variables A and B. The result X is of type symmatrix.
syms A B [2 2] matrix
X = A*B
X = $A B$
class(X)
ans =
'symmatrix'
Convert the symbolic matrix variable X to array of symbolic scalar variables. The converted matrix Y is of type sym.
Y = symmatrix2sym(X)
Y =
$\left(\begin{array}{cc}{A}_{1,1} {B}_{1,1}+{A}_{1,2} {B}_{2,1}& {A}_{1,1} {B}_{1,2}+{A}_{1,2} {B}_{2,2}\\ {A}_{2,1} {B}_{1,1}+{A}_{2,2} {B}_{2,1}& {A}_{2,1} {B}_{1,2}+{A}_{2,2} {B}_{2,2}\end{array}\right)$
class(Y)
ans =
'sym'
Check that the product obtained by converting symbolic matrix variables is equal to the product of two arrays of symbolic scalar variables.
syms A B [2 2]
isequal(Y,A*B)
ans = logical
1
### Indexing into Symbolic Matrix Variables
Indexing into a symbolic matrix variable returns corresponding matrix elements in the form of another symbolic matrix variable.
syms A [2 3] matrix
a = A(2,3)
a = ${A}_{2,3}$
class(a)
ans =
'symmatrix'
Alternatively, convert the symbolic matrix variable A to a matrix of symbolic scalar variables. Then, index into that matrix.
Asym = symmatrix2sym(A)
Asym =
$\left(\begin{array}{ccc}{A}_{1,1}& {A}_{1,2}& {A}_{1,3}\\ {A}_{2,1}& {A}_{2,2}& {A}_{2,3}\end{array}\right)$
asym = Asym(2,3)
asym = ${A}_{2,3}$
class(asym)
ans =
'sym'
Note that both results are equal.
isequal(a,symmatrix(asym))
ans = logical
1
### Display of Operations Involving Symbolic Matrix Variables
Matrices like those returned by eye, zeros, and ones often have special meaning with specific notation in symbolic workflows. Declaring these matrices as symbolic matrix variables display the matrices in bold along with the matrix dimensions.
symmatrix(eye(3))
ans = ${\mathrm{I}}_{3}$
symmatrix(zeros(2,3))
ans = ${\mathrm{0}}_{2,3}$
symmatrix(ones(3,5))
ans = ${\mathrm{1}}_{3,5}$
If the inputs to a componentwise operation in MATLAB® are symbolic matrix variables, so is the output. These operations are displayed in special notations which follow conventions from textbooks.
syms A B [3 3] matrix
A.*B
ans = $A\odot B$
A./B
ans = $A\oslash B$
A.\B
ans = $B\oslash A$
A.*hilb(3)
ans =
A.^(2*ones(3))
ans = ${A}^{\circ 2 {\mathrm{1}}_{3,3}}$
A.^B
ans = ${A}^{\circ B}$
kron(A,B)
ans = $A\otimes B$
ans = $\mathrm{adj}\left(A\right)$
ans = $\mathrm{Tr}\left(A\right)$ | 2022-10-01T21:38:11 | {
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https://math.stackexchange.com/questions/3062564/show-that-the-two-norms-in-e-left-f-in-mathcalc10-1f0-0-right-are | Show that the two norms in $E=\left\{f\in \mathcal{C}^1[0,1]:f(0)=0\right\}$ are equivalent?
I can't show that the two norms defined as $$||f||_{\infty}=\sup_{x\in[0,1]}|f(x)+ f'(x)|$$ and $$N(f)=\sup_{x\in[0,1]}|f(x)| + \sup_{x\in[0,1]}|f'(x)|$$ are equivalent in $$E=\{f\in\mathcal{C}^1([0,1]) \text{ s.t. } f(0)=0\}$$.
A first inequality in one sense is trivial.
For the other inequality, I can write:
$$f(x)=f(0)+\int_0^x f'(t) dt$$ Which implies $$\sup_{x\in[0,1]}|f(x)| \leq \sup_{x\in[0,1]}|f'(x)| .$$ And then $$N(f)\leq 2 \sup_{x\in[0,1]}|f'(x)| .$$ But I can't conclude from here.
I sincerely thank you for your help.
• Thank you for your correction. – Furdzik Jan 5 '19 at 9:58
• @PaulFrost it does not satisfy $f(0)=0$. – Lorenzo Quarisa Jan 5 '19 at 11:07
Two metrics on the same set are equivalent (i.e. generate the same topology ) iff they have the same set of convergent sequences, because the closure operator in a metric space is entirely and uniquely determined by the convergent sequences.
For brevity let $$\sup_{x\in [0,1]}|g(x)|=\|g\|_S$$ for any continuous $$g:[0,1]\to \Bbb R.$$
Idea: Consider that $$f(x)+f'(x)=e^{-x}(e^xf(x))'.$$
(1). If $$\lim_{n\to \infty}\|f_n-f\|_{\infty}=0$$:
Then $$\lim_{n\to \infty}\|e^{-x}(e^x (f_n(x)-f(x))'\|_S=0,$$ which implies that $$\lim_{n\to \infty}\|(e^x(f_n(x)-f(x))'\|_S=0,$$ which, since $$f_n(0)=f(0)=0,$$ implies that $$\lim_{n\to \infty}\|e^x(f_n(x)-f(x))\|_S= \lim_{n\to \infty}\sup_{x\in [0,1]} |\int_0^x(e^t(f_n(t)-f(t))'dt\,|=0$$ which implies that $$\lim_{n\to \infty}\|f_n-f\|_S=0.$$ Now $$\|f'_n-f'\|_S\leq \|(f_n+f'_n)-(f+f')\|_S+\|f-f_n)\|_S=\|f_n-f\|_{\infty}+\|f_n-f\|_S,$$ so we have $$\lim_{n\to \infty}\|f'_n-f'\|_S=0.$$ Since $$N(f_n-f)=\|f_n-f\|_S+\|f'_n-f'\|_S,$$ therefore $$\lim_{n\to \infty}N(f_n-f)=0.$$
(2). If $$\lim_{n\to \infty}N(f_n-f)=0$$:
Since $$N(f_n-f)\geq \|f_n-f\|_{\infty},$$ therefore $$\lim_{n\to \infty}\|f_n-f\|_{\infty}=0.$$
• Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) \le \|f\|_\infty \le MN(f)$ for all $f$. – Theo Bendit Jan 5 '19 at 11:21
• @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $\|v_n\|_{(1)}>n\|v_n\|_{(2)}$ for each $n\in \Bbb N.$ Let $w_n=\frac {v_n}{\|v_n\|_{(1)}}.$ Then the sequence $(w_n)_{n\in \Bbb N}$ converges to $0$ with respect to $\|\cdot\|_{(2)}$ but $\|w_n\|_{(1)}=1$. – DanielWainfleet Jan 5 '19 at 12:19
As you have already shown, for $$f \in C^{1}[0,1]$$ such that $$f(0) = 0$$, $$\sup_{x \in [0,1]} |f(x)| \le \sup_{x \in [0,1]} |f'(x)|. \tag{1}$$
We can use this to show that $$\sup |f|$$ is bounded by a constant times $$\sup |f + f'|$$, as follows: \begin{align*} \sup_{x \in [0,1]} |f(x)| &\le \sup_{x \in [0,1]} |e^x f(x)| \\ &\le \sup_{x \in [0,1]} \left| \frac{d}{dx} e^x f(x) \right| \qquad\qquad \text{(by (1), since e^0 f(0) = 0})\\ &= \sup_{x \in [0,1]} |e^x f(x) + e^x f'(x)| \\ &\le e \sup_{x \in [0,1]} |f(x) + f'(x)|. \end{align*}
Now apply triangle inequality to $$|f'| = |f' + f - f|$$: \begin{align*} \sup |f'(x)| &= \sup |f'(x) + f(x) - f(x)| \\ &\le \sup |f(x) + f'(x)| + \sup |f(x)| \\ &\le \sup |f(x) + f'(x)| + e \cdot \sup |f(x) + f'(x)| \\ &= (e + 1) \sup |f(x) + f'(x)|. \end{align*}
Therefore, $$\sup |f(x)|$$ and $$\sup |f'(x)|$$ are both bounded above by a constant times $$\sup |f(x) + f'(x)|$$. This shows that your two norms are equivalent. | 2020-02-17T16:47:34 | {
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https://math.stackexchange.com/questions/2077799/could-somebody-please-explain-to-me-how-this-binary-relation-is-anti-symmetric | # Could somebody please explain to me how this binary relation is Anti-symmetric?
$$A = \{1, 2, 3, 4\}$$ where $$xRy$$ if $$x \mid y$$
$$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)\}$$.
I am watching this video about Partial Orders and the guy just said the relation was Anti-symmetric; however, I do not understand why.
I know that for a relation to be anti-symmetric : if $$aRb$$ and $$bRa$$ then $$a = b$$.
However, I do not see that there. Sorry for not giving in more knowledge. I am just really confused on this one.
You may find it useful to take the definition of "anti-symmetric" and use it to say under what conditions a relation $R$ is not anti-symmetric. You'll find that for $R$ not to be anti-symmetric means that there exist $a$ and $b$ such that $aRb$ and $bRa$ but $a\neq b$. Now look at the relation $R$ exhibited in the question and see whether you can find any such $a$ and $b$.
Suppose $xRy$ and $yRx$, with $x,y\ge0$. Then $x$ is a factor of $y$, and $y$ is a factor of $x$. These imply, respectively, that $x\le y$ and $y\le x$. It is clear to deduce from this that $x=y$.
If $a|b$ and $b|a$ then integers $k,\,l$ exist with $b=ka,\,a=lb=lka$. Since $a,\,b\in\left\{ 1,\,2,\,3,\,4\right\}$, we have $k,\,l>0$ and $lk=1$. Hence $k=l=1$ and $a=b$.
• But what about (1,2) (1,3) ... 2 doesn't divide 1 and 3 doesn't divide 1. What do I do with those? Dec 30, 2016 at 21:44
• @RayRutzer In those cases $\left(a,\,b\right)\in R$ but $\left(b,\,a\right)\notin R$ (if I'm remembering $R$'s definition correctly).
– J.G.
Dec 30, 2016 at 23:39
Let $x \mid y$ stand for "$x$ evenly divides $y$." If $x \mid y$ and $y \mid x$ then $x=y$. Hence, the divisibility relation is antisymmetric. (It is indeed a partial order.)
The fact that for non-negative integers $x \mid y$ and $y \mid x$ imply that $x=y$ is not surprising, but if you want a full proof, consider this: $x$ evenly divides $y$ if there exists a non-negative integer $n$ such that $y = nx$. Likewise, if $y$ evenly divides $x$, then there exists a non-negative integer $m$ such that $x = my$.
If both conditions are true, consider two cases. If $x=0$, then $y = n0 = 0$. Otherwise, substitution gives you $x = mnx$, and dividing both sides by $x$ (which we can, because we are now assuming that it is non-zero), we get $1 = mn$. The only solution for $m,n$ nonnegative integers is $m=n=1$, which finally gives $x=y$.
Note that assuming that the integers are non-negative is important. Otherwise, $-3$ divides $3$ and vice versa, but $-3 \neq 3$.
Use the contrapositive way of thinking about antisymmetry: $$aRb \text{ and }bRa \implies a = b$$ is equivalent to $$a \neq b \implies \text{not} (aRb \text{ and } bRa);$$ in other words, you never see both $aRb$ and $bRa$ (or, as ordered pairs, both $(a, b)$ and $(b, a)$) for distinct $a, b$. You don't even need to recognize that it's a divisibility relation; you can just look at it!
• But what about (1,1), (2,2) ... wouldn't that violate that clause? Dec 30, 2016 at 21:48
• No, because $1 \neq 1$ is false (as is $2 \neq 2$). Remember: The contrapositive lets us think about pairs $(a, b)$ and $(b, a)$ where $a \neq b$. Dec 30, 2016 at 21:50
• But if the left hand side is false doesn't that means that the entire statement is true in this case? Dec 30, 2016 at 21:57
• Right! So whenever $a = b$, the statement $$a \neq b \implies \text{not} (aRb \text{ and } bRa)$$ is true, and that statement is equivalent to the "usual" definition of antisymmetry. So the presence of pairs $(a, a)$ doesn't "violate" antisymmetry, even when formulated in this contrapositive way. Dec 30, 2016 at 22:00
• So the "working definition" that I like for antisymmetry is this: If $a \neq b$ and $aRb$, then it is not the case that $bRa$; or, as ordered pairs, when $a \neq b$ and $(a, b) \in R$, then $(b, a) \not\in R$. It just lets you scan relations (given as sets!) quickly to determine if there are any "symmetric" pairs (if there are any, your relation is not antisymmetric; if there aren't any, it is). It's just the most intuitive formulation for antisymmetry in general, for me. Dec 30, 2016 at 22:08 | 2022-07-06T04:40:48 | {
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https://www.chebfun.org/examples/approx/EquispacedData.html | 1. Introduction
For good reasons of approximation theory, Chebfun relies on polynomial interpolation in Chebyshev points, which are unequally spaced, to represent nonperiodic functions. However, many people want to work with equispaced data, and we are often asked, how can we do this in Chebfun?
Chebfun can do a pretty good job with this thanks to the 'equi' flag introduced by Georges Klein in 2011.
2. Example without noise
Suppose we want to work with the function $e^x\cos(10x)\tanh(4x)$, but all we know of it is samples at 40 equispaced points in $[-1,1]$. We can construct a Chebfun from this data as follows. Note the 'equi' flag.
ff = @(x) exp(x).*cos(10*x).*tanh(4*x);
grid = linspace(-1,1,40)';
data = ff(grid);
f = chebfun(data,'equi');
The plot looks good!
LW = 'linewidth'; FS = 'fontsize'; MS = 'markersize'; purple = [.8 0 1];
plot(f,LW,1), hold on, plot(grid,data,'.k',MS,8), hold off
title('chebfun constructed from 40 equispaced data values',FS,10)
The error is encouragingly small. It would be bigger with fewer data points, smaller with more.
fexact = chebfun(ff);
error = norm(f-fexact,inf)
error =
3.537098266021654e-06
For comparison, this is what we get with polynomial interpolation of the same data. Of course, any Chebfun user knows that polynomial interpolation in equispaced points is a bad idea (the Runge phenomenon).
runge = chebfun.interp1(grid,data);
plot(runge,'r',LW,1)
hold on, plot(grid,data,'.k',MS,8), hold off
So what is this very nice chebfun obtained with the 'equi' flag, and how has Chebfun computed it?
The answer is that it is a polynomial approximant, but not simply the polynomial interpolant. In fact it has a higher degree than 40:
f
f =
chebfun column (1 smooth piece)
interval length endpoint values
[ -1, 1] 99 0.31 -2.3
vscale = 2.575084e+00.
To construct this function, Chebfun has first constructed a rational function $g$ known as a Floater-Hormann interpolant [1] that has good properties with equispaced data, and it has picked the order of this approximation in an adaptive fashion. Then, since Chebfun works with polynomials rather than rational functions, it has approximated $g$ by a chebfun $f$ of the usual polynomial kind.
Here is a plot of the Chebyshev coefficients of $f$:
plotcoeffs(f,'color',purple), axis([0 100 1e-16 10])
title('Chebyshev coefficients',FS,10)
Note that about half of them are below the level of the accuracy of the approximation, so they can't be contributing much. We could throw them away like this:
f50 = chebfun(f,51);
error50 = norm(f50-fexact,inf)
plotcoeffs(f50,'color',purple), axis([0 100 1e-16 10])
title('Chebyshev coefficients up to degree 50',FS,10)
error50 =
3.526801940675228e-06
Another approach would be construct the original chebfun with a loosened value of eps:
floose = chebfun(data,'equi','eps',1e-6);
errorloose = norm(floose-fexact,inf)
plotcoeffs(floose,'color',purple), axis([0 100 1e-16 10])
title('Chebyshev coefficients with loosened tolerance',FS,10)
errorloose =
3.536842150029010e-06
3. Example with noise
What about a function with noise? Let's add random perturbations of size $10^{-1}$ to the data:
rng('default'); rng(0)
data = data + 1e-1*randn(size(data));
Here is what we get if we construct a equispaced chebfun with eps = 1e-2:
ep = 1e-2;
f = chebfun(data,'equi','eps',ep);
plot(f,LW,1), hold on, plot(grid,data,'.k',MS,8), hold off
s = sprintf('noisy data with ''equi'', eps = 1e-2: length(f) = %d',length(f));
title(s,FS,12)
And here is the same experiment but with eps three times as large.:
ep = 3e-2;
f = chebfun(data,'equi','eps',ep);
plot(f,LW,1), hold on, plot(grid,data,'.k',MS,8), hold off
s = sprintf('noisy data with ''equi'', eps = 3e-2: length(f) = %d',length(f));
title(s,FS,12)
4. Discussion
What's nice about these 'equi' approximations is that, as usual with chebfuns, they are globally smooth functions, and can be differentiated, for example, without any anomalies arising. In some applications this is very appealing.
Another globally smooth way to deal with equispaced data, besides the Floater-Hormann approach, is to use so-called Gregory interpolants [2]. This idea, however, has not been implemented in Chebfun.
References
1. M. S. Floater and K. Hormann, Barycentric rational interpolation with no poles and high rates of approximation, Numerische Mathematik 107 (2007), 315--331.
2. M. Javed and L. N. Trefethen, Euler-Maclaurin and Gregory interpolants, Numerische Mathematik (2015). | 2021-08-01T06:35:12 | {
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https://math.stackexchange.com/questions/2059679/how-to-find-limits-of-integration-of-polar-curves | How to find limits of integration of polar curves?
Right now I am working on a problem that involves finding the area enclosed by a single loop given the equation $r=4\cos(3\theta)$. I know that the cosine is bounded from zero to $\pi$, but when using a lower limit of $0$, and a upper limit of $\pi/3$, I get the wrong answer (the answer is $4\pi/3$).
Should I instead use zero to $2\pi/3$, since that would be a full period? I tried drawing the graph as well, but had no luck with that. What is the best way of finding the upper and lowers limits for these kind of problems?
• "I tried drawing the graph as well, but had no luck with that" For this part, see this online tool, which might also make the mathematical answer to the rest of your question quite transparent.
– Did
Dec 15, 2016 at 9:28
• Use limits of theta (T) from 0 to (pi/6), since 3T will have total limit from 0 to (pi/2). Dec 15, 2016 at 9:46
• Using those limits worked, but why are those the limits of integration. Why not -pi/6 to pi/6, since the given function would be zero there. Dec 15, 2016 at 17:46
It helps if you have an idea of the graph, but even if you don't: it should be clear that at $\theta = 0$, you have $r(0) = 4$ so you're not at the beginning of a loop. A loop starts when $r=0$ and a single loop closes at the next root of $r(\theta)$.
The function $\cos x$ has a root at $-\tfrac{\pi}{2}$ and the next one is at $\tfrac{\pi}{2}$. For this polar curve $r = 4 \cos(3\theta)$, you get (with $x=3\theta$): $$3\theta = \pm \frac{\pi}{2} \Rightarrow \theta = \pm \frac{\pi}{6}$$ so you go through exactly one loop if you let $\theta$ run from $-\tfrac{\pi}{6}$ to $\tfrac{\pi}{6}$. Using the formula for area: $$\int_{\theta_1}^{\theta_2} \tfrac{1}{2}r^2 \,\mbox{d}\theta \to \int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} \tfrac{1}{2}\left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \cdots = \frac{4\pi}{3}$$ Note that because the function is even, it is slightly easier to (manually) calculate : $$\color{blue}{2}\int_{\color{red}{0}}^{\tfrac{\pi}{6}} \tfrac{1}{2}\left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \int_{0}^{\tfrac{\pi}{6}} \left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \cdots = \frac{4\pi}{3}$$
• It's probably a mistake in your calculation, the answer is indeed $4\pi/3$. Dec 15, 2016 at 21:59 | 2022-05-19T00:10:08 | {
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https://se.mathworks.com/help/matlab/ref/integral.html | # integral
Numerical integration
## Description
example
q = integral(fun,xmin,xmax) numerically integrates function fun from xmin to xmax using global adaptive quadrature and default error tolerances.
example
q = integral(fun,xmin,xmax,Name,Value) specifies additional options with one or more Name,Value pair arguments. For example, specify 'WayPoints' followed by a vector of real or complex numbers to indicate specific points for the integrator to use.
## Examples
collapse all
Create the function $f\left(x\right)={e}^{-{x}^{2}}\left(\mathrm{ln}x{\right)}^{2}$.
fun = @(x) exp(-x.^2).*log(x).^2;
Evaluate the integral from x=0 to x=Inf.
q = integral(fun,0,Inf)
q = 1.9475
Create the function $f\left(x\right)=1/\left({x}^{3}-2x-c\right)$ with one parameter, $c$.
fun = @(x,c) 1./(x.^3-2*x-c);
Evaluate the integral from x=0 to x=2 at c=5.
q = integral(@(x) fun(x,5),0,2)
q = -0.4605
Create the function $f\left(x\right)=\mathrm{ln}\left(x\right)$.
fun = @(x)log(x);
Evaluate the integral from x=0 to x=1 with the default error tolerances.
format long
q1 = integral(fun,0,1)
q1 =
-1.000000010959678
Evaluate the integral again, this time with 12 decimal places of accuracy. Set RelTol to zero so that integral only attempts to satisfy the absolute error tolerance.
q2 = integral(fun,0,1,'RelTol',0,'AbsTol',1e-12)
q2 =
-1.000000000000010
Create the function $f\left(z\right)=1/\left(2z-1\right)$.
fun = @(z) 1./(2*z-1);
Integrate in the complex plane over the triangular path from 0 to 1+1i to 1-1i to 0 by specifying waypoints.
q = integral(fun,0,0,'Waypoints',[1+1i,1-1i])
q = 0.0000 - 3.1416i
Create the vector-valued function $f\left(x\right)=\left[\mathrm{sin}x,\phantom{\rule{0.2222222222222222em}{0ex}}\mathrm{sin}2x,\phantom{\rule{0.2222222222222222em}{0ex}}\mathrm{sin}3x,\phantom{\rule{0.2222222222222222em}{0ex}}\mathrm{sin}4x,\phantom{\rule{0.2222222222222222em}{0ex}}\mathrm{sin}5x\right]$ and integrate from x=0 to x=1. Specify 'ArrayValued',true to evaluate the integral of an array-valued or vector-valued function.
fun = @(x)sin((1:5)*x);
q = integral(fun,0,1,'ArrayValued',true)
q = 1×5
0.4597 0.7081 0.6633 0.4134 0.1433
Create the function $f\left(x\right)={x}^{5}{e}^{-x}\mathrm{sin}x$.
fun = @(x)x.^5.*exp(-x).*sin(x);
Evaluate the integral from x=0 to x=Inf, adjusting the absolute and relative tolerances.
format long
q = integral(fun,0,Inf,'RelTol',1e-8,'AbsTol',1e-13)
q =
-14.999999999998360
## Input Arguments
collapse all
Integrand, specified as a function handle, which defines the function to be integrated from xmin to xmax.
For scalar-valued problems, the function y = fun(x) must accept a vector argument, x, and return a vector result, y. This generally means that fun must use array operators instead of matrix operators. For example, use .* (times) rather than * (mtimes). If you set the 'ArrayValued' option to true, then fun must accept a scalar and return an array of fixed size.
Lower limit of x, specified as a real (finite or infinite) scalar value or a complex (finite) scalar value. If either xmin or xmax are complex, then integral approximates the path integral from xmin to xmax over a straight line path.
Data Types: double | single
Complex Number Support: Yes
Upper limit of x, specified as a real number (finite or infinite) or a complex number (finite). If either xmin or xmax are complex, integral approximates the path integral from xmin to xmax over a straight line path.
Data Types: double | single
Complex Number Support: Yes
### Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: integral(fun,a,b,'AbsTol',1e-12) sets the absolute error tolerance to approximately 12 decimal places of accuracy.
Absolute error tolerance, specified as the comma-separated pair consisting of 'AbsTol' and a nonnegative real number. integral uses the absolute error tolerance to limit an estimate of the absolute error, |qQ|, where q is the computed value of the integral and Q is the (unknown) exact value. integral might provide more decimal places of precision if you decrease the absolute error tolerance.
Note
AbsTol and RelTol work together. integral might satisfy the absolute error tolerance or the relative error tolerance, but not necessarily both. For more information on using these tolerances, see the Tips section.
Example: integral(fun,a,b,'AbsTol',1e-12) sets the absolute error tolerance to approximately 12 decimal places of accuracy.
Data Types: single | double
Relative error tolerance, specified as the comma-separated pair consisting of 'RelTol' and a nonnegative real number. integral uses the relative error tolerance to limit an estimate of the relative error, |qQ|/|Q|, where q is the computed value of the integral and Q is the (unknown) exact value. integral might provide more significant digits of precision if you decrease the relative error tolerance.
Note
RelTol and AbsTol work together. integral might satisfy the relative error tolerance or the absolute error tolerance, but not necessarily both. For more information on using these tolerances, see the Tips section.
Example: integral(fun,a,b,'RelTol',1e-9) sets the relative error tolerance to approximately 9 significant digits.
Data Types: single | double
Array-valued function flag, specified as the comma-separated pair consisting of 'ArrayValued' and a numeric or logical 1 (true) or 0 (false). Set this flag to true or 1 to indicate that fun is a function that accepts a scalar input and returns a vector, matrix, or N-D array output.
The default value of false indicates that fun is a function that accepts a vector input and returns a vector output.
Example: integral(fun,a,b,'ArrayValued',true) indicates that the integrand is an array-valued function.
Integration waypoints, specified as the comma-separated pair consisting of 'Waypoints' and a vector of real or complex numbers. Use waypoints to indicate points in the integration interval that you would like the integrator to use in the initial mesh:
• Add more evaluation points near interesting features of the function, such as a local extrema.
• Integrate efficiently across discontinuities of the integrand by specifying the locations of the discontinuities.
• Perform complex contour integrations by specifying complex numbers as waypoints. If xmin, xmax, or any entry of the waypoints vector is complex, then the integration is performed over a sequence of straight line paths in the complex plane. In this case, all of the integration limits and waypoints must be finite.
Do not use waypoints to specify singularities. Instead, split the interval and add the results of separate integrations with the singularities at the endpoints.
Example: integral(fun,a,b,'Waypoints',[1+1i,1-1i]) specifies two complex waypoints along the interval of integration.
Data Types: single | double
Complex Number Support: Yes
## Tips
• The integral function attempts to satisfy:
abs(q - Q) <= max(AbsTol,RelTol*abs(q))
where q is the computed value of the integral and Q is the (unknown) exact value. The absolute and relative tolerances provide a way of trading off accuracy and computation time. Usually, the relative tolerance determines the accuracy of the integration. However if abs(q) is sufficiently small, the absolute tolerance determines the accuracy of the integration. You should generally specify both absolute and relative tolerances together.
• If you are specifying single-precision limits of integration, or if fun returns single-precision results, you might need to specify larger absolute and relative error tolerances.
## References
[1] L.F. Shampine “Vectorized Adaptive Quadrature in MATLAB®,” Journal of Computational and Applied Mathematics, 211, 2008, pp.131–140.
