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https://math.stackexchange.com/questions/605673/integrate-int-0-pi-2-frac11-tan-alphax-mathrmdx | Integrate $\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$
Evaluate the integral $$\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$$
• No restrictions on $\alpha$? – John Dec 13 '13 at 17:37
• Not as far as I know, but positive integers will probably suffice. – user85798 Dec 13 '13 at 17:38
• Playing around with Wolfram Alpha may help. I can't do the general case with the free version but it looks like there are separate forms for the solutions for even $\alpha$ and odd $\alpha$. – John Dec 13 '13 at 17:46
• In fact, unlike @RonGordon, mathematica spends a long time and returns the integral unevaluated. Yay! for humans. – Igor Rivin Dec 13 '13 at 18:11
• I'm actually surprised that this hasn't (AFAICT) shown up on the site before! This is a long-time contest problem; I've most often seen it with $\alpha=\sqrt{2}$. – Steven Stadnicki Dec 13 '13 at 18:12
Use the fact that
$$\tan{\left (\frac{\pi}{2}-x\right)} = \frac{1}{\tan{x}}$$
i.e.,
$$\frac1{1+\tan^{\alpha}{x}} = 1-\frac{\tan^{\alpha}{x}}{1+\tan^{\alpha}{x}} = 1-\frac1{1+\frac1{\tan^{\alpha}{x}}} = 1-\frac1{1+\tan^{\alpha}{\left (\frac{\pi}{2}-x\right)}}$$
Therefore, if the sought-after integral is $I$, then
$$I = \frac{\pi}{2}-I$$
and...
• That's brilliant. – John Dec 13 '13 at 18:12
• This is excellent! – user85798 Dec 13 '13 at 18:22
• And the answer doesn't even depend on $\alpha$! Way cool. – John Dec 13 '13 at 18:27
• This only works if $\alpha$ is a positive integer, very nice solution though. For other values all the answers to the integral are different. – Jeff Faraci Dec 13 '13 at 19:01
• @Jeff: why? What would be the value when $\alpha \lt 0$? – Ron Gordon Dec 13 '13 at 19:02
The answer given by Gordon is very good for computing the integral. But it does not provide much why it works. The following figure might help with that
This is the integral for $\alpha=\sqrt{2}$. It looks as if the area under the function is exactly half of the dashed rectangle... A good guess is therefore $$\int_0^{\pi/2} \frac{\mathrm{d}x}{1 + \tan(x)^\sqrt{2}} = \frac{1}{2}\left(\frac{\pi}{2}\right)$$ This striking symmetry can be shown true for any choice of $\alpha$, play around in geogebra or any ploting tool you fancy =)
The symmetry of such integrals is studied further in the paper Symmetri and Integration by Roger Nielsen Where it is shown that if $f(x) + f(a+b-x)$ is constant for all $x\in[a,b]$ - meaning it has this nice symmetric property.
Then the area can calculated as $$\int_a^b f(x) \mathrm{d}x = \frac{f(a)+f(b)}{2}(b-a) = f\left(\frac{a+b}{2}\right)(b-a)\,.$$ I will leave it to you to check that $f(x)+f(a+b-x)$ is constant.
Alternatively the integral can also be computed as follows \begin{align*} \int_0^{\pi/2} \frac{\mathrm{d}\theta}{(1 + (\tan \theta)^b} = \int_0^\infty \frac{\mathrm{d}x}{(1+x^2)(1+x^b)} = \frac{1}{2} \int_0^\infty \frac{\mathrm{d}x}{1 + x^2} = \frac{\pi}{4} \end{align*} Where the substitution $u \mapsto \tan \theta$ was used and that $$\int_0^\infty \frac{R(x)}{x^b+1}\mathrm{d}x = \frac{1}{2} \int_0^\infty R(x)\,\mathrm{d}x$$ Given that $R(x) = R(1/x)/x^2$, again check that this holds for $1/(1+x^2)$. A proof of the latter can be found on page $27$ here, section 1.9. More directly one has by the same methods \begin{align*} \int_0^\infty \frac{\mathrm{d}x}{(1+x^2)(1+x^b)} & = \int_0^1 \frac{\mathrm{d}x}{(1+x^2)(1+x^b)} + \int_1^\infty \frac{\mathrm{d}x}{(1+x^2)(1+x^b)} \\ & = \int_0^1 \frac{1}{(1+x^2)(1+x^b)} + \frac{x^b}{(1+x^2)(1+x^b)} \mathrm{d}x \\ & = \int_0^1 \frac{\mathrm{d}x}{1+x^2} = \frac{1}{2} \int_0^\infty \frac{\mathrm{d}x}{1+x^2} = \frac{\pi}{4} \end{align*} Where the substitution $x \mapsto 1/x$ was used in the last integral. These theorems and ideas are not the simplest way to attack the problem. But it might give you some broader insight further down the road.
• I'm not entirely sure what you mean by "it does not provide much why it works." My derivation, I think, could not be any more clear about why it works: the first line I wrote says it all. – Ron Gordon Dec 13 '13 at 20:22
• You use a "magical" substitution that in the end "happens" to work out. I try to justify the usage of $x \mapsto a+b-x$. Saying use this substitution here, does not give much explanation to as why you used it. Since $$\int_a^b f(x) \mathrm{d}x = \int_a^b f(a+b-x) \mathrm{d}x$$ Then the integral must me constant if $f(x)+f(a+b-x)$ is constant. If the latter is constant, it means that $f$ is symmetric about the point $(c,f(x))$, where $c=(a+b)/2$. – N3buchadnezzar Dec 13 '13 at 20:32
• Of course it "happens" to work out - that's why it's the tangent (or cotangent) function in the integrand and not, say, a log. Really, the plots are nice and numerical examples always help, but you really are making too much of this. The integral is beautiful because it is so simple...because of this unique property of the tangent. That's it. Nothing magical about it at all. – Ron Gordon Dec 13 '13 at 21:08
• It is simple because a first grader can look at the figure and say, oh it must be $\pi/4$ =) The substitution exploits symmetry which I showcased in my answer.There hundreds upon hundreds of integrals that have this intrinsic property – N3buchadnezzar Dec 13 '13 at 21:17
• Stop it guys, they were both good answers! – Bennett Gardiner Dec 15 '13 at 12:39
The integral can be solved by using the fact that $\tan\left(\dfrac{\pi}{2}-x\right)=\dfrac{1}{\tan x}$ and $$\int_a^b f(x)\;dx=\int_a^b f(a+b-x)\;dx.$$ Let $$I(\alpha)=\int_0^{\frac{\Large\pi}{2}} \frac{dx}{1+\tan^\alpha x},$$ then \begin{align} I(\alpha)&=\int_0^{\frac{\Large\pi}{2}} \frac{dx}{1+\tan^\alpha\left(\frac{\pi}{2}+0-x\right)}\\ &=\int_0^{\frac{\Large\pi}{2}} \frac{dx}{1+\dfrac{1}{\tan^\alpha x}}\\ &=\int_0^{\frac{\Large\pi}{2}} \frac{\tan^\alpha x}{1+\tan^\alpha x}dx. \end{align} Adding the two $I(\alpha)$'s yields \begin{align} 2I(\alpha)&=\int_0^{\frac{\Large\pi}{2}} \frac{1}{1+\tan^\alpha x}dx+\int_0^{\frac{\Large\pi}{2}} \frac{\tan^\alpha x} {1+\tan^\alpha x}dx\\ &=\int_0^{\frac{\Large\pi}{2}}\;dx\\ &=\frac{\pi}{2}\\ I(\alpha)&=\large\color{blue}{\frac{\pi}{4}}. \end{align}
P.S.
I did NOT copy Ron Gordon's answer since this answer is taken from the similar problem that I posted on Brilliant.org - Trigonometric Integral of The Year. A similar problem can also be found on Mathworld Wolfram - Definite Integral.
• What exactly was the point of bumping a 6 month year old question with an answer that is identical to the accepted answer then? – user85798 Jun 1 '14 at 19:20
• @Oliver And what is exactly the reason that I cannot answer this question? Is there a rule that restricts it? – Tunk-Fey Jun 1 '14 at 19:22
• @Tunk-Fey , if everybody is going to given exactly the same answer to a question then this is going to get very boring and ridiculous, evne more hwn you bump up a question 6 months old. This adds nothing to the asker nor to any other member. Furthermore, your claim that you didn't copy tRon's answer cannot be checked, and it indeed looks like you did copy his answer. – Timbuc Oct 28 '14 at 4:34
• This is not identical to the crowned answer, but it is not but a formalised formulation of it. – awllower Oct 28 '14 at 4:36
Set $$I(\alpha):=\int_0^{\pi/2} f(\alpha,x)dx$$ where $$f(\alpha,x):\mathbb{R}\times (0,\pi/2)\to \mathbb{R}$$ is defined as $$f(\alpha,x):=\frac{1}{1+\tan^\alpha(x)}$$.
Since both $$\frac{d}{d\alpha}f(\alpha,x)=-\frac{\tan^\alpha(x)\log(\tan(x))}{(1+\tan^\alpha(x))^2}$$ and $$f(\alpha,x)$$ are continuous on $$(0,\pi/2)\to \mathbb{R}$$ we can apply the Leibniz integral rule obtaining
$$I'(\alpha)=\frac{d}{d\alpha}\int_0^{\pi/2}f(\alpha,x)\,dx=\int_0^{\pi/2}\frac{d}{d\alpha}f(\alpha,x)\,dx=0$$ since $$\frac{d}{d\alpha}f(\alpha,\frac{\pi}{4}+x)=\frac{d}{d\alpha}f(\alpha,\frac{\pi}{4}-x)$$ for $$x\in[0,\frac{\pi}{4})$$.
Hence $$I(\alpha)$$ is constant on $$\mathbb{R}$$ and $$I(\alpha)=I(0)=\int_0^{\pi/2}\frac{dx}{1+1}=\color{red}{\frac{\pi}{4}}.$$ | 2021-06-19T09:24:05 | {
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https://la.mathworks.com/help/symbolic/sym.children.html | # children
Subexpressions or terms of symbolic expression
Starting in R2020b, the syntax subexpr = children(expr) for a scalar input expr returns subexpr as a nonnested cell array instead of a vector. You can use subexpr = children(expr,ind) to index into the returned cell array of subexpressions. For more information, see Compatibility Considerations.
## Description
example
subexpr = children(expr) returns a nonnested cell array containing the child subexpressions of the symbolic expression expr. For example, the child subexpressions of a sum are its terms.
example
subexpr = children(A) returns a nested cell array containing the child subexpressions of each expression in the symbolic matrix A.
example
subexpr = children(___,ind) returns the child subexpressions of a symbolic expression expr or a symbolic matrix A as a cell array indexed by ind.
## Examples
collapse all
Find the child subexpressions of the symbolic expression ${x}^{2}+xy+{y}^{2}$. The subexpressions are returned in a nonnested cell array. children uses internal sorting rules when returning the subexpressions. You can index into each element of the cell array by using subexpr{i}, where i is the cell index. The child subexpressions of a sum are its terms.
syms x y
subexpr = children(x^2 + x*y + y^2)
subexpr=1×3 cell array
{[x*y]} {[x^2]} {[y^2]}
s1 = subexpr{1}
s1 = $x y$
s2 = subexpr{2}
s2 = ${x}^{2}$
s3 = subexpr{3}
s3 = ${y}^{2}$
You can also index into each element of the subexpressions by specifying the index ind in the children function.
s1 = children(x^2 + x*y + y^2,1)
s1 = $x y$
s2 = children(x^2 + x*y + y^2,2)
s2 = ${x}^{2}$
s3 = children(x^2 + x*y + y^2,3)
s3 = ${y}^{2}$
To convert the cell array of subexpressions into a vector, you can use the command [subexpr{:}].
V = [subexpr{:}]
V = $\left(\begin{array}{ccc}x y& {x}^{2}& {y}^{2}\end{array}\right)$
Find the child subexpressions of the equation ${x}^{2}+xy={y}^{2}+1$. The child subexpressions of the equation are returned in a 1-by-2 cell array. Index into all elements of the cell array. The subexpressions of an equation are the left and right sides of that equation.
syms x y
subexpr = children(x^2 + x*y == y^2 + 1)
subexpr=1×2 cell array
{[x^2 + y*x]} {[y^2 + 1]}
subexpr{:}
ans = ${x}^{2}+y x$
ans = ${y}^{2}+1$
Next, find the child subexpressions of the inequality $\mathrm{sin}\left(x\right)<\mathrm{cos}\left(x\right)$. Index into all elements of the returned cell array. The child subexpressions of an inequality are the left and right sides of that inequality.
subexpr = children(sin(x) < cos(x))
subexpr=1×2 cell array
{[sin(x)]} {[cos(x)]}
subexpr{:}
ans = $\mathrm{sin}\left(x\right)$
ans = $\mathrm{cos}\left(x\right)$
Find the child subexpressions of an integral ${\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2777777777777778em}{0ex}}dx$. The child subexpressions are returned as a cell array of symbolic expressions.
syms f(x) a b
subexpr = children(int(f(x),a,b))
subexpr=1×4 cell array
{[f(x)]} {[x]} {[a]} {[b]}
V = [subexpr{:}]
V = $\left(\begin{array}{cccc}f\left(x\right)& x& a& b\end{array}\right)$
Find the Taylor approximation of the $\mathrm{cos}\left(x\right)$ function near $x=2$.
syms x
t = taylor(cos(x),x,2)
t =
$\mathrm{cos}\left(2\right)+\frac{\mathrm{sin}\left(2\right) {\left(x-2\right)}^{3}}{6}-\frac{\mathrm{sin}\left(2\right) {\left(x-2\right)}^{5}}{120}-\mathrm{sin}\left(2\right) \left(x-2\right)-\frac{\mathrm{cos}\left(2\right) {\left(x-2\right)}^{2}}{2}+\frac{\mathrm{cos}\left(2\right) {\left(x-2\right)}^{4}}{24}$
The Taylor expansion has six terms that are separated by $+$ or $–$ sign.
Plot the $\mathrm{cos}\left(x\right)$ function. Use children to separate out the terms of the expansion. Show that the Taylor expansion approximates the function more closely as more terms are included.
fplot(cos(x),[0 4])
hold on
s = 0;
for i = 1:6
s = s + children(t,i);
fplot(s,[0 4],'--')
end
legend({'cos(x)','1 term','2 terms','3 terms','4 terms','5 terms','6 terms'})
Call the children function to find the child subexpressions of the following symbolic matrix input. The result is a 2-by-2 nested cell array containing the child subexpressions of each element of the matrix.
syms x y
symM = [x + y, sin(x)*cos(y); x^3 - y^3, exp(x*y^2) + 3]
symM =
$\left(\begin{array}{cc}x+y& \mathrm{cos}\left(y\right) \mathrm{sin}\left(x\right)\\ {x}^{3}-{y}^{3}& {\mathrm{e}}^{x {y}^{2}}+3\end{array}\right)$
s = children(symM)
s=2×2 cell array
{1x2 cell} {1x2 cell}
{1x2 cell} {1x2 cell}
To unnest or access the elements of the nested cell array s, use braces. For example, the {1,1}-element of s is a 1-by-2 cell array of symbolic expressions.
s11 = s{1,1}
s11=1×2 cell array
{[x]} {[y]}
Unnest each element of s using braces. Convert the nonnested cell arrays to vectors using square brackets.
s11vec = [s{1,1}{:}]
s11vec = $\left(\begin{array}{cc}x& y\end{array}\right)$
s21vec = [s{2,1}{:}]
s21vec = $\left(\begin{array}{cc}{x}^{3}& -{y}^{3}\end{array}\right)$
s12vec = [s{1,2}{:}]
s12vec = $\left(\begin{array}{cc}\mathrm{cos}\left(y\right)& \mathrm{sin}\left(x\right)\end{array}\right)$
s22vec = [s{2,2}{:}]
s22vec = $\left(\begin{array}{cc}{\mathrm{e}}^{x {y}^{2}}& 3\end{array}\right)$
If each element of the nested cell array s contains a nonnested cell array of the same size, then you can also use the ind input argument to access the elements of the nested cell array. The index ind allows children to access each column of subexpressions of the symbolic matrix input symM.
scol1 = children(symM,1)
scol1=2×2 cell array
{[x ]} {[cos(y) ]}
{[x^3]} {[exp(x*y^2)]}
[scol1{:}].'
ans =
$\left(\begin{array}{c}x\\ {x}^{3}\\ \mathrm{cos}\left(y\right)\\ {\mathrm{e}}^{x {y}^{2}}\end{array}\right)$
scol2 = children(symM,2)
scol2=2×2 cell array
{[y ]} {[sin(x)]}
{[-y^3]} {[3 ]}
[scol2{:}].'
ans =
$\left(\begin{array}{c}y\\ -{y}^{3}\\ \mathrm{sin}\left(x\right)\\ 3\end{array}\right)$
## Input Arguments
collapse all
Input expression, specified as a symbolic number, variable, function, or expression.
Input matrix, specified as a symbolic matrix.
Index of child subexpressions to return, specified as a positive integer.
• If children(expr) returns a nonnested cell array of child subexpressions, then indexing with children(expr,ind) returns the ind-th element of the cell array.
• If children(A) returns a nested cell array of child subexpressions, where each cell element has the same size, then indexing with children(A,ind) returns the ind-th column of the nonnested cell array.
## Version History
Introduced in R2012a
expand all
Behavior changed in R2020b | 2022-06-29T08:00:01 | {
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https://brilliant.org/discussions/thread/the-buffalo-way/?sort=top | The Buffalo Way
Imgur
The Buffalo Way is a plug-and-bash method used to solve olympiad inequalities. It is usually applied to symmetric inequalities, where we can assume WLOG that the variables are in a specific order; that is, $x_1 \le x_2\le \cdots \le x_n$.
To illustrate this method, we shall prove $AM-GM$ for two variables using the method.
Prove that $\dfrac{x+y}{2}\ge \sqrt{xy}$ for non-negative reals $x,y$.
First, assume WLOG that $x\le y$. Thus, we can represent $x$ and $y$ by: $x=a$ $y=a+b$ where $a,b$ are non-negative reals. Make sure you see why this is true.
Thus, we want to prove $\dfrac{a+a+b}{2}\ge \sqrt{a(a+b)}\implies a+\dfrac{b}{2}\ge \sqrt{a^2+ab}$
Squaring both sides gives $a^2+ab+\dfrac{b^2}{4}\ge a^2+ab\implies \dfrac{b^2}{4}\ge 0$ which is true by the trivial inequality.
In general, if we have variables satisfying $x_1 \le x_2\le \cdots \le x_n$, then we substitute $x_1=y_1$$x_2=y_1+y_2$ $\vdots$ $x_n=y_1+y_2+\cdots +y_n$
Given that $a,b,c$ are non-negative reals such that $a\le b\le c$, then prove that $(a+b)(c+a)^2\ge 6abc$
We see that we already have the condition of $a\le b\le c$, so we can apply Buffalo's Way directly. Let $a=x$ $b=x+y$ $c=x+y+z$ where $x,y,z$ are non-negative reals. Thus, we want to prove $(2x+y)(2x+y+z)^2\ge 6x(x+y)(x+y+z)$
This expands to (told you Buffalo Way is a bash): $8x^3+12x^2y+8x^2z+6xy^2+8xyz+2xz^2+y^3+2y^2x+yz^2\ge 6x^3+12x^2y+6x^2z+6xy^2+6xyz$
Rearranging gives $2x^3+2x^2z+2xyz+2xz^2+y^3+2y^2z+yz^2\ge 0$ which is trivially true ($x,y,z$ are positive).
We can also see that this method gives an equality case: We must have $x=y=z=0$. Thus, $a=b=c=0$.
When actually solving Olympiad Inequalities in competitions, NEVER use this method, unless you have no idea now to do it otherwise. In cases where the inequality is relatively simple (no denominators) the Buffalo Way is almost guaranteed to work.
NOTE: I don't know where the name comes from. If you try to search it up, you won't get any results. I first heard of this method from the AoPS Community.
Note by Daniel Liu
5 years, 3 months ago
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Sort by:
2lazy2bash
- 5 years, 3 months ago
#?
- 5 years, 3 months ago
Yes #2lazy2bash
- 5 years, 3 months ago
Are there any Olympiad problems that can be solved with Buffalo?
- 5 years, 3 months ago
Well, try solving the second one without buffalo's.
But to answer your question, chances are there are no Olympiad question that can be only exclusively solved using the Buffalo way. This method is the least elegant method of olympiad inequalities ever, and it is almost guaranteed that there is a more elegant solution.
You might even be docked a point on the USA(J)MO if you used buffalo way, because of it's extreme inelegantness.
- 5 years, 3 months ago
"You might even be docked a point on the USA(J)MO if you used buffalo way, because of it's extreme inelegantness."
You might actually get a score of a point on that USA(J)MO (correct answer, bad proof)
- 5 years, 3 months ago
No, if you have a complete proof, no matter how inelegant, then you will get at least 5-6 points. It's impossible to get 1 point if you write a complete proof.
- 5 years, 3 months ago
We just have to make sure that all of our implications are reversible when using this method.
- 5 years, 3 months ago | 2019-10-14T02:17:16 | {
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https://www.physicsforums.com/threads/is-the-sign-in-this-integral-correct.948374/ | • I
In the table of integral, there is this formula
$$\int_0^{\infty}\frac{sin(ax)}{x}\,dx=\frac{\pi}{2}\text{sign} a$$
is sign a here is literally the sign of a, or it means something else?
Mark44
Mentor
In the table of integral, there is this formula
$$\int_0^{\infty}\frac{sin(ax)}{x}\,dx=\frac{\pi}{2}\text{sign} a$$
is sign a here is literally the sign of a, or it means something else?
Yes, it means the sign of a. ##\text{sign}(a)## and equals +1 if a > 0 and -1 if a < 0. So the integral above equals ##\pi/2## if a > 0, and ##-\pi/2## if a < 0.
EngWiPy
Yes, it means the sign of a. ##\text{sign}(a)## and equals +1 if a > 0 and -1 if a < 0. So the integral above equals ##\pi/2## if a > 0, and ##-\pi/2## if a < 0.
Thanks. I was looking for something similar but for cosine, and I found this
$$\int_0^{\infty}\frac{cos(ax)}{x+\beta}\,dx=-sin(a\beta)si(a\beta)-cos(a\beta)ci(a\beta)$$
where ##si(.)## and ##ci(.)## are the sine and cosine integrals, respectively. However, I want this for ##\beta=0##, which results in ##-ci(0)##. I am not sure, but it seems the result of this is ##-\infty##. Does this mean that ##cos(x)/x## is not integrable over ##[0, \infty)##?
Mark44
Mentor
Thanks. I was looking for something similar but for cosine, and I found this
$$\int_0^{\infty}\frac{cos(ax)}{x+\beta}\,dx=-sin(a\beta)si(a\beta)-cos(a\beta)ci(a\beta)$$
where ##si(.)## and ##ci(.)## are the sine and cosine integrals, respectively. However, I want this for ##\beta=0##, which results in ##-ci(0)##. I am not sure, but it seems the result of this is ##-\infty##. Does this mean that ##cos(x)/x## is not integrable over ##[0, \infty)##?
Right, because the graph of ##y = \frac{\cos(x)} x## has a non-removable discontinuity at 0, unlike the graph of ##y = \frac{\sin(x)} x##, which has a removable discontinuity that doesn't affect the integral.
From the graph here, ##Ci(0)## appears to be ##-\infty##, so I think you have an incorrect sign.
Thanks. Yes, right ##ci(0)=-\infty## not ##-ci(0)##. So, I cannot evaluate this integration? The original integration that I want to evaluate is
$$\int_0^{\infty}\frac{e^{-jtx}}{t}\,dt$$
where ##j=\sqrt{-1}##. Can I evaluate it in another way?
Svein
Ahlfors notes that using residues and some limit considerations, the Cauchy principal value of $\int_{-\infty}^{\infty}\frac{e^{jt}}{t}dt$ is $j\pi$. I would suggest substituting $u=x\cdot t$ in your integral and see where it leads.
Ahlfors notes that using residues and some limit considerations, the Cauchy principal value of $\int_{-\infty}^{\infty}\frac{e^{jt}}{t}dt$ is $j\pi$. I would suggest substituting $u=x\cdot t$ in your integral and see where it leads.
Thanks, but what about the integration limits? In my case. the lower limit is ##0## and not ##-\infty##. Even after the change of variables, the limits remains the same.
Svein
Thanks, but what about the integration limits? In my case. the lower limit is ##0## and not ##-\infty##. Even after the change of variables, the limits remains the same.
I haven't calculated the answer, just presented a tip. Next tip: split the integral in the middle and substitute -u for u in the negative part.
I haven't calculated the answer, just presented a tip. Next tip: split the integral in the middle and substitute -u for u in the negative part.
In my case, the exponent is negative, not just the limits. So, if I assume that ##u=-t##, I get the following integral:
$$-\int_{-\infty}^{\infty}\frac{e^{-ju}}{u}\,du = -\int_{-\infty}^0\frac{e^{-ju}}{u}\,du -\int_{0}^{\infty}\frac{e^{-ju}}{u}\,du = j\pi$$
I am interested in ##\int_{0}^{\infty}\frac{e^{-ju}}{u}\,du = -\int_{-\infty}^0\frac{e^{-ju}}{u}\,du -j\pi##, but then how to evaluate the integral in the RHS. It is the same problem as before.
Svein
I know that you may still not be satisfied, but $\int_{-\infty}^{0}\frac{e^{-ju}}{u}du=\int_{\infty}^{0}\frac{e^{ju}}{u}du=-\int_{0}^{\infty}\frac{e^{ju}}{u}du$. Add the two integrals: $\int_{0}^{\infty}\frac{e^{-ju}}{u}du-\int_{0}^{\infty}\frac{e^{ju}}{u}du=2\cdot j \int_{0}^{\infty}\frac{\sin(u)}{u}du$. | 2022-06-29T00:01:54 | {
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http://math.stackexchange.com/questions/125329/combinatorical-proof-rigor?answertab=oldest | # Combinatorical Proof Rigor [closed]
Many people feel that a combinatorical proof of a particular algebraic identity is not as rigorous as an analytic one. However, I'm not quite sure why people feel this way. So, I am curious as to what people on this site think of this style of proof. Is it rigorous enough? If it's rigorous, do they lack motivation?
-
I’ve never encountered such a prejudice; there is no inherent lack of rigor in combinatorial arguments. In my experience combinatorial proofs are frequently preferred, on the grounds that they offer more insight. – Brian M. Scott Mar 28 '12 at 5:39
Like @Brian said. You could provide references for, or examples of, the alleged prejudice. – Did Mar 28 '12 at 5:49
If you are talking about Isaac Solomon, he just likes to joke that way. He does know that combinatorics is a rigorous subject. – Daniel Montealegre Mar 28 '12 at 5:51
I am quoting one of the offered reasons for closure: "This question is not a good fit to our Q&A format. We expect answers to generally involve facts, references, or specific expertise; this question will likely solicit opinion, debate, arguments, polling, or extended discussion." Perfect fit - click... – Alex B. Mar 28 '12 at 6:01
@Alex: If the premise on which it’s based were correct, it would be a natural and reasonable question, deserving an answer. The OP seems to have believed the premise and asked the question in good faith; I see no reason to close it simply because the premise is in fact false. – Brian M. Scott Mar 28 '12 at 20:21
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## closed as not constructive by Alex B., Did, Hans Lundmark, Qiaochu YuanMar 28 '12 at 20:35
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or specific expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, see the FAQ for guidance.
I think that this question is based on a misapprehension: in several decades of doing and teaching mathematics I’ve never encountered a prejudice against combinatorial proofs of algebraic identities. On the contrary, I’ve found that they are often preferred, on the grounds that they offer more insight $-$ more sense of real understanding of why the result is true $-$ than purely computational arguments.
There is certainly no basis for such a preference: there is no inherent lack of rigor in combinatorial arguments. It’s true that by their nature they can often be stated somewhat informally and still be understood; this is in large part precisely because they tend to have a clear underlying idea, instead of being the result of a comparatively unmotivated string of computations. But this does not mean that they must be expressed informally: it is always possible to dot the i’s and cross the t’s just as meticulously as one would in any other argument. In practice there may be less need to do so, however, since $-$ the basic idea being relatively clear $-$ the reader who wishes to do so is likely to be able to fill in those details. | 2013-05-24T20:27:26 | {
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# In a rectangular coordinate system, which of the following
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In a rectangular coordinate system, which of the following points is intersected by the line connected by the coordinates (5,6) and (21,18)?
A. (9,9)
B. (12,12)
C. (13,13)
D. (12,13)
E. (16,15)
Originally posted by pzazz12 on 20 Oct 2010, 01:55.
Last edited by Bunuel on 22 Sep 2012, 09:03, edited 1 time in total.
Edited the question.
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Re: In a rectangular coordinate system ............ [#permalink]
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20 Oct 2010, 02:05
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pzazz12 wrote:
In a rectangular coordinate system ,which of the following points is intersected by the line connected by the coordinates (5,6) and(21,18)?
A.(9,9)
B.(12,12)
C.(13,13)
D.(12,13)
E.(16,15)
The slope of a line passing through the points (5,6) and (21,18) is "rise over run": $$\frac{18-6}{21-5}=\frac{3}{4}$$.
The slope of a line passing through the point (5,6) and one of the points from the answer choice must also be $$\frac{3}{4}$$ (as it's the same lin), so $$\frac{y-6}{x-5}=\frac{3}{4}$$.
Answer choice A works: $$\frac{9-6}{9-5}=\frac{3}{4}$$.
For more on these issues check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html
Hope it helps.
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20 Oct 2010, 02:15
pzazz12 wrote:
In a rectangular coordinate system ,which of the following points is intersected by the line connected by the coordinates (5,6) and(21,18)?
A.(9,9)
B.(12,12)
C.(13,13)
D.(12,13)
E.(16,15)
Equation of the line connecting the two points,(5,6) and (21,18);
y-6= {(6-18)/(5-21)}(x-5) or , 4y-3x=9
Only ( 9,9) is satisfying the equation.Hence answer is A.
Consider KUDOS if its helpful to u .Thanks.
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In a rectangular coordinate system, which of the following [#permalink]
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31 Dec 2014, 09:24
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I found the equation of the line to be $$y = \frac{3x}{4} + \frac{9}{4}.$$. Then, noticing that 3 of the options had the same value for x and y, I wanted to check at which point on the line the x and y values were equal. So I solved for x:
$$x = \frac{3x}{4} + \frac{9}{4}$$
$$\frac{x}{4} = \frac{9}{4}.$$
$$x = 9.$$
Therefore, (9,9) is a point on the line and our answer is A.
Further: If we would have found some other value (one that wasn't listed on the answer choices) then we at least could have eliminated 3 of the answer choices and then manually checked the other two.
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08 Mar 2017, 01:19
equation of line: y-6 = {(18-6)/(21-5)} (x-5)
rearranging we get 4y-3x = 9. Put the co ordinates, given in the option, in the equation.
we will get (9,9) as answer
Option A
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Re: In a rectangular coordinate system, which of the following [#permalink]
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10 Mar 2017, 09:35
pzazz12 wrote:
In a rectangular coordinate system, which of the following points is intersected by the line connected by the coordinates (5,6) and (21,18)?
A. (9,9)
B. (12,12)
C. (13,13)
D. (12,13)
E. (16,15)
Let’s first create the equation of a line for coordinates (5,6) and (21,18).
Slope = Δy/Δx = 12/16 = 3/4.
We can substitute in either ordered pair to calculate the y-intercept. Let’s use (5,6):
y = (3/4)x + b
6 = (3/4)(5) + b
6 = 15/4 + b
24/4 = 15/4 + b
9/4 = b
Thus, y = (3/4)x + 9/4
Let’s now substitute each set of coordinates to see which holds true in our equation.
A) (9,9)
9 = (3/4)(9) + 9/4 ?
9 = 27/4 + 9/4 ?
9 = 36/4 ?
9 = 9 ? → Yes!
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Re: In a rectangular coordinate system, which of the following [#permalink]
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31 Aug 2017, 23:56
Can someone solve this using graph?
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Re: In a rectangular coordinate system, which of the following [#permalink]
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29 Mar 2018, 15:37
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1
pzazz12 wrote:
In a rectangular coordinate system, which of the following points is intersected by the line connected by the coordinates (5,6) and (21,18)?
A. (9,9)
B. (12,12)
C. (13,13)
D. (12,13)
E. (16,15)
One approach is to first find the slope
Slope = (18 - 6)/(21 - 5) = 12/16 = 3/4
IMPORTANT: if the slope is 3/4, then we can start at (5,6) and move up 3 spaces and then move right 4 spaces to find other points on the line (with INTEGER values)
Start at: (5,6)
Move 3 units up and 4 units to the right to get to the point (9,9) (check answer choices - it's there!)
Cheers ,
Brent
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Re: In a rectangular coordinate system, which of the following &nbs [#permalink] 29 Mar 2018, 15:37
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# In a rectangular coordinate system, which of the following
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https://brilliant.org/discussions/thread/squares-and-primes/ | # Squares and primes
A few questions I've been wondering about regarding primes and squares that I thought the Brilliant community might have some insight on ...
(i) Are there an infinite number of perfect squares that are the averages of consecutive primes?
As examples, $4$ is the average of consecutive primes $3$ and $5$, and $9$ is the average of consecutive primes $7$ and $11$. $25$ is not such a perfect square as its "neighboring" primes are $23$ and $29$, which average to $26$.
The list of such perfect squares begins as
$4, 9, 16, 64, 81, 144, 225, 324, 441, 625, 1089, 1681, 2601, ...$.
It becomes less and less likely as the squares get larger that they will be the average of successive primes, but the notion that there is a largest such square seems unlikely. So the problem is to either prove that there is no such largest square, or prove that there must be one and in fact identify it.
(ii) Can every perfect square $n^{2} \gt 1$ be expressed as the average of two (not necessarily consecutive) primes?
While $25$ is not the average of two consecutive primes, it is the average of primes $19$ and $31$. $121$ is not the average of consecutive primes, but it is the average of primes $103$ and $139$.
(iii) How many perfect squares are the averages of two or more distinct pairs of primes?
$9$ is the average of both prime pairs $(7,11)$ and $(5,13)$. $64$ is the average of prime pairs $(61,67), (31,97)$ and $(19,109)$. So if we define $P(n)$ as the number of distinct prime pairs $(p,q)$ for which $n^{2} = \dfrac{1}{2}(p + q)$ then, for example, $P(3) = 2$ and $P(8) = 3$. With this definition, question (ii) becomes a matter of whether or not $P(n) \ge 1$ for all $n \gt 1$, and question (iii) becomes a matter of how many integers $n \gt 1$ there are such that $P(n) \ge 2$. Also, is there a maximum value for $P(n)$ over all integers $n$?
I'm not sure if these are open questions or have in fact been solved centuries ago, so I thought I might learn a few things by sharing them with the community. Enjoy!
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Identifying the largest square seems impossible , but I think we can prove that there are infinite squares ...
- 4 years, 2 months ago
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Very interesting. In the second problem, can we extend the problem to whether every(or an infinite sequence of numbers and if there is not an infinite sequence, then how many numbers can be expressed in that form) number can be expressed as an average of two primes?
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Yes, that extension is in essence the Goldbach conjecture, which is still an open question. As stated in the link, the conjecture is that every even integer $2n, n \ge 2,$ can be expressed as the sum of two primes $p,q$, i.e. that $2n = p + q \Longrightarrow n = \dfrac{p + q}{2}$. This last equation mirrors your extension, i.e., that every integer $n \ge 2$ can be expressed as the average of two (not necessarily distinct) primes. I believe that we can drop the "not necessarily distinct" qualifier if we restrict ourselves to $n \ge 4$. My function $P(n)$ is a measure of the number of Goldbach partitions. (Before I posted this note I had never paid any attention to the Goldbach conjecture, but now that I've made the connection I'm becoming aware of just how much research has been done on this problem and its many variations.)
Goldbach's weak conjecture has a still-to-be-verified proof via Harald Helfgott, (2013). So I guess we could could refer to my questions as a set of "Goldbach's perfect square conjectures". These are weaker than the original conjecture but stronger, (I think?), than the weak conjecture.
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http://mathematica.stackexchange.com/questions/8744/efficient-langevin-equation-solver/8747 | Efficient Langevin Equation Solver
This question is not about good algorithms for solving stochastic differential equations. It is about how to implement simple codes in Mathematica efficiently exploiting Mathematica's programming methodology. (Hopefully, this may be useful in a stochastic processes course, for instance).
A simple Langevin Eq. in a single random variable $X$ with additive noise reads $$\dot{X} = f(X) + \zeta(t)$$ where $f(X)$ is an arbitrary function and $\zeta(t)$ is a Gaussian white noise satisfying $$E(\zeta(t)) = 0, \qquad \text{and} \qquad E(\zeta(t) \zeta(t')) = \Gamma \delta(t-t')$$
To solve it we discretize time as $t = n dt$ and write $$X_{n+1} = X_{n} + f(X_n)dt + \sqrt{\Gamma dt}\xi_n$$ where $\xi_n \sim N(0,1)$.
Here is my best implementation thus far:
Langevin[x0_, f_, G_, tf_, n_, m_: 1] :=
With[{dt = N[tf/n], s = N[Sqrt[ tf G/n]], xx0 = Table[x0, {m}]},
Transpose@NestList[ # + dt f[#] + RandomVariate[NormalDistribution[0, s], m] &, xx0, n]];
It takes as input a initial condition $x_0$, a function $f[x]$, the spectral density $\Gamma$ (here written as $G$), the final integration time $t_f$ and the number of integration points $n$. The time step is then $dt = t_f/n$. It also takes an optional argument $m$ corresponding to the number of realisations.
The output consists of $m$ vectors $(X_0, X_1, X_2, \ldots,X_n )$ representing the stochastic realisations.
Here is this program applied to the famous bi-stable potential given by $V(x) = -\frac{x^2}{2} + \frac{x^4}{4}$, so that $f(x) = - V'(x) = x-x^3$. It simulates a cold ($\Gamma=0.1$, in data1) and a hot ($\Gamma=1$, in data2) condition:
First@AbsoluteTiming[
data1 = Langevin[0, -#^3 + # &, 0.1, 10, 10^3, 2000];
data2 = Langevin[0, -#^3 + # &, 1, 10, 10^3, 2000];]
0.317665
To analyse the steady state I discard some initial points (80% in this example). This shows how the particle remains distributed close to the potential minima when it's cold, but spread out when it's hot:
Show[
Histogram[{Flatten[data1[[All, 800 ;; 1000]]], Flatten[data2[[All, 800 ;; 1000]]]}, Automatic, "PDF"],
Plot[-z^2/2 + z^4/4, {z, -1.8, 1.8}, PlotStyle -> Red],
AxesOrigin -> {0, 0}, PlotRange -> {-0.3, 1.2}]
Now to the questions:
1. Any immediate improvements on this function?
2. Is there a better way through a different approach
3. Can I compile this function as is to gain speed?
A follow up would be to extend all this to systems of Langevin equations, replacing $X$, $f$ and $\zeta$ by vector valued functions. But, then, we loose the advantage of computing many realisations at once within the same NestList. I'll think more about this problem and if I come up with any ideas I'll update the question.
Thank you all in advance and I hope this may be of use to other researchers as well.
Note: here is an example using the idea of @R.M.: generate all random numbers at once and use an index through the iteration to move along:
LangevinBad[x0_, f_, G_, tf_, n_, m_: 1] := Block[{i = 1},
With[{dt = N[tf/n], r = RandomVariate[ NormalDistribution[0, N[Sqrt[ tf G/n]]], {n, m}], xx0 = Table[x0, {m}]},
Transpose@NestList[ # + dt f[#] + r[[i++]] &, xx0, n]]];
Maybe my coding is no good, but this version is really bad. Actually; Nest probably has an internal variable to keep track of what iteration step it is, but I have no idea if it is possible to access that.
ACL's version
@ACL came up with a really efficient code, which I copy here for completeness.
(* This was originally called l4 by ACL *)
LangevinACL[fn_] := With[{f = fn},
Compile[{{x0, _Real}, {G, _Real}, {tf, _Real}, {n, _Integer}},
Module[{dt, s, state, r}, dt = N[tf/n];
s = N[Sqrt[tf G/n]];
state = ConstantArray[0., n];
state[[1]] = x0;
r = RandomVariate[NormalDistribution[0, s], n];
Do[state[[nc]] = state[[nc - 1]] + dt*f@state[[nc - 1]] + r[[nc - 1]], {nc, 2, n}];
state], CompilationTarget -> "C"]]
Then to compile for a given function use
ll = LangevinACL[(# - #^3) &];
AbsoluteTiming[dat = Table[ll[0, .1, 10, 10^3], {2000}];]
This code is always faster then the originally posted and allows for easy parallelisation.
Vector Equations
In vector equations there are two possibilities; either all particles have the same fluctuating properties, in which case we usually write $E(\zeta_i(t) \zeta_j(t')) = \Gamma\delta_{i,j} \delta(t-t')$ for the components of the fluctuating vector; or each particle has a specific fluctuation: $E(\zeta_i(t) \zeta_j(t')) = \Gamma_{i,j} \delta(t-t')$, where $\Gamma_{i,j}$ are the entries of a covariance matrix.
Here are two implementations of the former (all equations with the same fluctuation).
The first is a simple variation of the original code, as suggested by @ACL, so that instead of computing several realisations at once, each function call evaluates only a single realisation, but for a vector system:
LangevinVec[x0_, f_, G_, tf_, n_] :=
With[{dt = N[tf/n], s = N[Sqrt[ tf G/n]], m = Length@x0},
NestList[ # + dt f[#] + RandomVariate[NormalDistribution[0, s], m] &, x0, n]];
Everything is exactly as in Langevin, except that on input $x_0$ should be an array of numbers. Note also that there is no failsafe to check if the function $f$ has the correct dimensionality! (it should be a mapping from $\mathbb{R}^m\rightarrow\mathbb{R}^m$, where $m$ is the length of $x_0$).
The second implementation is again motivated by @ACL's code:
LangevinVecACL[fn_] :=
With[{f = fn},
Compile[{{x0, _Real, 1}, {G, _Real}, {tf, _Real}, {n, _Integer}},
Module[{dt, s, state, r, m},
m = Length@x0;
dt = N[tf/n];
s = N[Sqrt[tf G/n]];
state = ConstantArray[0., {n, m}];
state[[1]] = x0;
r = RandomVariate[NormalDistribution[0, s], {n, m}];
Do[state[[nc]] = state[[nc - 1]] + dt*f@state[[nc - 1]] + r[[nc - 1]], {nc, 2,
n}];
state], CompilationTarget -> "C"]]
Now to applications. Here is a model of ferromagnetism reminiscent of the 1D Ising system. There are $m$ random variables in $\vec{x} = (x_1,x_2,\ldots,x_m)$ representing spins in a linear chain of atoms. The interaction potential is given by
$$V(\vec{x}) = - \sum_{i=1}^m (\frac{a x_i^2}{2} - \frac{b x_i^4}{4}) - c \sum_{i=1}^m x_i x_{i+1}$$ This refers to a bi-stable potential (as in the previous example) for each variable representing the magnetic order, plus a harmonic-type interaction between them. The corresponding force is $$f_i = a x_i - bx_i^3 + c(x_{i-1}+x_{i+1})$$
In matrix notation I can write $$f(\vec{x}) = A\vec{x} - b \vec{x}^3$$ where $\vec{x}^3$ stands for $(x_1^3,x_2^3,\ldots)$ and $A$ is an $m\times m$ tridiagonal matrix with of the form $$A = \left( \begin{array}{ccccc} a & c & 0 & 0 & c \\ c & a & c & 0 & 0 \\ 0 & c & a & c & 0 \\ 0 & 0 & c & a & c \\ c & 0 & 0 & c & a \end{array} \right)$$ Note that I am using periodic boundary conditions $x_{m+1}=x_1$ and thence the c's in the upper-right and lower-left corners.
Here is $f(x)$ in Mathematica
m = 100; a = 2.0; b = 3.0; c = 3;
A = SparseArray[{
{m, 1} -> c, {1, m} -> c,
Band[{1, 1}] -> a, Band[{2, 1}] -> c, Band[{1, 2}] -> c},
{m, m}];
f[x_] := A.x - b x^3
The choice of parameters is somewhat arbitrary and perhaps this definition of $f(x)$ is not the fastest due to the dot product.
I will use ACL's version LangevinVecACL, which is faster. So I first compile it
llvec = LangevinVecACL[f];
Here are two data sets for $\Gamma = 0.01$ (pretty cold) and $\Gamma = 10$ (pretty hot).
x0 = ConstantArray[0.0, m];
AbsoluteTiming[
data1 = llvec[x0, .01, 4, 10^4];
data2 = llvec[x0, 10, 4, 10^4];
]
The following code shows the steady-state distribution of a single realisation
GraphicsGrid[{Map[
ListPlot[#, PlotRange -> {{-1, m + 1}, {Floor@Min@#, Ceiling@Max@#}},
Filling -> Axis, Frame -> True, BaseStyle -> 14,
FrameLabel -> {"Position", "Magnetization"}] &,
{Last@data1, Last@data2}]}, ImageSize -> {600}]
As can be seen, at cold temperatures the system tends to divide itself into domains with all spins chunked either "up" or "down"; conversely, at high temperatures the domain configuration is clearly degraded.
The following function animates the time evolution of the system.
animateSpinChain[data_] := Animate[ListPlot[data[[i]],
PlotRange -> {{-1, m + 1}, {Floor[Min[data]], Ceiling[Max[data]]}},
Filling -> Axis], {i, 1, Length@data, Floor[Length@data/100]}]
-
That's a very nicely written question. Wish more were like this... – rm -rf Jul 26 '12 at 16:17
An immediate generalization to a system of equations would be nDLangevin[x0_, f_, covMat_, tf_, n_, m_: 1] := With[{nDim = Length[x0], mean = ConstantArray[0, Length[x0]], dt = N[tf/n], xx0 = Table[x0, {m}], nDf = Function[x, f[#] & /@ x]}, Transpose@NestList[# + dt nDf[#] + RandomVariate[MultinormalDistribution[mean, covMat], m] &, xx0, n]] but it's terribly slow. – b.gatessucks Jul 26 '12 at 16:49
I don't think it can be parallelized, because each step depends on the previous step. Parallelization works well when the steps are independent of each other (e.g. Map, Do, etc.). I think your NestList approach is very clean and efficient. An equivalent formulation would be using memoization and recursive functions, but that is about 2x slower in my tests. All the functions used are compilable, but I'm not sure how to handle the case of arbitrary f. If f is known in advance, then you could easily compile it for that f. – rm -rf Jul 26 '12 at 17:02
@R.M it can be parallelized if you calculate each realization independently (I ended up doing this when I was working on stochastic processes a few years back) – acl Jul 26 '12 at 17:17
You can use JIT-compilation, but my tests show that this only improves speed by about 10-15 percents. The running time seems to be dominated by the random number generation, which, for your example, is done in chunks large enough so that compilation does not bring much speed improvement. So, your function is pretty efficient, both because you generate random numbers in large enough chunks, and because NestList auto-compiles. It may make more sense to compile it if, for example, you would use it with small m argument in a loop. – Leonid Shifrin Jul 26 '12 at 17:29
The first thing I'd try is to compile this, and also split it up so each realization is done independently (to allow easy parallelization). Taking the mexican hat potential you mentioned in the question (see later for general potential):
l3 = Compile[
{
{x0, _Real},
{G, _Real},
{tf, _Real},
{n, _Integer}
},
Module[{dt, s, state, r}, dt = N[tf/n];
s = N[Sqrt[tf G/n]];
state = ConstantArray[0., n];
state\[LeftDoubleBracket]1\[RightDoubleBracket] = x0;
r = RandomVariate[NormalDistribution[0, s], n];
Do[
state\[LeftDoubleBracket]nc\[RightDoubleBracket] =
state\[LeftDoubleBracket]nc - 1\[RightDoubleBracket] +
dt*(# - #^3) &@
state\[LeftDoubleBracket]nc - 1\[RightDoubleBracket] +
r\[LeftDoubleBracket]nc - 1\[RightDoubleBracket],
{nc, 2, n}
];
state
],
CompilationTarget \[Rule] "C"
];
(just paste it and it will look better). Note that I produce the noise once, put it in a list, and then just access each element as needed, to avoid the overhead of calling RandomVariate many times (I may have messed something up here, I haven't checked if the moments are the same as yours for the same potential, but I think it's OK).
This seems slightly faster than the original version (without parallelization):
AbsoluteTiming[data1 = Langevin[0, -#^3 + # &, 0.1, 10, 10^3, 2000];]
(*0.245488*)
dat = Table[l3[1, .1, 10, 10^3], {2000}]; // AbsoluteTiming
(*{0.153759, Null}*)
and it can be easily parallelized (but I have not tried as my laptop has only two cores; you'd need a lot of cores for this to be worth it).
Generalizing to discretised partial SDEs (or systems of Langevin equations) is straightforard with this approach (I could provide code if you want). Obviously, you sacrifice the generality of accepting any f to be able to compile, which is a disadvantage.
Finally, I would suggest you avoid NestList for this. It's clean, but if you want to do this for a largish system of SDEs, and want eg 10^6 steps with eg 10^4 realizations, there's no way you'll be able to produce them at once, as you do. It simply doesn't scale. It's also more difficult to keep only the last N results, if you want to wait some time for the system to equilibrate.
EDIT: To compile this with arbitrary potential f, use
l4[fn_] := With[
{f = fn},
Compile[
{
{x0, _Real},
{G, _Real},
{tf, _Real},
{n, _Integer}
},
Module[{dt, s, state, r}, dt = N[tf/n];
s = N[Sqrt[tf G/n]];
state = ConstantArray[0., n];
state\[LeftDoubleBracket]1\[RightDoubleBracket] = x0;
r = RandomVariate[NormalDistribution[0, s], n];
Do[
state\[LeftDoubleBracket]nc\[RightDoubleBracket] =
state\[LeftDoubleBracket]nc - 1\[RightDoubleBracket] +
dt*f@state\[LeftDoubleBracket]nc - 1\[RightDoubleBracket] +
r\[LeftDoubleBracket]nc - 1\[RightDoubleBracket],
{nc, 2, n}
];
state
],
CompilationTarget \[Rule] "C"
]
]
and then
ll = l4[(# - #^3) &]
AbsoluteTiming[dat = Table[ll[0, .1, 10, 10^3], {2000}];]
(*{0.138822, Null}*)
However, Nest tries to automatically compile its first argument if it is nested over a number of times (100 by default) visible in SystemOptions[CompileOptions], so this may not be worth much effort for runs as small as these.
-
quizz: what package was used for figs 2 and 3 of this? – acl Jul 26 '12 at 18:15
acl, your timing is the same as mine i Think. In the first one you compute two simulations (data1 and data2) and in the second one you compute only one. Both have similar run times, and given the simplicity of NestList + the ability to accept any function, I don't really see much advantage. – Gabriel Landi Jul 26 '12 at 20:04
Yes you're right, I screwed up. Let me fix it later. The only advantage is that it is trivially parallelizable and it scales much easier (at least, for me). Try to solve the discretised KPZ equation (which is notorious for slow convergence) for large systems and thousands of runs and you will see what I mean. For obtaining intuition, though, I agree that the Nest approach is neater. In practice, anyway, for something as simple algorithmically as this, C is much faster and not very hard to code if performance is needed. – acl Jul 26 '12 at 20:21
@acl is there a prize...? :-) – sebhofer Jul 26 '12 at 20:47
@sebhofer a copy of the package :) – acl Jul 26 '12 at 20:51
I get a very small, but consistent improvement in time to calculate data1 and data2 by replacing With with Module:
Langevin[x0_, f_, G_, tf_, n_, m_: 1] := Module[{dt, s, xx0},
dt = N[tf/n];
s = N[Sqrt[tf G/n]];
xx0 = Table[x0, {m}];
Transpose@
NestList[# + dt f[#] +
RandomVariate[NormalDistribution[0, s], m] &, xx0, n]];
First@AbsoluteTiming[
data1 = Langevin[0, -#^3 + # &, 0.1, 10, 10^3, 2000];
data2 = Langevin[0, -#^3 + # &, 1 , 10, 10^3, 2000];]
but on my machine it only gives the difference between: 0.493533 and 0.436566
You can speed up the generation of the histograms using ParallelMap:
AbsoluteTiming[
histograms =
ParallelMap[
Histogram[#, Automatic, "PDF"] &, {Flatten[data1[[All, 800 ;; 1000]]],
Flatten[data2[[All, 800 ;; 1000]]]}];
Show[
histograms[[1]],
histograms[[2]],
Plot[-z^2/2 + z^4/4, {z, -1.8, 1.8},
PlotStyle -> Red],
AxesOrigin -> {0, 0},
PlotRange -> {-0.3, 1.2}
]]
but I don't think 3 or more than kernels will help over 2 (I'd love to be wrong on this).
I wonder if pre-calculating all of your RandomVariates might gain you a bit. I'll try something and update this answer if I can block out some time.
...
Responding to the OP's comment below and given that I may have missed what they meant in the comment did you mean something like the following when you wrote:
I thought about using a single RandomVariate. But I couldn't really figure out how to efficiently Nest that.:
Langevin2[x0_, f_, G_, tf_, n_, m_: 1] := Module[{dt, s, xx0, r},
dt = N[tf/n];
s = N[Sqrt[tf G/n]];
xx0 = Table[x0, {m}];
r = RandomVariate[NormalDistribution[0, s], m];
Transpose[
NestList[
# + dt f[#] + r &,
xx0,
n]
]
];
Faster, but given your comment and the original question I'm not certain it does what you need.
-
Hi Jagra. I thought about using a single RandomVariate. But I couldn't really figure out how to efficiently Nest that. – Gabriel Landi Jul 26 '12 at 17:36
I don't think so. This only generates the random numbers once. $r$ must be re-computed within each iteration. Ideally this would be done with r = RandomVariate[NormalDistribution[0,s],{m,n}]. But then you can't really nest this matrix; or at least I don't really know how. – Gabriel Landi Jul 26 '12 at 17:54
Why not r := RandomVariate[NormalDistribution[0, s], m] ? – b.gatessucks Jul 26 '12 at 17:59
@GabrielLandi You can probably use Fold then, and run an index through the pre-generated RandomVariate... – rm -rf Jul 26 '12 at 18:06
@GabrielLandi -- Have you thought about using FoldList and then used something like r = RandomVariate[NormalDistribution[0, s], {m,m}]? – Jagra Jul 26 '12 at 18:18 | 2013-12-06T06:56:46 | {
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https://math.stackexchange.com/questions/2318540/probability-of-a-survivor-in-chick-pecking-tournament | # Probability of a survivor in chick-pecking tournament
This is a follow up to this question: Suppose that $n$ chicks are arranged in a circle. Every chick randomly pecks either the chick to their right or the chick to their left. By the other question, the expected number of unpecked chicks is $n/4$. Instead of ending there, make a tournament out of it. Remove the pecked chicks from the circle and repeat the experiment with the remaining chicks. Iterate as long as possible. It is easy to see that the process ends with either $0$ or $1$ remaining chick.
Question: Let $p(n)$ denote the probability that the process ends with $1$ chick. What can be said about $\lim_{n \rightarrow \infty}p(n)$?
The following graph shows the result of Monte Carlo simulations which estimate $p(n)$ for all $n$ in the range $1$ to $1000$ (and using $1000$ tournaments for each $n$).
The wave-like nature of the graph is interesting. To get a better handle on it, we need the exact probabilities. The following is based on a nice formula by @6005 in the comments to this answer to the other question: $p(0) = 0$ and $p(1) = 1$. For any $n \geq 2$ we have:
$$p(n) = \begin{cases} \sum_{k=0}^\frac{n}{2} \left(\frac{\binom{n}{2k} + (-1)^k \binom{n/2}{k}}{2^{n-1}}\right) p(k) & \text{if n is even} \\ \sum_{k=0}^\frac{n-1}{2} \left(\frac{\binom{n}{2k}}{2^{n-1}}\right)p(k) & \text{otherwise}. \end{cases}$$
The following shows the graph of $p(n)$ from $n=1$ to $1000$:
The local maxima and minima appear approximately at powers of $2$ (sometimes shifted by $1$). The maxima are at powers of $2$ which are also powers of $4$ and the minima at the other powers of $2$ (hence of the form $2 \cdot 4^k$). This is somewhat intuitive given that the expected number of survisors in a single round is $\frac{n}{4}$. Furthermore, this expected value is also the most likely value (in the case that $n$ is a multiple of $4$). For example if you start with $128$, the first round could be expected to get you to $32$, the next round to $8$, from thence to $2$, which prompty peck each other, leaving you with $0$. This is a hueristic way of reasoning that becomes somewhat less plausible with each factor of $4$.
So the question: Does $\lim_{n\rightarrow \infty} p(n)$ exist, and, if so, to what?
My conjecture is that the observed oscillations get damped in the limit, and that the resulting limit is $0.5$, but I do not know how to compute such limits.
• My empirical calculations up to $2^{15}=32768$ suggest that your conjecture may be false, with the extremes possibly remaining outside $0.5434$ and $0.4037$ – Henry Jun 11 '17 at 17:55
As an expansion of my comment, on an empirical basis I suspect your conjecture may be false, from looking up to $2^{15}=32768$
You seem to have used R for your graphs, so here is my attempt:
c2a <- function(n,k){ exp( lchoose(n,2*k)-(n-1)*log(2) )}
c2b <- function(n,k){ (-1)^k * exp( lchoose(n/2,k)-(n-1)*log(2) )}
maxn <- 2^15
prob <- 1
for (n in 2:maxn){
if (n %% 2 == 0){
prob [n] <- sum( (c2a(n,1:(n/2)) + c2b(n,1:(n/2))) *
prob[1:(n/2)] )
}else{
prob [n] <- sum( (c2a(n,1:((n-1)/2))) *
prob[1:((n-1)/2)] )
}
}
A graph with a log scale looks like
plot(1:maxn, prob, log="x")
and looking at the values at powers of $2$ suggests that the peaks and troughs may have separate limits
> prob[4^(1:7)]
[1] 0.5000000 0.5571170 0.5468982 0.5442833 0.5436387 0.5434781 0.5434380
> prob[2*4^(1:7)]
[1] 0.3359375 0.3950479 0.4016141 0.4031632 0.4035449 0.4036400 0.4036638
while taking a weighted average of the probabilities across a cycle from $m$ through to $4m-1$ suggests a central figure something like $0.475$, which is marginally more than the average of the apparent limits for powers of $2$
wav <- numeric()
for (m in 1:(maxn/4)){
wav[m] <- sum(prob[m:(4*m-1)]/(m:(4*m-1))) / sum(1/(m:(4*m-1)))
}
plot(1:(maxn/4), wav, log="x", ylim=c(0.474, 0.476))
• Very nice (+1). The apparent fact that each peak and valley is slightly closer to each other than the previous peak/valley is what led me to conjecture to that there might be a common limit, though your evidence (especially the log-scale graph) suggests that if so it will only appear with truly enormous numbers. Perhaps the conjecture is wrong. It was more a hunch than anything. – John Coleman Jun 11 '17 at 23:32 | 2020-06-05T10:17:17 | {
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https://mathhelpboards.com/threads/finding-the-equation-of-a-tangent-line-at-a-given-point-tan-and-e-x-questions.6107/ | # Finding the equation of a tangent line at a given point (tan and e^x questions)
#### czzzar
##### New member
View attachment 1213
I don't need an answer, I'm just stumped as to how to properly turn these into point slope form to find the tangent, can anyone guide me through this? Help would be greatly appreciated.
I believe I understand the formulas that are used to solve problems such as these.
It starts by finding the derivative of the function.
After finding the derivative, I can plug in the given points and slope from the derivative into the point slope formula... but when i did this, my answer was wrong. I feel like I'm missing a key idea (or more) that is stopping me from finding this out
Last edited:
#### Chris L T521
##### Well-known member
Staff member
View attachment 1213
I don't need an answer, I'm just stumped as to how to properly turn these into point slope form to find the tangent, can anyone guide me through this? Help would be greatly appreciated.
I believe I understand the formulas that are used to solve problems such as these.
It starts by finding the derivative of the function.
After finding the derivative, I can plug in the given points and slope from the derivative into the point slope formula... but when i did this, my answer was wrong. I feel like I'm missing a key idea (or more) that is stopping me from finding this out
It seems to me that you have the right formula. The tangent line to $f(x)$ at $x=a$ is given by the equation $y=f(a)+f^{\prime}(a)(x-a)$. In the case of the first problem, $f(x)=\tan x$, and you're given the point $(a,f(a)) = \left(\frac{\pi}{4},1\right)$. With that, the equation of the tangent line is of the form
$y=f\left(\frac{\pi}{4}\right) +f^{\prime}\left(\frac{\pi}{4}\right) \left(x-\frac{\pi}{4}\right) = 1 + f^{\prime}\left(\frac{\pi}{4}\right) \left(x-\frac{\pi}{4}\right).$ I'm sure you can find $f^{\prime}\left(\frac{\pi}{4}\right)$ on your own, right?
The same goes for the other question: for $f(x)=(x-1)e^x$ at the point $(1,0)$, the equation of the tangent is of the form
$y=f(1)+f^{\prime}(1)(x-1)= f^{\prime}(1)(x-1).$
I would assume here that you can find $f^{\prime}(1)$ as well.
I hope this clarifies things!
#### czzzar
##### New member
It seems to me that you have the right formula. The tangent line to $f(x)$ at $x=a$ is given by the equation $y=f(a)+f^{\prime}(a)(x-a)$. In the case of the first problem, $f(x)=\tan x$, and you're given the point $(a,f(a)) = \left(\frac{\pi}{4},1\right)$. With that, the equation of the tangent line is of the form
$y=f\left(\frac{\pi}{4}\right) +f^{\prime}\left(\frac{\pi}{4}\right) \left(x-\frac{\pi}{4}\right) = 1 + f^{\prime}\left(\frac{\pi}{4}\right) \left(x-\frac{\pi}{4}\right).$ I'm sure you can find $f^{\prime}\left(\frac{\pi}{4}\right)$ on your own, right?
The same goes for the other question: for $f(x)=(x-1)e^x$ at the point $(1,0)$, the equation of the tangent is of the form
$y=f(1)+f^{\prime}(1)(x-1)= f^{\prime}(1)(x-1).$
I would assume here that you can find $f^{\prime}(1)$ as well.
I hope this clarifies things!
$f^{\prime}\left(\frac{\pi}{4}\right)$ i believe is 2 because the derivative of $f(x)=\tan x$ is $f(x)=\sec^2x$....
$y=1 + 2x - 2\frac{\pi}{4}.$ should pi/4 right be able to be multiplied? the answer to this question is 4x - 2y - pi +2 = 0.... It seems that I'm deviating away from the answer even more now...
#### Chris L T521
##### Well-known member
Staff member
$f^{\prime}\left(\frac{\pi}{4}\right)$ i believe is 2 because the derivative of $f(x)=\tan x$ is $f(x)=\sec^2x$....
$y=1 + 2x - 2\frac{\pi}{4}.$ should pi/4 right be able to be multiplied? the answer to this question is 4x - 2y - pi +2 = 0.... It seems that I'm deviating away from the answer even more now...
Yea, you should multiply them in order to simplify the equation. With that, you should have the tangent equation
$y=2x+1-\frac{\pi}{2}.$
However, your book goes the extra mile to get it in standard form (I don't know why they want you do get it into standard form, but what I have above is usually good enough). To get the book's answer, multiply both sides by 2 to get $2y=4x+2-\pi$ and then subtract $2y$ from both sides to get $4x-2y-\pi+2=0$.
I hope this makes sense! | 2020-10-20T00:06:42 | {
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https://math.stackexchange.com/questions/2555359/what-is-the-possibility-that-at-least-one-digit-will-not-show-up-in-a-20-digit | # What is the possibility that at least one digit will not show up in a 20-digit “code”?
A "code" is composed of 20 digits (numbers from 0 to 9), and we want to choose a number randomly. What is the possibility that at least one digit will not show up in the code?
What I did:
We have $10^{20}$ possibilities. Now, I want to choose 9 numbers out of the ten, and choose them randomly, so the possibility is:
$\frac{10\cdot9^{20}}{10^{20}} = \frac{9^{20}}{10^{19}}$
But when I put this in the calculator, I get $1.25\dots$ I thought maybe I need a more precise calculator, but even calculators I found in google returns the same answer.
The possibility isn't supposed to be above 1. What is the problem here?
• You're counting a lot of things twice. For example, you are counting the number 11111111111111111111 as "no twos", "no threes", "no fours" and so on. – 5xum Dec 7 '17 at 11:32
• Probability of a number not being $n$ is $9\over10$. Possibility in a 20-digit sequence is $\left({9\over10}\right)^{20}=0.12157...$ – MalayTheDynamo Dec 7 '17 at 11:33
• It's not your calculator. $9^{20} = 1.215\ldots \times 10^{19} > 10^{19}$. – Eric Towers Dec 7 '17 at 15:07
• This seems like a variant of the coupon collector's problem. – Necreaux Dec 7 '17 at 16:26
For any given digit, the probability the "code" does not contain that digit is indeed $(9/10)^{20}$. But computing the probability that any one of the ten digits is missing requires the inclusion/exclusion principle.
• Add $10×(9/10)^{20}$, like you did, for the probability that one digit is missing.
• Subtract $45×(8/10)^{20}$ for the probability that two digits are missing. 45 is the number of ways to choose two digits to omit.
• Add $120×(7/10)^{20}$ for the probability that three digits are missing. Again, 120 is the number of ways to omit three digits.
• Continue until adding $10×(1/10)^{20}$ for the probability that nine digits are missing.
The final, correct answer is $0.785262\dots$
• Doesn't the wording "..... a $certain$ digit will not show up ..." imply that a particular digit doesn't show up irrespective of other digits being present or absent ? – true blue anil Dec 7 '17 at 12:02
• @trueblueanil Meaning clarified. – Parcly Taxel Dec 7 '17 at 12:03
• Was that clarified by OP ?? – true blue anil Dec 7 '17 at 12:09
• I don't think it is right to edit OP's $question$. We can ask for clarifications, or mention that it is our interpretation. – true blue anil Dec 7 '17 at 12:12
• Well the answer was actually "a certain", but I think you answered for both "at least" and "a certain". So I got the other one for a bonus. Thanks :) – sheldonzy Dec 7 '17 at 12:18 | 2019-08-26T10:07:37 | {
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https://www.themathdoctors.org/10-calculus-says-so-or-not/ | # 1=0? Calculus Says So [or Not]
“False Proofs”, where seemingly good logic leads to nonsensical conclusions, can be a good way to learn the boundaries of reality — what to look out for when you are doing real math. We have a FAQ on the subject; there we discuss several well-known fallacies based in algebra, and have links to others. Today, I will look at some fallacies using differential and integral calculus.The FAQ quotes a summary I wrote:
Of course, these aren't really proofs, because they all have some error in them. What's important about these examples is that they show ways in which you can make a mistake in using math if you aren't careful enough. If you can understand where the error is, then you can look for the same kinds of errors in your own work, whether it's a proof for school or a calculation you make when you're designing a bridge. It also explains why mathematicians and scientists don't publish their results without first having others check them to make sure there isn't some subtle error in their calculations.