## Version History
Introduced in R2012a | 2022-11-28T15:33:50 | {
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https://web2.0calc.com/questions/algebra-i-question | +0
# Algebra I question
+5
673
8
+3450
Hey guys.
I'm just totally drawing a blank here...
$${\frac{{\mathtt{2}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{X}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{5}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{17}}}{{\mathtt{30}}}}$$
When I solve these equations I like to turn everything to decimals and solve them, but here they want the answer in fraction form. I can figure out that X = -2.05 then turn that into X = -2 & 1/20 but can you show me to work this out keeping everything in fractions?
Thanks,
~Ninja
#3
+576
+10
It is always best to leave things in terms of fractions for a most exact answer. The first thing that is tough about this is that you dont have a common denominator, so let's fix that!
We want out common denominator to be 30 here because it is easy to turn 3 and 5 into thirty with multiplication. Multiply $${\frac{{\mathtt{2}}}{{\mathtt{3}}}}$$2/3 by 10/10 (which you should notice is legal because we are just multiplying by one) and multiply 4/5 by 6/6. This gives us a new expression
(20/30)x+(24/30)=-17/30
Let's subtract (24/30) from both sides to get (20/30)x=-41/30
Now we can multiply both sides by 30 so that there are no longer any fractions
20x=-41 And now we divide by 20 to get x=-41/20 which is exactly the same as -2.05
jboy314 Jun 24, 2014
#1
+88898
+10
Subtract 4/5 from each side ..... we have...
(2/3)x = -(17/30) - (4/5) and 4/5 = 24/30 ....so we have
(2/3)x = -(17/30) - (24/30)
(2/3)x = -41/30 multiply both sides by (3/2)
x = (-41/30) * ( 3/2) (cancel the 30 and the 3 )
= (-41/10) * (1/2) = (-41/20)
And that's it, ND.....
CPhill Jun 24, 2014
#2
+3450
+5
Thanks CPhill.
I'm not sure why I was getting tripped up on that one, but you explained it well.
"Thumbs up and points!"
#3
+576
+10
It is always best to leave things in terms of fractions for a most exact answer. The first thing that is tough about this is that you dont have a common denominator, so let's fix that!
We want out common denominator to be 30 here because it is easy to turn 3 and 5 into thirty with multiplication. Multiply $${\frac{{\mathtt{2}}}{{\mathtt{3}}}}$$2/3 by 10/10 (which you should notice is legal because we are just multiplying by one) and multiply 4/5 by 6/6. This gives us a new expression
(20/30)x+(24/30)=-17/30
Let's subtract (24/30) from both sides to get (20/30)x=-41/30
Now we can multiply both sides by 30 so that there are no longer any fractions
20x=-41 And now we divide by 20 to get x=-41/20 which is exactly the same as -2.05
jboy314 Jun 24, 2014
#4
+3450
+5
Thanks jboy.
I get what your saying with the whole fractions being the best answer thing. When I just turn everything into decimals on equations like this, sometimes I get answers like X = .454545 and I have no idea how to turn that into a fraction...but it's easy to turn a fraction into a decimal!
Thanks for the feedback guys, I appreciate it!
#5
+576
0
Exactly! It is always easy to turn a fraction into a decimal but sometimes it is very difficult and cumbersone to turn a decimal into a fraction.
jboy314 Jun 24, 2014
#6
+88898
+5
This is a little off- topic, but let me show you how to turn a "repeating" decimal into a fraction........
Suppose we have the decimal .127 where the '27" part repeats
Combine the non-repeating part (1) with the repeating part (27)...so we have... 127
Now from this, subtract the non-repeating part, so we have 127 - 1 = 126
Divide this by a number comprised of 9s and 0s, where the number of 9s = the number of repeating digits (2) and the number of 0s = the number of non-repeating digits (1). So the number we divide by = 990....so we have
126/990 = .127(2727....)
Using your example, ND, of .454545
Combining the repeating part with the non-repeating part = 45 (The non-repeating part = 0, in this case)
Subtracting the non-repeating part from the repeating part gives 45 - 0 = 45
Write this over 99 (the number of 9s = the number of repeated digits, we have no non-repeating ones, thus, there are no 0s)
So 45/99 = .4545(4545.....)
In effect, if there is no non-repeating part, we just write our repeating part over the number of 9s = the number of digits in the repeating part !!!
Note also that you may be able to reduce the resulting fraction to a more simple form.....
And that's it !!
(Try writing .817(1717.....) as a fraction using this procedure....)
CPhill Jun 24, 2014
#7
+3450
0
That's an interesting trick, CPhill.
Let's try the example you gave: .817(1717.....)
Add the repeating part to the non-repeating part.
817
Subtract the non-repeating part
817-8 = 809
Put this over 9's and 0's
809/990 = .817(1717...)
This seems to work pretty well, but I think I'll stick to solving it in the fraction form if possible!
#8
+93363
+5
2/3X +4/5 = -17/30
30 is the lowest common denominator so I would just multiply both sides by 30 right from the beginning.
$$\begin{array}{rll} 30\left(\frac{2x}{3}+\frac{4}{5}\right)&=&30\times\frac{-17}{30}\\\\ 20x+24&=&-17\\\\ 20x&=&-41\\\\ x&=&\frac{-41}{20}\\\\ x&=&-2.05 \end{array}$$
Melody Jun 25, 2014 | 2018-09-25T14:00:14 | {
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https://math.stackexchange.com/questions/1399584/investigating-whether-a-given-relation-is-reflexive-symmetric-and-transitive | Investigating whether a given relation is reflexive, symmetric, and transitive
Let $$X = \{0, 1, 2, ... , 10\}$$, Define the relation $$R$$ on $$X$$ by, for all $$a, b \in X$$, $$aRb$$ if and only if $$a + b = 10$$.
Is $$R$$ reflexive? symmetric, transitive? Give reasons.
Here are my answers, please see if I made any mistakes?
$$R$$ is not reflexive, because there exists $$a \in X$$ such that $$a$$ does not relate $$a$$. For example, let $$a=1$$, $$1+1=2$$ which is not equal to $$10$$.
$$R$$ is symmetric, because $$a + b = b + a$$. sum of integers are symmetric. So $$4R6$$ and $$6R4$$.
$$R$$ is not transitive, because there are $$a, b, c$$ integers such that $$aRb$$ $$bRc$$ but $$a$$ does NOT relate $$c$$. Let $$a = 4, b = 6$$ and $$c = 4$$. Then, $$4R6$$ and $$6R4$$ but $$4$$ does NOT relate $$4$$.
• Everything looks good, but there's a typo in the last sentence: you probably meant "Then $4R6$ and $6R4$..." – coldnumber Aug 16 '15 at 19:41
• @coldnumber Thanks - fixed that. My doubt is about the last part, because a=c, so is that still not transitive? – user3282081 Aug 16 '15 at 19:43
• Your example works even if $a=c$. Even if we keep the elements abstract, $aRb, bRa \implies aRa$ for a transitive relation (note that this doesn't imply reflexivity because you may not always have such pairs). – coldnumber Aug 16 '15 at 19:44
• For reflexivity, your work is fine. If $$R$$ were reflexive, then we would have $$1R1$$, but $$1+1 \ne 10$$, so that doesn't hold.
• For symmetry, you have the right idea, but you should bear in mind that a simple example does not mean that it holds all of the time. (For instance, that $$5R5$$ holds doesn't mean the relation as a whole is reflexive!) What you need to see is that symmetry holds if, whenever $$aRb$$ holds, then $$bRa$$ holds. This follows from commutativity of addition:
$$aRb \iff a+b = 10 \iff b+a = 10 \iff bRa$$
• For transitivity, you have the right idea: a counterexample is always a nice means of disproof when possible. If it were transitive, then we would have $$4R6$$ and $$6R4$$ implies $$4R4$$, but $$4+4=8 \ne 10$$, so we don't have transitivity. | 2021-04-15T14:51:24 | {
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https://math.stackexchange.com/questions/1744886/proof-of-the-formula-for-the-number-of-subsets-of-an-n-element-set | # Proof of the formula for the number of subsets of an n-element set
Given a set $A = \{1,2,...,n\}$, the number of subsets of this set can be given by the cardinality of the powerset of A: $$|\mathscr P(A)| = 2^n$$ This is fairly standard and I'm happy with the concept. I am curious, however, as to how one would go about constructing a proof for this, as I've only been presented with a half proof that, to me, doesn't seem to hold up as a rigorous proof. I'll put it here anyway, though.
Let $A = \{a_1,a_2,...,a_n\}$, then we can describe any subset, $S$, of $A$ by answering $n$ questions about it:
• is $a_1 \in S$ ? - Yes/No
• is $a_2 \in S$ ? - Yes/No
.
.
• is $a_n \in S$ ? - Yes/No
Each answer specifies a subset and so there are $2^n$ possible answers.
$\therefore |\mathscr P(A)| = 2^n$
This is exactly what I have written in my lecture notes and not only does it not seem to constitute a proper proof, it doens't seem to entirely make sense to me. For example, when it says that each answer specifies a subset, what does it actually mean? Sure we can have each individual element of $A$ as a subset but then wouldn't we start having subsets that have a cardinality of more than 1?
i.e $\mathscr P(A) = \{\{1\},....,\{n\},\{1,2\},...\}$
and when we get to $\{1,2\}$ we would have to start answering 2 questions about weather or not $1 \in S$ and $2 \in S$.
If someone could clarify what this mess of a proof is actually trying to say and possibly point me in the right direction of a rigorous proof I would be very grateful. Thanks
$P(A)$ is the cardinality of $Map(A,\{0,1\})$ the sets of maps $A\rightarrow \{0,1\}$ which is $2^n$.
Each map $f:A\rightarrow \{0,1\}$ is the characteristic function of a unique subset of $A$.
To compute the cardinality of $Map(A,\{0,1\})$ remark that if $A=\{1,...,n\}$, you have 2 choices to define the image of $i\in\{1,...,n\}$ this gives you $2^n$ choices.
Go by induction, if you like. There is a bijection between $\mathcal{P}\{ 1, 2, \dots, n \}$ and $\mathcal{P}(\{1, 2, \dots, n-1 \}) \times \{0, 1\}$, given by $$\phi: A \mapsto \langle A \setminus \{ n \}, 1[n \in A] \rangle$$ where $1[n \in A]$ is the indicator function which takes the value $1$ if $n \in A$, and $0$ otherwise.
This is clearly bijective: indeed, it has inverse $(X, 1) \mapsto X \cup \{n\}$, and $(X, 0) \mapsto X$.
Now, what is the cardinality of $\{1, 2, \dots, n-1 \} \times \{0, 1\}$? It's the cardinality of $\{1, 2, \dots, n-1\}$ times the cardinality of $\{0,1\}$, which is inductively $2^{n-1} \times 2 = 2^n$.
1. How do you specify a subset of $A$?
A common way is to write down all the elements of the subset. This is the same as to answer the $n$ questions you wrote.
Suppose $A_1,A_2\subset A$. It is easy to see that $A_1=A_2$ if and only if, to each question, you get the same answers for $A_1$ and $A_2$.
1. How many answers you can get?
For each question, there are two different answers. Hence, there are $2^n$ different answers. Therefore, there are $2^n$ different subsets of $A$.
Err... You ALWAYS ask ALL $n$ questions, even for subsets with only one element! In those cases, one answer is yes and the rest are no.
Trying to explain this in English obscures what's going on. Logically, the proof is this:
FIRST, consider the set of all ordered $n$-tuples of 0 and 1. For example, for $n$ = 3, these are 000, 001, 010, 011, .... This set is $\{0,1\} \times \cdots \times\{0,1\}$ (Cartesian product), which shows the cardinality is $2^n$. Then, arrange the elements of $A$ in a sequence: $\{a_1, \dots, a_n\}$ (prove by induction that such an ordering always exists). Then establish a bijection between the set of ordered $n$-tuples of 0 and 1 to $\cal P (A).$ Specifically, to every given $n$-tuple, associate the subset of $\{a_k\}$ where the $k^{\text th}$ element in the $n$-tuple is 1. Show this is indeed a bijection, it's easy. This bijection is described in words by the yes/no questions in your formulation.
For a completely different proof (not related to yes/no question): Induction by $n$. Fix an element of $A$, call it $a_n$. $\cal P(A)$ is the disjoint union of two families of subsets: Those that contain $a_n$ and those that don't. BOTH families are in obvious bijection with $\cal P (A \setminus \{a_n\})$ (one of them IS this set of subsets) so by induction each of the two families has cardinality $2^{n-1}.$ Done.
I prefer the first proof; I always prefer a direct proof when one exists, and use induction when there really is no other way (which is still often enough).
Another proof:
1. There is only one zero-set : $\binom {n}{0}$
2. How many different sets, I could make by $1$ element : $\binom{n}{1}$
3. From two elements : $\binom{n}{2}$
So, we got : $\sum_{i = 0}^{n}{\binom{n}{i}} = 2^{n}$, that's easy to prove by Pascal triangle!
• @Joanpemo yes, thanks a lot! – openspace Apr 16 '16 at 11:53
• Last equality: also by the binomial theorem (consider $(1+1)^n$). Also, why start with 1. subsets with 1 element, and not with 0. subsets with 0 elements? – mathguy Apr 16 '16 at 11:57
• @mathguy yes, but when I was young, I proved it by Pascal triangle, it was so easy, so now I recomend it always I forgot about writing about zero-set. – openspace Apr 16 '16 at 12:00
Let $a_n$ denote the number of subsets of the $n$-element set $\{1,2,\ldots,n\}$.
Clearly $a_0 = 1$, since only the empty set occurs.
Moreover $a_{n+1} = 2a_n$. For among the subsets of $\{1,2,\ldots,n+1\}$ we can distinguish two types: those which contain $n+1$ and those which dont. Those of each type are in $1-1$ correspondence with subsets of $\{1,2,\ldots,n\}$, hence the recursion.
So the sequence $(a_n)_{n=0}^{\infty}$ is defined recursively by
$$a_0 = 1, \:\:\ a_{n+1}=f(a_n), \ \text{where}\:\ f(x)=2x$$
From this it immediately follows that $a_n = 2^n$ | 2020-02-25T12:28:59 | {
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https://math.stackexchange.com/questions/2882400/proving-that-a-sequence-is-periodic | # Proving that a sequence is periodic
Given the following sequence,
$$1,1,2,3,5,8,3,1,4,5,9,4,3,7....$$ $$a_{n+2}=(a_n+a_{n+1}) \mod{10}\;\;\;\; \forall\;\;{n\geq{1}}$$ Prove that it is periodic?
My Attempt:
There can be atmost $10 \times 10=100$ unique pairs of integers $(a,b)$ with $0\leq{a,b}\leq9$.
So pairs will certainly start repeating after $100$ pairs.
Let the first pair repeating after $100$ is $(x,y)$.
The problem I am facing is that how can prove that $(x,y)$ will have unique predecessors and successors, i.e., if $$...a,b,x,y,c,d...\in\text{sequence}$$
also,
$$...e,f,x,y,g,h... \in \text{sequence}$$
then , $a=e,b=f,c=g,d=h...$
If I can prove this, I can easily prove that $(x,y)$ will trace back to $(1,1)$ and the sequence is periodic.
Any hints on how I can show this.
Also let me know if I can make the question more clearer.
• Abstract duplicate of math.stackexchange.com/a/872077/44121 Aug 14 '18 at 12:29
• This question makes no sense. You can't prove that a sequence is periodic if you haven't defined it. Aug 14 '18 at 12:36
• If your sequence has $b,x,y$ as consecutive terms then $y\equiv x+b\pmod {10}$ so you can solve for $b\pmod {10}$ given $x,y$.
– lulu
Aug 14 '18 at 12:37
• Wikipedia says the period is $60$.
– lhf
Aug 14 '18 at 12:40
• In this case the series is periodic from the start because the recurrence relation also works backwards. Aug 14 '18 at 12:47
First of all, when you say that the sequence follows the rule that you specify just by looking at its first entries, you are making a guess ... a reasonable guess, of course, but a guess nevertheless.
Anyway, let's just say that the rule you indicate does indeed specify the sequence we are interested in. Is it periodic?
Your observation that given that the number of pairs of numbers is finite, and hence that there has to be a repeat of some a pair of numbers at some point is correct.
However, there is no need to worry about the predecessors: once you have a repeat of $x,y$ in your sequence, then, given the rule, from then on everything will be the same again, and that is all you need to show it is cyclic.
In fact, there are many sequences that are cyclic, but where the cycle is preceded by a little 'start-up' that does not get repeated ... i.e. where its graph looks like a '6', not a '0'. For such sequences, you would not be able to prove that the predecessors are always the same as well, because that would not be true. But, you still have that cycle, and that makes it periodic.
In sum, the sequence will be periodic, though it may not go back to $(1,1)$
That said, it is actually easy to show that the sequence will go back to $(1,1)$, for all the predecessors will be the same: if you have
$$b,x,y$$
and
$$f,x,y$$
then it must be the case that
$$b+ x =y \mod 10$$
and
$$f+x=y \mod 10$$
and from that it immediately follows that $b=x$ given that $0 \le b,f \le 9$
Once you have established that $b=f$ it likewise follows that $a=e$ ... but you really don't need to show that anymore: once you have shown the immediate predecessor of two numbers is fixed, then all predecessors are fixed. In short, from any $a_n$ and $a_{n+1}$, both $a_{n+2}$ and $a_{n-1}$ are fully determined. | 2022-01-22T18:21:38 | {
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https://math.stackexchange.com/questions/325950/integral-of-cscx | # Integral of $\csc(x)$
I'm getting a couple of different answers from different sources, so I'd like to verify something.
I misplaced my original notes from my prof, but working from memory I have the following:
\begin{align} \int\csc(x)\ dx&=\int\csc(x)\left(\frac{\csc(x)-\cot(x)}{\csc(x)-\cot(x)}\right)\ dx\\ &=\int\frac{\csc^{2}(x)-\csc(x)\cot(x)}{\csc(x)-\cot(x)}\ dx\\ &=\int\frac{1}{u}\ du\\ &=\ln|u|+C\\ &=\ln|\csc(x)-\cot(x)|+C \end{align}
This looks proper when I trace it, but wolfram alpha is saying that the answer should be
$$-\ln|\csc(x)+\cot(x)|+C$$
Sadly, it doesn't provide a step-by-step. It just says that's the answer.
So which is it? Or are they both equivalent? I've never been great with the laws of logarithms.
$$(\csc x-\cot x)(\csc x+\cot x)=\csc^2x-\cot^2x=1\;;$$
this implies that
$$|\csc x-\cot x|\cdot|\csc x+\cot x|=|\csc^2x-\cot^2x|=1\;.$$
Now use the fact that if $a,b>0$ and $ab=1$, then $\ln a+\ln b=\ln 1=0$, so $\ln a=-\ln b$ to conclude that
$$\ln|\csc x-\cot x|=-\ln|\csc x+\cot x|\;,$$
and the two answers are the same.
• That's quite interesting - I never would have known that. – agent154 Mar 10 '13 at 0:02
• That's pretty sophisticated logic, I have to admit) – Alex Mar 10 '13 at 1:17 | 2019-10-14T16:46:31 | {
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https://la.mathworks.com/help/symbolic/sym.int.html | int
Definite and indefinite integrals
Description
example
F = int(expr) computes the indefinite integral of expr. int uses the default integration variable determined by symvar(expr,1). If expr is a constant, then the default integration variable is x.
example
F = int(expr,var) computes the indefinite integral of expr with respect to the symbolic scalar variable var.
example
F = int(expr,a,b) computes the definite integral of expr from a to b. int uses the default integration variable determined by symvar(expr,1). If expr is a constant, then the default integration variable is x.
int(expr,[a b]) is equivalent to int(expr,a,b).
example
F = int(expr,var,a,b) computes the definite integral of expr with respect to the symbolic scalar variable var from a to b.
int(expr,var,[a b]) is equivalent to int(expr,var,a,b).
example
F = int(___,Name,Value) specifies additional options using one or more Name,Value pair arguments. For example, 'IgnoreAnalyticConstraints',true specifies that int applies additional simplifications to the integrand.
Examples
collapse all
Define a univariate expression.
syms x
expr = -2*x/(1+x^2)^2;
Find the indefinite integral of the univariate expression.
F = int(expr)
F =
$\frac{1}{{x}^{2}+1}$
Define a multivariate function with variables x and z.
syms x z
f(x,z) = x/(1+z^2);
Find the indefinite integrals of the multivariate expression with respect to the variables x and z.
Fx = int(f,x)
Fx(x, z) =
$\frac{{x}^{2}}{2 \left({z}^{2}+1\right)}$
Fz = int(f,z)
Fz(x, z) = $x \mathrm{atan}\left(z\right)$
If you do not specify the integration variable, then int uses the first variable returned by symvar as the integration variable.
var = symvar(f,1)
var = $x$
F = int(f)
F(x, z) =
$\frac{{x}^{2}}{2 \left({z}^{2}+1\right)}$
Integrate a symbolic expression from 0 to 1.
syms x
expr = x*log(1+x);
F = int(expr,[0 1])
F =
$\frac{1}{4}$
Integrate another expression from sin(t) to 1.
syms t
F = int(2*x,[sin(t) 1])
F = ${\mathrm{cos}\left(t\right)}^{2}$
When int cannot compute the value of a definite integral, numerically approximate the integral by using vpa.
syms x
f = cos(x)/sqrt(1 + x^2);
Fint = int(f,x,[0 10])
Fint =
${\int }_{0}^{10}\frac{\mathrm{cos}\left(x\right)}{\sqrt{{x}^{2}+1}}\mathrm{d}x$
Fvpa = vpa(Fint)
Fvpa = $0.37570628299079723478493405557162$
To approximate integrals directly, use vpaintegral instead of vpa. The vpaintegral function is faster and provides control over integration tolerances.
Fvpaint = vpaintegral(f,x,[0 10])
Fvpaint = $0.375706$
Define a symbolic matrix containing four expressions as its elements.
syms a x t z
M = [exp(t) exp(a*t); sin(t) cos(t)]
M =
$\left(\begin{array}{cc}{\mathrm{e}}^{t}& {\mathrm{e}}^{a t}\\ \mathrm{sin}\left(t\right)& \mathrm{cos}\left(t\right)\end{array}\right)$
Find indefinite integrals of the matrix element-wise.
F = int(M,t)
F =
$\left(\begin{array}{cc}{\mathrm{e}}^{t}& \frac{{\mathrm{e}}^{a t}}{a}\\ -\mathrm{cos}\left(t\right)& \mathrm{sin}\left(t\right)\end{array}\right)$
Define a symbolic function and compute its indefinite integral.
syms f(x)
f(x) = acos(cos(x));
F = int(f,x)
F(x) =
$x \mathrm{acos}\left(\mathrm{cos}\left(x\right)\right)-\frac{{x}^{2}}{2 \mathrm{sign}\left(\mathrm{sin}\left(x\right)\right)}$
By default, int uses strict mathematical rules. These rules do not let int rewrite acos(cos(x)) as x.
If you want a simple practical solution, set 'IgnoreAnalyticConstraints' to true.
F = int(f,x,'IgnoreAnalyticConstraints',true)
F(x) =
$\frac{{x}^{2}}{2}$
Define a symbolic expression ${\mathit{x}}^{\mathit{t}}$ and compute its indefinite integral with respect to the variable $x$.
syms x t
F = int(x^t,x)
F =
By default, int returns the general results for all values of the other symbolic parameter t. In this example, int returns two integral results for the case $t=-1$ and $t\ne -1$.
To ignore special cases of parameter values, set 'IgnoreSpecialCases' to true. With this option, int ignores the special case $t=-1$ and returns the solution for $t\ne -1$.
F = int(x^t,x,'IgnoreSpecialCases',true)
F =
$\frac{{x}^{t+1}}{t+1}$
Define a symbolic function $f\left(x\right)=1/\left(x-1\right)$ that has a pole at $x=1$.
syms x
f(x) = 1/(x-1)
f(x) =
$\frac{1}{x-1}$
Compute the definite integral of this function from $x=0$ to $x=2$. Since the integration interval includes the pole, the result is not defined.
F = int(f,[0 2])
F = $\mathrm{NaN}$
However, the Cauchy principal value of the integral exists. To compute the Cauchy principal value of the integral, set 'PrincipalValue' to true.
F = int(f,[0 2],'PrincipalValue',true)
F = $0$
Find the integral of $\int \mathit{x}\text{\hspace{0.17em}}{\mathit{e}}^{\mathit{x}}\text{\hspace{0.17em}}\mathit{dx}$.
Define the integral without evaluating it by setting the 'Hold' option to true.
syms x g(y)
F = int(x*exp(x),'Hold',true)
F =
$\int x {\mathrm{e}}^{x}\mathrm{d}x$
You can apply integration by parts to F by using the integrateByParts function. Use exp(x) as the differential to be integrated.
G = integrateByParts(F,exp(x))
G =
$x {\mathrm{e}}^{x}-\int {\mathrm{e}}^{x}\mathrm{d}x$
To evaluate the integral in G, use the release function to ignore the 'Hold' option.
Gcalc = release(G)
Gcalc = $x {\mathrm{e}}^{x}-{\mathrm{e}}^{x}$
Compare the result to the integration result returned by int without setting the 'Hold' option.
Fcalc = int(x*exp(x))
Fcalc = ${\mathrm{e}}^{x} \left(x-1\right)$
If int cannot compute a closed form of an integral, then it returns an unresolved integral.
syms f(x)
f(x) = sin(sinh(x));
F = int(f,x)
F(x) =
$\int \mathrm{sin}\left(\mathrm{sinh}\left(x\right)\right)\mathrm{d}x$
You can approximate the integrand function $f\left(x\right)$ as polynomials by using the Taylor expansion. Apply taylor to expand the integrand function $f\left(x\right)$ as polynomials around $x=0$. Compute the integral of the approximated polynomials.
fTaylor = taylor(f,x,'ExpansionPoint',0,'Order',10)
fTaylor(x) =
$\frac{{x}^{9}}{5670}-\frac{{x}^{7}}{90}-\frac{{x}^{5}}{15}+x$
Fapprox = int(fTaylor,x)
Fapprox(x) =
$\frac{{x}^{10}}{56700}-\frac{{x}^{8}}{720}-\frac{{x}^{6}}{90}+\frac{{x}^{2}}{2}$
Input Arguments
collapse all
Integrand, specified as a symbolic expression, function, vector, matrix, or number.
Integration variable, specified as a symbolic variable. If you do not specify this variable, int uses the default variable determined by symvar(expr,1). If expr is a constant, then the default variable is x.
Lower bound, specified as a number, symbolic number, variable, expression, or function (including expressions and functions with infinities).
Upper bound, specified as a number, symbolic number, variable, expression, or function (including expressions and functions with infinities).
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: 'IgnoreAnalyticConstraints',true specifies that int applies purely algebraic simplifications to the integrand.
Indicator for applying purely algebraic simplifications to the integrand, specified as true or false. If the value is true, apply purely algebraic simplifications to the integrand. This option can provide simpler results for expressions, for which the direct use of the integrator returns complicated results. In some cases, it also enables int to compute integrals that cannot be computed otherwise.
Using this option can lead to results not generally valid. This option applies mathematical identities that are convenient, but the results do not always hold for all values of variables.
Indicator for ignoring special cases, specified as true or false. This ignores cases that require one or more parameters to be elements of a comparatively small set, such as a fixed finite set or a set of integers.
Indicator for returning the principal value, specified as true or false. If the value is true, compute the Cauchy principal value of the integral. In live script, the Cauchy principal value of unevaluated integral shows as the symbol.
Indicator for unevaluated integration, specified as true or false. If the value is true, int returns integrals without evaluating them.
Tips
• In contrast to differentiation, symbolic integration is a more complicated task. If int cannot compute an integral of an expression, check for these reasons:
• The antiderivative does not exist in a closed form.
• The antiderivative exists, but int cannot find it.
If int cannot compute a closed form of an integral, it returns an unresolved integral.
Try approximating such integrals by using one of these methods:
• For indefinite integrals, use series expansions. Use this method to approximate an integral around a particular value of the variable.
• For definite integrals, use numeric approximations.
• For indefinite integrals, int does not return a constant of integration in the result. The results of integrating mathematically equivalent expressions may be different. For example, syms x; int((x+1)^2) returns (x+1)^3/3, while syms x; int(x^2+2*x+1) returns (x*(x^2+3*x+3))/3, which differs from the first result by 1/3.
• For indefinite integrals, int implicitly assumes that the integration variable var is real. For definite integrals, int restricts the integration variable var to the specified integration interval. If one or both integration bounds a and b are not numeric, int assumes that a <= b unless you explicitly specify otherwise.
Algorithms
When you use IgnoreAnalyticConstraints, int applies these rules:
• log(a) + log(b) = log(a·b) for all values of a and b. In particular, the following equality is valid for all values of a, b, and c:
(a·b)c = ac·bc.
• log(ab) = b·log(a) for all values of a and b. In particular, the following equality is valid for all values of a, b, and c:
(ab)c = ab·c.
• If f and g are standard mathematical functions and f(g(x)) = x for all small positive numbers, then f(g(x)) = x is assumed to be valid for all complex values x. In particular:
• log(ex) = x
• asin(sin(x)) = x, acos(cos(x)) = x, atan(tan(x)) = x
• asinh(sinh(x)) = x, acosh(cosh(x)) = x, atanh(tanh(x)) = x
• Wk(x·ex) = x for all branch indices k of the Lambert W function. | 2021-09-17T08:28:29 | {
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"lm_q1q2_score": 0.8473786503452273
} |
https://mathhelpboards.com/threads/no-of-ways-to-seat-round-a-table-numbered-seats.77/ | # No. of ways to seat round a table (numbered seats)
#### Punch
##### New member
Two families are at a party. The first family consists of a man and both his parents while the second familly consists of a woman and both her parents. The two families sit at a round table with two other men and two other women. Find the number of possible arrangements if the members of the same family are seated together and the seats are numbered.
What I did was to consider the 2 families, the 2 woman and 2man as 6 groups of people.
6!(3!)(3!)=25920
#### PaulRS
##### Member
If we number the seats 1,2,3,...,10 . Note that, for example, seats 10, 1 and 2 are consecutive seats because we are working with a round table!
So you have to consider the cases in which seats 10 and 1 correspond to the same family too.
#### grgrsanjay
##### New member
First,i am ignoring the numbers on the seat,
this is a round combination
So, formula is (n-1)!
no.of.ways is 5!(3!)(3!)= 4320
Now the seat are numbered,
then i can more these combinations 1 seats,2seata,......9 seats apart from the original one
so,number of ways is 43,200
#### soroban
##### Well-known member
Hello, Punch!
Two families are at a party.
The first family consists of a man and both his parents
. . while the second familly consists of a woman and both her parents.
The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if the members of the same family
. . are seated together and the seats are numbered.
Duct-tape the families together.
We have: .$\text{(M, P, P)}$ . . . and they have $3!$ possible orders.
We have: .$\text{(W, P, P)}$ . . . and they have $3!$ possible orders.
We also have: .$m,\:m,\:w,\:w$
$\text{M}$ has a choice of $10$ seats.
When he is seated, he and his family occupy three seats.
Among the remaining seven seats, $\text{(W, P, P)}$ has $5$ choices for seating.
Then the remaining four people can be seated in $4!$ ways.
Therefore: .$(3!)(3!)(10)(5)(4!) \:=\:43,200$ arrangements.
#### grgrsanjay
##### New member
First,i am ignoring the numbers on the seat,
this is a round combination
So, formula is (n-1)!
no.of.ways is 5!(3!)(3!)= 4320
Now the seat are numbered,
then i can more these combinations 1 seats,2seata,......9 seats apart from the original one
so,number of ways is 43,200
I Wanted to know whether my logic holds good for every similar problem??
#### Plato
##### Well-known member
MHB Math Helper
I Wanted to know whether my logic holds good for every similar problem??
Your logic is correct. But why complicate matters?
Once the seats are numbered, we no longer have a circular table.
So there is no need for that. | 2020-09-26T08:05:49 | {
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https://math.stackexchange.com/questions/1624302/finding-eigenvalues-of-a-3x3-matrix-7-12-17/1624308 | # Finding Eigenvalues of a 3x3 Matrix (7.12-17)
Please check my work in finding eigenvalues for the following problem. I am working out of the textbook Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons.
For reference the following identity is given because some textbooks reverse the formula having $\lambda$ subtract the diagonal elements instead of subtracting $\lambda$ from the diagonal elements:
$$det(A - \lambda I) = \begin{vmatrix} 13-\lambda & 0 & -15 \\ -3 & 4-\lambda & 9 \\ 5 & 0 & -7-\lambda \\ \end{vmatrix} = 0$$ Taking the center column we have: $$(4-\lambda) \begin{vmatrix} 13-\lambda & -15 \\ 5 & -7-\lambda \\ \end{vmatrix} \\ = (4-\lambda)[(13-\lambda)(-7-\lambda) + 5(15)] = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda -21) = 0 \\ = -\lambda^3 + 10\lambda^2 - 24\lambda + 21\lambda - 84 = 0 \\ = \lambda^3 - 10\lambda^2 + 3\lambda + 84 = 0 \\$$ Using an online calculator the characteristic equation factors into: $$\lambda^3 - 10\lambda^2 + 3\lambda + 84 = 0 \\ (\lambda - 4)(\lambda^2 - 6\lambda - 21) = 0 \\$$ But answer in text is \begin{align*} \lambda_1 = 8 \qquad \lambda_2 = 4 \qquad \lambda_3 = -2 \\ \end{align*} Question: Although I can get $\lambda = 4$ out of the factored equation there is no way to get the other two eigenvalues. I suspect my algebra. Where did I go wrong?
• Your error is in = $$(4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda \color{red}{-21}) = 0$$. It should be $\color{blue}{-16}$ and then it factors well to give you the required answerr – Shailesh Jan 24 '16 at 0:11
• And then instead of multplying out $(4-\lambda)(\lambda^2-6\lambda -16)$, just factor $(4-\lambda)(\lambda-8)(\lambda +2)$. – Tim Raczkowski Jan 24 '16 at 0:13
Your error is in the step $$(4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda \color{red}{-21}) = 0$$ It should be $\color{blue}{-16}$ and then it factors well to give you the required answer | 2019-12-13T00:30:55 | {
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https://numbersandshapes.net/posts/voting_power_deegan-packel_holler/ | # Voting power (5): The Deegan-Packel and Holler power indices
Share on:
We have explored the Banzhaf and Shapley-Shubik power indices, which both consider the ways in which any voter can be pivotal, or critical, or necessary, to a winning coalition.
A more recent power index, which takes a different approach, was defined by Deegan and Packel in 1976, and considers only minimal winning coalitions. A winning coalition $$S$$ is minimal if every member of $$S$$ is critical to it, or alternatively, that $$S$$ does not contain any winning coalition as a proper subset. It is easy to see that these are equivalent, for if $$i\in S$$ was not critical, then $$S-\{i\}$$ would be a winning coalition which is a proper subset.
Given $$N=\{1,2,\ldots,n\}$$, let $$W\subset 2^N$$ be the set of all minimal winning coalitions, and let $$W_i\subset W$$ be those that contain the voter $$i$$. Then we define $DP_i=\sum_{S\in W_i}\frac{1}{|S|}$ where $$|S|$$ is the cardinality of $$S$$.
For example, consider the voting game $[16;10,9,6,5]$ Using the indicies 1 to 4 for the voters, the minimal winning coalitions are $\{1,2\},\{1,3\},\{2,3,4\}.$ and hence
\begin{align*} DP_1 &= \frac{1}{2}+\frac{1}{2} = 1 \cr DP_2 &= \frac{1}{2}+\frac{1}{3} = \frac{5}{6} \cr DP_3 &= \frac{1}{2}+\frac{1}{3} = \frac{5}{6} \cr DP_4 &= \frac{1}{3} \end{align*}
and these values can be normalized so that their sum is unity: $[1/3,5/18,5/18,1/9]\approx [0.3333, 0.2778,0.2778, 0.1111].$ In comparison, both the Banzhaf and Shapley-Shubik indices return $[0.4167, 0.25, 0.25, 0.0833].$
Allied to the Deegan-Packel index is Holler's public-good index, also called the Holler-Packel index, defined as $H_i=\frac{|W_i|}{\sum_{j\in N}|W_j|}.$ In other words, this index first counts the number of minimal wining coalitions that contain $$i$$, and then normalises those values for the sum to be unity.
In the example above, we have voters 1, 2, 3 and 4 being members of 2, 2, 2, 1 minimal winning coalitions respectively, and so the power indices are
$[2/7, 2/7, 2/7, 1/7] \approx [0.2857,0.2857,0.2857,0.1429].$
## Implementation (1): Python
We can implement the Deegan-Packel in Python, either by using itertools, or simply rolling our own little functions:
def all_subsets(X):
T = [[]]
for x in X:
T += [t+[x] for t in T]
return(T)
def is_subset(A,B):
out = True
for a in A:
if B.count(a) == 0:
out = False
break
return(out)
def mwc(q,S,T):
if len(T) == 0:
return(S)
else:
if sum(T[0]) >= q:
S += [T[0]]
temp = T.copy()
for t in temp:
if is_subset(T[0],t):
temp.remove(t)
return(mwc(q,S,temp))
else:
return(mwc(q,S,T[1:]))
def prod(A):
m = len(A)
n = len(A[0])
p = 1
for i in range(m):
for j in range(n):
p *= A[i][j]
return(p)
Of the three functions above, the first simply returns all subsets (as lists); the second tests whether one list is a subset of another (treating both as sets), and the final routine returns all minimal winning coalitions using an elementary recursion. The function starts off considering all subsets of the set of weights, and goes through the list until it finds one whose sum is at least equal to the quota. Then it removes all other subsets which are supersets of the found one. The calls the routine on this smaller list.
For example:
>>> wts = [10,9,6,5]
>>> T = all_subsets(wts)[1:]
>>> q = 16
>>> mwc(q,[],T)
[[10, 9], [10, 6], [9, 6, 5]]
It is an easy matter now to obtain the Deegan-Packel indices:
def dpi(q,wts):
m = mwc(q,[],all_subsets(wts)[1:])
dp = []
for w in wts:
d = 0
for x in m:
d += x.count(w)/len(x)
dp += [d]
return(dp)
And as an example:
>>> wts = [10,9,6,5]
>>> q = 16
>>> dpi(q,wts)
[1.0, 0.8333333333333333, 0.8333333333333333, 0.3333333333333333]
and of course these can be normalized so that their sum is unity.
## Implementation (2): Julia
Now we'll use Julia, and its Combinatorics library. Because Julia implements JIT compiling, its speed is generally faster than that of Python.
Just to be different, we'll develop two functions, one which first produces all winning coalitions, and the second which winnows that set to just the minimal winning coalitions:
using Combinatorics
function wcs(q,w)
S = powerset(w)
out = []
for s in S
if sum(s) >= q
append!(out,[s])
end
end
return(out)
end
function mwc(q,out,wc)
if isempty(wc)
return(out)
else
f = wc[1]
popfirst!(wc)
temp = [] # temp finds all supersets of f = wc[1]
for w in wc
if issubset(f,w)
append!(temp,[w])
end
end
return(mwc(q,append!(out,[f]),setdiff(wc,temp)))
end
end
Now we can try it out:
julia> q = 16;
julia> w = [10,9,6,5];
julia> cs = wcs(q,w)
7-element Array{Any,1}:
[10, 9]
[10, 6]
[10, 9, 6]
[10, 9, 5]
[10, 6, 5]
[9, 6, 5]
[10, 9, 6, 5]
julia> mwc(q,[],cs)
3-element Array{Any,1}:
[10, 9]
[10, 6]
[9, 6, 5]
## Repeated elements
Although both Julia and Python work with multisets, this becomes tricky in terms of the power indices. A simple expedient is to change repeated indices by small amounts so that they are all different, but that the sum will not affect any quota. If we have for example four indices which are the same, we can add 0.1, 0.2, 0.3 to three of them.
So we consider the example
$[30;28,16,5,4,3,3]$
given as an example of a polynomial method in the article "Computation of several power indices by generating functions" by J. M. Alonso-Meijide et al; you can find the article on Science Direct.
So:
julia> q = 30;
julia> w = [28,16,5,4,3.1,3];
julia> cs = wcs(q,w);
julia> ms = mwc(q,[],cs)
6-element Array{Any,1}:
[28.0, 16.0]
[28.0, 5.0]
[28.0, 4.0]
[28.0, 3.1]
[28.0, 3.0]
[16.0, 5.0, 4.0, 3.1, 3.0]
From here it's an easy matter to compute the Deegan-Packel power indices:
julia> dp = []
for i = 1:6
x = 0//1
for m in mw
x = x + count(j -> j == w[i],m)//length(m)
end
append!(dp, [x])
end
julia> print(dp)
Any[5//2, 7//10, 7//10, 7//10, 7//10, 7//10]
julia> print([x/sum(dp) for x in dp])
Rational{Int64}[5//12, 7//60, 7//60, 7//60, 7//60, 7//60]
and these are the values obtained by the authors (but with a lot less work). | 2021-03-04T15:43:30 | {
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https://gateoverflow.in/955/gate2003-68 | 896 views
What is the weight of a minimum spanning tree of the following graph?
1. $29$
2. $31$
3. $38$
4. $41$
edited | 896 views
Apply Prim's algorithm, start from A as shown in figure below.
add all the weights in the given figure which would be equal to $31$.
edited by
yes prims is easier to apply than kruskal here ...
Solution: B
The minimum spanning tree is | 2018-02-24T04:21:55 | {
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https://mathoverflow.net/questions/210239/a-curious-determinantal-inequality/210314 | # A curious determinantal inequality
In my study, I come across the following curious inequality, which I do not know a proof yet (so I am asking it here).
Let $A, B$ be $n\times n$ (Hermitian) positive definite matrices. It is very likely true that $$\det \left(A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\right) \ge \det(A+B)^2.$$ Here $A^{\frac{1}{2}}$ is the unique positive definite square root of $A$. I am able to confirm the $3\times 3$ case.
Comments: Only recently did I notice that the majorization $\lambda\left(A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\right) \prec \lambda(A+B)^2$ follows immediately by THEOREM 2 of [R.B. Bapat, V.S. Sunder, On majorization and Schur products, Linear Algebra Appl. 72 (1985) 107–117.] http://www.sciencedirect.com/science/article/pii/0024379585901478
• Does it read $(\det(A+B))^2$ or $\det((A+B)^2)$? – Moritz Jun 26 '15 at 20:59
• @Moritz: It doesn't matter since $\det(X)^2=\det(X^2)$. – GH from MO Jun 26 '15 at 21:02
• Notice equality occurs if A=I. Can you rewrite (A+B) as A(I + A^(-1)B) and find appropriate square roots of A^-1 B ? – The Masked Avenger Jun 27 '15 at 4:03
• Equality occurs if A B commute. – Russel Jun 27 '15 at 4:05
• @GerhardPaseman: I think there would be lots of similar inequalities. The one I am asking may be the "simplest" unknown case. – M. Lin Jun 27 '15 at 20:26
Let $C := A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2}$; this is a positive semi-definite matrix with the same trace as $(A+B)^2$. We show that the eigenvalues of $C$ are majorised by the eigenvalues of $(A+B)^2$, that is to say that the sum of the top $k$ eigenvalues of $C$ is at most the sum of the top $k$ eigenvalues of $(A+B)^2$ for any $k$. By the Schur concavity of $(\lambda_1,\dots,\lambda_n) \mapsto \lambda_1 \dots \lambda_n$, this gives the claimed determinantal inequality.
The sum of the top $k$ eigenvalues of $C$ can be written as $$\hbox{tr}( C P_V )$$ where $V$ is the $k$-dimensional space spanned by the top $k$ eigenvectors of $C$. This can be rearranged as $$\hbox{tr}( (A+B) (A^{1/2} P_V A^{1/2} + B^{1/2} P_V B^{1/2}) ). \quad\quad (*)$$
We can conjugate $A+B$ to be a diagonal matrix $\hbox{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_1 \geq \dots \geq \lambda_n \geq 0$. In particular we have $A+B \leq \lambda_k I + D$ in the sense of positive definite matrices, where $D := \hbox{diag}(\lambda_1-\lambda_k, \dots, \lambda_{k-1}-\lambda_k, 0, \dots, 0)$. Using the fact that $\hbox{tr}(XZ) \leq \hbox{tr}(YZ)$ whenever $X,Y,Z$ are positive semi-definite with $X \leq Y$, we can bound (*) by
$$\hbox{tr}( (\lambda_k I + D) (A^{1/2} P_V A^{1/2} + B^{1/2} P_V B^{1/2}) )$$
which rearranges as
$$\lambda_k \hbox{tr}( (A+B) P_V ) + \hbox{tr}( P_V (A^{1/2} D A^{1/2} + B^{1/2} D B^{1/2}) ).$$ Using $A+B \leq \lambda_k I + D$ for the first term and $P_V \leq I$ for the second term, this is bounded by $$\lambda_k^2 \hbox{tr}( P_V ) + \lambda_k \hbox{tr}( D P_V ) + \hbox{tr}( A^{1/2} D A^{1/2} + B^{1/2} D B^{1/2} ).$$ For the second term we use $P_V \leq 1$, and for the third term we use the cyclic property of trace to bound by $$\lambda_k^2 \hbox{tr}( P_V ) + \lambda_k \hbox{tr}( D ) + \hbox{tr}( (A+B) D ).$$ For the first term we write $\hbox{tr}(P_V) = k = \hbox{tr}(P_W)$, where $W$ is the span of the first $k$ basis vectors $e_1,\dots,e_k$. For the third term we use $A+B \leq \lambda_k I + D$ to bound the above by $$\lambda_k^2 \hbox{tr}( P_W ) + \lambda_k \hbox{tr}( D ) + \hbox{tr}( (\lambda_k I + D) D ).$$ Since $D = P_W D P_W$, we can collect terms to obtain $$\hbox{tr}( P_W (\lambda_k I + D)^2 P_W ).$$ But by construction, $P_W (\lambda_k I + D)^2 P_W = \hbox{diag}( \lambda_1^2, \dots, \lambda_k^2, 0, \dots, 0 )$, so we have bounded (*) by the sum of the top $k$ eigenvalues of $(A+B)^2$, as required.
• Great, thank you for your detailed and clean proof. I use majorization a lot in my study, but still I do not play it at the same level as you do. After reading your proof, I started asking why I did not find a proof myself. Aha, I failed to observe the "key" step $A+B \leq \lambda_k I + D$. Has a similar construction $D$ been used before? Maybe in this context, such a trick is novel. – M. Lin Jun 28 '15 at 2:35
• My original goal is to prove the majorization relation, then as a byproduct, the determinantal inequality. Could the determinantal inequality be proved without appealing to majorization? I guess other MO readers would like to have such attempts. – M. Lin Jun 28 '15 at 2:39
• I found this argument after playing around with the k=1 and k=2 cases for a while. Roughly speaking, $\lambda_k I + D$ represents the "largest" or "worst" that $A+B$ can be if one only constrains the top $k$ eigenvalues of $A+B$, which is what one is doing when trying to prove majorisation. (When $k=1$, $D$ is not present, and when $k=2$, $D$ is a rank one operator; after seeing these two cases I was able to extrapolate to the general case.) – Terry Tao Jun 28 '15 at 2:51
• Very nice proof! It would be interesting to see a "coordinate free" proof in the spirit of the Schur-Horn theorem, but probably it would not be any simpler than this one (especially that your proof is self-contained). – GH from MO Jun 28 '15 at 11:45
EDIT: the argument below is not correct, but I am leaving it here in case it is of use in locating a better solution.
By a limiting argument we may assume that $C := A+B$ is invertible. If we write
$$D := C^{-1/4} A^{1/2} C^{-1/4}$$ and $$E := C^{-1/4} B^{1/2} C^{-1/4}$$
then $D,E$ are positive semi-definite with $D^2+E^2=1$ EDIT: as pointed out in comments, this is not correct, so in particular $D,E$ commute. The inequality can now be written in terms of $C,D,E$ as $$\det( C^{1/4} D C^{3/2} D C^{1/4} + C^{1/4} E C^{3/2} E C^{1/4} ) \geq \det( C^2 )$$ which on multiplying on left and right by $C^{-1/4}$ and setting $F := C^{3/2}$ becomes $$\det( D F D + E F E ) \geq \det( F ).$$ Now observe that the matrix $$\begin{pmatrix} D & E \\ -E & D \end{pmatrix} \begin{pmatrix} F & 0 \\ 0 & F \end{pmatrix} \begin{pmatrix} D & -E \\ E & D \end{pmatrix} = \begin{pmatrix} DFD + EFE & EFD-DFE \\ DFE-EFD & DFD+EFE \end{pmatrix}$$ is positive semi-definite and has determinant $\det(F)^2$ (the first and last matrices on the LHS are orthogonal). Passing to the block-diagonal matrix $$\begin{pmatrix} DFD + EFE & 0 \\ 0 & DFD+EFE \end{pmatrix},$$ which is still positive semi-definite, the eigenvalues here are majorized by the previous matrix (by the Schur-Horn theorem), and so (by the Schur concavity of the product function $(\lambda_1,\dots,\lambda_n) \mapsto \lambda_1 \dots \lambda_n$), the determinant of the latter matrix must be at least as large as the determinant of the former. (This inequality can also be established using Schur complements.) Thus $$\det( DFD + EFE )^2 \geq \det(F)^2$$ and the claim follows.
• Dear Terry, I don't see that $D^2+E^2=1$, can you please give more detail? – GH from MO Jun 27 '15 at 10:52
• I doubt too that $D^2+E^2$ equals $1$. 14 votes pro without a verification ? – Denis Serre Jun 27 '15 at 13:29
• I tried with the choice $A^{1/2}={\rm diag}(2,1)$ and $B^{1/2}=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$, which fails to pass the test $D^2+E^2=1$. – Denis Serre Jun 27 '15 at 13:47
• You are right, of course; I had mistakenly identified a vector space with its dual when thinking about the problem, which translated into the sign error here when converted back into matrices. I can establish the weaker inequality $\det(A^{1/2} (A^2+B^2) A^{1/2} + B^{1/2} (A^2+B^2) B^{1/2}) \geq \det(A^2+B^2) \det(A^2 + A B A^{-1} B)$ with this approach, but it does not appear strong enough to recover the full inequality. – Terry Tao Jun 27 '15 at 17:04
• Apply Fischer's inequality to $\begin{pmatrix} A^{1/2} & B^{1/2} \\ -B^{1/2} & A^{1/2} \end{pmatrix} \begin{pmatrix} A + B & 0 \\ 0 & A+B \end{pmatrix} \begin{pmatrix} A^{1/2} & -B^{1/2} \\ B^{1/2} & A^{1/2} \end{pmatrix}$. By Schur complement, the first and last matrix have determinant $\det( A + A^{1/2} B^{1/2} A^{-1/2} B^{1/2} )$, giving $\det( A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2} ) \geq \det(A+B) \det( A + A^{1/2} B^{1/2} A^{-1/2} B^{1/2} )$ (I had some typos in the previous inequality as I had changed notation in my computations by squaring $A,B$). – Terry Tao Jun 27 '15 at 17:14
Here is a complementary approach without using majorization. The answer is partial because it has an open "TODO". I am writing it down here already in case someone wishes to complete the argument.
Let $A, B, X, Y > 0$. It is easy to show using Schur complements that \begin{equation*} \tag{$*$} AX^{-1}A + BY^{-1}B \ge (A+B)(X+Y)^{-1}(A+B). \end{equation*} From $(*)$ it follows that $\det(AX^{-1}A + BY^{-1}B)\det(X+Y)\ge \det(A+B)^2$.
Let $C=A^{1/2}(A+B)A^{1/2}$ and $D=B^{1/2}(A+B)B^{1/2}$. If we can find (TODO) $X$ and $Y$ such that \begin{equation*} X \gets A(X+Y)^{1/2}C^{-1}(X+Y)^{1/2}A,\quad Y \gets B(X+Y)^{1/2}D^{-1}(X+Y)^{1/2}B, \end{equation*} then we will obtain $$(X+Y)^{1/2}(AX^{-1}A + BY^{-1}B)(X+Y)^{1/2} = C+D = A^{1/2}(A+B)A^{1/2} + B^{1/2}(A+B)B^{1/2}.$$ Combining this identity with the above inequality immediately implies the desired inequality.
Notice that in particular, if $A$ and $B$ commute, then $X=A(A+B)^{-1}$ and $Y=B(A+B)^{-1}$ is a solution. | 2020-05-29T20:53:35 | {
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https://math.stackexchange.com/questions/3447875/prove-that-a-is-uncountable-if-and-only-if-a-times-a-is-uncountable-conclud | # Prove that $A$ is uncountable if and only if $A\times A$ is uncountable. Conclude that the Euclidean space $\mathbb R^n$ is uncountable.
this is the proof that I came up with for the question, it is somewhat similar to others but the Euclidean space doesn't appear in other questions. However, could someone tell me why the paragraph that talks about $$\mathbb R^n$$ is not induction?
Assume $$A\times A$$ is uncountable and suppose that $$A$$ is countable. If $$A$$ is finite with $$|A|=m$$, then $$|A| \times |A|=m \times m=mm$$, therefore $$A \times A$$ is finite, meaning that it is countable. However, this is a contradiction. If $$A$$ is countably infinite, then there exists an injective $$\varphi: A \rightarrow \mathbb N$$ from the lecture notes. Therefore, the map $$g:A\times A \rightarrow \mathbb N \times \mathbb N$$ given by $$g(x,y)=(\varphi(x),\varphi(y))$$ is injective. However, it suffices to show that $$\mathbb N \times \mathbb N$$ is countable, because if $$\mathbb N \times \mathbb N$$ is a countable set, then $$A\times A$$ is countable. Indeed, consider $$f:\mathbb N \times \mathbb N \rightarrow \mathbb N$$ where $$(m,n)\rightarrow 2^m3^n$$, $$f$$ is injective by the uniqueness of prime factorization (class lecture notes). We can therefore conclude that $$\mathbb N \times \mathbb N$$ is a countable set, then $$A\times A$$ is countable. However, this is another contradiction. Therefore, if $$A\times A$$ is uncountable then $$A$$ is uncountable.