## Integration by parts
Let’s first look at a fallacy in integration, which teaches a very important lesson. Here is the question, from 2001:
1 = 0 Fallacy
Reading the Dr. Math pages - and especially the ones on 1 = 0 fallacies - I remembered a 'proof' we ran up against during high school (VWO in the Netherlands). It makes use of integral calculus.
We learned the following rule for 'partial integrating':
Int(f(x)*g(x))dx = f(x)*G(x) - Int(f'(x)*G(x))dx
with G(x): the primitive function of g(x) and f'(x): the derivative of f(x)
Now watch the following 'proof':
Int(1/x^2 * 2x)dx = 1/x^2 * x^2 - Int(-2/x^3 * x^2)dx (step 1)
this yields:
Int(2/x)dx = 1 - Int(-2/x)dx = 1 + Int(2/x)dx (step 2)
subtracting Int(2/x)dx on both sides yields:
0 = 1 (step 3)
Quite remarkable, I think!
We found two arguments that possibly explain the fallacy:
1) 1/x^2 * x^2 = 1 is an invalid step
2) we work with unbounded integrals. If we put lowerbound a
and upperbound b to the integral, we get for step 2:
Int(2/x)dx (a,b) = 1 (a,b) + Int(2/x)dx (a,b)
which yields:
Int(2/x)dx (a,b) = 1(b) - 1(a) + Int(2/x)dx (a,b)
because 1(b) = 1(a) = 1 we get:
Int(2/x)dx (a,b) = Int(2/x)dx (a,b)
which of course is true.
Which of these two arguments tackles the fallacy?
Recall that “integration by parts” uses the formula, $$\displaystyle\int u\ dv = uv\ – \int v du$$, or, equivalently, $$\displaystyle\int u\ v^\prime\ dx = uv\ – \int u^\prime\ v\ dx$$. It is explained here:
Choosing Factors When Integrating by Parts
What he has done here is to integrate$$\displaystyle\int \left(\frac{1}{x^2}\cdot 2x\right) dx$$ by parts, using $$u=\frac{1}{x^2}$$ and $$dv = 2x dx$$. (In his terms, $$f(x) = \frac{1}{x^2}$$, $$g(x) = 2x$$, so that $$G(x) = x^2$$.) Applying parts, we get:
$$\displaystyle\int \frac{1}{x^2}\cdot 2x dx = \frac{1}{x^2} \cdot x^2 – \int\frac{-2}{x^3} \cdot x^2 dx = 1 – \int\frac{-2}{x} dx = 1 + \int\frac{2}{x} dx$$
But the integral we started with simplifies to
$$\displaystyle\int \frac{2}{x} dx$$
Rather than evaluate this (getting $$2\ln{x}$$), we notice that these two ways of simplifying the integral imply that
$$\displaystyle\int \frac{2}{x} dx = 1 + \int\frac{2}{x} dx$$
Subtracting the integral from both sides, we have 0 = 1!
What went wrong? I replied,
Your second explanation is essentially right. I would say it is an "indefinite" integral, rather than "unbounded." When you work with indefinite integrals, you always have to keep in mind that an arbitrary constant can be added to the result, since differentiation of a constant yields zero. So what you really have is
Int(2/x)dx + C1 = 1 + Int(2/x)dx + C2
with a constant added to each side. This simplifies to
C1 = 1 + C2
which of course doesn't say much, since C1 and C2 could be anything. That eliminates the problem entirely.
Your method of turning the integrals into definite integrals amounts to the same thing; evaluating the constant at the limits makes it disappear, so you can ignore it.
This is a very important lesson to learn; for example, we often see students thinking that their answer for an integral is wrong because it doesn’t match the answer in the book, even after simplifying. The answer may be that the two answers differ by a constant. When this happens, the student ought to notice that the derivatives of the two answers are therefore the same (since the derivative of a constant is zero), so both are valid integrals. Here are some examples of this:
Calculus Constants
Constant Oversight
Putting it another way, by parts we got an answer of
$$1 + 2\ln{x} + C$$
while the direct method yielded
$$2\ln{x} + C$$
We just need different values of C to make them match.
## More about the constant of integration
We got a very similar question in 2003, with an even simpler integral:
Constant of Integration
Using integration by parts
integration of (1/x)dx = [x * (1/x)]+ integration of (1/x)dx
After simplifying by using the addition property of equality and multiplication, the answer would lead to 0 = 1, which should be wrong. The proof seems correct.
That is, taking $$u = \frac{1}{x}$$ and $$dv = dx$$, we get
$$\displaystyle\int \frac{1}{x}\ dx = x \cdot \frac{1}{x}\ – \int\frac{1}{x} dx$$
As before, subtracting the integral from each side, we are left with 0 = 1. Of course, we know the answer now; Doctor Jacques suggested trying the definite integral approach to clarify what was happening:
This looks like a paradox indeed, but try to see what you get if you evaluate the integral over an interval [a,b]...
Please feel free to write back if you are still stuck.
The student did that, but was not convinced about the indefinite form:
The problem is possible in the interval [a,b] but my teacher insists that there is a problem with the proof, while every one of us thinks the proof is correct.
Of course, a proof that 0 = 1 can’t be correct, unless all of math is wrong; presumably they just don’t see where the error is. So Doctor Jacques gave a brief explanation:
The problem is that an indefinite integral (antiderivative) is only defined up to an additive constant. More technically, it is not a single function, but an equivalence class of functions.
For example, INT(0 dx) = C, where C is any constant, since the derivative of a constant is 0.
In this case, we should have written:
(INT{dx/x} + C_1) = 1 + (INT{dx/x} + C_2)
and this merely shows that the constants must satisfy C_1 = 1 + C_2.
When you compute a "real" integral, i.e. between limits, these constants disappear.
“Defined up to an additive constant” means that answers may differ by that “$$+ C$$” that students are taught to write at the end by rote. So the real answer is not just the function you write, but all functions that can be obtained by using different numbers for the constant — an “equivalence class”.
This still left questions:
As I read your explanation I got confused with the constants. Isn't it that the constant of int(dx/x) on the left-hand side of the equation is equal to the constant at the right-hand side of the equation?
Writing “$$+ C$$” by rote leaves students not really understanding what it is! When I did that above ($$1 + 2\ln{x} + C$$ and $$2\ln{x} + C$$, using the same name C for the constant in each case), I had to consciously remind myself that C has to be different in each case. That is not obvious unless you think about it.
Doctor Jacques replied with a deeper explanation and a classic example:
These constants have no actual meaning - they are artificial.
When we write
INT{f(x)dx} = g(x)
we simply mean that the derivative of g(x) is f(x). Of course, the derivative of g(x) + C, where C is _any_ constant, is also f(x).
The particular constant that comes out depends on the method of integration. The whole point of the exercise is to show that different calculations can yield functions that differ by a constant.
We can even illustrate this with the function 1/x in another simpler way. We know that the "true" integral is ln(x).
Now, in INT{dx/x}, if we make the substitution ax = u, with a > 0, you will easily see that the result is
ln(ax) = ln(x) + ln(a) = ln(x) + constant
and, as we can take any positive number for a, we can make the constant ln(a) anything we wish.
There is no contradiction in writing:
INT{dx/x} = ln(x)
INT{dx/x} = ln(x) + C
because these are not true equalities between functions.
This is exactly the same as modular arithmetic. When we write
2 = 7 (mod 5)
the numbers 2 and 7 are not simple numbers. 2 represents all the numbers that are a multiple of 5 + 2, and 7 represents all the numbers that are a multiple of 5 + 7, and these sets of numbers are the same - that is what the equality means (in this case, we often use a special symbol instead of =, to mean congruence).
In a similar way, an expression like INT{f(x)dx} represents, not a single function, but the set of all functions whose derivative is f(x). An equality between integrals is an equality between sets of functions.
If you are not familiar with modular arithmetic, when we write $$2 \equiv 7 (\text{mod } 5)$$, it means that 2 and 7 are equivalent in the sense that their difference is a multiple of 5. They are both representatives of the same “equivalence class”, which consists of the numbers $$\{\dots , -8, -3, 2, 7, 12, \dots\}$$. The same idea applies to the indefinite integral: it is really the equivalence class of the function we write, meaning that we can add any constant to it and it will still be equivalent. That is what the “$$+ C$$” means.
## A fallacy in differentiation
Let’s move on to the other half of calculus. This question, from 2000, is a classic fallacy using differentiation:
Proof that 2 Equals 1 Using Derivatives
How can this be?
kx = x + x + ... + x (k-times) ......................[1]
xx = x + x + ... + x (x-times) ......................[2]
x^2 = x + x + ... + x (x-times) ......................[3]
dx(x^2) = 2x (diff. wrt x) ...........................[4]
dx(x + x + ... + x) = 1 + 1 + ... + 1 (x-times) ......[5]
so
2x = 1 + 1 + ... + 1 (x-times) {from eq. [4],[5]} ....[6]
so we have
2x = x ................................................[7]
so
2 = 1 (x <> 0) ......................................[8]
Thank you!
Akram starts with the (debatable!) fact that multiplication means repeated addition, letting x itself be the multiplier in order to get the square. Then he differentiates both sides of $$x^2 = x + x + \dots + x$$ to get $$2x = x$$, so that (dividing by x if it is non-zero), 2 = 1.
Alternatively, at the last step, one could “solve” $$2x = x$$ by subtracting x from both sides, yielding $$x = 0$$: that is, every number (since x was unspecified) is equal to zero. That would include 1 = 0.
What went wrong?
First, as I read it, when he differentiated (using a nonstandard notation “dx” apparently meaning “d/dx”) $$\underbrace{x + x + \dots + x}_{x\text{ times}}$$, he just differentiated each x to get 1, without considering that the number of terms is not constant. If you try to justify this by going to the definition of the derivative, you have to take the difference $$f(x + \Delta x)-f(x)$$, which here becomes the difference of sums of different numbers of terms. Doctor Rick took it from there:
In taking the difference, you forgot that not only has each term changed its value, but also the NUMBER of terms has changed. Let's put in some numbers to make this clear. Let x = 3 and delta(x) = 1. Then:
x^2 = 3 + 3 + 3
(x+delta(x))^2 = 4 + 4 + 4 + 4
delta(x^2) = 1 + 1 + 1 + 4
We still don't have the 2x that you expected; we've got 7 instead of 6. Why is this? You forgot something else. 2x is the DERIVATIVE of x^2 - the limit of delta(x^2)/delta(x) as delta(x) approaches zero. But the function as we have defined it (as a sum of x terms) has meaning only for integer values of x, so delta(x) can't be less than 1. The derivative is not defined. All we can define is a DIFFERENCE, as I have done (with delta(x) = 1, the smallest possible value), and this is not equal to 2x.
http://mathforum.org/dr.math/faq/faq.false.proof
At the bottom there is a link to "derivatives," an item in our archives that is directly related to your problem.
The reference at the bottom is to this answer by Doctor Rob:
Derivatives
Really, it’s hard to write something sensible when you try to see what is happening with specific numbers! The notation $$\underbrace{x + x + \dots + x}_{x\text{ times}}$$, for the specific numbers 3 and 4 has to mean $$\underbrace{3 + 3 + 3}_{3\text{ times}}$$ and $$\underbrace{4 + 4 + 4 +4}_{4\text{ times}}$$, respectively. But to take a derivative you have to be able to let x be 3.001, for example, as you take the limit; and it makes no sense to repeat something 3.001 times. The fact is that multiplication can only be thought of as repeated multiplication for whole numbers, so the foundation of the argument is faulty.
In the explanation you gave to show why the proof was wrong,
x^2 = 3 + 3 + 3
(x+delta(x))^2 = 4 + 4 + 4 + 4
delta(x^2) = 1 + 1 + 1 + 4
what is delta(x^2)?
There's a delta(x) = 1, and x = 3, but I thought delta(x) was a term of its own, not a function of x.
Doctor Rick clarified:
I took some shortcuts in my explanation, trying to correct the writer's notation without changing it too much. I'll go through it in a different way for you.
We're interested in finding the derivative of the function
f(x) = x^2
using the definition of x^2 as a sum of x copies of x. The claim was that you can differentiate
f(x) = x + x + x + ... + x (x times)
by taking the derivative of each term and adding:
df(x)/dx = 1 + 1 + 1 + ... + 1 (x times)
= x
In my explanation of why this is wrong, I can't really talk about derivatives, because the function has been defined only for integer values of x. Therefore I wrote in terms of finite differences: delta(x) is a finite (integer) change in x, and delta(x^2) is the change in x^2 due to this change in x. Normally we would use the Greek capital delta, and drop the parentheses around the x. You're right, it's not a function. Delta(x^2) is defined formally as follows:
delta(x^2) = f(x+delta(x)) - f(x)
That is, the derivative is defined as $$\displaystyle\frac{df(x)}{dx} = \lim_{x\rightarrow\Delta x}\frac{f(x + \Delta x)-f(x))}{\Delta x}$$. In this case, $$\displaystyle\frac{dx^2}{dx} = \lim_{x\rightarrow\Delta x}\frac{(x + \Delta x)^2-(x)^2)}{\Delta x}$$. Our $$\Delta x^2$$ is the numerator, $$(x + \Delta x)^2-(x)^2)$$.
You might prefer it if I talk in terms of independent variable x and dependent variable y:
[1] y = f(x) = x + ... + x (x times)
Let's remain general rather than choosing delta(x) = 1. For any value of x and any change in x, delta(x), we can evaluate f(x+delta(x)), which will differ from y = f(x) by an amount delta(y):
[2] y + delta(y) = (x+delta(x)) + ... + (x + delta(x)) (x+delta(x) times)
Subtract [1] from [2] to get delta(y):
delta(y) = delta(x) + ... + delta(x) (x times)
+ (x+delta(x)) + ... + (x+delta(x)) (delta(x) times)
It's that second line that the writer ignored. Applying the "definition" of (integer) multiplication, we get
delta(y) = x*delta(x) + delta(x)*(x+delta(x))
= 2x*delta(x) + delta(x)^2
Then
delta(y)/delta(x) = 2x + delta(x)
Written out, he has said that $$\Delta y = \underbrace{(x + \Delta x) + (x + \Delta x) + \dots + (x + \Delta x)}_{x + \Delta x\text{ times}} – \underbrace{(x + x + \dots + x)}_{x\text{ times}}$$ $$= \underbrace{\Delta x + \Delta x + \dots + \Delta x}_{x\text{ times}} + \underbrace{(x + \Delta x) + (x + \Delta x) + \dots + (x + \Delta x)}_{\Delta x\text{ times}}$$.
All this is done, of course, ignoring the fact that $$\Delta x$$ has to be an integer, so we can’t really take the limit.
If our function were defined on all real numbers rather than just integers, we would find the derivative by taking the limit of delta(y)/delta(x) as delta(x) approaches zero. The delta(x) term would go away, and we'd get the correct derivative. As it stands, you can see why (in my example in the original explanation) I got a difference of 7 instead of 6: there's an extra term delta(x)^2 = 1.
Another correspondent suggested that we "define" multiplication for non-integers like this (in my own notation):
x*y = x + ... + x ([y] times)
+ x*(y-[y])
where [y] is the greatest integer less than y. It's not very helpful, because it only defines multiplication by a number greater than 1 in terms of multiplication by a number less than 1 (namely, y-[y]). However, it does allow us to take delta(x) to zero. If you work through it, you'll find that the derivative works correctly.
Admittedly, much of this is really nonsense. But sometimes it is useful to examine nonsense to see why it is.
Incidentally, we have received at least two dozen questions equivalent to this one (going only by the number of times we referred to this answer). So this is not a rare issue!
### 1 thought on “1=0? Calculus Says So [or Not]”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-10-27T07:22:28 | {
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http://math.stackexchange.com/questions/209412/how-can-i-solve-this-limit | # How can I solve this limit?
$\lim_{x\to 1} \frac{\sin (x-1)}{x-1}$
I know the answer equals $1$ because $\lim_{x\to 0} \frac{\sin (x)}{x} = 1$ and in the following question $x-1$ gets arbitrary close to 0 so the same thing is happening. What I need is some steps to basically show that the question was not solved by a calculator.
I tried to use $\sin(A-B) = \sin A \mathrm{cos}B - \sin B \cos A$ but I had a $\frac {0}0$ which is obviously wrong. Any help/tip would be great.
-
Maybe try using sin series expansion? Or L'Hospital – Stefan Oct 8 '12 at 19:23
Try substitution $t=x-1$. – M. Strochyk Oct 8 '12 at 19:25
You could simply write down pretty much just what you’ve written here: it shows that you understand why the limit is $1$. However, a nice way is to make a substitution $y=x-1$; then clearly
$$\lim_{x\to 1}\frac{\sin(x-1)}{x-1}=\lim_{y+1\to 1}\frac{\sin y}y=\lim_{y\to 0}\frac{\sin y}y=1\;.$$ | 2016-05-26T14:43:48 | {
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https://dsp.stackexchange.com/questions/41466/fourier-transform-identities/41471 | # Fourier Transform Identities
We know the below,
$$\mathscr{F}\big\{x(t)\big\}=X(f) \tag{1}$$ $$\mathscr{F}\big\{x(-t)\big\}=X(-f) \tag{2}$$ $$\mathscr{F}\big\{x^*(t)\big\}=X^*(-f) \tag{3}$$
Now, if for some signal
$$x(-t)=x^*(t) \tag{4}$$
Then, is it safe to assume the following?
$$X(-f)=X^*(-f) \tag{5}$$
or does it depend on the type of signal?
You are correct. Your last equation is simply a fancy way of saying that $X(f)$ is real valued.
In general: if it's real in one domain, it's conjugate symmetric in the other.
Yes, if eqs. (2) and (3) hold for any "type of signal" (which they do), then (5) must hold.
Inserting (4) into (2) we get $$\mathscr{F}\big\{x^*(t)\big\} = X(-f)$$ and using (3) $$X(-f) = X^*(-f)$$
If we substitute $f = - g$ we get $$X(g) = X^*(g)$$ which, as Hilmar has already observed, means that $X(f)$ is real-valued. This is to be expected as, according to (4), $x(t)$ exhibits the conjugate complex symmetry.
The answers by @Deve and @Hilmar are technically perfect. I would like to provide some additional insights, with a few questions.
First, do you know of a signal satisfying this reversed-time/conjugate identity:
$$x(−t)=x^*(t)\,?$$
A first obvious idea is to choose among real and symmetric signals. A natural one in the Fourier framework is the cosine.
Now, let us get a little more complex (pun intented).
So second, what about the real sine? It is anti-symmetric. But if you remember that $$i^*=-i$$, the function $$t\to i.\sin t$$ also becomes a solution as well. So, by additivity, the function
$$t\to e^{i t}$$
(called complex exponential or cisoid) is also a solution. And its Fourier transform (as a generalized function) is indeed real (albeit somehow "infinite"). Going further, any linear combination of cisoids with real coefficients will do it.
Your question illustrates how Fourier duality is important, and how using it can simplify some issues. As seen in SYMMETRY OF THE DTFT FOR REAL SIGNALS:
In other terms, if a signal $$x(n)$$ is real, then its spectrum is Hermitian (conjugate symmetric'').
Here, your base signal $$x$$ is Hermitian, and the Fourier version is real. So to understand it better, just imagine $$t$$ is a frequency variable, and $$f$$ is its time dual. The standard representation is provided in Digital Analysis of Geophysical Signals and Waves/Complex Symmetry Properties.
It is also called the Heyser corkscrew/spiral. | 2020-01-23T15:16:26 | {
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https://math.stackexchange.com/questions/3060678/finite-difference-explicit-implicit-crank-nicolson-does-the-implicit-met/3060728 | # Finite difference - Explicit / Implicit / Crank Nicolson - Does the implicit method require the least memory?
Examine a dynamic 2D heat equation $$\dot{u} = \Delta u$$ with zero boundary temperature. A standard finite difference approach is used on a rectangle using a $$n\times n$$ grid. For the resulting linear systems a banded Cholesky solver is used. Compare the three methods explicit, implicit and Crank-Nicolson for the time stepping.
• Explicit method
$$\begin{gathered} \frac{u_{i+1,j} - u_{i,j}}{\Delta t} = \kappa \frac{u_{i,j-1} - 2u_{i,j} + u_{i,j+1}}{(\Delta x)^2} \\ u_{i+1,j} = u_{i,j} + \frac{\kappa \Delta t}{(\Delta x)^2}(u_{i,j-1} - 2u_{i,j} + u_{i,j+1})\\ \vec{u}_{i+1} = \vec{u}_i - \kappa \Delta t \textbf{A}_n \cdot \vec{u}_{i}\\ \end{gathered}$$
• Implicit method
$$\begin{gathered} \frac{u_{i+1,j} - u_{i,j}}{\Delta t} = \kappa \frac{u_{i+1,j-1} - 2u_{i+1,j} + u_{i+1,j+1}}{(\Delta x)^2} \\ \vec{u}_{i+1} = \vec{u}_i - \kappa \Delta t \textbf{A}_n \cdot \vec{u}_{i+1}\\ u_i = (\mathbb{I}_n + \kappa \Delta t \textbf{A}_n)^{-i}\cdot \vec{u}_0 \end{gathered}$$
• Crank-Nicolson
$$\begin{gathered} \frac{u_{i+1, j}-u_{i,j}}{\kappa \Delta t} = \frac{u_{i,j-1} - 2u_{i,j} + u_{i,j + 1}}{2(\Delta x)^2} + \frac{u_{i+1,j-1} - 2u_{i+1,j} + u_{i+1,j + 1}}{2(\Delta x)^2}\\ \vec{u}_{i+1} - \vec{u}_i = -\frac{\kappa \Delta t}{2}(\textbf{A}_n \cdot \vec{u}_{i+1} + \textbf{A}_n \cdot \vec{u}_i) \end{gathered}$$
Here is the question:
Does the implicit method require the least memory?
Here is my proposition for an answer:
I would say that the statement if FALSE
The implicit method requires more computational effort to give an answer, because the matrix $$\textbf{A}_n$$ needs to be inverted. However, I would say that this is not the reason why the statement is false: for the implicit method there is $$\textbf{no extra/less storage needed}$$ (compared to the explicit method for example) because there is no extra data generated from the computation (for example no other matrix generated). Is the answer correct as well as the reasoning behind it?
Following the comments of @zimbra314, I posted the question in Computational science beta
• The explicit methods should also be $u_i$ on the very right hand side of the bottom equation. then you can solve for $u_{i+1}$ explicitly. – tch Jan 3 at 16:02
• I think your reasoning is right. All the methods require you to store a current iterate and the matrix. It is possible that solving a linear system will require some additional memory, but that wouldn't mean the implicit memory uses less. Also, everything you do in the implicit method you need to do in the Crank-Nicholson method anyway, so there is no reason the implicit method would use less memory that Crank-Nicholson. – tch Jan 3 at 16:06
• Thanks for the comments @tch. I corrected the question according to your first comment. Regarding the second one, that's indeed the point which is not clear for me, but it seems we have the same intuition – ecjb Jan 3 at 16:11
• this question might be better suited for scicomp.stackexchange.com – piyush_sao Jan 3 at 16:21
It depends on how do you apply inversion. If you use a direct method such as LU factorization, the inverted matrix will have more non-zero entries due to fill-in. For $$n\times n$$ 2D grid, $$A$$ matrix will have $$\mathcal{O}(n^2)$$ entries while factored matrix will have $$\mathcal{O}(n^{2} \log n)$$ entries; so it would require more memory than explicit method.
• Thanks for the helpful comment @zimbra314. Then just to make the comparison straightforward. what would be the memory cost of explicit method in your example? – ecjb Jan 3 at 16:14
• $O(n^{2})$ on $n\times n$ 2D grid – piyush_sao Jan 3 at 16:18 | 2019-11-20T14:56:13 | {
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http://math.stackexchange.com/questions/880436/ratio-of-angles-in-a-right-triangle | Ratio of angles in a right triangle
P.S: I only want a hint,not the whole solution.
BdMO 2009 Problem 5 Secondary
In triangle ABC, $\angle A = 90$. M is the midpoint of BC. Choose D on AC such that AD = AM. The circumcircles of triangles AMC and BDC intersect at C and at a point P. What is the ratio of angles: $(\angle ACB)/( \angle PCB)$?
The diagram is extremely cluttered.I think the answer is $2:1$ based on some poorly drawn figures.
My try:
For our purposes,let the circle intersect $AB$ at Q.Then $QC=QA$ is the diameter of one of the circles.We can also get some angle equalities by angle chasing.The penultimate step seems to be showing that the arcs AP and PM are equal.I have listed out some of the possibilities down below,in increasing order of sophistication:
0)Show that BP=PD
1)Solve the question purely using angle-chasing.
2)Show that the arcs AP and PM are equal.
3)Construct an angle equal to $\angle ACP$ and then show it to be equal to $\angle PCB$.
4)Converse of Angle Bisector Theorem
None of the above have yielded anything good to work with.I am NOT looking for the whole solution,but merely a hint.Any help will be appreciated.
-
I hope that the solution is clear from the picture:
1. $\angle PAD=\angle PMB$ because $P$ lies on the circumcircle of $\triangle ACM$,
2. similar for $\angle ADP=\angle PBM$
3. $AD=BM(=AM)$ as $\angle A=90^\circ$.
Therefore $\triangle APD$ and $\triangle MPB$ are congruent.
-
Nice solution.Looks like my $0$th idea worked.A question though.What motivated you to look for congruent triangles in this messy figure?I mean,the figure was cluttered enough and trying to look for congruent triangles seemed like a mad thing to do.Your idea was quite ingenious.But may I know how you came up with it? – rah4927 Aug 2 '14 at 16:53
It's sorta backward reasoning, based on $AP=PM$ and $DP=PB$. Plus, I almost always try to use the fact that $BC=2AM$. – Quang Hoang Aug 2 '14 at 17:13
After giving this more thought, I see that the answer is in fact 2, and that this is the correct answer for any right triangle. The diagram provided by Quang Hoang is very well done, and Quang's observation about the congruent triangles is insightful, but I don't see how it settles the issue. Here is my most recent response to the original question:
First, there is a wonderful circle theorem that goes something like this: In a circle, if an inscribed angle A subtends arc K, then A = K/2. Let's call this the circle theorem.
After having drawn the specified figure, we can state that point P must lie at the intersection of the following two lines: The perpendicular bisector of chord AM, and the perpendicular bisector of chord DB. The perpendicular bisector of chord AM clearly cuts arc AM into two equal arcs, which are the arc PM and the arc PA. Note that inscribed angle PCA subtends arc PA. The circle theorem tells us that angle PCA is equal to half of arc PA. Similarly, inscribed angle PCM subtends arc PM, and so angle PCM is equal to half of arc PM. Because arcs PM and PA are equal, the angles that subtend them, specifically angles PCA and PCB, are equal to each other. It follows immediatly that angle ACB divided by angle PCB equals 2.
My earlier response to the question, which follows, is a brute-force approach. Nothing wrong with that, but it lacks elegance.
One approach is based on the assumption that you can state your answer in terms of various distances. For example, the distance between point C and point P, which can be stated as $\overline{CP}$. Furthermore, I also assume that you can use a trigonometry formula, in particular the law of cosines. If so, then you can simply forget about the circles and simply consider a point P inside $\triangle{ACB}$, which makes another triangle, which is $\triangle{PCB}$. You can also forget about right triangles, because the answer I have in mind works for any type of triangle.
The law of cosines can be stated as: $$c^2 = a^2 + b^2 - 2abCosC$$ Solving this for $CosC$ gives: $$CosC = \frac{a^2 + b^2 - c^2}{2ab}$$ Solving for C gives: $$C = Cos^{-1}\left( \frac{a^2 + b^2 - c^2}{2ab} \right)$$ You need to show the ratio of two angles, correct? You asked for a hint. This is it: Apply the third version of the law of cosines to your problem. You will need to apply it twice -- once for each angle as specified by the problem.
-
Use your eyes. Thanks to Quang Hoang's beautifully accurate drawing, anyone can see that DP and PM are not equal in length! In fact, DP is longer than PM – O.C.Riley Aug 17 '14 at 8:09
Ignore my last comment.I made a typo. It follows from his observations that $DP=PB$,which implies that $DB$ arc is bisected by $CP$. This implies in turn that ∠DCP=∠PCB and the conclusion follows. – rah4927 Aug 18 '14 at 3:23
In Quang Hoang's hint to you, he deliberately does not present prima facie evidence that DP = PB. You were supposed to fill in the details. – O.C.Riley Aug 20 '14 at 17:27
Yes I knew that but it was your comment ". . .but I don't see how that settles the issue" that made me write up the last comment. – rah4927 Aug 24 '14 at 17:08 | 2015-07-06T03:44:32 | {
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http://stat88.org/textbook/content/Chapter_02/04_Use_and_Interpretation.html | # 2.4. Use and Interpretation#
There are many situations in the law, medicine, and other fields, where Bayes’ Rule might help make decisions.
• Given the evidence, is the defendant guilty or not?
• Given the test results, does the patient have the disease, or not?
But not all medical or legal professionals have taken data science classes. So the calculations are sometimes misinterpreted or done incorrectly or simply not done at all.
Here is an example that demonstrates some of the issues that are involved.
## 2.4.1. Harvard Medical School Survey#
In 1978, Cascella, Schoenberger, and Grayboys asked 60 physicians, students, and house officers at the Harvard Medical school the following question:
“If a test to detect a disease whose prevalence is 1/1,000 has a false positive rate of 5 per cent, what is the chance that a person found to have a positive result actually has the disease, assuming that you know nothing about the person’s symptoms or signs?”
The answers ranged from about 2% to 95%, with 27 out of the 60 Medical School members surveyed answering 95%.
Let’s see what we think the answer should be. First, some background and terminology:
• Prevalence in the population is the percent of people who have the disease. It is also called the base rate of the disease.
• There is a test for the disease that has two possible results for a tested person. A positive result means that according to the test the person has the disease. A negative result means that according to the test the person doesn’t have the disease.
• The test can give a wrong result. The false positive rate is the proportion of positive results among people who don’t have the disease.
The true positive rate is the rate of positive results among those who do have the disease. The question doesn’t provide this rate. We will assume, as the people surveyed also did, that the test is good and the true positive rate is 100%.
Here are the data in a tree diagram.
The question asks for a chance but doesn’t say how the person is selected. As a first step, let’s assume that the person was selected at random from the population. Then the answer is a straightforward application of Bayes’ Rule:
$P(\text{disease} \mid \text{test positive}) ~ = ~ \frac{0.001 \times 1}{(0.001 \times 1) + (0.999 \times 0.05)} ~ \approx ~ 0.0196$
This calculation explains the answer of approximately 2% that was provided by 11 out of the 60 people surveyed.
## 2.4.2. Prior and Posterior Rates#
While the answer above is numerically correct, it is rather unsettling.
The medical test is pretty accurate. For a person who has the disease the test is error-free – it will always make the correct conclusion. For a person who doesn’t have the disease the test has an error rate of only 5%. For these people it will make the correct conclusion 95% of the time.
Yet for a randomly selected person who has tested positive, the calculation says the chance of disease is a mere 2%.
To understand what the 2% means, remember that it is a posterior probability of disease: the probability that the person has the disease, given that they tested positive.
This should be compared to the prior probability of disease: the probability we assigned to the person having the disease before we knew anything about test results. That probability was the base rate of 0.001, which is 0.1%.
Knowing that the person tested positive increased this probability by a factor of 20.
## 2.4.3. Effect of a Low Base Rate#
If 2% still seems small as an answer, look at the table below. It pretends that the population consists of 100,000 people, and displays the counts in the four branches of the tree. You can redo the table with a different population size if you wish, as long as you keep the proportions consistent with the question.
Test Positive
Test Negative
Disease
100
0
No Disease
4,995
94,905
As described in the question, 100 out of the 100,000 people (1/1000) have the disease. All these people test positive. Of the 99,900 people who don’t have the disease, 4,995 (5%) falsely test positive and the rest test negative.
If a randomly picked person has the disease, they are one of the people in the Test Positive column. This column illuminates an aspect of the calculation that must be understood in order to correctly interpret the result.
There are two groups of people who test positive: all those who have the disease, and also 5% of those who don’t have the disease.
The base rate is very small: the disease is rare in the population. So even though everybody with the disease tests positive, their number is small (100) in comparison to the group of people who don’t have the disease and falsely test positive (4,995). So the people who correctly test positive are a small fraction of all the people who test positive.
That is why when we are given that the test result of a randomly chosen person is positive, we conclude that there is a small chance that the person has the disease.
## 2.4.4. Base Rate Fallacy#
How could 27 out of the 60 Medical School members surveyed end up with an answer of 95% instead? It is because they made a common mistake known as the base rate fallacy. This error consists of simply ignoring the base rate and just focusing on the likelihoods.
Since the test had a 5% false positive rate, 27 people ignored all else and answered 95%, not noticing that they had provided $$P(\text{test negative} \mid \text{no disease})$$ when instead the question asked for $$P(\text{disease} \mid \text{test positive})$$.
The authors of the survey scathingly concluded that, “… formal decision analysis was almost entirely unknown and even common-sense reasoning about the interpretation of laboratory data was uncommon.”
Ouch. But the Med School members have plenty of company. The base rate fallacy occurs with depressing frequency in medicine, the law, and other fields. For some people who are not trained in data science, it is hard to keep track of several different rates at once and combine them appropriately.
## 2.4.5. Changing the Base Rate#
The most unsettling aspect of the Bayes’ Rule calculation above is the prior probability of 0.001. It is based on the assumption that the person was picked randomly from the population. But medical tests aren’t given to random members of the public. People get tested when they or their doctors think they might be at risk for the disease. In that case the base rate might not be the right figure to use for the prior probability of disease.
Suppose instead a person thinks that there is a small chance, say 10%, that they have the disease. If this person’s test result comes out positive, how should they update their probability of disease?
Since the test remains the same, all that has to change in the tree diagram is the prior probability of having and not having the disease.
The calculation changes to
$P(\text{disease} \mid \text{test positive}) ~ = ~ \frac{0.1 \times 1}{(0.1 \times 1) + (0.9 \times 0.05)} ~ \approx ~ 0.69$
The change in the prior probability of disease from 0.1% to 10% has a massive effect on the posterior probability of disease given a positive test result. That is now close to 70%.
The main lesson is that posterior probabilities are affected by the base rate as well as the likelihoods. Ignoring either of these can result in errors of calculation and interpretation. | 2023-01-28T13:48:01 | {
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https://math.stackexchange.com/questions/1407749/what-does-e-mean-in-this-expression | # What does $e$ mean in this expression?
I've seen this formula
$$1RM = \frac{100 \cdot w}{48.8 + 53.8 \cdot e^{-0.075 \cdot r}}$$
but I don't know what does the $e$ means.
The $w$ stands for weight. The $r$ for repetitions but I think the $e$ is from scientific notication but i'm not sure.
• $e$ is the base for the natural logarithm, for the exponential function, $e = \sum_{n = 0}^\infty \frac{1}{n!} \approx 2.7128$. – Daniel Fischer Aug 24 '15 at 10:00
• $e$ – Surb Aug 24 '15 at 10:01
• I had no idea it was a math constant. Thank you guys – Lothre1 Aug 24 '15 at 10:02
• By the way, when I search Google for "e maths", the first hit is mathsisfun.com/numbers/e-eulers-number.html which is correct, and gives you the string "euler's number" on which to search further. – Patrick Stevens Aug 24 '15 at 10:50
• @PatrickStevens When I google "e" the second link is the wikipedia one. – Surb Aug 24 '15 at 10:52
$\phantom{1233976327891467283146327864786231476892364983216468236874}$ $$\huge e$$
• Extracted sentences from the link: The number $e$ is an important mathematical constant that is the base of the natural logarithm. We have $$e= \sum_{n=0}^\infty \frac{1}{n!}=\lim_{n\to \infty}\Big(1+\frac{1}{n}\Big)^n\approx 2.71828$$ Sometimes called Euler's number after the Swiss mathematician Leonhard Euler. Also known as Napier's constant, but Euler's choice of the symbol $e$ is said to have been retained in his honor.The number $e$ is of eminent importance in mathematics. Like the constant $\pi$, $e$ is irrational: it is not a ratio of integers; and it is transcendental. – Surb Aug 24 '15 at 10:46
• I think that should probably be in the answer, not a comment. – Patrick Stevens Aug 24 '15 at 10:48
• @PatrickStevens I agree with you, but I like this idea to an answer with a single character (not often possible). However, I thought it is worth to extract a few informations from the link in case the link gets broken (although this is a wikipedia link which is unlikely to disappear). To cite Did: "OK, I confess: I always dreamt of reading/posting an answer with only one character... so I could not resist the occasion." – Surb Aug 24 '15 at 10:51
• Nicely done. Still, I hope this doesn't get as many upvotes as Did's one-letter answer, that would be a bad incentive for future answerers ;) – Daniel Fischer Aug 24 '15 at 10:57
$e\approx2.7182818284\dots$ is an irrational number. It has several representations. I'll give a few representation, along with a slight generalization.
\begin{align} e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\\ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n \end{align} This is the usual definition. The $\lim$ sign means, roughly, that you let $n$ get larger and larger. For example, $(1+\frac1{1000})^{1000}=2.7169\dots\approx e$, and you get better and better approximations as $n$ gets bigger. To prove the second thing from the first one, replace $n$ with $\frac nx$, and notice that $\frac nx\to\infty$ means the same thing as $n\to\infty$ (when $x$ is positive, at least. It works for negative $x$, too, though).
\begin{align} e&=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\dotsb\\ e^x&=\ 1\ +\ x\ +\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb \end{align} Remember that $0!=1$. You can try to prove this from the first set of equalities and the binomial formula. That's how Euler did it. (If you want to be rigorous, though, you need more effort. Euler was never very rigorous in this sort of thing.)
\begin{align} \int_1^e\frac1x\operatorname d\!x&=1\\ \int_1^t\frac1x\operatorname d\!x&=\log_e(t) \end{align} I need to explain the notation. The shape surrounded by the curve $y=\frac1x$, the vertical lines $x=1$ and $x=e$, and the $x$-axis has area $1$. More generally, the shape bounded by $\frac1x$, $x=1$, $x=t$, and the $x$-axis has area $\log_e(t)$. I won't prove this here.
And my favorite characterization: $e$ is the unique number that satisfies: $$e^x\ge x+1$$ for all $x$.
• +1 for your favorite characterization. While $e^x\geq x+1$ for every $x$ is clear to me, I'd be very interested in seeing a proof that $a^x\geq x+1$ for every $x$ implies $a=e$. – Surb Aug 25 '15 at 10:26
• @Surb Here's one: Notice that $x^{1/x}$ has a unique maximum at $x=e$ (Steiner's problem). It can be shown using just a few algebraic manipulations that $a^x\ge x+1$ implies that $a$ is a global maximum of $x^{1/x}$ (hint: substitute $x\mapsto\frac xa-1$ into the inequality), which means that $a=e$. – Akiva Weinberger Aug 25 '15 at 14:48 | 2019-05-21T08:43:08 | {
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https://byjus.com/question-answer/three-natural-numbers-are-taken-at-random-from-a-set-of-numbers-left-1-2/ | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
## The correct options are B $$\displaystyle \dfrac{^{89}\textrm{C}_{2}}{^{50}\textrm{C}_{47}}$$ C $$\displaystyle \dfrac{^{89}\textrm{C}_{87}}{^{50}\textrm{C}_{3}}$$$$n(S) =^{50}\textrm{C}_{3}$$As average value given $$30$$, mean sum of the numbers $$=90$$$$\therefore n(A) =$$ the number.of solution of the equation $$x_{1}+ x_{2}+ x_{3} = 90$$$$\Rightarrow x_{1}, x_{2}, x_{3}$$ each greater than $$1$$.$$=$$ Coefficient of $$x^{90}$$ in $$\left ( x+x^{2}+x^{3}\cdots \right )^{3}$$$$=$$ Coefficient of $$x^{87}$$ in $$\left ( 1+x+x^{2}+\cdots \right )^{3}$$$$=$$ Coefficient of $$x^{87}$$ in $$\left ( 1-x \right )^{-3}$$$$=$$ Coefficient of $$x^{87}$$ in $$\left ( 1+ ^{3}\textrm{C}_{1}x+^{4}\textrm{C}_{2}x^{2}+\cdots +^{89}\textrm{C}_{87}x^{87}+\cdots \right )$$$$=^{89}\textrm{C}_{87}=^{89}\textrm{C}_{2}$$$$\therefore$$Required probability $$=\displaystyle \dfrac{^{89}\textrm{C}_{87}}{^{50}\textrm{C}_{3}}=\dfrac{^{89}\textrm{C}_{2}}{^{50}\textrm{C}_{47}}$$Maths
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View More | 2022-01-18T14:41:20 | {
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http://math.stackexchange.com/questions/134410/an-infinite-cyclic-group-has-exactly-two-generators | # An Infinite cyclic group has exactly two generators.
Question: An infinite cyclic group has exactly two generators.
Answer: Suppose $G=\langle a\rangle$ is an infinite cyclic group. If $b=a^{n}\in G$ is a generator of $G$ then as $a\in G,\ a=b^{m}={(a^{n})}^{m}=a^{nm}$ for some $m\in Z$.
$\therefore$ We have $a^{nm-1}=e.$
(We know that the cyclic group $G=\langle a\rangle$ is infinite if and only if $0$ is the only integer for which $a^{0}=e$.) So, we have, $nm-1=0\Rightarrow nm=1.$ As $n$ and $m$ are integers, we have $n=1,n=-1.$
Now, $n=1$ gives $b=a$ which is already a generator and $n=-1$ gives $$H=\langle a^{-1}\rangle =\{(a^{-1})^{j}\mid j\in Z\} =\{a^{k}\mid k\in Z\}=G$$ That is $a^{-1}$ is also generator of $g$
My question is that am I approach this question correctly?
-
In your conclusion, I assume you mean to say $a^{nm-1} = e$? Other than that, you are absolutely right, not much to say here. – m_l Apr 20 '12 at 14:55
Just a tiny formatting thing: in the fourth line you mean $a^{nm - 1}$ rather than $a^{nm} - 1$. I couldn't edit it because it was too small an edit. – Tara B Apr 20 '12 at 14:55
snap =] $\hspace{1cm}$ – Tara B Apr 20 '12 at 14:56
You are writing your conclusions wrong. You don't want to conclude that $a^{-1}$ is also a generator (this is easy). You want to conclude that the only generators are $a$ and $a^{-1}$. So you want to show that if $a^n$ is a generator, then $n=1$ or $n=-1$. You are done once you get there. The rest is confused given what you are trying to show. Otherwise, the approach is fine. – Arturo Magidin Apr 20 '12 at 14:57
@Arturo: Your comment answers the question, so shouldn't it be an answer rather than a comment? – Tara B Apr 21 '12 at 8:01
I think you are getting confused along the way.
You want to show that if $G$ is an infinite cyclic group, then it has exactly two generators. This can be done by showing two things: that there are at most two generators, and then exhibiting two generators.
Exhibiting two generators is easy: if $G=\langle a\rangle$, then $a$ and $a^{-1}$ both generate; and $a\neq a^{-1}$, since $a$ has infinite order.
The bulk of your argument is an attempt at showing the other direction, namely you are trying to show:
If $a^n$ generates $G$, then $n=1$ or $n=-1$.
You analyse this correctly until the end of the paragraph that begins with a parenthetical remark. You successfully conclude $n=1$ or $n=-1$. So you are done.
But then you seem to be getting confused, and continue to argue; you are already done showing that there are at most two generators, so that's where the proof should end.
If the second part of the proof was meant to be what my first part was, then you are not clear in the first part. There should be an explicit statement where you say that your argument shows there are at most two generators. Finally, there is also the issue of noting that $a\neq a^{-1}$ (which is easy, but needs to be said).
- | 2014-08-30T03:01:22 | {
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https://acm.ecnu.edu.cn/problem/1166/ | # 1166. Simple Statistics
Given a set of numerical data, there are several ways in which to describe it. One way is the so-called 5-number summary. The five numbers used to describe the data set are the following: the minimum value, the first quartile, the median, the third quartile and the maximum value. The definition of the minimum and maximum values are obvious. The median of a set of numbers is the value of the number which would lie exactly in the middle of the set if it were sorted. For example, the median of the data set 7,-1, 9, 4, 1 would be 4. If there is an even number of values in the set, then the median is the average of the two values closest to the middle; if our set contained the values 7,-1, 9, 4, 1, 0 then the median would be (1 + 4)/2 = 2.5. The definition of the quartiles follows naturally from the definition of the median. If we take all the values that come before the median in the sorted list (in the case when we average two values for the median this set would include the lower of those two numbers) the first quartile is the median of this set.
The definition of the third quartile is identical except it uses those values that come after the original median. In our example above with 7,-1, 9, 4, 1, 0, the first quartile value would be 0 (the median of the value -1, 1 and 0 which are less than 2.5) and the third quartile would be 7. If the data set contained 1, 2, 2, 2, 3 (in any order), then the median would be 2, and the first and third quartiles would be 1.5 and 2.5, respectively. One special case is when there is only one element in the list, in which case the quartiles are equal to the median. One other way to characterize data is its skewness. A distribution is considered right-skewed whenever the maximum value is farther from the median than the minimum value, or when the maximum and minimum are equally distant from the median, but the third quartile is farther from the median than the first quartile. A left-skewed distribution is one with the opposite situation. For our purposes, a distribution which is neither left-skewed nor right-skewed is considered symmetric. Your task for this problem is to read in various sets of numbers and output the 5-number summary for each, along with the skewness of the data.
### 输入格式
There will be multiple input sets. Each input set will consist of a single line of the form n v1 v2 v3 . . . vn where n is the number of data values, and v1, . . . , vn are the values. All the values will be integers and the maximum value for n will be 100. A line which begins with 0 indicates end of input and should not be processed.
### 输出格式
For each input set, output the 5-number summary and skewness in the order minimum, first quartile,median, third quartile, maximum and skew, with a single space between each. Skew will either be the phrase right-skewed, left-skewed or symmetric.
### 样例
Input
6 7 -1 9 4 1 0
15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 0
0
Output
-1 0 2.5 7 9 right-skewed
1 4 8 12 15 symmetric
0 4 8 12 15 left-skewed
7 人解决，14 人已尝试。
7 份提交通过，共有 25 份提交。
7.3 EMB 奖励。 | 2022-12-01T23:08:59 | {
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http://math.stackexchange.com/questions/152467/parametric-form-of-a-plane | # Parametric form of a plane
Can you please explain to me how to get from a nonparametric equation of a plane like this:
$$x_1−2x_2+3x_3=6$$
to a parametric one. In this case the result is supposed to be
$$x_1 = 6-6t-6s$$ $$x_2 = -3t$$ $$x_3 = 2s$$
Many thanks.
-
Welcome to math.stackexchange!
A plane can be defined by three things: a point, and two non-colinear vectors in the plane (think of them as giving the plane a grid or coordinate system, so you can move from your first point to any other using them).
So first, we need an initial point: since there are many points in the plane, we can pick randomly. I'll just take $x_1=6,x_2=0$ so that $x_3=0$ and we see that the point $(6,0,0)$ solves the equation.
Now I need two vectors in the plane. I can do this by finding two other points in the plane, and subtracting them from this one (the difference of two vectors points from one to the other, so if both points are in the plane their difference will point along it). I'll take the points $(0,-3,0)$, and $(0,0,2)$. Notice the simple construction of all my points: set two variables to zero and find out what the third one should be. You can almost always do this, and it's probably the easiest way to go.
So my vectors are going to be these two points minus the original one I found. $$(0,-3,0)-(6,0,0)=(-6,-3,0)$$ $$(0,0,2)-(6,0,0)=(-6,0,2)$$ Now any vector in the plane, when scaled, is still in the plane. So I can define my plane like this: $$(6,0,0)+(-6,-3,0)t+(-6,0,2)s$$
I.e. start at the first point, and move $t$ amount in one direction and $s$ amount in another, where $t$ and $s$ range over the real numbers, so they cover the whole plane. Note that each of the scaled vectors, when plugged into the equation, give $0$. So for any point here, we're doing $6+0+0=6$, which solves the original equation. Splitting this up in terms of components $(x_1,x_2,x_3)$ instead of points, we get $$x_1=6-6t-6s$$ $$x_2=-3t$$ $$x_3=2s$$
There are infinitely many other parameterizations that could have worked, so your answer could look completely different while still being completely correct. But this is probably the logic they used, in case you were wondering.
-
Thank you very much indeed for your great answer and for welcoming me here! I hope I'll be able to learn how to write all the tags properly so you won't need to correct my questions again. Thanks. – Marco Greselin Jun 1 '12 at 14:29
Latex is quite easy to pick up. We're fortunate enough to have the folks over at tex.stackexchange.com helping us out if we need it. For this question there were only 3 things I did: surrounding things with single dollar signs makes them look nice inline. Using double dollar signs gives things their own line. And using an underscore within those dollar signs (called 'math mode') makes subscripts. If you want to see how any latex was made, just right click on it and go to Show math as -> tex commands. – Robert Mastragostino Jun 1 '12 at 14:33
@RobertMastragostino Is there a mistake here? Shouldn't the third point be (0, 0, 2) instead of (0, 2, 0)? – AndyG Nov 15 '13 at 18:55
@AndyG Yes, it seems it should be. I'll update, thanks – Robert Mastragostino Nov 18 '13 at 4:17
One way to do it is to let $x_1 = t$ and $x_2=s$ and then solve for $x_3$.
-
but how can i get to that result i wrote? – Marco Greselin Jun 1 '12 at 14:18
Look at $x_2$ and $x_3$ in your case, can you then somehow get $x_1$? – Sean Jun 1 '12 at 14:20
Many thanks for writing me back. But if I just follow what you wrote I get x3 = 6/3 +2/3 s- t/3 – Marco Greselin Jun 1 '12 at 14:25
Yip that's correct. This is equivalent to your solution in your question. – Sean Jun 1 '12 at 14:31
All right, thanks anyway! – Marco Greselin Jun 1 '12 at 14:32
There is more than one way to write any plane is a parametric way. To write a plane in this way, pick any three points $A$, $B$, $C$ on that plane, not all in one line. Then $$f(s, t) = A + (B-A)s + (C-A)t$$
- | 2014-04-18T05:50:57 | {
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https://thethong.wordpress.com/2009/09/17/probability-and-computing-chapter-1-exercises/ | ## Probability and Computing: Chapter 1 Exercises
Exercise 1.5:
Let $E_{n,m}$ be the event that Bob’s number is $n$ and Alice ‘s number is $m$; $E$ be the event that Alice wins.
(a): $E = Pr(E_{1,2}) + Pr(E_{1,4}) + Pr(E_{1,9}) + Pr(E_{6,9}) + Pr(E_{8,9}) =\frac{5}{9}$
(b): $E = Pr(E_{2,3}) + Pr(E_{2,5}) + Pr(E_{2,7}) + Pr(E_{4,5}) + Pr(E_{4,7}) =\frac{5}{9}$
(c): $E = Pr(E_{3,6}) + Pr(E_{3,8}) + Pr(E_{5,6}) + Pr(E_{5,8}) + Pr(E_{7,8}) =\frac{5}{9}$
Exercise 1.6:
Let $E_{k,n}$ be the event that there are k white balls (in a total off n balls) in the bin, B be the event that we pick out black ball, W be the event that we pick out white ball at one turn.
We will prove by induction the following statement:
$Pr(E_{1,n}) = Pr(E_{2,n})=...=Pr(E_{n-1,n})$
– The statement is true for $n= 2$
– Assume that the statement is true for some $n=k\ge 2$:
$Pr(E_{1,k}) = Pr(E_{2,k})=...=Pr(E_{k-1,k})$.
We will prove that it will aslo be true for $n = k + 1$, that is :
$Pr(E_{1,k + 1}) = Pr(E_{2,k + 1})=...=Pr(E_{k,k + 1})$
– Indeed, we have:
$Pr(E_{1, k + 1}) = Pr(B\mid E_{1,k}) = \frac{k-1}{k}.$
$Pr(E_{2,k + 1}) = Pr(B \mid E_{2,k}) + Pr(W \mid E_{1,k}) = \frac{k-2}{k} + \frac{1}{k} = \frac{k - 1}{k}.$
In general:
$Pr(E_{i,k + 1}) = Pr(B\mid E_{i,k}) + Pr(W\mid E_{i -1,k}) = \frac{k-i}{k} + \frac{i-1}{k}=\frac{k-1}{k}$
Thus,
$Pr(E_{1,k + 1}) = Pr(E_{2,k + 1})=...=Pr(E_{k,k + 1})=\frac{k-1}{k}$
Base on induction, the first statement is true.
### 22 responses to this post.
1. […] Webに転がってる解答 https://thethong.wordpress.com/2009/09/17/probability-and-computing-chapter-1-exercises/ を一部修正しただけですが． […]
2. Posted by Anonymous on September 8, 2011 at 2:57 pm
I have a different answer for this problem. when the white ball is just 1, so the probability for this situation is : 1/2 * 2/3 * 3/4……..*(n-1)/n = 1/n ,where n is the total numbers of bin. For the other sitution ,the answer is the same
3. Posted by Anonymous on May 25, 2012 at 11:49 pm
ex5 very simple but ínteresting. Alice always has choice to win
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Usually it is caused by the fire displayed | 2017-06-24T15:44:28 | {
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https://math.stackexchange.com/questions/4317065/is-the-map-sends-t-to-t-adjoint-of-t-surjective | Is the map sends $T$ to $T^*$ adjoint of $T$ surjective?
Let $$B(X)$$ denotes the set of all bounded linear operators from $$X$$ to $$X$$, where $$X$$ is a Banach space. Same is defined for the set $$B(X^*)$$, where $$X^*$$ denotes the set of all bounded linear functionals of $$X$$, that is all bounded linear functionals from $$X$$ to the field $$\mathbb F$$. In other words, $$X^*$$ is the dual space of $$X$$.
Let us now define a map $$\alpha :B(X) \to B(X^*)$$ by. $$\alpha(T)=T^*, \quad T \in B(X)$$ where $$T^*$$ is the Banach space adjoint of the operator $$T$$. That is, for Banach spaces $$X,Y$$ and $$T\in B(X,Y)$$ the adjoint operator $$T^*: Y^* \to X^*$$ is defined as $$T^*(y^*)(x)=y^*(T(x)),~~~y^* \in Y^*,~~x \in X.$$ Now note that, since $$\|T\|=\|T^*\|$$, the map $$\alpha$$ is an isometry and also injective. Can I show that $$\alpha$$ is a linear isomorphism? Linearity of $$\alpha$$ comes from the linearity of adjoint operators but I am not able show that $$\alpha$$ is surjective. One of my friends told me $$\alpha$$ may not be surjective but can't give any argument.
• So: the question is whether every member of $B(X^*)$ is the adjoint of some member of $B(X)$. Hint For a counterexample, you must take $X$ not reflexive. Nov 27, 2021 at 1:20
• Related post: math.stackexchange.com/questions/1832836/… Nov 27, 2021 at 14:58
In the case that $$X$$ is reflexive, I believe this is true, but I haven't checked it out; but I'm pretty sure that the map $$B(X^*)\to B(X^{**})$$ composed with the canonical isometric isomorphism $$B(X^{**})\cong B(X)$$ will give you the desired surjectivity.
However, the well-defined linear isometry $$B(X)\to B(X^*)$$, $$T\mapsto T^*$$ is not surjective in general. Take a non-reflexive Banach space $$X$$ (like $$c_0$$, or $$\ell^1$$) and let $$j:X\to X^{**}$$ be the canonical inclusion. Since $$j$$ is not surjective, let $$\chi\in X^{**}$$ be an element outside of $$j(X)$$. Fix a non-zero functional $$\psi_0\in X^*$$ and define an operator $$P:X^*\to X^*$$ by $$P(\phi):=\chi(\phi)\cdot\psi_0$$. This is a well-defined, bounded operator (and actually its range is one dimensional, but we dont care about it).
Now assume that $$P$$ is in the range of $$\alpha:B(X)\to B(X^*)$$, so there exists a bounded operator $$T\in B(X)$$ such that $$P=T^*$$, i.e. $$P(\phi)=T^*\phi=\phi\circ T$$ for all $$\phi\in X^*$$, so $$\phi(Tx)=\chi(\phi)\cdot\psi_0(x)$$ for all $$x\in X$$ and all $$\phi\in X^*$$. Since $$\psi_0$$ is non-zero, find $$x_0\in X$$ such that $$\psi_0(x_0)\ne0$$. We then have $$\phi(Tx_0)=\chi(\phi)\cdot\psi_0(x_0)$$ and this is true for all $$\phi\in X^*$$; set $$\psi_0(x_0):=\lambda\in\mathbb{C}\setminus\{0\}$$. We have just shown that $$\chi=\frac{1}{\lambda}j_{Tx_0}=j_{\frac{1}{\lambda}Tx_0}\in j(X)$$, which is a contradiction.
Note that this answer works for any non-reflexive space; in other words, if the claim in my first paragraph is true (which i think it is) we get the following:
Corollary: Let $$X$$ be a Banach space. The canonical linear isometry $$B(X)\to B(X^*)$$, $$T\mapsto T^*$$ is surjective if and only if $$X$$ is reflexive.
Comment: I've spent a few hours thinking about this, so I have to give some credit to GEdgar. I knew the counter-example would come from the non-reflexive world from the first moment I started on this, but I was trying to give an answer specifically for $$X=c_0$$; after seeing GEdgar's comment I realized there is no need to go in a specific space, the abstraction actually helps here.
• Nice answer! Lovely that you got a characterisation. Nov 27, 2021 at 10:20
• @QuantumSpace Thank you very much:) Nov 27, 2021 at 10:31
• @JustDroppedIn For reference, your corollary is true. I found this result in the book "Introduction to Functional Analysis" by Meise and Vogt, proposition 9.2. Dec 9, 2021 at 17:13
• @Mrcrg you mean $X$ is reflexive implies $B(X)\to B(X^*)$ is surjectve? this is obvious if you play around with the fact that the canonical embedding $X\to X^{**}$ is surjective Dec 9, 2021 at 19:24
• @Mrcrg but this is exactly what I prove: I prove $X$ not reflexive implies $T\mapsto T^*$ not surjective, which is equivalent to what you just wrote ($p\implies q$ is equivalent to negation of $q$ implies negation of $p$) Dec 9, 2021 at 19:47
This is an addition to the excellent answer of @JustDroppedIn, where I provide full details of the claim he sketches.
Theorem: If $$X$$ is reflexive, the map $$\varphi: B(X) \to B(X^*): T \mapsto T^*$$ is surjective.
Proof: Consider the canonical embedding $$j: X \to X^{**}: x \mapsto \operatorname{ev}_x$$. Since $$X$$ is reflexive, $$j$$ is an isometric isomorphism. As mentioned by @JustDroppedIn, this implies that we have a natural map $$\psi: B(X^*) \to B(X^{**}) \to B(X)$$ given by $$\psi(T) = j^{-1} T^* j, \quad T \in B(X^*).$$ We claim that $$\varphi \psi = \operatorname{id}_{B(X^*)}$$ and surjectivity of $$\varphi$$ will follow.
For this, fix $$T \in B(X^*)$$. We have to show that $$(j^{-1}T^* j)^* = T.$$ Fix $$f\in X^*$$. It suffices to show $$(j^{-1}T^*j)^*(f) = T(f) \in X^*$$ so fix $$x \in X$$, and note that it suffices to show $$((j^{-1} T j)^*(f))(x) = (T(f))(x).$$
From here, the proof is a simple calculation. Start by noting that $$\operatorname{ev}_x \circ T = \operatorname{ev}_y$$ for some $$y \in X$$. Hence, $$(T(f))(x) = \operatorname{ev}_x(T(f)) = f(y)$$ and thus \begin{align*}((j^{-1}Tj)^*(f))(x)&=f(j^{-1}T^*j(x))= f(j^{-1}(\operatorname{ev_x}\circ T))= f(y) = (T(f))(x)\end{align*} and we are done.
• Thanks for the confirmation, I appreciate it! Nov 27, 2021 at 10:45 | 2023-03-23T16:33:02 | {
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https://math.stackexchange.com/questions/2437485/are-the-vectors-linearly-dependent-or-independent | # Are the vectors linearly dependent or independent?
Problem: Let $\vec{v}=(5,9), \ \vec{u}=(3,-2)$ and $\vec{w}=(2,1)$. Determine the nature of their linear dependency.
Attempt: So, we are looking for constants $k_1,k_2,k_3$ such that $k_1\vec{v}+k_2\vec{u}+k_3\vec{w}=0.$ We can write this as
$$k_1\left[\begin{matrix} 5 \\ 9 \end{matrix}\right]+k_2\left[\begin{matrix} 3 \\ -2 \end{matrix}\right]+k_3\left[\begin{matrix} 2 \\ 1 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$$
which in canonical form is a system of linear equations in terms of $k_1,k_2$ and $k_3$.
$$M\vec{k}=\left[ {\begin{array}{cc} 5 & 3 & 1 \\ 9 & -2 & 1 \\ \end{array} } \right]\cdot \left[\begin{matrix} k_1 \\ k_2 \\ k_3 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right].$$
Row echelon form on $M$ gives
$$\left[ {\begin{array}{cc} 5 & 3 & 1 \\ 0 & -\frac{37}{5} & -\frac{4}{5} \\ \end{array} } \right],$$
This means that there are infinite solutions, but in order for them to be linearly independant, there should only exist one unique solution. Thus we have shown that the vectors $\vec{v},\vec{u},\vec{w}$ are linearly dependant. This means that $\text{span}\{\vec{v},\vec{u},\vec{w}\}=\mathbb{R}^2,$ because the addition of one of the vectors does not add another dimension to the span of the other two. The third vector that is linearly dependant of the other two, lies in their span.
Have I understood the concept of linear dependancy and span correctly? Any constructive input is very welcome!
• Note that at most $n$ vectors can be linearly dependant in $\mathbb R^n$, so for any $3$ vectors in $\mathbb R^2$, it is enough to check if they are pairwise dependant. Sep 20, 2017 at 14:03
• There is an almost identical example(s) provided here en.wikipedia.org/wiki/Linear_independence Sep 20, 2017 at 14:04
Yes, you understood correctly the concept of linear dependency. And you proof is correct too.
You have made a mistake, it should be:
$$M\vec{k}=\left[ {\begin{array}{cc} 5 & 3 & 2 \\ 9 & -2 & 1 \\ \end{array} } \right]\cdot \left[\begin{matrix} k_1 \\ k_2 \\ k_3 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$$
The solutions are of the form $\alpha\cdot \left[\begin{matrix} -7 \\ -13 \\ 37 \end{matrix}\right]$ for $\alpha \in \mathbb{R}$.
Thus, $$-7\vec{v}-13\vec{u}+37\vec{w} = 0$$
so $\{\vec{v}, \vec{u}, \vec{w}\}$ is linearly dependent in $\mathbb{R}^2$.
In fact, any three distinct vectors in a two-dimensional space must be linearly dependent.
Edit:
I found the solutions by reducing the matrix to row echelon form: $$\left[ {\begin{array}{cc} 5 & 3 & 2 \\ 9 & -2 & 1 \\ \end{array} } \right] \sim \left[ {\begin{array}{cc} 5 & 3 & 2 \\ -1 & -8 & -3 \\ \end{array} } \right] \sim \left[ {\begin{array}{cc} 1 & 8 & 3 \\ 5 & 3 & 2 \\ \end{array} } \right] \sim \left[ {\begin{array}{cc} 1 & 8 & 3 \\ 0 & -37 & -13 \\ \end{array} } \right] \sim \left[ {\begin{array}{cc} 1 & 0 & \frac{7}{37} \\ 0 & 1 & \frac{13}{37} \\ \end{array} } \right]$$
The first row gives $k_1 = -\frac{7}{37}k_3$, and the second row gives $k_2 = -\frac{13}{37}k_3$.
Thus, the solutions are given by $k_3\cdot \left[\begin{matrix} -\frac{7}{37} \\ -\frac{13}{37} \\ 1 \end{matrix}\right]$ for $k_3 \in \mathbb{R}$, which we can also write as $$\alpha\cdot \left[\begin{matrix} -7 \\ -13 \\ 37 \end{matrix}\right] \text{ for } \alpha\in\mathbb{R}$$ by setting $k_3 = 37\alpha$.
• How did you get the second line? the solutions are of the form...?` Sep 20, 2017 at 14:14
• @Parseval I have added it. Just reduce the matrix using row transformations to something more sparse. Sep 20, 2017 at 14:32
multiplying your second equation by $-2$ and addin g to the first we get $$-13k_1+7k_2=0$$ we get $$k_1=\frac{7}{13}k_2$$ therefore your vectors are dependent
• How do you conclude this based on your last formula? You're saying that $k_2$ is $k_1$ scaled by a factor of $\frac{7}{13}.$ How does this show that the vectors are dependant? Sep 20, 2017 at 14:16
• there exist other Solutions the $$k_1=k_2=k_3=0$$ choose for instance $$k_2=1,k_1=\frac{7}{13}$$ Sep 20, 2017 at 14:18
You worked well, but there's a little glitch.
The fact that the vectors are linearly dependent does not imply their span is $\mathbb{R}^2$: consider $(1,0)$, $(2,0)$ and $(3,0)$ to see why.
On the other hand, your matrix $M$ has two pivot columns, so indeed the span has dimension $2$ and therefore the three vector you were given do span $\mathbb{R}^2$.
For just establishing whether the three vectors are linearly dependent, it suffices to consider the fact that they're three and the space has dimension $2$: a set of $m$ vectors in a space of dimension $n$ is linearly dependent as soon as $m>n$; if $m\le n$ it can be linearly dependent or independent. | 2022-05-26T18:26:38 | {
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https://math.stackexchange.com/questions/2953600/how-to-map-interval-0-100-to-the-interval-100-350/2953802 | # How to map interval $[0, 100]$ to the interval $[100, 350]$?
I have an interval $$[0; 100]$$ and would like to map it to this new interval: $$[100;350]$$.
I thought about multiplying it by $$3.5$$, but that would give the interval $$[0;350]$$. And adding to each of these elements $$100$$ would give: $$[100;450]$$. Hence my question: is it possible to do what I want?
Note that I can settle for the interval $$[0;350]$$ : in my program, it will be enough if I exclude the numbers present in the interval $$[0;99]$$.
• how about multiplying by 2.5 instead? – Lord Shark the Unknown Oct 13 '18 at 8:41
To map from $$[a,b]$$ to $$[c, d]$$,
Consider the straight line that connects $$(a,c)$$ to $$(b,d)$$.
We have the slope $$m = \frac{d-c}{b-a},$$ we are able to recover $$m$$.
$$y=mx+C$$
To recover $$C$$, just substitute one of the value say $$(a,c)$$ and solve for $$C$$. For our example, we have $$a=0$$ and $$c=100$$.
Hence your transformation can be of the form of $$y=mx+100$$. Can you compute the $$m$$ to find what you want?
The ratio of the lengths of the intervals is $$2.5 :1,$$ the position of the left extremity is shifted by $$100.$$ So take the mapping $$f(x)=2.5x+100.$$
Since $$100\cdot t^0=100$$ for any positive $$t$$, we find $$t$$ such that $$100\cdot t^{100}=350\implies t=3.5^{0.01}$$. $$\boxed{y=100\cdot3.5^{0.01x}}$$ In general an exponential mapping from $$[a,b]$$ to $$[c,d]$$ is $$y=c\left(\frac dc\right)^{\frac{x-a}{b-a}}$$.
• Way too complicated; a simple affine transformation works. – saulspatz Oct 13 '18 at 15:21
• @saulspatz I know, since the linear transformation has already been said three times. I just wanted to give the next best approach. – TheSimpliFire Oct 13 '18 at 15:22
• But why the downvote? – TheSimpliFire Oct 13 '18 at 15:27
• I don't think the answer contributes anything positive to the discussion. – saulspatz Oct 13 '18 at 15:29
• @saulspatz My answer is an answer to the question: 'Is it possible to transform an interval into another?' What the OP specifically wanted has already been discussed many times, and I do not see any harm in adding an alternative method to perform such a transformation. – TheSimpliFire Oct 13 '18 at 15:31
You can consider $$x\mapsto ax + b\colon [0,100]\to [100,350]$$ such that $$0\mapsto 100$$ and $$100\mapsto 350$$.
Thus, $$a \cdot 0 + b = 100,\\ a\cdot 100 + b = 350,$$ and solving it gives $$a = \frac 52$$, $$b = 100$$.
In general, the same technique works for intervals $$[x_1,x_2]$$ and $$[y_1,y_2]$$:
$$ax_1 + b = y_1\\ ax_2 + b = y_2.$$
Solving it gives $$a = \frac{y_2 - y_1}{x_2 -x_1}$$ and $$b = y_1 - ax_1$$. All in all, it's a line $$y - y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1).$$ Looks familiar?
Another method that’s a bit more general and will come in handy if you want to map arbitrary curves is to parameterize your paths. That is, find a one-to-one mapping from your first path to $$[0,1]$$, $$(0,1]$$, or so on as appropriate. Then find a one-to-one mapping from $$[0,1]$$ to your second path. (The interval $$[0,1]$$ isn’t special, just convenient.) Finally, compose them.
Let’s say you want to map $$x^2$$ over the interval $$[0,4]$$ to $$\sin x$$ over the interval $$[0,2\pi]$$. A one-to-one mapping from the parabola to the line segment, $$t: [0,4] \to [0,1]$$, is $$t = \sqrt{x}/2$$, and a mapping from $$t \in [0,1]$$ to a sine wave over $$[0,2\pi]$$ is $$\sin {2\pi t}$$. Substituting, we get $$\sin {\left( \pi \sqrt{x}\right)}$$. | 2019-11-15T10:45:46 | {
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https://www.physicsforums.com/threads/a-convergence-question.285723/ | # A Convergence Question
1. Jan 18, 2009
### e(ho0n3
The problem statement, all variables and given/known data
Suppose {a_k} is a decreasing sequence of real numbers with a_k >= 0 for all k. Show that if a_1 + a_2 + ... converges, then lim k*a_k = 0. Is the converse true?
Relevant equations
Fact: If {a_k} is a sequence of real numbers such that a_1 + a_2 + ... converges, then lim a_k = 0.
The attempt at a solution
It seems to me that {a_k} would have to decrease faster than {1/k} in order for ka_k to converge to 0. For example, if {a_k} = {1/k}, lim k*a_k = 1. However, if {a_k} = {1/2^k}, then lim k*a_k = 0. This intuition has failed to lead me to the answer though. It has also come to my attention that if a_1 + a_2 + ... converges to A, that 0 <= lim k*a_k <= A, given that the limit exists. This fact however has also been unhelpful. Any tips?
2. Jan 18, 2009
### Dick
Suppose the limit k*a_k is not zero. That means that there is an e>0 such that k*a_k>e for an infinite number of different values of k. In particular you can find a subsequence k_i such that k_{i+1}>2*k_i for all i and k_i*A_{k_i}>e. Use that a_k is decreasing and think about how you can estimate the sum of the a_i for k_i<i<k_{i+1}.
3. Jan 18, 2009
### e(ho0n3
The sum of the a_i for k_i<i<k_{i+1} is greater than or equal to the sum a_i for k_i < i <= 2k_i and that sum is greater than or equal to k_i*a_{2*k_i}. Where do we go from here?
4. Jan 18, 2009
### Dick
There are more than k_{i+1}/2 terms in that group of terms and they are all greater than or equal to e/k_{i+1}. Hence?
5. Jan 18, 2009
### e(ho0n3
Actually, there are less than k_{i+1}/2 since k_{i+1}/2 > k_i. And where do you get that the are >= e/k_{i+1}?
6. Jan 18, 2009
### Dick
If k_{i+1}/2>k_{i} then k_{i+1}-k_{i}>k_{i+1}-k_{i+1}/2. I'm subtracting a LARGER number on the right side. And I picked a subsequence such that a_k_{i}*k_{i}>e and the a_k's are decreasing.
7. Jan 18, 2009
### e(ho0n3
OK, I understand now. But I don't know what you want me to conclude. For any k, we have that a_1 > a_2 > ... > a_k > e/k. What is so special about k_{i+1}?
8. Jan 18, 2009
### Dick
The point is that summing the terms between a_k_{i} and a_k_{i+1} gives you more than e/2. There are an infinite number of such intervals. In such a case the sum of the a_k must diverge. It's a proof by contradiction!
9. Jan 18, 2009
### e(ho0n3
Wow. I honestly did not see that. Let me ask you: What led you to figure that there would be an infinite number of intervals in the sum of a_k that are greater than e/2? Did you look at a particular example?
10. Jan 18, 2009
### Dick
It was the same intuition that made you say a_k must approach 0 faster than 1/k. The proof is roughly the same as the proof that a_k=1/k diverges. I looked at it and scratched my head for quite a while. It's not that easy. But no, particular examples aren't that helpful. Did you find the counterexample for the converse?
11. Jan 18, 2009
### e(ho0n3
I've never seen the proof that a_k = 1/k diverges. You wrote "for quite a while", which I found funny since I've been scratching my head since yesterday.
This should work as a counter-example: Let a_k = 1/k^{3/2} so that k*a_k = 1/sqrt(k), which converges to 0, but the sum diverges.
12. Jan 18, 2009
### Dick
Look up the proof, you'll see the resemblance. No, the sum of a_k=1/k^(3/2) does converge. It's a power series. You want something that diverges REALLY slowly. Slowly enough that the series is divergent but k*a_k still goes to zero.
13. Jan 18, 2009
### e(ho0n3
Oops. I was thinking about the sum of 1/sqrt(k) which diverges. My mistake. Let me ponder and tinker with this tonight. Hopefully I'll have something by tomorrow.
14. Jan 18, 2009
### Dick
BTW the elementary proof the harmonic series diverges is that the sum of 1/k from 1/2^k to 1/2^(k+1) is greater than 1/2. There are 2^k terms of size greater than 1/2^(k+1). Let me know what you find for the counterexample. There's a pretty easy one. It's just a LITTLE smaller than 1/k.
Last edited: Jan 19, 2009
15. Jan 19, 2009
### e(ho0n3
I did some googling and found that the sum of $$\sqrt[n]{2} - 1$$ diverges slowly and also $$\lim_{n \to \infty} (\sqrt[n]{2} - 1) = 0$$. I couldn't think of one as I don't really have a feel for these things yet.
16. Jan 19, 2009
### e(ho0n3
I just had a flash: Were you thinking about 1/(k ln k)? That should work too, I think.
17. Jan 19, 2009
### Dick
That's the one. An integral test shows it diverges, but k*(1/(k ln k)) goes to zero. | 2017-09-24T01:55:35 | {
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https://math.stackexchange.com/questions/1489366/proving-that-a-mathbbz-cap-b-mathbbz-operatornamelcma-b-mathbb-z/2124697 | # Proving that $a \mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb Z$ [duplicate]
Given $m \in \mathbb{Z}$, let $m\mathbb{Z}$ denote the set of integer multiples of $m$, i.e. $m\mathbb{Z} := \{mk\mid k \in \mathbb{Z}\}$. Now let $a,b \in \mathbb{Z}$ with $a,b$ not both $0$. Prove that $a\mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb{Z}$.
I am trying to write a proof for this, but I am unsure of what method to use. Also I am confused by $mk\mid k$, because wouldn't $m=1$ for this to be true.
## marked as duplicate by BCLC, Jendrik Stelzner, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 31 '18 at 15:03
• Yes, the $\mid$ symbol here is just a separator in a set definition. You could replace it with a colon character for clarity. – Thomas Andrews Oct 20 '15 at 16:51
• Why did you put the "equals" sign outside of MathJax in $a\mathbb Z \cap b\mathbb Z = \operatorname{lcm}(a,b)\mathbb Z$ in both of the places where you wrote that? There seem to be a fair number of people on m.s.e. who follow that incorrect usage. Is there some source instructing people to do it that way? ${}\qquad{}$ – Michael Hardy Oct 20 '15 at 16:59
• It doesn't say $\text{“}mk\mid m\text{''}$; it says $\text{“}\{\cdots \mid \cdots\cdots\}\text{''}$, where the vertical bar does not mean "divides", but rather means what it means in the context of expressions like $\text{“}\{\cdots \mid \cdots\cdots\}\text{''}$. ${}\qquad{}$ – Michael Hardy Oct 20 '15 at 17:03
Hint:
Step 1: Can you prove that $\operatorname{lcm}(a,b)\in a\mathbb{Z}\cap b\mathbb{Z}$?
Step 2: Can you prove that if $a$ and $b$ both divide $c$, then $c\in a\mathbb{Z}\cap b\mathbb{Z}$?
Step 3: How do steps 1 and 2, together, imply your result?
First of all, to clear up your confusion:
$$m\mathbb Z=\{mk\mid k\in\mathbb Z\}$$
does not mean that $mk$ divides $k$. The vertical line can be read as "so that" or "where". It means that the set $m\mathbb Z$ is the set of numbers in the form of $mk$, where $k$ is any element of $\mathbb Z$.
That said, to prove that $a\mathbb Z\cap b\mathbb Z=\operatorname{lcm}(a,b)\mathbb Z$, you need to prove that
1. if $x\in a\mathbb Z\cap b\mathbb Z$, then $x\in \operatorname{lcm}(a,b)\mathbb Z$.
2. if $x\in\operatorname{lcm}(a,b)\mathbb Z$, then $x\in a\mathbb Z\cap b\mathbb Z$.
• I changed what appeared to be an obvious typo, which said $m\mathbb Z\{mk\mid k\in\mathbb Z\}$, to $m\mathbb Z = \{mk\mid k\in\mathbb Z\}$, which seemed obviously intended. You changed it back. How does $m\mathbb Z\{mk\mid k\in\mathbb Z\}$ with no $\text{“}=\text{''}$ make sense? ${}\qquad{}$ – Michael Hardy Oct 20 '15 at 17:13
• @MichaelHardy Your change also changed the nice enumeration back into the inline enumeration, which is why I had to re-edit it. In doing so, I accidentally re-introduced a mistake that you correctly pointed out. Thanks for the edit. – 5xum Oct 20 '15 at 17:15
• I don't see how my edit could have affected the enumeration. I did nothing with the enumeration. ${}\qquad{}$ – Michael Hardy Oct 20 '15 at 17:17
• @MichaelHardy That's because while you were editing the mistakes in the top part, I edited the bottom one to improve enumeration. So first, my edit came in and fixed the enumeration, then your edit of the original post fixed everything else, but reverted the enumeration. Then, I wanted to re-fix the enumeration but ended up ruining everything else. – 5xum Oct 20 '15 at 17:21
I give another demonstration you might find useful, which allows you to reach the result without double implication, using only the definition of $lcm(a,b)$.
Definition: $d \in \Bbb Z$ is defined least common multiple of $a$ and $b$, in symbols $d := lcm(a,b)$ if:
1. $\exists m,n \in \Bbb Z$ such that $am = d = bn$, otherwise written as $a|d$ and $b|d$
2. If $c \in \Bbb Z$ is such that $a|c$ and $b|c$, then $d|c$
So we can write $$a\Bbb Z \cap b\Bbb Z =_{(1)} \left \{ x \in \Bbb Z : a|x, b|x \right \} =_{(2)} \left \{ x \in \Bbb Z : lcm(a,b)|x \right \} =_{(3)} lcm(a,b)\Bbb Z$$
Where:
• $=_{(1)}, =_{(3)}$ follows from the definition of $a\Bbb Z \cap b\Bbb Z$ and $lcm(a,b)\Bbb Z$
• $=_{(2)}$ follows from the point $2$ of the definition of $lcm(a,b)$
Well firstly, how do you define $\text{lcm}(a,b)$? I'll define it by the converse of a proposition in Artin Algebra (Prop 2.3.8)
Namely, we'll prove the converse of Prop 2.3.8 where $m:=\text{lcm}(a,b)$ is defined by the integer s.t.
(a) $m$ is divisible by both $a$ and $b$
(b) If $n$ is divisible by $a$ and $b$, then $n$ is divisible by $m$.
Pf: $(\subseteq)$
Let $n \in \mathbb Z m$. Then there is an integer $n_m$ s.t. $n_m = \frac n m$. Observe that $n_m = \frac{n_a}{m_a} = \frac{n_b}{m_b}$ where we define $n_a := \frac n a, m_a := \frac m a, m_b := \frac m b, n_b := \frac n b$. Observe that $m_a, m_b$ are integers by assumption (a) while we want to show that $n_a, n_b$ are integers because showing such is equivalent to showing $n \in \mathbb Za \cap \mathbb Zb$.
Now, $n_m m_a = n_a$ is a product of integers and hence an integer. The same is true for $n_m m_b = n_b$. Therefore, $n_a, n_b$ are integers and thus, $n \in \mathbb Za \cap \mathbb Zb$
$(\supseteq)$
This one is easier. Let $n \in \mathbb Za \cap \mathbb Zb$. Then $n_a, n_b$ as defined earlier are integers, i.e. $n$ is divisible by both $a$ and $by$. By assumption (b), $n$ is divisible by $m$.
QED | 2019-11-20T04:07:56 | {
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https://mathematica.stackexchange.com/questions/236274/how-can-i-plot-the-superimposition-of-2-functions-for-a-vibrating-circular-membr | # How can I plot the superimposition of 2 functions for a vibrating circular membrane?
I have only recently discovered Mathematica so I am quite weak with it, posting here as it seems to be the most comprehensive forum online.
My question is entirely based off a previous very high quality post namely
Circular membrane vibration simulation
I decided to make a new post for this since that post is nearly 6 years old now.
In the answers user 'Jens' describes the code for constructing the traditional Chladni Patterns for a circular membrane as the follows.
fXY[n_, k_][x_?NumericQ, y_?NumericQ] :=
BesselJ[n, k Sqrt[x^2 + y^2]] Cos[n ArcTan[y, x]]
wavePattern[n_, j_] :=
Module[{k0 = N[BesselJZero[n, j]]},
DensityPlot[fXY[n, k0][x, y], {x, -1.1, 1.1}, {y, -1.2, 1.1},
RegionFunction -> Function[{x, y}, x^2 + y^2 < 1],
ColorFunction ->
Function[{x},
Blend[{White, Darker@Brown, White}, 2 ArcTan[10 x]/Pi + .5]],
ColorFunctionScaling -> False, PlotPoints -> 100,
MaxRecursion -> 0,
Epilog ->
Inset[Grid[{{"n", n}, {"j", j}}, Frame -> All], {-.9, .9}],
BaseStyle -> {FontFamily -> "Arial"}]]
Show[GraphicsGrid@Table[wavePattern[m, n], {m, 0, 2}, {n, 1, 3}],
ImageSize -> 700]
which outputs the following
I want to know how I can be able to superimpose (by adding) the values of say wavePattern[2, 2] and wavePattern[3, 0]. And plot the values that occur, in a similar way. Which will displays the Chladni Pattern when both the modes are occurring simultaneously.
The reason I'm asking this question (not relevant to the question) : I recently saw a video lecture that claimed the seemingly non traditional (complex) Chladni patterns that are observed by modern experiments that use tone generators are actually the result of the superimposition of the basic modes that are predicted by the theoretical formula. This has me highly intrigued as I had seen nearly all experiments show the formation of Chladni patterns that don't match the "basic" modes. This seems to be a ingeniously simple solution which I seem to have missed and couldn't find anywhere else online. I want to verify it by using Mathematica, and since there already exists some great code I'm hoping someone here can help me tinker it in the form I am looking to do.
Something like that (adding two patterns)?
fXY[n_, k_][x_?NumericQ, y_?NumericQ] :=
BesselJ[n, k Sqrt[x^2 + y^2]] Cos[n ArcTan[y, x]]
addPattern[n1_, j1_, n2_, j2_] :=
Module[{k1 = N[BesselJZero[n1, j1]], k2 = N[BesselJZero[n2, j2]]},
DensityPlot[
fXY[n1, k1][x, y] + fXY[n2, k2][x, y], {x, -1.1, 1.1}, {y, -1.2,
1.1}, RegionFunction -> Function[{x, y}, x^2 + y^2 < 1],
ColorFunction ->
Function[{x},
Blend[{White, Darker@Brown, White}, 2 ArcTan[10 x]/Pi + .5]],
ColorFunctionScaling -> False, PlotPoints -> 100,
MaxRecursion -> 0,
Epilog ->
Inset[Grid[{{"n", n1, n2}, {"j", j1, j2}},
Frame -> All], {-.9, .9}],
BaseStyle -> {FontFamily -> "Arial"}]]
Example 1:
Show[addPattern[2, 2, 0, 3], ImageSize -> 250]
Example 2:
Show[addPattern[2, 2, 1, 3], ImageSize -> 250]
• Wow that's great I think this is exactly what I was looking for, plus its really easy to just add a third argument and make it add 3 instances at once too. Thanks a lot! Dec 11 '20 at 1:32 | 2022-01-24T10:36:58 | {
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https://stacks.math.columbia.edu/tag/07PG | Lemma 15.47.3. Let $R$ be a regular ring. Let $f \in R$. Assume there exists a derivation $D : R \to R$ such that $D(f)$ is a unit of $R$. Then $R[z]/(z^ n - f)$ is regular for any integer $n \geq 1$. More generally, $R[z]/(p(z) - f)$ is regular for any $p \in \mathbf{Z}[z]$.
Proof. By Algebra, Lemma 10.157.10 we see that $R[z]$ is a regular ring. Apply Lemma 15.47.2 to the extension of $D$ to $R[z]$ which maps $z$ to zero. This works because $D$ annihilates any polynomial with integer coefficients and sends $f$ to a unit. $\square$
Comment #3525 by Dario Weißmann on
Doesn't this work for any regular ring and any polynomial of the form $p+f$ where $p\in \mathbf{Z}[z]$ is regarded as a polynomial in $R[z]$?
A reference to the fact that a polynomial algebra over a regular ring is regular would be nice. I can't find it in the project so here is what I came up with (compiles fine for me but not as a comment?): Let $R$ be a regular ring. To show that $R[z]$ is a regular ring it suffices to show that the localization at all maximal ideals of $R[z]$ are regular local rings. Let $\mathfrak{m}$ be a maximal ideal of $R[z]$. Write $\mathfrak{p}=R\cap \mathfrak{m}$. Then $R[z]_{\mathfrak{m}}=R_{\mathfrak{p}}[z]_{\mathfrak{m}}$ and we can reduce to $R$ is a regular local ring, $\mathfrak{m}$ is a maximal ideal of $R[z]$ lying over $\mathfrak{m}_R$. Then $R[z]/\mathfrak{m}=\kappa[z]/\overline{\mathfrak{m}}$ is a field where $\overline{\mathfrak{m}}$ denotes the image of $\mathfrak{m}$ in $\kappa[z]$. Let $\overline{f}\in \kappa[z]$ be a generator of $\overline{\mathfrak{m}}$ and $f\in R[z]$ be a lift. Further let $f_1,\dots,f_r$ be a regular sequence in $R$ generating the maximal ideal. Then $f_1,\dots,f_r,f$ defines a regular sequence in $R[z]_{\mathfrak{m}}$ generating $\mathfrak{m}$.
Comment #3526 by Dario Weißmann on
OK, so it only didn't compile in the preview?
Comment #3529 by on
The preview and the final render are dealt with by different pieces of the software, but of course they should be as close together as possible. I'll see what I can do to improve this in the future, thanks for telling me about it.
Comment #3530 by on
Any smooth algebra over a regular ring is regular, see 10.157.10. The proof is essentially what you said in the case of polynomial rings.
Unfortunately, I don't understand your first comment. Can you clarify?
Comment #3531 by Dario Weißmann on
So the proof works by lemma 07PF for any polynomial mapping to a unit in R[z] under some derivation. We have $D(z^k)=kz^{k-1}D(z)=0$ by definition of $D$. By linearity the same is true for any polynomial $p$ with integer coefficients regarded as a polynomial in $R[z]$ via $\mathbf{Z}[z]\to R[z]$. So $D(p+f)=D(f)$ is a unit in $R[z]$.
We didn't need the $\mathbf{F}_p$-algebra structure and it holds for a larger class of polynomials.
Thank you for the reference!
Comment #3534 by on
OK, I understand now. Thank you. I will change the lemma when I next go through the comments.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2019-04-26T12:35:35 | {
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https://math.stackexchange.com/questions/1331461/does-the-series-sum-n-1-infty-frac2n1n2n12-converge | # Does the series: $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge?
does $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge? I think yes, it does, because the $a_n$ in the series converges to zero. but I'm trying to prove this by the help of the fact that:
$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$
any suggestions?
• If $n$ is very large, the individual terms are approximately $2n/n^4=2/n^3$. That suggest a suitable converging series for a comparison test. – Jyrki Lahtonen Jun 19 '15 at 13:08
• Is convergence sufficient, or do you need to know the limiting value? That it converges is established easily, since $\frac{2n+1}{(n+1)^2}\le1$, and since the series with $1/n^2$ converges. – Bernhard Jun 19 '15 at 13:09
• The fact, that the $a_n$ tend to $0$ is not enough, see the diverging series $$\sum_{j=1}^{\infty} \frac{1}{j}$$ – Peter Jun 19 '15 at 13:10
$$\frac{2n+1}{n^2(n+1)^2}=\frac{1}{n^2}-\frac{1}{(n+1)^2}$$ $$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}=\frac{1}{1}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}+....=1$$
• ...and hence the limiting value of the series is 1. – Bernhard Jun 19 '15 at 13:14
Notice that $$\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{n^2+2n+1-n^2}{n^2(n+1)^2},$$ so the series telescopes: $$\sum_{n=1}^m \frac{2n+1}{n^2(n+1)^2} = 1-\frac{1}{2^2} + \frac{1}{2^2} -\dotsb - \frac{1}{m^2} + \frac{1}{m^2} - \frac{1}{(m+1)^2} = 1- \frac{1}{(m+1)^2} \to 1$$ as $m \to \infty$
$$\sum_{n\geq1}\frac{2n+1}{n^{2}\left(n+1\right)^{2}}\leq3\sum_{n\geq1}\frac{1}{n^{3}}.$$
$$\frac{2n+1}{n^2(n+1)^2}=\frac{2n}{n^2(n+1)^2}+\frac{1}{n^2(n+1)^2}$$ $$=\frac{2}{n(n+1)^2}+\frac{1}{n^2(n+1)^2}$$
now it is very easy to compare the first term with $\frac{2}{n^3}$ and the second term with $\frac{1}{n^3}$
$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}<\sum_{n=1}^\infty \frac{2n+1}{n^4}=\sum_{n=1}^\infty \frac{2}{n^3} + \sum_{n=1}^\infty \frac{1}{n^4}$$
Another approach: Here $a_n=(2n+1)/n^2(n+1)^2$.
Take auxiliary series whose nth term $v_n =1/n^3$ ; which is convergent as $p>1$ in the power of n appearing in denominator.
Now, use Ratio test to observe that $a_n/v_n → a$ finite non zero number as $n→\infty$. Hence $a$_n also converges. | 2019-12-07T09:45:27 | {
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https://math.stackexchange.com/questions/607128/a-contest-question-on-probability | # A contest question on probability
Let A denote the set of those 6-digit positive integers in which each of the digits $1, 2, 3, 4, 5, 6$ appears exactly once. An integer in A, having standard decimal representation $d_1d_2 \cdots d_6$, is called nice if for each $k = 1, 2, 3, 4, 5$ the set {$d_1,.., d_k$} contains at least one digit strictly greater than $k$ (for example, 234561 is nice, whereas 321645 is not nice). From $A$, an integer $n$ is randomly picked. What is the probability that $n$ is nice?
My try:
Run the last digit from 1 through 6 and find the combinations by scanning the digits greater than the last digit across the first five positions in each of these scenarios. It is pretty cumbersome and I am wondering if there could be anything simpler by means of symmetry or something like that as it does not give us any pattern. :
$234561 - 5!=120$
$345612 - 4*4! =96$
$456123 - 3*4!+3*3! = 90$
$561234 - 2*4!+4*3!+6*2! = 84$
$612345 - 1*4!+3*3!+8*2!+13*1! = 71$
$123456 - =0$
Total number of ways = $6! = 720$
Required probability = $$\frac{120+96+90+84+71}{720} = \frac{461}{720}$$
A simpler solution would be very much appreciated after verifying the answer.
• It's late here, and I'm note sure what you did, but I just found $461$ instead of $446$, but perhaps I made a mistake ? – Xoff Dec 14 '13 at 23:30
• Could you show your working? Did you take the cue from me to work or is it something simpler than what I did? Let me know tomorrrow when you have a chance. It could be possible that I made a mistake too!! – Satish Ramanathan Dec 14 '13 at 23:34
• I made some recurrence relations and found the sequence oeis.org/A003319 which is indeed what you want. If needed, I'll give an explanation tomorrow. – Xoff Dec 14 '13 at 23:44
• The definition of nice integers reads like the following and nothing else: $d_5$ has to be 6-- since it's the only digit greater than 5. When $d_5=6$, $d_4=5$, .., $d_1=2$. And the only nice integer is 234561. something missing here. – ashley Dec 15 '13 at 0:14
• @satishramanathan I agree with the answer of obinna. – Xoff Dec 15 '13 at 8:27
To make the problem more general, let's say that we are interested in $M$ digit positive integers which contain numbers $1,2,3,...,M$.
A number is called nice if for each $k \in {1,2,3,...,(M-1)}$ , at least one digit out of $d_1,..,d_k$ is strictly greater than $k$.
We are interested in counting the numbers which are not nice.
Notice that if a number is not nice, it must have some index which is the first index that violates the nice property. i.e. there must be some minimum index $j+1$ such that $d_1,..,d_{j}$ has some $l \le j$, such that $d_l>j$, but $d_{j+1}$ has no such $l$.
Let's define $T(n)$ as the number of $n$ digit numbers for which the smallest digit that violates the nice property is $d_n$. It's easy to see that $T(M)$ gives the exact number of $M$ digit nice numbers.
Also, notice that since the smallest position, $i$ which violates the nice property has $d_i$ we see that $d_1,d_2,...,d_i$ must be permutations of $1,2,3,...,i$ for which $d_1,d_2,...,d_{i-1}$ obey the nice property but $d_{i}$ doesn't. This is because if we had some integer greater than $i$ among $d_1,d_2,...,d_{i}$, then $d_i$ must also obey the nice property.
Also note that we can count the number of nice numbers by considering the smallest position which violates the nice property over all $n$ digit numbers.
In particular once we know what this minimum violating index is, we can simply permute the remaining numbers. This leads us to the following recurrence relation
$$T(n) = n! - \sum_{i=1}^{n-1} T(i)(n-i)!$$
$$T(1)=T(2)=1$$
Note that $\sum_{i=1}^{n-1} T(i)(n-i)!$ is the total number of numbers that are not nice.
Going back to the original question, the probability is simply
$$\frac{T(M)}{M!}$$ where $M=6$.
We see that $T(6)=461$ and hence the answer is $\frac{461}{720}$
Xoff pointed out in a comment above that $T(n)$ has an OEIS entry
• Nice solution, This is exaxtly what I was looking for. Thanks to Xoff too!!. – Satish Ramanathan Dec 15 '13 at 15:09
• @Xoff,obinna: I found the extra 5 counts missing and you guys are right on the answer. – Satish Ramanathan Dec 15 '13 at 19:31 | 2019-10-21T14:59:13 | {
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http://math.stackexchange.com/questions/275171/simple-calculus-function/275174 | # Simple Calculus Function
I am doing a course in calculus and I was given this problem :
Given that $f(x)=3x^4−6x^3+4x^2−7x+3$, evaluate $f(−2)$.
The answer is meant to be 129 according to the tutor, but no matter how many times I try and work it out I can't see how they got that answer.
-
Please tell us what you tried, and what answer your attempted solution gave. – Andrew Uzzell Jan 10 '13 at 14:40
The answer is correct, and it's just substituting values, what problem are you having? – MyUserIsThis Jan 10 '13 at 14:42
My second term was -48 when it had to be positive. – 8BitSensei Jan 10 '13 at 14:51
It appears your problem is with order of operations.
Let's look at an easier function: $$g(x) = 6x^4$$
What is $g(-2)$?
$$g(-2) = 6(-2)^4$$
Remember--you evaluate exponents before multiplication: $$g(-2) = 6(16)$$ $$g(-2) = 96$$
If the positive/negative thing is still hard, just turn the exponent into multiplication: $$g(-2) = 6[(-2)(-2)(-2)(-2)]$$ $$g(-2) = 6[(-2)(-2)(4)]$$ $$g(-2) = 6[(-2)(-8)]$$ $$g(-2) = 6(16)$$ $$g(-2) = 96$$
### EDIT:
From comments, it appears the error wasn't with raising the negative to an even power, but rather with the second term. The answer has been dealt with in other responses, but I'll include it here for archive purposes:
$$f(x)=3x^4−6x^3+4x^2−7x+3$$ $$f(-2)=3(-2)^4−6(-2)^3+4(-2)^2−7(-2)+3$$ $$f(-2)=3(16)−6(-8)+4(4)−7(-2)+3$$ (Note the minus signs in front of the six and seven. I now multiply out those negatives, which makes them positive.) $$f(-2)=3(16)+6(8)+4(4)+7(2)+3$$ $$f(-2)=48+48+16+14+3$$ $$f(-2)=129$$
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I found that my problem was that my final abstract was 48-48+16+3, when it should have been, 48+48+16+3, still not entirely sure why the second term becomes positive though. – 8BitSensei Jan 10 '13 at 14:54
Oh! ok. I added that part to my answer now. – apnorton Jan 10 '13 at 16:09
Thank you, the tutorial I was using did not show any working out with the answer, that made it difficult to work out where I was going wrong. – 8BitSensei Jan 10 '13 at 16:12
Observe the magic of color:$$f(\color{#C00}{x}) = 3\color{#C00}{x}^4 - 6\color{#C00}{x}^3 + 4\color{#C00}{x}^2 - 7\color{#C00}{x} + 3$$Now, we essentially have to replace our colored expressions with numbers.$$f(-2) = 3(-2)^4 - 6(-2)^3 + 4(-2)^3 - 7(-2) + 3$$Now use the order of operations to simplify. For example, $3(-2)^4 = 3\times(-2)^4 = 3\times16 = {48}$.
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Ummm... it's $f(-2)$ The downvote is not mine, but that's probably the reason you got it... – apnorton Jan 10 '13 at 14:44
Thank you, turns out that my hyphen key was broken and I was not looking at what I was typing. :-) – Parth Kohli Jan 10 '13 at 14:45
$$f(-2)=3\cdot(-2)^4-6\cdot(-2)^3+4\cdot(-2)^2-7\cdot(-2)+3=48+48+16+14+3=129$$
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(−2)3 = 16, so -6⋅16 = -48, how does it become positive? – 8BitSensei Jan 10 '13 at 14:46
The first and third term have an even power, so everything will be nonnegative. The second and fourth term have odd powers, but we're subtracting them, so we're subtracting a negative number which makes it positive. – Clayton Jan 10 '13 at 14:47
Also, $(-2)\cdot 3=-6$, not $16$, and $(-2)^3=-8$. – Clayton Jan 10 '13 at 14:52
the 3 is a exponent of (-2), not very clear in my comment. – 8BitSensei Jan 10 '13 at 14:58
I thought that might be the case, and that was why I wanted to point out $(-2)^3=-8$. – Clayton Jan 10 '13 at 14:59
$$f(x)=3x^4−6x^3+4x^2−7x+3=(x+2)(3x^3-12x^2+28x-63)+129$$
$$\implies f(-2)=0(3x^3-12x^2+28x-63)+129=129$$
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Horner's method? $$3 + x \cdot ( -7 + x \cdot (4 + x \cdot (-6 + 3\cdot x)))=$$ $$=3 + (-2 )\cdot ( -7 + (-2) \cdot (4 + (-2) \cdot (-12)))=$$ $$=3 + (-2) \cdot ( -7 + (-2) \cdot (4 + 24))=$$ $$=3 + (-2) \cdot ( -7 + (-2) \cdot 28)=$$ $$=3 + (-2) \cdot ( -7 - 56)=$$ $$=3 + (-2) \cdot (-63)=$$ $$=129$$
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Why do you want to make it so difficult......... – Applied mathematician Jan 10 '13 at 14:52 | 2016-02-11T00:32:04 | {
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https://khan-academy.appspot.com/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/8th-slope-intercept-form/v/linear-equations-in-slope-intercept-form?_escaped_fragment_= | Linear equations and functions
# Slope-intercept form
## More examples of constructing linear equations in slope-intercept form
Linear Equations in Slope Intercept Form
Back
## More examples of constructing linear equations in slope-intercept form
Discussion and questions for this video
I don't understand why we divide by three at about 7:50. help?
He is basically simplifying the problem 5/6 multiplied by -3/1, -3 and 6 both are divisible by 3 so he simplifies the equation making it 5/2 multiplied by -1 which then results in -5/2
how do you change from point-slope form to slope-intercept?
y - y1 = m(x - x1)
=> y - y1 = mx - mx1
=> y = mx - mx1 + y1
then simplify.
With an example containing actual numbers, you see it's much easier than the steps above imply:
y - 7 = 4(x - 2)
=> y - 7 = 4x - 8
=> y = 4x - 8 + 7
=> y = 4x - 1
1 Comment
Why do we use X as the common variable in algebra? X is also the horizontal axis in graphs.
That's not a coincidence, as you will see when you start graphing algebraic equations.
1 Comment
What is the difference between f(x) and y? Sal said that he assumed they were equal, but can there be cases where they aren't? How would that work?
They're the same thing, but in higher math classes you use f(x) because you graph functions and y can be used as a variable. Say you have a question that says
f(x)=3x + 4, what is f(2), you would simply plug in 2 for x and get 10. It's much more useful in more advanced math.
You can also watch this video.
At first when you are learning functions, you would think that they are silly and you can use y instead. But later on, you will realize that y is silly.
I hope this helps!
1 Comment
at 3:27 shoudnt it be y2-y1 over x2-x1 instead of y1-y2 over x1-x2
Believe it or not, they are the same thing. Try it.
Why is 6/-3 equal to -6/3 or -2?
6/-3 means 6 ÷ -3
A positive divided by a negative is a negative, and 6 ÷ 3 = 2.
Therefore 6/-3 = -2
-6/3 means -6 ÷ 3
A negative divided by a positive is a negative, and 6 ÷ 3 = 2
Therefore -6/3 = -2
-2 = -2
Since 6/-3 = -2, -6/3=-2, and -2=-2 (i.e. they are all equal to the same thing), they must also be equal to each other.
This same process could be done for any numbers meaning that
(-x)/y = x/(-y) = -(x/y)
or in words: A negative in a fraction can go in front, in the numerator, or in the denominator, but not both.
1 Comment
to find the slope sal subtracts (at 3:00) 6-0/2-5. check. next problem i paused, tried to figure it out myself and subtracted (at 5:54) 5-0/3-(-3) to find the slope. sal however reversed the order and did 0-5/-3-3. we ended up with very different answers. how do i know which point to subtract from which point? he did them in two different orders.
Ramona, it looks like you are on the right track. Continuing from your set up, you should get this result:
slope = (5-0)/(3-(-3)) = 5/(3+3) = 5/6
That's the same answer Sal got to in the video.
To answer your question about knowing which point to subtract from which point - either way is correct! It is _your choice_ which is the 'first' point and which is the 'second' - just make sure you make the _same_ choice for both the x and y coordinates. In the video, Sal chose to set up the calculation 'green minus orange' and you set it up 'orange minus green'. But either way, once all the arithmetic is done, you should have the same answer. (Why? Note what happens at 7:01: _"...the negatives cancel out."_
At 8:37, if b in the equation y = mx+b is 5/2 or 5 halves, how would it be plotted on the coordinate plane in fraction form as the y intercept? Isn't the y intercept supposed to be a whole number, so that it can be plotted on the coordinate plane?
The y-intercept can be any number, it need not be a whole number (usually is not). You just plot 2.5 as best you can, half-way between 2 and 3.
This isn't exactly related to the topic, but why does 6-0/2-5 become -6/-3 ?
Why isn't it just 6/-3 ? I never understood this and it always stuffs me up.
Tomm0,
Your mistake is that (6-0) is +6, not -6
If you had a line that included the points (6,2) and (0,5), the slope would be (change in y)/(change in x) which would be (6-0)/2-5)
(6-0) is +6 , not -6
(2-5) is -3
So the fraction is 6/-3
which reduces to -2/1 = -2
I hope that helps make it click for you.
1 Comment
How Would You Work This Type Of Problem When It's Set Up In A Table???
do you mean if its like
x-2/3/4/5/6
y-4/6/8/10/12 you have to find a pattern (in this case y is twice the amount of x) so y=2x
I don't get it. At 3:39 is confusing. Why can't we use the slope equation to get the slope and then the slope would be 2 not -2. Am I missing something? He said if it's switch then it would be be negative?
I think I understand what you are trying to say, if you are talking about the equation y2-y1/x2-x1. So you would have 0-6/5-2. First, if we think about it, the slope will be the same no matter if it is from (2,6) to (5,0) or (5,0) to (2,6). Just know the slope is the same, so we can do it Sal's way of 6-0/2-5 or the way 0-6/5-2. You probably thought that 0-6 was just 6, but 0-6 is -6. Then you get 5-2 which is 3, so -6/3. If you thought 0-6 was 6, you would have got 2 as the slope. I think that is what you are asking
Comment if you get it or if you meant something else
write th equation of the line in slope- intercept form if the slope is 1/5 and the y-intercept is -9
when you put an equation in y intercept form you always start with y= the slope goes with the x and the intercept is the next term.
y = mx +b
m is the slope, b is the intercept
y= 1/5x -9
What if the slope is undefined, but you are given a point. What do you do?
Hannah,
If you have a linear equation where the slope is undefined, then the denominator of the slope must be 0 since anything divided by 0 is undefined.
The denominator of the slope is the change in x
So the change in x is zero.
So if your point was (1,3) and the slope is undefined, you know that some other point has a change in y but x changes by zero,
So x would always by 1.
Points (1,3), (1,4), (1,5) would all be on your line.
The equation of this line is x=1+0y or x=1
If you graphed it you would have a vertical line. going through the x axis at the point given as the x value in your original point.
I hope that makes it click for you.
if you have 2lbs of jellybeans for $9, what of .5 lbs? 3lbs? 4 lbs? I would solve this using proportions: if you have the proportion$9 per 2 lbs., you can divide both by 2 to get $4.50 for 1 lb. From here you can multiply both by 3, 4, or 1/2 to get your answers. I'll leave that up to you ;D at 1:48 how did he find the y-intercept of the equation if it was not written in standard form ? 2 Votes 1 Comment He knew that line goes through the point (4/5, 0), right? 4/5 is an x value and 0 is an y value of this point. Also he knew that slope (m) is equal to -1. Therefore by substituting the values to the formula of y = mx + b we get 0 = -1 * 4/5 + b which is an equation with one variable and just like Sal did - 0 = -4/5 + b therefore b = 4/5. what do i do when the equation is Y = X it means that whatever your X value is,that will also be your Y value. when X=1 Y=1 when X=5 Y=5.and therefore the slope is 1(1/1). it follows the usual equation of a line Y=mX + b.when X is 0 so is Y, that's why the Y intercept (b) isn't in the equation, its 0. AT 2:23 I don't get how he got -x Try watching the video again. How do I write the slope-intercept form of the equation of the line through: (3,-2), if the slope is undefined? Hi, If a slope is undefined, it means that the x value does not change. Remember that slope is calculated as the change in the y value divided by the change in the x value. If x does not change, then we are dividing by zero, which is undefined. Therefore, an undefined slope is a vertical line through the point you were given. You cannot really write it in a correct slope/intercept form. Usually this is written as: x=whatever the x value of your point is. In your case, the equation would be x=3 And the graph would be a vertical line running through the 3 on the x axis. Hope that helps :-) I didn't understand the equation f(1.5)=-3,f(-1)=2? This is because our function here is *f(x)=(x)-2* So each time we put in a value of x we multiply it by -2 Ex. f(1.5)=(1.5)*-2=-3 f(-1)=(-1)*-2=2 how do you find a y intercept in a graph i don't understand what is going on when you have to solve for b, can i please have some help? When solving for b we put the values (x,y) of the given point (which sit on the line) in the the equation of the line in its general form: y=mx+b, where m is our slope (in this example -1) Than we separate b on one side of the equation and numbers on another. how is it " m=delta y over delta x? " why is it not delta x over delta y? slope can also be though of in terms of rise over run. since delta y is how far the graph is rising and delta x is how far the graph "runs", thats why the slope is delta y over delta x. It is just a mathematical convention that slope is measured that way. Is it m=y2-y1 over x2-x1 or m=y1-y2 over x1-x2? I need help. Please and Thank You. It really doesn't matter. As long as you have the y's over the x's, you'll get the slope of your line. In fact, the two ratios are equal anyways because y1-y2=-(y2-y1), annd the same for the x's, so multiply one way by -1/-1, and you'll get the other. About the substitution of the points. What if you don't have a zero? Do you just go with the points that are closest to one? @ 1:07} Do you have to use m to represent the slope or do you choose any variable? Well technically, you have to use m to represent the slope or else it wouldn't be right. How do you make it easier to convert from a problem such as -5x + 2y = 6 without getting confused over negative numbers? Tommy, Typically to isolate y (so you can get into the form y = mx + b) you would start this problem by adding 5x to both sides (to get rid of the -5x on the left). This first step would give you: 2y = 5x + 6 Bingo. Your negative sign is gone. Now a you divide both sides by 2: y = 5/2x + 3 ... and you're done! at 3:53 , why does Sal put a negative when 6 - 0 is not negative? because he went the other way around just to prove you would get the same value for the slope. how do i find an equation of the slope line? I do not get your question. it is y1 -y2 over x1- x2 How do you solve a problem when it asks you, " Write the equation that describes each line in slope-intercept form." and the problem is [slope= -2/7, (14,-3) is on the line]? PLEASE ANSWER SOON OR PUT A VIDEO UP ON KHAN ACADEMY ABOUT MY QUESTION!!!!!!! well look at it this way: slope= -2/7, (14,-3), look at in the video 10: 11 through the rest of the video, I hoped this helped you Sophia. And when graphing these can if you can switch the 6/-3 around will it change the graph points Stephanie, If I understand your question correctly, you are asking how to find slope. If you are given points, (example: 1,2, 3,4) then you can find your slope using the equation: Y2 - Y1 over X2-X1. In this case, Y2 is 4 because this is the second y in the two points, Y1 is 2, X2 is 3, and X1 is 1. We now have: 4-2 over 3-1 which is 2 over 2 which can be simplified to 1. The slope of the equation is 1. Hope this helped! Isn't -2-1=3 ? Sal said that it equals to 2?! The answer is not 2. Probably just a simple mistake 2 Votes 1 Comment what is the *slope-intercept formula* and x and y are points on the graph. At 6:39, i understand that Sal started with the second set of coordinates (-3,0) first but, i thought it didn't matter. I later tried doing the equation with the first set of coordinates first (3,5) and it doesn't work because you end up dividing 5 by 0. This doesn't work. Can someone explain how to know which coordinate set to use first? I am a little confused. Thank you, Elena You were initially right: the order doesn't matter when finding slope. (5 -0)/(3--3) = 5/6 (0-5)/(-3-3) = -5/-6 = 5/6 Both give the same answer. 3 Votes 1 Comment At 6:11, how come its 0-5 instead of 5-0? Is that crucial to get the correct answer or does it not matter, because you are still finding the change? (y2 - y1)/(x2-x1) works either way. Just make sure that whatever point you started with for the difference in y in the numerator is the same point you start with for the difference of x in the denominator. 3 Votes 1 Comment i really do not understand this video. how would you write in a equation form: Jan wants to buy maps and atlases. the maps cost$2 each and the atlases cost each $5. If she buys 3 atlases and spends$25, how many maps can she buy?
Can you plze explain how to write an equation with this example?
An equation for this is
t = 2m + 5a
where t = total, m = maps, a = atlases
You're also told that a = 3 and t = 25, so
25 = 2m + 5 * 3
Solve for m.
Will you please show how to find the slope-intercept form for the line satisfying the following:
x-intercept 3, y-intercept 2/3?
Well remember x represent the horizontal line and so your first point will be (3,0) while the vertical line (y) represent a point in y lower than 1 (dividing 1 in thirth so you take 2 of them and you have 2/3). Finally your second point is (0,2/3) and join the dots.
m= 3 - 0/0 - 2/3= -9/2 = -4 1/2... y= -9/2x+2/3 (mx+b)
Sorry if I was a little obvious but my intention was be clear!
How do you write the the equation if its a word problem?
Write down what you know and what you don't know and what it is asking for clearly. Try to find a relationship. Remember that all lines have a slope (m), which is basically how steep it is, which is simply a measure of how far it moves along the x axis over how far it moves along the y axis, and a point of origin where it crosses a known point. In normal notation, this is the y intercept (b), which occurs when x = 0.
what is the x for ?
Its the value of the number on the x axis.
Oh yeah, - x is just a variable.
If we can plot an equation in which no x is defined ( e.g. 2y=-8 where we assume 0x) why can we not graph an equation in which no y is defined (e.g. 4x=-8)
You can graph 4x=-8. When you divide both sides by 4 you get x=-2. This is simply a vertical line which crosses the x intercept at -2. That vertical line includes all points in which x is equal to -2.
Is there a particular reason why Sal used less common variables in this video? Or did he just want to through in a variety?
Well, if you are talking about him using the variables: m, x, y, and b, he is using the equation of slope-intercept form, which is *y = xm + b*. "b" is the y-intercept and "m" is the slope of the line, and x + y are points on the line, of course. This equation makes a straight line. If you want to know more, just look here: https://www.khanacademy.org/math/algebra/linear-equations-and-inequalitie/equation-of-a-line/v/graphing-a-line-in-slope-intercept-form.
Also, the f at around 10:32 is a symbol for function.
at 2:57 in the videos sal used these triangular shaped things in the equation what are they and what do they represent
They are the Greek symbol delta. They are accepted in math and science as the word "change".
1 Comment
How do you write an equation in slope intercept form?
A linear equation in slope-intercept form is y = mx + b. Where "m" is your slope and "b" is your y-intercept.
How to graph y=x
The y-intercept would be zero and the slope would be 1.It would have a constant increase and go in a straight equal line.
Hi you have done that in a problem
Line contain points (2,6) and (5,0)
m=y1-y2/x1-x2
but the formula is
http://cs.selu.edu/~rbyrd/math/slope/slo_eq1.gif
Since (x1,y1) and (x2,y2) could be either of the two points, it it true that is doesn't matter which way the slope formula is written. It is important that you use your original point for both the numerator and denominator or your slope will have the opposite sign it should have.
Around 4:12, Sal had a problem which said,
"The line contains points (2,6) and (5,0)"
In this problem, Sal subtracts this way, "6-0/2-5"
In the next problem 6:48 Sal does his problem "The line contains points (3,5) and (-3,0)" and SWITCHES AROUND compared to his problem before, by subtracting "0-5/3 - -3".
I'm not a mathematical genius compared to Sal so please help me see where I'm going wrong. I thought the problem would turn out like this,
"5-0/ -3-3"
Sal's Equation: "0-5/ 3- -3"
Christian, that's ok! You see, when you want to do these problems, you usually have two sets of points, right? Let's call them (X1,Y1) and (X2, Y2). Now, in school, usually the general formula taught for finding slope is (Y2 - Y1) divided by (X2 - X1). But, as you'll learn later on (if you aren't taking higher level math yet) that the equation is actually "CHANGE OVER Y" divided by "CHANGE OVER X". The order in which you subtract doesn't matter, as long as you subtract the Ys and divide them by the subtraction of the Xs. If you'll notice, your equation gives you the answer "5/-6," which is "-5/6". Sal's answer is also "-5/6".
Hope that helps!
what does delta mean?
Delta is a Greek letter: http://en.wikipedia.org/wiki/Delta_(letter).
Mathematicians use the letter delta to represent a change in value of a variable. For example, ∆x is the change in the x variable. So when you see ∆y/∆x you know you're dealing with slope as it's the change in y compared to (or over) the change in x. This is also called rise over run: rise is how far you change vertically. Run is how far you change horizontally.
what is linear equation
Yogi,
A linear equation is an equation which graphs to form a line.
It usually takes the form of y = ax+b where a and b are constants such as y=2x+1.
When you graph it, it forms a strait line, so they call it a linear equation.
How are you supposed to find the fraction
aka the answer?! I need help.. any1?
IM THE SHORT GUY PEOPLE PLZ HELP!.... and thx and i wonder who else needs help ..
TCulton,
The slope intercept form is y=mx+b
You have the form x+2y=5
In the slope intercept form, the y is all by itself on the left. So convert
x+2y=5 to a from that has y by itself on the left side of the equation
x+2y=5 First get rid of the x on the left by subtracting x from both sides.
x-x+2y=5-x so
2y=5-x Now get rid of the 2 by dividing each side by 2
2y/2 = 5/2 - x/2 so
y=5/2 - x/2 Now change the order on the left
y=(-1/2)x+5/2
So it is now in slope y-intercept form. The slope is -1/2 and the y-intercept is the point (0,5/2)
At 7:20 why did he choose the second point, what if the first piont??
It doesn't matter which point you select. Sal probably picked the 2nd point because the zero usually makes some of the math easier to do. If you were to use the 1st point to find "b", you would get the same value for "b". Here it is:
5 = 5/6(3) + b
5 = 15/6 + b
5 = 5/2 + b
10/2 - 5/2 = b
5/2 = b
how do i make 6x - 3y = -9 in slope intercept form
Isolate y:
6x - 3y = -9, add 3y to both sides
6x = -9 + 3y, then add 9 to both sides
6x + 9 = 3y, then divide everything by 3
2x + 3 = y
you can always enter this kind of stuff into wolfram alpha and they will solve it and graph it for you: http://www.wolframalpha.com/
how do i solve 6x - 3y = -9 for y
check my previous answer, but be aware that the value of y depends on x. Every solution is a point that is on the line 6x - 3y = -9
How would you put a problem like 2x - 5y = 15 into Slope intercept form?
To be in slope intercept form, you need "y" to be by itself on one side.
So, in your example, move the 2x (subtract it from both sides:
-5y = -2x + 15
Now, you need to make -5y into only y, so divide entire equation by -5:
y = 2x/5 - 3
1 Comment
i dont understand why negatives cancel out
A negative is the opposite of a number. The opposite of 2 is -2. The opposite of -2 is 2. Two negatives means 'the opposite of the neg. number', which is positive. Therefore --2=2. Three negatives (---2) is the opposite of the opposite of the opposite of 2.
The thing that throws me off about slope-intercept form is that when I see the y variable, my brain wants to make that the y-axis. Does any one else find this confusing? Why is it like this? Is it because the coordinates are also written (x,y)?
y is a variable, and most often (at this level of math) it is the _dependent_ (sometimes also called the _output_) variable, and it _*is*_ associated with the y axis. Given a value of x (the _independent_ or _input_) variable, what pops out is y. The ordered pair of coordinates (x, y) can be thought of as (input, output) or (independent, dependent). So when you see (1, 5) what that means is there is a function, or expression, that when 1 is input, that is, when x=1, then what is output is 5, or y=5.
I assure you that you will grow accustomed to this representation over time. It is very common, and very useful.
Why does he subtract the two points when one is greater in x and y values than the other?
Humza,
For some reason, I can't access the video right now, but what Sal was probably trying to do was to calculate the slope of the line from two points that are on the line. In order to do that, the formula is m=(y2-y1)/(x2-x1). Basically that means that you call one of the points, "point 1" and the other "point 2". Then you subtract point 1's y value from point 2's, and divide that by point 2's x value minus point 1's x value.
i cant figure out what (34,87) (51,25) can u plaese help me
I know that the slope is the change in y divide by the change in x. But what are you getting, if anything, if you divided the change in x by the change in y?
George,
If you take the *negative* of change in x over change in y, you would have the slope of a line perpendicular to the original line.
I thought it was y2 - y1 over x2 - x1.
and it is, the short way to write it like you did is just to put the little triangle before the x and the y. tha just means change!
At 3:28 why does he put the first coordinate first? the equation is y>2- y>1
x>2- x>1
You can pick either one to be honest. you can start with the first coordinates or the second coordinates it doesn't matter the order.
he says change in y but the way i learned is rise over run so which one is it
Those mean the same thing. "Rise" corresponds to "change in y", and "run" corresponds to "change in x". See, if we denote "change in y" by ∆y and "change in x" by ∆x, "rise over run" refers to ∆y / ∆x, which is the slope.
1 Comment
how would you determinr the the rate of change is constant for tables given.
Spencer,
If the
(change in y) / (change in x) is always the same, then the rate of change is constant.
If you table has the following
(3,5)
(4,7)
(6,11)
(7,14)
You could calculate (change in y)/(change in x)
For the first two terms (7-5)/4-3) = 2/1 = 2
The second & third term (11-7)/(6-4) = 4/2 = 2
So far the change is constant
The third and fourth term (14-11)/(7-6) = 3/1 = 3
This last point in our table shows we have a different ratio for
(change in y)/(change in x) so the table does not have a constant rate of change.
I hope that helps make it click for you.
what does f(x) - f of x mean?
f=function. If you have an equation like y = 3x + 7 then you can also say that y is a function of x, or f(x). f(x) = 3x +7.
Sal, you kept saying that the points listed MUST satisfy the equation, what if the points do not satisfy the equation, I'm just wondering?
Points that dont' satisfy the equation cannot be on the line.
Sorry, could someone explain to me why you multiply x by the slope?
When x increases by 1, y increases by the slope m.
Consider the points (0,b) and (x,y).
Here the first coordinate increases by x and therefore the second coordinate should increase by m times x:
y = m*x + b
When finding the slope from the points could you just stack and subtract? Would that work everytime for points?
that was the way i was taught. it never gave me any trouble before. that math is much easier to do in your head anyway and maybe only need to write one or two numbers down till you get the answer.
what if all you have are two points? how would you find the y-intercept? I still don't understand!!
Two points define a line, and if you define a line, then you define the y-intercept. If you have two points, you can calculate the slope of a line between them. Then you can write y = mx + b, and you know x. Plug in the y and x of one point, and solve for b. That's the y -interecept.
Please explain how to graph the equation and state the slope of the line if the slope existed Example x=-8
x = -8 is a vertical line, because it says no matter what the y value is, x will always be -8.
The slope can be said to be "infinite", but since infinity is not a number, the slope in this case is "undefined", and that's what you should put down when taking a test.
The graph, once again, is just a vertical line extending in both directions forever, at x = -8.
Hope I helped!
I don't get this?
Help!
Well, It'd help if you told us *what* you don't get..
Just a suggestion.
- tempest | 2015-07-28T01:26:53 | {
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http://math.stackexchange.com/questions/49435/what-is-a-free-parameter-in-a-computational-model/49503 | What is a “free parameter” in a computational model?
In many articles regarding computational models of some particular phenomenon, there seems to be a consensus: "the smaller the number of 'free parameters' in the model, the better". So, what is meant by "free parameter", and why is it less desirable to model something with such a parameter?
Thanks!
-
A free parameter is one that can be adjusted to make the model fit the data. If I make a model that says $A$ is proportional to $B$, there is one free parameter, the proportionality constant. If my model has a specific value of the proportionality constant, there are no free parameters.If I say that $A$ is a quadratic function of $B$, there are three free parameters, $a,b,c$ in $A=aB^2+bB+c$. That makes it easier to fit the data, even if my model is not correct, so it is a less impressive.
-
The following marvellous quote from Freeman Dyson's "A meeting with Enrico Fermi" contains both an example for your first question and an answer to your second.
[Enrico Fermi] delivered his verdict in a quiet, even voice. . . . "To reach your calculated results, you had to introduce arbitrary cut-off procedures that are not based either on solid physics or on solid mathematics."
In desperation I asked Fermi whether he was not impressed by the agreement between our calculated numbers and his measured numbers. He replied, "How many arbitrary parameters did you use for your calculations?" I thought for a moment about our cut-off procedures and said, "Four." He said, "I remember my friend Johnny von Neumann used to say, with four parameters I can fit an elephant, and with five I can make him wiggle his trunk." With that, the conversation was over. . . .
- | 2015-02-01T16:41:20 | {
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https://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/hilberts-hotel-busses-injectivity-surjectivity-27001.html?s=6cdbac81e3fd803e8822e972d4366152 | # Thread: Hilbert's hotel and busses : Injectivity - Surjectivity
1. Hey!!
I look at the problem with hilbert's hotel and the busses. First one bus with infinitely many guests, then four busses with infinitely many guests and then infinitely many busses with infinitely many guests.
At each case we move all the guests that are already in the hotel to the odd room numbers.
So the even room numbers are free for the new guests.
In the first with one bus we use the formula $n = 2(i-1)+2$ where $i\in \mathbb{N}\setminus\{0\}$ is the number of place in the bus and $n$ is the even room number that this guest will get.
In the second case with the four busses we use the following:
\begin{equation*}\begin{matrix} & 1. \text{ Bus } & 2. \text{ Bus } & 3. \text{ Bus } & 4. \text{ Bus } \\ 1. \text{ Person } & \text{ Room } 2 & \text{ Room } 4 & \text{ Room } 6 & \text{ Room } 8 \\ 2. \text{ Person } & \text{ Room } 10 & \text{ Room } 12 & \text{ Room } 14 & \text{ Room } 16 \\ 3. \text{ Person } & \text{ Room } 18 & \text{ Room } 20 & \text{ Room } 22 & \text{ Room } 24 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ i. \text{ Person } & \text{ Room } 8(i-1)+2 & \text{ Room } 8(i-1)+4 & \text{ Room } 8(i-1)+6 & \text{ Room } 8(i-1)+8\end{matrix}\end{equation*}
In the last case with the infinitely many busses we use the formula $2^{i}\left (2(j-1)+1\right )$ where $i$ is the place in the bus and $j$ is the number of the bus.
Using this we get \begin{equation*}\begin{matrix} \text{Place}/\text{Bus} & 1. \text{ Bus} & 2. \text{ Bus} & 3. \text{ Bus} & 4. \text{ Bus} & 5. \text{ Bus} & \ldots \\ 1 & 2 & 6 & 10 & 14 & 18 & \ldots \\ 2 & 4 & 12 & 20 & 28 & 36 & \ldots \\ 3 & 8 & 24 & 40 & 56 & 72 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \end{matrix}\end{equation*}
I want to describe the injectivity and the surjectivity of the maps, and if the maps for the old and the new guests are injective and surjective.
The map for the old guests is injective, since it means that the new room thet they get either had no previous guest or one.
The same holds also for the map for the new guests. Their new rooms either had no previous guests or one.
As for the surjectivity. As for the rooms of the old guests, only the odd room umbers have a preimage and so the map is not surjective. As for the rooms of the new guests, only the even room umbers have a preimage and so the map is not surjective.
Is that correct?
2.
3. mathmari:
Quote:
At each case we move all the guests that are already in the hotel to the odd room numbers.
i woulndn't. For example, in tne second case where, after all rooms are full, four new buses each containing an infinite (we really need "countably infinite") number of guests come I would move guests in room "n" to room 5n. Then put the guest from the first new bus in rooms 5n+ 1, from the second bus in 5n+ 2, etc.
For the last problem where a (countably) infinite number of busses, each containing a (countably) infinite number of guest, consider the set of rational numbers. That set is countable so we can "number" each rational number. We can think of the original set of guests as correponding to the rational numbers 1/n then the guests from the first new bus 2/n, etc. Put each guest into the integer numbered room corresponding to that rational number. I think that is basically what you have done.
Originally Posted by HallsofIvy
mathmari:
i woulndn't. For example, in tne second case where, after all rooms are full, four new buses each containing an infinite (we really need "countably infinite") number of guests come I would move guests in room "n" to room 5n. Then put the guest from the first new bus in rooms 5n+ 1, from the second bus in 5n+ 2, etc.
So is the idea of my post wrong? Isn't it similar to yours?
Originally Posted by HallsofIvy
For the last problem where a (countably) infinite number of busses, each containing a (countably) infinite number of guest, consider the set of rational numbers. That set is countable so we can "number" each rational number. We can think of the original set of guests as correponding to the rational numbers 1/n then the guests from the first new bus 2/n, etc. Put each guest into the integer numbered room corresponding to that rational number. I think that is basically what you have done.
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I haven't really understood the last part:
"Put each guest into the integer numbered room corresponding to that rational number."
What exactly do you mean?
5. Originally Posted by mathmari
I want to describe the injectivity and the surjectivity of the maps, and if the maps for the old and the new guests are injective and surjective.
The map for the old guests is injective, since it means that the new room they get either had no previous guest or one.
Hey mathmari!!
We don't look at just the new rooms do we?
Nor does it matter whether a room had a previous guest or not, does it?
I think we are talking about the set of old guests, the set of newly arriving guests, and the set of rooms yes?
If so, then initially all rooms were occupied by exactly 1 old guest.
So the initial map of old guests to rooms is both injective and surjective.
After the old guests were moved to new rooms, each room had either an old guest, or no guest.
So the new map of old guests to rooms is injective.
Or do you have different sets or maps in mind?
Originally Posted by mathmari
The same holds also for the map for the new guests. Their new rooms either had no previous guests or one.
Shouldn't it be that each room either had 1 new guest, or not a new guest?
Does it matter that a room had a previous guest or not?
I guess we might also look at the map of all guests to the rooms.
Afterwards all rooms are occupied by exactly 1 guest, so the map is injective.
Originally Posted by mathmari
As for the surjectivity. As for the rooms of the old guests, only the odd room umbers have a preimage and so the map is not surjective. As for the rooms of the new guests, only the even roomn umbers have a preimage and so the map is not surjective.
Is that correct?
The initial map of old guests to rooms is surjective, isn't it?
After they have moved, the map of old guests to rooms is indeed not surjective.
And the map of new guests to rooms is indeed not surjective either.
I wonder if the map of all guests to rooms is surjective.
Originally Posted by Klaas van Aarsen
I think we are talking about the set of old guests, the set of newly arriving guests, and the set of rooms yes?
If so, then initially all rooms were occupied by exactly 1 old guest.
So the initial map of old guests to rooms is both injective and surjective.
After the old guests were moved to new rooms, each room had either an old guest, or no guest.
So the new map of old guests to rooms is injective.
I guess we might also look at the map of all guests to the rooms.
Afterwards all rooms are occupied by exactly 1 guest, so the map is injective.
The initial map of old guests to rooms is surjective, isn't it?
After they have moved, the map of old guests to rooms is indeed not surjective.
And the map of new guests to rooms is indeed not surjective either.
Ah ok!!
Originally Posted by Klaas van Aarsen
I wonder if the map of all guests to rooms is surjective.
In this case all the rooms are occupied and that would mean that the map of all guests to rooms is surjective, or not?
7. Originally Posted by mathmari
In the last case with the infinitely many busses we use the formula $2^{i}\left (2(j-1)+1\right )$ where $i$ is the place in the bus and $j$ is the number of the bus.
Using this we get \begin{equation*}\begin{matrix} \text{Place}/\text{Bus} & 1. \text{ Bus} & 2. \text{ Bus} & 3. \text{ Bus} & 4. \text{ Bus} & 5. \text{ Bus} & \ldots \\ 1 & 2 & 6 & 10 & 14 & 18 & \ldots \\ 2 & 4 & 12 & 20 & 28 & 36 & \ldots \\ 3 & 8 & 24 & 40 & 56 & 72 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \end{matrix}\end{equation*}
Originally Posted by mathmari
In this case all the rooms are occupied and that would mean that the map of all guests to rooms is surjective, or not?
I was looking at your formula for infinitely many buses.
It makes me wonder if every even room gets a guest assigned to it.
Originally Posted by Klaas van Aarsen
I was looking at your formula for infinitely many buses.
It makes me wonder if every even room gets a guest assigned to it.
How can we check that?
9. Originally Posted by mathmari
How can we check that?
For surjectivity we have to verify if every even number can be written in the form of your formula, so that there is at least 1 guest that gets the room with that number.
Btw, for injectivity we should also verify that it cannot happen that 2 guests are assigned to the same room with an even number.
Each number can be written uniquely as a power of 2 multiplied by an odd number can't it?
It follows from the fundamental theorem of arithmetic, also known as the unique factorization theorem, doesn't it?
Does it mean that we can always find a bus passenger for an even room? | 2020-01-18T07:20:22 | {
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https://math.stackexchange.com/questions/3030557/are-there-any-two-numbers-such-that-multiplying-them-together-is-the-same-as-put | # Are there any two numbers such that multiplying them together is the same as putting their digits next to each other?
I have two natural numbers, A and B, such that A * B = AB.
Do any such numbers exist? For example, if 20 and 18 were such numbers then 20 * 18 = 2018.
From trying out a lot of different combinations, it seems as though putting the digits of the numbers together always overestimates, but I have not been able to prove this yet.
So, I have 3 questions:
1. Does putting the digits next to each other always overestimate? (If so, please prove this.)
2. If it does overestimate, is there any formula for computing by how much it will overestimate in terms of the original inputs A and B? (A proof that there's no such formula would be wonderful as well.)
3. Are there any bases (not just base 10) for which there are such numbers? (Negative bases, maybe?)
• If $B$ has $b$ digits then $B < 10^b$ but $AB > 10^b A$. This argument applies in any (positive) base. Dec 8, 2018 at 0:43
• Yes. I admit that was a little unclear given that you're using it to mean concatenation. Dec 8, 2018 at 0:48
• @TobyMak that's still overestimating, just like 20 * 18 < 2018 Dec 8, 2018 at 0:48
• Just a side note: as multiplication of numbers is commutative, those we would need A=B, because otherwise the concatenations A.B and B.A would yield different results whereas AB and BA are equal. Dec 8, 2018 at 18:31
• @PhilippImhof Not necessarily, since you're not asking that the same holds for the reversed order. For instance change a bit the quest and suppose to look for $A, B$ such that $2A\cdot B=A*B$, where * is the concatenation and $\cdot$ the multiplication. Then your reasoning should apply again, but $A=3, B=6$ is a solution.
– Del
Dec 9, 2018 at 11:16
I have two natural numbers, $$A$$ and $$B$$, such that $$A \times B = AB$$.