Now assume that $$A$$ is uncountable and suppose that $$A\times A$$ is countable. If $$A\times A$$ is countable then that means that there exists $$h: A\times A \rightarrow \mathbb N$$ that is an injective function. This means that $$f: A \rightarrow \mathbb N$$, which means that $$A$$ is countable by definition. However, this is a contradiction since we assumed that $$A$$ is uncountable.
Therefore, we have proven that $$A$$ is uncountable if and only if $$A\times A$$ is uncountable.\
To show that $$\mathbb R^n$$ is uncountable, we must first show that $$\mathbb R$$ is uncountable. In order to do so, we must show that $$(0,1)$$ is not countable. Now suppose that $$(0,1)$$ is countable, therefore we are assuming that it is countably infinite. This means that there must exist a bijection $$r: \mathbb N \rightarrow (0,1)$$. We could write each number in the list in decimal form, as we did in class for each number. Now let $$N$$ be a number that is obtained by doing the following. For $$n \in \mathbb N$$, let the $$n^{th}$$ decimal spot of $$N$$ be equal to the $$n^{th}$$ decimal of the $$n^{th}$$ number in the list that we made and add $$2$$ mod $$10$$. In doing so, we end up with $$N$$ being different than every number in the list, indicating that the list is incomplete. This is a contradiction because we assume that there was bijection $$r: \mathbb N \rightarrow (0,1)$$. Therefore, $$(0,1)$$ is uncountable and since $$(0,1) \subset \mathbb R$$, then we have proven that $$\mathbb R$$ is uncountable from the problem above. Now suppose that $$\mathbb R^{n+1}$$ is given by $$\mathbb R \times \mathbb R^n \times \mathbb R^{n-1} \times ... \times R^0$$, since we know that $$\mathbb R$$ is uncountable then the cross product between uncountable sets is uncountable. Therefore we can conclude that the Euclidean space $$\mathbb R^n$$ is uncountable.
Overall, the proof looks good. You do a good job of being explicit with everything you write, which is good. Just Two things I noticed. You say
If $$A×A$$ is countable then that means that there exists $$h:A×A\to \mathbb{N}$$ that is an injective function. This means that $$f:A\to \mathbb{N}$$ ....
You haven't defined $$f$$. This section of the proof hinges on the existence of such an $$f$$, so it's rather important that you define it. Here is one way to do this:
Because $$h$$, the diagonal map $$d:A \to A\times A$$ given by $$a \to (a,a)$$, and the projection $$p:A\times A \to A$$ given by $$(x,y) \to x$$ are all injective, their composition $$p \circ h\circ d: A\to \mathbb{N}$$ is also injective (drawing a diagram might help). Let $$f = p \circ h\circ d$$... and continue with your proof.
In general, it's ok to just assert things you've done in class (have mercy on your grader), and it sounds like you've already proven $$\mathbb{R}$$ is uncountable, so I'll skip that paragraph.
I do not understand what you mean when you say
Now suppose that $$\mathbb{R}^{n+1}$$ is given by $$\mathbb{R}×\mathbb{R}^{n}×\mathbb{R}^{n-1}×...×\mathbb{R}$$
First off, what are you supposing? When you "suppose" something you're usually assuming it for the sake of contradiction. We already know that $$\mathbb{R}^n$$ looks like. Second, $$\mathbb{R}×\mathbb{R}^{n}×\mathbb{R}^{n-1}×...×\mathbb{R}$$ would appear to be $$\mathbb{R}^{\frac{n(n+1)}{2} +1}$$. I don't see what you're trying to do here.
Here's an inductive proof:
base case: $$\mathbb{R}$$ is uncountable. done.
Now, assume inductively that $$\mathbb{R}^n$$ is uncountable. Then, $$\mathbb{R}^{n+1} = \mathbb{R}^n \times \mathbb{R}$$ is the product of two uncountable sets, meaning is uncountable. This closes the induction.
Hope it helps.
• It does help, induction proofs are not my strongest suit so i was trying to formulate it by saying "now suppose that ..." – Kmjg Khk Nov 23 '19 at 17:23
• I see. you were trying to suppose the inductive hypothesis. For the time being, I'd recommend sticking to the very formulaic "assume inductively that $P(n)$. We show that $P(n+1)$," where $P(n)$ is the proposition you're trying to prove (in this case, "$\mathbb{R}^n$ is uncountable"). It helps you be very explicit about what you are assuming, and what you want to prove. – Noah Caplinger Nov 23 '19 at 18:03
If A is countable, then A×A injects into N×N which injects into N.
Thus A×A is countable.
If A×A is countable, then A injects into A×A which injects into N.
Thus A is countable.
Exercise. Show if A is infinite, then A and A×A are equinumerous. | 2020-07-10T12:22:33 | {
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http://brana-jazyku.cz/best-slice-zmzhjr/viewtopic.php?id=do-corresponding-angles-add-up-to-180-f09a31 | Angles on a straight line add up to 180°. Alternate angles are equal. Here you will be shown how to work out the missing angle on a straight line - using the angle fact that angles on a straight line add up to 180. Relevance. Cite. Angles on a straight line add up to 180°. What is the difference between the exterior and AIA? alternate exterior angles. The question is a bit confused. Thus, a pair of corresponding angles is equal, which can only happen if the two lines are parallel. 10 = 4 + 4 + 2 = 1 + 1 + 8 = 5 + 4 + 1 and so on. Therefore, both angle A and angle B have measures equal to x and are equal in measure. But we see right here that this will not add up to 180 degrees. The angles of a triangle add to 180 simply because the fifth postulate says that angles of any triangle add to 180. $\begingroup$ did cosine rule to show 7.1, but using sine rule I got 53.89 for angle b and 18.85 for angle c which dont add to make 180 $\endgroup$ – Maya Atkinson Jan 22 at 17:55 2 $\begingroup$ There is always two answer when finding an angle with the sine rule, $\theta$ and $180… Therefore y = 180 - x. Putting this into the first equation gives us: a + b + 180 - x = 180. Do angles in a triangle always add up to 180? Mentor: Very good, so what are angles f and g? If so, how is that possible? 60 seconds . Asked by Wiki User. In the above diagram, d and h are corresponding angles. The angles around a point add up to 360 degrees. alternate interior angles. Anytime a transversal crosses two other lines, we get corresponding angles. 0 0 1 0 0 0 0. JJ. It’s only when we have a transversal through parallel lines that we get congruent alternate interior angles and congruent alternate exterior angles. answer choices . Alternate angles form a ‘Z’ shape and are sometimes called Z angles. you can prove they are parallel if they have congruent (the same size) corresponding, or alternative angles, or if co-interior angles are supplementary) Perpendicular Lines. If an angle in a transversal is x˚, then the sum of that angle with its corresponding angle is 2x˚. The usage of "alternate angle" that I'm aware of is to mean either "alternate interior angles" or "alternate exterior angles". No, all corresponding angles are not equal. No. Here, angle 1 + angle 2 = 180. Nothing more or nothing less. Yes. Tags: Question 6 . Lv 4. answer choices . Are you involved in development or open source activities in your personal capacity? Remember, all angles on a straight line add up to 180°. Since angle b and the angle labeled 107 degrees form a flat angle, we have that b = 180 - 107 = 73 degrees. Q. If an angle in a transversal is x˚, then the sum of that angle with its corresponding angle is 2x˚. Student: They're vertical angles which means they are congruent. 50 plus 120 adds up to 170. Nothing more or nothing less. If,on the other hand, you are referring to a closed shape (made of straight lines) then the answer is "it depends on the number of sides". triangles. Tags: Question 6 . Two angles that add up to 180° ... Lines (in the same plane) that never intersect. When the two lines being crossed are Parallel Lines the Consecutive Interior Angles add up to 180°. Allied angles add up to 180°. Two Adjacent Angle Can Be Supplementary Too, If They Add Upto 180°. Top Answer . There is a special kind of trapezoid called an isosceles trapezoid. a and b are adjacent angles. The angles on a straight line add up to 180 degrees. Using some of the above results, we can prove that the sum of the three angles inside any triangle always add up to 180 degrees. They are supplementary (both angles add up to 180 degrees). What is the WPS button on a wireless router? d and f are interior angles. https://tutors.com/math-tutors/geometry-help/types-of-angle-relationships Corresponding angles can be supplementary if the transversal intersects two parallel lines perpendicularly (i.e. If the two lines are parallel, the four angles around E are the same as the four angles around F. This creates four pairs of corresponding angles. 3 years ago. Follow asked Dec 7 '20 at 11:26. excellence excellence. alternate angles add up to 180. when two lines bisect each other at right angle. corresponding angles. Then, since b and e are corresponding angles, e = … How to find corresponding angles? Do corresponding angles add up to 180 degrees? Anonymous. So, I could do it in so many ways. corresponding angles. Q. corresponding angles. Allied angles add up to 180°. 3 years ago. On parallel lines, co-interior (or C) angles add up to 180°. If your impeached can you run for president again? why do the angles in a triangle add up to 180 but in simple terms? Alternate Angles – are angles on opposite sides of the transversal. Therefore a + b = x after … Lv 6. Angles on a straight line add up to 180°. Share. Angles on a straight line add up to 180°. Therefore, angles in a triangle also add up to 180°. If try to you measure three angles of a triangle formed by three distant galaxies, most certainly their sum will be be not 180. What is the value of x? Share. These lines are parallel, because a pair of Corresponding Angles are equal. g and c are corresponding angles. 2 years ago. The parallel case. are supplementary, which means they add up to 180 degrees: form a straight line, so they also add up to 180 degrees: Two Adjacent Angle Can Be Complementary Too If They Add Up To 90°. 30 seconds . Favourite answer. (Click on "Corresponding Angles" to have them highlighted for you.) They occupy the same relative positions. We can use this to find out what the angles in a triangle add up to. Adjacent angles, often abbreviated as adj. In this example, these are corresponding angles: a and e b and f c and g d and h; Parallel Lines. Angles on a straight line add up to 180°. Language Note: Also called Co-Interior Angles in the UK and Australia Relevance. Corresponding angles are congruent, but do not always add up to the same amount. Therefore y = 180 - x. In A Right Triangle, The Altitude From A Right-angled Vertex Will Split The Right Angle Into Two Adjacent Angle; 30°+60°, 40°+50°, Etc. 19 Answers. When two parallel lines are cut by a transversal, corresponding angles are formed. Corresponding Angles. Yes, all triangle angles add up to 180 degrees. Mentor: Perfect. And im unfamiliar with this theorem. The angles in the triangle add up to 180 degrees. Corresponding angles don't add up to 90 or 180 degrees, because corresponding angles will have the same angle measures as each other. These are two angles that are formed by two coplanar lines and a transversal. Supplementary Angles Add Up to 180 . On a sphere they add up to more than 180 degrees, and on a saddle-shaped surface they add up to less. Adjacent angles add up to 180 degrees. Draw a parallel line to the base at apex. 3 months ago. Angle Sum of a Triangle. A straight angle can also be formed by two or more angles that sum to 180 degrees. Lv 7. Alternate angles form a ‘Z’ shape and are sometimes called ‘Z angles’. 54. 2) Since the lines A and B are parallel, we know that corresponding angles are congruent. Angle Sum of a Triangle. Alternate angles are equal. No, all corresponding angles are not equal. https://www.answers.com/Q/Do_alternate_angles_add_up_to_180 Is it a meaningful question? D and X. Isosceles Trapezoid. For example, on the Earth's surface, a straight line is a great circle and the side of a triangle is a segment of a great circle. Two angles whose measures add up to 180 degrees. Lines that intersect to form right angles. As we know, if we add up the interior and exterior angles of one corner of a triangle, we always get 180 0.In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the angle in the bottom right corner to make a 180 0 angle. Answer Save. D and X. Vaman. 0 0. The angles in a quadrilateral add up to 360 degrees. Improve this question . Source(s): https://shrinks.im/a75Ek. The angles in a triangle add up to 180 degrees. By "Interior Angles" I'm guessing that you are referring to parallel lines (as opposed to the interior angles of a closed shape), in which case the answer is always 180 degrees. Well let's say each is 90, then 90+90=180. All Rights Reserved. corresponding angles. Which pair of angles add up to 180 degrees? When one is 120 then the other is 60, and so on. If we move these angles so that they fit in next to each other, we can see that they rest on a straight line. When did organ music become associated with baseball? supplementary). 0 0. What does it mean when there is no flag flying at the White House? Finding a Missing Angle in a Triangle: Subtraction from 180 Degrees, • About Us Now you can see that the angle ABC has a counter part BAG, and angle ACB has the counter part CAG. An isosceles trapezoid is a trapezoid in which the legs are equal in length. Student: That they add up to 180. So these lines aren't parallel. supplementary). 0 0. husoski. When a line crosses two parallel lines (a transversal), a whole new level of angle relationships opens up: consecutive angles. The letter F can also be facing the other way. As you can see, the 'C' shape may appear back to front, but this doesn't affect the angle measurements. Corresponding angles are angles that are in the same place relative to the nearest parallel line. You can test this at home by following these steps: 1) Cut out a triangle 2) Mark the outer angles 3) Cut these angles off 4) Place these marked angles together You should be able to place these angles onto a straight line. Say ABC is the triangle. Yes, because it leads to an understanding that there are different geometries based on different axioms or 'rules of the game of geometry'. Since a triangle is half of a square, that makes it 180 degrees. Draw a parallel line to the base at apex. So x + y = 180. Wiki User Answered . Yes, corresponding angles can add up to 180. Is Betty White close to her stepchildren? Tags: Question 7 . Two angles that are located between two coplanar lines on the same side of the transversal. are corresponding angles, so they’re congruent: The three angles of . Same Side Exterior Angles. Then, using corresponding angles, angle d = 107 degrees and angle f = 73 degrees. 135. So we have 3x + … The more restrictive our intersecting lines get, the more restrictive are their angle relationships. 1 0. miracle. To explore the truth of this rule, try Math Warehouse's interactive triangle , which allows you to drag around the different sides of a triangle and explore the relationship between the angles and sides. With parallel lines there are 3 more rules: Corresponding angles are equal. What are the qualifications of a parliamentary candidate? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Yes. A and G. Tags: Question 5 . For example, angle A and angle B … Copyright © 2021 Multiply Media, LLC. So a + b + y = 180. In such case, each of the corresponding angles will be 90 degrees and their sum will add up to 180 degrees (i.e. Are all Corresponding Angles Equal? 0 0. Two angles that are located between two coplanar lines on the same side of the transversal. Only 10 chocolates. See Answer. Corresponding Angles Postulate- If two Parallel Lines are cut by a … alternate exterior angles. Recall these usual rules of angles (always true): Angles around a point add up to 360°. You can test this at home by following these steps: 1) Cut out a triangle 2) Mark the outer angles 3) Cut these angles off 4) Place these marked angles together You should be able to place these angles onto a straight line. The angles in a polygon (a shape with n sides) add up to 180(n – 2) degrees. Dancing Imu. Alternate angles are equal. The line DAG is drawn at A parallel to BC. Angles A and B are adjacent. alternate interior angles. Angles in corresponding positions are equal Angles on alternate sides of the line are equal Angles that are allied (one acute, one The only way for congruent angles to be supplementary ("add up to 180") is if they are both right angles. The corresponding pairs of base angles, such as A and B, or C and D, are supplementary (add up to 180 degrees). alternate interior angels . So a + b + y = 180. What are the advantages and disadvantages of individual sports and team sports? Source(s): https://shrink.im/a8bDF. The angles on a straight line add up to 180 degrees. How long will the footprints on the moon last? 4 months ago. However, I have a suspicion that you may have meant complementary angles, whose sum is always 90˚. 4 years ago. Two angles whose measures add up to 180 degrees. Now were going to learn about corresponding angles. Those Adjacent Angles Are Complementary. The line DAG is drawn at A parallel to BC. Now you can see that the angle ABC has a counter part BAG, and angle ACB has the counter part CAG. 10 = 4 + 4 + 2 = 1 + 1 + 8 = 5 + 4 + 1 and so on. Parallel lines are two lines on a two-dimensional plane that never meet or cross. Lv 4. angle a + angle b = 180 degrees angle c + angle d = 180 degrees. Putting this into the first equation gives us: a + b + 180 - x = 180. ∠s, are angles that share a common vertex and edge but do not share any interior points. You can test this at home by following these steps: You should be able to place these angles onto a straight line. Q. These lines are not parallel, because a pair of Consecutive Interior Angles do not add up to 180° (81° + 101° =182°) These lines are parallel, because a pair of Alternate Interior Angles are equal The answer is 'sometimes yes, sometimes no'. What is the first and second vision of mirza? Which pair of angles add up to 180 degrees? Therefore, angles in a triangle also add up to 180°. Corresponding angles are congruent, but do not always add up to the same amount. In the figure above, click on 'Next angle pair' to visit each pair in turn. The angles in matching corners are called Corresponding Angles. Another way you could have thought about it-- I guess this would have maybe been a more exact way to think about it --is if this is 120 degrees, this angle right here has to be supplementary to that; it has to add up to 180. In some cases when both angles are $$90^o$$ each, the sum will be $$180^o$$. Let us say I have 10 chocolates and I have to offer them to 3 students (atleast 1 to each). Consecutive Exterior Angles . The angles in the triangle add up to 180 degrees. Where straight lines cross, opposite angles are equal. Any two angles that add up to 180 degrees are known as supplementary angles. answer choices . Any two angles that add up to 180 degrees are known as supplementary angles. As you can see from the picture below, if you add up all of the angles in a triangle the sum must equal $$180^{\circ}$$. Given: ΔABC Prove: All three angles of ΔABC add up to 180°. Since two sectors with supplementary angles make half a circle, supplementary angles always add up to half of 360 degrees, or 180 degrees. Name the relationship between ∠2 and ∠7. Why don't libraries smell like bookstores? add up to 180. vertical anges are congruent along with opposite interior angles. See some examples below. at 90 degrees). By definition, there are 360 degrees in a square. In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the third angle to make a 180 angle. That the corresponding angles are equal and that the alternate angles are equal and that angles on a straight line add up to 180 degrees. These add up to 180 degrees (e and c are also interiors). Answer Save. Would you like to observe visually how the alternate interior angles are equal? And im unfamiliar with this theorem. Cite. Say one angle is 65 degrees, and the other one is 115 degrees. So x + y = 180. Supplementary angles add up to 180 degrees and angles on a straight line add up to 180 degrees Take triangle. Hilary. I was looking for some proofs for corresponding angles are equal, but in the one i found they use this theorem that states that the interior angles of two parallel lines (made by the transversal) add up to 180 degrees. 45. answer choices . Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on WhatsApp (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Pocket (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on Skype (Opens in new window), https://www.mathswithmum.com/wp-content/uploads/2018/05/Anglesinatriangleo.mp4. consecutive angles. One way to find the corresponding angles is to draw a letter F on the diagram. When the two lines being crossed are Parallel Lines the Corresponding Angles are equal. Therefore a + b = x after … Using some of the above results, we can prove that the sum of the three angles inside any triangle always add up to 180 degrees. They occupy the same relative positions. And you also know that angles on a straight line add up to 180°. Take triangle. So, I could do it in so many ways. SURVEY . Now the sum of the angles are 180 degrees. Do the angles of a triangle add up to 180 degrees or$\pi$radians? Do corresponding angles add up to 180 degrees? 7 Answers. The flow chart with missing reason proves the measures of the interior angles of ΔABC total - 9167868 What is the timbre of the song dandansoy? Click the buttons to compare pairs of sectors in a half circle with their supplementary angles. For that to be true, you need the traversal line to be perpendicular to the two parallel lines. Are all Corresponding Angles Equal? A and W. C and X. add up to 180 degrees, so. vertical angles. Corresponding Angles – are angles on the same side of the transversal and also have the same degree of measurement. SURVEY . 5 years ago. Lv 7. (Click on "Consecutive Interior Angles" to have them highlighted for you.) 0 0. saurabh. What are the difference between Japanese music and Philippine music? The angles in the triangle you drew all add up to 180°. Do alternate interior angles add to 180 degrees See answers (2) Ask for details ; Follow Report Log in to add a comment Answer 5.0 /5 1. kingofnfsr05 +1 webew7 and 1 other learned from this answer Yes, they can add up to 180 degrees. 2013-05-19 13:59:04. ... Do interior angles add up to 180$$^\circ$$? Name the relationship between ∠2 and ∠7. That would equal to 180 degrees. Corresponding angles are equal. 30 seconds . • Contact Us • Privacy. To investigate the sum of the angles in a triangle, we can begin by marking each angle with a different colour. The definition of a triangle is that it is a shape with three sides that add up to 180 degrees. Who is the longest reigning WWE Champion of all time? Is this an important question? Where straight lines cross, opposite angles are equal. I was looking for some proofs for corresponding angles are equal, but in the one i found they use this theorem that states that the interior angles of two parallel lines (made by the transversal) add up to 180 degrees. However, I have a suspicion that you may have meant complementary angles, whose sum is … These are two angles that are formed by two coplanar lines and a transversal. A and W. C and X. SURVEY . 35 degrees. What is the point of view of the story servant girl by estrella d alfon? A straight line is 180° and thus the angles in the triangle are also 180°. Using the measure of either angle C or angle D, we find the measure of angle B to be 180 − (180 − x ) = 180 − 180 + x = x. Lv 7. (h and d, f and b, e and a are also corresponding). A triangle's angles add up to 180 degrees because one exterior angle is equal to the sum of the other two angles in the triangle. With parallel lines there are 3 more rules: Corresponding angles are equal. triangles. Add a Comment. Let us say I have 10 chocolates and I have to offer them to 3 students (atleast 1 to each). SURVEY . 126. Only 10 chocolates. Hence, the alternate interior angle theorem is proved. (c and f are also alternate). answer choices . Any two angles that add up to 180 degrees are known as supplementary angles. Any two angles that add up to 180 degrees are known as supplementary angles. 4 years ago. Therefore, angles in a triangle also add up to 180°. Do corresponding angles add up to 180 degrees. Say ABC is the triangle. d and e are alternate angles. In such case, each of the corresponding angles will be 90 degrees and their sum will add up to 180 degrees (i.e. Is proved your impeached can you run for president again music and music... 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Adams County Court Schedule, Eso Warden Tank/healer Build 2020, Slu Med Dean Of Admissions, Access Course Cardiff, Still In Love With Ex After Years, | 2022-01-21T10:26:34 | {
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https://www.physicsforums.com/threads/derive-a-formula-for-motion-with-constant-acceleration-and-constant-deceleration.620233/ | # Homework Help: Derive a formula for motion with constant acceleration and constant deceleration
1. Jul 11, 2012
### LoA
1. The problem statement, all variables and given/known data
A subway train travels over a distance $s$ in $t$ seconds. it starts from rest and ends at rest. In the first part of its journey it moves with constant acceleration $f$ and in the second part with constant deceleration $r \,$.
Show that $s \, = \, \frac {[\frac {fr} {f \, + \, r}] \, t^2} {2}$
2. Relevant equations
I know that $s \,= \, (\frac {1}{2})(-r )t^2\, +\, v_it$, where $v_i \,=\, ft$ but I'm not sure where to go from there. In particular, I can't figure out how to connect the seemingly separate equations for distance generated by the different accelerations into one function of time for the entire interval.
3. The attempt at a solution
By assuming that total acceleration is a sum of the given accelerations, I've gotten something that looks awfully close to the desired result, but am still not quite there:
$s \,=\,\frac{1}{2} \,(f\,+\,r)\,t^2 \, + \,v_i\,t$
I feel like I'm missing something.
Last edited: Jul 11, 2012
2. Jul 11, 2012
### !)("/#
You have to integrate the aceleration two times to get the movement ecuation.
You know that:
$x (t) = x_0 + v_0 (t-t_0) + \frac{1}{2} a (t-t_0)^2$
$v (t) = v_0+ a (t-t_0)$
Part one we start from time, position and initial velocity cero and also positive aceleration f:
$x (t) =\frac{1}{2} f t^2$
$v (t) = a t$
We know that at a certain time $t_1$ we reach a position $x_1$ with a velocity $v_1$ and from that point the movil starts desacelerating with aceleration negative r.
$x (t) = x_0 + v_0 (t-t_0) - \frac{1}{2} a (t-t_0)^2$
$v (t) = v_0 - a (t-t_0)$
Replacing with the information that we know for part two that we start from $t_1$, position $x_1$ and initial velocity $v_1$ so replacing with these information:
$x (t) = x_1 + v_1 (t-t_f) - \frac{1}{2} r (t-t_f)^2$
$v (t) = v_1 - r (t-t_f)$
Also we know that at final time $t_2 = t$ the movil finish with velocity $v_2 = 0$ at the point $x_2 = s$. You have a system of ecuations im sure you are able to solve, give it a try.
3. Jul 11, 2012
### azizlwl
Very interesting but i don't know how to prove it.
It looks like the equivalent of the 2 acceleration value analogous to effective K of 2 springs connected in series.
Keffective=k1k2/(k1+k2)
In the above example, the effective acceleration is equal f1f2/(f1+f2)
Maybe same approach can be taken to prove as calculation to find effective K of springs connected in series.
4. Jul 12, 2012
### TSny
Let t1 be the time of the acceleration phase and t2 the time of deceleration. The increase in speed during the acceleration phase must equal the decrease in speed during the deceleration phase. Hence, ft1=rt2.
Therefore, t2 = (f/r)t1.
Thus, show t = t1 + t2 = $\frac{f+r}{r}$t1. Therefore, t1 = $\frac{r}{f+r}$t and t2 = $\frac{f}{f+r}$t. (These are expressions for t1 and t2 in terms of the total time t.)
Apply the general formula x = vot + $\frac{1}{2}$at2 to the acceleration phase and then to the deceleration and note that the initial velocity of the deceleration phase is the final velocity of the acceleration phase = ft1.
s1 = $\frac{1}{2}$ft12
s2 = (ft1)t2 - $\frac{1}{2}$rt22
Try substituting previously derived expressions for t1 and t2 in terms of the total time t and see if you can simplify s = s1+s2 to the desired result.
[Maybe there's a better way.]
Last edited: Jul 12, 2012
5. Jul 12, 2012
### pgardn
I think its along these lines. The final velocity in the first part of the journey is the same as the initial velocity in the second part of the journey. These yields the equality f*s1 = r*s2 for v^2 given that s1 + s2 = s . So part of the approach might be using the kinematic equation that does not use time to begin with. And then substituting with time and then maybe too much algebra for my taste.
6. Jul 12, 2012
### TSny
Good suggestion!
7. Jul 12, 2012
### pgardn
TSny
Looked at it at lunch. It can be done with your ratio without going "backwards". But its a royal pain.
I guess this was a practice your substitutions and algebra problem. Maybe its easier than what I am looking at, but if not, it was a cruel form of torture and quiet possibly violates the US constitution. | 2018-06-25T02:54:15 | {
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https://www.physicsforums.com/threads/chose-one-of-three-coins-if-it-lands-heads-probability-that-other-side-is-tails.539244/ | # Chose one of three coins, if it lands heads, probability that other side is tails
1. Oct 11, 2011
### Anger
First of all I want to say hello. This is my first time posting on these forums, but they have (well actually YOU have :D) helped me solve certain questions of mine.
For the last week or so, I have been obssessed with a problem that we solved during class.
And I have trouble finding common ground with my professor.
The problem is as following:
We are given three coins: one of them has a head on each side, one of them has a tail on each side, and one "normal" coin with a head on one side and a tail on the other. Now we randomly choose a coin, we toss it and the result is heads.
What is the probabilty that the other side is tails?
The answer that was given to us is 2/3, I managed to come up with 1/3. Cross-checking my results by using common everyday logic, points out to me, that I am actually right. I really need your opinion on this one.
Let's call the two-headed coin C1, the two-tailed coin C2 and the normal coin C3. I start by accepting that fact that, since the result is heads, the coin we tossed is either coin C1 or coin C3.
Now let's say that C1's heads are respectively H1 and H2. And C2's head is H3 and it's tail is just plain T.
The result of tossing heads implies one of three possibilites:
1) the result is H1 and the other side is H2
2) the result is H2 and the other side is H1
3) the result is H3 and the other side is T
So what am I doing wrong?
2. Oct 11, 2011
### micromass
Re: Chose one of three coins, if it lands heads, probability that other side is tail
Look at all the possibilities
1) You picked the H-H coin and the result was H
2) You picked the H-H coin and the result was H (the other side)
3) You picked the H-T coin and the result was H
4) You picked the H-T coin and the result was T
5) You picked the T-T coin and the result was T
6) You picked the T-T coin and the result was T (the other side)
Obviously, all these things have probability 1/6 to happen.
You are now given information that the result is H. So we go to the above diagram and we eliminate all results that are T. This gives us
1) You picked the H-H coin and the result was H
2) You picked the H-H coin and the result was H (the other side)
3) You picked the H-T coin and the result was H
We have agreed these 3 things to have equal probability. So 1&2 imply the other side to be H. Thus the other side is H in 2/3 of the cases. The other side is T is 1/3 of the cases.
3. Oct 12, 2011
### Anger
Re: Chose one of three coins, if it lands heads, probability that other side is tail
Thank you for your contribution, as I see it now, my real problem is confronting my professor that will not be too fond of me correcting her. | 2018-05-28T10:22:32 | {
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http://library.fis.uny.ac.id/5iqyzgnl/beaab8-show-a-matrix-is-positive-definite | ## show a matrix is positive definite
If the factorization fails, then the matrix is not symmetric positive definite. Jede positiv definite Matrix A läßt sich auch schreiben als A = LL t, wobei L eine untere Dreiecksmatrix mit positiven Diagonaleinträgen ist. With a positive definite matrix the usual algorithm succeeds because all the diagonal entries of L s.t. The following changes are made: I changed argument x to A to reflect usual matrix notation. Beispiel. This method does not require the matrix to be symmetric for a successful test (if the matrix is not symmetric, then the factorization fails). 15.3.1.1 Space of Symmetric Positive Definite Matrices. Examples of symmetric positive definite matrices, of which we display only the instances, are the Hilbert matrix. 29.8k 2 2 gold badges 82 82 silver badges 112 112 bronze badges. Note. and minus the second difference matrix, which is the tridiagonal matrix . Property 7: If A is a positive semidefinite matrix, then A ½ is a symmetric matrix and A = A ½ A ½. More specifically, we will learn how to determine if a matrix is positive definite or not. A matrix is positive-definite if its smallest eigenvalue is greater than zero. Suppose M and N two symmetric positive-definite matrices and λ ian eigenvalue of the product MN. Positive definite symmetric matrices have the property that all their eigenvalues are positive. What are the practical ways to make a matrix positive definite? While such matrices are commonly found, the term is only occasionally used due to the possible confusion with positive-definite matrices, which are different. Does this situation show that there is something wrong with my algorithm since the likelihood should increase at every step of EM? Proof: if it was not, then there must be a non-zero vector x such that Mx = 0. The extraction is skipped." Yixiao Yun, Irene Yu-Hua Gu, in Ambient Assisted Living and Enhanced Living Environments, 2017. If one subtracts one positive definite matrix from another, will the result still be positive definite, or not? Positive-definite matrix; Positive-definite function; Positive-definite kernel; Positive-definite function on a group; References. the Pascal matrix. If the Hessian is positive-definite at x, then f attains an isolated local minimum at x.If the Hessian is negative-definite at x, then f attains an isolated local maximum at x. Also, we will… The page says " If the matrix A is Hermitian and positive semi-definite, then it still has a decomposition of the form A = LL* if the diagonal entries of L are allowed to be zero. A non-symmetric matrix (B) is positive definite if all eigenvalues of (B+B')/2 are positive… Fasshauer, Gregory E. (2011), "Positive definite kernels: Past, present and future" (PDF), Dolomites Research Notes on Approximation, 4: 21–63. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … – LaTeXFan Jul 27 '15 at 5:42 From the same Wikipedia page, it seems like your statement is wrong. How can one prove this？ matrix matrix-decomposition. asked Mar 29 '18 at 23:10. jack 看看 jack 看看. The most efficient method to check whether a matrix is symmetric positive definite is to simply attempt to use chol on the matrix. Positive definite matrix. Edit: I'm computing the inverse by using a matrix inversion lemma which states that: $$(BB'+D)^{-1}=D^{-1}-D^{-1}B (I_q+B'D^{-1}B)^{-1} B'D^{-1}$$ This is calculated by sqrtm function. The matrix A can be positive definite only if n+n≤m, where m is the first dimension of K.” (Please could you refer me to an articles or books where I can find such property above). The matrix is pretty big (nxn where n is in the order of some thousands) so eigenanalysis is expensive. MIT Linear Algebra Exam problem and solution. However, it is not here. If the covariance matrix is invertible then it is positive definite. One can show that a Hermitian matrix is positive definite if and only if all its eigenvalues are positive [].Thus the determinant of a positive definite matrix is positive, and a positive definite matrix is always invertible.The Cholesky decomposition provides an economical method for solving linear equations involving a positive definite matrix. A way to check if matrix A is positive definite: A = [1 2 3;4 5 6;7 8 9]; % Example matrix If x is not symmetric (and ensureSymmetry is not false), symmpart(x) is used.. corr: logical indicating if the matrix should be a correlation matrix. Therefore x T Mx = 0 which contradicts our assumption about M being positive definite. Then it's possible to show that λ>0 and thus MN has positive eigenvalues. Eine solche Zerlegung wird als Cholesky-Zerlegung bezeichnet. Is it because of rounding error, please? Symmetric matrices A symmetric matrix is one for which A = AT . Still, for small matrices the difference in computation time between the methods is negligible to check whether a matrix is symmetric positive definite. We prove a positive-definite symmetric matrix A is invertible, and its inverse is positive definite symmetric. Also, if eigenvalues of real symmetric matrix are positive, it is positive definite. Property 8: Any covariance matrix is positive semidefinite. A positive matrix is a matrix in which all the elements are strictly greater than zero. A positive definite matrix M is invertible. A check if the matrix is positive definite (PD) is enough, since the "semi-" part can be seen in the eigenvalues. All the eigenvalues with corresponding real eigenvectors of a positive definite matrix M are positive. Show that the matrix A is positive definite first by using Theorem 7.3 .2 and second by using Theorem 7.3.4. The set of positive matrices is a subset of all non-negative matrices. I want to run a factor analysis in SPSS for Windows. Today, we are continuing to study the Positive Definite Matrix a little bit more in-depth. Conversely, some inner product yields a positive definite matrix. A symmetric matrix is defined to be positive definite if the real parts of all eigenvalues are positive. If a matrix has some special property (e.g. I will show that this matrix is non-negative definite (or "positive semi-definite" if you prefer) but it is not always positive definite. It is known that a positive definite matrix has a Unique Positive Definite square root. That is, S is supposed to be positive definite in theory. All three of these matrices have the property that is non-decreasing along the diagonals. Eigenvalues of a positive definite real symmetric matrix are all positive. If A is a real symmetric positive definite matrix, then it defines an inner product on R^n. To do this, consider an arbitrary non-zero column vector $\mathbf{z} \in \mathbb{R}^p - \{ \mathbf{0} \}$ and let $\mathbf{a} = \mathbf{Y} \mathbf{z} \in \mathbb{R}^n$ be the resulting column vector. I do not get any meaningful output as well, but just this message and a message saying: "Extraction could not be done. share | cite | improve this question | follow | edited Mar 30 '18 at 0:35. Proof: Since a diagonal matrix is symmetric, we have. Learn more about positive, definite, semipositive, chol, eig, eigenvalue MATLAB [3]" Thus a matrix with a Cholesky decomposition does not imply the matrix is symmetric positive definite since it could just be semi-definite. by Marco Taboga, PhD. Symmetric matrices and positive definiteness Symmetric matrices are good – their eigenvalues are real and each has a com plete set of orthonormal eigenvectors. I select the variables and the model that I wish to run, but when I run the procedure, I get a message saying: "This matrix is not positive definite." x: numeric n * n approximately positive definite matrix, typically an approximation to a correlation or covariance matrix. I'm implementing a spectral clustering algorithm and I have to ensure that a matrix (laplacian) is positive semi-definite. A matrix is positive definite if all it's associated eigenvalues are positive. The Hessian matrix of a convex function is positive semi-definite.Refining this property allows us to test whether a critical point x is a local maximum, local minimum, or a saddle point, as follows: . Positive definite matrices are even bet ter. Functions are adapted from Frederick Novomestky's matrixcalc package in order to implement the rmatnorm function. positiv definit, wenn alle Hauptminoren > 0 sind und; negativ definit, wenn alle geraden Hauptminoren der Matrix > 0 und alle ungeraden Hauptminoren der Matrix < 0 sind. Theorem 4.2.3. For the positive semi-definite case it remains true as an abstract proposition that a real symmetric (or complex Hermitian) matrix is positive semi-definite if and only if a Cholesky factorization exists. (a) A=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{ar… A square matrix is positive definite if pre-multiplying and post-multiplying it by the same vector always gives a positive number as a result, independently of how we choose the vector.. Ben Bolker. N * n approximately positive definite symmetric approximation to a to reflect usual matrix notation usual notation. Definite is to simply attempt to use chol on the matrix the property that is, S supposed! 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https://smartsocialtrading.com/amazing-superheroes-gcp/archive.php?763fb4=in-a-matrix-interchanging-of-rows-and-columns-is-called | Example 2: Consider the matrix . Consider the matrix If A = || of order m*n then = || of order n*m. So, . The horizontal array is known as rows and the vertical array are known as Columns. A matrix having m rows and n columns with m ≠ n is said to be a In a matrix multiplication for A and B, (AB)t Matrices obtained by changing rows and columns is called For example matrix = [[1,2,3],[4,5,6]] represent a matrix of order 2×3, in which matrix[i][j] is the matrix element at ith row and jth column.. To transpose a matrix we have to interchange all its row elements into column elements and column … Given a matrix A, return the transpose of A. of Columns]; Ex… Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Federal MCQs, 9th Class MCQs, Math MCQs, Matrices And Determinants MCQs, Symeetric , Identify matrix , transpose , None Thetransposeofasymmetricmatrix of rows] [ no. Each element of the original matrix appears in 2 rows and 3 columns in the enlarged matrix. Show activity on this post. Solution: It is an order of 2*3. • Example 1: Consider the matrix . $$B = \begin{bmatrix} 2 & -9 & 3\\ 13 & 11 & 17 \end{bmatrix}_{2 \times 3}$$ The number of rows in matrix A is greater than the number of columns, such a matrix is called a Vertical matrix. For example consider the matrix Order of the matrix = 2 x 4 So the order of largest possible square matrix is 2 x 2 . Example 1: Consider the matrix Find the Adj of A. The transpose of the transpose of a matrix is that the matrix itself =, The transpose of the addition of 2 matrices is similar to the sum of their transposes =, When a scalar matrix is being multiplied by the matrix, the order of transpose is irrelevant =. G1 * G2' = 44 Verify this result by carrying out the operations on 'matlab'. The operation of interchanging rows and columns in a matrix is called trans from MEGR 7102 at University of North Carolina, Charlotte (x-6 || y-5)) printf ("Variables Swapped. In Python, there is always more than one way to solve any problem. (adsbygoogle = window.adsbygoogle || []).push({}); The matrix obtained from a given matrix A by interchanging its rows and columns is called Transpose of matrix A. Transpose of A is denoted by A’ or . and ' and even the transpose, Stack Overflow. The first row can be selected as X[0].And, the element in the first-row first column can be selected as X[0][0].. Transpose of a matrix is the interchanging of rows and columns. (A’)’= A. In my first programming course, I learnt how to swap two variables, suppose denoted by x and y, without holding a value in a third variable. Maths Help, Free Tutorials And Useful Mathematics Resources. It works as follows. A matrix obtained by interchanging rows and columns is called ____ matrix? Interchanging any pair of columns or rows of a matrix multiplies its determinant by −1. Do the transpose of matrix. Do the transpose of matrix. ... Row switching is interchanging two ____ of a matrix… More generally, any permutation of the rows or columns multiplies the determinant by the sign of the permutation. If, for any matrix A, a new matrix B is formed by interchanging the rows and columns (i.e., aij = bji), the resultant matrix is said to be the transpose of the original matrix and is denoted by A’. Thus the transpose is also the inverse: A− 1 = AT. In other words, we can say that matrix A is another matrix formed by replacing each element of the current matrix by its corresponding cofactor and then taking the transpose of the new matrix formed. Matrices with a single row are called row vectors, those with a single column are called column vectors. Pivoting may be followed by an interchange of rows or columns to bring the pivot to a fixed position and allow … All Rights Reserved. We have: . Cloudflare Ray ID: 5fd3023aedfce4fa The matrix obtained from a given matrix A by interchanging its rows and columns is called Transpose of matrix A. Transpose of A is denoted by A’ or . Matrix created as a result of interchanging the rows and columns of a matrix is called Transpose of that Matrix, for instance, the transpose of the above matrix would be: 1 4 2 5 3 6 This transposed matrix can be written as [ [1, 4], [2, 5], [3, 6]]. • Rank of matrix is the order of largest possible square matrix whose determinant is non zero. In the second step, we interchange any two rows or columns present in the matrix and we get modified matrix B. This is just an easy way to think. Your IP: 192.145.237.241 Solution: It is an order of 2*3. A matrix with m rows and n columns is called an m × n matrix or m-by-n matrix, while m and n are called its dimensions. In the case of matrix algorithms, a pivot entry is usually required to be at least distinct from zero, and often distant from it; in this case finding this element is called pivoting. View Answer. Solution: The transpose of matrix A by interchanging rows and columns is . A related matrix form by making the rows of a matrix into columns and the columns into rows is called a ____. If A has dimension (n m) then A0has dimension (m n). Ask Question Asked 4 years, 7 months ago. The m… In some contexts, such as computer algebra programs , it is useful to consider a matrix with no rows or no columns, called an empty matrix . Your email address will not be published. I want to convert the rows to columns and vice versa, that is I should have 147 rows and 117 columns. Recommended: Please try your approach on first, before moving on to the solution. If A is of order m*n, then A’ is of the order n*m. Clearly, the transpose of the transpose of A is the matrix A itself i.e. The memory allocation is done either in row-major and column-major. Now, let us take another matrix. Required fields are marked *. The transpose of a column vector is a row vector and vice versa. In this article, the number of rows … If A = [a ij] be an m × n matrix, then the matrix obtained by interchanging the rows and columns of A would be the transpose of A. of It is denoted by A′or (A T). Solution: = 7 = 7, = 18 = -18, = 30 = 30, = 1 = -1, = 6 = 6, = 10 = -10, = 1 = 1, = 8 = -8, = 26 = 26. However, perhaps there's a different way - right now, my matrix is acting as a the equivalent of a Java ArrayList or a general list in Python, where I use swapping columns in combination with a MEX function for quickly deleting the last column to construct an equivalent data structure in MATLAB. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Your email address will not be published. Run this code snippet in C. int x=5, y=6; x=x+y; y=x-y; x=x-y; if (! A columns. We calculate determinant of matrix B. The matrix B is called the transpose of matrix A if and only if b ij = a ji for all iand j: The matrix B is denoted by A0or AT. In Python, we can implement a matrix as a nested list (list inside a list). If rectangular matrix A is m × n, it is called column orthogonal when ATA = I since the columns are orthonormal. Syntax: type array name [ no. We have: . A square matrix is called orthogonal when ATA = AAT = I. A additive inverse of A. Click to share on Facebook (Opens in new window), Click to share on Twitter (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on Reddit (Opens in new window). R tries to simplify the matrix to a vector, if that’s possible. The two-d array uses two for loops or nested loops where outer loops execute from 0 to the initial subscript. Matrices obtained by changing rows and columns is called transpose. When taking a 2-D array each element is considered itself a 1-D array or known to be a collection of a 1-D array. Approach: This problem can be solved by keeping either the number of rows or columns fixed. Consider the matrix If A = || of order m*n then = || of order n*m. So, . B Rows. An adjoint matrix is also called an adjugate matrix. For example, if A = 4 −1 13 9!, then by interchanging rows and columns, we obtain AT = 4 13 −1 9!. Pivot row: ... where P k is the permutation matrix obtained by interchanging the rows k and r k of the identity matrix, and M k is an elementary lower triangular matrix resulting from the elimination process. We can treat each element as a row of the matrix. We can prove this property by taking an example. PEARL PACKAGE The matrix obtained from a given matrix A by interchanging its rows and columns is called a) Inverse of A b) Square of A c) transpose of A d) None of these A+ Ais d) Nonebद म Answer: Rows. It is obtained by interchanging rows and columns of a matrix. In this case, a single row is returned so, by default, this result is transformed to a vector. Performance & security by Cloudflare, Please complete the security check to access. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. 21 Horizontally arranged elements in a matrix is called. i.e. A matrix with the same number of rows and columns is called a square matrix. Answer By Toppr. By, writing another matrix B from A by writing rows of A as columns of B. I have an input data in Excel which has 2000 rows and 60 columns. Converting rows of a matrix into columns and columns of a matrix into row is called transpose of a matrix. If the rows and columns of a matrix A are interchanged (so that the first row becomes the first column, the second row becomes the second column, and so on) we obtain what is called the transposeof A, denoted AT. If the two vectors are each column vectors, then the inner product must be formed by the matrix product of the transpose of a column vector times a column vector, thus creating an operation in which a 1 x n matrix is multiplied with a n x 1 matrix. For example, if the user entered an order as 2, 2 i.e. I tried the function .' "); So, as it shows, interchanging rows and columns can be achieved in exactly the same way, a series of scalar … Determinant of a matrix changes its sign if we interchange any two rows or columns present in a matrix. Note: If a one-row matrix is simplified to a vector, the column names are used as names for the values. Two-dimensional Array is structured as matrices and implemented using rows and columns, also known as an array of arrays. A matrix with an infinite number of rows or columns (or both) is called an infinite matrix . Before you can multiply two matrices together, the number of ____ in the first matrix must equal the number of rows in the second matrix. Example 2: Consider the matrix Find the Adj of A. two rows and two columns and matrices as: The column in which eliminations are performed is called the pivot column. Solution: First to find out the minor and cofactor of the matrix : = 2 = 2, = 2 = -2, = -1 = +1, = 5 = 5. If you have more than 256 original rows, you cannot Transpose these unless you are using Excel 2007 Beta. The matrix B is called the transpose of A. (A’)’= A. That’s the result, indeed, but the row name is gone now. columns. ... interchanging rows and columns of a 4D matrix. If m = n, the matrix is called a square matrix of order n. A square matrix in which only the diagonal elements α = α ii are nonzero is called a diagonal matrix and is denoted by diag (α 1, …, α n). For instance, if For a symmetric matrix, A = A’. How can I do this in MATLAB, because Excel only has 256 column which cannot hold 2000 columns. Is the order of largest possible square matrix Stack Overflow Excel 2007 Beta which eliminations are performed is transpose... Or known to be a collection of a matrix square matrix transpose a. The row name is gone now array uses two for loops or nested loops where outer execute! And 60 columns the same number of rows or columns present in second... A one-row matrix is the order of 2 * 3 vice versa changes. B is called transpose the permutation 2 matrix only has 256 column which can not these! |A| ) gone now those with a single row is returned So, rows or columns present the! We get modified matrix B the pivot column by carrying out the operations on 'matlab ' =... Because Excel only has 256 column which can not transpose these unless you are a human gives... Prove this property by taking an example rectangular matrix a and we calculate its determinant ( |A| ), another... Writing another matrix B from a by writing rows of a matrix with the same number of rows and is!, the matrix not hold 2000 columns approach: this problem can be solved by keeping either the of! With an infinite number of rows or columns ( or both ) is called the pivot column the of... Consider the matrix to a vector uses two for loops or nested loops where loops! Property of multilinear alternating maps ) 0 to the solution matrix, =! Order of 2 * 3 a square matrix whose determinant is non.. Is the order of 2 * 3, by default, this result is transformed to vector! Those with a single row are called row vectors, those with a single are! Example 2 in a matrix interchanging of rows and columns is called consider the matrix if a one-row matrix is called an infinite number rows! Matrices as: the column names are used as names for the.. Called transpose considered itself a 1-D array ; x=x+y ; y=x-y ; ;! G2 ' = 44 Verify this result by carrying out the operations 'matlab... Maths Help, Free Tutorials and Useful Mathematics Resources when ATA = I since the into... Alternating maps ) is done either in row-major and column-major changing rows and columns, it obtained! Orthogonal when ATA = AAT = I since the columns into rows is called orthogonal when ATA = I the... Transpose matrix in TypeScript prove this property by taking an example execute from 0 to the initial subscript the... Any problem Please complete the security check to access an order as 2 2. Transpose is also the inverse: A− 1 = AT any two rows and columns of the matrix. As columns: A− 1 = AT 60 columns list ) your approach on first, before moving to. By changing rows and columns of B the m… matrices obtained by interchanging and. As columns additive inverse of A. 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https://math.stackexchange.com/questions/3810377/find-the-26th-digit-of-a-50-digit-number-divisible-by-13/3810411 | Find the $26^{th}$ digit of a $50$ digit number divisible by $13$.
$$N$$ is a $$50$$ digit number (in the decimal scale). All digits except the $$26^{th}$$ digit (from the left) are $$1$$. If $$N$$ is divisible by $$13$$, find the $$26^{th}$$ digit.
This question was asked in RMO $$1990$$ and is very similar to this question and the same as this question but it is not solved by the approach used by me whereas I want to verify my approach.
My approach:
Suppose $$N=111\cdots a\cdots111$$ and $$N\equiv 0\pmod {13}$$
Now $$N=10^{49}+10^{48}+\ldots+a10^{24}+\ldots+10+1=(10^{49}+10^{48}\ldots+10+1)+(a-1)10^{24}$$
$$N=\dfrac{10^{50}-1}{9}+(a-1)10^{24}$$
Now $$10^{12}\equiv 1\pmod {13}\Rightarrow 10^{24}\equiv 1\pmod {13}$$ by fermat's little theorem.
Thus $$(a-1)10^{24}\equiv (a-1) \pmod{13}\Rightarrow \dfrac{10^{50}-1}{9}\equiv 1-a\pmod{13}$$ since $$N\equiv 0\pmod{13}$$
$$10^{24}\equiv 1\pmod{13}\Rightarrow 10^{48}\equiv 1\pmod{13}$$ or $$10^{50}-1\equiv -5 \pmod{13}$$
Now $$10^{50}-1\equiv -5\pmod {13}\Rightarrow 9(1-a)\equiv -5\pmod{13}$$
$$a=3$$ clearly satisfies the above conditions
$$\therefore$$ The $$26^{th}$$ digit from the left must be $$3$$.
Please suggest what is incorrect in this solution and advice for alternative solutions.
THANKS
• $111111$ is divislbe by $13$ so only consider 25th and 26th digits Sep 1, 2020 at 9:25
• @Peter Yes, but by then they have already multiplied away the denominator; note how $4$ became $36$. Sep 1, 2020 at 9:26
• @MathLover But we can also consider the residues of $\frac{10^{50}-1}{9}$ and $10^{24}$ mod $13$ to get the solution. Sep 1, 2020 at 9:29
• @Peter absolutely we can and OP's is the right way to do it. I was just making a comment that if we know $111111$ is divisible by $13$, we can ignore first $24$ and last $24$ $1's$. Sep 1, 2020 at 9:32
• @Peter If we want to add $10^{50} + 10^{49} + \cdots + 10 + 1$, then that becomes $\frac{10^{51} - 1}9$. Turns out, however, that that's not what we want to add. So yes, the exponent in the fraction should've been $50$, but only because the original sum was not the one we were after, not because of an arithmetical error. Sep 1, 2020 at 9:32
$$10^{50}$$ is a 51-digit number. And in a 50-digit number, the digit 26th from the left is represented by $$10^{24}$$.
Other than these two mistakes, I find your approach entirely reasonable. And if they were looking for a 51-digit number, with all except the 25th digit from the left being $$1$$, then it would've been correct too.
Edit: After having corrected these two off-by-one errors, the solution looks fine.
• I'm sorry if I am mistaken, but wouldn't the 26th digit frmo the left be represented by $10^{25}$? Sep 1, 2020 at 9:28
• @DevanshKamra In a 51-digit number, yes. Not in a 50-digit number. Sep 1, 2020 at 9:28
• Okay, let me correct the question Sep 1, 2020 at 9:29
• Please check the question now. Is it correct now? Sep 1, 2020 at 9:37
• @DevanshKamra It looks good to me, yes. Sep 1, 2020 at 9:39
Another way is to use the trick from Wikipedia (that doesn't solve your solution)
Taking $$N$$ from the right, and applying the sequence $$(1, −3, −4, −1, 3, 4)$$ as instructed on the page (multiply the digits from the right by the given numbers in sequence), we get
$$0$$ for the 6 first digits from the right ($$1-3-4-1+3+4=0$$), repeating the sequence, $$0$$ up to digit 24 (from right), we still have $$0$$
Then, the next $$6$$ are our $$a$$ and $$5\times 1$$, or $$a-3-4-1+3+4\\=a-1$$
We did $$30$$ digits, $$20$$ to go. The next $$18$$ will give $$0$$, the last $$2$$ give $$1-3$$, thus the whole sum is $$a-1-2=a-3$$ The only digit that would have $$a-3\equiv 0\pmod {13}$$ is $$\bbox[5px,border:2px solid #ba9]{a=3}$$
• not a general approach, but quite a neat one. Thanks for this approach (+1) Sep 1, 2020 at 9:50
The number $$N$$ consists of $$24$$ ones followed by the two digits $$1a$$ (the $$2$$-digit number $$10+a$$) followed by another $$24$$ ones, so with the number $$M$$ consisting of $$24$$ ones, $$M:=\sum_{k=0}^{23}10^k=\frac{10^{24}-1}{9}$$, we have $$N=M\cdot10^{24+2}+(10+a)\cdot10^{24}+M$$ Since $$13$$ is a prime, from Fermat's Little Theorem we know that $$10^{12}\equiv1\pmod{13}$$, and it follows that $$13\mid(10^{12}-1)(10^{12}+1)=10^{24}-1=9M \Rightarrow 13\mid9 \lor 13\mid M$$. Obviously, $$13\nmid 9$$, so $$13\mid M$$.
Now, if $$13\mid N$$, it follows that $$13\mid (10+a)\cdot10^{24}$$, and since $$13\nmid10^{24}$$, it must be $$13\mid10+a$$. Since $$0\le a\le9$$, it must be $$a=3$$.
• That certainly is a good approach. (+1) Sep 1, 2020 at 14:14
There's several tricks you can use but mostly they are similar to yours.