Do any such numbers exist? For example, if $$20$$ and $$18$$ were such numbers then $$20 \times 18 = 2018$$.
Lets put aside the trivial answer $$A=0$$ and $$B=0$$ and consider both $$A, B>0$$.
You want numbers such that $$A\times B = A\times10^k + B$$ where $$k$$ is the number of digits of $$B$$, that is with $$10^{k-1}\leq B < 10^k$$. So you need $$B=10^k+\dfrac{B}{A}$$ with $$B<10^k$$. From which results $$B>10^k$$ and $$B<10^k$$.
So if there's no mistake there, the answer is no.
• This is really nice. Dec 8, 2018 at 1:34
There is the pathological example $$A=B=0$$.
For the rest:
Let $$B$$ have $$m$$ digits. We have $$AB= A*10^m+B$$ We want $$AB=A* B$$,
We have $$A*10^m+B-A*B = A*(10^m-B)+B>B,$$ because $$10^m>B$$ as $$B$$ has $$m$$ digits. So you always overestimate by at least $$B$$.
From this its also clear, that the result is independent of the chosen base.
Let me generealise a bit:
If we allow $$A$$ to be negative, we need to change the condition to $$AB=A*10^m-B$$.
This leads to $$A=\frac B {10^m-B},$$ but then the right hand side is positive and again we get a contradiction.
So the last possibility is $$B<0$$. But then we first need to define $$AB$$.
A natural way to do this would be $$A(-B)$$.
Then we have for $$A>0$$: $$AB=A*10^m-B$$ which implies $$A=\frac B {10^m-B}.$$ This time, there is no contradiction because of sign issues.
However, now $$-B>0$$, hence $$|10^m-B|>|B|$$, so the right hand side is no integer.
The analogous argument gives also a contradiction for $$A,B>0$$.
So even in the more generalised setting, the answer is no.
If $$B$$ has $$n$$ digits then $$10^{n-1} \le B <10^n$$ and we want $$AB = 10^nA + B$$ or
But $$B<10^n$$ so $$AB < 10^nA \le 10^nA + B$$
So 1) Yes over compensation always
2) by $$10^nA + B - AB = 10^{[\log_{10}(B)]+1}A + B - AB$$
3) The same argument applies to any base $$> 1$$
• By AB for part (2), do you mean A*B? Dec 8, 2018 at 7:55
• Also, I believe (1) should include the case where A = B = 0 Dec 8, 2018 at 8:02
• I don't think we should include $A=B=0$ because $00$ isn't a valid expression. I took it as a given that $A \ge 1$. But if you wish to include that as a case you may. And yes $AB$ is the product of $A$ and $B$. Dec 8, 2018 at 16:59
• It's perfectly consistant. I always used AB to mean A times B and i never used it to mean anything else. I never used any notation for concatenation. Dec 9, 2018 at 5:01
• Of course. The left hand side is $AB$ is multiplying the two digits. And $10^nA + B$ the right hand side is concatinating. Dec 10, 2018 at 2:22
I will use $$*$$ for your concatenation operation, and $$\cdot$$ for true multiplication.
1) If you allow for leading zeroes to be ingored, then $$0*0=00=0$$. Notice that for all pairs of non-zero natural numbers, $$a\cdot b\leq a\cdot10^{\lceil\log_{10}(b)\rceil}$$ but that $$a*b>a\cdot10^{\lceil\log_{10}(b)\rceil}$$. If $$a=0$$ and $$b\neq 0$$ and we ignore leading zeroes, then we are still an overestimation, and if $$a\neq0$$ and $$b=0$$ then we are still an overstimation.
2) Based on my crude estimates above, yes, there is a way of putting bounds on the size of the overestimation, but I don't know if you can do much better honestly.
3) Also based on my crude estimations, if you replace the logarithms with different bases I'm fairly sure that this shows no base greater than 1 works. Base 1 itself actually has both types of behaviour; $$1*11=111>11$$, but $$111*111=111111<111111111$$. Additionally, you have an example of equality in $$11*11=1111$$. There is of course the issue that $$0$$ is a strange object to try and work with in base 1, so let's just ignore that for now...
I can't muster the strength of will to try and prove anything for negative bases. I suspect that negative bases less than $$-1$$ will fail, but it is easy to see that in base $$-1$$ there are trivial representations that will also give you equality; $$11*11=1111$$ where everything in sight is $$0$$ in base 10.
• Those are not base-1 numbers, as in any base b, the allowed digits are from 0 to b-1 Dec 9, 2018 at 3:45
• @BenVoigt: In unary numeral system, you can use any arbitrary symbol for tallying. If you insist, you can use $0$ as repeated symbol. So zero would be represented with an empty string (not very convenient...), one with $0$, two with $00$ and four with $0000$. Dec 9, 2018 at 16:13
• @EricDuminil: That's a fine system and "unary" is a good name for it, but not "base-1", since it has no relationship to positional number systems. It does not qualify as an answer to OP's final question "Are there any bases for which there are such numbers?" Dec 10, 2018 at 0:31
So it was impossible with natural numbers (see other answers) But if you willing to bend the rules a bit, by making that when decimal numbers are involved $$A.a \times B.b = AB.ab$$
then you can find
$$x.99999999999... \times 9.9999999999... = x9.99999999999...$$
or more compactly
$$x.(9) \times 9.(9) = x9.(9)$$
this is possible because $$x.(9) = x + 1$$
and for an example if $$x=4$$
$$5 \times 10 = 50 \Leftrightarrow 4.(9) \times 9.(9) = 49.(9)$$
(1) A slightly different method:
Let $$k>1$$ be an an arbitrary base. We know that $$\lceil\log_kB\rceil$$ is the number of digits in $$B$$ in base $$k$$.
We want to prove if there exists a solution to $$A*B=A*k^{\lceil log_{k}B\rceil}$$+B.
Isolate $$A$$ and $$B$$, $$\:\:1-\frac{1}{A}=\frac{k^{\lceil log_{k}B\rceil}}{B}$$. For positive $$A$$, we have that $$1-\frac{1}{A} < 1$$.
Recall that by the definition of logarithm, $$k^{log_kB}=B$$. $$\:$$ Since$$\lceil x\rceil\geq x$$, we know $$\frac{k^{\lceil log_{k}B\rceil}}{B}\geq1$$. Thus, along with the other answers, we come to a contradiction. One side of the equation is less than 1, the other at least 1.
(3) Now what if $$k<0$$?
I will first point out that there do exist integers $$A$$ and $$B$$ for which $$A * B=AB$$.
We will use the following definition of negative base 10:
A number of the form $$\ldots d_2d_1d_0$$ with numerical value $$\ldots+ d_2(-10)^2+d_1(-10)^1+d_0(-10)^0$$ where every $$0\leq d_i\leq9.$$
Now consider $$A=-4_{10}=-4_{-10}$$ and $$B=8_{10}=8_{-10}$$. Discarding the negative sign, $$AB=48$$.
$$A*B=-32_{10}=4(-10)^1+8(-10)^0=48_{-10}$$.
Another example:
$$A=-9_{10}=-9_{-10}, B=90_{10}=90_{-10}$$. $$AB =990$$.
$$A*B=-810_{10}=9(-10)^1+9(-10)^1+0(-10)^0=-990_{-10}$$. This time we must include the negative sign, but regardless the digits are the same.
So for your third question, provided that we let the negative sign in the concatenation be optional, we can find examples where multiplying two numbers will yield the concatenation of their digits.
Here's my approach: take two natural numbers $$n,m$$ with $$x,y$$ number of digits respectively. Then in particular we can bound $$n \cdot m \leq (9 \cdots \ \text{(x times)} \ \cdots 9) \cdot (9 \cdots \ \text{(y times)} \ \cdots 9) = (10^{x} - 1)(10^y - 1),$$ so $$n \cdot m \leq 10^{x+y} - 10^x - 10^y + 1$$. Now, we can also bound $$nm \geq (10 \cdots \ \text{(x-1 zeros)} \cdots \ 0)(10 \cdots \ \text{(y-1 zeros)} \cdots \ 0) \geq 10 \cdots \ \text{(x+y-1 zeros)} \ \cdots 0,$$ so that $$nm \geq 10^{x+y-1}$$.
This is as far as I've got, there are already better answers around.
• Absolutely right - fixed it now. Probably would've been more standard to swap the roles of $x,y$ and $n,m$ but meh shrugs Dec 9, 2018 at 21:08 | 2022-05-25T23:50:22 | {
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https://math.stackexchange.com/questions/3311137/number-of-ways-to-split-n-up-into-k-baskets-such-that-different-arrangements/3311148 | # Number of ways to split $N$ up into $k$ baskets such that different arrangements of the $k$ baskets are considered equivalent?
I've been considering the problem of integer partitions and while there have been some answers for related questions, I haven't came across a solution for my following problem.
Suppose you have $$N$$ balls and wish to throw it into $$k$$ indistinguishable baskets. Find the number of ways to do this.
Then $$S_1+S_2+...+S_k= N$$ where each $$S_i$$ ca n only take on integer values. So if $$k=3$$ and $$N=5$$, then something like $$(1,1,3)$$ will be equivalent to $$(1,3,1)$$ and $$(3,1,1)$$.
I've thought about generating polynomials, and if I wanted the number of non-distinct ways to do this, I would take the coefficient of $$x^5$$ in the expansion of $$(x^1+x^2+x^3)^3$$, which also can be evaluated by the multinomial coefficient formula to give $$6$$. It makes sense as the only sets of values $$(S_1,S_2,S_3)$$ can take are $$(1,1,3)$$ and $$(1,2,2)$$, both of which can be permuted $$3$$ times.
There was another solution to a related problem, and it involved the number of ways to split $$N$$ up into $$N$$ integers or less such that no two numbers are the same. For our problem, it would be the sum of the number of ways to split $$5$$ into $$1$$ number, split $$5$$ into $$2$$ numbers, split $$5$$ into $$3$$ numbers... such that $$S_i \neq S_j, \forall i \neq j$$. In this case, integer partitions of $$5$$ into $$3$$ number will not be considered, since both $$(1,1,3)$$ and $$(1,2,2)$$ contain repetitions. The $$3$$ ways that this can be done are $$(5,0), (4,1), (3,2)$$.
But obviously this is not what I want as it doesn't count $$(1,1,3)$$ and $$(1,2,2)$$.
Is there a formula to do this? A related question is here, but no explicit algorithm/formula is given.
EDIT: @marcelgoh said that Stirling numbers of the second kind would work. I have a follow-up question:
Is there a way to iterate through permutations of numbers making up $$N$$, but in a 'Stirling' sense?
For instance, if I wanted to express:
$$\frac{20!}{(2*1+1)!(2*1+1)!(2*3+1)!} + \frac{20!}{(2*1+1)!(2*3+1)!(2*1+1)!} + \frac{20!}{(2*3+1)!(2*1+1)!(2*1+1)!} + \frac{20!}{(2*2+1)!(2*2+1)!(2*1+1)!} + \frac{20!}{(2*2+1)!(2*1+1)!(2*2+1)!} + \frac{20!}{(2*1+1)!(2*2+1)!(2*2+1)!}$$
I could use: $$\sum_{i+j+k=5, i,j,k\geq 1}\frac{20!}{(2i+1)!(2j+1)!(2k+1)!}$$
But what if I just wanted:
$$\frac{20!}{(2*1+1)!(2*1+1)!(2*3+1)!} + \frac{20!}{(2*2+1)!(2*2+1)!(2*1+1)!}$$
Could I use something like:
$$\sum_{i+j+k=5, 1\leq i\leq j\leq k}\frac{20!}{(2i+1)!(2j+1)!(2k+1)!}$$
Or is there some less messy notation for the same concept?
• I'm having trouble understanding the follow-up question. My guess is that you're fixing $N = 5$ and you want to iterate through each of the $\big\{ {5\atop k} \big\}$ combinations of summands equalling $5$, for $k = 1,2,3,4$. Is this correct? – marcelgoh Aug 2 '19 at 7:02
• More of I want to iterate through the summands for $n=5, k=3$. Also I checked Stirling numbers of the second kind out and it is useful to $n$ labelled elements, but what if my elements are unlabelled, say identical balls? So if let's say $n=6, k=4$, I would like to iterate through $(1,1,1,3), (1,1,2,2)$. If $n=7, k=3$, then the iteration would be $(1,1,5), (1,2,4), (1,3,3), (2,2,3)$. – Yip Jung Hon Aug 2 '19 at 7:05
• Ahh okay. I believe this is $p_k(n)$, the number of partitions of $n$ into exactly $k$ parts. I'll edit my answer. – marcelgoh Aug 2 '19 at 7:11
I believe that the Stirling numbers of the second kind $$\big\{{n\atop k}\big\}$$ are what you need. This is the number of ways to partition $$n$$ labelled elements into $$k$$ unlabelled non-empty subsets.
EDIT: If we're trying to partition $$n$$ unlabelled elements into $$k$$ subsets, then the function we're actually looking to use is $$p_k(n)$$. According to Wikipedia, this function satisfies the recurrence relation $$p_k(n) = p_k(n-k) + p_{k-1}(n-1),$$ with initial conditions $$p_0(0) = 1$$ and $$p_k(n) = 0$$ if either of $$n$$ or $$k$$ is non-positive. | 2020-03-31T23:40:59 | {
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https://dsp.stackexchange.com/questions/54109/scale-image-by-using-fft | # Scale Image by using FFT
Can we change the scale of image by using FFT? I mean, how should i do process on frequency domain of image to upscale or downscale the orginal image?
The other question of mine is that how can changing scale of image affect on fft of this image?
• What do you mean by 'scale'? Resolution? – lxop Dec 13 '18 at 12:58
• @Ixop yes, I mean upscaling or downscaling the resolution of image. – Mert Ege Dec 13 '18 at 13:01
Yes image scaling can also be performed in frequency domain using DFT/FFT.
For example, given an $$N \times M$$ image, you can implement the interpolation by integer $$K$$ in the frequency domain by enlarging the $$N \times M$$ point DFT/FFT of the original image into $$K \cdot N \times K \cdot M$$ size where new DFT/FFT samples are all zero; except at the four corners; effectiveley you are zero padding the original DFT/FFT in frequency domain. The effect in time domain wil be interpolation.
The following is an example of 2x image zooming using FFT.
K = 2;
S = size(I);
F = fft2(I);
F2 = repmat( K*K*F , K , K);
Mask(end-S(1)/2 : end , 1:S(2)/2+1) = 1;
Mask(1:S(1)/2+1 , end-S(2)/2 :end ) = 1;
Mask(end-S(1)/2 : end , end-S(2)/2 :end ) = 1; | 2021-06-12T20:49:16 | {
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https://math.stackexchange.com/questions/516219/finding-out-the-area-of-a-triangle-if-the-coordinates-of-the-three-vertices-are/1414021 | # Finding out the area of a triangle if the coordinates of the three vertices are given
What is the simplest way to find out the area of a triangle if the coordinates of the three vertices are given in $x$-$y$ plane?
One approach is to find the length of each side from the coordinates given and then apply Heron's formula. Is this the best way possible?
Is it possible to compare the area of triangles with their coordinates provided without actually calculating side lengths?
What you are looking for is called the shoelace formula:
\begin{align*} \text{Area} &= \frac12 \big| (x_A - x_C) (y_B - y_A) - (x_A - x_B) (y_C - y_A) \big|\\ &= \frac12 \big| x_A y_B + x_B y_C + x_C y_A - x_A y_C - x_C y_B - x_B y_A \big|\\ &= \frac12 \Big|\det \begin{bmatrix} x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1 \end{bmatrix}\Big| \end{align*}
The last version tells you how to generalize the formula to higher dimensions.
PS. Another generalization of the formula is obtained by noting that it follows from a discrete version of the Green's theorem:
$$\text{Area} = \iint_{\text{domain}}dx\,dy = \frac12\oint_{\text{boundary}}x\,dy - y\,dx$$
Thus the signed (oriented) area of a polygon with $$n$$ vertexes $$(x_i,y_i)$$ is
$$\frac12\sum_{i=0}^{n-1} x_i y_{i+1} - x_{i+1} y_i$$
where indexes are added modulo $$n$$.
You know that AB x AC is a vector perpendicular to the plane ABC such that |AB x AC|= Area of the parallelogram ABA’C. Thus this area is equal to ½ |AB x AC|.
From AB= $$(x_2 -x_1, y_2-y_1)$$; AC= $$(x_3-x_1, y_3-y_1)$$, we deduce then
Area of $$\Delta ABC$$ = $$\frac12[(x_2-x_1)(y_3-y_1)- (x_3-x_1)(y_2-y_1)]$$
• This is the best answer, as it covers the 6005 and the Jailcu answers. Vector cross product. Aug 24 '16 at 21:09
• By this way can area come as negative value? I am trying to understand this problem. geeksforgeeks.org/orientation-3-ordered-points Dec 19 '17 at 13:39
• What has been taken is the absolute value of the cross product. Dec 20 '17 at 14:31
• Yes it can be negative. The sign depends on the orientation of the three vertices. Feb 9 '18 at 7:03
• Looking at the parallelogram, why not drop a perpendicular from C to AB. Now there are two right triangles whose area is easily found (You KNOW some coordinates.) No need to divide by 2 b/c the two right triangles cover the triangle under observation. Jul 12 '19 at 18:29
Heron's formula is inefficient; there is in fact a direct formula. If the triangle has one vertex at the origin, and the other two vertices are $(a,b)$ and $(c,d)$, the formula for its area is $$A = \frac{\left| ad - bc \right|}{2}$$
To get a formula where the vertices can be anywhere, just subtract the coordinates of the third vertex from the coordinates of the other two (translating the triangle) and then use the above formula.
The simplest way to remember how to calculate is by taking $\frac{1}{2}$ the value of the determinant of the matrix $$\begin{bmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{bmatrix}$$
• – Zaz
Aug 17 '16 at 16:21
if $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the vertices of a triangle then its area is given by:-
$$\frac{ 1}{2}|(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))|$$
The area of a triangle $P(x_1, y_1)$, $Q(x_2, y_2)$ and $R(x_3, y_3)$ is given by $$\triangle= \left|\frac{1}{2}(x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}))\right|$$ If the area of triangle is zero, it means the points are collinear.
If we code this in Python3, it will look like
def triangle_area(x1, y1, x2, y2, x3, y3):
return abs(0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2)))
If we code in js, it will look like this
function triangle_area(x1, y1, x2, y2, x3, y3){
return Math.abs(0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2)))
}
Now here's a solution which works in any vector space with an inner product: take the half of the root of the Gram-Determinant of two sides of the triangle, that is $$\frac12\sqrt{\det\begin{pmatrix}\langle b-a,b-a\rangle& \langle b-a,c-a\rangle\\ \langle b-a,c-a\rangle & \langle c-a,c-a\rangle \end{pmatrix}}.$$
For fun, I'll just throw out the really long way that I learned in 3rd grade, only because it hasn't been submitted. I don't endorse this, the Shoelace/Surveyor's formula is way better.
1. Determine the distance between two of the three points, say $\big( x_{1}, y_{1} \big)$ and $\big( x_{2}, y_{2} \big)$. $d = \sqrt{ \big(x_{2} - x_{1}\big) ^{2} + \big(y_{2} - y_{1}\big) ^{2} }$
2. Determine the slope $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ and y-intercept, $a = y_{1} - \big(m \times x_{1}\big)$, of the line between $\big( x_{1}, y_{1} \big)$ and $\big( x_{2}, y_{2} \big)$.
3. Determine the slope of the a line perpendicular the line from $\big( x_{1}, y_{1} \big)$ and $\big( x_{2}, y_{2} \big)$, which is the negative reciprocal of the first slope. $n = \frac{-1}{m}$
4. Determine the equation of the line parallel to this second line, that passes through the third point $\big( x_{3}, y_{3} \big)$, by finding the y-intercept in $y = n*x+b$, since you already have the slope and a point on the line. $y_{3} - \big(n \times x_{3}\big)=b$.
5. Determine where this new line intersects the line between $\big( x_{1}, y_{1} \big)$ and $\big( x_{2}, y_{2} \big)$, by solving the system of equations of the new line and the original line: $y = m*x+b$ and $y = n*x+b$. Call this point $\big( x_{4}, y_{4} \big)$. I won't write this out, I'll leave it as an "exercise for the reader".
6. Determine the distance between $\big( x_{3}, y_{3} \big)$ and the new point$\big( x_{4}, y_{4} \big)$. $c = \sqrt{ \big(x_{4} - x_{3}\big) ^{2} + \big(y_{4} - y_{3}\big) ^{2} }$
7. If $d$ is the base of the triangle, and $c$ the height, the area is $A = \frac{1}{2} c*d$.
8. Realize you've spent several minutes solving a trivial problem... cry silently.
• You learnt this in third grade?! Jan 12 '19 at 0:34
The area $A$ of the triangle two of whose vertices lie on the axes, with coordinates $(a, 0)$, $(0, b)$, and a third vertex $(c, d)$ is obtained from previous formula by a mere horizontal axis shift of -a units as $$A = \frac{|-ad + b(a - c)|}{2}$$
• What "previous formula" do you mean? May 25 '15 at 11:41
• Sorry to be vague about that ! I meant the formula A = |ad - bc|/2 for triangle with coordinates (a, b) , (c, d) and origin. May 27 '15 at 4:56 | 2021-09-20T03:00:23 | {
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http://www.xn--mamuki-6ib.pl/madhya-pradesh-rfzfeh/vampire-weekend---this-life-a42a3e | the definition only tells us a bijective function has an inverse function. The function f: ℝ2-> ℝ2 is defined by f(x,y)=(2x+3y,x+2y). Let f: A → B. Now we much check that f 1 is the inverse … I've got so far: Bijective = 1-1 and onto. Then f has an inverse. The codomain of a function is all possible output values. A bijective group homomorphism $\phi:G \to H$ is called isomorphism. Proof. The Attempt at a Solution To start: Since f is invertible/bijective f⁻¹ is … Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. A bijection of a function occurs when f is one to one and onto. is bijective, by showing f⁻¹ is onto, and one to one, since f is bijective it is invertible. Let f 1(b) = a. Bijective. it's pretty obvious that in the case that the domain of a function is FINITE, f-1 is a "mirror image" of f (in fact, we only need to check if f is injective OR surjective). Let f : A !B be bijective. Show that f is bijective and find its inverse. The domain of a function is all possible input values. https://goo.gl/JQ8NysProving a Piecewise Function is Bijective and finding the Inverse Let f : A !B be bijective. I think the proof would involve showing f⁻¹. 1.Inverse of a function 2.Finding the Inverse of a Function or Showing One Does not Exist, Ex 2 3.Finding The Inverse Of A Function References LearnNext - Inverse of a Bijective Function … This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. Yes. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. In mathematical terms, let f: P → Q is a function; then, f will be bijective if every element ‘q’ in the co-domain Q, has exactly one element ‘p’ in the domain P, such that f (p) =q. Let’s define $f \colon X \to Y$ to be a continuous, bijective function such that $X,Y \in \mathbb R$. If we fill in -2 and 2 both give the same output, namely 4. So x 2 is not injective and therefore also not bijective and hence it won't have an inverse.. A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached. Since f is surjective, there exists a 2A such that f(a) = b. Click here if solved 43 The range of a function is all actual output values. Accelerated Geometry NOTES 5.1 Injective, Surjective, & Bijective Functions Functions A function relates each element of a set with exactly one element of another set. A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. It means that each and every element “b” in the codomain B, there is exactly one element “a” in the domain A so that f(a) = b. Bijective Function Examples. it doesn't explicitly say this inverse is also bijective (although it turns out that it is). Theorem 1. Please Subscribe here, thank you!!! A function is called to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. Since f is injective, this a is unique, so f 1 is well-de ned. In order to determine if $f^{-1}$ is continuous, we must look first at the domain of $f$. 1. An example of a function that is not injective is f(x) = x 2 if we take as domain all real numbers. 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https://physics.stackexchange.com/questions/652007/addition-of-velocities-vs-addition-of-forces | # Addition of velocities vs. Addition of forces
Imagine two strings tied to a box.
Case 1: Two strings are pulled with the same $$u$$ velocity. The box will also move with velocity $$u$$.
Case 2 : Tension along $$\text{a}$$ string is $$T$$. Therefore total force acting on the box is $$T+T=2T$$. (Box is accelerating)
I think my problem is obvious. Both velocity and force are vectors. But why we can not get the velocity of the box in the first case as $$u+u=2u$$? (This is obviously wrong, but why?)
• Link to a similar question - physics.stackexchange.com/questions/594941/… Jul 20, 2021 at 1:43
• One way to think about it: The velocity is in parallel, while the force is in series. Jul 20, 2021 at 14:46
## 7 Answers
I'll start my answer by departing from the specific example, to provide a more general answer. In the end, I'll summarize how the general discussion applies to the specific example.
Let's focus on your observation that
Both velocity and force are vectors.
True. Both are vectors, and an intrinsic characterization of vectors is that there is a sum of vectors. However, although the vector addition is a well definite mathematical concept (in essence, the parallelogram rule), the application of vectors in physics cannot avoid the additional step of carefully identifying the physical meaning of the sum.
When we identify the physical entity force with vectors, we implicitly or explicitly have to provide an operative meaning of all the vector operations (sum and multiplication by a scalar). In classical mechanics, if we identify the presence of a force $${\bf F}$$ by the resulting acceleration of a test particle, the sum of two forces applied to the same particle and product of a force by a scalar, are directly related to the corresponding addition of accelerations and multiplication of acceleration by a scalar.
Notice that an important ingredient of the force addition concept is adding only forces applied to the same body. Missing that, we would go into trouble if we try to sum an action-reaction pair of forces. In a more mathematical flavor, we could say that forces on different bodies are in different vector spaces and therefore cannot be summed.
A similar discussion can be done on velocities. Displacements of a point-like object in a time $$\Delta t$$, can be represented by vectors. How do we know that? We simply define the sum of two displacements of the same body as the resulting displacement. With this definition, it is a non-trivial physical finding that the order of the two displacements does not matter (sum is commutative), that there is a zero-displacement. There is an opposite displacement for every displacement such that the sum of both is equivalent to the zero displacement. Moreover, it is possible to define multiplication by a scalar, by using displacements in the same direction. Such a multiplication fulfills all the corresponding axioms in the definition of a vector space.
The key point is that the sum of displacements as vectors has the physical meaning of combining different displacements of the same body. Whatever can be said about displacements, can be said about velocities, of course.
To summarize, what can or cannot be done when summing entities called velocities or forces depends on the physical meaning we give to the mathematical concepts. It is not enough to have vector quantities to sum them without analyzing what kind of vectors are.
Let's now come to the example. Forces on the same body can be summed and this would result in an acceleration which is the sum of the accelerations present if only one of the forces at a time were present. Summing velocities of two different points of the same rigid body is meaningless because the body (the box) position is identified by one point only.
Notice that clarifying the concepts behind vector addition of velocities is an important prerequisite to avoid confusion with the laws of transformation of velocities in different reference frames in Relativity.
• Considering this answer is already accepted: one very helpful analogy for OP would be the classic person on a train. This is the example where adding velocities makes sense indeed, and is clearly different from his example of the box with two strings.
– AnoE
Jul 19, 2021 at 13:58
• @AnoE I considered this example. But I decided not to include it. The reason is implicitly given by the last sentence. If one takes for granted that the relative velocity is always a vector sum, the so-called addition of velocities in SR becomes ununderstandable. Jul 19, 2021 at 15:55
• Fair enough, @GiorgioP. I think OP is firmly in the nonrelativistic realm so far, but it's your answer. ;)
– AnoE
Jul 20, 2021 at 6:43
Two strings are pulled with same velocity $$u$$. The box will also move with velocity $$u$$.
You don't "pull" a box with a velocity. You can pull it by applying a force that results in a change of velocity. Imagine you have two forces acting on the box, one that gives it a velocity $$\vec v$$ and the other gives it a velocity $$\vec w$$. Then it is okay to say that the resulting velocity is $$\vec v +\vec w$$.
Tension along a string is $$T$$. Therefore total force acting on the box is $$T+T=2T$$.(Box is accelerating)
In this case, you are applying two forces to the box, and so they add to give you a resultant force. The first case doesn't really make sense physically, but the second case does.
Both velocity and force are vectors
Yes they are, and mathematically we can add any two vectors, but when adding vectors that represent physical quantities, we need to be sure what these physical quantities represent.
In your first example, you cannot add different velocities at different points on the same body since its velocity is defined by the translational motion of one point (usually the center) of the body (assuming the body is rigid).
The second example with forces works fine since the net force on an object is the vector sum of all forces acting on it, as you have done.
• Yes, we can't pull a box with a velocity. It's purely fault of my word choice. What I wanted to imply is similar to the question in this post.
– ACB
Jul 21, 2021 at 7:29
• Ok. And did you find your answer? Jul 21, 2021 at 7:35
• Yes, I got the answer. Thanks
– ACB
Jul 21, 2021 at 7:36
• Excellent! Cheers. Jul 21, 2021 at 7:39
You are walking down the street with your friend. Now you hold hands. Are you now moving twice as fast?
If addition of velocity worked like that, you would get total nonsense. Whenever two moving objects would stick to each other, they would move faster.
Another, more mathematical way of looking at it. The energy of an object of mass $$m$$ moving at velocity $$v$$ is $$E = \frac12 mv^2.$$ Now mentally imagine dividing this object in two objects of masses $$m_1$$ and $$m_2$$ (like person = body + head). We obviously have $$m = m_1+m_2.$$ Now imagine these two parts are moving with velocities $$v_1$$ and $$v_2$$ respectively. But since mentally dividing an object cannot change its energy, we must have $$\frac12 mv^2 = \frac12\left(m_1v_1^2 + m_2v_2^2\right).$$ Solving this will lead you to $$v=v_1=v_2$$
This is an insightful question.
Just like not all things with legs are tables, not all things represented by vectors (in a mathematical sense) are the same.
In classical mechanics, there are two classes of vector quantities, each with some common properties. The nomenclature below isn't standardized, as different authors have used different names for the same concepts below.
• Axis Vector - A unique vector that conveys the direction and magnitude of a quantity belonging to the body that represents a line (or axis) in space.
Some examples on a rigid body are
• The momentum vector is a single vector describing the translational momentum state of a rigid body. Regardless of how a body rotates, it is always defined as $$\boldsymbol{p} = m\, \boldsymbol{v}_{\rm C}$$ where C is the center of mass. The line associated with momentum is called the axis of percussion.
• The force vector is a single vector describing the loading a body is under. Force is the time derivative of momentum and the total load is thus $$\boldsymbol{F} = m \, \boldsymbol{a}_C$$ The line associated with force is called the line of action.
• The Rotational velocity vector is a common quantity shared by all particles of a body. Any point on the body (or the general extended frame) is said to rotate by $$\boldsymbol{\omega}$$ relative to each other point. The line associated with rotation is called the rotation axis.
• Moment Vector - A vector that varies by location and is defined by taking the moment of an axis vector. This requires the right-hand rule convention in the form of a cross product to define the direction of the vector. This defines a vector field around the lines mentioned above. A vector field is a vector that changes direction and magnitude by location.
Some examples are at some arbitrary point A are:
• The velocity vector depends on the location of the particle measured, with the general formula $$\boldsymbol{v}_A = \boldsymbol{r}_A \times \boldsymbol{\omega}$$ where $$\boldsymbol{r}_A$$ is the location of the rotation axis relative to the particle.
• The torque vector depends on the location of the particle where torque is summed with the general formula $$\boldsymbol{\tau}_A = \boldsymbol{r}_A \times \boldsymbol{F}$$ where $$\boldsymbol{r}_A$$ is the location of the force line of action relative to the summation point.
• The angular momentum vector depends on the location where it is summed with the general formula $$\boldsymbol{L}_A = \boldsymbol{r}_A \times \boldsymbol{p}$$ where $$\boldsymbol{r}_A$$ is the location of the axis of percussion relative to the summation point.
Now for the vector algebra part of the question. How do we add two forces or two velocities and how does this process differ.
• Two Forces add up to the loading state of the body, by sliding the vectors along their line of action until they meet at a common point. At the common point use the trapezoidal rule to find out the direction, magnitude and location of the resultant force.
The result is the vector addition of the axis vectors (Forces) and the moment vectors (Torques) component by component.
• Two Velocities do not add up to change the state of the body (unless the body is purely translating, a special case). The two velocities are just expressions of the same rotational velocity at different points
To describe the kinematics of a body relative to another body you need to add up both the axis vectors (rotational velocity) and the moment vectors (translational velocity) expressed at a common point, just like the forces above need to be added at a common point.
The geometry of this situation leads to the relative centre of rotation theorem.
It is worth reading the following article from 1901
• SLATE, F. The Use of “;Axis-vectors”. Nature 64, 54–55 (1901). DOI, download pdf
Additionally, read this answer here about the nature of torque and the moment vectors that are defined using the cross product.
In summary, the common quantities in mechanics are interpreted as follows $$\begin{array}{r|l|l} \text{concept} & \text{value} & \text{moment}\\ \hline \text{rotation axis} & \text{rot. velocity}, \boldsymbol{\omega} & \text{velocity}, \boldsymbol{v} = \boldsymbol{r}\times \boldsymbol{\omega} \\ \text{line of action} & \text{force}, \boldsymbol{F} & \text{torque}, \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} \\ \text{axis of percussion} & \text{momentum}, \boldsymbol{p} & \text{ang. momentum}, \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} \end{array}$$ The stuff under the value column are fundamental quantities that give us the magnitude of something (as well as the direction). The stuff under the moment column are secondary quantities that depend on where they are measured and give use the relative location of the fundamental quantities. Hence the terms torque = moment of force, velocity = moment of rotation and angular momentum = moment of momentum. All that means is that these quantities are $$\boldsymbol{r} \times \text{(something fundamental)}$$ and they describe the moment arm to this something.
Sorry for my poor english. French is my native language.
To define a vector, it is necessary to specify the vector space on which it is defined. In general, for a manifold, we have a tangent vector space at each point. In classical physics, space has an affine flat space structure and we define at each point a tangent vector space which, for an affine flat space, can all be identified with each other.
It is in this tangeant vector space that the addition of two vectors is defined. One thus defines at each point the vector space of the displacements. The vector sum of two displacements from a point is a displacement. By dividing by time, we go to velocity vectors.
On the other hand, there are difficulties in defining the derivative of the velocity : we compare vectors at different points. To be able to do this, we have to define a parallel transport and connection that allows a vector to be transported from one point to another. It is very easy in the case of the affine space of classical physics. More complicated in a variety: it is necessary to introduce a covariant derivative.
So even the addition of two forces at different points is not a simple consequence of the structure of vector space. We have to transport the vectors. And this is only simple for an affine flat space.
The fact that velocity and force are vectors is only secondary here. The true distinction here is whether the quantities in question are intensive or extensive, that is, how they scale with the size of the system.
With scalar values, we could ask exactly the same thing about temperature and mass. Consider two identical objects with temperature $$T$$ and mass $$m$$. When put together, their temperature will not increase to $$T$$, but will still remain at $$T$$, because temperature is an intensive property. However, mass is an extensive property, so the combined mass will increase to $$2m$$.
Combining two unequal extensive properties reduces to a sum, while with intensive properties it is a weighted average over some extensive property of a system: in this example it would be the total heat capacity of each of the bodies.
• Your argument is correct. But how do you express intensive or extensive properties for velocity or force? Furthermore I think velocity and force are not properties of an object. They are quantities that describe the motion.
– ACB
Jul 20, 2021 at 4:51
• Force is certainly not a property of an object, but velocity can be treated as such. So can linear momentum, which is clearly extensive. It seems to me that the intensive*/*extensive category can be naturally extended to many physical quantities that are not properties. I guess I need to think about it some more. Jul 20, 2021 at 11:05
In case (1) constant velocity means the acceleration is zero. In the absence of gravity, there is no net force. Only relative velocities add.
• I am unable to get your last point. Here we can't say $u+u=2u$ and it's truely incorrect. Then think of a horizontal projection( horizontal is for simplicty). Using equations of motion we calculate horizontal($v_x$) and vertical($v_y$) velocities seperatedly. Then using vector addition, we calculate net velocity of the projectile($v_{net}=\sqrt{v_x^2 +v_y^2}$). Here, are horizontal and vertical velocities relative?(This example should be asked on the original question, but I think it's not suitable to edit it now.)
– ACB
Jul 20, 2021 at 2:59
• Also I understand vector addition and relative velocity are two different things.
– ACB
Jul 20, 2021 at 3:12
• A relative velocity problem is one that comes to mind in which you might have a reason to add velocity vectors. In this case, the two velocity vectors represent the velocities of two different points within the object (it is not rotating). Jul 20, 2021 at 13:45 | 2022-08-11T02:26:26 | {
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http://sahincetinkaya.net/simple-lemon-ftmq/03ndx.php?tag=matlab-check-if-spd-dbe578 | The input argument M to ispd is Hermitian and therefore theoretically its determinant is real. MATLAB operators that contain a period always work element-wise. A matrix is positive definite if all it's associated eigenvalues are positive. \n \n ', '0: The matrix is not positive definite. Check the value of the BeingDeleted property to verify that the object is not about to be deleted before querying or modifying it. It is often required to check if a given matrix is positive definite or not. ISO 26262 classifies functions in automotive safety integrity levels (ASIL) from A to D – that is, from least to most stringent. part that is then overwritten with the result. I need help creating a function using for loops! So that is why I used $\mathbf{A}=nearestSPD(\mathbf{X})$ to get a SPD matrix. The following Matlab project contains the source code and Matlab examples used for kmeans spd matrices. Introduction. I prefer my one-line solution below, which also checks if the matrix is symmetric. (abbreviated SPD), we have that the SVD and the eigen-decomposition coincide A=USUT =EΛE−1 withU =E and S =Λ. tf = det(M)>0 && (length(M)>=1 && ispd(M(1:end-1,1:end-1))); Yes I edited by hand and made a mistake. Description [xv,lmb,iresult] = sptarn(A,B,lb,ub,spd,tolconv,jmax,maxmul) finds eigenvalues of the pencil (A – λB)x = 0 in interval [lb,ub]. I'm asking because of the thread in the first link that Ameer posted way earlier in this thread that shows that the chol function may not be reliable in determining if a matrix is SPD (repeating the link here for easy access): https://www.mathworks.com/matlabcentral/answers/101132-how-do-i-determine-if-a-matrix-is-positive-definite-using-matlab?s_tid=srchtitle. Auf welche Kauffaktoren Sie zuhause beim Kauf Ihres Check wlan achten sollten. (A matrix of linear polynomials A ij – λB ij, A – λB, is called a pencil.). And since Fix now. (I have not tried it myself. The Wikipedia page does talk about the non-Hermitian case in the “Extension ...” section. Simulink Check™ automates checking for both guidelines. My function script needs to. The code incorrectly tells me that. then the input matrix is symmetric positive definite and the factorization was successful." But do non-symmetric positive definite matrices have any useful properties? Decimal point: 102.5543. Continental provides a generic VCU device as well as basic software and a MATLAB ... Legal limits for CO 2, higher level fuel cost, and more strictly legislated countries are pushing down on the OEM to find a new sustainable way to meet these needs, causing the OEM to look at hybridization further. A and B are sparse matrices.lb and ub are lower and upper bounds for eigenvalues to be sought. I am supposed to write code using for loops and the det() function. Based on your location, we recommend that you select: . It is easy to check that (SPD(n),(,! Examples . I have changed my code a little, so now it reads as follows: M = [ 1 2 3 4 ; 5 6 7 8 ; 9 10 11 12 ; 13 14 15 16]. spmd, statements, end defines an spmd statement on a single line. Estimation & Control Library for Guidance, Navigation and Control Applications - PX4/ecl Discover what MATLAB ® can do for your career. You may receive emails, depending on your. $\begingroup$ @ Rodrigo, I asked that question yesterday and my take away from the comments was that in MATLAB, a matrix $\mathbf{X}$ is not PSD just because the way it is constructed. Though the best method to check PD IMO is using incomplete cholesky factorization for full and EIGS for sparse matrix, and not using DET at all. M is not definite postive because v'*M*v = -1 < 0, that violates the definition, “Your matrix is not positive definite first of all because it is not symmetric.”. Reload the page to see its updated state. The answer is. Description: Three or more periods at the end of a line continues the current command on the next line.If three or more periods occur before the end of a line, then MATLAB ignores the rest of the line and continues to the next line. has a positive determinant, yet is certainly not positive definite, even though it is symmetric. Reload the page to see its updated state. return the result to the calling program - 1 if positive definite, 0 otherwise. $\endgroup$ – Suvrit Jan 25 '11 at 16:38 $\begingroup$ But how does matlab do it? We provide a MATLAB implementation of the KSVD-SPD and Online-SPD algorithms described in these two papers. Welcome to Battleship 2.0, a Matlab special. Opportunities for recent engineering grads. Opportunities for recent engineering grads. Maybe I’m just parsing words here, but you don’t need the symmetry assumption to apply those criteria. Or do you want to write the algorithm from scratch? tic,Uj = nearestSPD(U);toc Elapsed time is 0.008964 seconds. I'm creating a reduced-order model for the heat equation in 1-dimension based on the finite element method. $\endgroup$ – … M = [ 1 2 3 4 ; 5 6 7 8 ; 9 10 11 12 ; 13 14 15 16]; Opps, for the second line, it should read "x = NAME_lab01 (M)". useful properties compared to SPD matrixes. Uses: Line continuation. I think Sepehr is implying that the "p" output of chol() is returning 0, implying that chol thinks it, positive definite. Why is my code not running through all determinates? A co-diagonal submatrix is a square matrix of any size, contained within the original matrix, that shares the diagonal with the original matrix.". So the change made will be essentially insignificant, EXCEPT that MVNRND will work after the perturbation made by nearestSPD. I need to come up with a pattern for the matrices to be check. You need to first symmetrize the matrix M = M + M' before checking determinant of principal minors. Wir begrüßen Sie zuhause auf unserer Webseite. LOG-EUCLIDEAN MEANS 497 A and B are sparse matrices.lb and ub are lower and upper bounds for eigenvalues to be sought. NOTE: CHOL expects its input matrix to be symmetric and only looks at the upper triangular portion of the matrix. These additional checks, together with the existing ISO 26262 checks in Simulink Check, ensure that the MATLAB implementation generates code suitable for high-integrity applications. If the input matrix is not positive definite, then "p" will be a positive integer: The CHOL function will return an error if it is only provided with a single output argument, and is also given a matrix that is not positive definite. These data could come … I see the question come up every once in a while, so I looked in the file exchange to see what is up there. Comprehensive user guidance through analysis results as well as the repair and improvement process effectively ensures ISO 26262 standard compliance for your software models. MATLAB: Eigenvectors of an SPD matrix being saved as complex doubles. What's the scoop on chol's undocumented p output? It sounds like you are not sure. Find the treasures in MATLAB Central and discover how the community can help you! \n \n'). $\endgroup$ – Bill Greene Feb 16 '17 at 15:55 $\begingroup$ Ok,if as a new question, i were to check a matrix is positive definite , then i need to check for positive definite and i am searching a way to code it efficiently! This person who gives you this "defnition" must assume M is symmetric, or makes a mistake. Community Treasure Hunt. (A matrix of linear polynomials A ij – λB ij, A – λB, is called a pencil.). command should take a single argument (the matrix whos determinant you want to calculate). nearestSPD will be able to convert U into something that is indeed SPD, and for a 100 by 100 matrix, do it quickly enough. Ah well, if Golub and Van-Loan says so, then fine. Find the treasures in MATLAB Central and discover how the community can help you! ). 2 Application The Cholesky factorization is used to solve the linear system Ax = y when A is SPD: Meaning my code is correct? Sure enough, the gt operator only considers the real part of a complex operand. Chol returns zero if the matrix is positive semi-definite not positive definite. Excuse me, i did not understand you.My matlab knowledges are not so good. Jetzt bei Amazon.de bestellen! I'm currently working on a lab where I need to check if a square matrix is positive and definite. WindRose for Matlab 14 March 2015 Page 2 of 29 Data We start from some simple data which we want to be represented in a wind rose. Just like the classic Battleship game created in the mid-1900’s, our version of Battleship will ask you to place battleships on the board and choose your attack coordinates; however, rather than playing a friend, you play the computer, the treacherous pirate of … ". Based on your location, we recommend that you select: . If chol returns a second argument that is zero, then MATLAB (and mvnrnd) will be happy! Perhaps you misunderstand the meaning of positive definite? However, your code is by no means correct. $\begingroup$ matlab can do this for you; the restriction to SPD is not necessary. However, due to roundoff errors the computed value of det(M) could have a small imaginary part, which got me thinking about what det(M) > 0 means if det(M) returns a complex number. Please help me write some code! Code Generation. Cholesky decomposition is an efficient method for inversion of symmetric positive-definite matrices. it seems like you need the symmetry assumption to apply Sylvester's criterion, the eigenvalue criterion, and pretty much every other property I've seen used in practice. Accelerating the pace of engineering and science. For previous releases, read below for any additional information: Rather than using the EIG function to obtain the eigenvalues in order to determine positive definiteness, it is more computationally efficient to use the CHOL function. I believe that I updated my code to just take one argument, I will attatch it below: fprintf('1: The matrix is positive definite. https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_997999, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998035, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998044, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998077, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998092, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998095, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998101, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998104, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998107, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998113, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998116, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998122, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998125, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998128, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998131, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998137, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998140, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998143, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998146, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998149, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998200, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998230, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998263, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998272, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998302, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998329, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998341, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998344, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998350, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998359, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998374, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998533, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#answer_490375, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#answer_490549, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998764, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998770, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_998779, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_1001848, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_1002061, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_1003048, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#comment_1003075, https://de.mathworks.com/matlabcentral/answers/589435-how-do-i-create-a-function-script-to-check-the-positive-definiteness-of-a-a-square-matrix-of-any-siz#answer_490609. | 2021-04-23T05:17:19 | {
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https://www.emathhelp.net/notes/calculus-2/applications-of-integrals/volume-of-solid-of-revolution-and-method-of-cylinders/ | # Volume of Solid of Revolution and Method of Cylinders
Sometimes it is very hard to use Method of Disks/Rings to obtain volume of solid of revolution.
But that's not a problem. We can use any shape for the cross section as long as it can be expanded or contracted to completely cover the solid.
In this note we introduce method of cylindrical shells.
First suppose that we rotate about y-axis. Imagine a hollow cylinder whose wall has width Delta x. In other words, imagine a rectangle of with Delta x rotated around y-axis. Radius of i-th cylinder is r(x_i^**) (it depends on x_i^**)where x_i^** lies in interval [x_(i-1),x_i] . On the figure radius is r(x_i^**)=x_i^**.
Imagine this shell to be cut and flattened. The resulting rectangular slab has length 2pir(x_i^**) (length of circle with radius r(x_i^**)), width Delta x and height h which depends on x_i^** (on figure height is f(x_i^**)).
So, the volume of the shell is 2pir(x_i^**)h(x_i^**)Delta x.
If we do this for every subinterval and add the results, we get an approximation to the volume of the solid: V~~sum_(i=1)^n 2pir(x_i^**)h(x_i^**)Delta x.
This approximation improves as increases, so V=lim_(n->oo)sum_(i=1)^n 2pir(x_i^**)h(x_i^**)Delta x.
We recognize in this limit of Riemann sum definite integral, therefore V=int_a^b 2pir(x)h(x)dx.
Volume of solid rotated around y-axis is V=int_a^b 2pi r(x)h(x)dx.
As can be seen the formula for area of the cross section in case of cylindrical shells is A=2pi(radius)(height).
It is always a good practice to draw a sketch of problem in order to correctly determine radius and height.
In case we rotate around x-axis, radius and height will depend on y.
Example 1. Find the volume of the solid obtained by rotating about the y-axis the region bounded by the curve y=4(x-1)^2(x-3)^2 and x-axis.
First of all note that we can't use method of disks here. Since we rotate about y-axis we need function in terms of y, but it is very hard to express x in terms of y.
Now we draw a sketch. In this case radius of cylinder is x and height is y(x)=4(x-1)^2(x-3)^2.
Bounds of integration are 1 and 3.
Therefore, V=int_1^3 2pi x*4(x-1)^2(x-3)^2dx=
=8pi int_1^3 (x^4-8x^3+22x^2-24x+9)dx=
=8pi (1/5 x^5-2x^4+22/3 x^3-12x^2+9x)|_1^3=
=8pi (1/5*3^5-2*3^4+22/3*3^3-12*3^2+9*3)-8pi (1/5*1^5-2*1^4+22/3*1^3-12*1^2+9*1)=(128pi)/15.
Next examples can be solved using method of disks, but we will solve them using cylindrical shells.
Example 2. Find the volume of the solid obtained by rotating about the y-axis the region bounded by the curves y=sqrt(x) and y=x/2.
First we draw a sketch. In this case radius of cylinder is x and height is difference between function values at x: h(x)=sqrt(x)-x/2.
Bounds of integration are points of intersection of y=sqrt(x) and y=x/2, i.e. 0 and 4.
Therefore, V=int_0^4 2pi x*(sqrt(x)-x/2)dx=2pi int_0^4 (x^(3/2)-1/2 x^2)dx=
=2pi (2/5 x^(5/2)-1/6 x^3)|_0^4=2pi (2/5 *32 -64/6)=(64pi)/15.
Example 3. Find the volume of the solid obtained by rotating about the line x=-1 the region bounded by the curves y=sqrt(x) and y=x/2.
First we draw a sketch. In this case radius of cylinder is not x, it is distance between line x=-1 and edge of cylinder: 1+x. Height is y(x)=sqrt(x)-x/2.
Bounds of integration are points of intersection of y=sqrt(x) and y=x/2, i.e. 0 and 4.
Therefore, V=int_0^4 2pi (1+x)*(sqrt(x)-x/2)dx=
=2pi int_0^4 (x^(3/2)+x^(1/2)-1/2 x^2 -1/2 x)dx=
=2pi (2/5 x^(5/2) +2/3 x^(3/2) -1/6 x^3-1/4 x^2)|_0^4=2pi (64/5+16/3-64/6-16/4)=(104pi)/15.
Now let's see how to handle situations when we rotate about x-axis.
Example 4. Find the volume of the solid obtained by rotating about the x-axis the region bounded by the curve y=x^2 on interval [0,1].
Since we rotate about x-axis, we need function in terms of y: x=+-sqrt(y). Since we are in first quadrant (interval [0,1]) then we choose positive value x=sqrt(y).
Now we draw a sketch. In this case radius of cylinder is y and height is distance from 1 to function, i.e. h(y)=1-sqrt(y).
Bounds of integration are 0 and 1.
Therefore, V=int_0^1 2pi y*(1-sqrt(y))dy=
=2pi int_0^1 (y-y^(3/2))dy=2pi (1/2 y^2 -2/5 y^(5/2))|_0^1=2pi(1/2-2/5)=pi/5.
Example 5. Find the volume of the solid obtained by rotating about the x-axis the region bounded by the curves y=x^2 and y=x.
Since we rotate about x-axis, we need functions in terms of y: x=sqrt(y) and x=y.
Now we draw a sketch. In this case radius of cylinder is y and height is difference between outer and inner function at y: h(y)=sqrt(y)-y.
Bounds of integration are 0 and 1.
Therefore, V=int_0^1 2pi y(sqrt(y)-y)dy=
=2pi int_0^1 (y^(3/2)-y^2)dy=2pi (2/5 y^(5/2)-1/3 y^3)|_0^1=2pi (2/5-1/3)=(2pi)/15.
Example 6. Find the volume of the solid obtained by rotating about horizontal line y=2 the region bounded by the curves y=x^2 and y=x.
We again need functions in terms of y: x=sqrt(y) and x=y.
Bounds of integration are points of intersection of x=sqrt(y) and x=y, i.e. 0 and 1.
Now we draw a sketch.
At y radius of cylinder is 2-y and height is difference between values of outer and inner functions at y: sqrt(y)-y.
Therefore, V=int_0^1 2pi (2-y)(sqrt(y)-y)dy=2pi int_0^1 (2y^(1/2)+y^2-2y-y^(3/2))dy=
=2pi (4/3 y^(3/2)+1/3 y^3-y^2-2/5 y^(5/2))|_0^1=2pi(4/3 +1/3 -1-2/5)=(8pi)/15. | 2020-07-08T03:53:48 | {
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http://www.artofproblemsolving.com/wiki/index.php/2007_USAMO_Problems/Problem_2 | # 2007 USAMO Problems/Problem 2
## Problem
(Gregory Galperin) A square grid on the Euclidean plane consists of all points $(m,n)$, where $m$ and $n$ are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least 5?
## Solutions
### Solution 1
Lemma. Among 3 tangent circles with radius greater than or equal to 5, one can always fit a circle with radius greater than $\frac{1}{\sqrt{2}}$ between those 3 circles.
Proof. Descartes' Circle Theorem states that if $a$ is the curvature of a circle ($a=\frac 1{r}$, positive for externally tangent, negative for internally tangent), then we have that $$(a+b+c+d)^2=2(a^2+b^2+c^2+d^2)$$ Solving for $a$, we get $$a=b+c+d+2 \sqrt{bc+cd+db}$$ Take the positive root, as the negative root corresponds to internally tangent circle.
Now clearly, we have $b+c+d \le \frac 35$, and $bc+cd+db\le \frac 3{25}$. Summing/square root/multiplying appropriately shows that $a \le \frac{3 + 2 \sqrt{3}}5$. Incidently, $\frac{3 + 2\sqrt{3}}5 < \sqrt{2}$, so $a< \sqrt{2}$, $r > \frac 1{\sqrt{2}}$, as desired. $\blacksquare$
For sake of contradiction, assume that we have a satisfactory placement of circles. Consider 3 circles, $p,\ q,\ r$ where there are no circles in between. By Appolonius' problem, there exists a circle $t$ tangent to $p,\ q,\ r$ externally that is between those 3 circles. Clearly, if we move $p,\ q,\ r$ together, $t$ must decrease in radius. Hence it is sufficient to consider 3 tangent circles. By lemma 1, there is always a circle of radius greater than $\frac{1}{\sqrt{2}}$ that lies between $p,\ q,\ r$. However, any circle with $r>\frac 1{\sqrt{2}}$ must contain a lattice point. (Consider placing an unit square parallel to the gridlines in the circle.) That is a contradiction. Hence no such tiling exists.
### Solution 2
It is not possible. The proof is by contradiction. Suppose that such a covering family $\mathcal{F}$ exists. Let $D(P,\rho)$ denote the disc with center $P$ and radius $\rho$. Start with an arbitrary disc $D(O,r)$ that does not overlap any member of $\mathcal{F}$. Then $D(O,r)$ covers no grid point. Take the disc $D(O,r)$ to be maximal in the sense that any further enlargement would cause it to violate the non-overlap condition. Then $D(O,r)$ is tangent to at least three discs in $\mathcal{F}$. Observe that there must be two of the three tangent discs, say $D(A,a)$ and $D(B,b)$ such that $\angle AOB\leq 120^\circ$. By the Law of Cosines applied to triangle $ABO$, $$(a + b)^2\leq (a + r)^2 + (b + r)^2 + (a + r)(b + r),$$ which yields $$ab\leq 3(a + b)r + 3r^2,$$ and thus $$12r^2\geq (a - 3r)(b - 3r).$$ Note that $r < 1/\sqrt{2}$ because $D(O,r)$ covers no grid point, and $(a - 3r)(b - 3r)\geq (5 - 3r)^2$ because each disc in $\mathcal{F}$ has radius at least 5. Hence $2\sqrt{3}r\geq 5 - 3r$, which gives $5\leq (3 + 2\sqrt{3})r < (3 + 2\sqrt{3})/\sqrt{2}$ and thus $5\sqrt{2} < 3 + 2\sqrt{3}$. Squaring both sides of this inequality yields $50 < 21 + 12\sqrt{3} < 21 + 12\cdot 2 = 45$. This contradiction completes the proof.
Remark: The above argument shows that no covering family exists where each disc has radius greater than $(3 + 2\sqrt{3})/\sqrt{2}\approx 4.571$. In the other direction, there exists a covering family in which each disc has radius $\sqrt{13}/2\approx 1.802$. Take discs with this radius centered at points of the form $\left(2m + 4n + \frac{1}{2}, 3m + \frac{1}{2}\right)$, where $m$ and $n$ are integers. Then any grid point is with $\sqrt{13}/2$ of one of the centers and the distance between any two centers is at least $\sqrt{13}$. The extremal radius of a covering family is unknown.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
• <url>viewtopic.php?t=145844 Discussion on AoPS/MathLinks</url>
2007 USAMO (Problems • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 All USAMO Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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http://bootmath.com/proving-gcd-leftfracagcd-abfracbgcd-abright1.html | # Proving $\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$
How would you go about proving that $$\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$$
for any two integers $a$ and $b$?
Intuitively it is true because when you divide $a$ and $b$ by $\gcd(a,b)$ you cancel out any common factors between them resulting in them becoming coprime. However, how would you prove this rigorously and mathematically?
#### Solutions Collecting From Web of "Proving $\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$"
Very simply it can be done like this: $\gcd(a,b)=d$.
Now we ask can: $\gcd(\frac{a}{d},\frac{b}{d})=e$ for $e>1$?
Well, this implies $e\mid\frac{a}{d},e\mid\frac{b}{d} \Rightarrow em=\frac{a}{d}, en=\frac{b}{d} \Rightarrow dem=a,den=b \Rightarrow de$ is a common divisor of $a,b$ which is greater than $d$, thus a contradiction as $d$ by definition was supposed as the $\gcd$. Hence, $e=1$.
Let $d=\gcd(a,b)$. Let $a=md$ and $b=nd$. If some $k\gt 1$ divides $m$ and $n$, then $kd$ divides $a$ and $kd$ divides $b$, contradicting the fact that $d$ is the greatest common divisor of $a$ and $b$.
This is a special case of the GCD distributive law ($4$ proofs of it are below). Namely
$$\color{#c00}{\bf c} = (a,b) = (a/c,b/c)\color{#c00}{\bf \,c}\overset{\rm\large cancel\ \color{#c00}{\bf c}}\Longrightarrow 1 = (a/c,b/c)\qquad\qquad$$
Below are sketches of four proofs of the gcd Distributive Law $\rm\:(ax,bx) = (a,b)x\:$ using various approaches: Bezout’s identity, the universal gcd property, unique factorization, and induction.
First we show that the gcd distributive law follows immediately from the fact that, by Bezout, the gcd may be specified by linear equations. Distributivity follows because such linear equations are preserved by scalings. Namely, for naturals $\rm\:a,b,c,x \ne 0$
$\rm\qquad\qquad \phantom{ \iff }\ \ \ \:\! c = (a,b)$
$\rm\qquad\qquad \iff\ \: c\:\ |\ \:a,\:b\ \ \ \ \ \ \&\ \ \ \ c\ =\ na\: +\: kb,\ \ \$ some $\rm\:n,k\in \mathbb Z$
$\rm\qquad\qquad \iff\ cx\ |\ ax,bx\ \ \ \&\ \ \ cx = nax + kbx,\ \,$ some $\rm\:n,k\in \mathbb Z$
$\rm\qquad\qquad { \iff }\ \ cx = (ax,bx)$
The reader familiar with ideals will note that these equivalences are captured more concisely in the distributive law for ideal multiplication $\rm\:(a,b)(x) = (ax,bx),\:$ when interpreted in a PID or Bezout domain, where the ideal $\rm\:(a,b) = (c)\iff c = gcd(a,b)$
Alternatively, more generally, in any integral domain $\rm\:D\:$ we have
Theorem $\rm\ \ (a,b)\ =\ (ax,bx)/x\ \$ if $\rm\ (ax,bx)\$ exists in $\rm\:D.$
Proof $\rm\quad\: c\ |\ a,b \iff cx\ |\ ax,bx \iff cx\ |\ (ax,bx) \iff c\ |\ (ax,bx)/x\ \ \$ QED
The above proof uses the universal definitions of GCD, LCM, which often served to simplify proofs, e.g. see this proof of the GCD * LCM law.
Alternatively, comparing powers of primes in unique factorizations, it reduces to the following
$$\begin{eqnarray} \min(a+x,\,b+x) &\,=\,& \min(a,b) + x\\ \rm expt\ analog\ of\ \ \ \gcd(a \,* x,\,b \,* x)&=&\rm \gcd(a,b)\,*x\end{eqnarray}\qquad\qquad\ \$$
The proof is precisely the same as the prior proof, replacing gcd by min, and divides by $\,\le,\,$ and using the universal property of min instead of that of gcd, i.e.
$$\begin{eqnarray} {\rm employing}\quad\ c\le a,b&\iff& c\le \min(a,b)\\ \rm the\ analog\ of\quad\ c\ \, |\, \ a,b&\iff&\rm c\ \,|\,\ \gcd(a,b) \end{eqnarray}$$
Then the above proof translates as below, $\$ with $\,\ m(x,y) := {\rm min}(x,y)$
$c \le a,b \!\iff\!c\!+\!x \le a\!+\!x,b\!+\!x\!$ $\!\iff\! c\!+\!x \le m(a\!+\!x,b\!+\!x)\!$ $\!\iff\!\! c \le m(a\!+\!x,b\!+\!x)\!-\!x$
Theorem $\ \$ If $\ a,b,x\$ are positive naturals then $\ (ax,bx) = (a,b)x$
Proof $\$ By induction on $\color{#0a0}{{\rm size}:= a\!+b}.\,$ If $\,a=b,\,$ $\,(ax,bx) = (ax,ax) = (a)x = (a,b)x\,$ hence it is true. Else $\,a\neq b;\,$ wlog, by symmetry, $\,a > b\,$ so $\,(ax,bx) = (ax\!-\!bx,bx) = \color{}{((a\!-\!b)x,bx)}\,$ with smaller $\rm\color{#0a0}{size}$ $\,(a\!-\!b) + b = a < \color{#0a0}{a\!+b},\,$ therefore $\,((a\!-\!b)x,bx)\!\underset{\rm induct}=\! (a\!-\!b,b)x = (a,b)x$.
Let $a,b$ have the following prime factorisations:
$$a= \prod_{n=1}^\infty p_n^{\alpha_n} ,b=\prod_{n=1}^\infty p_n^{\beta _n}.$$
(Here $(p_n)$ is the ascending sequence of prime numbers). We then have $$\gcd(a,b)=\prod_{n=1}^\infty p_n^{\min(\alpha_n,\beta_n)},$$ and consequently $$\frac{a}{\gcd(a,b)}=\prod_{n=1}^\infty p_n^{\alpha_n-\min(\alpha_n,\beta_n)},\frac{b}{\gcd(a,b)}=\prod_{n=1}^\infty p_n^{\beta_n-\min(\alpha_n,\beta_n)}.$$
Now ask yourself, could it be that one of the factors $p_n$ could have a strictly positive exponent, in both of the last products simultaneously?
Every other answer is doing part of this, but since you asked for rigorous, here is rigorous:
Let $A$ be the multiset of prime factors of a, $B$ be the multiset of prime factors of $b$. The multiset of prime factors of $1$ is the empty set.
$$\text{GCD}\left(\frac{a}{\text{GCD}(a,b)}, \frac{b}{\text{GCD}(a,b)}\right) = 1$$
Standard conversion to multisets:
$$\bigg(A – (A\cap B)\bigg) \cap \bigg(B – (A \cap B)\bigg) = \{\}$$
Standard conversion to Boolean Algebra ($x \in A$ becomes $A$):
$$\bigg(A \land \lnot (A \land B)\bigg) \land \bigg(B \land \lnot (A \land B)\bigg) = \bot$$
Standard reduction (or you could use truth table):
$$\bigg(A \land (\lnot A \lor \lnot B)\bigg) \land \bigg(B \land (\lnot A \lor \lnot B)\bigg) = \bot$$
$$\bigg(A \land \lnot B\bigg) \land \bigg(B \land \lnot A\bigg) = \bot$$
$$\bot = \bot$$
$$\top$$
Assume WLOG that $a, b \geq 1$. Let $m = \dfrac{a}{\gcd(a,b)}$, and $n = \dfrac{b}{\gcd(a,b)}$, and let $c = \gcd(m,n)$. Then $c \mid m$, and $c \mid n$. This means: $(c\cdot \gcd(a,b)) \mid a$, and $(c\cdot \gcd(a,b)) \mid b$. So $(c\cdot \gcd(a,b)) \mid \gcd(a,b)$. but $\gcd(a,b) \mid (c\cdot \gcd(a,b))$. Thus: $c\cdot \gcd(a,b) = \gcd(a,b)$, and this means $c = 1$.
A Bezout-centric approach. Let $d=\text{gcd}(a,b)$. Then, there are integers $m$ and $n$ such that $d=ma+nb$. But then $m$ and $n$ are also integers such that
$$1=m\left(\frac{a}{d}\right)+n\left(\frac{b}{d}\right)$$
so $\text{gcd}(\frac{a}{d},\frac{b}{d})=1$.
Suppose $d \mid \dfrac a{\gcd(a,b)}$ and $d\mid \dfrac b{\gcd(a,b)}$.
Then $nd = \dfrac a {\gcd(a,b)}$ and $md\mid \dfrac b {\gcd(a,b)}$.
So $nd\cdot\gcd(a,b) = a$ and $md\cdot\gcd(a,b) = b$.
So $d\cdot\gcd(a,b)$ is a common divisor of $a$ and $b$.
Since $\gcd(a,b)$, is the greatest of those, we must have $d\not>1$. | 2018-07-16T20:20:24 | {
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https://themaximalist.org/category/integral-calculus/ | ## logarc
In the back of my calculus book there is a table of famous integrals. Here’s integral number 21 in that table:
From Maxima integrate(), I get
What’s going on?
Both forms give a workable antiderivative for the original integrand:
Furthermore, we believe that both forms are correct because of this helpful identity for hyperbolic sine:
${\rm asinh}(z)=\ln(z+\sqrt{1+z^2}).$
Turns out (thanks to a Barton Willis for pointing me in the right direction) there’s a variable logarc that we can set to make Maxima return the logarithmic form instead of hyperbolic sine:
I haven’t yet encountered cases where this would be a bad idea in general, but I’ll update this if I do.
## logabs
In the first week of my differential equations course, we study methods of direct integration and separation of variables. I like to emphasize that the absolute values can lend an extra degree of generality to solutions with antiderivatives of the form
$\int \frac{1}{u}\;du = \ln |u|+ C.$
As an example, for the initial value problem
$y' = \frac{x}{1-x^2}$, $y(0)=1$,
it is convenient for treating all possible initial conditions ($x \ne \pm 1$) in one step to use the antiderivative
$y = -\frac{1}{2} \ln | 1-x^2 | + C.$
However, Maxima omits the absolute values.