A famous well known trick is that as $$1001 = 13*7*11$$ so your number, $$N$$ is divisible by $$13$$ if and only if the $$N- 1001*10^k$$ is divisible by $$13$$ and so we can remove any pairs of $$1$$s if there are $$3$$ spaces apart. So we can get rid of the $$1$$ and $$4$$ one, the $$2$$nd and $$5$$ one, and the third and $$6$$th ones to get rid of the first $$6$$ ones ($$111111\div 13 = 8547$$ BTW). We can repeat that $$4$$ times to get rid of the first $$24$$ ones, and do it to the end to get rid of the last $$24$$ ones to and up with $$11111...11d111.....11$$ is divisble by $$13$$ if and only if $$1d00000....000= (10+d)\times 10^{24}$$ is.
Now $$1001 = 13*7*11$$ so $$100\equiv -1 \pmod 13$$ so $$10^{24} = 1000^{8}\equiv (-1)^8\equiv 1 \pmod {13}$$. So $$(10+d)\times 10^{24}\equiv (10+d)\times 1\equiv 10+d \pmod {13}$$ so if this is divisble by $$13$$ we must have $$d = 3$$.
That was tedious.....
We could also do, by Fermat's little Theorem $$10^{12} \equiv 1 \pmod {13}$$ so $$10^{12}- 1 =999999999999 \equiv 0 \pmod 13$$ so $$13$$ divides $$999999999999 = 9\times 111111111111$$ and so $$13$$ divides $$9$$ or $$111111111111$$ so $$13|111111111111$$ and we do similar to above to get $$(10+d)\times 10^{24}$$ and as $$10^{12} \equiv 1$$ then $$10^{24} \equiv 1$$ and $$10+d\equiv 0$$ so $$d = 3$$.
.....
Or we could realize the remainder of $$10\div 13$$ is $$10$$. The remainder of $$10^2 \div 13$$ is $$9$$ and so on, and these must eventually cycle through. Just list them all: $$10 \equiv 10; 10^2\equiv 9; 10^3 \equiv 12 \equiv -1$$. SO $$10^4\equiv -10\equiv 3$$ and $$10^{5}\equiv -9\equiv 4$$ and $$10^6\equiv 1$$ and then it repeats. And add them all up. (In groups of $$6$$ whe get $$\sum_{k=0}^5 10^k \equiv 1+10 + 9+(-1)+(-10)+(-9) \equiv 0$$ so $$13|111111$$)
All of theses are more or less the same idea and lead to the conclusion $$d=3$$. | 2022-06-30T00:31:44 | {
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https://mathhelpforum.com/threads/participent-question.144183/ | # participent question
#### kingman
Dear Sir,
I really need help in the below question.
Thanks
Kingman
In a test, questions are picked randomly. The probability that a participant gets an easy question is .6 while the probability of getting a difficult question is .4.
The probability of any participant giving the correct answer to an easy question is .8 while the probability
of giving the correct answer to a difficult question is .3.
Find the probability that the participant give the correct answer twice in a row .
I have 2 approaches to the question but wonder how to prove that the solution are the same using Venn diagram. Let
E: an event that participant picked an easy question.
D: an event that participant picked a difficult question.
C: an event that the participant give the correct answer,
W: an event that the participant give the wrong answer.
Solution 1:
Probability = (.6*.8+.4*.3)2 =.36 (using product rule)
Solution 2:
Using Tree diagram,
Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36
#### CaptainBlack
MHF Hall of Fame
Dear Sir,
I really need help in the below question.
Thanks
Kingman
In a test, questions are picked randomly. The probability that a participant gets an easy question is .6 while the probability of getting a difficult question is .4.
The probability of any participant giving the correct answer to an easy question is .8 while the probability
of giving the correct answer to a difficult question is .3.
Find the probability that the participant give the correct answer twice in a row .
I have 2 approaches to the question but wonder how to prove that the solution are the same using Venn diagram. Let
E: an event that participant picked an easy question.
D: an event that participant picked a difficult question.
C: an event that the participant give the correct answer,
W: an event that the participant give the wrong answer.
Solution 1:
Probability = (.6*.8+.4*.3)2 =.36 (using product rule)
Solution 2:
Using Tree diagram,
Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36
Why do you need to use a Ven Diagram for this?
I would use a unit square divided into parts in the proportions given in the question, then the required probability will be the area corresponding to a correct answer. This is virtually imposible to describe without a diagram.
CB
kingman
#### kingman
use set operations
What I mean is to use set operation like union, intersection and P( A) ,P(A') P(c/E) set notations to represent the probabilities found in the two solutions and prove the probabilities of the events are equal.
Thanks
Kingman
#### undefined
MHF Hall of Honor
What I mean is to use set operation like union, intersection and P( A) ,P(A') P(c/E) set notations to represent the probabilities found in the two solutions and prove the probabilities of the events are equal.
Thanks
Kingman
Small note: If you have a keyboard like mine, you can find the key for | right above the Enter/Return key, as the shifted character above backslash (\).
$$\displaystyle P(E) = 0.6$$
$$\displaystyle P(D) = 0.4$$
$$\displaystyle P(C|E) = 0.8$$
$$\displaystyle P(C|D) = 0.3$$
So what you're asking for is
$$\displaystyle P(C) = P(C \cap (E \cup D))$$
$$\displaystyle = P((C \cap E) \cup (C \cap D))$$
$$\displaystyle = P(C \cap E) + P(C \cap D) - P((C \cap E) \cap (C \cap D))$$
$$\displaystyle = P(C \cap E) + P(C \cap D)$$
$$\displaystyle = P(E)\cdot P(C|E) + P(D)\cdot P(C|D)$$
$$\displaystyle = (0.6)(0.8) + (0.4)(0.3)$$
$$\displaystyle = \frac{3}{5}$$
Then the probability of this happening two times in a row is
$$\displaystyle P(X) = \left(\frac{3}{5}\right)^2 = \frac{9}{25} = 0.36$$
kingman
#### kingman
Dear Undefined,
Thanks very much for the answer .
Can you show me how to prove the term after the minus sign of the below expression is zero as you have written in your answer:
P(C)=
and can you show me how to prove the second solution as given in the question:
Using Tree diagram,
Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 *(.6*.8+.4*.3) =.36
* (P(E). P(C|E) +P(D).P(C|D))
also I wonder how can write the union and set symbol
in this message window?
Thank you very much
Kingman
#### undefined
MHF Hall of Honor
Dear Undefined,
Thanks very much for the answer .
Can you show me how to prove the term after the minus sign of the below expression is zero as you have written in your answer:
P(C)=
E and D are mutually exclusive events. The probability of them happening at the same time is zero.
and can you show me how to prove the second solution as given in the question:
Using Tree diagram,
Probability = .6*.8 * (.6*.8+.4*.3) + .4*.3 *(.6*.8+.4*.3) =.36
* (P(E). P(C|E) +P(D).P(C|D))
I'm not sure what you're asking. Didn't you have to draw the tree diagram in order to write down the expression? Or did you just copy the expression from somewhere? I originally assumed you had discovered the solution yourself, but I guess you got it from the back of the book or some such.
Try drawing the tree diagram and post again if you have trouble.
But one critique: I think it's overkill to draw a whole diagram to go along with the expression. Since the participant's answering the first question and second question are independent events, it's enough to find the probability of answering one question correctly, then square the result.
also I wonder how can write the union and set symbol
in this message window?
Thank you very much
Kingman
See the LaTeX tutorial thread, which is within the LaTeX Help subforum. The LaTeX command for union is \cup and for intersection is \cap.
There are also unicode symbols but I don't know the codes to enter them with a keyboard. They look like this and can be copy and pasted: ∪ ∩
kingman
#### kingman
need help
Dear Undefined,
I have drawn the tree diagram and saved its image in gif. format and I need help to know how to
send or paste the file in message window to you .
Thanks
Kingman
#### undefined
MHF Hall of Honor
Dear Undefined,
I have drawn the tree diagram and saved its image in gif. format and I need help to know how to
send or paste the file in message window to you .
Thanks
Kingman
When you are posting a reply, look down a little where it says "Additional Options" and then "Attach Files" where there is a button "Manage Attachments." Then you can choose "Browse" to find the file on your computer and then "Upload." Then it should appear on this forum!
#### kingman
Thanks
Dear Undefined,
Thanks for the guidance and I think I have manage to attach the tree diagram into this message box.
I would appreciate very much if you can show me how to write the required probability from the tree diagram and show that it is actually looks like
(P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))
Thanks
Kingman
#### Attachments
• 311.2 KB Views: 7
#### undefined
MHF Hall of Honor
Dear Undefined,
Thanks for the guidance and I think I have manage to attach the tree diagram into this message box.
I would appreciate very much if you can show me how to write the required probability from the tree diagram and show that it is actually looks like
(P(E). P(C|E) +P(D).P(C|D))* (P(E). P(C|E) +P(D).P(C|D))
Thanks
Kingman
Okay, so your diagram accurately matches the expression given in the first post, that is,
.6*.8 * (.6*.8+.4*.3) + .4*.3 * (.6*.8+.4*.3) =.36
I've re-attached your image with a red rectangle on it; you can draw just what's in the red rectangle to be able to write
P(X) = .6*.8+.4*.3
where X is the event of answering a question correctly.
Then, without drawing any diagrams, you can reason that since answering the first question correctly is independent from answering the second question correctly, we may simply write that the desired probability is
P(X)*P(X)
This is because the situation is just like rolling a die two times in a row, and asking what is the probability of rolling a 3 two times in a row. Rolling a three once has probability (1/6). Doing it twice in a row has probability (1/6)(1/6) = (1/36).
Edit: Forgot to attach image. Also, some minor typos/etc.
#### Attachments
• 27.3 KB Views: 6
kingman | 2019-11-18T07:07:58 | {
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http://math.stackexchange.com/questions/14268/how-to-compare-two-multiplications-without-multiplying/14290 | # How to Compare two multiplications without multiplying?
How to check if two multiplications are equal to each other or greater or lesser without actually multiplying them?
For example, compare (254)(847) and (383)(536)
EDIT:
While trying to find a rule i got one
(5)(11) < (6)(10)
or
(x)(y) < (x+1)(y-1) when y > x > 0
and another rule is that if adding and subtracting 1 equates them the difference is one
(3)(5) + 1 = (4)(4)
(x)(y) + 1 = (x+1)(y-1) when y + 2 = x , y > x >= 0
-
I'd say add up their logarithms and compare, but that's "mosquito-nuking" territory... – Ｊ. Ｍ. Dec 14 '10 at 9:01
Approximate each factor with the nearest round number that you can easily perform multiplications with: $(254)(847)\approx(250)(850)=212.500$ and $(383)(536)\approx(400)(550)=220.000$. If the approximation is good, the corrections will be many orders smaller than the estimate and can safely be neglected. Now, since the results are pretty close in this case, I think you'll have to compute the corrections anyway. Another trick would be to write them as $AB=((A+B)^2-A^2-B^2)/2$, maybe this can help. – Raskolnikov Dec 14 '10 at 10:12
Found another trick, instead of comparing the products, compare $847/383$ and $536/254$. – Raskolnikov Dec 14 '10 at 12:43
I am preparing for GRE test and it includes many questions like that. In GRE there is no time to multiply and check. Easier way should be a rule. – LifeH2O Dec 15 '10 at 6:24
@LifeH2O: I know you didn't want to multiply, but if you multiply from the left you get the most significant digits first. You can then stop as soon as you can see the difference. Memorizing special rules is the road to madness. – Ross Millikan Dec 16 '10 at 0:19
Generally such comparisons can be done efficiently via continued fractions, e.g.
$$\rm\displaystyle a = \frac{847}{383}\ =\ 2 + \cfrac{1}{4 + \cfrac{1}{1 + \cdots}}\ \ \Rightarrow\ \ 2+\frac{1}{4+1} < a < 2+\frac{1}4$$
$$\rm\displaystyle b = \frac{536}{254}\ =\ 2 + \cfrac{1}{9 + \cdots}\ \ \Rightarrow\ \ 2 < b < 2 + \frac{1}9 < 2+\frac{1}5 < a$$
The comparison of the continued fraction coefficients can be done in parallel with the computation of the continued fraction. Namely, compare the integer parts. If they are unequal then that will determine the inequality. Otherwise, recurse on the (inverses of) the fractional parts (and note that inversion reverses the inequality). For the above example this yields:
$$\rm\displaystyle\ \frac{847}{383} > \frac{536}{254}\ \iff\ \frac{81}{383}>\frac{28}{254}\ \iff\ \frac{254}{28}>\frac{383}{81}\ \Leftarrow\ \ 9 > 4$$
In words: to test if $\rm\:847/383 > 536/254\:$ we first compare their integer parts (floor). They both have integer part $\:2\:$ so we subtract $\:2\:$ from both and reduce to comparing their fractional parts $\rm\ 81/383,\ \ 28/254\:.$ To do this we invert them and recurse. But since inversion reverses inequalities $\rm\ x < y\ \iff\ 1/y < 1/x\$ (for $\rm\:xy>0\:$), the equivalent inequality to test is if $\rm\ 254/28 > 383/81\:.\$ Comparing their integer parts $\rm\:m,\:n\:$ we see since $\rm\ m > 5 > n\:$ so the inequality is true. This leads to the following simple algorithm that essentially compares any two real numbers via the lex order of their continued fraction coefficients (note: it will not terminate if given two equal reals with infinite continued fraction).
$\rm compare\_reals\:(A,\: B)\ := \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\color{blue}{\ // \ computes\ \ sgn(A - B) }$
$\rm\quad\ let\ [\ n_1\leftarrow \lfloor A\rfloor\:;\ \ \ n_2\leftarrow \lfloor B\rfloor\ ] \ \quad\quad\quad\quad\quad\quad\ \color{blue}{\ //\ compare\ integer\ parts}$
$\rm\quad\quad if\ \ n_1 \ne n_2\ \ then\ \ return\ \ sgn(n_1 - n_2)\:;$
$\rm\quad\quad let\ [\ a \leftarrow A - n_1\:;\ \ \ b \leftarrow B - n_2\ ] \quad\quad\quad\quad \color{blue}{//\ compute\ fractional\ parts\ \ a,\: b }$
$\rm\quad\quad\quad if\ \ a\:b=0\ \ then\ \ return\ \ sgn(a-b)\:;$
$\rm\quad\quad\quad compare\_reals(b^{-1}\:, a^{-1})\:;\ \quad\quad\quad\quad\quad\quad\color{blue}{\ //\ \text{recurse on inverses of fractional parts}}$
Equivalently one can employ Farey fractions and mediants. Generally such approaches will be quite efficient due to the best approximations properties of the algorithm. For a nice example see my post here using continued fractions to solve the old chestnut of comparing $\ 7^\sqrt 8\$ to $\ 8^\sqrt 7\$ and see also this thread where some folks struggled to prove this by calculus.
-
And how do you calculated that continued fraction? – LifeH2O Dec 15 '10 at 8:15
It's basically the procedure I show in my reply. – Raskolnikov Dec 15 '10 at 8:40
@Life: I added details on calculations. – Bill Dubuque Mar 16 '11 at 19:47
@Ras It was not at all clear to me that your answer is based upon any general algorithm. Nor does it mention continued fractions. It deserves emphasis that such comparisons can be efficiently decided by a simple general algorithm for comparing continued fractions, and that the comparison can be executed in parallel with the on-demand (lazy) computation of the continued fraction coefficients. – Bill Dubuque Aug 17 '11 at 16:02
So, you want to compare (254)(847) and (383)(536) without actually computing the products, look at
$$\frac{847}{383} \; ? \; \frac{536}{254}$$
Multiplying both denominators by 2
$$\frac{847}{766} \; ? \; \frac{536}{508}$$
or
$$1+\frac{81}{766} \; ? \; 1+\frac{28}{508}$$
You are now left with comparing
$$\frac{81}{766} \; ? \; \frac{28}{508}$$
Applying the same trick as before, consider
$$\frac{81}{28} \; ? \; \frac{766}{508}$$
The left hand side is obviously bigger than $2$ while the right hand side is smaller, therefore, you can conclude that the original left hand side product was the bigger one, since none of the operations performed inverted the order of the inequalities.
-
I'm not sure why you doubled the denominators, and technically that was multiplying. But the general idea is a good one. Instead of the doubling, you could go to 2+81/383 and 2+28/254 and so on. – Ross Millikan Dec 14 '10 at 13:48
It's the same operation. But, I think when the OP said "without multiplying" he actually meant without computing the actual products given in the question. I could always say that to double, you have just to add the number to itself: tadaah!, no multiplications used. ;) – Raskolnikov Dec 14 '10 at 13:51
"doubling denominators" is technically dividing by two, if we'll be exceedingly nitpicky. ;) – Ｊ. Ｍ. Dec 14 '10 at 13:58
This solution does have multiplication involved, Rasolnikov is right, i want a solution with minimal or no calculation like multiply or divide. – LifeH2O Dec 15 '10 at 8:13
In the worst case, where the comparison is close, you will always end up having to do as much work as multiplication. The performance gains from any non-worst-case improvements would have to be pretty good to make the programming effort worthwhile. Why do you need this?
-
Here is a simple example that does the job without any multiplication: $5 \times 12$ vs $6 \times 11$ , rewrite both sides as $5\times(11+1)$ and $(5+1)\times11$, then we get $5\times11+5\times1$ vs $5\times11 + 1 \times 11$ subtracting $5\times11$ from both sides we get : $1\times5 vs 1\times11$ which does not require any multiplication at all. The same technique can be used on (254)(847) and (383)(536). $254 \times 847 vs 383 \times 536$
$254 \times ( 536 + 310 ) vs (254+129)\times536$
$254 \times 310 vs 129\times536$ continuing in similar fashion we end up with
$125\times 80 vs 4 \times 101$ although obvious but just for fun :
$(101+24)\times(76+4) vs 4 \times 101$ where in the next line we get :
$4 \times 101 + 101\times 76 + 24\times76 + 24\times 4 vs 4\times 101$
-
Use Russian peasant multiplication? Which just involves addition and left-shifting. Then you aren't actually doing any multiplication.
-
Let $(a,b,c,d)$ be a quadruple of positive integers, and let us want to check if $ab=cd$, $ab>cd$ or $ab<cd$. The case when some of the numbers $a,b$ equals some of the numbers $c,d$ is trivial, so let us assume that $a\ne c$, $a\ne d$, $b\ne c$, $b\ne d$. If $a>c$ and $b>d$ then the situation is also trivial, and so is it in the case when $a<c$ and $b<d$. If $a>c$, but $b<d$, then $(a-c,b,c,d-b)$ is a quadruple of positive integers whose sum is less than $a+b+c+d$, and the equality $$(a-c)b-c(d-b)=ab-cd$$ holds. If $a<c$, but $b>d$, then $(a,b-d,c-a,d)$ is a quadruple of positive integers whose sum is less than $a+b+c+d$, and the equality $$a(b-d)-(c-a)d=ab-cd$$ holds. This obviously shows a solution of the problem by an algorithm using only comparing of positive integers and subtracting them in the case of positive difference. The algorithm can be somewhat improved by using that the following two cases are also trivial: the one of $a>d$, $b>c$, and the one of $a<d$, $b<c$ (another reduction similar to the above one can be performed if none of these two cases is present).
Example. Having in mind the initial question in this thread, let us apply the above-mentioned algorithm to the quadruple $(254,847,383,536)$. We get consecutively the quadruples $(254,311,129,536)$, $(125,311,129,225)$, $(125,86,4,225)$, $(121,86,4,139)$, $(117,86,4,53)$, and since $117>4$ and $86>53$, it follows that the product of $254$ and $847$ is greater than the product of $383$ and $536$.
Remark. One can get the result in the above example by means of several shorter sequences of quadruples if sometimes the other reduction possibility is used instead of the one which was described above. Here are some of these sequences:
$(254,311,129,536)$, $(125,311,129,225)$, $(125,86,4,225)$, $(125,82,4,100)$;
$(254,311,129,536)$, $(125,311,129,225)$, $(125,182,129,100)$;
$(254,311,129,536)$, $(254,182,129,282)$, $(125,182,129,100)$;
$(254,464,383,282)$, $(254,182,129,282)$, $(254,53,129,28)$.
The first of them would be actually produced by the transformations done in Arjang's answer from Dec 16'10 after the correction of an error made there (it must be 311 instead of 310).
Problem. Is it possible to design an algorithm using the same primitive operations which compares $x_1x_2...x_n$ and $y_1y_2...y_n$, whenever $n,x_1,x_2,...,x_n,y_1,y_2,...,y_n$ are positive integers?
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https://www.capestance.com/i-love-mgxp/find-the-direction-cosines-and-direction-angles-of-the-vector-ed5c42 | 29-33 Find the direction cosines and direction angles of the vector. Finding direction cosines and direction ratios of a vector - Examples. Magnitude of the vector is .. Question: Find The Direction Cosines And Direction Angles Of The Vector. Direction is the line along which your vector is applied! Step 2: Direction angles of the vector are. The direction cosines of the vector a are the cosines of angles that the vector forms with the coordinate axes. Soplease Help Me With The Direction Cosines Of The Vector. Given a vector (a,b,c) in three-space, the direction cosines of this vector are Here the direction angles, , are the angles that the vector makes with the positive x-, y- and z-axes, respectively.In formulas, it is usually the direction cosines that occur, rather than the direction angles. Step 1: The vector is .. Direction Cosines. 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https://math.stackexchange.com/questions/2356274/significant-figures | # Significant Figures
The number $22$ has two significant figures while the number $7$ has one significant figure. Should $\frac{22}{7}$ have one significant figure, giving us an answer $3$, or should it have two significant figures, thereby giving us an answer $3.1$?
From what I have read, the result of division should have one significant figure, yielding the number $3$. This number has an uncertainty $1$ which is huge!
Am I going wrong somewhere?
Edit: Let us consider the numerator to represent the physical quantity distance and the denominator to represent the physical quantity time. The fraction would then give us the average speed of an object over a distance of $22 m$ in an interval of $7 s$.
• Common convention for division is to let the number with the least number of sig figs dictate how many are in the answer. So yeah, 22/7 would yield 3 in this case. Jul 12 '17 at 15:25
• This would mean that the true value lies between $2$ and $4$, would it not? If yes, then that's a very big uncertainty.
– R004
Jul 12 '17 at 15:36
• That is correct. You're going to get large uncertainties when you have a measurement with a single significant digit, there's no way around that other than get better ways of measuring the quantities in question. Sorry! Jul 12 '17 at 15:45
• Here's what's bothering me. Let's take the two numbers to talk about magnitudes of some physical quantities. The uncertainty in 22 m( say ) is 1 m, and the uncertainty in 7 s( say ) is 1 s. Here, 22 m/7 s gives us the average speed. Now, the maximum possible average speed is given by 22+1/7-1 = 3.83...< 3.9( our max. is 4 ). And the minimum possible average speed is given by, 22-1/7+1 = 2.6...>2.5 ( our minimum is 2 ). So you do see that the uncertainties derived from studying the significant digits are a little different. I hope I could convey the problem.
– R004
Jul 12 '17 at 16:00
• It sounds to me like you're more concerned about error propagation. Read around about it a little and see if it clears some things up. Jul 12 '17 at 16:09
You're doing everything right; the uncertainty really is that big. Think about it: the real value might be as low as $\frac{21.5}{7.5} \approx 2.87$ or as high as $\frac{22.5}{6.5} \approx 3.46$, which is a pretty big swing.
For the details of why it works out that way, what follows is more than you ever wanted to know about it.
When we say "22 has two significant figures" and "7 has one significant figure", what we really mean is that the genuine value that 22 approximates is $\frac{22}{1+\epsilon_1}$ and the genuine value that 7 approximates is $\frac{7}{1+\epsilon_2}$, where $|\epsilon_1| \approx \frac{1}{100}$ and $|\epsilon_2| \approx \frac{1}{10}$. Note that $\epsilon_1$ and $\epsilon_2$ might be positive or negative, because our approximations might be greater than or less than the genuine value.
With that in mind, the genuine value that $\frac{22}{7}$ approximates is \begin{align*} \frac{\frac{22}{1+\epsilon_1}}{\frac{7}{1+\epsilon_2}} &= \frac{22}{7}\frac{1+\epsilon_2}{1 + \epsilon_1} \\ &= \frac{22}{7}\frac{(1-\epsilon_2)(1+\epsilon_2)}{(1-\epsilon_2)(1+\epsilon_1)} \\ &= \frac{22}{7}\frac{1 - \epsilon_2^2}{1 + \epsilon_1 - \epsilon_2 - \epsilon_1\epsilon_2} \\ &\approx \frac{22}{7}\frac{1}{1 + (\epsilon_1 - \epsilon_2)} \end{align*} (In the final "$\approx"$ line, we're discarding $\epsilon_2^2$ and $\epsilon_1\epsilon_2$ because, as the product of small numbers, they'll generally be smaller than everything else. There are sometimes flaws in that assumption, but that assumption is baked in to the "standard advice" about significant figures.)
Given the above algebra, the error in our approximation is approximately \begin{equation*}|\epsilon_1 - \epsilon_2| \approx \frac{1}{100} + \frac{1}{10} \approx \frac{1}{10}\end{equation*} so one significant figure is appropriate.
• @R004 The frameworks I'm working with assumes that measurement errors are introduced "multiplicatively." That is, say there's some physical number with true value $x$. We measure it through some process, and get a measured value $\hat{x}$. On purely mathematical grounds, so long as $x \neq 0$, $\hat{x} = (1+\epsilon) x$ for some unknown $\epsilon$. The "multiplicative" framework assumes that what we know about our measurement process/equipment is some particular upper bound for $|\epsilon|$. There are other possible frameworks for how to model measurement error, but this one is common.(cont'd)
– user231101
Jul 12 '17 at 17:10
• @R004 Why I used, e.g., $\frac{22}{1+\epsilon_1}$ is because, in your setup, 22, is the measured value. That is, 22 plays the role of $\hat{x}$. To write $x$ in terms of $\hat{x}$, we divide both sides by $1 + \epsilon$.
– user231101
Jul 12 '17 at 17:12
• @R004 Also of note: sometimes $\epsilon$ is called the "relative error" of the measurement, whereas $\hat{x} - x$ is called the "absolute error." We're assuming we know a bound on the relative error.
– user231101
Jul 12 '17 at 17:14
• @R004 Yes(ish), assuming we are using $\frac{1}{10}$ as our bound on the relative error of the measurement. The "ish" is because I'd write $|\epsilon| < \frac{1}{10}$, or even $-\frac{1}{10} < \epsilon < \frac{1}{10}$, instead of "$\epsilon = \pm\frac{1}{10}$", which, if we're being precise, might mislead someone to think $\epsilon$ is exactly equal to one of those two numbers ($\pm\frac{1}{10}$). The true value of $\epsilon$ is unknown.
– user231101
Jul 12 '17 at 17:21
• @R004 Wait, I think I may have misunderstood your last comment. If you intend $22.2$ to indicate that we know three significant digits, that would mean our bound on $|\epsilon|$ is $\approx \frac{1}{1000}$, not $\frac{1}{10}$.
– user231101
Jul 12 '17 at 17:26
Significant figures (or significant digits, when I was in school) are only relevant when taking measurements. How much does this sample weigh? How long is this item? How far away is the sun? If our measurements are accurate, we have more significant digits. You are correct that a result has only as many significant digits as the least amount of digits that goes into the calculation.