For this case, we could consider only the needed interval $-1, but still…
Turns out we can set the Maxima variable logabs to make integrate() include absolute values in cases like this:
But then later in the course, I saw that logabs also impacts the Ordinary Differential Equation solver ode2(). I encountered an example for which Maxima, in particular solve() applied to expressions involving absolute value, didn’t do what I wanted with logabs:true
For the logistic equation
$\frac{dP}{dt} = kP\left ( 1-\frac{P}{P_c} \right )$, $P(0)=P_0$
we expect that by separating variables we can obtain the solution
$P(t) = \frac{P_0 P_c}{(P_c-P_0)e^{-kt}+ P_0}.$
Here’s what happens when we use ode2() with and without logabs:true:
## Path Integrals
For my multivariable calculus class, I wanted an easy-to-call suite of symbolic integrators for path integrals of the form
$\int_C f(x,y) dr$,
$\int_C {\bf F}(x,y)\cdot d{\bf r} = \int_C \langle P(x,y), Q(x,y) \rangle\cdot \langle dx, dy \rangle$, or
$\int_C {\bf F}(x,y,z)\cdot d{\bf r} = \int_C \langle P(x,y,z), Q(x,y,z), R(x,y,z) \rangle\cdot \langle dx, dy, dz \rangle$.
My overarching design idea was that the input arguments needed to look the way they do when I teach the course:
• a scalar field $f(x,y):R^2 \rightarrow R$ or a vector field ${\bf F}(x,y): R^2 \rightarrow R^2$ or ${\bf F}(x,y,z): R^3 \rightarrow R^3$
• a curve $C$ defined by a vector-valued function ${\bf r}(t): R \rightarrow R^n, a\le t \le b$ where $n=2,3$ as appropriate.
It took me a while to work out how to evaluate the integrand along the path within my function. Things that worked fine on the command line failed when embedded into a batch file to which I passed functions as arguments. I ended up using subst, one variable at a time. I’d like to be able to do this in a single command which can detect whether we’re in 2 or 3 dimensions so that I don’t need separate commands.
For now, here’s what I came up with along with some illustrative examples taken from Paul’s online math notes, that show how to call these new commands I, I2 and I3.
/* path integral of a scalar integrand f(x,y) on path r(t) in R^2, t from a to b */
I(f,r,t,a,b):=block(
[f1,f2,dr,Iout],
f1:subst(x=r[1],f),
f2:subst(y=r[2],f1),
dr:sqrt(diff(r,t).diff(r,t)),
Iout: integrate(f2*dr,t,a,b),
Iout
);
/* path integral of a vector integrand F(x,y) on path r(t) in R^2, t from a to b */
I2(H,r,t,a,b):=block(
[H1,H2,I],
H1:subst(x=r[1],H),
H2:subst(y=r[2],H1),
I: integrate( H2.diff(r,t),t,a,b),
I
);
/* path integral of a vector integrand F(x,y,z) on path r(t) in R^3, t from a to b */
I3(H,r,t,a,b):=block(
[H1,H2,H3,I],
H1:subst(x=r[1],H),
H2:subst(y=r[2],H1),
H3:subst(z=r[3],H2),
I: integrate( H3.diff(r,t),t,a,b),
I
);
Here’s an update: a related maxima function for evaluating a complex integral
$\int_\Gamma f(z) dz$
where $f: C \rightarrow C$ and the curve $\Gamma$ is given by $r: R \rightarrow C$.
/* path integral of a complex integrand f(z): C --> C, on path z(t): R --> C, t from a to b */
IC(f,r,t,a,b):=block(
[f1,dz,Iout],
f1:subst(z=r,f),
dz:diff(r,t),
Iout: integrate(f1*dz,t,a,b),
Iout
);
## Two Maxima Functions for Riemann Sums
Two early attempts at programming Maxima functions. I’d love to hear your comments about how to make these work better. Many thanks to the online community from whom I learned how to get this far!
The script is given at the bottom of this post. Copy and paste into a file (in your working directory) called Riemann.mac, and then load into Maxima using
/* Two Maxima functions for introductory integral calculus
(c) 2016, themaximalist.org */
/* RiemannSum
fn is the function to be integrated on the interval [a,b]
n rectangles
opt specifies:
0: left endpoints
1: midpoints
2: right endpoints
*/
RiemannSum(fn,a,b,n,opt):=
block([xx,s],
xx(i,n):= a+(i+opt/2)*(b-a)/n,
s: sum(ev(fn,x=xx(i,n)),i,0,n-1)*(b-a)/n,
float(s)
)$/* RiemannRectangles to Draw rectangles fn is the function to be integrated: on the interval [a,b] n rectangles opt specifies: 0: left endpoints 1: midpoints 2: right endpoints */ load(draw)% RiemannRectangles(fn,a,b,n,opt):= block([xi,xx,rects,wxd], xi(i,n):= a+i*(b-a)/n, xx(i,n):= a+(i+opt/2)*(b-a)/n, rects(n):=makelist(rectangle([xi(k,n),0], [xi(k+1,n),ev(fn,x=xx(k,n))]),k,0,n-1), wxd(n):=apply(wxdraw2d, append([xrange=[a-(b-a)/4, b+(b-a)/4],color=blue], rects(n), [transparent=true, explicit(fn,x,a, b)] ) ), wxd(n) )$ | 2021-02-28T22:30:08 | {
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https://math.stackexchange.com/questions/562119/difference-of-two-binomial-random-variables | # Difference of two binomial random variables
Could anyone guide me to a document where they derive the distribution of the difference between two binomial random variables. So $X \sim \mathrm{Bin}(n_1, p_1)$ and $Y \sim \mathrm{Bin}(n_2, p_2)$, what is the distribution of $|X-Y|$.
thank you.
(Also $X$ and $Y$ are independent)
This question is more tricky than it sounds. To solve it, I will use here a combination of both manual methods and automated methods, in particular computer algebra tools [the mathStatica package (of which I am an author) for Mathematica and the latter itself].
If I may change the notation slightly:
The Problem
Let $X_1$ ~ $\text{Binomial}(n,p)$ and $X_2$ ~ $\text{Binomial}(m,q)$ be independent.
Find the pmf of $|X_1-X_2|$
Given: Due to independence, the joint pmf of $(X_1, X_2)$, say $f(x_1,x_2)$, is:
Solution
Let $Y=X_1-X_2$ and $Z=X_2$. Then, the joint pmf of $(Y,Z)$, say $g(y,z)$, is:
where Transform is a mathStatica function that derives the joint pmf using the Method of Transformations. Deriving the domain of support of $Y$ and $Z$ is a bit more tricky. To make things clearer, here is a rough plot that illustrates the (smoothed continuous version of) the domain of support:
This suggests two cases:
• Case 1: When $y \ge 0$: $0 \le z \le n-y$
• Case 2: When $y < 0$: $-y \le z \le m$
The density of $Y=X_1-X_2$ is then obtained by summing out $Z$ in each part of the domain:
Finally, to find the pmf of $|Y|$, the pmf for strictly positive values will be:
and when $Y=0$:
Summary
The pmf of $|X_1-X_2|$, say $\phi(y)$ is:
with domain of support $Y$ = {0, 1, ..., max$(m,n)$}.
All done.
Monte Carlo check
It is always a good idea to check ones work using Monte Carlo methods. Here, for instance, are 100,000 pseudo-random drawings from each of $X_1$ and $X_2$, given some parameter assumptions:
x1data = RandomVariate[BinomialDistribution[12, .1], 100000];
x2data = RandomVariate[BinomialDistribution[ 7, .9], 100000];
Next, compare the empirical distribution of $|X_1-X_2|$ (red triangles) to the theoretical density $\phi(y)$ (blue dots) derived above, given the same parameter assumptions:
Looks good :)
• For the special case in which $p=q=1/2$ and $n=m$, we get: $Pr[Y=y] = p^{2n} \cdot \text{Binomial}[n,y] \cdot \text{Hypergeometric2F1}[-n,-n+y,1+y,1]$ – Erel Segal-Halevi Feb 3 '15 at 21:11
I doubt there is a special name for the distribution in general. There is one special case of interest: $p_1 = 1 - p_2$. Note that $n_2 - Y \sim {\text Bin}(n_2, 1-p_2)$, and so in this special case $X - Y + n_2 \sim {\text Bin}(n_1 + n_2, p_1)$.
• This explains the distribution of X−Y (in the special case). How do you get from it to the distribution of |X-Y|? – Erel Segal-Halevi Feb 4 '15 at 6:17
• $P(|X-Y|=z) = P(X-Y=z) + P(X-Y=-z)$ for nonnegative integers $z$. – Robert Israel Feb 4 '15 at 6:29
I can give you an answer for the pmf of X-Y. From there |X - Y| is straightforward.
$$X \sim Bin(n_1, p_1)$$
$$Y \sim Bin(n_2, p_2)$$
We are looking for the probability mass function of $$Z=X-Y$$
First note that the min and max of the support of Z must be $$(-n_2, n_1)$$ since that covers the most extreme cases ($$X=0$$ and $$Y=n_2$$) and ($$X=n_1$$ and $$Y=0$$).
Then we need a modification of the binomial pmf so that it can cope with values outside of its support.
$$m(k, n, p) = \binom {n} {k} p^k (1-p)^{n-k}$$ when $$k \leq n$$ and 0 otherwise.
Then we need to define two cases
1. $$Z \geq 0$$
2. $$Z \lt 0$$
In the first case
$$p(z) = \sum_{i=0}^{n_1} m(i+z, n_1, p_1) m(i, n_2, p_2)$$
since this covers all the ways in which X-Y could equal z. For example when z=1 this is reached when X=1 and Y=0 and X=2 and Y=1 and X=4 and Y=3 and so on. It also deals with cases that could not happen because of the values of $$n_1$$ and $$n_2$$. For example if $$n_1 = 4$$ then we cannot get Z=1 as a combination of X=5 and Y=4. In this case thanks to our modified binomial pmf the probablity is zero.
For the second case we just reverse the roles. For example if z=-1 then this is reached when X=0 and Y=1, X=1 and Y=2 etc.
$$p(z) = \sum_{i=0}^{n_2} m(i, n_1, p_1) m(i+z, n_2, p_2)$$
Put them together and that's your pmf.
$$f(z)= \begin{cases} \sum_{i=0}^{n_1} m(i+z, n_1, p_1) m(i, n_2, p_2),& \text{if } z\geq 0\\ \sum_{i=0}^{n_2} m(i, n_1, p_1) m(i+z, n_2, p_2), & \text{otherwise} \end{cases}$$
Here's the function in R and a simulation to check it's right (and it does work.)
https://gist.github.com/coppeliaMLA/9681819
• A great answer from Simon. Can you tell me the inverse cumulative for X - Y? – sjb-sjb Sep 19 '18 at 22:48 | 2019-04-24T13:59:49 | {
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http://lincoln-marti.net/daniel-doheny-msmesob/xzc9s78.php?5668b7=sss-postulate-examples | In this mini-lesson, we will learn about the SSS similarity theorem in the concept of the SSS rule of congruence, using similar illustrative examples. This states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent. Example of Postulate. If we can show that two sides and the included angle of one triangle are congruent to two sides and the included angle in a second triangle, then the two triangles are congruent. Asked By: Ayada Lugo | Last Updated: 7th January, 2020. Teacher’s Activity Students’ Activity Yes Adrian Very good! The following postulate, as well as the SSS and SAS Similarity Theorems, will be used in proofs just as SSS, SAS, ASA, HL, and AAS were used to prove triangles congruent. Examples: Geometric Postulates Show Answer. [AAS postulate.] How to pronounce postulate. Section 4. How to use the Pythagorean Theorem? Teacher’s Activity Students’ Activity For your better understanding, here is now the exact statement for SSS Congruence Postulate. This geometry video tutorial provides a basic introduction into triangle congruence theorems. Example What postulate would you use to prove the triangles congruent? So we will give ourselves this tool in our tool kit. var vidDefer = document.getElementsByTagName('iframe'); Topics. In this lesson, we will examine this postulate, see how and why it works, and put it to use in various examples. Like any field, the present system of accounting has certain underlying axioms which form the basis of … The Parallel Postulate - Through a given external point, there is at most one line parallel to a given line. Hammond’s postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. And they were able to do it because now they can write "right angle," and so it doesn't form that embarrassing acronym. Given : In ΔABC, AD is a median on BC and AB = AC. Hl or hypotenuse leg for right triangles only. 14 Votes) SAS Postulate. We had the SSS postulate. AB ¯ = DE ¯ [Given.] Postulates are also called as axioms. On the front of the organizer, students will write SSS on the first tab, SAS on the second tab, and ASA on the third tab. Discussion. I’m confident that after watching this lesson you will agree with me that proving triangles congruent is fun and straightforward. This is the only postulate that does not deal with angles. ΔABC and ΔDBC are two isosceles triangle. Angle-Angle (AA) Similarity Postulate If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. Can you can spot the similarity? Chapter 4. Listen to the audio pronunciation in the Cambridge English Dictionary. Postulate 20. Use the SAS Similarity Theorem to determine if triangles are similar. Name the postulate, if possible, that makes the triangles congruent. Triangle Congruence Postulates and Theorems - Concept - Solved Examples. Example 2: If ¯PN ⊥ ¯MQ and ¯MN ~= ¯NQ as shown in Figure 12.4, write a two-column proof that ΔPNM ~= ΔPNQ. In cat below. How to say postulate. We can say that two triangles are congruent if any of the SSS, SAS, ASA, or AAS postulates are satisfied. 6 Check It Out! Hence sides AB and CD are congruent, and also sides BC and DA are congruent. AC = 8x + 1 = 33 EF = 2(x + 1) = 10 When x = 4, the triangles are similar by the SSS Similarity Theorem. The ASA Postulate was contributed by Thales of Miletus (Greek Name two sides and the included angle between the sides. 4 ways of proving that triangles are congruent. Step: 4. This means that the pair of triangles have the same three sides and the same three angles (i.e., a total of six corresponding congruent parts). SSS Similarity. Introduction to triangle congruency lesson. As Math is Fun accurately states, there only five different congruence postulates that will work for proving triangles congruent. SSS Congruence Postulate If the three sides of a traingle are conrresponding and congruent to the three sides of the other triangle, th the two triangles are congruent. For a list see Congruent Triangles. Explain your reasoning. This is the only postulate that does not deal with angles. Each triangle postulate has a clear example with pictures, want to see? Heather Z. Oregon State University. https://www.onlinemath4all.com/side-side-side-congruence-postulate.html Stay Home , Stay Safe and keep learning!!! 2. Prove: $$\triangle ABC \cong \triangle EFC$$ Side Angle Side Example Proof. In a square, all four sides are congruent. State the postulate or theorem. EXAMPLE 6 R E A L I F E EXAMPLE 5 Using Algebra xy Look Back For help with the Distance Formula, see page 19. Congruent Triangles. For your better understanding, here is now the exact statement of the SSS Congruence Postulate. Category: medical health lung and respiratory health. There's the Side-Angle -Side postulate, or SAS. Proof 1. Examples. 8. Learn more. -> If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent. Video Examples: The five postulates of Euclidean Geometry. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent. // Last Updated: January 21, 2020 - Watch Video //. Introduction. SSS Congruence Postulate. Play this game to review Geometry. We can use the SSS postulate (which has no A's—unlike your geometry tests). if(vidDefer[i].getAttribute('data-src')) { Sas triangle congruence postulate explained youtube. You must be signed in to discuss. How to prove triangles congruent sss, sas, asa, aas rules. NOT CONGRUENT The Congruence Postulates SSS ASA SAS AAS SSA AAA Name That Postulate SAS ASA SSS SSA (when possible) Name That Postulate (when possible) ASA SAS AAA SSA Name That Postulate (when possible) SAS SAS SAS Reflexive Property Vertical Angles Vertical Angles Reflexive Property SSA Let’s Practice Indicate the additional information needed to … Now we have the SAS postulate. 1). Practice Proofs. Examples: Determine if the lengths represent the sides of an acute, right, or obtuse triangle if a triangle is possible. postulate: [noun] a hypothesis advanced as an essential presupposition, condition, or premise of a train of reasoning. The other triangle LMN will change to remain congruent to the triangle PQR. As you will quickly see, these postulates are easy enough to identify and use, and most importantly there is a pattern to all of our congruency postulates. Teacher’s Activity Students’ Activity Yes Adrian Very good! Example 1. In an exothermic reaction, the transition state is closer to the reactants than to the products in energy (Fig. ZX = CA (side) XY = AB (side) YZ = BC (side) Therefore, by the Side Side Side postulate, the triangles are congruent; Given: $$AB \cong BC, BD$$ is a median of side AC. Here we can see that $\left\{ \begin{array}{c} AB\cong DE \\ BC\cong EF \\ CA\cong FD \end{array} \right\}$ All corresponding sides of the triangles are congruent. You could view this as angle-side-side. Examples: Determine if the triangles are congruent. To prove that these triangles are congruent, we use SSS postulate, as the corresponding sides of both the triangles are equal. Check out the interactive simulation to explore more congruent shapes and do not forget to try your hand at solving a … Holt McDougal Geometry Triangle Similarity: AA, SSS, SAS Example 1: Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. You know you have to prove the triangles congruent, and one of the givens is about angles, so SAS looks like a better candidate than SSS (Side-Side-Side) for the final reason of the proof. From side side side postulate to Postulates of Congruent Triangle. © and ™ ask-math.com. The following postulate, as well as the SSS and SAS Similarity Theorems, will be used in proofs just as SSS, SAS, ASA, HL, and AAS were used to prove triangles congruent. Learn more. Geometry › SSS Postulate. Example How can you use congruent triangles to prove j Q @ j D Since QWE ≅ DVK by AAS, you know that ∠ The Area Postulate - To every polygonal region there corresponds a unique positive real number. Glacial-interglacial sea surface temperature changes across the subtropical front east of New Zealand based on alkenone unsaturation ratios and foraminiferal assemblages In ΔABC, AD is a median on BC and AB = AC. 5 Example 1 Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. Step: 1 . In another lesson, we will consider a proof used for right triangl… STUDENT HELP y 1 x 1 A( 7, 5) C( 4, 5) B( 7, 0) H(6, 5) G(1, 2) F(6, 2) 216 Chapter 4 Congruent Triangles 1.Sketch a triangle and label its vertices. Is it true that ∆ ABC ≅ ∆ ADC? Share . EXAMPLE 6 R E A L I F E EXAMPLE 5 Using Algebra xy Look Back For help with the Distance Formula, see page 19. And here, they wrote the angle first. How do you know? So that actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. function init() { Perhaps the easiest of the three postulates, Side Side Side Postulate (SSS) says triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle. This video is provided by the Learning Assistance Center of Howard Community College. congruent. The two triangles also have a common side: AC. The Multiplication Postulate: If x = y, then x * 3 = y * 3 . Learn the triangle congruence postulates which are the SAS, ASA, SSS, AAS, and HL. Prove: $$\triangle ABD \cong \triangle CBD$$ Every single congruency postulate has at least one side length known! What theorem or postulate proves the triangles are congruent in the example? Postulate is used to derive the other logical statements to solve a problem. Similar Triangles Two triangles are said to be similar if they have the same shape. Example of Postulate. If the are, write a similarity statement. Step: 6. Two geometric figures are similar if one is a scaled version of the other. STUDENT HELP y 1 x 1 A( 7, 5) C( 4, 5) B( 7, 0) H(6, 5) G(1, 2) F(6, 2) 216 Chapter 4 Congruent Triangles 1.Sketch a … Or, if we can determine that the three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. EXAMPLE 1 Use the AA Similarity Postulate Determine whether the triangles are similar. SAS; SSS; ASA; AAS; Yep, you guessed it. Definition and examples of sas congruency postulate | define sas. Congruent trianglesare triangles that have the same size and shape. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. This means that the corresponding sides are equal and the corresponding angles are equal. This is one of them (SSS). SSS Postulate - Every SSS correspondence is a congruence. ... Side-Side-Side (SSS) Congruence Postulate. Solution : The game plan is to make use of the SAS Postulate. Example 1. This is called the Side Angle Side Postulate or SAS. postulate meaning: 1. to suggest a theory, idea, etc. Accounting Postulate: A fundamental assumption in the field of accounting. Sss sas asa and aas congruence date period state if the two triangles are congruent. All Rights Reserved. SSS Congruence Postulate. Title: SideSideSide SSS Congruence Postulate 1 Side-Side-Side (SSS) Congruence Postulate If three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent. Glencoe Geometry. The global warming postulate is based almost entirely on models, and today's models are deliberately biased to support global warming. There's no other one place to put this third side. Side Side Side Postulate -> If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent. 97 examples: They are also postulated to stimulate other cells for granuloma formation… Postulate 17. Because if we can show specific sides and/or angles to be congruent between a pair of triangles, then the remaining sides and angles are also equal. The first two postulates side angle side sas and the side side side sss focus predominately on the side aspects whereas the next lesson discusses two additional postulates which focus more on the angles. ASA Postulate Example Angle-Angle-Side Whereas the Angle-Angle-Side Postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then … 8. window.onload = init; © 2021 Calcworkshop LLC / Privacy Policy / Terms of Service. Using the Angle Addition Postulate and definition of. Five-Minute Check (over Lesson 4–3) CCSS Then/Now New Vocabulary Postulate 4.1: Side-Side-Side (SSS) Congruence Example 1: Use SSS to Prove Triangles Congruent Example 2: Standard Test Example: SSS on the Coordinate Plane Postulate 4.2: Side-Angle-Side (SAS) Congruence Example 3: Real-World Example: Use SAS to Prove Triangles are Congruent Example 4: Use SAS or SSS in Proofs Over Lesson … It is the only pair in which the angle is an included angle. Postulate 18. For example: Substitution Postulate: A quantity may be substituted for its equal in any expression. Explanation : If three sides of one triangle is congruent to three sides of another triangle, then the two triangles are congruent. We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. Everybody read! Proving Congruence SSS, SAS. Top Geometry Educators. Covid-19 has affected physical interactions between people. Try this Drag any orange dot at P,Q,R. Congruent Triangles. But there is a warning; we must be careful about identifying the accurate side and angle relationships! Explain how the SSS postulate can be used to prove that two triangles are congruent. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher). 10) These two angles are linear pair angles and they are supplementary, 13) AP is the perpendicular bisector of BC, 13) By definition of perpendicular bisector and from (9) and (12). SSS Postulate First, there's the side-side-side postulate, or SSS . In order to prove that triangles are congruent, all the angles and sides have to be congruent. Two triangles are said to be congruent if one can be superimposed on the other such that each vertex and each side lie exactly on top of … CCSS.MATH.CONTENT.7.G.A.2 Draw (freehand, with ruler and protractor, and with technology) geometric shapes with given conditions. Use SSS. Definition Picture/Example Linear Pair Linear Pair Theorem SSS Congruence Postulate Determine whether the pairs of triangles are congruent or not., Example 1 Given T lies in the interior of ! Examples, solutions, videos, worksheets, games, and activities to help Geometry students learn triangle congruence by side-side-side (SSS) and side-angle-side (SAS). It also discusses the CPCTC theorem, to draw further conclusions from congruency. Postulate 19. Congruence Postulate SSS. What if we aren't given any angles? Example $$\triangle ABC \cong \triangle XYZ$$ All 3 sides are congruent. The Pythagorean Theorem can be used when we know the length of two sides of a right triangle and we need to get the length of the third side. Methods of proving triangle congruent mathbitsnotebook(geo. 7, 24, 25; 5, 12, 16; 6, 8, 9; 3, 5, 9 ; Show Video Lesson. A X B C Y Z . i) ΔABD ≅ ΔACD ii) AP is the perpendicular bisector of BC. And as seen in the image to the right, we show that trianlge ABC is congruent to triangle CDA by the Side-Side-Side Postulate. They are called the SSS rule, SAS rule, ASA rule and AAS rule. Relationships Within Triangles. We discuss what the abbreviations stand for and then students identify which postulate can be used to prove the triangles from the Do Now are congruent. Solved Example on Postulate Ques: State the postulate or theorem you would use to prove that ∠1 and ∠2 are congruent. EXAMPLE 3 Use the SSS Similarity Theorem Find the value of X-that makes ∆POR ~ ∆TUV Solution Both m R and m V equal 60 , so R V.Next, find the value of x that makes the About Cuemath. For example: Since the sum of 3 and 8 are both 8, we can substitute each expression with 8 and they will still equal to one another. A few examples were shown for a better understanding. If what you postulate is true, then both the lawyer and DeLay are too incompetent to enter into a courtroom to begin with. AAS Postulate Example. 4.6/5 (14 Views . We refer to this as the Side Side Side Postulate or SSS. If all three sides in one triangle are the same length as the corresponding sides in the other, then the triangles are congruent. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. Get access to all the courses and over 150 HD videos with your subscription, Monthly, Half-Yearly, and Yearly Plans Available, Not yet ready to subscribe? are similar. 8. Example St. Francis Preparatory School. Addition Postulate: If equal quantities are added to equal quantities, the sums are equal. Side-angle-side (sas) triangle: definition, theorem & formula. SAS Postulate Example Side-Side-Side Or, if we can determine that the three sides of one triangle are congruent to three sides of another … SSS postulate. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. Prove that $$\triangle LMO \cong \triangle NMO$$ Advertisement. In which pair of triangles pictured below could you use the Angle Side Angle postulate (ASA) to prove the triangles are congruen. Did you know that there are five ways you can prove triangle congruency? Postulate is a true statement, which does not require to be proved. And as seen in the image, we prove triangle ABC is congruent to triangle EDC by the Side-Angle-Side Postulate. ... (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions. So we already know, two triangles are congruent if they have the same size and shape. 14 December 2020 . Determine which congruent triangle postulate is likely to be the ticket for proving the triangles congruent. In this lesson, we will consider the four rules to prove triangle congruence. If the two angles and the non included side of one triangle are congruent to the two angles and the non included side of another triangle, then the two triangles are congruent. In today’s geometry lesson, we’re going to tackle two of them, the Side-Side-Side and Side-Angle-Side postulates. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. Students then glue the diagrams onto the back of the correct tab (see the file “SSS, SAS, and ASA Activity”). pagespeed.lazyLoadImages.overrideAttributeFunctions(); called a linear pair. Given: 1) point C is the midpoint of BF 2) AC = CE. More About Postulate. Now, I’ll group you into 4 groups and form a circle with your group. MC Megan C. Piedmont College. Example 1 Solution Because they are both right angles, B and E are congruent. Covid-19 has led the world to go through a phenomenal transition . SSS Postulate. Examples of postulate in a sentence, how to use it. Answer. How to Prove Triangles Congruent? Thankfully we don’t need to prove all six corresponding parts are congruent… we just need three! Solving sas triangles. Those are the Angle-Side-Angle (ASA) and Angle-Angle-Side (AAS) postulates. 1) In triangle ABC, AD is median on BC and AB = AC. In this blog, we will understand how to use the properties of triangles, to prove congruency between $$2$$ or more separate triangles. } } } Side Angle Side Practice Proofs. Triangles ABC has three sides congruent to the corresponding three sides in triangle C… Take Calcworkshop for a spin with our FREE limits course. 3) By definition of median. Statement of the Aim *complete the congruent marks to illustrate that the triangles are congruent through SSS Congruence Postulate; *match the given sides of triangles to show that the triangles are congruent through SSS Congruence Postulate; *illustrate the importance of being part of a group by citing an example. Theorem or postulate? Solution to Example 2 1. Pair four is the only true example of this method for proving triangles congruent. Everybody read! Solutions. AB ... •Example: because of HL. Side-Side-Side (SSS) 1. Is it true that ∆ABC ≅ ∆XYZ? Some of the worksheets below are Geometry Postulates and Theorems List with Pictures, Ruler Postulate, Angle Addition Postulate, Protractor Postulate, Pythagorean Theorem, Complementary Angles, Supplementary Angles, Congruent triangles, Legs of an isosceles triangle, … So SAS-- and sometimes, it's once again called a postulate, an axiom, or if it's kind of proven, sometimes is called a theorem-- this does imply that the two triangles are congruent. Advertisement. SAS Congruence Postulate. Congruence is defined as agreement or harmony. You’ll quickly learn how to prove triangles are congruent using these methods. The Division Postulate: If x = y, then x / 7 = y / 7 . Figure 12.4 ¯PN ⊥ ¯MQ and ¯MN ~= ¯NQ. As a consequence, their angles will be the same. So we need to learn how to identify congruent corresponding parts correctly and how to use them to prove two triangles congruent. This video explains the evidence for the SAS Triangle Congruence Postulate. 2 Use the SSS Congruence Postulate Example 1 Solution It is given that and _____. for (var i=0; i | 2022-01-19T11:49:16 | {
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https://www.physicsforums.com/threads/area-between-curves.154524/ | # Area between curves
1. Feb 3, 2007
### tony873004
1. The problem statement, all variables and given/known data
Find the area of the region bounded by:
y=cosx, y=sin2x, 0, x=pi/2
2. Relevant equations
3. The attempt at a solution
I made a graph. I believe I'm trying to find the area I shaded.
red=cos(x), blue=sin(2x)
I need to find the intersection point so I will know the limits of my 2 integrals.
cosx = sin2x
But I don't know how to do this. There should be an infinate number of intersections, but I am only interested in the one that appears to happen around x=1/2 and the next one at what appears to be pi/2.
I can verify with my calculator that cos(pi/2) and sin(2pi/2) both equal 0, and that the right limit given by the problem is indeed the intersection, but that is not the case for the 1st intersection. How do I solve this? And what if the book gave the right limit as x=2. My method of eyeballing it and verifying my guess with the calculator would fail.
Assuming I find the intersection point, the next thing I was going to do is:
$$\left( {\int_0^{???} {\cos x} \,dx\, - \,\int_0^{???} {\sin 2x} \,dx} \right)\, + \,\,\left( {\int_{???}^{\pi /2} {\sin 2x\,dx} - \int_{???}^{\pi /2} {\cos x\,dx} } \right)$$
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Feb 3, 2007
### Gib Z
sin(2x)=2 sinx cos x
cos x= 2sinx cos x
1=2 sin x
sin x =1/2
So the solution is arcsin (1/2). What value of sin x gives us 1/2? pi/6 of course.
Thats the systematic way of doing it, you may also have done cos x= sin (pi/2 -x) so equate them, sin (pi/2 - x)=sin 2x, 2x=pi/2 - x, 3x=pi/2, x=pi/6.
The 2nd way is only useful for angles less than pi/2, which means it can be used here.
Now that you have both intersection points just integrate as how you were going to do next, knowing the ???'s is pi/6.
3. Feb 3, 2007
### tony873004
sin(2x)=2 sinx cosx
I guess that's just a trig identity I forgot? Playing with it on my calculator, I can see that it works for all values of x.
cosx = 2 sinx cosx
Playing with this on my calculator, it does not work for all values of x. Is this a trig identity, or specific to this problem?
If specific to this problem, from sin(2x)=2 sinx cosx i get
cosx = sin(2x)/2sin x
How did you get cosx = 2 sinx cosx ?
4. Feb 3, 2007
### Gib Z
Sin 2x = 2sinx cos x is a identity you just forgot.
Cos x= 2sinx cos x is specific to the problem. When we want to find points of intersection, we set the 2 functions to be equal, then solve. The functions we have are sin 2x and cos x. Since sin 2x=2sinx cos x for all x, we sub that in, then we have to solve cos x=2sinx cos x.
I dont know why you got cos x= sin(2x)/2 sin x, thats not needed.
cos x = 2sin x cos x
Divide both sides by 2 cos x, sin x=1/2. Solving for that, arcsin 1/2=pi/6
5. Feb 3, 2007
### Gib Z
The Indentity can be easily seen from expanding the formula for sin(x+y), letting y=x. | 2017-11-23T08:28:33 | {
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https://math.stackexchange.com/questions/2687074/using-long-division-to-divide-one-polynomial-by-another-of-a-higher-degree | # Using long division to divide one polynomial by another of a higher degree
My friend and I were finding the series expansion of the function $\frac{2x}{x^2+1}$. I naturally went for the Maclaurin expansion method, using the differentials to find the series expansion. The answer I got (which is correct) is $$\sum_{n=0}^\infty (-1)^n 2x^{(2n+1)}$$ The problem is, this took me a while to differentiate enough to see the pattern. My friend, however did something much faster - he used long division! $$\require{enclose} \begin{array}{r} \end{array} {x^2+1} \enclose{longdiv}{2x}$$ Apparently, instead of using the $x^2$ (the highest power of x) as the subject for his division, he used 1 (the lowest power of $x$), and he got the answer almost immediately. Example: 1 goes into $2x$ $2x$ times, and $(2x)({x^2+1}) = ({2x^3+2x})$. Subtract this from $2x$ and continue.
I couldn't find any evidence of this type of long division anywhere, and I'm having trouble understanding the logic behind it. How can he start subtracting something larger than $2x$ from $2x$? What sort of long division is this?
Could someone please help me understand?
• There is an easy series for $1/(1+t)$ that works as long as $|t| < 1.$ Analogous for $1/(1 + x^2)$ as long as $|x| < 1.$ You may multiply that by the numerator $2x$ – Will Jagy Mar 11 '18 at 23:10
• Your objection is that $2x^3 + 2x$ is "larger" than $2x$, but that depends on what you mean by "large". In the ring of formal power series $\Bbb{R}[[x]]$, it makes sense to order things so that $x^m + \ldots \le x^n + \ldots$, when $m \ge n$. With this notion of ordering, your friend's idea for calculating the power series representing a rational function works. – Rob Arthan Mar 11 '18 at 23:10
## 5 Answers
The reason we usually go with the leading term in polynomial division is because this will make it so that we cancel out the leading term in the numerator. This makes the total degree of what we're working with go down and we stop when we can't go any further.
But, this only works if the leading term of the denominator is smaller than the leading term of the numerator. So we shouldn't be able to apply it to something like $\frac{2x}{x^2+1}$.
But, note that the final term in the denominator has smaller degree than the final term in the numerator. This means that we can always make the final terms equal. Doing it like this will make the total degree of the output larger, hence we get the series expansion for the rational function.
So, in your example, $2x=2x(x^2+1)-2x^3$. Here we're just making the $+1$ part of $x^2+1$ equal to $2x$ and then subtracting out whatever is leftover (some may say the remainder). With this, you can write
$$\frac{2x}{x^2+1} = 2x - \frac{2x^3}{x^2+1}$$ Note that the power in the numerator increased, so we can do this again, but with $-\frac{2x^3}{x^2+1}$. In this case, $-2x^3=-2x^3(x^2+1) + 2x^5$, and so overall we have $$\frac{2x}{x^2+1} = 2x - 2x^3 + \frac{2x^5}{x^2+1}$$ And, again, the power in the numerator increased. Since we can increase this without bound, we can continue on to obtain the power series expansion.
This kind of long division is used all over places.
For example $$\frac {1}{1-x}=1+x+x^2+x^3+....$$ $$\frac{1}{1+x}=1-x+x^2-x^3+....$$ $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+....$$
It is much easier to get your series this way than going through lots of differentiation.
If $f,g$ are two polynomials such that $\deg f\geq\deg g$, then long division of $f$ by $g$ always terminates with a unique quotient polynomial and a constant remainder (this is the division algorithm).
If, however $\deg f\lt\deg g$, then the algorithm never terminates and the infinite series you generate as the quotient becomes the power series of $f/g$. Of course, convergence of this power series must be verified with $f/g$ to ensure that the power series obtained is the correct one.
Think of it analogously as dividing $1$ by $3$ using long division to obtain the decimal representation of $1/3$: you always get the remainder $10$ which is $\gt 3$ and the division algorithm never terminates.
By long division we obtain
$$\frac{1}{x^2+1}=\frac{x^2+1-x^2}{x^2+1}=1-\frac{x^2}{x^2+1}=1-\frac{x^4+x^2-x^4}{x^2+1}=1-x^2+\frac{x^4}{x^2+1}=$$
and so on.
As an alternative note that
$$\frac{2x}{x^2+1}=2x(1+x^2)^{-1}$$
then use binomial expansion for $(1+x^2)^{-1}$.
So you basically want to understand just the long division.
How many times does $x^2$ go into $2x$?
Let $k=\text{amount of times }x^2 \text{ goes into }2x$
$$kx^2=2x$$ $$k=\dfrac{2x}{x^2}$$
Therefore, $x^2$ goes into $2x$ $\dfrac {2x}{x^2}$ times.
Your friend would have continued by doing $1\times \dfrac{2x}{x^2}=\dfrac{2x}{x^2}$, to get a quotient of $\dfrac{2x}{x^2}+\dfrac{-\frac{2x}{x^2}}{2x}$, which he would have simplified. | 2019-10-14T13:27:41 | {
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https://www.physicsforums.com/threads/what-is-the-work-done-in-this-case.908239/ | # What is the work done in this case?
1. Mar 19, 2017
### vcsharp2003
1. The problem statement, all variables and given/known data
A spring (k = 500 N/m) supports a 400 g mass in a vertical position. The spring is now stretched 15 cm and maintained at the stretched position. How much work is done in stretching the spring by 15 cm? Is this work dependent on the magnitude of force applied while stretching the spring?
I am not sure if the work done would be just 0.5kx2 = 0.5 x 500 x 0.15[SUP}2[/SUP] = 5.625 J or one should use a more rigorous approach outlined under 3. The attempt at a solution.
2. Relevant equations
For a spring,
F= kx
PEspring = 0.5kx2
Work energy Theorem
ΔKE + Σ ΔPE = Work done
3. The attempt at a solution
Initial stretch of spring by 400 g mass = (0.400 x 9.8) / 500 = 0.00784 m
The initial stretch is used to determine initial spring PE.
Since the initial and final positions are at rest, so ΔKE = 0.
Σ ΔPE = ΔPEspring + ΔPEgravity = (0.5 x 500 x (0.15 + 0.00784)2 - 0.5 x 500 x (0.00784)2) + (0 - 0.400 x 9.8 x 0.15 )
= (6.23 - 0.015) + (- 0.588)
= 5.627 J
∴ Work done = ΔKE + Σ ΔPE = 0 + 5.627 = 5.627 J
This work would be independent of the magnitude of force applied since work done depends on only ΔKE and Σ ΔPE, both of which are independent of force applied.
2. Mar 19, 2017
### PetSounds
With only one significant figure in your original measurements, I don't think the difference in your answers will matter.
3. Mar 19, 2017
### TJGilb
So, the work in this context is force applied dotted with the displacement. You already recognize the force of the spring is $F=kx$. We know that the displacement is 15 cm. Since the force changes as we stretch it further and further, we need to take the integral of the force with respect to x. When you do this, you'll end up with $W=\frac 1 2 kx^2$. | 2017-08-19T09:37:39 | {
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https://math.stackexchange.com/questions/3430974/characteristic-of-a-field-is-a-minimal-number/3430998 | # Characteristic of a field is a minimal number
I have a an issue with certain terminology of field properties.
Characteristic is defined as as the minimal number such as $$1+1+1+1...=0$$
is there a possiblity that, there exists any other number than 1 that satisfies this condition?
suppose I have : $$\mathbb{Z}/7$$ then $$1+1+1+1+1+1+1=0$$ I can't think of any other minimal number that makes it equal.
• Do you mean to ask whether a field can have more than one characteristic at the same time? Or do you mean to ask why we are adding together copies of $1$ rather than some other number? – Arthur Nov 11 '19 at 12:05
• Chiming in with Arthur. In the latter case do observe that $$a+a+\cdots+a=a(1+1+\cdots+1)$$ for all the elements $a$ of the field. – Jyrki Lahtonen Nov 11 '19 at 12:08
• yes. Can it have anything other than 1? – user6394019 Nov 11 '19 at 12:08
There is a very good reason for unsing $$1$$ when defining the characteristic: The advantages of using $$1$$ specifically is that
1. it gives the largest characteristic among elements of the ring
2. any other element is going to yield "characteristic" which is a divisor of the $$1$$-characteristic
3. $$1$$ is the one element we can count on existing in the ring
For instance, in the ring $$\Bbb Z/7\Bbb Z$$, we get that $$1+1+1+1+1+1+1 = 0$$, but no other smaller sum of $$1$$'s is equal to $$0$$. Now, it also turns out that we get the same answer for any other non-zero number. For instance, $$2+2+2+2+2+2+2 = 0$$, but no other smaller sum of $$2$$'s is going to give $$0$$ (apart from the empty sum). In fields, you could use whatever non-zero number you want. This is where point 3 above comes in, though: In $$\Bbb Z/7\Bbb Z$$, using the number $$5$$ works, but in $$\Bbb Z/5\Bbb Z$$ it doesn't. Declaring that we will use $$1$$ means we aviod this issue.
As a different example, note that characteristic is also defined for non-field rigs (at least as long as they are commutative and unital). For instance, $$\Bbb Z/6\Bbb Z$$ has characteristic $$6$$, because $$1+1+1+1+1+1 = 0$$. However, if we use $$2$$ then we get $$2+2+2 = 0$$ and if we use $$3$$ we get $$3+3 = 0$$. So different numbers give different characteristics.
However, no matter which element you take, the "characteristic" that you get will be a divisor of $$6$$, and no matter which element you take, adding $$6$$ of them together will give you $$0$$, and no smaller number than $$6$$ works for all elements simultaneously. So the number $$6$$ still very much characterises the additive structure of the ring. So that's what we use.
• I understand. Can you also explain why char(Q)=char(R)=char(C)=0? – user6394019 Nov 11 '19 at 13:45
• @user6394019 That's just a convention. There is no way to write $0$ as a finite, non-empty sum of $1$'s in those fields, so they do not have positive finite characteristic the way $\Bbb Z/7\Bbb Z$ does, for instance. It could have been called $\operatorname{char}(\Bbb Q) = \infty$, but the established convention is to call it $\operatorname{char}(\Bbb Q) = 0$. – Arthur Nov 11 '19 at 13:50 | 2020-02-25T01:44:21 | {
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https://math.stackexchange.com/questions/1037738/proof-that-an-involutory-matrix-has-eigenvalues-1-1 | # Proof that an involutory matrix has eigenvalues 1,-1
I'm trying to prove that an involutory matrix (a matrix where $A=A^{-1}$) has only eigenvalues $\pm 1$.
I've been able to prove that $det(A) = \pm 1$, but that only shows that the product of the eigenvalues is equal to $\pm 1$, not the eigenvalues themselves.
Does anybody have an idea for how the proof might go?
Thanks.
Let $\lambda$ a eigenvalue of A and $x \neq 0$ respective eigenvector, then
$Ax = \lambda x \Leftrightarrow A^{-1}A x= \lambda A^{-1} x \Leftrightarrow x = \lambda A x \Leftrightarrow x = \lambda^2 x \Leftrightarrow (1-\lambda^2)x = 0$
then $\lambda =\pm 1$
• Thanks. I feel like such an idiot for trying to use determinants for hours. – spc38 Nov 25 '14 at 6:40
• @spc38 I tried determinants just now. See answer. – BCLC Oct 24 '18 at 5:57
Another approach is to note that, since $A^2 = I$, the minimal polynomial of an involutory matrix will divide $x^2 - 1 = (x-1)(x+1)$. The cases where the minimal polynomial is $(x-1)$ or $(x+1)$ correspond to the "degenerate" cases $A = I$ and $A = -I$. Here, the eigenvalues are all $1$ and all $-1$ respectively. All other cases result in $A$ having a mix of both $-1$ and $1$ eigenvalues, recognizing of course that there's no distinction between $-1$ and $1$ when $A$ is over a base field of characteristic two.
More generally, for a complex base field, this approach can be used to show that the set of eigenvalues of a matrix $m$-involution $A$ (for which $A^m=I$ for an integer $m>1$) belongs to the set of $m$-th roots of unity.
Here's another approach with diagonalisation. Let $$A=S\Lambda S^{-1}$$, where $$S$$ has the eigenvectors of $$A$$ as its columns and $$\Lambda$$ is the matrix with eigenvalues on its main diagonal. Then $$A^2=S\Lambda^2S^{-1}=I$$, so $$S\Lambda^2=S$$ and $$\Lambda^2=I$$. Since the diagonal entries of $$\Lambda^2$$ are the eigenvalues squared, then $$\lambda_i^2=1$$ by comparing the entries of $$\Lambda^2$$ and $$I$$. So $$\lambda_i=\pm1$$.
For a proof with determinants:
$$0 = \det(A- \lambda I) = 0$$
$$= \det(A- \lambda (A^2))$$
$$= \det(A (I- \lambda A))$$
$$= \det(A) \det(I- \lambda A)$$
Hence
$$0 = \det(A) \det(I- \lambda A)$$
$$\iff 0 = (\pm1) \det(I- \lambda A)$$
$$\iff 0 = \det(I- \lambda A)$$
$$\iff 0 = \det((- \lambda)(\frac{1}{- \lambda}I+ A))$$
$$\iff 0 = (- \lambda)^n \det(\frac{1}{- \lambda}I+ A) \tag{1}$$
$$\iff 0 = \det(\frac{1}{- \lambda}I+ A)$$
$$\iff 0 = \det(-\frac{1}{ \lambda}I+ A)$$
$$\iff 0 = \det(A -\frac{1}{ \lambda}I)$$
Therefore, $$\lambda$$ is an eigenvalue of $$A$$ if and only if it equals its reciprocal assuming hopefully, that I didn't make a logical error and actually conclude only that $$\lambda$$ is an eigenvalue of $$A$$ if and only if its multiplicative inverse is too. If I did make an error, then I hope someone can tell me how to proceed.
QED
$$(1)$$ I don't know what $$\lambda$$ is, but I know what $$\lambda$$ isn't: $$\lambda$$ is nonzero because $$\det(A)$$ is nonzero because $$\det(A) = \pm 1$$
• Indeed, you proved that if $\lambda$ is an eigenvalue then $1-\lambda$ also is an eigenvalue, but not that they're equal. The key fallacy is in that $\det(A-\lambda I)=\det(A-(1/\lambda)I)$ does not imply $A-\lambda I=A- (1/\lambda)I$. – YiFan Oct 24 '18 at 6:04
• @user496634 (there are 2 sentences in the parenthetical remark) well ok so eigenvalues occur in pairs of reciprocals. Any suggestions on how to proceed? – BCLC Oct 24 '18 at 6:21
• Unfortunately I can't think of a way to proceed with the determinant approach. Good try though! – YiFan Oct 24 '18 at 9:32
You can easily prove the following statement:
Let $$f: V\to V$$ be an endomorphism. If $$\lambda$$ is an eigenvalue of $$f$$, then $$\lambda^k$$ is an engeinvalue of $$\underbrace {f\ \circ\ ...\ \circ f}_{k \text{ times}}$$
In this case, let $$A$$ be a matrix of an endomorphism $$f$$ such that $$f\circ f = I$$. This means that $$A$$ is an involutory matrix (Because $$AA=I$$). So if $$\lambda$$ is an eingenvalue of $$f$$, then $$\lambda ^2$$ is an eigenvalue for $$f \circ f = I$$. The only eigenbalue of the identity funcion is $$1$$, so $$\lambda^2 = 1$$ meaning that $$\lambda = \pm1$$.
Suppose $$\lambda$$ is an eigenvalue of $$A$$ then we know that $$1/\lambda$$ is an eigenvalue of $$A^{-1}$$. But here, $$A= A^{-1}$$. So, for every $$\lambda$$ in $$A$$, $$\lambda=1/\lambda$$. And hence $$\lambda^2 = 1$$. And hence $$\lambda= \pm 1$$.
• This reasoning is incorrect. You have argued that if $\lambda$ is an eigenvalue of $A$ then so is $\lambda^{-1}$. But this does not mean that $\lambda = \lambda^{-1}$. – user279515 Dec 13 '18 at 19:15 | 2021-08-03T09:32:59 | {
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http://math.stackexchange.com/questions/15415/prove-the-existence-of-a-largest-integer-less-than-or-equal-to-a-rational-number | # Prove the existence of a largest integer less than or equal to a rational number
Prove that there is a largest integer $n$ such that $n \le x$ for any fixed rational $x$.
What about any fixed real number $x$?
-
Where did you get this problem? This looks like it could be a homework problem. Did you try it and are stuck? If you show your working, people can help you better. – Aryabhata Dec 24 '10 at 7:52
Oh, it's not a homework problem. I teach myself, for me there is no such thing as "homework". :) It's just a sentence in an analysis textbook with no proof given. – qed Dec 24 '10 at 12:54
As Christian Blattner says, in questions like this it is difficult to know what can be used and what can not in the proof. It is "obviously true" (not denigrating the question), because so much of what we do depends on it. But if you are constructing the numbers, there are different routes, which lead to different answers. – Ross Millikan Dec 24 '10 at 15:24
The answer to your question depends on the set of facts you take for granted. In terms of elementary number theory I would argue as follows: Let $x=p/q$ with $q>0$. Then the $q$ consecutive numbers $p-q+1$, $p-q+2$, $\ldots$, $p$ form a complete residue system mod$\thinspace q$, so there is exactly one among them, say $r$, that is divisible by $q$. It follows that $n:=r/q$ is the largest integer $≤x$.
-
I like this answer because it doesn't make use of the extra properties which $\mathbb{R}$ has. It seems preferable to me to avoid special properties of $\mathbb{R}$ to prove this fact which certainly remains true if we totally ignore the existence of $\mathbb{R}$. – Zach Conn Dec 24 '10 at 17:44
Thank you, this proves it in a beautiful way. – qed Dec 26 '10 at 6:20
(Sorry to post this as an answer, but I can't comment yet). Do you mean that given a fixed rational number $x$, the set $\lbrace n: n\leq x, n \in \mathbb Z\rbrace$ has a maximum? (Hint: Use the well-ordering principle)
-
Yeah. That is what I meant. – qed Dec 24 '10 at 12:49
I got your point. Consider $A=\{m:m>x,\; m\in Z\}$ , then A is bounded below and has a least element. Let $m_{0}$ be the least element of A, then $m_{0}-1$ is not in A, hence $m_{0}-1 \le x$, and it is the greatest integer satisfying that condition. – qed Dec 24 '10 at 13:29
But I'll wait for a different proof. Especially a proof using Archimedean property, since the textbook says this result is based on it. – qed Dec 24 '10 at 13:31
First, as a clarification pointed out by Dactyl, you probably want a proof of:
For any rational (or real) $x$, there is a largest integer $n$ such that $n \leq x$.
The reversal of words makes the claim false because it says that there is a fixed largest integer $n$ less than or equal to arbitrarily large $x$.
Also for Dactyl's hint, you have to be careful when $x$ is negative because the well-ordering principle is for Natural numbers. Specifically, not every nonempty subset of the integers has a smallest element (e.g., the integers themselves have no least element). However, every nonempty subset of the integers that has a lower bound has a least element, and you can prove this by induction on the absolute value of an integral lower bound.
Of course, $n$ is typically used to denote a Natural number so maybe you're restricting yourself to nonnegative $x$ anyway.
Edit (in response to request for different proof):
Consider $\{n \in \mathbb{N}| |x| < n\}$. The Archimedean property tells us that for any Real number, we can find a Natural number that's larger so this set is non-empty. Then by the well-ordering of the Natural numbers, there exists a least element in this set, call it $n_0$. If $x \geq 0$, verify that $n_0 - 1$ is the desired integer, and if $x < 0$, verify that that $1 - n_0$ is the one you want if $x$ is an integer and $-n_0$ otherwise.
-
Since the problem is tagged "real analysis", I'd just use the supremum of the set $\{ n \in \mathbf{Z} : n \le x \}$. Of course, you need to prove that this set is not empty and that the supremum is actually a maximum.
- | 2015-05-25T10:25:33 | {
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https://math.stackexchange.com/questions/998459/how-to-proceed-have-i-correctly-transformed-the-question-to-an-equation | # how to proceed ? Have I correctly transformed the question to an equation?
Amongst Teaching staff of ABC university the ratio of men and women is $5:2$. Amongst the women $\large\frac{3}{7}$ are not married . If the number of married women teacher is $56$ then the total number of teacher is ?
What I have tried :
$$M:W=5:2$$
$$W_{\text{unmarried}}= \frac{3}{7}W$$
$$W_{\text{married}}= 56$$
now what to do?
$$W=W_{\text{unmarried}}+W_{\text{married}}$$
$$W = \frac{3}{7}W + 56$$ $$W= 98$$
• You can only accept one answer so choose wisely. Later you'll earn privilege to vote on questions and answers. – user171358 Oct 30 '14 at 16:30
Yes You're right
Let $x$ and $y$ be number of Male and Female teachers $$2x=5y\tag{1}$$
$\large\frac{3}{7}$ Females are unmarried that means , $\large\frac{4}{7}$ Females are married and number of married Females is also $56$
$$56=\frac{4y}{7}\tag{2}$$ After Solving $(1)$ and $(2)$ simultaneously we get $$x=245 \quad, \quad y=98$$
and $$\text{Total Number of Teachers} =343$$
$\dfrac{M}{W}=\dfrac{5}{2}=t \implies W=2t$
$\dfrac{4}{7}W=56 \implies W=98$
$\therefore t=49 \implies M=490$
Hint: If $T$ is the total number of women, you know that $\frac{3}{7}T$ are not married and therefore $\frac{4}{7}T$ are married. So, $\frac{4}{7} T = 56$, so $T =\,...$ Can you finish?
The trick is to maintain units, and not just work with numbers.
5 men : 2 wom : 7 prof
3 wu : 4 wm : 7 wom
56 wm * 7 wom / 4 wm * 7 prof / 2 wom = 56 /8 *7 *7 * wm/wm * wom/wom * prof
= 343 prof | 2020-02-27T18:23:18 | {
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https://math.stackexchange.com/questions/2938417/determining-which-vector-is-linearly-dependent-in-a-set | # Determining which vector is linearly dependent in a set
I'm confused with a particular problem when given something like this to solve:
I am told to consider sets of vectors in $$\mathbb R^3$$ and $$\mathbb R^4$$ respectively. I am then told to construct a subset of this set that will span the subspace $$U = Span \ (A)$$, so basically I have to snub out any linearly dependent vectors in these two sets:
I'll merely look at one of the examples. The solution to this involved bringing the vectors together to create a matrix with the columns as the vectors, and then row-reduce to echelon form. Once doing so, analyzing the system of equations included something I find concerning. In my own solution, I wrote the column vectors as row vectors, and row-reduced. I'll compare our row reduced matrices.
$$\begin{bmatrix} 2 & -1 & 2 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$
Which lead me to believe that the fourth vector was linearly dependent, as the row referring to it is a zero row. Thus, I stated that the basis that will span $$U$$ is $$B = \{(2,-1,2), (0,3,4), (0,0,-1)\}$$. However, my lecturer's answer was much different. Instead, they did what I stated in that they just kept the column vectors as columns for their matrix and then converted it to echelon form:
$$\begin{bmatrix} 1 & -1 & -1 & 1\\ 0 & 1 & 2/3 & 2/3 \\ 0 & 0 & 1 & 7 \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta\\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$
Where $$\alpha, \beta, \gamma, \delta \in F$$. This led to the conclusion:
$$\alpha - \beta - \gamma + \delta = 0$$ $$\beta + (2/3) \ \gamma + (2/3) \ \delta = 0$$ $$\gamma - 7 \ \delta= 0$$
The lecturer used $$\delta = 1$$ and was able to show that, by determining the other constants with this statement, a $$\{(4,-1,3)\}$$ was able to be expressed as a linear combination of the vectors preceding it.
My questions are two-fold:
a) Why was my method, and echelon form matrix, not valid for this purpose?
b) Is any initialization of the free parameter $$\delta$$ viable? If so, I did not succeed in showing this for $$\delta = 3$$, as I could not represent the fourth vector as a linear combination of vectors like the lecturer did. If only $$\delta = 1$$ is the only viable option to get the correct answer, how is one supposed to know what to set $$\delta$$ to to arrive at the correct answer in this case? Just for clarity, I mean "correct answer" in the context of it being the correct initialization of $$\delta$$ such that the fourth vector can be shown to be linearly dependent.
• Comment only since I haven't read your whole answer. If a list of vectors is dependent there's no way to single out one of them as "the" dependent vector. Whenever you have a linear combination that's $0$ any vector with a nonzero coefficient is linearly dependent on the others. – Ethan Bolker Oct 1 '18 at 19:58
• Short answer to a): Your method found a basis for the span, but you can’t identify which vectors of the original set are redundant. – amd Oct 1 '18 at 20:01
• @amd why can’t I? – sangstar Oct 1 '18 at 20:07
• The zero rows are always at the bottom of the matrix and the process might reorder rows. There’s no direct correspondence between the rows of the echelon form and the rows of the original matrix as there is for the columns. Try it with $\small\begin{bmatrix}0&1&1\\0&-1&-1\\1&0&0\end{bmatrix}$. Your conclusion would be that $(1,0,0)$ is a linear combination of the other two rows, which is patently impossible. – amd Oct 1 '18 at 20:10
If you perform Gaussian elimination on the matrix with the given vectors as columns, then the columns with no leading $$1$$ can be removed.
\begin{align} \begin{bmatrix} 2 & 1 & 0 & 4 \\ -1 & 1 & 1 & -1 \\ 2 & 3 & 1 & 3 \end{bmatrix} &\to \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 3/2 & 1 & 1 \\ 0 & 2 & 1 & -1 \end{bmatrix} && \begin{aligned} R_1&\gets\tfrac{1}{2}R_1 \\ R_2&\gets R_2+R_1 \\ R_3&\gets R_3-2R_1\end{aligned} \\&\to \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 1 & 2/3 & 2/3 \\ 0 & 0 & -1/3 & -7/3 \end{bmatrix} && \begin{aligned} R_2&\gets\tfrac{2}{3}R_2 \\ R_3&\gets R_3-2R_2\end{aligned} \\&\to \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 1 & 2/3 & 2/3 \\ 0 & 0 & 1 & 7 \end{bmatrix} && R_3\gets -3R_3 \end{align} At this point we see that the fourth vector can be removed. If we also compute the RREF \begin{align} \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 1 & 2/3 & 2/3 \\ 0 & 0 & 1 & 7 \end{bmatrix} &\to \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 5 \end{bmatrix} && R_2\gets -\frac{2}{3}R_3 \\&\to \begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 7 \end{bmatrix} && R_1\gets -\frac{1}{2}R_2 \end{align} We see that $$\begin{bmatrix} 4 \\ -1 \\ 3 \end{bmatrix}= 4\, \begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}\, -4\, \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}\, +7\, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$
You may have taken a transpose where not appropriate. In both cases, four columns, the vector created by summing $$w c_1 + x c_2 + y c_3 + z c_4$$ is exactly the result of multiplying the matrix on the left times the column vector $$V = \left( \begin{array}{c} w \\ x \\ y \\ z \end{array} \right)$$ Finding the reduced row echelon form of the rectangular matrix leads to a way to find a null vector. For the first matrix, 3 by 4, it is guaranteed there is at least one nonzero vector $$V$$ that is a null vector. Write out $$V,$$ any nonzero entry is a column in the rectangular matrix that can be written in terms of the others.
In the 4 by 4 case, it is possible that the only $$V$$ that works is the zero vector. Not sure. If there is a nonzero $$V,$$ same as before
$$V = \left( \begin{array}{c} 4 \\ -4 \\ 7 \\ -1 \end{array} \right)$$ or any nonzero multiple of that, so that any one of the four columns can be written in terms of the other three. Put the other way, with the four columns in the 3 by 4 matrix, we have $$4 c_1 - 4 c_2 + 7 c_3 - c_4 = 0$$ | 2020-11-30T07:42:05 | {
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https://dsp.stackexchange.com/questions/27014/discrete-time-biquad-filter-relation-between-peak-frequency-and-pole-frequency?noredirect=1 | Discrete-time Biquad filter: Relation between peak frequency and pole frequency
This question arose from a discussion in the comments to this question and its answer. The author of that question discussed a discrete-time second order filter described by the following difference equation:
$$y[n]=b_0x[n]+b_1x[n-1]+b_2x[n-2]-a_1y[n-1]-a_2y[n-2]\tag{1}$$
with $b_0=b_2=(1-r^2)/2$, $b_1=0$, $a_1=-2r\cos(2\pi f_0 T)$, and $a_2=r^2$.
The transfer function corresponding to the difference equation $(1)$ is
$$H(z)=\frac{1-r^2}{2}\frac{1-z^{-2}}{1-2r\cos(\omega_p)z^{-1}+r^2z^{-2}}\tag{2}$$
where I use the abbreviation $\omega_p=2\pi f_0 T$.
In the original question quoted above, $f_0$ was called "resonant frequency", and the discussion inspired by that question was about whether or not $f_0$ equals the peak frequency of the filter's frequency response.
So the questions I pose are:
1. What is the meaning of $r$ and $f_0$ as used in the coefficients of Eqs. $(1)$ and $(2)$?
2. Does $f_0$ equal the filter's peak frequency (i.e., the frequency where the magnitude of the filter's frequency response attains its maximum)?
• i've up-arrowed both question and answer. i guess maybe i should ask you exactly what you mean by "pole frequency"? if you have complex conjugate poles, is it the angle of the pole location against the $\Re\{z\}$-axis? keep in mind that, if the $Q$ is low enough, there might be no complex conjugate poles. Nov 13, 2015 at 8:29
• okay, you answered my question. still reading. Nov 13, 2015 at 8:31
According to the Audio EQ Cookbook the transfer function of a second-order band-pass filter with 0dB peak gain is given by
$$H(z)=\frac{\alpha-\alpha z^{-2}}{(1+\alpha)-2\cos(\omega_0)z^{-1}+(1-\alpha)z^{-2}}\\=\frac{\alpha}{1+\alpha}\cdot\frac{1-z^{-2}}{1-\frac{2}{1+\alpha}\cos(\omega_0)z^{-1}+\frac{1-\alpha}{1+\alpha}z^{-2}}\tag{1}$$
where $\omega_0$ is the angular frequency where the peak occurs ("peak frequency"), and $\alpha=\sin(\omega_0)/(2Q)$ is a parameter determined by $\omega_0$ and the desired quality factor $Q$.
If we express the same transfer function $H(z)$ in terms of its poles we get
$$H(z)=k\frac{1-z^{-2}}{\left(1-re^{j\omega_p}z^{-1}\right)\left(1-re^{-j\omega_p}z^{-1}\right)}= k\frac{1-z^{-2}}{1-2r\cos(\omega_p)z^{-1}+r^2z^{-2}}\tag{2}$$
where $k$ is some gain constant to be determined, $0<r<1$ is the pole radius, and $\omega_p=2\pi f_p$ is the pole angle, corresponding to the pole frequency $f_p$. Comparing the denominator of $(2)$ to the filter coefficients given in this question (and shown under Eq. $(1)$ in the question above), it is clear that the frequency $f_0$ in that question equals the pole frequency $f_p$ in our formulation. And $r$ in the question above is obviously the pole radius.
Comparing $(1)$ and $(2)$ we can express $\alpha$ and $\omega_0$ of Eq. $(1)$ in terms of the pole radius $r$ and the pole angle $\omega_p$ used in Eq. $(2)$:
\begin{align}&\frac{1-\alpha}{1+\alpha}=r^2 \Longrightarrow \alpha=\frac{1-r^2}{1+r^2},\qquad k=\frac{\alpha}{1+\alpha}=\frac{1-r^2}{2}\\ &\frac{2}{1+\alpha}\cos(\omega_0)=(1+r^2)\cos(\omega_0)=2r\cos(\omega_p)\end{align}\tag{3}
The second line of Eq. $(3)$ shows that the pole angle $\omega_p$ and the peak frequency $\omega_0$ are not equal, and that the following relationship holds:
$$\frac{1+r^2}{2r}\cos(\omega_0)=\cos(\omega_p)\tag{4}$$
The following Matlab/Octave example illustrates the above. We choose $\omega_0=0.2\pi$ and $Q=1$ and design the corresponding biquad band-pass filter according to Eq. $(1)$. The figure below shows the magnitude of the resulting frequency response, and the locations of the peak frequency $\omega_0$ (red line) and of the pole angle $\omega_p$ (green line), both normalized by $\pi$.
w0 = .2*pi;
Q = 1;
al = sin(w0)/(2*Q);
b = [al,0,-al]/(1+al); % numerator coeffs
a = [1+al,-2*cos(w0),1-al]/(1+al); % denominator coeffs
p = roots(a); % poles
r = abs(p(1)); % pole radius
r2 = sqrt((1-al)/(1+al)); % pole radius according to formula
abs(r-r2) % ans = 0
wp = abs(angle(p(1))); % pole angle: wp = 0.17877*pi
wp2 = acos((1+r^2)/(2*r)*cos(w0)); % pole angle according to formula
abs(wp-wp2) % ans = 0
[H,w]=freqz(b,a,1024);
plot(w/pi,abs(H))
hold on, plot([w0,w0]/pi,[0,1],'r',[wp,wp]/pi,[0,1],'g'), hold off
EDIT:
In Eq. $(2)$ I assumed that there are two complex conjugate poles, because in the original question, which triggered this question and answer, the filter coefficients were chosen in such a way that only complex conjugate poles were possible (cf. the coefficients under Eq. $(1)$ of the question above).
However, the general transfer function in Eq. $(1)$ can also have a double real-valued pole or two different real-valued poles. It can be shown that a double real-valued pole occurs when the quality factor $Q$ is chosen as $Q=\frac12$. This case is still covered by Eq. $(2)$ with $\omega_p=0$ (i.e., a positive real-valued pole) or $\omega_p=\pi$ (a negative real-valued pole). For $Q<\frac12$ we get two different real-valued poles, and Eq. $(2)$ is not valid anymore. Instead, the transfer function becomes
$$H(z)=k\frac{1-z^{-2}}{(1-\beta_1z^{-1})(1-\beta_2z^{-1})}= k\frac{1-z^{-2}}{1-(\beta_1+\beta_2)z^{-1}+\beta_1\beta_2z^{-2}}\tag{a}$$
where $\beta_1$ and $\beta_2$ are the two real-valued poles. Comparing this transfer function to the transfer function given in Eq. $(1)$, we get the following relationships between $\alpha$, $\omega_0$, $k$ and $\beta_1$ and $\beta_2$:
\begin{align}\alpha&=\frac{1-\beta_1\beta_2}{1+\beta_1\beta_2}\\ \cos(\omega_0)&=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2}\\ k&=\frac{1-\beta_1\beta_2}{2}\end{align}\tag{b}
Of course, the pole angles are now either zero or $\pi$, because both poles are real-valued. Note that the peak frequency $\omega_0$ is not necessarily $0$ or $\pi$. The peak frequency $\omega_0$ can in fact take any value, and the two pole angles will always be either $0$ or $\pi$ as long as the quality factor $Q$ satisfies $Q<\frac12$.
• this appears to be completely correct. i withdraw my previous dispute with you, Matt. you are correct and i was mistaken. what do you do about $f_p$ when the $Q < \frac12$? and, BTW, that $Q$ (from the cookbook) still is applicable only for the Bilinear Transform mapping of $s$ to $z$. the mapping of $Q$ to $r$ will be different for Impulse Invariant. Nov 13, 2015 at 8:35
• @robertbristow-johnson: I'm glad we finally agree. It would be great if you could also mention that fact in the comments to the original question. Not for me, but for the OP and future users. Then we can also clean up the mess over there ... Nov 13, 2015 at 8:39
• @robertbristow-johnson: Good question about low Q, I'll add that case to my answer as soon as I have the time to do so. Nov 13, 2015 at 9:55
• @robertbristow-johnson: I added some information about $Q<\frac12$. Since we get real-valued poles, the pole angles are of course either $0$ or $\pi$, and $\omega_0$ can still take any value. Nov 13, 2015 at 12:33
• I can confirm your formula for the relationship. I got the same by solving for the zero of the derivative of the magnitude frequency response. Sep 29, 2016 at 18:39 | 2022-05-18T06:45:36 | {
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http://rnr.rs/eau-de-acpo/e3d36a-are-the-two-diagonals-of-a-rectangle-equal-why | # are the two diagonals of a rectangle equal why
The diagonal of the rectangle is the hypotenuseof these triangles.We can use Pythagoras' Theoremto find the length of the diagonal if we know the width and height of the rectangle. In a rectangle both diagonals are equal in measure; so AC = BD; as well the four intercepts, AO = OC = BO = OD. toppr. A rectangle has two diagonals as it has four sides. NCERT RD Sharma Cengage KC Sinha. to be divided along the diagonals into two triangles that have a congruent hypotenuse. Opposite sides of a rectangle are equal; so, AB = DC and AD = BC. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. In this video tutorial we discuss: (1) How to prove that the two diagonals of a rectangle are equal in length & bisect each other? Videos: 1 Duration: 00:03:47 Language: English. In ΔADB and ΔBCD: AD = BC (Opposite sides of a rectangle are equal.) Similarly we can prove that PC = PA . His goal is to help you develop a better way to approach and solve geometry problems. "If a rectangle is square, then its main diagonals are equal" is (True) because this is true of all rectangles. Biology. Geometry doesn't have to be so hard! Answer. b:Rectangle. Geometry answers, proofs and formulas for solving geometry problems, and useful tips for how to approach these problems. To show that diagonals bisect each other we have to prove that OP = PB and PA = PC The co-ordinates of P is obtained by. Now, since a rectangle is a parallelogram, its opposite sides must be congruent and it must satisfy all other properties of parallelograms. We can also prove this from scratch, repeating the proofs we did for parallelograms, but there’s no need. IF the quadrilateral is a rectangle, then the two diagonals are equal in length. Real World Math Horror Stories from Real encounters. A rectangle is a parallelogram with 4 right angles. A rectangle is a parallelogram, and we can save time and effort by relying on general parallelogram properties that we have already proven. Rectangles are a special type of parallelogram. EASY. 00:03:47 undefined. a:Square. If side MN = 12 and side ML = 5, what is the length of the other two sides? Don’t worry if you didn’t see this immediately, or if you chose other triangles – it is easy to prove this property of rectangles using other combinations of triangles, such as ΔABC and ΔDCB or ΔDAB and ΔCBA; any two pairs will do, as long as the diagonals are the corresponding sides in each triangle. Write a two column-proof given: fghi is a rectangle. 직사각형의 두 대각선은 길이가 같고 서로 다른 것을 이등분한다. Thank you! Books. Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. (1)The diagonals of a parallelogram are equal. d:Rhombus. Physics. ii) From O, draw a perpendicular OP to meet AB at P. In triangle … Rectangles are a special type of parallelogram. Like many other geometry problems where we need to prove that two line segments are of equal lengths, we will turn to triangle congruency as our go-to tool. Explain why a rectangle is a convex quadrilateral. prove: the diagonals of a rectangle have equal lengths. $$\angle SZT$$ and $$\angle SZA$$ are supplementary angles, Answered By . Maths. In this lesson, we will show you two different ways you can do the same proof using the same rectangle. Therefore, x = 30 °. (Two diagonals of a rectangle are equal in length and bisect each other.) OP = OB . e:Trapezium etc..... New questions in Math. If a diagonal bisects a rectangle, two congruent right triangles are obtained. [00:03:47] S. Login/Register to track your progress. are congruent NO is 5 and lO is 12. It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. Chemistry . Therefore $$\angle SZA = 120°$$. If you remember your Pythagorean theorem, you should be able to see why. By the Pythagorean theorem, we know that. Therefore, SZ = AZ, making SZA isosceles and $$\angle$$ZSA$$\angle$$ZAS, being base angles of an isosceles triangle. ∴ The diagonals of a rectangle bisects each other and equal . Diagonals of rectangle bisect each other. It's easy to prove that the diagonals of a rectangle with the Pythagorean theorem. Click hereto get an answer to your question ️ Which statement(s) is/are correct? Books. Click hereto get an answer to your question ️ Prove that the diagonals of a rectangle divide it in two congruent triangles. Here, we don’t even need to construct any special triangles, because the diagonals themselves have defined the triangles. Prove that the diagonals of a rectangle are congruent. Again, we can use the Pythagorean theorem to find the hypotenuse, NL. Video Explanation. parallelograms; class-8; Share It On Facebook Twitter Email. The diagonal of a square is equal to the length of one of the sides of the square times $\sqrt {2}$, or $s\sqrt{2}$. Click hereto get an answer to your question ️ Prove logically that the diagonals of a rectangle are equal The two diagonals are not necessarily equal in a (a)rectangle (b)square (c) rhombus (d) isosceles trapezium ← Prev Question Next Question → 0 votes . 343 views. Get Instant Solutions, 24x7. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. CD = CD (Common) Click to learn more... By accessing or using this website, you agree to abide by the Terms of Service and Privacy Policy. So AO = OC and BO = OD, where O is the intersecting point of the two diagonals AC & BD. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. To find MZ, you must remember that the diagonals of a parallelogram bisect each other. Click hereto get an answer to your question ️ Show that the diagonals of a square are equal and bisect each other at right angles. Copyright © 2020. Haroldescorcia2527 is waiting for your help. (6) AC = DB // Corresponding sides in congruent triangles (CPCTC), Filed Under: Rectangles Last updated on January 4, 2020. A rectangular field is 15 m long and 10 m wide, Another rectangular field having the sameperimeter has its sides in the ratio 4 : 1. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Note also that the diagonals are equal and cut each other in half at right angles. Interactive simulation the most controversial math riddle ever! What is the value of x in rectangle STAR below? LMNO and divide along the diagonal MO into two right triangles. Length of a diagonal of a square = √2 x Length of a Diagonal of a Rectangle Similar to a square, the length of both the diagonals in a rectangle are the same. … 0%. What is the formula of collinear? In order to prove that the diagonals of a rectangle are congruent, consider the rectangle shown below. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. … Triangle MLO is a right triangle, and MO is its hypotenuse. (2)The diagonals of a square are perpendicular to each other. Related Questions. To Prove that the two Diagonals of a Rectangle Are of Equal Length. If we divided the rectangle along diagonal NL, we would create triangle LNO. 1 A square is a rectangle.alwayssometimesnever2 The diagonals of a rhombus are perpendicular. Free Algebra Solver ... type anything in there! In rectangle STAR below, SA =5, what is the length of RT? We also know that AB= CD as they are opposite sides in a parallelogram. A Quadrilateral has two diagonals. c:Parallelogram. Since the diagonals of a rectangle are congruent, RT has the same length as SA. alwayssometimesnever4 The diagonals of a trapezoid are equal. (4)Every quadrilateral is either a trapezium or a parallelogram or a kite. Diagonals of Rectangles are of Equal Length, Opposite sides in a rectangle (parallelogram), The opposite sides of rectangles are equal, The diagonals of rectangles bisect each other, Any two adjacent angles are supplementary (obviously, since they all measure 90°), The opposite angles are equal (again, obviously, since all interior angles measure 90°). You can contact him at [email protected]. Rectangles are a special type of parallelogram, in which all the interior angles measure 90°. The two diagonals are equal in length. NCERT RD Sharma Cengage KC Sinha. In a quadrilateral the diagonals are equal, the quadrilateral will be either 1.A square or 2.A rectangle In a parallelogram the diagonals bisect each other and they never equal 00:02 CBSE class 9 maths NCERT Solutions chapter 10 Quadrilaterals q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle. Next, remember that the diagonals of any parallelogram bisect each other and the diagonals of a rectangle are congruent. Click here to see the proof. Please email us at [email protected]. The length of a diagonal (d) of a rectangle whose length is l and whose breadth is b is calculated by the Pythagoras theorem. Which element of a text best helps the reader determine the central idea? New questions in Mathematics. I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree. A rectangle and a crossed rectangle are quadrilaterals with the following properties in common: Opposite sides are equal in length. The formula to find the length of the diagonal of a rectangle is: (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties), If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13. In the figure above, click 'reset'. 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https://mathematica.stackexchange.com/questions/83118/hightlight-all-the-self-intersections-of-a-lissajous-figure | # Hightlight all the self-intersections of a Lissajous figure
This graph–also known as a Lissajous figure–contains so many self-intersections.How can I highlight them?
ParametricPlot[{Sin[100 t], Sin[99 t]}, {t, 0, 2 π},
PlotRange -> All]
• Too tired to tackle this now, but a hint: for general $n$ (this is case $n=100$) the number of nodes is $2n^2-4n+1$. They are distributed on a $2n-1$ by $2n-3$ grid, occupying a diagonal pattern (like the white squares on a chessboard). The key is going to be to find the positions of the lines of the grid. – 2012rcampion May 11 '15 at 3:27
• Looks like they have x-coordinates Cos[Pi Range[2 (n - 1) - 1]/(2 (n - 1))] and y-coordinates Cos[Pi Range[2 (n - 1) + 1]/(2 (n - 1) + 2)]. – 2012rcampion May 11 '15 at 3:34
• A quick idea: extract the Line[] objects from a plot of the curve (Cases[] is useful here), split any polylines present into simple lines of the form Line[{pt1, pt2}], and then use a line intersection algorithm on the lines produced. Polishing with FindRoot[] is optional. – J. M. will be back soon May 11 '15 at 5:02
Manipulate[
ParametricPlot[({Sin[n t1], Sin[(n - 1) t1]}), {t1, 0, 2 Pi},
Epilog -> {Red, PointSize[Large],
Table[If[OddQ[i + j],
Point[{Cos[Pi i/(2 (n - 1))], Cos[Pi j/(2 (n))]}]], {i,
2 n - 3}, {j, 2 n - 1}]}], {{n, 5}, 2, 20, 1}]
## General Case I
We can generalize to the Lissajous curve specified by the two non-negative integers $a$ and $b$:
$$x = \sin at \\ y = \sin bt \\ t \in [0,2\pi)$$
Without loss of generality, I will assume $b<a$ and $b\nmid a$.
We can start by making a table of small cases:
Column[Row /@
Table[ParametricPlot[({Sin[a t], Sin[b t]}), {t, 0, 2 Pi},
Epilog -> {}, PlotLabel -> {a, b}, Axes -> False,
ImageSize -> Tiny], {a, 10}, {b,
Select[Range[a - 1], CoprimeQ[#, a] &]}], Alignment -> Center]
When both $a$ and $b$ are odd, we get a degenerate curve that traces itself twice. I'll handle those cases later.
It looks like each self-intersection occurs on a horizontal and vertical line shared with several other solutions. We can make a table mapping $a$ and $b$ to the number of horizontal and vertical grid lines (ignoring $b=1$ as a special case for now):
$$3,2\to 3,5\\ 4,3\to 5,7\\ 5,2\to 3,9\\ 5,4\to 7,9\\ 6,5\to 9,11$$
It's fairly evident that the number of grid lines is merely:
$$2b-1,\,2a-1$$
The spacing of the grid lines looks mathematically like it might be more difficult. However, the spacing looks familiar to me: like the spacing of points in an airfoil .dat file:
Graphics[Point@
Rest[Import[
I remember from AE311 (incompressible flow) that this spacing follows the transformation:
$$x\mapsto \frac{c}{2}\left(1-\cos(\theta)\right)$$
with the points evenly spaced in $\theta$. Could it really be that simple?
Manipulate[
ParametricPlot[({Sin[a t], Sin[b t]}), {t, 0, 2 Pi},
GridLines -> {Cos[Pi Range[2 b - 1]/(2 b)],
Cos[Pi Range[2 a - 1]/(2 a)]}, PlotLabel -> {a, b},
Axes -> False], {{a, 5}, 2, 20, 1}, {{b, 4},
Select[Range[a - 1], CoprimeQ[#, a] &]}]
Heck yeah it is: lucky guess!
Note that only every other grid node contains an intersection; they form a sort of checkerboard pattern. This accounts for the seeming fewer number of grid lines when $b=1$: only every other line is occupied, so there are twice (plus one) as many grid lines as intersections.
We can also take a look at the odd-odd special cases:
We can see that they follow a double-size checkerboard pattern, with adjacent intersections two diagonals apart.
With all this in mind, we can now extend the code from the original example:
Manipulate[
ParametricPlot[
{Sin[a t], Sin[b t]}, {t, 0, 2 Pi},
GridLines -> {Cos[Pi Range[2 b - 1]/(2 b)], Cos[Pi Range[2 a - 1]/(2 a)]},
PlotLabel -> {a, b}, Axes -> False,
Epilog -> {Red, PointSize[Large],
Table[If[
If[OddQ[a] && OddQ[b],
EvenQ[i] && Divisible[i + j + a + b + 2, 4],
OddQ[i + j]],
Point[Cos[Pi/2 {i/b, j/a}]]
],
{i, 2 b - 1}, {j, 2 a - 1}]}
],
{{a, 5}, 2, 20, 1}, {{b, 4}, Select[Range[a - 1], CoprimeQ[#, a] &]}
]
## General Case II
We can follow a similar procedure for phased Lissajous curves. Without loss of generality, we can apply a phase $\phi$ to the $x$-coordinate:
$$x = \sin(at +\phi) \\ y = \sin(bt) \\ t \in [0,2\pi)$$
If we apply phases $\phi_a$ and $\phi_b$ to the $x$ and $y$-coordinates, respectively, this is equivalent to a curve with $\phi=\phi_a-\frac a b \phi_b$ and $t'=t+\frac{\phi_b}b$.
First we'll take a look at what's going on:
Manipulate[
ParametricPlot[{Sin[a t + ϕ], Sin[b t]}, {t, 0, 2 Pi},
PlotLabel -> {a, b}, Axes -> False], {{a, 5}, 2, 20, 1}, {{b, 4},
Select[Range[a - 1], CoprimeQ[#, a] &]}, {ϕ, 0, 2 Pi}]
I like to visualize this as the projection of a pattern on the surface of a vertical cylinder, rotating about its axis:
A little bit of work transforms the original solution to follow the intersections of the cylinder pattern:
Note that we're missing half of the intersections now! The missing intersections are where lines from the 'front' half of the cylinder overlap the 'back' half. We can get those via a similar process, treating the pattern as a projection from the surface of a horizontal cylinder. In the image above, we essentially want to reflect the 'missing' intersections across the diagonal:
This gives us our final result:
Manipulate[
With[{gcd = GCD[a, b]},
With[{a = a/gcd, b = b/gcd},
ParametricPlot[
{Sin[a t + ϕ], Sin[b t]}, {t, 0, 2 Pi},
PlotLabel -> {a, b}, Axes -> False,
Epilog -> {
PointSize[Large],
Red,
Table[
If[EvenQ[i + j],
Point[{Sin[2 Pi (i + a)/(2 b) + ϕ], Cos[Pi j/a]}]
],
{i, 2 b}, {j, a - 1}
],
Orange,
Table[
If[EvenQ[i + j],
Point[{Cos[Pi i/b], Sin[2 Pi (j + b)/(2 a) - b/a ϕ]}]
],
{i, b - 1}, {j, 2 a}
]
}
]
]
],
{{a, 6}, 1, 20, 1}, {{b, 13}, 1, 20, 1}, {{ϕ, Pi/10}, 0, 2 Pi}
]
(Note that for some values of ϕ, you will see repeated intersections or intersections at the edge of the curve. This happens when the curve becomes degenerate and overlaps itself.)
• Neat! and really fast as well! (+1) – MarcoB May 11 '15 at 3:52
• @Marco Much faster if you use 2.. – 2012rcampion May 11 '15 at 3:53
• @2012rcampion You code comes first , but it's a pity that it becomes slow when n is large. – WateSoyan May 12 '15 at 11:02
• @Wate Actually the answer to that one is quite simple: the number of intersections for $a$ and $b$ both odd is $(a-1)(b-1)/2$ and the number of intersections in the other cases is $2ab-(a+b)$. The number of regions is simply the number of intersections plus one (every time you close a curve by intersection, you split a previous region into two, and you start with one region). – 2012rcampion May 12 '15 at 15:25
• N.B. These intersection points are related to what are now termed Padua points. – J. M. will be back soon Jul 19 '16 at 14:08
One way (whew, there are a lot of intersections! -- here's a shorter version):
sol = NSolve[{Sin[10 t], Sin[9 t]} == ({Sin[10 t], Sin[9 t]} /. t -> s) &&
0 <= t < s < 2 Pi, {t, s}];
ParametricPlot[{Sin[10 t], Sin[9 t]}, {t, 0, 2 π},
Epilog -> {Red, PointSize[Large], Point[{Sin[10 t], Sin[9 t]} /. sol]}]
({Sin[100 t], Sin[99 t]} will take a lot longer.)
General solution via Mathematica that is not too slow
Solve returns solutions in the form ConditionalExpression, and the condition can be used to generate the values {m, n} for each point of intersection (via Solve inside Block).
gensols = Cases[
Solve[(-a t + a s == 2 Pi m || a t + a s == Pi + 2 Pi m) &&
(b t - b s == 2 Pi n || b t + b s == Pi + 2 Pi n) &&
a > b > 0 && {a, b, m, n} ∈ Integers &&
0 <= t < s < 2 Pi, {s, t}],
HoldPattern[t -> t0_] :> {t -> t0}, 2];
Block[{a = 100, b = 99},
pts = Flatten[
Hold[{Sin[a t], Sin[b t]} /. Solve[Last[t], {m, n}]] /. gensols // ReleaseHold,
1]
] // Length // AbsoluteTiming
(* {1.55879, 19601} *)
ParametricPlot[{Sin[100 t], Sin[99 t]}, {t, 0, 2 Pi},
PlotStyle -> {Black, Thickness[0.0015]}, PlotPoints -> 3000,
Epilog -> {GraphicsComplex[N@pts, {Red, PointSize[0.003], Point[Range@Length@pts]}]}]
• 19,601 intersections if my math is correct. – 2012rcampion May 11 '15 at 3:56
• It's a pity that NSolve takes a very long time. – WateSoyan May 11 '15 at 10:45
• @WateSoyan Well, 20K points are a lot to track down and NSolve might be converting the trig. eqns. to polynomial ones -- ouch. The difference between solving (this answer) and having been solved (rcampion2012's) is to be expected. You must be making a pretty big poster to show all those points! :) – Michael E2 May 11 '15 at 12:31
• @Michael E2 I have posted my answer below,purely based on Solve. – WateSoyan May 11 '15 at 13:43
• @WateSoyan To me it's a pity that I preferred to figure out how to get Mathematica to solve the problem, than to solve it mathematically myself, like rcampion2012. (But then again, this site is about how to use M.) – Michael E2 May 11 '15 at 15:20
GraphicsMeshMeshInit[];
eps = 1/1000000;
pp = ParametricPlot[{Sin[10 t], Sin[9 t]}, {t, eps, 2 π}];
intersections = GraphicsMeshFindIntersections[pp];
Show[pp, Epilog -> {Red, PointSize[Large], Point@intersections}]
GraphicsMeshFindIntersections[ParametricPlot[{Sin[100 t], Sin[99 t]},
{t, eps, 2 π}]] // Length // Timing
{0.078125, 20330}
Row[Show[plt = ParametricPlot[{Sin[# t], Sin[(# - 1) t]}, {t, 0, 2 π}],
ImageSize -> 300, Epilog -> {Red, PointSize[.2/#],
Point@GraphicsMeshFindIntersections[plt]}] & /@ {5, 10, 20, 50}]
• Now I remember that function! +1. – Michael E2 May 11 '15 at 17:24
• @Michael, thank you for the vote. – kglr May 11 '15 at 17:39
• Glad to see my idea works. :) – J. M. will be back soon May 11 '15 at 21:44
• @J.M, sorry I was distracted by Cases[...] in your comment above and failed to click the link you provided. Yes, it does work; and directly on the graphics input without the extra need to extract the lines. – kglr May 11 '15 at 21:58
• @WateSoyan, it came up as one of the search results from ??**Intersections*. – kglr May 12 '15 at 11:11
I find a workaround which can find all that exact self-intersections:
sol = Solve[(100 (t1 - t2) == 2 k1 \[Or]
100 (t1 + t2) == (2 k1 + 1)) && (99 (t1 - t2) == 2 k2 \[Or]
99 (t1 + t2) == (2 k2 + 1)), {t1, t2}];
Flatten[({t1, t2} /. # /. Solve[(0 <= t1 < t2 < 2) /. #, {k1, k2}, Integers]) & /@ sol, 1]
• You have slightly more than twice as many solutions as rcampion2012. I think you want Flatten[({t1, t2} /. # /. Solve[(0 <= t1 < t2 < 2) /. #, {k1, k2}, Integers]) & /@ sol, 1]` to get each solution exactly once. (I was working along similar lines, but I was trying to get the solution even faster.) – Michael E2 May 11 '15 at 13:52
• @Michael E2 Yeah,I forgot to check it. – WateSoyan May 11 '15 at 14:46 | 2019-11-14T07:14:00 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/83118/hightlight-all-the-self-intersections-of-a-lissajous-figure",
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https://math.stackexchange.com/questions/719825/using-the-polar-form-of-1-i-and-sqrt3-i-to-deduce-cos-frac-pi12 | # Using the polar form of $1 + i$ and $\sqrt3 + i$ to deduce $\cos (\frac{\pi}{12}), \sin(\frac{\pi}{12})$
I have been beating my head against the following problem and would like a gentle nudge in the right direction.
The question states, by writing $1 + i$ and $\sqrt3 + i$ in polar form, deduce that
$$\cos (\frac{\pi}{12}) = \frac{\sqrt3 + 1}{2\sqrt2}, \sin(\frac{\pi}{12}) = \frac{\sqrt3 - 1}{2\sqrt2}$$
so I have written them in polar form
EDIT (as polar forms were incorrect): $$1 + i = \sqrt2e^{i\pi/4}, \sqrt3 + i = 2e^{i\pi/6}$$
another part of the question also asks you to put $\frac{1 + i}{\sqrt3 + i}$ into form $x + yi$ which I figured is
$$\frac{\sqrt3 + 1}{4} + \frac{1 - \sqrt3}{4}i$$
I just can't seem to connect it all together unfortunately so any help would be greatfully received.
• $1+i = \sqrt{2}e^{\frac{i\pi}{4}}$ and $\sqrt{3}+i = 2e^{\frac{i\pi}{6}}$. – John Habert Mar 20 '14 at 14:01
• Your polar forms are wrong. For example: $1+i=\sqrt{2}e^{i\frac{\pi}{4}}$. – Michael Hoppe Mar 20 '14 at 14:02
• I don't understand the downvote on this. Writing a wrong expression in an answer deserves a downvote, in a question the expressions can be wrong. That is why the OP is asking! – Guy Mar 20 '14 at 14:10
Hint: $\frac14-\frac16=\frac{3}{12}-\frac{2}{12}=\frac{1}{12}$.
• This is great, I felt this might be the case but as above I had gotten the polar forms wrong which confused me. I am accepting as this is core to the problem. – Matt Martineau Mar 20 '14 at 14:06
Notice
$$1 + i = \sqrt2e^{i\frac{\pi}{4}}$$ $$\sqrt{3} - i = 2e^{-i\frac{\pi}{6}}$$
Multiply both and we get
$$(1 + i)(\sqrt{3} - i) = \sqrt2e^{i(\frac{\pi}{4} - \frac{\pi}{6})}$$ $$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt{2}e^{i\frac{\pi}{12}}$$ $$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt2\left(\sin{\frac{\pi}{12}} + i\cos\frac{\pi}{12}\right)$$ $$\frac{\sqrt3 + 1}{2\sqrt2} + \frac{\sqrt3 - 1}{2\sqrt2}i = \cos{\frac{\pi}{12}} + i\sin\frac{\pi}{12}$$
Now, compare both real and imaginary parts to deduce:
$$\sin{\frac{\pi}{12}} = \frac{\sqrt3 + 1}{2\sqrt2}$$ $$\cos{\frac{\pi}{12}} = \frac{\sqrt3 - 1}{2\sqrt2}$$
• Awesome, very thorough solution to the problem. I used this to check once I had given it a stab myself. – Matt Martineau Mar 20 '14 at 14:40
• Hope it helped you, this appeared on my homework as well! – Yiyuan Lee Mar 20 '14 at 14:44 | 2021-07-24T16:21:13 | {
"domain": "stackexchange.com",
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https://money.stackexchange.com/questions/132385/how-do-i-calculate-monthly-payments-given-a-certain-loan-amount-apr-and-loan-t | How do I calculate monthly payments given a certain loan amount, APR, and loan term in years?
I have a homework assignment to create a calculator that calculates a monthly payment, given a certain loan amount, APR, and loan term (in years). What is the formula I need to use?
• Have you searched for "loan payment formula? I get many results that give the fairly simple formula. Oct 30 '20 at 17:02
Using the following variables
``````s = principal
r = periodic rate
n = number of periods
d = periodic payment
``````
Standard loan equation - formula derived by induction
$s=\sum_{k=1}^{n}\frac{d}{(1+r)^k}=\frac{d-d(r+1)^{-n}}{r}$
$\therefore d=rs(\frac{1}{(1+r)^n-1}+1)$
``````d = r s (1/((1 + r)^n - 1) + 1)
``````
If your APR is quoted in the USA it will be a nominal rate. For a loan with monthly repayments and a "nominal APR compounded monthly" the periodic rate `r = APR/12`
E.g.
``````APR = 5% = 0.05
r = 0.05/12 = 0.00416667
``````
If your APR is quoted in Europe it will be an effective annual rate. To convert to a periodic monthly rate `r = (1 + APR/100)^(1/12) - 1`
E.g.
``````APR = 5% = 0.05
r = (1 + 0.05)^(1/12) - 1 = 0.00407412
``````
See Investopedia - Present Value of an Ordinary Annuity for more insight into how this is calculated.
For example, the Investopedia loan illustration, with 5 annual payments and APR of 5% (nominal compounded annually is the same as effective annual rate.)
Using the same example of five \$1,000 payments made over a period of five years, here is how a present value calculation would look. It shows that \$4,329.58, invested at 5% interest, would be sufficient to produce those five \$1,000 payments.
``````s = 4329.58
r = 0.05
n = 5
d = r s (1/((1 + r)^n - 1) + 1) = 1000
``````
• Just to check my math. Given s=\$4329.58, r=0.05/12 n = 5*12 = 60 months, then d = \$81.70. Is that correct? Oct 30 '20 at 18:14
• @moonman239 Hi, yes correct for a nominal APR compounded monthly. The type of nominal rate should always be quoted, because 5% APR compounded quarterly is not the same as 5% compounded monthly, or daily. The correct answer \$81.70 compares OK to the ballpark estimate \$1000/12 = \$83.33, just as a rough check. Oct 30 '20 at 18:22 | 2021-09-19T19:49:42 | {
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http://mathoverflow.net/questions/157795/number-of-partitions-whose-blocks-form-arithmetic-progressions | # Number of partitions whose blocks form arithmetic progressions
As is known, the set $\{1,\ldots,n\}$ has $2^n$ many subsets and $B_n$ (the $n$th Bell number) many partitions, where clearly $B_n<2^{2^n}$ and it is actually known that $B_n<n^n$ for large $n$.
A napkin calculation suggests that $\{1,\ldots,n\}$ has about $n^2\log n$ many subsets that form arithmetic progressions, i.e., are evenly spaced''. Example: the only subsets of $\{1,2,3,4\}$ that are not evenly spaced are $\{1,2,4\}$ and $\{1,3,4\}$.
It seems harder to determine the number $A_n$ of partitions of $\{1,\ldots,n\}$ in which each block is evenly spaced. So my question is: what can be said about the growth rate of $A_n$?
-
I suggest to add this statistic to findstat.org – Per Alexandersson Feb 17 '14 at 8:42
Here are some values for $A_n$ computed with a modern napkin: $1,2,5,13,37,111,359,1211$, which is oeis.org/A053732 (though the page doesn't contain any new information). @PerAlexandersson It's not really a statistic on a combinatorial collection, is it? – Christian Stump Feb 17 '14 at 9:18
blocks of size 1 and 2 count as arithmetic progressions? – Gerry Myerson Feb 17 '14 at 10:57
@GerryMyerson yes – Bjørn Kjos-Hanssen Feb 17 '14 at 14:33
@ChristianStump: This is a subset of set partitions right? So, one can have a statistic that is 1 if all blocks are arithmetic progressions and 0 otherwise. – Per Alexandersson Feb 17 '14 at 15:59
I will prove that $$A_n = \Big(\frac{n}{e}\Big)^{n/2} \exp(O(\sqrt{n})).$$ With more effort, one could probably even get an asymptotic formula. Moreover the argument suggests that a typical such set partition consists of about $C_1\sqrt{n}$ singleton sets, about $C_2\sqrt{n}$ sets of size $3$, a bounded number of sets of size four, no sets of size five or more, and the rest comprising of doubleton sets. (See the Additional Remarks section for a rough sketch of what I have in mind.) This is in contrast to a typical set partition where the sets tend to have size about $\log n$.
Let $C(n)$ denote the number of ways of partitioning an $n$-element set into doubleton sets. Clearly $C(n)=0$ if $n$ is odd and $C(n)$ equals $(n-1)\cdot (n-3)\cdot \ldots \cdot 1$ if $n$ is even. Using Stirling's formula $C(n)\sim \sqrt{2} (n/e)^{n/2}$ if $n$ is even.
Clearly $A_n$ is at least the number of ways of partitioning $\{1,\ldots, n\}$ into just singleton and doubleton sets. Note that $k$ singleton sets may be chosen in $\binom{n}{k}$ ways, and the remaining doubleton sets in $C(n-k)$ ways (and we may assume that $n$ and $k$ have the same parity). Thus $$A_n \ge \sum_{k} \binom{n}{k} C(n-k)$$ and a calculation using our asymptotic for $C(n-k)$ shows that this is $$\sim C \Big( \frac{n}{e}\Big)^{n/2} e^{\sqrt{n}},$$ for some positive constant $C$. This gives the lower bound part of the claimed behavior for $A_n$.
Now let's turn to the upper bound. Suppose that the set partition has $k_1$ singletons, $k_3$ sets of size $3$, $k_4$ sets of size $4$ and so on. Note that $k_1+3k_3+4k_4+\ldots$ must be at most $n$ and have the same parity as $n$. The number of choices for the $k_1$ singletons is $\binom{n}{k_1} \le n^{k_1}/k_1!$. Consider now the choices for the three element sets. There are at most $n^2/4$ ways of picking a three term progression in $\{1,\ldots, n\}$ (pick the starting point $a$ and then there are at most $(n-a)/2$ choices for the common difference $d$), and therefore the number of ways of picking $k_3$ three term sets is at most $(n^2/2)^{k_3}/k_3!$. Similarly the number of $j$-term progressions in $\{1,\ldots, n\}$ is at most $n^2/(2(j-1))$, and so the number of ways of picking $k_j$ sets with $j$ elements is at most $(n^2/(2(j-1)))^{k_j}/k_j!$. Thus we find that $$A_n \le \sum_{k_1, k_3, k_4, \ldots} \frac{n^{k_1}}{k_1!} \prod_{j\ge 3} \frac{1}{k_j!} \Big(\frac{n^2}{2(j-1)}\Big)^{k_j} C(n-k_1-3k_3-\ldots).$$ Now we use above that $C(n-k_1-3k_3 -\ldots) = O((n/e)^{(n-k_1-3k_3-\ldots)/2})$. Thus we get that $$A_n = O\Big( \Big(\frac{n}{e}\Big)^{n/2} \sum_{k_1, k_3, k_4, \ldots} \frac{(en)^{k_1/2}}{k_1!} \prod_{j\ge 3} \frac{1}{k_j!} \Big( \frac{e^{j/2}n^{2-j/2}}{2(j-1)}\Big)^{k_j} \Big). \tag{1}$$ Simply extend the sums over $k_1$, $k_3$, $\ldots$ to run over all non-negative integers. We find that $$A_n = O\Big( \Big(\frac{n}{e}\Big)^{n/2} \exp\Big( \sqrt{en} + \sum_{j\ge 3} \frac{e^{j/2}n^{2-j/2}}{2(j-1)}\Big) \Big) = O\Big( \Big(\frac{n}{e}\Big)^{n/2} \exp(O(\sqrt{n}))\Big).$$
Additional remarks. If one inspects the upper bound argument above, then one can show that terms with $k_1+3k_3+4k_4+\ldots \ge 100\sqrt{n}$ contribute an amount $\le (n/e)^{n/2}$. To see this, multiply the bound in (1) by $\exp((k_1+3k_3+\ldots)-100\sqrt{n})$ which is $\ge 1$ for such terms, and then extend $k_j$ to all natural numbers and bound as before. So certainly the bulk of the contribution will come from terms with $k_1+3k_3+\ldots \le 100\sqrt{n}$, so that already most sets are doubletons. In this range we can replace $C(n-k_1-3k_3-\ldots)$ by $\sim \sqrt{2} (n/e)^{n/2} n^{-(k_1+3k_3+\ldots)/2}$. This leads to a better bound in (1) with the $e^{k_1/2}$ and $e^{jk_j/2}$ terms removed. Now carrying out our argument gives the more precise bound $$A_n = O\Big( \Big(\frac{n}{e}\Big)^{n/2} \sum_{k_1,k_3,\ldots} \frac{n^{k_1/2}}{k_1!} \prod_{j\ge 3} \frac{1}{k_j!} \Big(\frac{n^{2-j/2}}{2(j-1)}\Big)^{k_j} \Big), \tag{2}$$ which leads to the sharper upper bound $$A_n = O\Big( \Big(\frac{n}{e}\Big)^{n/2} \exp\Big( \frac{5\sqrt{n}}4\Big)\Big). \tag{3}$$ I expect this bound to be tight, and that $A_n$ is really asymptotic to some constant times this expression (reason below). Granting this for the moment, what (2) tells us is that we should view $k_1$ as being roughly speaking Poisson with parameter $\sqrt{n}$, $k_3$ is Poisson with parameter $\sqrt{n}/4$, $k_4$ is Poisson with parameter $1/6$, and note that $k_5$ and above are Poisson with a very small parameter so that usually they are likely to be zero. This is the reasoning behind my comment on the structure of a typical set partition of the type considered here -- $k_1$ is likely to be about $\sqrt{n}$, $k_3$ of size $\sqrt{n}/4$, $k_4$ of constant size, and $k_5$ and above are likely to be zero.
Lastly let me explain why there should be a corresponding lower bound of the same size as in (3) (this wouldn't be too hard to prove; a precise asymptotic would be more challenging). In the upper bound argument we just sampled arithmetic progressions of length $j$ in $\{1,\ldots,n\}$ repeatedly, ignoring the fact that the progressions we pick should be disjoint. The point is that if we pick only a small number of progressions (which is true when $k_1+3k_3+\ldots \le 100\sqrt{n}$), then there is a positive probability that sampling repeatedly will lead to a choice with disjoint sets. In other words, there is a lower bound of the same shape as (2) in this range of parameters, and that is enough to give the lower bound (3). Another way to say this is that if we just consider set partitions into blocks of size $1$, $2$ or $3$ (and these being APs) then we'd get about the quantity in (3) -- first pick about $\sqrt{n}/4$ disjoint three term progressions, then pick about $\sqrt{n}$ singleton sets, and then use doubletons for the rest.
-
@BjørnKjos-Hanssen: I added some more remarks on this. Please do let me know if something is unclear. – Lucia Feb 19 '14 at 1:04
Lucia, thanks, that helps. Great answer. – Bjørn Kjos-Hanssen Feb 19 '14 at 6:31 | 2015-03-30T00:02:26 | {
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# The temperature of a certain cup of coffee 10 minutes after
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The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the formula F=120(2^-at) + 60, where F is in degrees Fahrenheit and a is a constant. Then the temperature of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?
A. 65
B. 75
C. 80
D. 85
E. 90
OPEN DISCUSSION OF THIS QUESTION IS HERE: the-temperature-of-a-certain-cup-of-coffee-10-minutes-after-98165.html
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The temperature of coffee 10 minutes after it was poured (120 F) will help in solving the constant “a”.
120 = 120(2 ^ 10a) + 60
2 ^ -1 = 2 ^ 10a
a = -1/10
The temperature of coffee 30 minutes after it was poured is:
F = 120 ( 2 ^ -30/10) + 60
F = 120 * 1/ 8 + 60
F = 15 + 60
F = 75
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You're right - OA is B.
When I see a problem that looks this complex, and I'm being timed, I think my brain freezes. Seriously, it's hard to think straight and just plow through it. I'm worried about how much time it will take to solve, and should I spend that much time, etc. Less than a week to go, I really need to get my speed up.
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Robin in NC wrote:
The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the formula F=120(2^-at) + 60, where F is in degrees Fahrenheit and a is a constant. Then the temperature of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?
A. 65
B. 75
C. 80
D. 85
E. 90
We have the formula for calculating temperature: $$F=120*2^{-at}+60$$ and the value for $$t=10$$ --> $$F(t=10)=120=120*2^{-10a}+60$$ --> $$\frac{1}{2^{10a}}=\frac{1}{2}$$ --> $$10a=1$$ --> $$a=0.1$$. So formula is $$F=120*2^{-0.1t}+60$$
Now for $$t=30$$ the temperature in Fahrenheits will be $$F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75$$.
OPEN DISCUSSION OF THIS QUESTION IS HERE: the-temperature-of-a-certain-cup-of-coffee-10-minutes-after-98165.html
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Re: The temperature of a certain cup of coffee 10 minutes after [#permalink] 26 Sep 2014, 00:10
Display posts from previous: Sort by | 2017-10-18T09:50:01 | {
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10 May 2017, 12:17
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In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Assuming a constant rate of consumption, how much gas would be consumed by 7 cars in an 8-hour race?
A. 560
B. 654.5
C. 768
D. 864
E. 896
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10 May 2017, 12:22
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E.
6(x)*5= 480
X = 16
Plug in and get 896
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Re: In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Ass [#permalink]
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10 May 2017, 13:30
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In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Assuming a constant rate of consumption, how much gas would be consumed by 7 cars in an 8-hour race?
A. 560
B. 654.5
C. 768
D. 864
E. 896
I suspect this method may be a slightly unorthodox, but I often set up equations wrong on rate questions, so I drew a table:
Attachment:
CONSUMPTION - RATE - TABLE.jpg [ 35.95 KiB | Viewed 883 times ]
Because this is a rate question that resembles the "multiple workers" at unknown individual rate (where total work is analogous to total number of gallons consumed, as in pool-draining rate questions), I drew a table.
TOTAL gallons consumed / [total # of cars * total hrs gas is consumed] = RATE of individual car's consumption (see second column)
$$\frac{480}{6*5}$$= $$\frac{480}{30}$$ = 16 gallons of gas per hour for INDIVIDUAL car
Now we have individual rate of consumption, such that (# of cars)*(rate of consumption)*(hours spent consuming) = TOTAL gallons consumed (see second row)
7*16*8 = 896
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In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Ass [#permalink]
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10 May 2017, 14:05
In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Assuming a constant rate of consumption, how much gas would be consumed by 7 cars in an 8-hour race?
A. 560
B. 654.5
C. 768
D. 864
E. 896
480/(6*5)=16 gallons per car per hour
16*7*8=896 gallons
E
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Re: In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Ass [#permalink]
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10 May 2017, 14:12
30(5 cars*6 hours) car hours consumes 480 gallons of fuel.
We need to find how much 56 car hour consume? (7 cars * 8 hours)
30 ---- 480
56 ---- x
Cross multiplying and solving for x,
30x = 480*56
x = 16*56 = 896(Option E)
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Re: In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Ass [#permalink]
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15 May 2017, 17:32
1
In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Assuming a constant rate of consumption, how much gas would be consumed by 7 cars in an 8-hour race?
A. 560
B. 654.5
C. 768
D. 864
E. 896
Since 6 cars consume 480 gallons, one car consumes 480/6 = 80 gallons in 5 hours. Thus, the consumption rate for each car is 80/5 = 16 gallons per hour.
Since there are 7 cars, they will consume a total of 16 x 7 = 112 gallons per hour. Since the race takes 8 hours, all 7 cars will consume 112 x 8 = 896 gallons of gas.
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Re: In a 5-hour race, 6 cars consumed 480 gallons of gas between them. Ass [#permalink] 15 May 2017, 17:32
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# How many terminating zeroes does 200! have?
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How many terminating zeroes does 200! have?
(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
[Reveal] Spoiler: OA
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15 Apr 2005, 07:39
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You have 40 multiples of 5, 8 of 25 and 1 of 125. This will give 49 zeros
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15 Apr 2005, 08:40
You have the correct answer.
Please explain how did you go for 5, 25, 125 what is the reasoning for your approach to this soln.
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16 Apr 2005, 00:15
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in order to determine the zeros you need to know how many 2 and 5 200! contains. Apparently, 2 are much more than 5 since every second number is a multiple of 2. That is why when you find the multiples of 5 you will get the number of zeros. Now every 5-th number is a 5 multiple, where every 25-th has two fives, and 125 has 3 fives. So you have 40 times 5, 8 times 5x5 and 1 time 5x5x5. But note that 25 and 125 already contain one five from the fives multiples, and you will have 40+8+1 times 5 which is 49.
Hope that helps
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Re: How many terminating zeroes does 200! have? [#permalink]
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To find number of terminating zero's count the number of 5's
$$\frac{200}{5}$$ = 40
$$\frac{200}{25}$$ = 8
$$\frac{200}{125}$$ = 1
If you add them its 49 zero's
Explanation of this concept on this thread in detail...
question-on-terminating-zero-s-154440.html
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Re: How many terminating zeroes does 200! have? [#permalink]
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25 Feb 2015, 08:28
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To solve this question, it would be good to know the formular: n!/5^x and in this example: 200/5=40 200/25=8 and 200/125=1 and then the sum is 49, Hence C
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How many terminating zeroes does 200! have? [#permalink]
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25 Feb 2015, 12:36
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TS wrote:
How many terminating zeroes does 200! have?
(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
Nice question. Answer is C.
To answer a question about trailing zeros in a factorial, we must realize that the trailing zeros depend on the number of 10s (i.e. 10s or its multiples) in the factorial. 10=5*2, so we are looking for 5s and 2s. By virtue of being the more frequent number, we don't need to count the number of 2s, because there will always be equal or more 2s than 5s in a factorial. So, essentially we are looking for 5s and its exponents.
Start with finding 5s. 200/5= 40. So we have 40 fives (and 40 twos) which means we'll have 40 zeros. If the answer was a decimal, such as 201/5, round down.
Now look for two 5s, that is 25. So, 200/25= 8, so we have eight more zeros.
Lastly, 5^3= 125, and we have one such number in 200!, which gives us one more trailing zero. Therefore, we have 40+8+1=49 trailing zeros.
Hope this helps.
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Re: How many terminating zeroes does 200! have? [#permalink]
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25 Feb 2015, 12:47
TS wrote:
How many terminating zeroes does 200! have?
(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.
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Re: How many terminating zeroes does 200! have? [#permalink]
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25 Feb 2015, 16:45
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Hi All,
It looks like everyone who posted in this thread understands the correct answer. For anyone who doesn't quite get why the math works the way that it does, here are some examples to prove the point.
200! = (200)(199)(198)(197)....(2)(1) so we know that it's a gigantic number. The reason why it will end in a "string" of 0s is because of all of the multiples of 5 involved.
When multiplying integers, there are two ways to get a number that ends in a 0:
1) multiply a multiple of 5 by an even number
2) multiply a multiple of 10 by an integer.
(5)(2) = 10 so we get one 0 for every multiple of 5
However, 25 = (5)(5). It has TWO 5s in it, so there will be two 0s.
eg (25)(4) = 100
With 125 = (5)(5)(5), we have THREE 5s, so there will be three 0s.
eg (125)(8) = 1,000
So when we divide 200 by 5.....200/5 = 40 multiples of 5. SOME of those multiples of 5 are multiples of 25 (or 125) though. Each of those special cases has to be accounted for.
25, 50, 75, 100, 125, 150, 175 and 200 are all multiples of 25, so they each include one extra 0 (and 125 includes two extra 0s).
So we have 40 + 8 + 1 = 49 zeroes.
[Reveal] Spoiler:
C
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Re: How many terminating zeroes does 200! have? [#permalink]
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25 Jan 2017, 10:11
1) To paraphrase this question, we need to find the number of prime factor pairs 2*5 in the product 200!
2) In a range of consecutive integers, there will always be more prime factors 2, than prime factors 5, so we only need to calculate the number of prime factors of 5.
3) 200/5=40; 200/25=8; 200/125=1
4) 40+8+1=49
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Re: How many terminating zeroes does 200! have? [#permalink]
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08 Mar 2018, 17:44
TS wrote:
How many terminating zeroes does 200! have?
(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
To determine the number of trailing zeros in a number, we need to determine the number of 5-and-2 pairs within the prime factorization of that number. Note that each 5-and-2 pair creates a “10,” which contributes one trailing zero.
Since we know there are fewer 5s in 200! than there are 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.
To determine the number of 5s within 200!, we can use the following shortcut in which we divide 200 by 5, then divide the quotient of 200/5 by 5 and continue this process until we no longer get a nonzero quotient.
200/5 = 40
40/5 = 8
8/5 = 1 (we can ignore the remainder)
Since 1/5 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 5 within 200!.
Thus, there are 40 + 8 + 1 = 49 factors of 5 within 200!. This means that there are 49 5-and-2 pairs in 200!; therefore, there are 49 terminating zeros in 200!.
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Re: How many terminating zeroes does 200! have? [#permalink] 08 Mar 2018, 17:44
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-03-18T07:52:31 | {
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https://math.stackexchange.com/questions/4661090/is-the-trace-of-a-constant-simply-that-constant | # Is the trace of a constant simply that constant?
I can't seem to find an answer to this online, basically what I am asking is
$$\mathrm{Tr} ((4)_{1\times 1})=4?$$ or in general is $$\mathrm{Tr} ((k)_{1\times 1})=k?$$ Where $$k$$ is a constant.
I think the answer is yes, since a $$1\times 1$$ matrix is simply a number (or constant) so the sum of the diagonal elements is just that element.
But I would like to see some confirmation of this if possible.
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Mar 17 at 18:05
• The answer is indeed yes Mar 17 at 18:49
If I have guessed correctly what you mean, I would suggest that you should write $$\mathrm{Tr} ((4)_{1\times 1})$$ which causes less confusion. As the answer for your question, the trace of a $$1\times 1$$ matrix $$(a)$$ is indeed $$a$$. What do you think is ambiguous? If we define the trace of a $$1\times 1$$ matrix $$(a)$$ as $$a$$, then the properties of traces for $$n\times n$$ $$(n\ge 2)$$ matrices still holds for $$1\times 1$$ matrix. For example $$a$$ is the sum of all of its eigenvalues (since $$0=\det (\lambda (1)-(a))=\det (\lambda -a)\Longleftrightarrow \lambda=a$$, i.e. $$a$$ is the unique eigenvalue of $$(a)_{1\times 1}$$.) | 2023-03-31T18:48:50 | {
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"openwebmath_perplexity": 117.34810355761188,
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https://ivyglobal.com/study/explanations/438 | ## 1.
The Correct Answer is (B) — In scientific notation, the decimal place comes after the first digit of the number, which is multiplied by a power of 10. The only answer choice that gives the number stated and follows this rule properly is B.
## 2.
The Correct Answer is (H) — To answer this question, it may be helpful to work backwards. To take the average of 2 numbers, you add them together and divide by 2. To find one of the original numbers, multiply the average by 2 and subtract the known number: (184 × 2) — 170 = 198.
## 3.
The Correct Answer is (E) — To find the GCF, multiply together all of the prime factors that the numbers have in common. $72=2\times2\times2\times3\times3$ and $88=2\times\2\times2\times11$ and $120=2\times2\times2\times3\times5$. For these three numbers, the common factors multiplied together give a product of $2\times2\times2=8$. In this instance, however, it may be faster to guess and check.
## 4.
The Correct Answer is (F) — The questions equates two expressions that are both in terms of z. First, isolate for z. Then, collect like terms and solve.
\begin{align*} 6z+2&=5-11z \\ 6z+11z&=5 \\ 17z&=3 \\ z&=\frac{3}{17} \end{align*}
## 5.
The Correct Answer is (E) — You know that the area of a rectangle is length multiplied by width, so to find the missing length, divide the area by the known width: $\frac{320000}{400}=800$.
To find the perimeter, add all the lengths together: $400+400+800+800=2400$.
If you chose (A), you probably found the missing length and stopped there.
## 6.
The Correct Answer is (J) — To answer this question, it may be easier to convert the ratio into a decimal: $\frac{\frac{2}{5}}{\frac{1}{8}}=3.2$. The only ratio from the answer choices that gives this same ratio is J.
You can also multiply both sides of the ratio by $\frac{5}{8}$ to get 16:5, or J.
## 7.
The Correct Answer is (D) — Since this is a function, simply substitute x=2 and solve. \begin{align*} f(x)&=(a^4b^3)^{2x+1} \\ f(x)&=(a^4b^3)^{2(2)+1} \\ f(x)&=(a^4b^3)^5 \\ f(x)&=a^{20}b^{15} \end{align*}
If you chose (A), you probably added the exponents instead of multiplying them.
## 8.
The Correct Answer is (K) — First, determine how many times the recipe can be completed with the given quantities of ingredients. $\frac{7.5}{1.25}=6$ iterations of the recipe. This means $6\times5=30$ cupcakes.
If you chose (G), you probably found the number of times the recipe could be done and stopped there.
## 9.
The Correct Answer is (D) — The number can be odd or even, so (A) and (B) are incorrect. Doubling a number makes the number even, so (C) is incorrect. Tripling a number can make the product odd or even depending on the original number, so (E) is incorrect. (D) is the correct answer: doubling the number will always result in an even number, and adding 1 to an even number will always result in an odd number.
## 10.
The Correct Answer is (H) — To find the angle, first find the proportion of the circle graph that homework occupies on the graph. $\frac{24-8-8-1-3}{24}= \frac{4}{24}= \frac{1}{6}$. Multiply this proportion by 360° to get the angle measure, or 60°.
## 11.
The Correct Answer is (B) — Expand and simplify the expression using the FOIL technique: \begin{align*} &=(x-5)(x+2) \\ &=x^2-5x+2x-10 \\ &=x^2-3x-10 \end{align*}
## 12.
The Correct Answer is (H) — Translating the question from words into equations, $a= \frac{1}{2}b$, and $3a=c$, . Rewrite the latter as $a= \frac{1}{3}c$ and substitute this into the first expression:
$\frac{1}{3}c= \frac{1}{2}b$ and $c= \frac{3}{2}b$
b is greater than c by a factor of $\frac{3}{2}$.
## 13.
The Correct Answer is (B) — Working through |x – 5| = 3, gives you: x – 5 = 3, which simplifies to x = 8, and –(x – 5) = 3, which simplifies to x = 2. Possible values of x are therefore 8 and 2. Substituting x = 8 into the second equation yields 33, which is not an answer choice. Substituting x = 2 into the second expression yields 9.
## 14.
The Correct Answer is (H) — The ratio of the longer side of the floor to the longer side of the blueprint is 15:10, or 3:2. This means that the floor is 1.5 times as large as the blueprint. Cross multiply to find the length of the floor: $\frac{6}{10}=\frac{x}{15}, x=\frac{6\times15}{10}=\frac{90}{10}=9$.
## 15.
The Correct Answer is (C) — Since you know two of the angles in triangle CDE, you can find angle DCE, which is 55°. Since CDE and BCE are supplementary angles, you can find the value of BCE by subtracting 55° from 180°, which yields 125°. Since there are 180° in a triangle, subtracting 125° from 180° gives the value of the remaining two angles. x + y = 55°.
## 16.
The Correct Answer is (K) — Solving for x given the first equation gives x = 4. Substituting this into the second expression gives $4^2+4^3=16+64=80$.
## 17.
The Correct Answer is (D) — To solve this question, drawing a diagram may help.
Since the distance from the Sun to Jupiter is 65 cm and from the Sun to Mars is 20 cm, the distance from Mars to Jupiter must be 45 cm. To find the distance from Mercury to Mars, subtract the distance from Mercury to Jupiter, 60 cm, from the distance from Mars to Jupiter, 45 cm. 60 — 45 = 15 cm.
## 18.
The Correct Answer is (H) — From the question, if y is the larger number, then the smaller number is equal to 2y — 6. Subtracting this from the larger number gives y — (2y — 6). The question tells you this expression is equal to 1, therefore the equation is y — (2y — 6) = 1.
## 19.
The Correct Answer is (D) — You can factor $x^4$ out of the expression under the exponent, leaving $x^2 - x$ inside the brackets. Since the entire expression is to the exponent $\frac{1}{2}$, when $x^4$ is factored out from the exponent $\frac{1}{2}$ (or the square root), it becomes $x^2$. The final expression is $x^2\sqrt{x^2-x}$.
## 20.
The Correct Answer is (F) — To answer this question, substitute the known values using $a^2 + b^2 = c^2$ from the Pythagorean Theorem, and solve for the unknown height. \begin{align*} 6^2&=4^2+x^2 \\ 36&=16+x^2 \\ x^2&=20 \\ x&=\sqrt{20}=2\sqrt{5} \end{align*}
## 23.
The Correct Answer is (B) — This question is most easily solved by the process of elimination. You know that the line slopes downwards, which means the slope must be negative. This eliminates answer choices (C), (D), and (E). Of the two remaining answer choices, the slopes are $-2$ and $\frac{-1}{2}$. The line is not steep enough for a slope of $-2$ , therefore $\frac{-1}{2}$ must be the slope and (B) is the answer. This can be verified by solving for the slope: $\frac{0-2}{3-(-1)}=\frac{-2}{4}=\frac{-1}{2}$.
## 24.
The Correct Answer is (F) — The chimpanzee eats up to 63 bananas per week, or an average of $\frac{63}{7} = 9$ bananas per day. If the chimpanzee eats up to than 63 bananas a week, the average daily consumption must be less than or equal to 9 bananas. The only answer choice that represents this inequality is choice (F).
## 25.
The Correct Answer is (E) — To solve this question, simplify the expression by factoring the bottom and top separately, then cancelling out like terms when possible: \begin{align*} &=-(\frac{z^2-4}{16z^3-64z}) \\ &=-(\frac{z^2-4}{16z(z^2-4)}) \\ &=-\frac{-1}{16z} \end{align*}
## 26.
The Correct Answer is (J) — Set up a ratio between the height and the length of the door stopper: $\frac{2}{5}=\frac{3}{x}$. Cross multiply and solve for the length of the door stopper: $x = \frac{3\times5}{2}=\frac{15}{2}=7.5$
## 27.
The Correct Answer is (E) — Remember to apply the exponents first, according to the order of operations:
\begin{align*} &=(8x^2 y^3 )^2\times(-2x)^3 \\ &=64x^4 y^6×-8x^3 \\ &=-512x^7 y^6 \end{align*}
All of the answer choices other than (E) have a + or – y term at the end, which is incorrect.
## 28.
The Correct Answer is (F) — Set the two expressions on the right-hand side of both equations equal to each other, and then solve: \begin{align*} x+1&=-3x-3 \\ 4x&=-4 \\ x&=-1 \end{align*}
Substitute $x=-1$ into either of the original equations: $y=-1+1=0$.
The solution is the point that lies on both lines: (—1,0).
## 29.
The Correct Answer is (A) — The difference in votes, ab, must be greater than or equal to 30 votes. The only answer choice that represents this inequality is (A).
## 30.
The Correct Answer is (G) — You can rewrite 32 as $2^5$, so the expression becomes $(2^{5})^{\frac{2}{5}}$. Using exponent laws: $(2^{5})^{\frac{2}{5}}=2^2$.
## 31.
The Correct Answer is (C) — You can see from the right triangle formed at the right of the parallelogram that the height of the parallelogram is 4. This is because the right triangle is a 3-4-5 triangle. Alternatively, you can use the Pythagorean Theorem to solve for the missing length. It may, however, be faster to memorize some Pythagorean triples. To find the area of a parallelogram, multiply the height of the parallelogram by the base. (5+3)(4)=32.
If you chose (A), you probably forgot to add 3 to 5 when finding the length of the base.
## 32.
The Correct Answer is (H) — This is an arithmetic sequence because each new row has two more than the row before it. You can write out the number of circles for the first 9 rows and sum them together:
$5+7+9+11+13+15+17+19+21=117$
Or, recognize that each "pair" found by starting from the outside edges and working in, such as 5 and 21, 7 and 19, and so on sums to 26. There are four pairs that sum to 26 plus the remaining row of 13, so 4×26 + 13 = 117. If you are familiar with the Gaussian method of adding arithmetic sequences, recognize that the pattern of this sequence is $5+(n-1)\times2$, where n is the 9th row. 9th row $= 5+(n-1)2=5+(9-1)2=5+16=21$. The $9^{th}$ row will have a total of $\frac{n}{2}(5+21)=\frac{9}{2}(26)=117$ circles.
## 33.
The Correct Answer is (E) — Remember that $\sin a = \frac{\text{opposite}}{\text{hypotenuse}}$. You can sub in the relevant values from the question to get $\sin a=\frac{6}{12}=\frac{1}{2}$.
## 34.
The Correct Answer is (G)$\frac{60}{4}=15$ students participate in musical theater. Of those 15 students, $\frac{15}{3}=5$ students do not play an instrument. Therefore, 10 students participate in musical theater and play an instrument.
## 35.
The Correct Answer is (E) — To average $100 per day for a week, the fundraiser needs to raise a total of$700. Sum up the money they have made in the first 6 days: 52+84+106+66+94+98=500. $200 more needs to be raised. Each cookie is$2, therefore 100 cookies need to be sold.
## 36.
The Correct Answer is (H) — The outputs of the function increase at regular intervals of 3. For each increase by 1 in the x column, the output of the function increases by 3 times as much. Therefore, the slope of the function, or the value m is 3.
Alternatively, you could plug any two points into the slope formula to get an m value of $\frac{7-4}{1-0}=3$.
## 37.
The Correct Answer is (C) — You know that angle BFC is 90°, and that the triangle BCF is an isosceles triangle. Therefore, the other two angles, including angle FBC, of the triangle must each be 45°.
## 38.
The Correct Answer is (H) — The equation of the first line is $y = \frac{1}{2}x$. Since it passes through the origin, the y intercept must be 0. Using the given point of intersection and the origin, you can deduce that the slope is $\frac{1-0}{2-0}=\frac{1}{2}$.
The slope of the second line is perpendicular to the first line, so it is the negative reciprocal: —2. To find the y-intercept, substitute the known point of (2,1) into the equation of the second line and solve. \begin{align*} 1 &= (—2)(2) + b \\ 1 &= —4 + b \\ 5 &= b \end{align*}
The y intercept is 5, therefore the line intersects the y axis at (0,5).
## 39.
The Correct Answer is (B) — This problem requires you to work backwards from what you are given. You know there are 5 times as many footballs sold as baseballs; if the store sold 720 footballs, it sold $\frac{720}{5}=144$ baseballs. Additionally, there are 3 times more baseballs sold than hockey sticks: $\frac{144}{3}=48$ hockey sticks sold.
## 40.
The Correct Answer is (J) — The area of the larger square is $(2+3)^2=25$. To find the side length of the smaller square, use $a^2 + b^2 = c^2$ from the Pythagorean Theorem. $2^2+3^2=c^2,c^2=13, c=\sqrt{13}$. The area of the smaller triangle is $(\sqrt{13})^2=13$. Now, divide the area of the smaller square by the larger square and multiply by 100 to get the percentage of the larger square that is occupied by the smaller square: $\frac{13}{25}\times100 = 52\%$.
## 41.
The Correct Answer is (B) — Simply plug in the given expressions for c and d, and reduce.
\begin{align*} \frac{c}{d}&=\frac{2x^2-10x-28}{4x+8} \\ \frac{c}{d}&=\frac{2(x^2-5x-14)}{4(x+2)} \\ \frac{c}{d}&=\frac{2(x-7)(x+2)}{4(x+2)} \\ \frac{c}{d}&=\frac{x-7}{2} \end{align*}
## 42.
The Correct Answer is (F) — The shaded area is above the line $y=\frac{1}{2}x-\frac{1}{2}$, indicating greater than $y=\frac{1}{2}x-\frac{1}{2}$. The shaded area is also above $3x+2$, indicating greater than $3x+2$. The only answer choice that combines both of these inequalities is (F).
## 43.
The Correct Answer is (C) — Expanding and simplifying this expression gives:
\begin{align*} &=i^2-2i+2i-4 \\ &=i^2-4 \\ &=-1-4=-5 \end{align*}
## 44.
The Correct Answer is (G) — For this question, treat the vectors as equations, where i and j are variables. Substitute the expressions u and v and solve. \begin{align*} &=2u — 3v \\ &=2(7i + 4j) \\ &=14i + 8j — 9i + 6j \\ &=5i + 14j \end{align*}
## 45.
The Correct Answer is (B) — This question requires knowledge of special triangles. The easiest way to solve this question is to be familiar with the trigonometric ratios of special triangles. Alternatively, you can draw out the triangle to help you deduce the value of θ.
The sides of the triangle are $1-2-\sqrt{3}$. The longest side of the triangle must be opposite the greatest angle. Since 2 is opposite the right angle, 60° must be opposite $\sqrt{3}$ and 30° must be opposite 1. Therefore, the value of theta is 30°.
## 46.
The Correct Answer is (J) — Since the perimeter of the square is 16 inches, you know that $\frac{16}{4}=4$ is the length of one of the sides. From the diagram, you can see that a single slant height of the trapezoid (the side lying along the square) is 2 inches long. To find the two bases of the trapezoid, use the Pythagorean Theorem with the legs of each triangle being formed by the square. The total slant lengths of the trapezoid of colored glass is $2+2=4$. The long base of the trapezoid measures $4^2+4^2=c^2,c^2=32,c=4\sqrt{2}$. The short base of the trapezoid measures $2^2+2^2=c^2,c^2=8,c=2\sqrt{2}$. Adding all of these lengths together gives $4+4\sqrt{2}+2\sqrt{2}=4+6\sqrt{2}$.
## 47.
The Correct Answer is (A) — If you rewrite the equation as a logarithm, you get $\log_x a = log_x \text{?}$. From this equation, you can deduce that ? = a.
## 48.
The Correct Answer is (H) — Since the company wants to cover just the top and bottom of each bar, you only need to sum the area of the two rectangular bases of the gold bar: (7)(20)+(6)(20)=140+120=260.
## 49.
The Correct Answer is (C) — To find the number of gold bars that the company can make, divide the total amount of gold that they have by the volume of one gold bar. $\frac{100,000}{1040}=96.15$, or 96 bars can be made – round down, since you cannot make a fraction of a gold bar. Now, find the amount of gold left over by subtracting the total volume of all the gold bars from the total amount of gold. 100,000-(1,040×96)=100,000-99,840=160.
## 50.
The Correct Answer is (K) — Substitute the given values into the equation given for the purity of gold and solve: $\frac{12.52}{12.55}\times100=99.76\doteq99.8$
## 51.
The Correct Answer is (B) — All of the other answer choices are infinite, except for a line segment, or (B).
## 52.
The Correct Answer is (F) — If only v can change, v is raised to the power of 2, and F is doubled, the only factor of multiplication that will give 2F is $\sqrt{2}$. Choices (G) and (H ) would increase F by too much, and choices (J) and (K) would decrease F.
## 53.
The Correct Answer is (A) — Isolate the first equation for v in terms of u, and isolate the second equation for w in terms of v.
$v=\frac{5u}{3}, w=\frac{4}{v}$
Substitute the value of v into the second equation and solve:
\begin{align*} w&=\frac{4}{\frac{5u}{3}} \\ w&=4\times\frac{3}{5u} \\ w&=\frac{12}{5u} \end{align*}
## 54.
The Correct Answer is (K) — The length of the base is 2(1-0)=2, as seen by the top length of the triangle on the left. Therefore, a=2. Since (a,b) lies on the x axis, b = 0. $\frac{0}{2}+2=0+2=2$.
## 55.
The Correct Answer is (B) — Recognize from the diagram that the lengths form a right triangle. To find the value of d, you can use the Pythagorean Theorem:
\begin{align*} 60^2+50^2&=d^2 \\ d^2&=6100 \\ d&=\sqrt{6100} \\ d&\doteq78 \end{align*}
You could also use the cosine law, which is stated in the note below the diagram, by substituting the given values and calculating the value of c. However, there is a lot of potential for error when using the cosine law, and it is a much more time consuming.
## 56.
The Correct Answer is (K) — Choices (F) and (J) directly contradict the statement in the question. Choices (G) and (H) are not necessarily true: the statement original statement does not dispute that, for example, a yellow wagon can be painted. The only choice that must be true is choice (K).
## 57.
The Correct Answer is (E) — The minimum volume displaced by the 3 people is, at least, the volume required to make the tank full. As you can see from the diagram, the empty volume of the tank before the 3 people enter is: (30)(15)(0.5)=225 cubic feet.
## 58.
The Correct Answer is (J) — The easiest way to answer this question is by process of elimination. The domain of the three functions is given in the question. Since you know that all the functions are less than or equal to 2, the filled dots must be on the right side of the graph; this eliminates (H) and (K). Of the remaining choices, the parabola is in the same place for all of them, meaning that you have to focus on functions g(x) and f(x), the linear graphs. At x = -1, g(x) = 0, eliminating (G). At x = 1, f(x) = 0, eliminating (F). The answer is (J). You may have incorrectly checked the endpoints of the functions if you didn't notice that the domain is -2 < 2x ≤ 2, which simplifies to -1 < x ≤ 1.Of the remaining 3 choices, (J) is the only one that correctly graphs g(x) and h(x).
## 59.
The Correct Answer is (B) — To answer this question, it may help to draw a diagram.
Drawing a perpendicular line from the center of the table to the wall and a line from the center of the table to the point where the table meets the wall gives a right triangle with a hypotenuse of 40 centimeters (since the radius is 40 centimeters) and a height of 32 centimeters. Solve for the missing edge by using the Pythagorean Theorem: $40^2=32^2+c^2,c^2=576,c=24$. Don’t forget to double this value, since the question asks for the entire length of the flat edge of the table. Doubling 24 gives 48.
If you chose (A), you probably found half of the length of the flat edge and stopped there.
## 60.
The Correct Answer is (G) — Changing a changes the height of the tides, or the amplitude. Changing c changes the base height of the tides around which they oscillate, or translates the function up or down. Changing x changes the time, or the input value. Changing y changes the height of the tide at any given time, or the output value. Changing k changes the time it takes for a tide to complete its interval, or the period. Therefore, changing k would change the time between high tides. | 2022-08-14T18:02:22 | {
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https://www.physicsforums.com/threads/find-the-values.254575/ | # Find the values:
1. Sep 8, 2008
### Euler_Euclid
Find all values of p and q if p and q are prime numbers, $$p^2 + q^2 + 7pq$$ is a perfect square.
2. Sep 8, 2008
### CRGreathouse
(p, p), (3, 11), (11, 3) all work. I'd suggest trying the equation mod 4 or 36 to see if you can rule out others.
3. Sep 8, 2008
### Euler_Euclid
please explain the whole solution. It was an Olympiad question and all of my friends couldn't do it.
4. Sep 8, 2008
### praharmitra
$$p^2 + 7pq + q^2$$
For p = q, it reduces to $$9p^2 = (3p)^2$$. It therefore is a perfect square.