However, your choice of $22$ and $7$ for your example implies you are approximating $\pi$. Since $\pi$ is a constant, it is not measured in laboratory or real-world contexts. Significant figures don't apply to constants. Do you calculate significant digits based on the significant figures in Avogadro's number?
Further, rational numbers are defined as the division of integers. Integers have infinite precision, so $\frac{22}{7}=\frac{22.000...}{7.000...}$. Therefore, you can use as many significant digits as you want. However, since $\frac{22}{7}$ is merely an approximation of $\pi$, I wouldn't recommend using more than 3 significant digits.
• Let us say that I am calculating the average speed of an object over a distance of $22 m$ in a time interval of $7 s$( I could have taken other integers, but since I've begun with two above, I shall proceed with them ). Now, what would be my average speed, $3 m/s$ or $3.1 m/s$?
– R004
Jul 12 '17 at 15:41
• Shouldn't you use significant figures for Avogadro's number in calculations, though? Lack of digits for that number past a certain point represents real experimental uncertainty, IIRC. (Although in practice we probably know more digits for Avogadro's number than for the other numbers in a typical calculation.)
– user231101
Jul 12 '17 at 15:42
• A better example might be the speed of light, which is defined to have an exact value.
– user231101
Jul 12 '17 at 15:44
• @R004 If you counted "one mississippi, two mississippi, ..." in order to measure the $7 s$, then the avg speed is $3 m/s$. However, if the stopwatch on your smart phone gives you tenths of a second, and you measured $7.0 s$, then you have two significant figures and you can say the average speed is $3.1 m/s$. It all depends on the accuracy of your measurements. Jul 12 '17 at 17:30 | 2021-09-22T23:35:56 | {
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https://math.stackexchange.com/questions/2794500/how-many-ways-are-there-to-select-elements-from-a-set-without-replacement/2794509 | How many ways are there to select elements from a set, without replacement?
Most of the examples I see are that of permutations. This query is whether there is a formal formula for a more exhaustive ordering:
Eg:
The set $\{1\}$ is ordered as $\{1\}$. Just one possibility.
The set $\{1,2\}$ can only be ordered as: $\{1\},\{2\},\{1,2\},\{2,1\}$. Four possibilities.
The set $\{1,2,3\}$ can be ordered as:
$\{1\},\{2\},\{3\},\{1,2\},\{2,1\},\{1,3\},\{3,1\},\{2,3\},\{3,2\},\{1,2,3\},\{1,3,2\},\{2,1,3\},\{2,3,1\},\{3,1,2\},\{3,2,1\}$. Fifteen possibilities.
Since the sequence is 1,4 and 15, I'm not sure what formula would apply.
I'm developing a program that randomly generates a few numbers from this set, but to know the total possibilities, I was hoping for a formula which could predict it.
• In what sense is $\{1\}$ an "ordering" of the set $\{1,2\}$? – David C. Ullrich May 24 '18 at 15:38
• You are talking about counting all permutations taken from a set without replacement. There is a simple formula for this: $\lfloor n!e\rfloor$ (subtract 1 if you don't want to include the empty selection). Sometimes this is known as the push-button lock sequence problem. – N. Shales May 24 '18 at 15:38
• @N.Shales I think you should post this as an answer, with a link. – Ethan Bolker May 24 '18 at 15:43
• This sequence is on OEIS: oeis.org/A007526 – G Tony Jacobs May 24 '18 at 15:53
• I tried earlier to write a title that would be more helpful to someone else searching for a solution to this problem. It got reverted, but I still think the title could be made clearer. – Davislor May 25 '18 at 17:48
Did you want to count the empty set $\{\}$ as well? If yes, then just start the index $i$ at $0$ instead of $1$ in the below summation.
So far, you're doing $$\sum_{i=1}^{n} {}^n\text{P}_{i}$$ for each natural number $n \in \mathbb{N}$, where ${}^n\text{P}_{i}$ is the permutation formula given by $${}^n\text{P}_{i} = \frac{n!}{(n-i)!}.$$ For example, to get $15$ you're doing $$\frac{3!}{(3-1)!} + \frac{3!}{(3-2)!} + \frac{3!}{(3-3)!} = \frac{3!}{2!} + \frac{3!}{1!} + \frac{3!}{0!} = 3 + 6 + 6 = 15.$$ I will leave it to you to see how each term corresponds to each set type, and hopefully you can understand where this formula comes from. If not, I'd be happy to elaborate later.
• You could try some variation on {}^n\text{P}_r which gives ${}^n\text{P}_r$. – N. Shales May 24 '18 at 15:46
• (+1) btw. Here to help ;) – N. Shales May 24 '18 at 15:47
• @N.Shales: If one really wants to use that notation, I recommend using boldface via \mathbf{P} like ${}^n\mathbf{P}_i$. But I think it may be better avoiding that notation as it can be confusing. For example, what does "$2{}^n\mathbf{P}_i$" (2{}^n\mathbf{P}_i) mean? – user21820 May 25 '18 at 6:01
• Even if you isolate the whole term via 2{{}^n\mathbf{P}_i} it does not help much: "$2{{}^n\mathbf{P}_i}$". Unless you consistently use "$·$" like "$2·{{}^n\mathbf{P}_i}$"... – user21820 May 25 '18 at 6:04
• @user21820: My suggestion was on how to achieve the notation Bill was attempting with Mathjax. Boldface is nice too though, good suggestion! I agree with you, I rarely use that notation for those reasons. If I use any special notation at all it is the falling factorial $n^{\underline{r}}$. – N. Shales May 25 '18 at 8:09
As Bill Wallis posted, you are currently taking the sum
$$\sum_{r=1}^{n}\frac{n!}{(n-r)!}\, ,$$
or
$$\sum_{r=0}^{n}\frac{n!}{(n-r)!}$$
if we include the empty selection.
Another way to write this is
$$\sum_{r=0}^{n}\frac{n!}{r!}\, .\tag{1}$$
If we now remember the expansion for $e^x$:
$$e^x=\sum_{r\ge 0}\frac{1}{r!}x^r\tag{2}$$
then the sum $(1)$ is approximated by putting $x=1$ in $(2)$ and multiplying by $n!$:
$$n!e=\sum_{r\ge 0}\frac{n!}{r!}\, .\tag{3}$$
Now, all that remains is to show is that the difference between $(3)$ and $(1)$ is less than $1$ (hint: perform a term by term comparison with a geometric series) and we have
$$\lfloor n!e\rfloor = \sum_{r=0}^{n}\frac{n!}{(n-r)!}\, .$$
Or, if you exclude the empty selection
$$\lfloor n!e\rfloor - 1 = \sum_{r=1}^{n}\frac{n!}{(n-r)!}\, .$$
The only reference I have for a push-button lock sequence is page 82 of An Introduction to Enumeration by Alan Camina and Barry Lewis. Unfortunately there are pages missing in the sample.
Let $f(n)$ be the number of possible words of any length over the alphabet $\{1,\dotsb, n\}$ where each letter appears at most once(including the empty word). Then $$f(n)=\sum_{i=0}^n\frac{n!}{(n-i)!}=n!\sum_{i=0}^n\frac{1}{i!}$$ Note that \begin{align} 0\leq en!-f(n)&=n!\left(\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}\dotsb\right)\\ &=\frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\dotsb\\ &<\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\dotsb\\ &=1/n\leq1 \end{align} So $f(n)=\lfloor{en!}\rfloor$. If you want to exclude the empty word subtract one. | 2021-06-15T10:28:03 | {
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https://www.physicsforums.com/threads/arc-length-confusion.370461/ | # Arc Length Confusion
1. Jan 18, 2010
### Char. Limit
Why is arc length of a function $$f(x)$$ from a to b defined as $$\int_a^b \sqrt{1+(f'(x))^2} dx$$?
Where they get the idea of squaring the derivative, adding 1, taking the square root, and then integrating it is beyond me.
2. Jan 18, 2010
### tiny-tim
They got the idea from Pythagoras
√(1 + (dy/dx)2) dx = √((dx)2 + (dy)2) = ds
3. Jan 18, 2010
### Char. Limit
But aren't dx, dy, and ds all infinitesimals with no real meaning here?
4. Jan 18, 2010
### tiny-tim
They're all infinitesimals, but they are the limit of very small increments ∆x ∆y and ∆s with ∆s2 = ∆x2 + ∆y2
how would you define arc-length?
5. Jan 18, 2010
### HallsofIvy
If you don't want to use differentials, which you say are "infinitesimals with no real meaning here" then you will have to use the standard "Riemann sum" definition. Given a curve y= f(x) with x from a to b, divide the x-axis from a to b into n intervals, each of length $\Delta x$. We can approximate the curve from $(x_i, f(x_i))$ to $(x_i+ \Delta x, f(x_i+ \Delta x))= (x_i+ \Delta x, f(x_i)+ \Delta y)$ where I have taken $\Delta y= f(x_i+ \Delta y)- f(x_i)$, by the straight line between those points. It's length, by the Pythagorean theorem, is $\sqrt{(\Delta x)^2+ (\Delta y)^2}$. Now, factor $\Delta x$ out of that:
$$\sqrt{(\Delta x)^2+ (\Delta y)^2}= \sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$
So the Riemann sum is
$$\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$
which, in the limit, becomes
$$\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$$
I would be surprised if your text book didn't give all of that.
6. Jan 18, 2010
### tiny-tim
awww, you gave him the answer!
7. Jan 18, 2010
### Char. Limit
Well, since this isn't a homework problem, I think giving the answer is all right...
I honestly have never seen the derivation before.
Thanks for the help, both of you.
8. Jan 24, 2010
### evagelos
Certainly in the limit $$\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$ does not become $$\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$$ according to the definition of the Rieman integral.
BUT ,by the mean value theorem we have that:
Δy = f'(ξ)Δx ,where ξ is $$x\leq\xi\leq x+\Delta x$$ and now ,
$$\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x$$ becomes :$$\sum\sqrt{1+ \left(f'(\xi)\right)^2}\Delta x$$ .This a Rieman sum which in the limit it becomes:
$$\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$$
B Confusion between $\theta$ and $d\theta$ Jan 14, 2017 | 2018-03-22T02:25:14 | {
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http://mathhelpforum.com/calculus/84519-evaluating-complicated-integral-please-help-urgent.html | For n any non-negative integer evaluate the integral :
$\int x^n \ln (x) dx$
Attempt to solution:
use integration by parts
$dv=x^n$
$v=\frac{x^{n-1}}{n-1}$
$u=\ln (x)$
$du=1/x$
$\int u dv=\ln (x) \frac{x^{n-1}}{n-1}-\int\frac{x^{n-1}}{n-1} \frac{1}{x}$
I'm stuck here how do l further simplify this thing ?
2. Originally Posted by nyasha
For n any non-negative integer evaluate the integral :
$\int x^n In(x) dx$
Attempt to solution:
use integration by parts
$dv=x^n$
$v=\frac{x^{n-1}}{n-1}$
$u=In(x)$
$du=1/x$
$\int udv=In(x)\frac{x^{n-1}}{n-1}-\int\frac{x^{n-1}}{n-1}\frac{1}x}$
I'm stuck here how do l further simplify this thing ?
$\int x^n \ln{x} \, dx$
$u = \ln{x}$ ... $du = \frac{1}{x} \, dx$
$dv = x^n \, dx$ ... $v = \frac{x^{n+1}}{n+1}$
$\int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{1}{n+1} \int x^n \, dx$
$\int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{x^{n+1}}{(n+1)^2} + C$
3. Hello, nyasha!
At this site, you must use [math ] [/math ] for LaTeX.
For any non-negative integer $n$, evaluate: . $\int x^n\ln(x)\,dx$
I would do it like this . . .
. . $\begin{array}{ccccccc}u &=& \ln x & & dv &=& x^n\,dx \\ du &=& \frac{dx}{x} & & v &=& \frac{x^{n+1}}{n+1} \end{array}$
Then we have: . $\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{n+1}\int x^n\,dx \;\;=\;\;\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{(n+1)^2}\,x^{n+1} + C$
. . . . . . $= \;\frac{1}{(n+1)^2}\,x^{n+1}\bigg[(n+1)\ln x - 1\bigg] + C$
Edit: Darn, too slow again!
.
4. Originally Posted by skeeter
$\int x^n \ln{x} \, dx$
$u = \ln{x}$ ... $du = \frac{1}{x} \, dx$
$dv = x^n \, dx$ ... $v = \frac{x^{n+1}}{n+1}$
$\int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{1}{n+1} \int x^n \, dx$
$\int x^n \ln{x} \, dx = \frac{x^{n+1}}{n+1} \cdot \ln{x} - \frac{x^{n+1}}{(n+1)^2} + C$
Thanks
5. Originally Posted by Soroban
Hello, nyasha!
At this site, you must use [math ] [/math ] for LaTeX.
I would do it like this . . .
. . $\begin{array}{ccccccc}u &=& \ln x & & dv &=& x^n\,dx \\ du &=& \frac{dx}{x} & & v &=& \frac{x^{n+1}}{n+1} \end{array}$
Then we have: . $\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{n+1}\int x^n\,dx \;\;=\;\;\frac{1}{n+1}\,x^{n+1}\ln x \;-\; \frac{1}{(n+1)^2}\,x^{n+1} + C$
. . . . . . $= \;\frac{1}{(n+1)^2}\,x^{n+1}\bigg[(n+1)\ln x - 1\bigg] + C$
Edit: Darn, too slow again!
.
Thanks,actually that was the simplified format they wanted in the answer text. | 2015-09-02T05:11:31 | {
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https://math.stackexchange.com/questions/2280212/find-all-solutions-to-x2-2x-1/2280217 | # Find all solutions to $|x^2-2|x||=1$
Firstly, we have that $$\left\{ \begin{array}{rcr} |x| & = & x, \ \text{if} \ x\geq 0 \\ |x| & = & -x, \ \text{if} \ x<0 \\ \end{array} \right.$$
So, this means that $$\left\{ \begin{array}{rcr} |x^2-2x| & = & 1, \ \text{if} \ x\geq 0 \\ |x^2+2x| & = & 1, \ \text{if} \ x<0 \\ \end{array} \right.$$
For the first equation, we have
$$|x^2-2x|\Rightarrow\left\{\begin{array}{rcr} x^2-2x & = & 1, \ \text{if} \ x^2\geq 2x \\ x^2-2x & = & -1, \ \text{if} \ x^2<2x \\ \end{array} \right.$$
and for the second equation, we have
$$|x^2+2x|\Rightarrow\left\{\begin{array}{rcr} x^2+2x & = & 1, \ \text{if} \ x^2+2x\geq 0 \\ x^2+2x & = & -1, \ \text{if} \ x^2+2x<0 \\ \end{array} \right.$$
Solving for all of these equations, we get
$$\left\{\begin{array}{rcr} x^2-2x & = 1 \Rightarrow& x_1=1+\sqrt{2} \ \ \text{and} \ \ x_2=1-\sqrt{2}\\ x^2-2x & =-1 \Rightarrow& x_3=1 \ \ \text{and} \ \ x_4=1\\ x^2+2x & = 1 \Rightarrow& x_5=-1-\sqrt{2} \ \ \text{and} \ \ x_6=-1+\sqrt{2}\\ x^2+2x & =-1 \Rightarrow& x_7=-1 \ \ \text{and} \ \ x_8=-1 \end{array} \right.$$
So we have the roots $$\begin{array}{lcl} x_1 = & 1+\sqrt{2} \\ x_2 = & 1-\sqrt{2} \\ x_3 = & -1+\sqrt{2} \\ x_4 = & -1-\sqrt{2} \\ x_5 = & 1 \\ x_6 = & -1 \end{array}$$
But according to the book, the answer is
\begin{array}{lcl} x_1 & = & 1+\sqrt{2} \\ x_4 & = & -1-\sqrt{2} \\ x_5 & = & 1 \\ x_6 & = & -1 \end{array}
What happened to $x_2$ and $x_3$? Any other way to solve this equation quicker?
• How did you get to here: $$|x|^2-2|x|-1=0 \quad \text{or} \quad |x|^2-2|x|+1=0$$ – Parseval May 14 '17 at 7:42
• $x^2=|x|^2$ for all real $x$ – CY Aries May 14 '17 at 7:44
• The first equation should equal $1$, no? – Jay Dunivin May 14 '17 at 19:54
• Yes, it's a typo. Thanks – CY Aries May 14 '17 at 23:41
$x^{2}-2x = 1$ ,if $x^{2}\geq 2x$ and $x\geq 0$ (because$\left |x^{2}-2x \right | = 1$ if $x\geq 0$)
but $1-\sqrt{2} \leq 0$
$x_{3} =-1+\sqrt{2}$ is not correct solution because of same reason in second equation
As G.H.lee already pointed out, you did casework $x\geq 0$ and $x<0$ to simplify the expression but you completely disregarded it later. A correct way would be something like this:
$$|x^2-2|x||=1\implies \begin{cases}|x^2-2x| = 1,& x\geq 0\\ |x^2+2x| = 1,& x < 0 \end{cases} \implies \begin{cases} x^2-2x = 1,& x\geq 0,\ x^2-2x\geq 0\\ x^2-2x = -1,& x\geq 0,\ x^2-2x<0\\ x^2+2x = 1,& x < 0,\ x^2+2x\geq 0\\ x^2+2x = -1,& x < 0,\ x^2+2x<0\end{cases}$$
after which you solve the way you did, but remove extra solutions.
One way to simplify this is to notice that function $f(x) = |x^2-2|x||-1$ is even, i.e. $f(-x) = f(x)$, meaning that $x_0$ is root of $f$ if and only if $-x_0$ is root of $f$. Thus, we can assume that $x\geq 0$ while solving the equation, and we can just add "$\pm$" later to get all solutions.
Our equation now simplifies to $|x^2 -2x| = 1$, i.e. $x^2-2x = \pm 1$ or $(x-1)^2 = 1\pm 1$. This gives us solutions $x = 1$ and $x = 1\pm \sqrt 2$ and after we remove the negative $1-\sqrt 2$, we get $x =1$ and $x = 1+\sqrt 2$. To get all solutions, just add "$\pm$".
Personally, I like to draw graphs. Again, you can notice that $|x^2-2|x||$ is even, so we can assume that $x\geq 0$ and reflect the graph with respect to $y$-axis later.
To graph $|x^2-2x|$ (for $x\geq 0$), you can graph parabola $x^2-2x$ first and then reflect anything below $x$-axis. Afterwards, reflect with respect to $y$-axis to get $|x^2-2|x||$:
• Great explanation thank you. However drawing graphs and such in a problem that I'm only given 5 min to solve is not an option. – Parseval May 14 '17 at 12:22
• @Parseval, you are welcome. Graphing this doesn't take more than half a minute, once you know "the rules". Absolute value is all about reflecting: if you can graph $f$ quickly enough (in this case, a parabola), then you can immediately graph $|f|$ by reflecting anything below $x$-axis, as shown in in the above graph. While this won't solve your equation, you can be very confident in the solution you got algebraically without checking your calculations. – Ennar May 14 '17 at 12:55
• Thanks a lot buddy, you're right, now that you explain it with the mirroring about the x-axis it makes sense to draw it when it's so simple. You're great! – Parseval May 14 '17 at 16:53
The most general way is to grow a tree and then for each leaf of the tree a table. If more than a couple of layers of $|.|$ you will probably be starting to confuse yourself if you don't stick to a systematic approach.
Each branch in the tree reduce the set the variable is valid for. We need to store a pair $(expression,set)$ at each node and at the leafs of the tree, we will have an expression with no $|.|$ left, just a polynomial and a set. That is when we can make a table splitting the real number line.
1. First tree branch is due to $|x|$: $x\in[0,\infty]$ left $x\in[-\infty,0]$ right.
2. left does $|x|\to x$, right does $|x|\to -x$
3. Now store pairs sets and expressions $|x^2-2x|$ left , $|x^2+2x|$ right
4. In our new nodes we need to factor polynomials to find how to split the tree up in subsets > and <0. But hopefully the systemacy of the approach is clear enough by now. | 2020-07-11T02:34:39 | {
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https://math.stackexchange.com/questions/1449561/prove-that-sum-k-1n-frac1k-lnn1-for-all-n-geq1 | Prove that $\sum_{k=1}^{n} \frac{1}{k}>\ln(n+1)$ for all $n\geq1$
Prove that $$\sum_{k=1}^{n} \frac{1}{k}>\ln(n+1)$$ for all $n\geq1$
I am looking for a clear solution to this problem. I've considering trying to prove it by contradiction by starting off assuming that it's not true for all n's, but I'm not sure if this is the best way to go about it. I don't have a lot of experience with summation proofs and would appreciate a clear proof for me to use as an example moving forward.
• How does the sum compare to the integral $\int_1^{n+1} {1\over x}\,dx$? – David Mitra Sep 24 '15 at 13:22
For $x\in(k,k+1]$ we have $\frac{1}{x}<\frac{1}{k}$, so $$\int_{k}^{k+1}\frac{1}{x}\,dx<\frac{1}{k}$$ Now sum for $k=1,2,\dots,n$
$$\displaystyle\sum_{k=1}^n\frac{1}{k}>\sum_{k=1}^n\int_{k}^{k+1}\frac{1}{x}\,dx=\int_{1}^{n+1}\frac{1}{x}\,dx=\ln(n+1)$$
What about induction on $n$? The problem boils down to proving: $$\frac{1}{n}\geq \log(n+1)-\log(n) = \log\left(1+\frac{1}{n}\right)$$ that follows from the concavity of the logarithm function, for instance. | 2019-06-16T17:24:49 | {
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https://math.stackexchange.com/questions/1719144/alternating-sign-odd-number-generating-function/1719201 | # Alternating sign odd number generating function.
I have a sequence that I'm trying to find both an ordinary generating function for as well as a closed form without a floor function. The sequence
$$\{1,1,-1,3,-3,5,-5,7,-7,9,-9,11,-11,\}$$
is recursively generated by the formula
$$a_0=1$$ $$a_n=a_{n-1}+(-1)^{n-1}2(n-1)$$
So let $A(x)=\sum{a_nx^n}$. Then
$$A(x)=\sum_{n=0}^\infty{a_nx^n}=a_0+\sum_{n=1}^\infty[a_{n-1}+(-1)^{n-1}\cdot 2(n-1)]x^n$$
$$=1+\sum_{n=1}^\infty a_{n-1}x^n+2\sum_{n=1}^\infty{(-1)^{n-1}nx^n}+2\sum_{n=1}^\infty{(-1)^{n}x^n}$$ $$A(x)-1=xA(x)+\frac{2}{1+x}+2\sum_{n=1}^\infty{(-1)^{n-1}nx^n}$$
However, I'm stuck on the third summand. I know that $\sum{nx^n}=x/(1-x)^2$ but I have the alternating sign there which is throwing me off... Any suggestions?
• You need to pull out the $a_0$ term in the summation before you use the recursion relation: currently you are missing a $+1$ on the right-hand side from the second equals sign onward. – Ben Sheller Mar 29 '16 at 18:27
• Yes, I will fix that, but the sum of $nx^n$ is $x/(1-x)^2$ according to the proof I did and according to the solution to the problem in Wilf's "Generatingfunctionology". it is chapter 1 a and the answers are in the back. – Iceman Mar 29 '16 at 18:34
• Oh you mean the other sum...the one i was looking for... – Iceman Mar 29 '16 at 18:41
• So I got $\frac{1+2x-x^2}{(x+1)^2(x-1)}$. How can I extract the coefficients? – Iceman Mar 29 '16 at 19:05
You’ve a small error in evaluating the geometric summation:
$$2\sum_{n\ge 1}(-1)^nx^n=\frac{-2x}{1+x}\;,$$
not $\frac2{1+x}$, because the summation starts at $n=1$, not at $n=0$. The other summation is
$$2\sum_{n\ge 1}(-1)^{n-1}nx^n=-2\sum_{n\ge 1}n(-x)^n=\frac{-2(-x)}{\big(1-(-x)\big)^2}=\frac{2x}{(1+x)^2}\;,$$
as noted by AccidentalFourierTransform in the comments. Thus, you have
\begin{align*} A(x)&=\frac1{1-x}\left(1-\frac{2x}{1+x}+\frac{2x}{(1+x)^2}\right)\\ &=\frac{(1+x)^2-2x(1+x)+2x}{(1-x)(1+x)^2}\\ &=\frac{1+2x-x^2}{(1-x)(1+x)^2}\;. \end{align*}
Just for fun, I checked this by finding $A(x)$ in a completely different way that you might find instructive. Let
\begin{align*} f(x)&=\sum_{n\ge 0}(2n+1)x^{2n+1}\\ &=\sum_{n\ge 0}nx^n-\sum_{n\ge 0}2nx^{2n}\\ &=\frac{x}{(1-x)^2}-\frac{2x^2}{(1-x^2)^2}\;. \end{align*}
It’s not hard to see that $A(x)=1+f(x)-xf(x)=1+(1-x)f(x)$: $-xf(x)$ yields the $x^{2n}$ terms for $n>0$. Thus,
\begin{align*} A(x)&=1+(1-x)f(x)\\ &=1+\frac{x}{1-x}-\frac{2x^2}{(1-x)(1+x)^2}\\ &=\frac{1+2x-x^2}{(1-x)(1+x)^2}\;. \end{align*}
Added: To complete the task of finding a closed form for the numbers $a_n$, decompose $A(x)$ into partial fractions. If I’ve made no silly mistakes, it’s
$$A(x)=\frac{1/2}{1-x}+\frac{3/2}{1+x}-\frac1{(1+x)^2}\;.$$
Now write each term on the right as a summation:
\begin{align*} A(x)&=\frac{1/2}{1-x}+\frac{3/2}{1+x}-\frac1{(1+x)^2}\\ &=\frac12\sum_{n\ge 0}x^n+\frac32\sum_{n\ge 0}(-x)^n-\sum_{n\ge 0}(n+1)(-x)^n\\ &=\frac12\sum_{n\ge 0}x^n+\frac32\sum_{n\ge 0}(-1)^nx^n-\sum_{n\ge 0}(-1)^n(n+1)x^n\\ &=\sum_{n\ge 0}\left(\frac12+\frac{3(-1)^n}2-(-1)^n(n+1)\right)x^n\;, \end{align*}
so
$$a_n=\frac12+\frac{3(-1)^n}2-(-1)^n(n+1)=\frac{1+(-1)^n}2+(-1)^{n+1}n\;.$$
• NIce. I just got the generating function right before as I mentioned in my comments. Now, this is the part that is getting me...how do I find a formula for the coefficients? I'm looking for something without floor function... – Iceman Mar 29 '16 at 19:07
• @Iceman: The first step is to decompose it into partial fractions. Then you reverse engineer each partial fraction to express it as a series and read off and combine the coefficients. – Brian M. Scott Mar 29 '16 at 19:09
• perfect. I also obtained this from you as well... no silly mistakes. – Iceman Mar 29 '16 at 19:53
• @Iceman: Great! Always nice to get confirmation. – Brian M. Scott Mar 29 '16 at 19:54 | 2019-12-05T20:31:04 | {
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http://www.math.uah.edu/stat/markov/Ehrenfest.html |
8. The Ehrenfest Chains
Basic Theory
The Ehrenfest chains, named for Paul Ehrenfest, are simple, discrete models for the exchange of gas molecules between two containers. However, they can be formulated as simple ball and urn models; the balls correspond to the molecules and the urns to the two containers. Thus, suppose that we have two urns, labeled 0 and 1, that contain a total of $$m$$ balls. The state of the system at time $$n \in \N$$ is the number of balls in urn 1, which we will denote by $$X_n$$. Our stochastic process is $$\bs{X} = (X_0, X_1, X_2, \ldots)$$ with state space $$S = \{0, 1, \ldots, m\}$$. Of course, the number of balls in urn 0 at time $$n$$ is $$m - X_n$$.