You can put p = 2 and show that only q = 2 gives a feasible solution, which is covered in p = q. Thus, we conclude that p,q are both odd, and hence the value of the expression is odd
For other values
$$p^2 + 7pq + q^2 = (p+q)^2 + 5pq = k^2$$
$$5pq = k^2 - (p+q)^2 = (k - p - q)(k + p + q)$$
Thus, since p and q are primes, the only possible solutions are
$$1.k-p-q = 5, k+p+q = pq$$
From the above equations we get k = 5+p+q and hence
$$5+2p+2q = pq => 5 + 2q = p(q-2) => p = \frac{2q+5}{q-2} = 2 + \frac{9}{q - 2}$$
$$=> q - 2 = 1, 3, or 9$$
This gives (p,q) = (11,3) or (5,5) or (3,11)
The other possibilites are
$$2. k-p-q = 5p, k+p+q = q$$(no solution)
$$3. k+p+q = 5, k-p-q = pq$$(no solution)
$$4. k+p+q = 5p, k-p-q = q$$(gives p = q)
Thus, the only possible solutions are (p,p), (3,11) and (11,3)
5. Sep 9, 2008
### snipez90
Nice solution praharmitra. I saw the decomposition as well but didn't complete the solution since I didn't analyze the small cases at first. I have to remember that even/odd parity analysis is always a good first step in number theory equations.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 2017-12-18T11:46:08 | {
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http://math.stackexchange.com/questions/825911/probability-of-getting-the-sum-of-two-numbers-selected-from-two-ranges-as-odd | # Probability of getting the sum of two numbers selected from two ranges as odd.
Consider two integers N and M.
One number is selected from a range of 1 to N.
Another number is selected from a range of 1 to M.
What is the probability that the sum of the two numbers is odd?
Note: I figured out that, whenever either N or M is even, the probability will be 1/2. But what if both or any one of them is odd.
-
As you correctly pointed out, if either $N$ or $M$ is even, then the probability is $1/2$, by a symmetry argument. Suppose both $M,N$ are odd. Then there are two cases:
1. We pick exactly $M$ from $[1,M]$
2. We pick anything else from $[1,M]$.
In case 1, the probability that the sum is odd is $\frac{(N-1)/2}{N}= \frac{N-1}{2N}$, slightly less than half, because slightly less than half of the numbers in $[1,N]$ are even. In case 2, the probability that the sum is odd is 1/2 again, because this is equivalent to picking from $[1,M-1]$. Hence the combined answer is
$$\frac{1}{M}\frac{N-1}{2N} + \frac{M-1}{M}\frac{1}{2}=\frac{MN-1}{2MN}$$
A slightly more compact way to write the answer, which handles the cases of $M,N$ even as well, is $$\left \lfloor \frac{MN}{2}\right\rfloor \frac{1}{MN}$$ where $\lfloor\cdot\rfloor$ denotes the floor function.
-
$A \equiv \text{Number selected from the range$1\ldots N$is odd}\\ B \equiv \text{Number selected from the range$1\ldots M$is odd}$
Now $P(A) = \frac 1 N\left\lceil\frac N 2\right\rceil,\ P(\bar{A}) = \frac{1}{N}\left\lfloor \frac N 2 \right\rfloor$
and $P(B) = \frac 1 M\left\lceil\frac M 2\right\rceil,\ P(\bar{B}) = \frac{1}{M}\left\lfloor \frac M 2 \right\rfloor$
Then the probability that the sum of the two numbers is odd is
$P(A)P(\bar{B}) + P(\bar{A})P(B)\\ = \dfrac 1 N\left\lceil\dfrac N 2\right\rceil \dfrac{1}{M}\left\lfloor \dfrac M 2 \right\rfloor + \dfrac{1}{N}\left\lfloor \dfrac N 2 \right\rfloor \dfrac 1 M \left\lceil\dfrac M 2\right\rceil\\ = \dfrac{1}{MN}\left( \left\lceil\dfrac N 2\right\rceil \left\lfloor \dfrac M 2 \right\rfloor + \left\lfloor \dfrac N 2 \right\rfloor \left\lceil\dfrac M 2\right\rceil \right)$
-
Odds and evens form a checkerboard.
Write the numbers from 1 to M down one side, and the numbers from 1 to $N$ along the top. Put $i+j$ in column $i$, row $j$. Colour $i+j$ red if even, blue if it is odd.
You can cover the checkerboard with dominos. Each domino covers a blue and a red, and there will be one square left over because there is an odd number of squares.
All four corners are even : 1+1, 1+N, 1+M, M+N - so it makes sense that the extra square is even.
There are $(NM+1)/2$ even squares and $(NM-1)/2$ odd squares, making a total of $NM$ squares.
- | 2016-02-06T16:31:35 | {
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https://math.stackexchange.com/questions/3532084/logarithms-solve-for-m | # Logarithms. Solve for m
Solve $$x y^m = y x^3$$ for $$m$$.
I know this has to do with logarithms, but I'm not able to figure out how it relates to logs.
• Welcome to Mathematics Stack Exchange. Can you take the logarithm of both sides of the equation? – J. W. Tanner Feb 2 '20 at 20:46
• Can you divide both sides by $xy$? – Narasimham Feb 2 '20 at 20:51
$$xy^m=yx^3\implies \log x + m \log y=\log y + 3 \log x \implies (m-1)\log y=2\log x$$
$$\implies m-1=\dfrac{2\log x}{\log y}\implies m=\dfrac{2\log x}{\log y}+1$$
• Thank you. Exactly what I needed to understand it. – B.Bart Feb 2 '20 at 22:34
Rewrite $$xy^m = yx^3$$ as
$$y^{m-1}=x^2$$
Take log of both sides,
$$(m-1)\ln y = 2\ln x$$
$$m= 2\log_y x +1$$ | 2021-05-18T13:05:08 | {
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http://edugeetha.com/jcx/properties-of-isosceles-triangle.php | # Properties of isosceles triangle
The altitude is a perpendicular distance from the Isosceles Triangle Properties. Triangle has three vertices, three sides and three angles. b. Properties of Triangle. Also, since triangle ABD is isosceles, ray AM bisects angle BAD, so angle BAM = angle DAM. An equilateral triangle has all three sides equal in length. Our objectives for this learning packet are: -Defining an isosceles triangle - Establishing the different parts of an isosceles triangle and their properties - Solving problem; triangle; isosceles; isosceles triangle; angles; angle; missing angle In this tutorial, you'll learn about the properties of a polygon, see the names of the ISOSCELES TRIANGLE AREA, HEIGHT AND SECTION PROPERTIES Isosceles triangle area, altitude (height), distances from the centroidal axes to the An isosceles triangle is a triangle with two sides of the same length. We need to prove that the angles opposite to the sides AC and BC are equal, that is, ∠CAB = ∠CBA. 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The above isosceles trapezoid property calculator is based on the provided equations and does not account for all mathematical limitations. Then we did the proof of the base angles theorem. Exterior Angle The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. Theorems Theorem 4. Equilateral Triangle An equilateral triangle has all sides and all angles are equal in size. (4) Hence the altitude drawn will divide the isosceles triangle into two congruent right triangles. Lessons 4. Answer Wiki. Triangles Scalene Isosceles Equilateral Use both the angle and side names when classifying a triangle. Since the total degrees in any triangle is 180°, an obtuse triangle can only have one angle that measures more than 90°. An obtuse triangle has only one inscribed square. The bisector of the vertex angle is the perpendicular bisector of the base. We draw a perpendicular line from A down to BD, which intersects at point C. Objectives: 1. Properties of Isosceles Triangles. Improve your math knowledge with free questions in "Proofs involving isosceles triangles" and thousands of other math skills. In this tutorial, see how identifying your triangle first can be very helpful in solving for that missing measurement. e 180° ). Properties of a The property In a right triangle, the median drawn to the hypotenuse, has the measure half the hypotenuse. Since this is an isosceles right triangle, the only problem is to find the unknown hypotenuse. 206 CHAPTER 4 Discovering and Proving Triangle Properties Imagination is built upon knowledge. There are several properties that are true of every isosceles triangle. 2. An isosceles triangle 6. Isosceles and scalene triangles can be right triangles; all isosceles triangles have the additional useful property of being able to be split into two right triangles. The figure has four right angles and four congruent sides. If the legs are congruent we have what is called an isosceles trapezoid. Enjoy a range of interesting triangle facts for kids and have fun learning about the 3-sided polygon. We know that each of the lines which is a radius of the circle (the green lines) are the same length. If all three side lengths are equal, the triangle is also equilateral. The upper base angles are congruent. Of TrianglePropertiesT- 1-855-694-8886Email- info@iTutor. Symbols If AB&*cAC&*, then aC caB. The angle formed by the legs is the vertex angle. If you're like me, you probably don't weigh exactly what you'd like to weigh. LEGS The _____ sides of an isosceles triangle opposite the base angles. The three vertices of the triangle are denoted by A, B, and C in the figure below. ' The third and unequal side of an isosceles triangle is known as the 'base. All angles in a triangle add up to 180˚ so 180 - (50+50) = 80˚ So an isosceles triangle has only got two sides of equal length and two angles the same. Given below is an example of obtuse/oblique angle triangle. In the picture on the left, the shaded angle is the obtuse angle that distinguishes this triangle. The Consider the right most B and corresponding triangle CBF. Like the 30-60-90 triangle, the lengths of the sides of a 45-45-90 triangle follow a specific pattern that you should know. Scalene Isosceles Equilateral Acute 7 11 Geometry calculator for solving the simiperimeter of a isosceles triangle given the length of sides a and b. Let’s assume ABD is an equilateral triangle with each side = 2. The circumcenter of a triangle is always inside the triangle. Q. 21 If two times the square of the diameter of the circumcircle of a triangle is equal to the sum of the squares of its sides then prove that the triangle is right angled. 1 and 4. Properties Of Triangle 2. Show that the triangle is isosceles. c. The angles opposite the Activity: The Isosceles (45-45) Right Triangle. The Triangle and its Properties Triangle is a simple closed curve made of three line segments. Which properties belong to all isosceles triangles? Check all that apply. Properties of an isosceles trapezium; 12. An equilateral triangle is also an equiangular triangle. Properties The unequal side of an isosceles triangle is usually referred to as the 'base' of the triangle. (S). x = 19, the measure of angle ABC = 4(19) - 12 = 64. The Calabi triangle is a special isosceles triangle with the property that the other two inscribed squares, with sides collinear with the sides of the triangle, are of the same size as the base square. Triangle questions account for less than 10% of all SAT math questions. 2 words related to isosceles: symmetric, symmetrical. Also the Pythagorean theorem can be used for non ACT Math Help » Geometry » Plane Geometry » Triangles » Isosceles Triangles » Acute / Obtuse Isosceles Triangles Example Question #1 : Acute / Obtuse Isosceles Triangles What is the area of an isosceles triangle with a vertex of degrees and two sides equal to ? Isosceles triangle calculator computes all properties of an isosceles triangle such as area, perimeter, sides and angles given a sufficient subset of these properties. The Properties of Scalene Triangles. Isosceles triangle has two sides, b and c, and the two angles opposite them <B and <C equal. It is known that if a and b are two given line segments, then their third Calculates the other elements of an isosceles triangle from the selected elements . 3) The . Two sides of an isosceles triangle are equal which means two of its angles are also equal. An isosceles triangle has two sides of equal length. The most important rule is that this triangle has one right angle, and two other angles are equal to 45°. Example: Find the values of the variables x 41° y Synonyms for isosceles in Free Thesaurus. Angles opposite to equal sides of an isosceles triangle are also equal. It has at least two congruent sides. Angles opposite to equal sides in an isosceles triangle are always of equal measure. ” Think of it as a “same-sided” triangle. Area of Isosceles Triangle Formula. Bisector via angles (isosceles triangle) Activity. If two angles of a triangle are Isosceles triangles are special and because of that there are unique relationships that involve their internal line segments. The sides a, b and c will be chords of the circle. These worksheets explain equilateral, isosceles, and scalene triangles. The triangle in primary-school geometry: how children learn about equilateral, scalene, isosceles and right-angled triangles in KS2. An isosceles triangle is a polygon having two equal sides and two equal angles adjacent to equal sides. What are synonyms for isosceles? The Isosceles Trapezoid and its Dissecting Similar Triangles Larry Hoehn Abstract. We call this little statement the Angle Sum Theorem for triangles. Sometimes it An isosceles triangle is a triangle that has (at least) two equal side lengths. If any two angles and a sides of one Properties of the Triangles: The sum of the three angles of the a triangle is two right angles (i. (3) Perpendicular drawn to the third side from the corresponding vertex will bisect the third side. Isosceles Triangle. Isosceles triangle is a polygon with three vertices (corners) and three edges (sides) two of which are Given the properties of an isosceles triangle, students can be asked to draw their own isosceles triangle. Since S is the midpoint of ¯PQ , ¯PS≅¯QS . Equilateral Triangle Properties In geometry, where an equilateral polygon is a polygon which has all sides are same and the equal of the same length in equilateral triangle. Quiz on properties of quadrilaterals; 11. Computed angles, perimeter, medians, heights, centroid, inradius and other properties of this triangle. In this article isosceles triangle properties In this page we have NCERT Solutions for Class 7 Maths Chapter 6: Triangle and Its Properties All Exercises 6. The lower base angles are congruent. '. Intensify practice with this compilation of area of a triangle worksheets featuring skills like finding the area of scalene, isosceles and equilateral triangles, find the missing base or height, find the area with measures offered as integers, decimals, fractions and algebraic expressions to mention just a few. In geometry, an isosceles triangle is a triangle that has two sides of equal length. Thank you Triangles. A right triangle with the two legs (and their corresponding angles) equal. Altitude The altitude of a triangle is the perpendicular from the base to the opposite vertex. The third angle, <A, could be acute or obtuse. Here we have on display the majestic isosceles triangle, D U K. select elements base and height base and hypotenuse base and base angle hypotenuse and height hypotenuse and base angle height and base angle area and base area and height area and hypotenuse area and base angle height and vertex angle A triangle's height is the length of a perpendicular line segment originating on a side and intersecting the opposite angle. Use the Pythagorean theorem to calculate the hypotenuse of a right triangle. This worksheet contains problems where students must apply the properties and theorems of isosceles triangles. Move the vertices of the triangle around. One of the properties of such Find information related to equilateral triangles, isosceles triangles, scalene triangles, obtuse triangles, acute triangles, right angle triangles, the hypotenuse, In this lesson students explore and use the properties of isosceles triangles to solve real world problems. The two angles adjacent to the base are called base angles. The shortest distance from a vertex to the opposite side is the altitude to that side. The angles, however, HAVE to all equal 60°. This implies that in every isosceles triangle, the angles opposite to the equal sides are also equal. Students will identify and reason with relationships between lines and angle properties in triangles to develop a system of Properties of Isosceles Triangle ,Triangles - Get topics notes, Online test, Video lectures, Doubts and Solutions for CBSE Class 9 on TopperLearning. (ii) acute angled triangle (iii) isosceles triangle (vi) equilateral triangle (v) 60° Triangle. When an isosceles triangle has exactly two congruent sides, these two sides are the legs. 4. That will only happen in an equilateral triangle. If in a triangle the median has the measure half the length of the side it is drawn to, then the triangle is a right triangle. This isosceles triangle calculator can help in your geometry problems, finding area, height, angles, perimeter or many Properties of the isosceles triangle:. If you are, that knowledge can help you. If you look at the picture below, angles w and x are congruent and angles z and y are congruent. Lesson Notes In Lesson 23, students study two proofs to demonstrate why the base angles of isosceles triangles are congruent. One of the properties of an isosceles triangle is that the height to the base bisects the apex angle. (The base may need to be extended). Triangles, as I'm sure you know, are three sided, 2D shapes. We know thatArea of triangle = 1/2 × Base × HeightHere,Base = BC = bHeight = ADFinding heightNow,In an isosceles triangle,Median & Altitude are the sameSo, D is mid-point of BC∴ BD = DC = b/2Find area of triangle ABCWe know thatArea of triangle = 1/2 × Base × HeightHere,Base = BC = b = 4 cmHeigh Notes 4-9: Isosceles and Equilateral Triangles What is an isosceles triangle? _____ The congruent sides are called the legs. If another triangle can be divided into two right triangles (see Triangle), then the area of the triangle may be able to be determined from the sum of the two constituent right triangles. The base angles of an isosceles triangle are always equal. Discover a relationship between the base angles of an isosceles triangle. Triangle properties. The equal sides of an isosceles triangle are known as the ' legs. Isosceles = at least two equal sides Equilateral = three equal sides What are Isosceles triangles have two equal sides and two equal angles. On the triangle below, classify and appropriately label each side and angle using the terms vertex angle, base angle, leg, base. Theorems include: measures of interior angles of a triangle sum to 180 degrees; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. The Obtuse Triangle has an obtuse angle (an obtuse angle has more than 90°). In the above XYZ, the side XY=XZ and XYZ = XZY, so XYZ is classified as an Isosceles triangle. An isosceles triangle is a triangle with two equal side lengths and two equal angles. What is the value of the angle at the top of this Isosceles triangle? Ans w er . Of course, by itself, the equilateral triangle is not a right triangle, but we can cut it in half and get a right triangle. The longest side of an obtuse triangle is the one opposite the obtuse angle vertex. An isosceles triangle has two sides congruent. There are various types of triangles with unique properties. The interior angle measure of an isosceles trapezoid will equal 360 degrees and the adjacent angles are supplementary. Lesson 23: Base Angles of Isosceles Triangles Student Outcomes Students examine two different proof techniques via a familiar theorem. This type of triangle is also known as an isosceles right triangle, since it’s both isosceles and right. The sum of all three interior angles of a triangle is equal to 180°. Any lower base angle is supplementary to any upper base Isosceles Triangle Theorem. The hypotenuse length for is called Pythagoras's constant. A triangle with all three equal sides is called equilateral. An equilateral triangle has all three sides of equal length and all three angles of equal measure. Sometimes it is specified as having two and only two sides of equal length, and sometimes as having at least two sides of equal length, the latter version thus including the equilateral triangle as a special case. The sides opposite to equal angles of a triangle are also equal. The angles opposite the congruent sides are called the base angles. In ∆XYZ, XY = XZ. Therefore each of the two triangles is isosceles and has a pair of equal angles. In this write-up, we had chance to investigate some interesting properties of the orthocenter of a triangle. The third side is called the base. To solve a triangle means to know all three sides and all three angles. For example, the sum of all interior angles of a right triangle is equal to 180°. the triangle as a function of h, where h denotes the height of the triangle. No homework assignment given. equilateral triangle – isosceles triangle with all three sides are congruent. By Reflexive Property , The converse of the Isosceles Triangle Theorem is also true. In an equilateral triangle, all the three sides and three angles will be equal and each angle will measure 60 °. The chart below shows an example of each type of triangle when it is classified by its sides and angles. Calculates the other elements of an isosceles triangle from the selected elements. The congruent sides of the isosceles triangle are called the legs. And the corresponding angles of the equal sides will be equal. 1) 7 x 7 2) 6 x 6 3) 6 x 6 4) 4 x 4 5) 40° x 70 ° 6) x 75° 75 ° 7) 54° x 72 ° 8) x 75° 30 ° 9) 65° x 80 ° 10) 28° x 56 °-1- The sum of the measures of the angles of a triangle is 180. Therefore, ∠XYZ = ∠ XZY Calculator to find sides, perimeter, semiperimeter, area and altitudes of Isosceles Triangles. In our calculations for a right triangle we only consider 2 known sides to calculate the other 7 unknowns. an isosceles triangle. A triangle is an important basic geometry shape. No, matter where the apex or the peak points, it is still going to be an isosceles triangle. Three equal sides Three equal angles, always 60° Isosceles Triangle . One such property is the sum of any two sides of a triangle is always greater than the third side of the triangle. Properties of the Triangles: The sum of the three angles of the a triangle is two right angles (i. Thus, we see that angle BDE must be the angle x in the problem, which must be 180-(50+50) = 80. In this post, we will discuss the isosceles triangle formula and its area and the perimeter. The page presents a triangle where the user can drag any vertex. 20 If r 1 = r + r 2 + r 3 then prove that the triangle is a right angled triangle. The calculator has been provided with educational purposes in mind and should be used accordingly. 4) Every median is also an altitude and a bisector. A right triangle is a type of isosceles triangle. Properties of an Isosceles Triangle Definition : A triangle is isosceles if two of its sides are equal. " The answer from the key is A(h) = (piR^2) - (h times the square root of (2Rh - h^2)). 5 Properties of Trapezoids and Kites 399 Using Properties of Isosceles Trapezoids The stone above the arch in the diagram is an isosceles trapezoid. ” HOMEWORK: Lesson 4. Let M denote the midpoint of BC (i. like a cross has at least two congruent sides. All three sides are congruent. i have no idea what consecutive means, sorry. Hope you like them and do not forget to like , social share and comment at the end of the page. Just Example 1: Use the Distance Formula to find the distance between the points with coordinates (−3, 4) and (5, 2). Then there's the awkward cousin to equilateral and isosceles triangles: the scalene triangle. It therefore also has #color(blue)(" two equal sides. Definition: A triangle is isosceles if two of its sides are equal. A triangle is a closed figure made of three line segments. How to check whether a triangle is equilateral, scalene or isosceles triangle in C programming. ” o Dividing an isosceles triangle from the vertex between the equal angles results in two equal halves. When the 3rd angle is a right angle, it is called a "right isosceles triangle". To Construct a Triangle when Two of its Sides and the included Angles are given. In the case of an isosceles triangle, the median and the perpendicular is the same, when drawn from the vertex which joins the two equal sides. Find m∠K, m∠M, and m∠J. a right isosceles triangle 2. Property 1: In an isosceles triangle the notable lines: Median, Angle Bisector, Altitude and Perpendicular Bisector that are drawn towards the side of the BASE are equal in segment and length . having two sides of equal length 2. The Isosceles Triangle Properties An isosceles triangle is a triangle that has at least two congruent sides. Based on the length of the sides, triangles are classified into three types: Scalene Triangle; Isosceles Triangle; Equilateral Triangle Equilateral Triangle . The hash mark in the figure denotes the congruency. 3 VOCABULARY TIP Isos- means “equal,” and -sceles means “leg. I would like to know how the answer was derived, given that we only know that R is a constant, and we know the properties of an isosceles triangle. Students complete proofs involving properties of an isosceles triangle. The altitude is a perpendicular distance from the PROPERTIES OF ISOSCELES TRIANGLE ABC. a) Medians of a triangle: A line drawn from any vertex to the mid point of its opposite side is called a median with respect to that vertex. You can draw one yourself, using D U K as a model. All three angles are congruent. Please enter two properties of the isosceles triangle. Of course, the main property of isosceles triangles is their two congruent sides. all of them have there own properties, some that can be figured out by their name, some that are not so easy to figure out. Triangle. The two angles opposite to the equal sides are congruent to each other. An isosceles right triangle therefore has angles of , , and . The answer is 80˚. Converse also true: If two angles of a triangle are congruent, then the sides opposite those angles are congruent. The next day we did isosceles and equilateral triangles. Basic Properties. The isosceles triangle is an important triangle within the classification of triangles, so we will see the most used properties that apply in this geometric figure. e. 2, If all three angles of ABC are acute, the circumcenter will lie inside the triangle. The third side of an isosceles triangle which is unequal to the other two sides is called the base of the isosceles triangle. Notice that the opposite of vertex A is side a, opposite to vertex B is side B, and opposite to vertex C is side c. Use symbols: a, b, h, T, p, A, B, C, r, R. Scalene Triangles Isosceles Triangle Formulas An Isosceles triangle has two equal sides with the angles opposite to them equal. · Characteristics of an isosceles triangles: o Isosceles Triangle Theorem (ITT) states: “The angles opposite the equal sides are equal. isosceles triangle definition: 1. Def. On a diagram, equal sides of a triangle have one small line or dash drawn on each side. An isosceles triangle therefore has both two equal sides and two equal angles. a) Triangle ABM This property is equivalent to two angles of the triangle being equal. Identifying the vertex angle of an isosceles triangle is important. If just two of a triangle’s sides are congruent, then it is called an isosceles triangle. Properties of an isosceles triangle (1) two sides are equal (2) Corresponding angles opposite to these sides are equal. Example 2: A triangle has vertices A(12,5), B(5,3), and C(12, 1). The properties of the isosceles trapezoid are as follows: The properties of trapezoid apply by definition (parallel bases). Write a C program to input sides of a triangle and check whether a triangle is equilateral, scalene or isosceles triangle using if else. Students will be working with a partner to complete questions and short written paragraph. Isosceles triangle properties are used in many proofs and problems where the student must realize that, for example, an altitude is also a median or an angle bisector to find a missing side or angle. isosceles synonyms, isosceles pronunciation, isosceles translation, English dictionary definition of isosceles. Student Help H J K Now, by angle sum property, ∠P + ∠Q + ∠R = 180° x + 90° + x = 180° 90° + 2x = 180° 2x = 180° − 90° 2x = 90° x = (90°)/2 x = 45° Find angle x In ∆XYZ, XY = YZ (Given) Therefore, ∠Z = ∠X (Angles opposite to equal sides are equal) x = ∠X ∠X = x Now, by angle sum property, Isosceles Triangle. Free PDF download of NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and Its Properties solved by Expert Teachers as per NCERT (CBSE) Book guidelines. Jump on to the statements directly. This Demonstration shows that in an isosceles tetrahedron, the three plane angles at each vertex sum to . If one sides of the triangle is produced , the exterior angle formed is equal to the sum The internal bisectors of ∠Q and ∠R of δPQR intersect at A. An equilateral triangle can be considered a special case of isosceles triangle, having all three sides equal. Isosceles triangle, 70 degrees of a rectangle Plane of symmetry Prime factors Prime numbers Prisms Probability of a single event Properties of 3D shapes Students struggled with it for the rest of the year. Isosceles Triangle –at least two sides have the same length. Theorem: Let ABC be an isosceles triangle with AB = AC. Some other properties of triangle. Equilateral Triangle. 20 as a triangle with two equal sides. This means triangle ABC must be an isosceles triangle such that AB = AC. I had noticed that whenever I built a fire, Ajor outlined in the air before her with a forefinger an isosceles triangle, and that she did the same in the morning when . The unequal side length of an isosceles triangle is called the base. An isosceles triangle with three equal sides is called an equilateral triangle. a triangle with two sides of equal length 2. Fold your triangle so that the two legs coincide. When it comes to angles: acute triangles (all angles are acute), right triangles (one right angle), obtuse triangles (one obtuse angle), and equiangular triangles (you guessed it; have all equal angles). The bases in an isosceles triangle are parallel, and the legs are congruent. We have XY=XZ. Here we will solve some numerical problems on the properties of isosceles triangles Find x° from the given figures. Properties of Obtuse Triangles. A triangle may be classified An isosceles triangle may be right, obtuse, or acute (see below). . Find the supplementary of the largest angle. 1. Solution: We are given a triangle ABC but we don’t know what kind of a triangle it is. Moreover, triangle centers, one from each similar triangle, form the vertices of a centric triangle which has special properties. If the three angles measure 60 then it is an equilateral triangle. Note that the inradius is 1 3 the length of an altitude, because each altitude is also a median of the triangle. A B C THEOREM 4. Equilateral triangle properties: 1) All sides are equal. This is called as isosceles triangle theorem. Sometimes you will need to draw an isosceles triangle given limited information. , M is the point on BC for which MB = MC). Properties of a triangle 1. A right triangle has all the properties of a general triangle. In the triangle on the left, the side corresponding to 1 has been multiplied by 6. The two angles touching the base (which are congruent, or equal) are called base angles. An isosceles tetrahedron is also called a disphenoid. REMYA S 13003014 MATHEMATICS MTTC PATHANAPURAM 3. Properties of Isosceles Triangles by – Marco A. In an equilateral triangle, all sides are congruent AND all angles are congruent. In Δ ABCSides: AB, BC and CAAngles: ∠BAC, ∠ABC and ∠BCAVertices: A, B and CThe side opposite to the vertex A is BC. The angle bisector is also a median. ' Think about yourself on a scale. The interior angles of a triangle are given as 2x + 5, 6x and 3x – 23. Its three angles are also equal and they are each 60º. 2 Properties of Isosceles Triangles Recall from Chapter 1 that an isosceles triangle is a triangle with at least two congruent sides. The each side is 60 degree. A number of properties apply only to specific kinds of triangles. So, the altitude divides the base into two equal segments. The sum of all the three angles of a triangle is 180°. Isosceles Triangle: An isosceles triangle has two sides that are equal in length, called legs and the third side is known as base. In maths exam papers there are two or three 2. The Triangle and its PropertiesTriangle is a simple closed curve made of three linesegments. Properties of right triangles By the definition, a right triangle is a triangle which has the right angle. Isosceles triangle calculator. Snapshots. A triangle having two equal sides is called an isosceles triangle. Let ABC be an isosceles triangle with <B=<C, and <A at the top. Gonzalez Activity overview In this activity and by using the Nspire handhelds, students will discover the different properties and attributes of Isosceles Triangles. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. (as DE is BD Types of Triangles - Cool Math has free online cool math lessons, cool math games and fun math activities. 3. supplementary: two angles that together add up to 180 degrees. Isosceles trapezoids are dissected into three similar triangles and re-arranged to form two additional isosceles trapezoids. All isosceles right triangles are similar since corresponding angles in isosceles right triangles are equal. The diagonals divide each other in the same ratio. Students must use the Isosceles Triangle Theorem to find missing values in triangles and to complete two-column proofs. Properties of isosceles triangle ABC 1. 2) Angles of every equilateral triangle are equal to 60° 3) Every altitude is also a median and a bisector. You'll also learn the theorem of isosceles triangles. Label the properties of quadrilateral ACA'C'. An isosceles triangle also has two angles of the same measure, namely the angles opposite to the two sides of the same length; this fact is the content of the isosceles triangle theorem, which was known by Euclid. List the conjectures about the relationships you observe in the angles of your isosceles If all sides are equal, the triangle is called equilateral. So if we have three equal angles we are going to take that and divide it by three. An equilateral triangle is always isosceles. Two equal sides Two equal angles Scalene Triangle . Therefore, the distance from the altitude to the base vertex of the isosceles triangle is half of the length of the base length. Logic to classify triangles as equilateral, scalene or isosceles triangle if sides are given in C program. Isosceles Triangle; Properties; Isosceles Triangle Theorem; Converse; Converse Proof; Isosceles Triangle. The parallel sides are called bases while the nonparallel sides are called legs. isosceles isosceles triangle adj. Define isosceles. Statement 1: The bisector of ?A is a median in triangle ABC. The angle sum property of a triangle: The total measure of the three angles of a triangle is 180°. Students will investigate the properties of isosceles triangles. Properties Of Triangle 1. The name Sal proves that the base angles in isosceles triangles are congruent, and conversely, that triangles with congruent Theorems concerning triangle properties. In the given isosceles triangle, if AB = AC then ∠B = ∠C The two interior angles that are opposite these sides are equal to each other. Here we give an interesting property of isosceles triangles. Isosceles triangle is a triangle whereTwo sides are equalAngles opposite to equal sides are equalHere,a is the side which is equal, AB = ACb is the base, BC = bAngles opposite to equal sides are equal, ∠B = ∠C (Proof - link)PerimeterPerimeter of isosceles triangle = Sum of all sides= a + a + b= 2a + If all sides are equal, the triangle is called equilateral. Thes properties result in shortcuts that make it easy to find unknown measures of parts of an isosceles right triangles A triangle having three unequal sides is called a scalene triangle. isosceles: two sides are equal. The hypotenuse is the side of the triangle opposite the right angle. In an isosceles triangle, the altitude to the base bisects the base and the vertex angle. But we have no idea about the measure of angle A. When the unequal angle of an isosceles triangle is 90 degrees, it is referred to as a right isosceles triangle. Isosceles Triangles. Free practice questions for ACT Math - How to find the area of an acute / obtuse isosceles triangle. an acute scalene triangle 4. In this math lesson plan we will discuss triangles and their properties. An isosceles triangle has two sides that are congruent. 120 degrees - 10 degrees - 50 degrees is a scalene triangle since all the angle measures are different. The remaining side is called a base. The legs are congruent by definition. Let us discuss further how to calculate the area, perimeter, and the altitude of an isosceles triangle. Thank you for the help. ELIZABETH STUART PHELPS L E S S O N4. The Pythagorean theorem can only be used with isosceles triangles that are right triangles. adj 1. 4,6. Triangle is a basic shape which has several properties based on its sides, interior angles and exterior angles. If you know the side lengths, base, and altitude, it is possible to do this with just a ruler and compass (or just a compass, if you are given line segments). Isosceles trapezoids have some additional properties: Pythagoras Theorem applied to triangles with whole-number sides such as the 3-4-5 triangle. and perpeindicular are two lines that meet at a point to make a right angle. Trying to find a missing interior angle measurement in a triangle? See if you're working with a special type of triangle such as an equilateral or isosceles triangle. The other side is called the base and the angles between The conjectures focus on triangle congruence and on the special case of isosceles triangles, with additional study of sums of the measures of interior angles of a triangle. An isosceles triangle has two sides of equal length and the angles opposite those sides are of equal measure. In the figure AB = AC, so triangle ABC is isosceles. Illustrated definition of Isosceles Triangle: A triangle with two equal sides. Properties of a square; 4. Sal proves that the base angles in isosceles triangles are congruent, and conversely, that triangles with congruent base angles are isosceles. Properties of basic quadrilaterals; 10. Using rulers and protractors, students will sketch a triangle that should be an isosceles triangle. 1,6. It implies that two sides - legs - are equal in length and the hypotenuse can be easily calculated. com 2. However, some properties are applicable to all triangles. But in every isosceles right triangle, the sides are in the ratio 1 : 1 : , as shown on the right. comBy iTutor. The other interesting properties of the 45 45 90 triangles are: It's the only possible right triangle that is also an isosceles triangle This interactive web-based resource demonstrates isosceles triangles (two sides with the same length). The relationship between the lateral side $$a$$, the based $$b$$ of the isosceles triangle, its area A, height h, inscribed and circumscribed radii r and R respectively are give by: Problems with Solutions Problem 1 A triangle with two sides of equal length is an isosceles triangle. The triangle on the left is scalene because it has three different angles. For an isosceles right triangle with side lengths , the hypotenuse has length , and the area is . Antonyms for isosceles. The following properties of triangles shall make the concept more clear to you: 1. We aren't gullible enough to believe everything we hear. and. Equilateral triangle. GCSE Maths section looking at the Properties of Triangles and Quadrilaterals the angle sum of a triangle and a quadrilateral and identifying quadrilaterals by their geometric properties. We can now apply the Transitive Property to show that ?TUS and ?TSU are congruent. If two of its sides are equal, a triangle is called isosceles. The Pythagorean Theorem. A triangle with equal side ℓ and isosceles angle θ has area ℓ2sin2θ 2, as the vertex angle has measure 180∘−2θ, and sin (180∘−2θ)=sin2θ. Introduction : Isosceles triangle is one of the important shapes in geometry. Triangle Angle-Sum Theorem The sum of the measures of the angles of a triangle is 180. Includes full solutions and score reporting. Properties, convex, cyclic. Any lower base angle is supplementary to any upper base The properties of the isosceles trapezoid are as follows: The properties of trapezoid apply by definition (parallel bases). Notes: ISOSCELES AND EQUILATERAL TRIANGLES Geometry Unit 4 – Relationships w/in Triangles Page 229 TERM DEFINITION EXAMPLE ISOSCELES TRIANGLE A triangle with at least one pair of _____ sides. Equi- refers to things that are the “same” or “equal”, and lateral means “sided. If any two angles and a sides of one Identifying isosceles triangles. In case of isosceles tetrahedron, the associated parallelepipied is a cuboid, parallelepapiped with orthogonal faces and edges. Here are online calculators, generators and finders with methods to generate the triples, to investigate the patterns and properties of these integer sided right angled triangles. Some properties of a trapezoid: Two adjacent angles of a trapezoid are supplementary. Like an isosceles triangle, isosceles trapezoids have base angles that are congruent. If we have a quadrilateral where one pair and only one pair of sides are parallel then we have what is called a trapezoid. All The Triangle and Its Properties Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. When the altitude to the base of an isosceles triangle is drawn, two congruent triangles are formed, proven by Hypotenuse - Leg. Some isosceles triangles can be equilateral if all three sides are congruent. F is midpoint of AC. Thus CF is half CB which is half CA. the isosceles trapezoid (like an isosceles triangle) Properties of the sides of an isosceles An isosceles triangle is defined in I. The median is parallel to the bases. (These are the angles that are adjacent to the base. A Special Triangle & Its Properties (I) Activity. In an isosceles triangle XYZ, two sides of the triangle are equal. General properties. Triangle Facts for Kids. . 7 Use Isosceles and Equilateral Triangles THEOREMS For When does Triangle ABC obtain its maximum area? What is the relationship between its sides? What is the relationship between the base and height? Construct a formal or informal proof as to when the maximum area of an isosceles triangle is obtained. Properties of triangles. Given 2 unequal known sides you can find the unknowns of the Oct 16, 2015 Types of triangles: properties of isosceles, scalene and right angled triangles. Schwartzman's The Words of Mathematics explain the etymology (the origins) of the words. an obtuse scalene triangle 7. In Euclidean geometry, the isosceles triangle theorem states that the angles opposite to the two equal sides of an isosceles triangle are equal. a right scalene triangle 3. In isosceles (and equilateral) triangles, a segment drawn from the vertex angle to the opposite side is the altitude, angle bisector and median. The third side is the base of the isosceles triangle. That being said, you still want to get those questions right, so you should be prepared to know every kind of triangle: right triangles, isosceles triangles, isosceles right triangles—the SAT could test you on any one of them. 2/1-10. The altitude is a perpendicular distance from the In this article, we will state two theorems regarding the properties of isosceles triangles and discuss their proofs. In general, it is necessary to know the properties of the isosceles triangle. Each equal angle in a right isosceles triangle is equal to 45 degrees. Draw and cut out an isosceles triangle. A triangle is said to be equilateral if each one of its sides is of the same length and each of one its angles measures. The properties of different kinds of triangles are listed below. Like every triangle a scalene has three vertices. We will practice using these theorems to help us solve the following exercises. We’ve learned that you can classify triangles in different ways. If AB BC, then May 12, 2013 This post contains the proof of the isosceles triangle theorem. Properties of a kite; 9. An isosceles triangle is a triangle with two congruent sides. Properties of a rhombus; 7. ) Area of a Triangle Worksheets. The median's length is half the sum of the bases. In an equilateral triangle, like S U N below, each height is the line segment that splits a side in half and is also an angle bisector of the opposite angle. The smallest angle is across from the smallest side The sum of the length of any two sides of a triangle is greate… A triangle that has 2 equal sides. Properties of Isosceles and Equilateral Triangles Mark the triangles to reflect the properties discovered. Isosceles Triangle: Theorems. The geometry lesson plan on triangles will also talk about the procedure for teaching triangles in the classroom. an obtuse isosceles triangle . The altitude to the base of an isosceles triangle bisects the base. Mathematics Having at least two equal sides: an isosceles triangle. By the Distance Formula, Because AB = BC, triangle ABC is isosceles. ")# What's so special about these angles, though? Well, one of the most important properties of triangles is the fact that the sum of the three inner angles is always 180°. In addition, by Reflexive Property, \overline{BD} \cong \overline{BD} . The diagonals are congruent. In ∆XYZ, XY ≠ YZ ≠ ZX Is this isosceles triangle or not? 2. The base angles of an isosceles triangle Reflexive Property (A quantity is congruent to itself. Every tetrahedron is associated with (inscribable into) a parallelepiped. An equilateral triangle has all three sides congruent. The altitude creates the needed right triangles, the congruent legs of the triangle become the congruent hypotenuses, and the altitude becomes the shared leg, satisfying HL. Basic Properties. Classify the quadrilateral ACA'C', formed from the composite reflections of an isosceles right triangle. Thus A is on m. This is a 90-60-30 triangle. An isosceles triangle also has two angles of the same measure; namely, the angles opposite to the two sides of the same length; this fact is the content of the Isosceles triangle theorem. Geo-Activity Properties of Isosceles Triangles Base Angles Theorem Words If two sides of a triangle are congruent, then the angles opposite them are congruent. The difference between the lengths of any two sides is smaller than the length of the third side. DAILY LEARNING GOAL: I can identify the parts and properties of isosceles triangles. If two sides of a triangle are congruent, then the angles opposite those sides are congruent. all sides are of equal length What are the basic properties of a 45-45-90 triangle? so this is an isosceles triangle. Classification of triangles based on angles. Tim Brzezinski. ACA'C' is a square. an acute equilateral triangle 6. Finally, by the Isosceles Triangle Theorem, we know that the sides Proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal Equilateral triangle properties: 1) All sides are equal. In addition, all isosceles triangles also have congruent base angles. Isosceles triangles In this lesson, you'll learn how an isosceles triangle's sides and angles make it unique. All equilateral triangles are similar to each other, and have 60 degree with internal angles. The triangle on the right is NOT scalene because it has two angles of the same size. Note that this is 2 3 the length of an altitude, because each altitude is also a median of the triangle. Scalene means unequal, but it doesn't sound like unequal. Similarly, if 2 angles of a triangle are congruent, then their opposite sides are congruent. Let’s start with an equilateral triangle, another magically symmetrical shape. Any lower base angle is supplementary to any upper base angle. an acute isosceles triangle 5. It sounds like 'scale. a triangle that has two sides of equal length and…. Properties of triangles: Classification of triangles by sides. 1) What do you notice about line segments AB and AC? *A right triangle can be a scalene triangle, but an isosceles triangle cannot because it has two equal sides. Take a look! Tetrahedron is isosceles if its opposite edges are pairwise equal. Isosceles and Equilateral Triangles Date_____ Period____ Find the value of x. Isosceles Triangle Equations Formulas Calculator - Semiperimeter Geometry AJ Design Geometry calculator for solving the inscribed circle radius of a isosceles triangle given the length of sides a and b. 5) Every bisector is also an altitude and a median. A triangle having three equal sides is called an equilateral triangle. Properties of a parallelogram; 6. Base Angle Converse (Isosceles Triangle) Section 7. So let’s look at the rules. Consider isosceles triangle ABC &n. Isosceles trapezoids are special types of trapezoids that have the pair of of non-parallel legs being congruent to each other. Students using Anglegs to explore properties of isosceles triangles (angle bisectors, perpendicular bisectors, etc). And the corresponding angles of the equal sides will be We know that a triangle adds up to 180 degrees. In an isosceles triangle, the angle between the two congruent Define isosceles triangle. The 30-60-90 Triangle. A scalene triangle is a triangle with no equal sides or angles. Properties of isosceles triangle inscribed in a circle, Hint: use Pythagoras' theorem twice, then eliminate CE between the following to find r: OE2+CE2=r2(OE+r)2+CE2=AC2. An Isosceles right triangle is a special triangle with several special properties. This calculator calculates any isosceles triangle specified by two of its properties. 5. Angle BAM = angle BAC and angle DAM = angle DAC (same rays) properties of ∆ACA' apply to the reflected triangle. Can you find an isosceles triangle whose base is an integer, whose perimeter is the cube of an integer, and whose area is the cube of an integer? Is the isosceles triangle with these properties unique? Free Isosceles Triangle Area & Perimeter Calculator - Calculate area, perimeter of an isosceles triangle step-by-step 4. Therefore, a net of an isosceles tetrahedron is a triangle similar to the faces of the solid, and the sides of the net are twice the sides of the faces. In an isosceles triangle, if the side length is a, then the base of the triangle (or the hypotenuse) is equal to sqrt(2) a. Property of the lengths of sides of a triangle: The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Well, some of these types of triangles have special properties! Isosceles Triangle. What are the different types of triangle? The sum of angles in any triangle is 180°. The angle opposite the base is called the vertex angle. The two sides opposite the base angles are congruent. Make certain the triangle is a right triangle. A triangle with no two of its sides congruent is called a scalene triangle and is shown below. To Construct a Triangle whose Three Sides are given. An obtuse triangle may be either isosceles (two equal sides and two equal angles) or scalene (no equal sides or angles). Is the triangle possible, if sides of the triangle are 5 cm,12cm and 6cm? A triangle is scalene if all of its three sides are different (in which case, the three angles are also different). Theorem 1: Angles opposite to the equal sides of an isosceles triangle are also equal. Isosceles Triangle An isosceles triangle is a triangle with two equal sizes. After learning the basics of isosceles triangles they Isosceles triangle definition is - a triangle in which two sides have the same length. Therefore, the two base angles measure 64°. ” So, isosceles means equal legs. Â In an isosceles triangle, the angles opposite the equal sides are equal. Every triangle has three sides, three angles, and three vertices. Triangle has three vertices, three sides and three angles. Properties of a rectangle; 13. One hundred and eighty divided by three is equal to sixty. (Back to triangle at top) i. Exercise 1 An isosceles triangle is a triangle with at least two sides of the same length. Worksheet on Triangle. 13 13 10 - Acute isosceles triangle, area=60. In a right triangle, the median drawn to the hypotenuse divides the triangle in two isosceles triangles. Hash marks show sides ∠ D U ≅ ∠ D K, which is your tip-off that you have an isosceles triangle. Proof: Consider an isosceles triangle ABC where AC = BC. There are 4 different types, as illustrated on the left; there is an equilateral, an isosceles, a right angled and a scalene triangle. In an isosceles triangle, the altitude to the base is the perpendicular bisector of the base; it is also the angle bisector of the vertex angle, line of symmetry of the triangle, and the median from the apex to the base as well. These two sides are called legs. The first two The triangle ABD is isosceles. Then, also recall that the altitude of an isosceles triangle is the median of the base. Find information related to equilateral triangles, isosceles triangles, scalene triangles, obtuse triangles, acute triangles, right angle triangles, the hypotenuse, angles of a triangle and more. By the symmetry properties of the isosceles triangle, the line AM is the perpendicular bisector of BD = m. An isosceles triangle has two equal sides (or three, technically) and two equal angles (or three, technically). PQR is an isosceles,inscribed in a circle with centre O, such that PQ=PR=13cm and QR=10cm. A triangle has three sides and is made of straight lines. Concepts Isosceles Triangles, Vertex Angle, Base Angles, legs & base of the Isosceles Triangle Teacher preparation Properties of Isosceles Triangle - Get Get topics notes, Online test, Video lectures & Doubts and Solutions for ICSE Class 9 Mathematics on TopperLearning The Triangle and its Properties 3 21. Classification of Triangles by Sides PROPERTIES OF ISOSCELES TRIANGLES If 2 sides of a triangle are congruent, then their opposite angles are congruent. I purchased the isosceles and equilateral triangles from Lisa Davenport on TPT. 18 = 108 0 A triangle which has one angle that measures more than 90 0 is an obtuse angle triangle. Besondere Punkte/Special Points. Since there are three possible bases, there are also three possible altitudes. Thus triangle ACE is isosceles, thus angle CAE = 20. A 45-45-90 triangle is a triangle with two 45º angles and one right angle. List the properties below each triangle. 9 – Isosceles Triangle Theorem If 2 sides of a triangle are congruent, then the angles opposite those sides are congruent. Along with more informal methods, rigid motions and their properties are introduced as ways to establish triangle congruence criteria. An equilateral triangle is equiangular. The above is an isosceles triangle. ). The area of an equilateral is (√3)/4 × s 2, where s is the side length. · If two shapes are same in size, they’re called congruent. Included Angle or Vertex Angle Included angle is the angle subtended by two sides at the vertex of the triangle. A triangle with angles 45, 45, 90 degrees is called as an isosceles right triangle. An Isosceles triangle has got two sides of equal length and 2 angles equal. These special properties of the isosceles triangle allow you to calculate the area from just Sep 9, 2014 Introduction. Here ∠P = ∠Q = ∠PRS. A triangle with base b and isosceles angle θ has area b2tanθ 4, as the length of the altitude to the base is btanθ 2. The circumradius of an equilateral triangle is s3√ 3. select elements base and height base and hypotenuse base and base angle hypotenuse and height hypotenuse and base angle height and base angle area and base area and height area and hypotenuse area and base angle height and vertex angle Angles of the two equal sides of an isosceles triangle are the same or equal. We can split the triangle in two by drawing a line from the centre of the circle to the point on the circumference our triangle touches. Therefore, each of these angles have to measure 60 degrees. base b and a arm a. 2 Triangle Sum Properties & Properties of Isosceles Triangles-Classify triangles and find measures of their angles. Examples of Properties of Triangle. Not that they had trouble applying it, but they had trouble identifying when they could apply it. The two equal sides of an isosceles triangle are known as ‘legs’ whereas the third or unequal side is known as the ‘base’. Angle bisectors, perpendicular bisectors, midpoints, and medians are also examined in this lesson. The Triangle and Its Properties. A side that is not equal to the other sides is called the base of the triangle. 1/1-9 . Classification of Triangle. This endows isosceles An equilateral triangle is also a special isosceles triangle. Properties of a trapezium; 8. If one sides of the triangle is produced , the exterior angle formed is equal to the sum of the interior opposite angles. Thus provides the calculation of all parameters of the triangle if you enter two of its parameters eg. Which is the largest side of a right angle triangle? (A) Median (B) Hypotenuse (C) Arm (D) Altitude 22. The equal sides are called legs, and the third side is the base. The angles in a triangle add up to . Share a scenario where this would be applicable in real life. Zoltán Kovács. The line joining one vertex of a triangle to the midpoint of the opposite side is called a median of the triangle. Alternatively, if the apex angle has measure α, the area is b2 4cotα 2. 3,6. I can sketch a right equilateral triangle. S. Find the radius of the circle. No equal sides No equal angles Triangles and Trigonometry Properties of Triangles Reveal All Steps Let’s start simple: a triangle is a closed shape that has three sides (which are line segments ) and three vertices (the points where the sides meet). 2,6. All equilateral triangles are also isosceles triangles since every equilateral triangle has at least two of its sides congruent. Solution: The interior angles of a triangle sum up to : 180 0 (2x + 5) + (6x) + (3x – 23) = 18011x – 18 = 18011x = 198x = 18 0 The largest angle is 6x = 6 . Then. The Triangle and its Properties. The inradius of an equilateral triangle is s3√ 6. An Isosceles Triangle has the following properties: Two sides are congruent to each other. Base Angle Theorem (Isosceles Triangle) If two sides of a triangle are congruent, the angles opposite these sides are congruent. Learn more. In geometry, an isosceles triangle is a triangle that haes twa sides o equal lenth. As a consequence, all angle are congruent, and they each measure 60 degrees. If each angle is less than 90°, then the triangle is called an acute-angled triangle. The altitude is a perpendicular distance from the Properties The unequal side of an isosceles triangle is usually referred to as the 'base' of the triangle. An equilateral triangle is equiangular, so each angle would have to measure 60° because there are 180° in a triangle. 6) If the length of a side is a the area of the equilateral triangle is ¼a 2 √ 3 The properties of the isosceles trapezoid are as follows: The properties of trapezoid apply by definition (parallel bases). The properties of parallelograms can be applied on rhombi. Because of the highly symmetrical properties of the equilateral triangle, the segment AC (a) forms a right angle at the base, (b) bisects the angle at A, and (c) bisects the base BD. AB and AC BASE ANGLES An isosceles triangle is a special case of a triangle where 2 sides, a and c, are equal and 2 angles, A and C, are equal. This means that the trapezoid appears symmetrical, and that the diagonals are equal in length. Children learn to classify triangles as equilateral, isosceles, scalene or right-angled in KS2 geometry. The sides can measure anything as long as they are all the same. Topic: Isosceles Triangle Theorems - Worksheet 4 1. In the above triangle sides AC and BC are equal and therefore angles A and B are also equal. First of all, let’s review the definition of the orthocenter of a triangle. Every equilateral triangle is also an isosceles triangle, but not every isosceles triangle is an equilateral triangle. An isosceles trapezoid is a trapezoid whose legs are congruent. “The orthocenter of a triangle is the point at which the three altitudes of the triangle meet. Therefore, an equilateral triangle is also an equiangular triangle. When all angles are congruent, it is called equiangular. As a vertex is dragged, the others move automatically to keep the triangle isosceles. In an isosceles triangle, the base is usually taken to be the unequal side. properties of isosceles triangle
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https://greasaggedet.firebaseapp.com/455.html | # Find taylor series of sinx centered at pi
How to evaluate sinxx using a taylor series expansion quora. If we wish to calculate the taylor series at any other value of x, we can consider a variety of approaches. By using this website, you agree to our cookie policy. An applied approach mindtap course list you have been observing the local democratic party in a large city and have compiled some. Taylor series of sinx at pi2 math, calculus, taylor series showme. This could be its value at mathx 0math as is considered a popular interview questions, i. In this section we will discuss how to find the taylormaclaurin series for a function. This is very useful information about the function sinx but it doesnt tell the whole story. Find the taylor series for 4442172 find the taylor series for fx, centered at the given value of a. Finding the domain and range of a function in exercises 16, find the domain and range of the function. Find the taylor series expansion for fx cos x, at x pi4.
Find the taylor series for fx centered at the given value. Operations on power series related to taylor series in this problem, we perform elementary operations on taylor series term by term di. Find the taylor series for fx, centered at the given value of a. A calculator for finding the expansion and form of the taylor series of a given function. We can then differentiate this 3 times in order to find fx, fx and fx. Find the taylor series for fx centered at the given. The maclaurin series is just a taylor series centered at a 0. Apr 25, 2011 find the taylor series of 1 x centered at c 1. Apr 03, 2014 observe that the first term has even powers of x pi 3, while the second term has odd powers of x pi 3. Find the taylor series for fx, centered at the given. Find the taylor series for fx centered at the given value of a and its radius of convergence do not need to show that the limit of the remainders is 0.
If there is no c in the series representation, then the function is centered at 0. Taylor and maclaurin power series calculator emathhelp. Free taylor series calculator find the taylor series representation of functions stepbystep. Operations on power series related to taylor series. Jul 20, 2005 hi, im trying to get the taylor series for f x sin x, centered at xpi2, but i am seem to be getting an incorrect taylor series, any help. Answer to find the taylor series for fx sinx centered at c pi 2. The period of sinx is 2 how is this series related to the number 1. We also derive some well known formulas for taylor series of ex, cosx and sinx around x0. This will require use of the chain rule and the product rule. Thus both series are absolutely convergent for all x.
Generally speaking, the interval of convergence for the representing taylor series may be dierent from the domain of the function. For both series, the ratio of the nth to the n1th term tends to zero for all x. Apr 24, 2016 then use this series and taylor series for sin x to find the quartic taylor polynomial about 0 for the function f x asked by jay on april 24, 2016. Many properties of the cosine and sine functions can easily be derived from these expansions, such as. If you are unsure of an answer, you can use a graphing utility to compare the original function to the rst few. In this video, i show how to find the power series representation for sinx using a taylor maclaurin series expansion. This article shows how to derive the maclaurin expansion series for sin.
So, the n 0 term with n being even comes from only the first term. Remember, the maclaurin series is just the taylor series centered at zero. Find the taylor series for eq\sin x eq centered at eq\pi. Answer to find the taylor series centered at a pi 6 for f x sin x. Answer to find the taylor series for fx sinx centered at c pi2. Find the taylor series for fx centered at the given value of a. Find the taylor series for fx, centered at the given value. Voiceover lets see if we can find the maclaurin series representation of f of x, where f of x is equal to x to the third times cosine of x squared. Then use this series and taylor series for sin x to find the quartic taylor polynomial about 0 for the function fx asked by jay on april 24, 2016.
What this tells you is that the even numbered derivativesie. Answer to find the taylor series centered at a pi6 for fx sin x. Maclaurin expansion of sinx the infinite series module. Trigonometrypower series for cosine and sine wikibooks. We might choose a taylor series centered at x e rather than at x 1. Stack overflow for teams is a private, secure spot for you and your coworkers to find and share information.
We also derive some well known formulas for taylor series of e x, cos x and sin x around x0. Taylor expansions are very similar to maclaurin expansions because maclaurin series actually are taylor series centered at x 0. Find the taylor series for eq\ sin x eq centered at eq\ pi. Fx sin x, centered at x pi 2 question physics forums. This will work for a much wider variety of function than the method discussed in the previous section at the expense of some often unpleasant work. Have your program read x and the desired accuracy some small positive number. For example, its hard to tell from the formula that sinx is periodic. We need to compute a bunch of derivatives of fx cosx. In this section we will discuss how to find the taylor maclaurin series for a function. Ken bube of the university of washington department of mathematics in the spring, 2005. For easy angles such as 45 degrees pi4, we know exactly what sine is, thanks to the unit circle. Thus, a taylor series is a more generic form of the maclaurin series, and it can be centered at any xvalue. Mathematics stack exchange is a question and answer site for people studying math at any level and professionals in related fields. The calculator will find the taylor or power series expansion of the given function around the given point, with steps shown.
1387 409 376 1097 1598 252 100 1385 367 64 596 613 966 499 245 1401 1329 138 1087 777 950 1381 38 485 996 859 849 356 518 966 293 474 1327 95 285 163 | 2022-10-05T14:36:49 | {
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http://math.stackexchange.com/questions/904093/limit-of-a-sequence-of-averages-three-variables | # Limit of a sequence of averages (three variables)
Let $a_0 = 0$, $a_1 = 0$, $a_2=1$ and for $n>2$, $a_n = \dfrac{a_{n-1}+a_{n-2}+a_{n-3}}{3}$. Consider $\lim\limits_{n \to +\infty} a_n$.
Using a python script I found that $a_n$ tends to $\frac{1}{2}$ as $n \to +\infty$. However I don't know how to prove this...I tried the approach from this question that I asked here a while back, but I couldn't anything like it to work because the differences between terms kept altering between positive and negative in an unpredictable fashion (i.e. -+-++-++-++-+--+--...) and the absolute value of the differences did not always decrease as $n$ increased ($|a_7-a_8| < |a_8-a_9|$). (Source: the python code, which I can include if that helps...)
Another approach I tried was to use induction to try to prove the general case that if the first $k$ terms $a_0,\cdots,a_{k-1}$ are given where $a_0,\cdots,a_{k-2} = 0$ and $a_{k-1} = 1$ and for $n>k-1$,
$a_n = \dfrac{\sum\limits_{i=1}^k a_{n-i}}{k}$
then that $\lim\limits_{n \to +\infty} a_n = \dfrac{2}{k+1}$ (this is just a guess after the first five or six terms), but I also made no progress there.
Could anybody help point me in the right direction? How do I prove this? (The $k=3$ case is enough; proving the general case was just an idea.)
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One way is to solve the recurrence explicitly, by using the characteristic polynomial. – André Nicolas Aug 20 '14 at 14:33
See j.c.'s answer here for a proof and a more general approach! – Silenttiffy Aug 20 '14 at 14:36
There is a way to solve this problem, but it requires time and work. The recurrence relation for this sequence is similar to that of the Fibonnaci sequence, in general, these types of sequences have closed forms in terms of a sum of exponentials of n.
Here is how you can solve the problem: by the linearity of the recurrence relation, if $s_n$ is a sequence which satisfies the relation, then $ks_n$ also does ( $k$ is any value). If also, say $t_n$ satisfies the relation, then the sum $$s_n+t_n$$ also does. Hence, to find the closed form of the sequence, you must find three distinct sequences which satisfy the recurrence relation. Then, your sequence is of the form $$k_1 s_n + k_2t_n + k_3u_n$$ where $s_n$, $t_n$ and $u_n$ are the 3 sequences you found.
To find the 3 sequences, assume they are of exponential form: $$s_n= x^n$$ for some constant $x$. Substituting in the recurrence relation and simplifying, youre left with:
$$3x³=x²+x+1$$ I leave you to solve that ( hint, one solution is easy by inspection, then factor the cubic and solve the final quadratic)
You will find that two of the roots are complex and one is real. Then, find your $k_1$, $k_2$, $k_3$ mentioned before by substituting the the values for $n=0,1,2$
Whew. ( I know it's long) Finally, take the limit as $n \to \infty$. You will find two of the exponentials vanish ( hint: what is their modulus?) and the third is trivial. That will yield your limit to be $\frac{1}{2}$.
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The calculation that you have very nicely described works smoothly and quickly. – André Nicolas Aug 20 '14 at 14:45
Thank you Andre Nicolas. It can get quite tedious sometimes. – Assaultous2 Aug 20 '14 at 14:46
Use a telescoping sum to get $$3a_n+2a_{n-1}+a_{n-2}=3a_2+2a_1+a_0.$$ Now it is easy to find the answer, by taking $n\to\infty$ in the above equation: $${3\cdot 1+2\cdot0+1\cdot0\over3+2+1}=\frac12.$$ Actually, we first have to prove that $\lim_{n\to\infty}a_n$ exists.
With this method, you can also prove the general case.
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$\lim_{n\to\infty}(3a_n+2a_{n-1}+a_{n-2})=6a_{\infty}$. Since $3a_2+2a_1+a_0=3+0+0$, we obtain $a_{\infty}=1/2$. Nice results! – mike Aug 20 '14 at 14:38
@mike How do I show the step of taking the limit more rigorously? – Thornshadow17432 Aug 20 '14 at 14:49
Please take a look at the following MSE question and Answer. math.stackexchange.com/questions/829721/… – mike Aug 20 '14 at 14:53 | 2015-01-29T06:42:00 | {
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https://math.stackexchange.com/questions/2373639/closed-set-mapped-to-itself-in-a-compact-hausdorff-space | # Closed set mapped to itself in a compact Hausdorff space
Let $X$ be a compact Hausdorff space and let $f:X\to X$ be continuous. Prove that there exists a non-empty closed set $C$ such that $f(C) = C$.
Proving by contradiction seems hard to me, since if $f(C)\neq C$ for all closed sets $C$, then we can have either $f(C)\subset C$, $f(C)\cap C\neq\varnothing$ or $f(C)\cap C = \varnothing$. The only thing I might have thought about is to set $C = f(X)$. Since $X$ is compact, $f(X)$ is compact and since $X$ is Hausdorff, $f(X)$ is closed. Then we have that $f(C)\subseteq C$ by definition. I was not able to prove that $C\subseteq f(C)$, namely that $f(X) \subseteq f(f(X))$. It is not obvious for me that in general it is even true, so maybe the choice $C = f(X)$ does not help.
• Your guess will not work. Take $X$ to be the closed unit interval $[0,1]$ and $f$ to be the multiplication with $\frac{1}{2}$. – tiefi Jul 27 '17 at 14:07
• Hint: Take the intersection of all nonempty closed subsets $A\subset X$ such that $f(A)\subset A$. – Moishe Kohan Jul 27 '17 at 14:14
• Hint. Define $K_0 = X$ and $K_{n+1} =f(K_n)$. Then $K_0 \supset K_1 \supset K_2 \supset \cdots$ is a sequence of non-empty nested compact sets. What can you say about the intersection $C=\cap_{n\geq 0} K_n$? – Sangchul Lee Jul 27 '17 at 14:23
• @MoisheCohen, How do you know that the intersection is non-empty? For instance, consider $$X=\{x\in\mathbb{R}^2:1\leq|x|\leq2\}, \qquad f(x) =-x.$$ Then you can create two disjoint closed sets $A_1$, $A_2$ for which $f(A_i)\subseteq A_i$ holds for each $i=1,2$. – Sangchul Lee Jul 27 '17 at 14:29
• @tiefi, For instance, you can take $A_1=X\cap(\mathbb{R}\times\{0\})$ and $A_2=X\cap(\{0\}\times\mathbb{R})$ as the intersection of $X$ with the horizontal axis and the vertical axis, respectively. You can check that in fact these sets are invariant under the antipodal map $f$. – Sangchul Lee Jul 27 '17 at 14:37
Indeed define $K_0 = X$. Then define $K_{n+1} = f[K_n]$ for all $n \ge 0$. By induction all $K_n$ are compact subspaces of $X$ and for all $n$: $K_{n+1} \subseteq K_n$:
Both clearly hold for $n=0$. And if $K_n$ is compact and $K_{n+1} \subseteq K_n$, then $K_{n+1} = f[K_n]$ is compact as the continuous image of a compact set, and $K_{n+2} = f[K_{n+1}] \subseteq f[K_n] = K_{n+1}$, as function images preserve inclusions.
Finally define $K=\cap_n K_n$. (it doesn't mattter whether we start from $n=0$ or $n=1$ as $K_0 = X$ has no effect.) Then $K$ is closed as an intersection of closed (compact in Hausdorff implies closed) sets. It is also non-empty as a decreasing intersection of non-empty closed sets in a compact space.
Clearly $$K = \bigcap_{n\ge 1} K_n = \bigcap_{n \ge 1} f[K_{n-1}] \supseteq f[\bigcap_{n \ge 1} K_{n-1}] = f[\bigcap_{n \ge 0} K_n] = f[K]$$
So $f[K] \subseteq K$. To see that the reverse holds:
Let $x \in K$ then for all $n$, $x \in K_{n+1}$ so that $F_n = K_n \cap f^{-1}[\{x\}]$ is non-empty and all $F_n$ are also closed (Hausdorff implies singleton sets are closed) thus compact. So the sets $F_n$ are also a decreasing family of non-empty closed sets and so $$\cap_n F_n \neq \emptyset$$
And note that a $p \in \cap_n F_n$ has the property that $p \in K$ and $f(p) = x$, so that $K \subseteq f[K]$.
• Any reference for the claim "a decreasing intersection of closed sets in compact space has a non-empty intersection"? – Joshhh Jul 30 '17 at 14:17
• @Joshhh non-empty closed subsets. If the intersection were empty, the complements would be an open cover without a finite subcover, contradicting compactness. – Henno Brandsma Jul 30 '17 at 14:51
The answer to your question is given in the comments. I would like to just add the following as an extension to the answers of SangchulLee and MoisheCohen.
Since $X$ is compact, any closed subset $A\subseteq X$ is compact. Since $f$ is continuous, for any compact subset $C\subseteq X$, we have that $f(C)$ is a compact subset of $X$. Finally, as $X$ is Hausdorff, any compact subset of $X$ is closed in $X$.
Now consider the collection $\beta=\{A\subseteq X:A \hspace{2mm}\mbox{closed in} \hspace{2mm} X,\, A\neq\varnothing, \,f(A)\subseteq A\}$. We seek an element $A\in\beta$ such that $f(A)=A$. By the previous paragraph, all elements of $\beta$ are compact and closed subsets of $X$, and for all $B\in\beta$, we have $f(B)\in\beta$.
The idea of MoisheCohen is to take the intersection of all elements of $\beta$. However, this may not give an element of $\beta$, since $\beta$ may contain a pair of disjoint sets. Nevertheless, this idea can be made to work. Suppose we could find $A\in\beta$ such that: if $B\in\beta$ satisfies $B\subseteq A$, then $B=A$. In other words, $A$ is a minimal element of $\beta$. Then by the previous paragraph we have $f(A)\in\beta$ and $f(A)\subseteq A$, whence $f(A)=A$ as required. If we used Zorn's Lemma, we could find such an $A$ and be done. However, we can obtain the desired result by generalising the argument of SangchulLee instead (since here compactness serves as a kind of 'Noetherian' condition).
Claim: If $C\subseteq X$ is closed and satisfies $f(C)\subseteq C$, then we can find $A\subseteq C$ which is also closed in $X$ such that $f(A)=A$.