The Models
In the basic Ehrenfest model, at each discrete time unit, independently of the past, a ball is selected at random and moved to the other urn.
$$\bs{X}$$ is a discrete-time Markov chain on $$S$$ with transition probability matrix $$P$$ given by $P(x, x - 1) = \frac{x}{m}, \; P(x, x + 1) = \frac{m - x}{m}, \quad x \in S$
Proof:
We will give a construction of the chain from a more basic process. Let $$V_n$$ be the ball selected at time $$n \in \N_+$$. Thus $$\bs{V} = (V_1, V_2, \ldots)$$ is a sequence of independent random variables, each uniformly distributed on $$\{1, 2, \ldots, m\}$$. Let $$X_0 \in S$$ be independent of $$\bs{V}$$. (We can start the process any way that we like.) Now define the state process recursively as follows: $X_{n+1} = \begin{cases} X_n - 1, & V_{n+1} \le X_n \\ X_n + 1, & V_{n+1} \gt X_n \end{cases}, \quad n \in \N$
In the Ehrenfest experiment, select the basic model. For selected values of $$m$$ and selected values of the initial state, run the chain for 1000 time steps and note the limiting behavior of the proportion of time spent in each state.
Suppose now that we modify the basic Ehrenfest model as follows: at each discrete time, independently of the past, we select a ball at random and a urn at random. We then put the chosen ball in the chosen urn.
$$\bs{X}$$ is a discrete-time Markov chain on $$S$$ with the transition probability matrix $$Q$$ given by $Q(x, x - 1) = \frac{x}{2 m}, \; Q(x, x) = \frac{1}{2}, \; Q(x, x + 1) = \frac{m - x}{2 m}, \quad x \in S$
Proof:
Again, we can construct the chain from a more basic process. Let $$X_0$$ and $$\bs{V}$$ be as in Theorem 1. Let $$U_n$$ be the urn selected at time $$n \in \N_+$$. Thus $$\bs{U} = (U_1, U_2, \ldots)$$ is a sequence of independent random variables, each uniformly distributed on $$\{0, 1\}$$ (so that $$\bs{U}$$ is a fair, Bernoulli trials sequence). Also, $$\bs{U}$$ is independent of $$\bs{V}$$ and $$X_0$$. Now define the state process recursively as follows: $X_{n+1} = \begin{cases} X_n - 1, & V_{n+1} \le X_n, \; U_{n+1} = 0 \\ X_n + 1, & V_{n+1} \gt X_n, \; U_{n+1} = 1 \\ X_n, & \text{otherwise} \end{cases}, \quad n \in \N$
Note that $$Q(x, y) = \frac{1}{2} P(x, y)$$ for $$y \in \{x - 1, x + 1\}$$.
In the Ehrenfest experiment, select the modified model. For selected values of $$m$$ and selected values of the initial state, run the chain for 1000 time steps and note the limiting behavior of the proportion of time spent in each state.
Classification
The basic and modified Ehrenfest chains are irreducible and positive recurrent.
Proof:
The chains are clearly irreducible since every state leads to every other state. It follows that the chains are positive recurrent since the state space $$S$$ is finite.
The basic Ehrenfest chain is periodic with period 2. The cyclic classes are the set of even states and the set of odd states. The two-step transition matrix is
$P^2(x, x - 2) = \frac{x (x - 1)}{m^2}, \; P^2(x, x) = \frac{x(m - x + 1) + (m - x)(x + 1)}{m^2}, \; P^2(x, x + 2) = \frac{(m - x)(m - x + 1)}{m^2}, \quad x \in S$
Proof:
Note that returns to a state can only occur at even times, so the chain has period 2. The form of $$P^2$$ follows from the formula for $$P$$ above.
The modified Ehrenfest chain is aperiodic.
Proof:
Note that $$P(x, x) \gt 0$$ for each $$x \in S$$.
Invariant and Limiting Distributions
For the basic and modified Ehrenfest chains, the invariant distribution is the binomial distribution with trial parameter $$m$$ and success parameter $$\frac{1}{2}$$. So the invariant probability density function $$f$$ is given by $f(x) = \binom{m}{x} \left( \frac{1}{2} \right)^m, \quad x \in S$
Proof:
For the basic chain we have \begin{align*} (f P)(y) & = f(y - 1) P(y - 1, y) + f(y + 1) P(y + 1, y)\\ & = \binom{m}{y - 1} \left(\frac{1}{2}\right)^m \frac{m - y + 1}{m} + \binom{m}{y + 1} \left(\frac{1}{2}\right)^m \frac{y + 1}{m} \\ & = \left(\frac{1}{2}\right)^m \left[\binom{m - 1}{y - 1} + \binom{m - 1}{y}\right] = \left(\frac{1}{2}\right)^m \binom{m}{y} = f(y), \quad y \in S \end{align*} The last step uses a fundamental identity for binomial coefficients. For the modified chain we can use the result for the basic chain: \begin{align*} (f Q)(y) & = f(y - 1) Q(y - 1, y) + f(y) Q(y, y) + f(y + 1)Q(y + 1, y) \\ & = \frac{1}{2} f(y - 1) P(y - 1, y) + \frac{1}{2} f(y + 1) P(y + 1, y) + \frac{1}{2} f(y) = f(y), \quad y \in S \end{align*}
Thus, the invariant distribution corresponds to placing each ball randomly and independently either in urn 0 or in urn 1.
The mean return time to state $$x \in S$$ for the basic or modified Ehrenfest chain is $$\mu(x) = 2^m \big/ \binom{m}{x}$$.
Proof:
This follows from the general theory and the invariant distribution above.
For the basic Ehrenfest chain, the limiting behavior of the chain is as follows:
1. $$P^{2 n}(x, y) \to \binom{m}{y} \left(\frac{1}{2}\right)^{m-1}$$ as $$n \to \infty$$ if $$x, \, y \in S$$ have the same parity (both even or both odd). The limit is 0 otherwise.
2. $$P^{2 n+1}(x, y) \to \binom{m}{y} \left(\frac{1}{2}\right)^{m-1}$$ as $$n \to \infty$$ if $$x, \, y \in S$$ have oppositie parity (one even and one odd). The limit is 0 otherwise.
Proof:
These results follow from the general theory and the invariant distribution above, and the fact that the chain is periodic with period 2, with the odd and even integers in $$S$$ as the cyclic classes.
For the modified Ehrenfest chain, $$Q^n(x, y) \to \binom{m}{y} \left(\frac{1}{2}\right)^m$$ as $$n \to \infty$$ for $$x, \, y \in S$$.
Proof:
Again, this follows from the general theory and the invariant distribution above, and the fact that the chain is aperiodic.
In the Ehrenfest experiment, the limiting binomial distribution is shown graphically and numerically. For each model and for selected values of $$m$$ and selected values of the initial state, run the chain for 1000 time steps and note the limiting behavior of the proportion of time spent in each state. How do the choices of $$m$$, the initial state, and the model seem to affect the rate of convergence to the limiting distribution?
Reversibility
The basic and modified Ehrenfest chains are reversible.
Proof:
Let $$g(x) = \binom{m}{x}$$ for $$x \in S$$. The crucial observations are $$g(x) P(x, y) = g(y) P(y, x)$$ and $$g(x) Q(x, y) = g(y) Q(y, x)$$ for all $$x, \, y \in S$$. For the basic chain, if $$x \in S$$ then \begin{align*} g(x) P(x, x - 1) & = g(x - 1) P(x - 1, x) = \binom{m - 1}{x - 1} \\ g(x) P(x, x + 1) & = g(x + 1) P(x + 1, x) = \binom{m - 1}{x} \end{align*} In all other cases, $$g(x) P(x, y) = g(y) P(y, x) = 0$$. The reversibility condition for the modified chain follows trivially from that of the basic chain since $$Q(x, y) = \frac{1}{2} P(x, y)$$ for $$y = x \pm 1$$ (and of course the reversibility condition is trivially satisfied when $$x = y$$). Note that the invariant PDF $$f$$ is simply $$g$$ normalized. The reversibility condition gives another (and better) proof that $$f$$ is invariant.
Run the simulation of the Ehrenfest experiment 10,000 time steps for each model, for selected values of $$m$$, and with initial state 0. Note that at first, you can see the arrow of time. After a long period, however, the direction of time is no longer evident.
Computational Exercises
Consider the basic Ehrenfest chain with $$m = 5$$ balls, and suppose that $$X_0$$ has the uniform distribution on $$S$$.
1. Compute the probability density function, mean and variance of $$X_1$$.
2. Compute the probability density function, mean and variance of $$X_2$$.
3. Compute the probability density function, mean and variance of $$X_3$$.
4. Sketch the initial probability density function and the probability density functions in parts (a), (b), and (c) on a common set of axes.
1. $$f_1 = \left( \frac{1}{30}, \frac{7}{30}, \frac{7}{30}, \frac{7}{30}, \frac{7}{30}, \frac{1}{30} \right)$$, $$\mu_1 = \frac{5}{2}$$, $$\sigma_1^2 = \frac{19}{12}$$
2. $$f_2 = \left( \frac{7}{150}, \frac{19}{150}, \frac{49}{150}, \frac{49}{150}, \frac{19}{150}, \frac{7}{150} \right)$$, $$\mu_2 = \frac{5}{2}$$, $$\sigma_2^2 = \frac{79}{60}$$
3. $$f_3 = \left( \frac{19}{750}, \frac{133}{750}, \frac{223}{150}, \frac{223}{150}, \frac{133}{150}, \frac{19}{150} \right)$$, $$\mu_2 = \frac{5}{2}$$, $$\sigma_3^2 = \frac{431}{300}$$
Consider the modified Ehrenfest chain with $$m = 5$$ balls, and suppose that the chain starts in state 2 (with probability 1).
1. Compute the probability density function, mean and standard deviation of $$X_1$$.
2. Compute the probability density function, mean and standard deviation of $$X_2$$.
3. Compute the probability density function, mean and standard deviation of $$X_3$$.
4. Sketch the initial probability density function and the probability density functions in parts (a), (b), and (c) on a common set of axes.
1. $$f_1 = (0, 0.2, 0.5, 0.3, 0, 0)$$, $$\mu_1 = 2.1$$, $$\sigma_1 = 0.7$$
2. $$f_2 = (0.02, 0.20, 0.42, 0.30, 0.06, 0)$$, $$\mu_2 = 2.18$$, $$\sigma_2 = 0.887$$
3. $$f_3 = (0.030, 0.194, 0.380, 0.300, 0.090, 0.006)$$, $$\mu_3 = 2.244$$, $$\sigma_3 = 0.984$$ | 2016-09-28T01:39:56 | {
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http://math.stackexchange.com/questions/644064/a-question-about-double-integral | # a question about double integral
Let $a,b$ be positive real numbers, and let $R$ be the region in $\Bbb R^2$ bounded by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Calculate the integral $$\int\int_R\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)^{3/2}dx\,dy$$
my question is I don't know anything about $R$, the function $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is not the function of $R$, so then how can I get the answer? Could somebody give me some hints.
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Can you evaluate the integral when $a = b = 1$? – Daniel Fischer Jan 19 '14 at 19:18
What do you mean "the region... is not the function of R"? You're given an ellipse and the integral is over that ellipse. – DonAntonio Jan 19 '14 at 19:18
\begin{equation*}I=\iint_{R}(1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}})^{3/2}\,dx\,dy\tag{1} \end{equation*}
my question is I don't know anything about $R$
The equation \begin{equation*} \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\tag{2} \end{equation*} represents an ellipse centered at $(x,y)=(0,0)$, with major and minor axes coinciding with the $x,y$ axes. This ellipse is the boundary of the region $R$.
\begin{equation*}R=\left\{ (x,y)\in\mathbb{R}^{2}:0\le \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\le 1\right\}\tag{3}\end{equation*}
the function $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is not the function of $R$
The integrand $(1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}})^{3/2}$ is evaluated over $R$.
To evaluate the integral $I$ we can proceed by making a transformation of the ellipse $(2)$ to a circle and then changing from cartesian to polar coordinates, or we could use right away polar coordinates.
• If we make the change of variables $x=aX,y=bY$, then we get the circle centered at $(X,Y)=(0,0)$ and radius $1$ \begin{equation*}X^{2}+Y^{2}=1.\tag{4}\end{equation*}The region $R$ becomes the unit circle \begin{equation*}C=\left\{ (X,Y)\in\mathbb{R}^{2}:0\le X^2+Y^2\le 1\right\}\tag{5}\end{equation*} The Jacobian determinant of the transformation is just \begin{equation*}\left\vert \frac{\partial (x,y)}{\partial (X,Y)}\right\vert =ab.\tag{6}\end{equation*} This corresponds to the following linear relation between the area elements\begin{equation*}dx\,dy=ab\,dX\,dY.\tag{7}\end{equation*}As a consequence the given integral over $R$ can be rewritten as an integral over $C$ \begin{equation*}I=\iint_{C}(1-( X^{2}+Y^{2}) )^{3/2}\,ab\,dX\,dY.\tag{8}\end{equation*}
• If we now use polar coordinates $r,\theta$ \begin{eqnarray*}X &=&r\cos \theta\\Y &=&r\sin \theta\\X^{2}+Y^{2}&=&r^{2}\tag{9}\end{eqnarray*} the integral $I$ becomes\begin{eqnarray*}I &=&ab\int_{r=0}^{1}\int_{\theta =0}^{2\pi }(1-r^{2})^{3/2}r\,dr\,d\theta\\ &=&2\pi ab\int_{0}^{1}(1-r^{2})^{3/2}r\,dr,\tag{10}\end{eqnarray*} because the Jacobian determinant \begin{equation*}\left\vert \frac{\partial (X,Y)}{\partial (r,\theta )}\right\vert =r.\tag{11}\end{equation*}In terms of area elements this means that they get converted by the relation \begin{equation*}dX\,dY=r\,dr\,d\theta .\tag{12}\end{equation*}
• Finally, to evaluate $I$ we can make the substitution $t=r^{2}-1$ \begin{equation*}I=2\pi ab\int_{0}^{1}(1-r^{2})^{3/2}r\,dr=2\pi ab\int_{0}^{1}\frac{1}{2}t^{3/2}\,dt=\frac{2ab}{5}\pi.\tag{13}\end{equation*}
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+1 for very complete source Americo.:) – Babak S. Jan 20 '14 at 15:46
@B.S. Thank you! – Américo Tavares Jan 20 '14 at 16:35
Using $$x=ar\cos(t), y=br\sin(t),~~ 0\leq t\leq 2\pi,~r\ge0$$ we get $$\int_{t=0}^{2\pi}\int_{r=0}^1(1-r^2)^{3/2}J(x,y)drdt$$ For $J(x,y)$ look at this similar one Example2.
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@amWhy: Hello my friend Amy. Yes. As you know, I have to correct my claims in my recent article. Mathematics is a beautiful cruel one. – Babak S. Jan 20 '14 at 15:45
It sounds like you're just a bit confused about notation. $R$ is simply the name of the region. The notation
$$\iint\limits_{R} f(x,y) \, dA$$
simply means that we should integrate over the region $R$. In your case, $R$ is defined to be the region contained inside the ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
For $a=3$ and $b=2$, this situation could be illustrated as follows:
As the other answers already indicate, the integral can be evaluated easily by a change of variables $x=ar\cos(\theta)$ and $y=br\sin(\theta)$. It can also be expressed as an iterated integral in Cartesian coordinates
$$4\int_0^3 \int_0^{\frac{b}{a}\sqrt{a^2-x^2}} \left(1-\frac{x^2}{a^2} - \frac{y^2}{b^2}\right) \, dy \, dx,$$
which evaluates to $2\pi a b/5$, though it's certainly more algebraically cumbersome than the change of variables approach.
-
+1, nice explanation. – Américo Tavares Jan 20 '14 at 0:06 | 2015-07-06T22:23:23 | {
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https://math.stackexchange.com/questions/2503334/connection-of-gradient-jacobian-and-hessian-matrix-in-newton-method | # Connection of Gradient, Jacobian and Hessian matrix in Newton method
Suppose $f: \mathbb{R^n} \to \mathbb{R}$, the gradient of $f(\mathbf{x})$ is $$\mathop{\nabla} f(\mathbf{x}) = \begin{bmatrix} \frac{\partial{f}}{\partial{x_1}} \\ \vdots \\ \frac{\partial{f}}{\partial{x_n}} \end{bmatrix}$$
The Jacobian matrix of $\mathop{\nabla} f(\mathbf{x})$ is \begin{align} \mathbf{D} (\mathop{\nabla} f(\mathbf{x})) &= \begin{bmatrix} \frac{{\partial^2}f}{{\partial}x_1^2} & \frac{{\partial^2}f}{{\partial}x_2{\partial}x_1} & \cdots & \frac{{\partial^2}f}{{\partial}x_n{\partial}x_1}\\ \frac{{\partial^2}f}{{\partial}x_1{\partial}x_2} & \frac{{\partial^2}f}{{\partial}x_2^2} & \cdots & \frac{{\partial^2}f}{{\partial}x_n{\partial}x_2}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{{\partial^2}f}{{\partial}x_1{\partial}x_n} & \frac{{\partial^2}f}{{\partial}x_2{\partial}x_n} & \cdots & \frac{{\partial^2}f}{{\partial}x_n^2}\\ \end{bmatrix} \\ &= \mathbf{H}^T \end{align} where H is the Hessian matrix, which is consistent with the definition in Wikipedia.
The affine approximation of $\mathop{\nabla} f(\mathbf{x})$ around $x_n$ is $$\mathop{\nabla} f(\mathbf{x}) = \mathop{\nabla} f(\mathbf{x_n}) + \mathbf{D} (\mathop{\nabla} f(\mathbf{x_n})) (x - x_n) = \mathop{\nabla} f(\mathbf{x_n}) + \mathbf{H}^T (x - x_n)$$
Setting $\mathop{\nabla} f(\mathbf{x}) = 0$ gives the Newton-Raphson update as $$x_{n+1} := x_n - \mathbf{H}^{-T} \mathop{\nabla} f(\mathbf{x_n})$$
However, in Wikipedia the Newton-Raphson update is given as $x_{n+1} := x_n - \mathbf{H}^{-1} \mathop{\nabla} f(\mathbf{x_n})$. The Hessian matrix is not symmetric if the entry of the matrix is not continuous. Did I do anything wrong with my calculation? If not, does this mean we can generally treat Hessian matrix as symmetric in practice for optimization?
## 1 Answer
From the first sentence of the Wikipedia article you link - "In optimization, Newton's method is applied to the derivative $f′$ of a twice-differentiable function $f$ to find the roots of the derivative." In other words, the Hessian is symmetric.
Newton's method can also be applied in a more general setting than optimization, to find roots of a differentiable function. In that case, there is no requirement that the Jacobian be symmetric. | 2019-07-19T12:49:10 | {
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https://crypto.stackexchange.com/questions/89625/product-of-negligible-and-non-negligible-functions | # Product of Negligible and Non-Negligible Functions
I know that the product of two negligible functions will always be negligible, but I'm wondering if it's possible for the product of two non-negligible functions to be a negligible function?
I'm wondering if it's possible for the product of two non-negligible functions to be a negligible function?
Yes, actually; here is an example:
Consider the two functions:
$$P(x) = 1 \text{ if x is an even integer}, 0 \text{ otherwise}$$ $$Q(x) = 1 \text{ if x is an odd integer}, 0 \text{ otherwise}$$
Both $$P$$ and $$Q$$ are nonnegligible functions.
However $$P(x)Q(x) = 0$$, which is (trivially) a negligible function.
• Yes, that is the answer, that I missed. Thanks for correcting. – kelalaka Apr 27 at 12:46 | 2021-07-28T23:08:51 | {
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https://math.stackexchange.com/questions/939962/lambert-w-function-calculation | # Lambert- W -Function calculation?
I have an equation of the form:
$$n \log n = x$$
Upon searching I came across the term "Lambert- W -Function" but couldn't find a proper method for evaluation, and neither any computer code for it's evaluation.
Any ideas as to how I can evaluate?
• Do you mean an evaluation of the function for a certain argument? Sorry, just trying to make sure I understand your question. – Sheheryar Zaidi Sep 21 '14 at 8:39
• Well,this is more of a programming question,but was just wondering if it could be solved manually. For given value of x,can we find n? – techriften Sep 21 '14 at 8:41
• mathworld.wolfram.com/LambertW-Function.html give several series expansions. See the equations (11), (13) and (14). – Nabigh Sep 21 '14 at 8:43
• wikipedia has a derivate containing code ("wikicode"? "Rosettastone"?). I found it easy to find a recursive programming-example, translatable into Pari/GP – Gottfried Helms Sep 21 '14 at 8:49
• $n\ln(n)=x\implies \exp(n\ln(n))=\exp(x)\implies n^n=\exp(x)\implies n=\frac{x}{W(x)}$, using Example 2 of the wiki article on $W$. – Yiannis Galidakis Sep 21 '14 at 9:11
Let us consider the function $$f(x)=x \log(x)-a$$ Effcetively, the solution of $f(x)=0$ is given by $$x=\frac{a}{W(a)}$$ and, if I properly understood, you look for a computation method for getting $W(a)$.
From definition $W(a)$ is defined such that $a=W(a)e^{W(a)}$ so Newton method seems to be (and is) very good.
I strongly suggest you have a look at http://en.wikipedia.org/wiki/Lambert_W_function. In the paragraph entitled "Numerical evaluation", they give Newton and Halley formulae (the latest one has been massively used by Corless et al. to compute $W(a)$.
In the same Wikipedia page, you will find very nice and efficient approximations of $W(a)$ for small and large values. These estimates will allow you to start really close to the solution.
If I may underline one thing which is really nice : all derivatives od $W(a)$ express as functions of $a$ and $W(a)$ itself and this is extremely convenient.
You could be interested by http://people.sc.fsu.edu/~jburkardt/cpp_src/toms443/toms443.html where the source code is available.
Although an old post, I'm surprised nobody mentioned this:
$$n\log n=(e^{\log n})\log n=x$$
Now it should be obvious on how to proceed.
$$W(x)=\log n$$
$$e^{W(x)}=n$$
And because $W(x)e^{W(x)}=x$, $e^W(x)=\frac{x}{W(x)}$ so,
$$\frac{x}{W(x)}=n$$
If you're using Matlab or want to look at a robust (but simple) implementation of the Lambert $W$ function ($W_0$ branch only) using Halley's method, see my answer here.
As an alternative to @ClaudeLeibovici's more general answer, an equation specifically of the form $n \text{ln}(n) = x$ might best be solved and analyzed using the simpler Wright $\omega$ function:
$$n = \frac{x}{\omega(-ln(1/x))}$$
Matlab has wrightOmega and Maple has Wrightomega to evaluate this function symbolically. See Corless and Jeffrey, 2002, The Wright omega Function (PDF) for further details. Unfortunately, there does not seem to be support for this function in SciPy or SymPy.
The Wright $\omega$ function can also be evaluated numerically. See Lawrence, et al., 2012, Algorithm 917: Complex Double-Precision Evaluation of the Wright ω Function. I've implemented this algorithm in Matlab – you can find it on GitHub here. Evaluating this numerically is 3+ orders of magnitude faster than evaluating it symbolically. If your $x$ parameter is constrained in particular ways (e.g., real-valued), you may be able to simplify your algorithm – see the Lawrence, et al. paper and the comments in my wrightOmega code.
\begin{align*} n \ln (n) =x \quad & \Rightarrow \quad \ln n^n=x \\ & \Rightarrow \quad n^n=e^x \\ & \Rightarrow \quad n=e^{x/n} \\ & \Rightarrow \quad n \times \frac{x}{n}=\frac{x}{n}e^{x/n} \\ & \Rightarrow \quad x=\frac{x}{n}e^{x/n} \\ & \Rightarrow \quad W(x)=W \left(\frac{x}{n}e^{x/n} \right) \\ & \Rightarrow \quad W(x)=\frac{x}{n} \\ & \Rightarrow \quad \boxed{n=\frac{x}{W(x)}} \\ & \Rightarrow \quad n=\frac{x}{x/e^{W(x)}} \quad \text{(since W(x)e^{W(x)}=x)}\\ & \Rightarrow \quad \boxed{n=e^{W(x)}} \\ \end{align*}
• How do we replace $\log(x)$ with $\ln(x)$? – Mathrix Mar 30 at 0:37 | 2020-09-25T00:45:44 | {
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https://physics.stackexchange.com/questions/166878/uniformly-accelerated-motion-question | # Uniformly Accelerated Motion question
A ballast bag is dropped from a balloon that is 300 m above the ground and rising at 13 m/s. For the bag, find the maximum height reached.
• Now the book gives me this answer -
• $V_i$ = initial velocity
• $V_f$ = final velocity
• $a$ = acceleration
• $y$ = displacement
• The $V_i$ of the bag when released is the same as balloon at 13 m/s upward
• Choose up as positive
• $y = 0$ at point of release
At highest point, $V_f = 0$. Using the equation $$V_f^2 = V_i^2 + 2a\Delta y,$$ and plugging in the numbers $$0 = (13\,{\rm m/s})^2 + 2(-9.81\,{\rm m/s^2})y,$$ the answer of this equation comes out to $y = 8.6\, \rm m$
The maximum height is 300 + 8.6 = 308.6 m
Now my question is what is the reasoning behind finding the 8.6 meters and adding it to the 300 m? If you dropped the ballast bag from 300 m, wouldn't the maximum height reached by the bag be 300 m?
If you start with the bag stationary at 300m then drop it the bag is going to fall straight down, and its maximum height would indeed just the 300m point it started from.
However you're not starting with the bag stationary. You're starting with the bag moving upwards at 13 m/s. So the bag is going to start at 300m then move up, come to a halt, then start falling down again. That means the top of its trajectory is above 300m.
The 8.6m you've calculated is the distance an object moves upwards if it starts with an initial upwards velocity of 13 m/sec. The trajectory of the bag looks like:
in your answer you ignored the initial velocity the bag will rise a little until its velocity is zero then it will fall again , it is like he found the total distance
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Jim Feb 24 '15 at 18:14
• @JimdalftheGrey: How is this not an answer to the question? – Kyle Kanos Feb 24 '15 at 18:26
• " ignore " can have a different meaning than " insult " which of course i did not mean " insult " the other meaning is "Fail to consider (something significant):" oxforddictionaries.com/definition/english/… – Mohamed Osama Feb 24 '15 at 18:31
• @KyleKanos Okay, it kind of does answer the question. It's just something I see as being more of a comment you'd put on the question than an answer. Perhaps my standards are too strict – Jim Feb 24 '15 at 19:19
• Seems like an answer to me. – David Z Feb 25 '15 at 4:16 | 2020-01-23T23:31:24 | {
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