Proof: We can assume $C\in\beta$ (since for $C=\varnothing$ the result is trivial). Let $C_{1}=C$ and for all $n\geqslant1$ define $C_{n+1}=f(C_{n})$. Observe that $$C_{1}\supseteq C_{2}\supseteq C_{3}\supseteq \cdots$$is a descending chain of non-empty closed sets. Let $A=\cap_{n\geqslant1}C_{n}$.
If $A=\varnothing$, then by taking complements we see that $X=\cup_{n\geqslant1}(X\setminus C_{n})$. By compactness of $X$, we can pass to a finite subcover $X=\cup_{n=1}^{m}(X\setminus C_{n})=X\setminus C_{m}$. But then we have the contradiction that $C_{m}=\varnothing$.
Therefore, we can conclude that $A\neq\varnothing$. Now we just show that $f(A)=A$. It is easy to show from the definitions of the $C_{n}$ that we have $f(A)\subseteq A$.
Now suppose $a\in A$. We seek $b\in A$ such that $f(b)=a$. Note that as $f$ is continuous and $X$ is Hausdorff, $K_{n}:=C_{n}\cap f^{-1}(\{a\})$ is closed in $X$ for all $n$. Each $K_{n}$ is non-empty because $a\in C_{n+1}=f(C_{n})$ for all $n\geqslant1$. Thus, just like the $C_{n}$, the $K_{n}$ form a descending chain of non-empty closed sets. Hence, by a similar compactness argument to when we showed that $A\neq\varnothing$, we see that $K:=\cap_{n\geqslant1}K_{n}\neq\varnothing$. Also $K\subseteq A$, and so for any $b\in K$ we have $a=f(b)\in f(A)$. Therefore $f(A)\supseteq A$. $\hspace{1cm}\square$ | 2019-10-21T21:13:42 | {
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https://math.stackexchange.com/questions/2603199/sum-sum-n-2-infty-frac-1nn2n-2/2603205 | # sum $\sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$
I have the following series:
$\sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$
I am not able to do the telescoping process in the above series. I converted that into following partial fraction: $\sum _{n=2}^{\infty}\frac{(-1)^n}{(n+2)(n-1)}$ but nothing seems to cancel as usually happens in telescoping method. How can I do the above series. Is there any other method to do the above problem?
Writing the $n$th term of your sum in a form that telescopes can be avoid altogether by converting the problem to a double integral as follows.
Noting that $$\frac{1}{n - 1} = \int_0^1 x^{n - 2} \, dx \qquad \text{and} \qquad \frac{1}{n + 2} = \int_0^1 y^{n + 1} \, dy,$$ the sum can be rewritten as \begin{align*} \sum_{n = 2}^\infty \frac{(-1)^n}{n^2 + n - 2} &= \sum_{n = 2}^\infty \frac{(-1)^n}{(n - 1)(n + 2)}\\ &= \int_0^1 \int_0^1 \sum_{n = 2}^\infty (-1)^n x^{n - 2} y^{n + 1} \, dx dy \tag1\\ &= \int_0^1 \int_0^1 \frac{y}{x^2} \sum_{n = 2}^\infty (-xy)^n \, dx dy\\ &= \int_0^1 \int_0^1 \frac{y}{x^2} \cdot \frac{x^2 y^2}{1 + xy} \, dx dy \tag2\\ &= \int_0^1 \int_0^1 \frac{y^3}{1 + xy} \, dx dy\\ &= \int_0^1 y^2 \Big{[} \ln (1 + xy) \Big{]}_0^1 \, dy\\ &= \int_0^1 y^2 \ln (1 + y) \, dy\\ &= \frac{1}{3} \ln (2) - \frac{1}{3} \int_0^1 \frac{y^3}{1 + y} \, dy \tag3\\ &= \frac{1}{3} \ln (2) - \frac{1}{3} \int_0^1 \left (y^2 - y + 1 - \frac{1}{1 + y} \right ) \, dy \tag4\\ &= \frac{1}{3} \ln (2) - \frac{1}{3} \left [\frac{y^3}{3} - \frac{y^2}{2} + y - \ln (1 + y) \right ]_0^1\\ &= \frac{2}{3} \ln (2) - \frac{5}{18}. \end{align*}
Explanation
1. Interchanging the summation with the double integration.
2. Summing the series which is geometric.
3. Integration by parts.
4. Partial fraction decomposition.
• Hello, omegadot. I liked the way you solved this sum problem. I am thinking whether this method of summing can be applied in other situations also. – RAHUl JHa Jan 30 '18 at 7:35
• Whenever I see linear factors in the denominator of a convergent sum and a numerator of either $1$ or $(-1)^n$ I immediately think of trying to write the $n$th term of the sum in terms of an integral. After interchanging the integral sign with the summation, summing the resultant series which is geometric can be readily done. One hopefully can then perform the remaining integration. I have recently done this for a number of examples which you can follow here. – omegadot Feb 5 '18 at 23:30
• It is just another tool to add to your toolbox. The only potential problem the method may throw up is integrals which are either tedious or difficult to find. – omegadot Feb 6 '18 at 3:53
• Yes. But remember the sum this can be done on must first converge (try the technique on $\sum_{n =1}^\infty 1/n$ as see what happens) and the number of linear factors appearing in the denominator should not be too many since as their number grows the number of integrations that need to be performed also grows and can become very tedious. – omegadot Feb 6 '18 at 4:04
• And as an example of a sum that leads to a more difficult integral to find (unless you are familiar with special functions that is or other more advanced techniques for finding improper integrals) is $\sum_{n = 1}^\infty 1/n^2$. See what you end up with if you let $1/n = \int_0^1 x^{n - 1} \, dx$ and $1/n = \int_0^1 y^{n - 1} \, dy$. – omegadot Feb 6 '18 at 4:06
You need to fully expand the partial fraction.
\begin{align} (-1)^n \over (n+2)(n-1) &= {((n+2) - (n-1)) \times (-1)^n \over 3\times (n+2)(n-1)} \\ &= {(n+2) \times (-1)^n \over 3\times (n+2)(n-1)} - {(n-1) \times (-1)^n \over 3\times (n+2)(n-1)} \\ &= {(-1)^n \over 3\times (n-1)} - {(-1)^n \over 3\times (n+2)} \end{align}
Now this can be handled using Alternating Harmonic Series.
The final result, as Mathematica calculated, is $\frac {-5 + 12 \log 2} {18}$.
$$\sum_{k=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}=\frac{1}{3}\sum_{k=2}^{+\infty}(-1)^n\left(\frac{1}{n-1}-\frac{1}{n+2}\right)=$$ $$=\frac{1}{3}\left(1-\frac{1}{4}-\frac{1}{2}+\frac{1}{5}+\frac{1}{3}-\frac{1}{6}-\frac{1}{4}+\frac{1}{7}+...\right)=$$ $$=-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)+\frac{2}{3}\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}=\frac{2}{3}\ln2-\frac{5}{18}.$$ | 2019-09-18T12:02:10 | {
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http://math.stackexchange.com/questions/373918/what-is-261531-21-326-1531-21-3/373935 | # What is $(26+15(3)^{1/2})^{1/3}+(26-15(3)^{1/2})^{1/3}$?
$$(26+15\cdot\sqrt3)^{1/3}+(26-15\cdot\sqrt3)^{1/3}$$
I'm trying to get the result of this number. Through some algebra I found that it is close to $52^{1/3}$. Through some observation I found that it is a root of this cubic equation $x^3-3x-52=0$ and I found that the only real solution of that eq. is $4$ so now I know that the number I'm looking for is $4$ (close to my first try $52^{1/3}$). My question is, is there any algebraic process to get from the original expression and simplify it to $4$?
Thank you all very much in advance.
-
What exactly do you want help with? How to obtain that cubic equation for $x$? Or how to solve that cubic equation for $x$? – TMM Apr 26 '13 at 22:44
I want to know how to get from that expression to 4. – Eesu Apr 26 '13 at 22:53
$26 + 15 \sqrt{3} = 8 + 12 \sqrt{3} + 18 + 3 \sqrt{3} = 2^3 + 3 \cdot 2^2 \sqrt{3} + 3 \cdot 2 \sqrt{3}^2 + \sqrt{3}^3 = (2 + \sqrt{3})^3.$
$26 - 15 \sqrt{3} = 8 - 12 \sqrt{3} + 1 8- 3 \sqrt{3} = 2^3 - 3 \cdot 2^2 \sqrt{3} + 3 \cdot 2 \sqrt{3}^2 - \sqrt{3}^3 = (2 - \sqrt{3})^3.$
$(26 + 15 \sqrt{3})^{\frac{1}{3}} + (26 - 15 \sqrt{3})^{\frac{1}{3}} = 2 + \sqrt{3} + 2 - \sqrt{3} = 4.$
+1 How did you have this idea? Did you try first something like \begin{equation*} 26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3\left( a^{2}b+b^{3}\right) \sqrt{3} \end{equation*} and solved the system \begin{equation*} \left\{ \begin{array}{c} a^{3}+9ab^{2}=26 \\ a^{2}b+b^{3}=5 \end{array} \right. \end{equation*} a solution of which is $a=2,b=1$? – Américo Tavares Apr 26 '13 at 23:27 | 2014-04-17T01:46:13 | {
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http://math.stackexchange.com/questions/241692/how-to-solve-this-recurrence-relation-tn-4-cdot-t-sqrtn-n?answertab=votes | # How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$
I was trying to solve this recurrence $T(n) = 4T(\sqrt{n}) + n$. Here $n$ is a power of $2$.
I had try to solve like this:
So the question now is how deep the recursion tree is. Well, that is the number of times that you can take the square root of n before n gets sufficiently small (say, less than $2$). If we write: $$n = 2^{\lg(n)}$$ then on each recursive call $n$ will have it's square root taken. This is equivalent to halving the above exponent, so after $k$ iterations we have: $$n^{1/2^{k}} = 2^{\lg(n)/2^{k}}$$ We want to stop when this is less than $2$, giving:
\begin{align} 2^{\lg(n)/2^{k}} & = 2 \\ \frac{\lg(n)}{2^{k}} & = 1 \\ \lg(n) & = 2^{k} \\ \lg\lg(n) & = k \end{align} So after $\lg\lg(n)$ iterations of square rooting the recursion stops. For each recursion we will have $4$ new branches, the total of branches is $4^\text{(depth of the tree)}$ therefore $4^{\lg\lg(n)}$. And, since at each level the recursion does $O(n)$ work, the total runtime is: $$T(n) = 4^{\lg\lg(n)}\cdot n\cdot\lg\lg(n)$$
But appears that this is not the correct answer...
Edit:
$$T(n) = \sum\limits_{i=0}^{\lg\lg(n) - 1} 4^{i} n^{1/2^{i}}$$
I don't know how to get futher than the expression above.
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Does it help to simplify $4^{\lg\lg n}$ as $(\lg n)^2$? – alex.jordan Nov 21 '12 at 2:16
Can you explain "For each recursion we will have 4 new branches" to me? I don't understand this, but I have little to no experience with computational complexity. – alex.jordan Nov 21 '12 at 2:18
The first call of T(n) will generate 4 branches, each one of this branches will call 4 new branches and so on – dreamcrash Nov 21 '12 at 2:26
That's what I don't understand. If I call $T(16)$, then what I see in your recursion makes me call $T(4)$, and that's it. I'm only seeing "one branch" as opposed to a recursion like $T(n)=T(\sqrt{n})+T(n/4)$, where I would see one call as branching into two. – alex.jordan Nov 21 '12 at 3:07
You are right. I just edit. – dreamcrash Nov 21 '12 at 14:55
For every $k\geqslant0$, let $U(k)=T(2^k)$, then $U(k)=4U(k/2)+2^k$ hence $U(k)\geqslant2^k$ for every $k$.
Choose $C\geqslant2$ so large that $U(k)\leqslant C 2^k$ for every $k\leqslant5$. Let $k\geqslant6$. If $U(k-1)\leqslant C2^{k-1}$, then $U(k)\leqslant 4C2^{k/2}+2^k$. Since $k\geqslant3$, $2^{k/2}\leqslant 2^k/8$ hence $U(k)\leqslant (C/2+1)2^k$. Since $C/2+1\leqslant C$, the recursion is complete.
Finally, $U(k)=\Theta(2^k)$.
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We want to find a simple upper bound for $$T(n) = \sum_{i=0}^{\lg\lg{n}-1} 4^i n^{1/2^i}.$$
Note that the first summand is $n$ and the last summand is less than $$4^{\lg\lg{n}} n^{1/2^{\lg\lg{n}}} = \log^2{n} \cdot n^{1/\lg{n}}.$$
The first term is significantly larger and probably doing most of the work, so we will aim to show that $T(n) = O(n)$. Indeed, \begin{align*} T(n) &= \sum_{i=0}^{\lg\lg{n}-1} 4^i n^{1/2^i}\\ &\le n + \sum_{i=1}^{\lg\lg{n}} 4^i n^{1/2^i}\\ &\le n + \sum_{i=1}^{\lg\lg{n}} 4^i \sqrt{n}\\ &\le n + \sqrt{n}\lg\lg{n} + \sum_{i=1}^{\lg\lg{n}} 4^i\\ &\le n + \sqrt{n}\lg\lg{n} + \frac{4\lg^2{n} - 1}{4 - 1}\\ &= O(n). \end{align*}
Also, clearly $T(n) \ge n$, so we have $T(n) = \Theta(n)$.
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https://www.physicsforums.com/threads/integral-question.254229/ | Integral question
Homework Statement
I have a question dealing with this integral: $$\int\frac{x}{\sqrt{x^2-4}}$$. I did the trig substitution to check my other method of substitution and got two different answers: $$\sqrt{x^2-4}$$ and $$\sqrt{x^2-4}/2$$. The latter is from trig substitution and the former from regular substitution.
The Attempt at a Solution
Ok, here is my work for the trig substitution
x=2sec@
dx=2sec@tan@
(x^2-4) = 4tan^2@
I finally get the integral of sec^2@ which gives me tan@ and looking from the triangle I drew its answer is $$\sqrt{x^2-4}/2$$. Why don't these two answers agree? Is there some rule suggesting not to use trig substitution or one way over the other?
Thank You
Related Calculus and Beyond Homework Help News on Phys.org
Oh! I made a mistake with a coefficient, they agree with each other! Sorry
HallsofIvy
Homework Helper
Homework Statement
I have a question dealing with this integral: $$\int\frac{x}{\sqrt{x^2-4}}$$. I did the trig substitution to check my other method of substitution and got two different answers: $$\sqrt{x^2-4}$$ and $$\sqrt{x^2-4}/2$$. The latter is from trig substitution and the former from regular substitution.
The Attempt at a Solution
Ok, here is my work for the trig substitution
x=2sec@
dx=2sec@tan@
(x^2-4) = 4tan^2@
I finally get the integral of sec^2@ which gives me tan@ and looking from the triangle I drew its answer is $$\sqrt{x^2-4}/2$$. Why don't these two answers agree? Is there some rule suggesting not to use trig substitution or one way over the other?
Thank You
You skipped over the critical step! Yes, if you let $x= 2 sec(\theta)$, then dx= 2sec(\theta)tan(\theta)d\theta[/itex] (you dropped the "$d\theta$" just as you dropped the "dx" in the original integral- bad habit.) and $x^2- 4= 4tan^2(\theta)$ so $\sqrt{x^2- 4}= 2 tan^2(\theta)$. Did you forget that the"4" in $4 tan^2(\theta)$ became "2" when you took the square root?
Now, the part you skipped- putting all that into the integral.
$$\int\frac{x dx}{\sqrt{x^2- 4}}= \int \frac{[2 sec(\theta)][2sec(\theta)tan(\theta)d\theta}{2 tan(\theta)}$$
$$= 2\int sec^2(\theta)d\theta= 2 tan(\theta)+ C= 2tan(sec^{-1}(\frac{x}{2})+ C$$
Notice the "2" still in there?
Now, if $sec(\theta)= x/2$, then we can represent that as a triangle with angle $\theta$, near side= 2, and hypotenuse= x. By the Pythagorean theorem, the opposite side has length $\sqrt{x^2- 4}$ and so $tan(\theta)= \sqrt{x^2- 4}/2$. But because of the "2" multiplying the integral,
$$\int \frac{x dx}{\sqrt{x^2- 4}}= \sqrt{x^2- 4}+ C$$
exactly as you would get if you make the substitution u= x2- 4.
Excellent Halls,
I found my mistake but thanks for showing me the extended steps, :D | 2020-12-03T18:29:07 | {
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http://math.stackexchange.com/questions/305001/how-can-the-real-numbers-be-a-field-if-0-has-no-inverse | # How can the real numbers be a field if $0$ has no inverse?
I'm reading a linear algebra book (Linear Algebra by Georgi E. Shilov, Dover Books) and the very start of the book discusses fields. 9 field axioms discussing addition and multiplication are given then the author goes on to discuss common sets of numbers.
The integers are identified as being a set of numbers which is not a field because there does not exist a reciprocal element for every integer (axiom # 8 in this book states the existence of a reciprocal element $B$ for a number $A$ such that $AB=1$). The author goes on to call the real numbers a field, and asserts that an axiomatic treatment can be had by supplementing the field axioms with the order axioms and the least upper bound axiom.
My understanding consists of the following statements I believe are facts:
1. zero is a member of the reals
2. there exists no reciprocal element of zero that is a real number
Given those two facts it seems to me that the reals fail the same test for being a field that the author states the integers fail. Yet the author is calling the real numbers a field. To my mind this is a contradiction.
Is there a resolution to this apparent contradiction? I'm a total beginner at this sort of math (I'm an engineer by training, not a mathematician!) and would appreciate any assistance!
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Read the field axioms again. – Chris Eagle Feb 15 '13 at 19:09
@Paul Read where it says Existence of additive inverses and multiplicative inverses and check rschwieb's answer. – Git Gud Feb 15 '13 at 19:14
You guys are quite correct... my failing is in taking inadequate notes and then referring back to those rather than the original. Regaining my mental acuity after joining the corporate world is one of my main motivations in reading this book... looks like it's coming along well thanks to the people on this board! – Paul Feb 15 '13 at 19:46
It's great that you are taking an active approach to learning the material and asking questions when things don't make sense to you. I wish more of the students I teach took such an approach! – Michael Joyce Feb 15 '13 at 20:43
Zero is always excluded from having a reciprocal.
(The axioms should say that every nonzero element of the field has a reciprocal.)
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It does say non-zero. Did my notes say that? Of course not! Thank you very much! – Paul Feb 15 '13 at 19:41
Many things would be nicer if we could assume that everything in a field, including 0, had a multiplicative inverse. Unfortunately, that is not possible in any useful way.
In any structure that satisfies the field axioms, it must hold for any $a$ that $$0\cdot a = (0+0)\cdot a = 0\cdot a + 0\cdot a$$ and canceling one $0\cdot a$, we see that $0\cdot a = 0$ for all $a$.
If $0$ had an inverse, then we would have $a = 1\cdot a = (0\cdot 0^{-1})\cdot a = 0\cdot(0^{-1}\cdot a) = 0$, and so this is only possible if every element in the field equals $0$, which is to say that $0$ is the only element of the field.
The set $\{0\}$ with $0\cdot 0=0$ and $0+0=0$ is indeed a field according to some formulations of the field axioms. But it is not a very interesting field, and actually it tends to be such a cumbersome special case that by convention it is not considered a "field" at all. Formally this convention is expressed by requiring as an axiom that $1\ne 0$.
-
Name, if possible, three things that would be nicer... I'm interested! – rschwieb Feb 15 '13 at 22:03
@rschwieb: Um... Matrices with entries in a field would always have inverses. You could cancel factors in equations and fractions without worrying that the factors might be 0. ("Such-and-such must be nonzero" assumptions disappear from theorems all over). Every polynomial function would have a scalar multiple that was monic of any higher degree, so the quotient in long polynomial division could always be a scalar. – Henning Makholm Feb 16 '13 at 15:12
Here it is from Google Books:
That "$\ne 0$" in there is the thing you are missing.
- | 2016-02-13T03:20:39 | {
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https://www.khanacademy.org/math/calculus-all-old/taking-derivatives-calc/exponential-functions-differentiation-calc/v/derivatives-of-sin-x-cos-x-tan-x-e-x-and-ln-x | If you're seeing this message, it means we're having trouble loading external resources on our website.
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# Derivatives of sin(x), cos(x), tan(x), eˣ & ln(x)
Learn the derivatives of several common functions. Created by Sal Khan.
## Want to join the conversation?
• Lets say I have an equation sin x = 1/2. Then clearly, x = 30 degrees or pi/6 radians.
Now if I differentiate both sides of the equation with respect to x (because both are equal, their derivatives should also be equal), I will have cos x = 0, right?
But that means x = 90 degrees, which is obviously not the solution! Could somebody please explain why this is happening?
• Once you say that "x = some number", x is no longer a variable but rather it is a constant -- and you must then treat it like a constant.
Since x = π/6, both "x" and "sin(x)" are constants (because sin(π/6) = 1/2) and the derivative of a constant is zero.
Does this clear things up for you?
• Think of e like pi - it's just an accepted name for a number that's useful in a bunch of formulas. It's equal to around 2.71828, so plug that into wherever you see e and you'll be good, or you can often leave things in terms of e.
• What is the explanation of why the derivative of sec (x) is sec (x) tan(x)?
It comes up in the questions for this topic on Khan Academy in the Special derivatives exercise.
• If we accept that d/dx (cos x) = − sin x, and the power rule then:
sec x ≡ 1/cos x
Let u = cos x, thus du = − sin x dx
sec x = 1/u
(1/u) = (u⁻¹)
By the power rule:
derivative of (u⁻¹) = −u⁻² du
Back substituting:
= −(cos x)⁻² ( − sin x) ∙ dx
= [sin x / (cos x)²] ∙ dx
= [(sin x / cos x) ∙ (1/cos x)] ∙ dx
= [tan (x) ∙ sec (x)] ∙ dx
• I think I understand how to do the derivatives of the trig functions. But What if instead of Sin(x,
there was Sin( and equation. Like, Sin(x^2+2). How would you take the derivative when it's something like that, and not just Sin or Cos?
• You always have to multiply the outer derivative with the inner derivative. That's true even for sin(x), it's just that the inner derivative is 1. (d/dx x = 1)
d/dx sin(x) = cos(x) * 1 = cos(x)
d/dx sin(2x) = cos(2x) * 2 = 2 cos(2x)
d/dx sin(x^2) = cos(x^2) * 2x = 2x cos(x^2)
d/dx sin(x^2 + 2) = cos(x^2 + 2) * 2x = 2x cos(x^2 + 2)
• Where i can find the proof of these derivatives?
• The website https://proofwiki.org has proofs for many mathematical identities and theorems, including various common derivatives.
• Is there a way of proving the derivative of e^x without using the derivative of ln(x)? If yes, can you please provide a link to it?
• Yes, I can prove it:
derivative of e^x , by definition of derivative:
= lim h→0 { e^(x+h) - e^x} / h
= lim h → 0 {e^x(e^h) -e^x}/h
= lim h → 0 e^x{(e^h) -1}/h
= (e^x) lim h → 0 {(e^h) -1}/h
By definition of e:
e = lim h→ 0 (1+h)^(1/h)
And so,
e^h = lim h→ 0 (1+h)^(h/h)
= lim h→ 0 (1+h)
Substituting that in:
= (e^x) lim h → 0 {(1+h) -1}/h
= (e^x) lim h → 0 { h/h}
= (e^x) lim h → 0 { 1 }
= e^x * (1)
= e^x
• What are the proofs of sin, cos & tan derivatives? Are there any videos proving them?
• Is there a video that goes over the intuition behind integrals of trig functions (sec(x) and cot(x) included ideally)?
• Are there videos of proving the derivations of trig functions? I've only seen the sine limit approaching zero proof where he proves it using sandwhich / squeeze theorem, (that is proof that limit of sinx/x is equal to 1 as x approaches zero). Also, could I use squeeze function and 2 parabolas to prove sine function instead of the overly complicated way he's proving it? And another thing, how can derivative of function tanx be cosecant when he's clearly written 1/cos^2(x) which is cosecant squared?
(1 vote)
• Unfortunately there's no proof currently on Khan of the derivatives of sine, cosine, or tangent.
Also, the derivative of tangent is secant squared.
`1/cos x = sec x`
`d/dx (tan x) = 1/cos^2 x = sec^2 x`
As for proofs, here's a good proof of the derivative of sine:
https://proofwiki.org/wiki/Derivative_of_Sine_Function/Proof_2
Using the proof for sine, you can easily prove cosine using the equality
`cos(x) = sin(x+π/2)` and the chain rule.
Using the derivative of sine and the derivative of cosine, you can use the definition of tangent
`tan(x) = sin(x)/cos(x)` and the quotient rule to prove the derivative of tangent. | 2023-02-03T23:43:59 | {
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https://brilliant.org/discussions/thread/polynomial-expansions-useful-formulas/ | ×
# Polynomial Expansions (useful formulas)
Let $$\mathrm{A}$$ be the set of all "$$\color{Blue}{\text{useful}}$$" things.
Let $$\mathrm{B}$$ be the set of all "$$\color{Green}{\text{Awesome}}$$" things.
Let $$\mathrm{C}$$ be the set of all "$$\color{Red}{\text{fascinating}}$$" things.
Let $$\mathrm{D}$$ be the set of all "$$\color{Brown}{\text{easily understandable}}$$" things.
What this note contains is an element of $$\mathrm{A \cap B \cap C \cap D}$$....
$\color{Blue}{\textbf{Polynomial Expansions}}$
$$\mathbf{1.}\quad \displaystyle \dfrac{1-x^{m+1}}{1-x} = 1+x+x^2+...+x^m = \sum_{k=0}^m x^k$$
$$\mathbf{2.} \quad \displaystyle \dfrac{1}{1-x} = 1+x+x^2+... = \sum_{k=0}^\infty x^k$$
$$\mathbf{3.}\quad \displaystyle (1+x)^n = 1+\binom{n}{1} x + \binom{n}{2}x^2+...+\binom{n}{n}x^n = \sum_{k=0}^n \dbinom{n}{k} x^k$$
$$\mathbf{4.}\quad \displaystyle (1-x^m)^n = 1-\binom{n}{1}x^m+\binom{n}{2}x^{2m}-...+(-1)^n\binom{n}{n} x^{nm}$$
$$\quad \quad \quad \quad\quad \displaystyle = \sum_{k=0}^n (-1)^n \dbinom{n}{k}x^{km}$$
$$\mathbf{5.}\quad \displaystyle\dfrac{1}{(1-x)^n} = 1+ \binom{1+n-1}{1} x + \binom{2+n-1}{2} x^2+...+\binom{r+n-1}{r} x^r+......$$
$$\quad \quad \quad \quad \quad \displaystyle =\sum_{k=0}^\infty \dbinom{k+n-1}{k} x^k$$
$$\color{Purple}{\text{Tremendously useful}}$$ in calculating the $$\color{Blue}{\text{co-efficient}}$$ of any term in specially generating functions that we come across, in many combinatorics problems...
Taken From - Alan Tucker's "Applied Combinatorics"
Good luck problem solving !
3 years, 2 months ago
Sort by:
I know that the second one only works for $$x<1$$. But do any of the others work for only $$x>1$$. Also, thank you so much for this note, it's very useful.
- 3 years, 2 months ago
-1<x<1 actually.
- 3 years, 2 months ago
Thank you, but do u know which ones only work for this case
- 3 years, 2 months ago
Can you add parts of this page into the algebra wiki? I think that Algebraic Identities and Algebraic Manipulation - Identities, would be suitable places to add them.
Staff - 3 years ago
what is the derivation of 5th one.
- 3 years, 2 months ago
Standard result, it's related to "Newton's generalised Binomial theorem", but if you do want the derivation please see it here
- 3 years, 2 months ago
Does the last one work for 0<x<1???
- 3 years, 2 months ago
Definitely 2.
- 3 years, 2 months ago
set contains Everything
- 3 years, 2 months ago
Is this for nerds like you?
- 3 years, 2 months ago
Dude don't use nerds in a derogatory manner please. If you don't like nerds or aren't one yourself, you should remove yourself from Brilliant.org. Have a nice day.
- 3 years, 2 months ago
@Finn Hulse , thanks for helping here, really !
img
I truly like your comment, by these many likes :-
img
- 3 years, 2 months ago
Haha, anytime dude. :D
- 3 years, 2 months ago
Agreed. BTW, some nerds can be good at sports as well.
- 3 years, 1 month ago
BTW sharky, participate in JOMO 8, we miss your submission.
@Sharky Kesa , JOMO 8 starts $$\color{Red}{\textbf{TOMORROW}}$$ and has some good questions I made.
- 3 years, 1 month ago
# (#Sharky_Surprises )
- 3 years, 1 month ago
This is for using in generating functions we design for combinatorics problems.... For example, see the set "vegetable combinatorics".... (type in search bar simply).... That's for all who want to learn, nothing high-figh technique or anything, just formulas to get co-efficient of a specific term in a generating function. @Jack Daniel Zuñiga Cariño
- 3 years, 2 months ago | 2017-10-18T20:30:47 | {
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http://mathhelpforum.com/discrete-math/160437-help-proof-sets-print.html | # Help with Proof of Sets
• October 20th 2010, 06:59 PM
JSB1917
Help with Proof of Sets
Prove $(A \cup B) - (A \cap B) = (A-B) \cup (B-A)$
$(A \cup B) - (A \cap B) \subseteq (A-B) \cup (B-A)$
Let $x \in (A \cup B) - (A \cap B)$. Then $x \in A \cup B$ and $x \notin A \cap B$. Then $x \in A$ or $x \in B$ but $x \notin A$ and $x \notin B$. Since $x \in A$ and $x \notin B$ then $x \in (A-B)$. Also since $x \in B$ and $x \notin A$ then, $x \in (B-A)$. Now $x \in (A-B)\cup(B-A)$.
Therefore, $(A \cup B) - (A \cap B) \subseteq (A-B) \cup (B-A)$
Do I have the general idea right? What can I fix? I'm sure I did something wrong. Yes, I also realized I skipped the other part of the proof.
• October 20th 2010, 09:25 PM
Soroban
Hello, JSB1917!
I used a two-column proof . . .
Quote:
$\text{Prove: }\;(A \cup B) - (A \cap B) = (A-B) \cup (B-A)$
$\begin{array}{ccccccc}
1. & (A \cup B) - (A \cap B) && 1.& \text{ Given} \\ \\
2. & (A \cup B) \cap (\overline{A \cap B}) && 2. & \text{d{e}f. Subtraction} \\ \\
3. & (A \cup B) \cap (\overline A \cup \overline B) && 3. & \text{DeMorgan's Law} \\ \\
4. & (A \cap \overline A) \cup (A \cap \overline B) \cup (B \cap \overline A) \cup (B \cap \overline B) && 4. & \text{Distributive Prop.} \\ \\
5. & \emptyset \;\;\cup (A \cap \overline B) \cup (B \cap \overline A) \cup \;\;\emptyset && 5. & S \cap S \:=\:\emptyset \\ \\
6. & (A \cap \overline B) \cup (B \cap \overline A) && 6. & S \cup \emptyset \:=\:S \\ \\
7. & (a - B) \cup (B - A) && 7. & \text{d{e}f. Subtraction}
\end{array}$ | 2015-01-29T00:59:25 | {
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http://mathhelpforum.com/advanced-algebra/175046-compute-change-basis-matrix.html | # Math Help - Compute the change of basis matrix
1. ## Compute the change of basis matrix
I wish to compute the change of basis for the following:
B = { $\begin{bmatrix} 1\\ 1 \end{bmatrix},\begin{bmatrix} 1\\ -1 \end{bmatrix}$}
C = { $\begin{bmatrix} 0\\ 1 \end{bmatrix},\begin{bmatrix} 2\\ 1 \end{bmatrix}$}
How can I compute the change of basis matrix $P_{C \leftarrow B}$?
There are supposedly (at least) two ways to do this.
I know how to do this by expressing the vectors in B in terms of the basis C; when I do that I get the result that:
$P_{C \leftarrow B}$ = $\begin{bmatrix} 1/2 & -3/2\\ 1/2 & 1/2 \end{bmatrix}$
Is this correct at all? Although even if it is correct I'd like to be able to do it using a different approach I got described.
Supposedly I first have to transform the B coordinates to the canonical coordinates and thereafter the canonical coordinates to C coordinates, this should give $P_{C \leftarrow B}$ as $P_C^{-1}P_B$
Could someone show me how to do this for this example?
2. Originally Posted by posix_memalign
I wish to compute the change of basis for the following:
B = { $\begin{bmatrix} 1\\ 1 \end{bmatrix},\begin{bmatrix} 1\\ -1 \end{bmatrix}$}
C = { $\begin{bmatrix} 0\\ 1 \end{bmatrix},\begin{bmatrix} 2\\ 1 \end{bmatrix}$}
How can I compute the change of basis matrix $P_{C \leftarrow B}$?
There are supposedly (at least) two ways to do this.
I know how to do this by expressing the vectors in B in terms of the basis C; when I do that I get the result that:
$P_{C \leftarrow B}$ = $\begin{bmatrix} 1/2 & -3/2\\ 1/2 & 1/2 \end{bmatrix}$
Is this correct at all? Although even if it is correct I'd like to be able to do it using a different approach I got described.
It is correct and hopefully you understand what you did here: you express each element of the new
basis as a lin. combination of the new basis and the matrix we're looking for is the transpose of the
coefficient matrix obtained in the first step.
This is, as far as I can tell, the easiest and quickest method.
Supposedly I first have to transform the B coordinates to the canonical coordinates and thereafter the canonical coordinates to C coordinates, this should give $P_{C \leftarrow B}$ as $P_C^{-1}P_B$
Could someone show me how to do this for this example?
As I can see this is exactly the same as before: to express the basis B in terms of the canonical
coordinates is just to take the matrix B formed with the elements of B, and then...etc. Exactly the same!
Tonio
3. Originally Posted by tonio
As I can see this is exactly the same as before: to express the basis B in terms of the canonical
coordinates is just to take the matrix B formed with the elements of B, and then...etc. Exactly the same!
Tonio | 2014-12-25T22:54:06 | {
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https://math.stackexchange.com/questions/4036241/deck-of-cards-probability-one-card-missingrandomly-probability-its-a-jack/4036277#4036277 | # Deck of Cards Probability: One card missing(randomly), Probability it's a jack?
The question is:
A card is missing from a deck of 52 cards. You draw a card, What's the probability that it's a jack?
I been struggling with probability and would need help understanding this question. What I got so far is that I need to consider two scenerios.
A: Jack is one of the missing cards.
B: Jack is not one of the missing cards.
P(A) = 4/52
P(B) = 48/52
To continue I feel like I might be on the right direction but I could be getting my probabilities wrong and that's what is confusing me and I don't know what do after. Any suggestion are welcomed. Thank you.
• So far you're correct. Now you just have to figure the probability that the card is a Jack in each of the two cases, and apply the law of total probability. Feb 23 at 1:45
You are correct so far. There are two cases to consider and they are disjoint: the missing card is or is not a jack. If the jack is missing, then there are only $$3$$ possible jacks to draw from the remaining $$51$$-card deck. Similarly, if the missing card is not a jack, then there are $$4$$ possible jacks to draw from the remaining $$51$$-card deck.
Remember how conditional probability works: $$P(A \cap B) = P(B) \, P(A \mid B)$$.
Try to calculate this yourself, but click below to reveal a spoiler when you're ready.
\begin{align*}\def\j{\text{draw jack}}\def\jm{\text{jack missing}}\def\njm{\text{jack not missing}}P(\j) &= P(\j \cap \jm) + P(\j \cap \njm) \\&= P(\jm) \, P(\j \mid \jm) \\&\qquad\qquad {}+ P(\njm) \, P(\j \mid \njm) \\&= \frac{4}{52}\frac{3}{51} + \frac{48}{52}\frac{4}{51} \\&= \frac{1}{13}\frac{3}{51} + \frac{12}{13}\frac{4}{51} \\&= \frac{51}{13 \cdot 51} \\&= \frac{1}{13} \end{align*}
• I got what you got, I remember about conditional probability and the Law of Total Probability. Thank you for the tips. Feb 23 at 2:32
Use the events, $$A:$$ the missing card is a Jack, $$B:$$ the missing card is not a Jack, and $$C:$$ a Jack is drawn.
You have $$\mathsf P(A)=4/52$$ and $$\mathsf P(B)=48/52$$
The Law of Total Probability says: $$\mathsf P(C)=\mathsf P(A)\,\mathsf P(C\mid A)+\mathsf P(B)\,\mathsf P(C\mid B)$$
$$\mathsf P(C\mid A)$$ is the probability for drawing a Jack when three are left among the fifty-one remaining cards.
$$\mathsf P(C\mid B)$$ is the probability for drawing a Jack when all four are among the fifty-one remaining cards.
$$~\\~\\~$$
What do you notice about the result? Is this surprising? | 2021-12-07T08:25:16 | {
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https://math.stackexchange.com/questions/1486464/find-all-five-solutions-of-the-equation-z5z4z3z2z1-0/1486476 | # Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$
$z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
• Doesn't the polynomial look a lot like a partial sum of a familiar series? Oct 18 '15 at 20:49
• Whilst the given solutions are neater, you should also learn that the factoring you did gave a quadratic in $z^2$ that you know how to solve - so nothing more clever is actually required here. Oct 19 '15 at 0:09
Note that $(z-1)(1+z+z^2+z^3+z^4+z^5) = z^6-1$. Solve $z^6-1=0$ and discard the $z=1$ solution (which comes from the $z-1$ factor).
• this is clearly the best way :)
– Ant
Oct 18 '15 at 20:57
• Although I should have seen the solution the other ways, this will allow me to utilize the method I used previously. Oct 18 '15 at 21:01
Hint:
Since $$z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)$$ Solve $z^2+z+1=0$ and $z^2-z+1=0$ by using the quadratic formula: $$z^2+z+1=0\implies z_{1,2}=\frac{-1\pm\sqrt{1-4}}{2}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$ $$z^2-z+1=0\implies z_{3,4}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$$
Here is another way to pursue this (just for fun!).
Recall that for $|x| < 1$ we have:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$
Multiplying both sides by $x^6$ yields:
$$\frac{x^6}{1-x} = x^6 + x^7 + x^8 + x^9 + \cdots$$
Now we subtract the former line from the latter:
$$\frac{x^6 - 1}{1-x} = -(1 + x + x^2 + x^3 + x^4 + x^5)$$
We now multiply both sides by $-1$ to obtain your original expression:
$$\frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$
In one way, this is no different than the suggestion in your accepted answer. But I think (hope) seeing the connection arise in a different manner is of interest. Anyway, at this point you seek to find the roots of $x^6 - 1 = (x^3 + 1)(x^3 - 1)$, and you can now factor the latter two parenthetical expressions as the sum and difference of cubes, respectively, to find the roots.
(Noting that/why $x\neq 1$.)
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
False.
It also shows the remaining four solutions are roots of $z^4 + z^2 + 1 = 0$.
• Which is a quadratic equation in $x=z^2$. In fact, it's one of my favorite equations.
– Joel
Oct 19 '15 at 10:15
$$z^5+z^4+z^3+z^2+z+1 = 0 \Longleftrightarrow$$ $$(z+1)(z^2-z+1)(z^2+z+1) = 0 \Longleftrightarrow$$ $$z+1=0 \vee z^2-z+1=0 \vee z^2+z+1=0 \Longleftrightarrow$$ $$z=-1 \vee z=\frac{1\pm\sqrt{-3}}{2} \vee z=\frac{-1\pm\sqrt{-3}}{2} \Longleftrightarrow$$ $$z=-1 \vee z=\frac{1\pm i\sqrt{3}}{2} \vee z=\frac{-1\pm i\sqrt{3}}{2}$$
\begin{align}z^6&=1\\ z^6-1&=0\\ (z-1)(\underbrace{z^5+z^4+z^3+z^2+z+1}_{f(z)})&=0 \end{align} Hence the solutions to $f(z)$ are \begin{align}z^6&=1=e^{i2\pi} &&(z\neq1)\\ z&=e^{i\frac {2n\pi}6}=e^{i\frac {n\pi}3}&& (n=1,2,3,4,5)\quad\blacksquare\end{align}
I quote: Factoring it into (z+1)(z4+z2+1)(z+1)(z4+z2+1) didn't do anything but show -1 is one solution. Put $z^2=u$ then you will find the solutions. | 2021-09-29T03:09:21 | {
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https://web2.0calc.com/questions/help_18263 | +0
# help
0
32
2
+529
All the positive integers greater than 1 are arranged in five columns (A, B, C, D, E) as shown. Continuing the pattern, in what column will the integer 800 be written?
Logic Dec 7, 2018
#1
+14579
+3
A C E are only odd nos.
E = 5 + 8(n) would put 797 there .....800 would be 3 more to the left in column B
ElectricPavlov Dec 7, 2018
#2
+20683
+5
All the positive integers greater than 1 are arranged in five columns (A, B, C, D, E) as shown.
Continuing the pattern, in what column will the integer 800 be written?
We rearrange:
$$\small{ \begin{array}{|cccc|cccc|cccc|cccc|cccc|c|} \hline B&C&D&E& D&C&B&A& B&C&D&E& D&C&B&A& B&C&D&E& \ldots\\ \hline 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 & 11 & 12 & 13 &14 & 15 & 16 & 17 &18 & 19 & 20 & 21 & \ldots \\ \hline \end{array} }$$
$$\begin{array}{|r|r!c|c|c|c|l|} \hline \text{Position} ~&(& 1 & 2 & 3 & 4\text{ or }0& ) \\ \hline \text{Row odd} ~&(& B & C & D & E &) \\ \text{Row even} ~&(& D & C & B & A &) \\ \hline \end{array}$$
Formula:
$$\begin{array}{|rcll|} \hline \text{Row} &=& 1 + \left\lfloor \dfrac{n-2}{4} \right\rfloor \quad & \quad \left\lfloor\ldots \right\rfloor \text{Function floor or Integerpart} \\\\ \text{Position} &=& (n-1) \pmod{4} \quad & \quad \text{If the position is zero, so the position is 4}\\ \hline \end{array}$$
$$\text{Row} =\ ?, ~n = 800:$$
$$\begin{array}{|rcll|} \hline \text{Row} &=& 1 + \left\lfloor \dfrac{800-2}{4} \right\rfloor \\ &=& 1 + \left\lfloor \dfrac{798}{4} \right\rfloor \\ &=& 1 + \left\lfloor 199.5 \right\rfloor \\ &=& 1 + 199 \\ &=& 200 \quad & \quad \text{the row is even!} \\ \hline \end{array}$$
The number 800 is in row = 200. And the row is an even number.
$$\text{Position} =\ ?, ~n = 800:$$
$$\begin{array}{|rcll|} \hline \text{Position} &=& (800-1) \pmod{4} \\ &=& 799 \pmod{4} \\ &=& 3 \\ \hline \end{array}$$
$$\text{Column} =\ ?, ~\text{Row is even}:$$
$$\begin{array}{|rcll|} \hline \hline \text{Position} ~&(& 1 & 2 & \color{red}3 & 4\text{ or }0& ) \\ \hline \text{Row even} ~&(& D & C & \color{red}B & A &) \\ \hline \end{array}$$
The integer 800 will be written in column B
heureka Dec 7, 2018
edited by heureka Dec 7, 2018 | 2018-12-17T11:12:24 | {
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https://www.physicsforums.com/threads/de-involving-limits.715250/ | # DE involving limits
1. Oct 8, 2013
### ehabmozart
1. The problem statement, all variables and given/known data
Find the general solution of the given differential equation, and use it to determine how
solutions behave as t→∞.
a) y' − 2y = t2e2t
2. Relevant equations
DE
3. The attempt at a solution
After doing the linear DE steps I end up getting y(t)= t3e2t/3 + ce2t. Now I have a problem in getting the limit. The book says the 'It is
evident that all solutions increase at an exponential rate.' Now, my doubt is how would it increase. Or what makes us sure that it will increase? The first part indeed goes to infinity but the second part has a c which can either be positive or negative. So we can end up getting infinity - infinity. Can someone help me?
2. Oct 8, 2013
### UltrafastPED
Assuming that you have found the general solution, note that both terms have exponentials with positive exponents for t>0 - the exponential dominates the t^3, so the solution increases exponentially as t-> infinity.
3. Oct 9, 2013
### ehabmozart
Excuse me sir, but I found this an irrelevant reply. I kept on mentioning the constant C. With examples, consider C to be -5 the limit would then be NEGATIVE INFINITY as t gets larger. But in cases such as C being 6 the answer goes to POSITIVE INFINITY. WHy is that the book mentions only the shooting UP!
4. Oct 9, 2013
### UltrafastPED
Spend some time looking at the expression that you have ...
y(t)=(t^3/3 +C)exp(2t)
Does the value of C matter at all?
5. Oct 9, 2013
### ehabmozart
Well, if C is NEGATIVE, then yes it will matter. The first half definitely goes to positive infinity. But the second half would go to Negative inifinity since exp(2t) shoots up, this negative C can simply reflect this on the x axis and it is negative infinity???
6. Oct 9, 2013
### UltrafastPED
What is the limit as t-> infinity for this expression: (t^3/3 +C)?
7. Oct 9, 2013
### ehabmozart
In this case it is infinity because C comes after a plus sign unsupported with any exponential or something. What would be the limit then for the expression (t^3/3 + Ct^3/3). Consider the case where C is negative.
8. Oct 9, 2013
### UltrafastPED
Your assertion is incorrect - are you clear how limits work?
9. Oct 9, 2013
### Ray Vickson
His answer is not irrelevant; it is 100% accurate and gives you exactly all you need. Perhaps you do not understand the issues, but that is a separate matter, and one that YOU need to address.
For example, suppose C = -5. Then $C + t^3/3 = t^3/3 - 5 \geq 1$ if $t \geq 18^{1/3} \doteq 2.6207$; that is, if $t \geq 2.6207$, $y(t) \geq e^{2t}$. Does this go to $+\infty$ as $t \to + \infty$?
Suppose C = -5,000,000. Then $C + t^3/3 = t^3/3 - 5000000 \geq 1$ if $t \geq 100 \times 15^{1/3} \doteq 246.6212$. So, if $t \geq 246.6212$, $y(t) \geq e^{2t}$. Does this go to $+\infty$ as $t \to + \infty$?
Last edited: Oct 9, 2013 | 2017-10-19T21:13:57 | {
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https://math.stackexchange.com/questions/3967582/doubt-in-open-dense-sets-of-a-metric-space | # Doubt in open dense sets of a metric space.
Let $$B_d(x,r) \cap G_1$$ where $$G_1$$ is an open dense set. Now this intersection is non empty. Let $$x_1 \in B_d(x,r) \cap G_1$$.Then, $$B_d(x_1,r_1) \subset G_1$$ .Now my doubt is that will there exists a $$r_2 > 0$$ such that $$B_d(x_1,r_2) \subset G_1$$ and $$B_d(x_1,r_2) \subset B_d(x,r)$$.
I think it will be true. However I was wondering if there is a contradicting example.
Whether $$B_d(x_1,r_1)\subseteq G_1$$ depends on $$r_1$$. What is true is that there is an $$r_1>0$$ such that $$B_d(x_1,r_1)\subseteq G_1$$. As for your question, $$B_d(x,r)\cap G_1$$ is an open set containing $$x_1$$, and the open balls are a base for the topology, so necessarily there is an $$r_2>0$$ such that $$B_d(x_1,r_2)\subseteq B_d(x,r)\cap G_1$$, which of course implies that $$B_d(x_1,r_2)\subseteq B_d(x,r)$$ and $$B_d(x_1,r_2)\subseteq G_1$$.
• What about the closure of $B_d(x_1,r_2)$ ?Will it be contained in $G_1$? Dec 31 '20 at 2:13
• @GuriaSona: Not automatically, but a metric space is regular, so you can always choose $r_2$ small enough so that $\operatorname{cl}B_d(x_1,r_2)\subseteq B_d(x,r)\cap G_1$. Dec 31 '20 at 2:28
• I am sorry if it sounds stupid but I'm unable to prove the statement that such a $r_2$ will exist such that closure is contained in the ball?Is it because of the fact that a closed ball of radius $r_2<r$ will be contained in $B_d(x,r)$.(If this is the reason then I can probably see it in $\mathbb{R}^k$) .how do I conclude it in case of any general metric space (say discrete metric space)? Dec 31 '20 at 7:28
• @GuriaSona: In any metric space $\langle X,d\rangle$ it’s true that $\operatorname{cl}B_d(x,r)\subseteq\{y\in X:d(x,y)\le r\}$, but since $x_1$ and $x$ are in general different points, that fact doesn’t help here. As I said, use regularity: there is an open set $U$ such that $$x_1\in U\subseteq\operatorname{cl}U\subseteq B_d(x,r)\cap G_1\,.$$ And $U$ is an open nbhd of $x_1$, so there is an $r_2>0$ such that $B_d(x_1,r_2)\subseteq U$, and hence $$\operatorname{cl}B_d(x_1,r_2)\subseteq\operatorname{cl}U\,.$$ Dec 31 '20 at 7:52 | 2022-01-20T14:31:38 | {
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https://math.stackexchange.com/questions/1286924/find-the-area-of-the-face-of-the-clock | # Find the area of the face of the clock.
The minute hand of a clock is $10$ cm long .Find the area of the face of the clock described by the minute hand between $9$ am and $9:35$ am.
\begin{align}&\color{green}{a.)183.3\quad cm^2}\\ &b.)366.6\quad cm^2\\ &c.)244.4\quad cm^2\\ &d.)188.39\quad cm^2\\ \end{align}
I tried
$Area=\dfrac{30}{360} \times \pi\times 10^2=26.17\quad cm^2$
what is the mistake ?
• What's the angle? Definitely not $30^\circ$... – b00n heT May 17 '15 at 21:04
What makes you think it is $\frac{30}{360}$ ?
The correct rate is $\frac{35}{60}$, since 35 minutes pass and one full circle is 60 minutes.
• Oh ok i see it now. thnx – R K May 17 '15 at 21:06
When the minute hand moved from 9 am to 9:35 am,
It moved $\frac7{12}$ cicrle,that is,
it moved $\frac{7}{12}\cdot360=210$ degree NOT $30$ degree.
In a 12 hour clock, we know that $$1 \space minute=\frac{360}{60}=6^o=\frac{6\times \pi}{180}=\frac{\pi}{30} \space radian$$
Hence, the area swept by the minute hand between 9 am to 9:35 am is $$=\frac{1}{2}\times(\text{total angle swept})\times(\text{length of minute hand})^2=\frac{1}{2}\times\left(35\times \frac{\pi}{30}\right)\times(10)^2=\frac{175\pi}{3}\approx183.2595715 \space cm^2$$
The fraction of full angle in minutes time =$\dfrac{35}{60}$
The fraction of swept area =$\dfrac{35}{60} \pi 10^{2} \approx 183.26.$ | 2020-04-07T04:24:31 | {
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http://mathhelpforum.com/statistics/32197-counting-permutations-combinations-print.html | Counting, permutations and/or combinations
• Mar 26th 2008, 08:30 PM
tony351
Counting, permutations and/or combinations
How many ways can 8 people, 4 men and 4 women, be seated at a round table if people sitting opposite each other must be the opposite gender?
I know this is a circular permutation but I can't seem to come up with this answer.
I thought it would be 7!/2
• Mar 27th 2008, 06:28 AM
awkward
Hi Tony351,
Let's designate one of the men to be Fred and specify the seating relative to Fred's place at the table, numbering the seats counterclockwise as 1,2,3,4,5,6,7,8 with Fred in seat 1.
We can choose the woman opposite Fred in one of 4 ways. There are now 6 people left.
We can choose the person in seat 2 in one of 6 ways and the person opposite in one of 3 ways. There are now 4 people left.
We can choose the person in seat 3 in one of 4 ways and the person opposite in one of 2 ways. There are now 2 people left.
We can choose the person in seat 4 in one of 2 ways and the person opposite in only 1 way. We're done.
Putting it all together, the number of possible arrangements is
4 * (6 * 3) * (4 * 2) * (2 * 1) = 1152.
• Mar 27th 2008, 09:26 AM
iknowone
There are 8 choices for the first person's seat (male or female). Once they sit the opposite chair must be filled by the opposite gender so there are 4 people meeting that criteria, choose 1 to sit.
6 choices for third person (male or female), they sit and there are 3 remaining of the oposite gender one of which must fill the opposite seat. etc.
4 choices for 5th person, 2 choices
2 choices for the 7th person, last one is forced.
8*4*6*3*4*2*2.
The difference here from the proposed answer is the factor of 8 (this counts the 8 rotations of the group around the table for each ordering). You don't count this if the rotation of the table doesn't change the "seating", but the problem isn't entirely clear about that. Since you have the answer as 1152 I suppose they define seating arrangment to be invariant of rotation (true of say a necklace with 4 different black and 4 different white beads) but if you're at dinner I would argue that your actual position (not just your relative position to everyone else at the table) matters. But this is just being nit-picky, the idea is the same.
• Mar 27th 2008, 10:30 AM
tony351
Thank you
Thank you for the detailed answer, cleared up mu confusion :)
• Mar 27th 2008, 05:44 PM
Soroban
Hello, Tony!
Another approach . . .
It's rather primitive, but that's the way my brain works.
Quote:
How many ways can 4 men and 4 women, be seated at a round table
if people sitting opposite each other must be the opposite gender?
Let's seat the men first.
The first man can sit anywhere.
The second man has a choice of the other 7 seats,
. . but must not sit opposite the first man.
Hence, he has 6 choices.
The third man has a choice of the other 6 seats,
. . but he must not sit opposite the first two men.
Hence, he has 4 choices.
The fourth man has a choice of the other 5 seats,
. . but he must not sit opposite the first three men.
Hence, he has 2 choices.
There are: . $6\cdot4\cdot2 \:=\:{\color{blue}48}$ ways for the men to be seated.
The four women can occupy the remaining four seats in $4!= {\color{blue}24}$ ways.
Therefore, there are: . $48 \times 24 \:=\:\boxed{1152}$ seating arrangements. | 2017-05-27T19:56:59 | {
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http://math.stackexchange.com/questions/310499/analysis-convergence | # Analysis convergence
This is a past exam question which im not sure how to solve..
${For \ n=1,2,3,... consider \ f_n: [0,2] \rightarrow \mathbb{R} \ given \ by}$
$f_n(x)=$ $\left\{\begin{array}{l l}nx & \quad \text{0$\leq x <\dfrac{1}{n} $}\\2-nx & \quad\text{$\dfrac{1}{n} \leq x < \dfrac {2}{n}$} \\0 & \quad\text{otherwise} \end{array} \right.$
$a)Determine \ the \ pointwise \ limit \ f(x)=\displaystyle\lim_{n\to \infty}f_n(x)$
$b)Does \ f_n \rightarrow f \ with \ respect \ to \ the \ 1-norm ||g||_1 = \displaystyle\int^2_0 |g(x)|\ dx?$
$c)Does f_n \rightarrow f \ uniformly?$
-
Try sketching the graphs of the first few $f_n$ and see what you can surmise. – David Mitra Feb 21 '13 at 20:01
@DavidMitra I have drawn the graph for the few few $f_n$ but im not sure how this helps. – Mathsstudent147 Feb 21 '13 at 20:32
Can you see that the pointwise limit $0$, to begin with? – 1015 Feb 21 '13 at 20:33
Look at them. Note the graph of $f_n$ is $0$ most everywhere, except for a spike of height $1$ centered at $x=1/n$. As $n$ grows larger, the spike "heads towards" the $y$-axis. So what is the pointwise limit? For b), what do the areas under the graphs tend to? For c), is it true that for large $n$, $f_n$ is uniformly close to the pointwise limit? – David Mitra Feb 21 '13 at 20:35
@DavidMitra The areas under the graph tend to 0 c) For large n, $f_n$ is uniformly close to the pointwise limit, which is 0. – Mathsstudent147 Feb 21 '13 at 20:44
At a) $f_n(0)=0$ for every $n$. For $x>0$ there exists $n_0$ such that $\frac{1}{n}<\frac{1}{n_0}<x$ for every $n>n_0$. Consequently $f_n(x)\to 0$ for every $x\in [0,2]$.
At b) $f_n$ is a triangle with height $1$ and width $2/n$. Consequently $||f_n||=1/(4n)\to 0=||f||$
At c) no since $\max\{f_n(x)-f(x)\}=1$ which is arbitrary large
For c), the $\max$ is $1$. – David Mitra Feb 21 '13 at 20:48 | 2015-01-26T18:29:36 | {
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https://math.stackexchange.com/questions/1146363/what-is-the-probability-of-losing-in-the-taiwainese-imo-teams-game | # What is the probability of losing in the Taiwainese IMO team's game?
Evan Chen's recount of the Taiwanese IMO team's journey recorded a game the team members played at their free time, which runs as the following:
There are $n$ team members (in the actual case $n=6$ but here we simply take $n\geq2$)Every team member points at another team member (whom must be different from him/herself) and thus we obtain a (directed) graph with $n$ vertices and $n$ edges. It has one more edge than a tree and therefore must contain a cycle. Any member who is a vertex of any cycle in this graph loses the game. (So it is possible that everyone loses but it's impossible that no one loses.)
Assume everyone chooses the person s/he points at randomly, what is the probability of a player losing the game?
Reference: Evan Chen's recount
It is: $$p_{1}+\cdots+p_{n-1}$$ where: $$p_{k}:=\frac{n-2}{n-1}\frac{n-3}{n-1}\cdots\frac{n-k}{n-1}\frac{1}{n-1}$$ is the probability of losing together with $k$ mates that are in the same (losing) cycle.
Let's start with denoting the player who starts with $P_0$ and the player he points at with $P_1$. Inductively denote the player pointed at by $P_{n-1}$ with $P_n$. Then $P_0$ will lose with exactly $k$ mates in the same cycle if $P_0,\dots,P_{k-1}$ are distinct and $P_k=P_0$.
If e.g. $P_0$ loses together with $3$ mates $P_1$ must point at $P_2\neq P_0$ (probability $\frac{n-2}{n-1}$), $P_2$ must point at $P_3\notin\{P_0,P_1\}$ (probability $\frac{n-3}{n-1}$) and finally $P_3$ must point at $P_0$ (probability $\frac1{n-1}$).
This leads to $p_3=\frac{n-2}{n-1}\frac{n-3}{n-1}\frac{1}{n-1}$.
• This looks awesome but can you please provide an outline of where the $p_k$'s come from or at least point me to a reference? – Zhipu 'Wilson' Zhao Feb 13 '15 at 11:28
Equivalently to the other answers so far, it's $\displaystyle\sum_{k=2}^n\frac{(k-1)!\binom{n-1}{k-1}}{(n-1)^k}$, or to make the sum work out nicely (adding the missing first term and shifting the index $k$ by one): $$-\frac{1}{n-1}+\displaystyle\sum_{k=0}^{n-1}\frac{k!\binom{n-1}{k}}{(n-1)^{k+1}}$$
I'd just like to note that this can be simplified via using generating functions. We start with the simple generating function $$(1+x)^{n-1}=\displaystyle\sum_{k=0}^{n-1}\binom{n-1}{k}x^k.$$ Note that for $y$ positive, $\displaystyle\int_0^{\infty}x^ke^{-x/y}dx=y^{k+1}\displaystyle\int_{0}^{\infty}z^ke^{-z}dz=k!y^{k+1}$ (using the substitution $x=yz$).
This is a convenient tool allowing us to get the desired $k!$ in our generating function. In particular, we find that
$$\displaystyle\int_0^{\infty}(1+x)^{n-1}e^{-x/y}dx=\displaystyle\sum_{k=0}^{n-1}\binom{n-1}{k}\displaystyle\int_0^{\infty}x^ke^{-x/y}dx=\displaystyle\sum_{k=0}^{n-1}k!\binom{n-1}{k}y^{k+1}.$$
Now the idea is clear, as we have that subsituting $y=\frac{1}{n-1}$, $$\displaystyle\sum_{k=0}^{n-1}\frac{k!\binom{n-1}{k}}{(n-1)^{k+1}}=\displaystyle\int_0^{\infty}(1+x)^{n-1}e^{-(n-1)x}dx=\displaystyle\int_0^{\infty}((1+x)e^{-x})^{n-1}dx,$$ so we have that the probability with $n$ people is $$P_n=-\frac{1}{n-1}+\displaystyle\int_0^{\infty}((1+x)e^{-x})^{n-1}dx.$$ This integral formulation is nice in that it allows us to evaluate this asymptotically, as the integral is $$\displaystyle\int_0^{\infty}e^{(n-1)(\ln(1+x)-x)}dx\approx\displaystyle\int_0^{\infty}e^{(n-1)(-x^2/2)}dx$$ for large $n$, which is a Gaussian integral and evaluates to $\sqrt{\frac{\pi}{2(n-1)}}$. The $\frac{1}{n-1}$ term can thus be ignored and we get an asymptotic result of $\sqrt{\frac{\pi}{2(n-1)}}$.
Let's play this game with $n$ people. I'll call a graph coming from a play of this game a round. Since each person has $n-1$ choices of whom to point to, there are $(n-1)^{n}$ possible rounds.
Let us calculate the number of rounds in which Person $1$ loses.
Of course in every round, every vertex has out-degree exactly $1$. It follows that if Person $1$ loses in a given round, then the cycle they are on is a directed cycle and, moreover, they are in exactly one directed cycle.
If Person $1$ loses because they are in a directed $k$-cycle for some $k\geq 2$, then there are ${n-1\choose k-1}$ choices of people that they can be involved in a $k$-cycle with, and $(k-1)!$ different directed cycles involving just these people. So there are $(k-1)!{n-1\choose k-1}(n-1)^{n-k}$ such rounds.
Note that we are not over-counting anywhere here since it cannot be the case that Person $1$ is in more than one cycle.
Thus the probability of Person $1$ losing is $$\frac{\sum_{k=2}^n(k-1)!{n-1\choose k-1}(n-1)^{n-k}}{(n-1)^n}$$
For $n=6$, this amounts to a probability of a little over a half ($0.50208$). | 2020-10-28T20:42:39 | {
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https://math.stackexchange.com/questions/2995751/proof-matrix-a-beginbmatrix-1-1-0-1-endbmatrixn-text-when-n | # Proof matrix $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^n , \text{ when } n \in \mathbb{N}$ [duplicate]
## Problem
Find generalitazion for matrix A exponents, when $$n\in\{1,2,3,\dots\},n \in \mathbb{N}$$
$$A^n=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^n , \text{ when } n \in \mathbb{N}$$
Proof generalization by induction.
## Attempt to solve
By computing a set of $$A$$ exponent's $$n\in \{\ 1,2,3,4 \}$$. It is possible to form generalization that is applicable for set $$n\in \{1,2,3,4 \}$$
$$A^1=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix},A^2=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix},A^3=\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix},A^4=\begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix} \dots A^n =\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$$
### Induction proof
Induction hypothesis
Assume expression is valid when $$n=k$$
$$A^k = \begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix}$$
Base case
When $$n=1$$ $$A^1=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ which is valid by definition.
Induction step
When $$n=k+1$$
$$A^{k+1} = \begin{bmatrix} 1 & k+1 \\ 0 & 1\end{bmatrix}$$
$$A^{k+1}=A^kA^1$$
By utilizing induction hypothesis we have
$$\implies A^{k+1}=\begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ By utilizing matrix multiplication we have $$\implies A^{k+1} \begin{bmatrix} 1\cdot 1 + k\cdot 0 & 1 \cdot 1 + 1 \cdot k \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot 1 + 1 \cdot 1 \end{bmatrix}$$ $$\implies A^{k+1}=\begin{bmatrix} 1 & k+1 \\ 0 & 1 \end{bmatrix}$$
$$\tag*{\square}$$
## EDIT
The point of posting this was to have comment on if my solution seems correct or not. If you can notice something that doesn't look right, let me know !
## marked as duplicate by Mark McClure, ArsenBerk, Chris Custer, user10354138, Lee David Chung LinNov 13 '18 at 2:00
• It is correct, just that it is confusing the way you "define " $A$ in the first equation as you define $A$ to be what you later call $A^n$. – Surb Nov 12 '18 at 19:36
Your proof looks good to me. There is an alternative proof: Note that $$A=I+N$$, where $$N=\pmatrix{0&1\\ 0&0}$$ squares to zero. Therefore, by binomial expansion, $$A^n=(I+N)^n=I+nN+\binom{n}{2}N^2+\cdots+\binom{n}{n}N^n=I+nN=\pmatrix{1&n\\ 0&1}.$$ | 2019-07-20T16:25:54 | {
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https://byjus.com/question-answer/if-d-1-left-begin-array-r-r-r-1-x-y-z-1-y/ | Question
If $$D _ { 1 } = \left| \begin{array} { r r r } { 1 } & { x } & { y z } \\ { 1 } & { y } & { z x } \\ { 1 } & { z } & { x y } \end{array} \right|$$ and $$D _ { 2 } = \left| \begin{array} { c c c } { 1 } & { 1 } & { 1 } \\ { x } & { y } & { z } \\ { x ^ { 2 } } & { y ^ { 2 } } & { z ^ { 2 } } \end{array} \right|$$ then find relation between $$D_1$$ and $$D_2$$.
Solution
$$D_1=\begin{vmatrix} 1& x& yz \\ 1& y& zx\\ 1& z& xy\end{vmatrix}$$multiply 1st row by $$x$$ and divide 2 nd row by $$y$$by 3rd row by $$z$$$$=\dfrac{1}{xyz}\begin{vmatrix} x& x^2& xyz\\ y& y^2& xyz \\ z& z^2& xyz\end{vmatrix}$$Take $$xyz$$ from 3rd column$$=\dfrac{xyz}{xyz}\begin{vmatrix} x& x^2& 1 \\ y & y^2 & 1 \\ z & z^2& 1\end{vmatrix}$$[column are changed to rows but determinant will not change]$$=\begin{vmatrix} x& y& z\\ x^2& y^2& z^2\\ 1& 1& 1\end{vmatrix}$$Interchanged $$C_3$$ and $$C_1$$ and again $$C_2$$ and $$C_3$$$$=\begin{vmatrix} 1& 1& 1\\ x^2& y^2& z^2\\ x& y& z\end{vmatrix}$$$$C_3\leftrightarrow C_1$$$$=(-1)(-1)\begin{vmatrix} 1 & 1 & 1\\ x& y& z\\ x^2& y^2& z^2\end{vmatrix}\Rightarrow D_1=D_2$$[When you interchange two column/rows determinant will change its sign]Mathematics
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"url": "https://byjus.com/question-answer/if-d-1-left-begin-array-r-r-r-1-x-y-z-1-y/",
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https://julesjacobs.com/2019/03/15/leapfrog-verlet.html | Leapfrog and Verlet are the same method
Leapfrog and Verlet are two popular methods to integrate Newton’s equations of motion in physics simulations and games. These methods occupy a sweet spot between Euler’s method (the simplest method) and higher order methods. They are almost as simple as Euler’s method and use only one force calculation per time step, yet they have crucial advantages: they are second order accurate (compared Euler’s method which is of order one), and they are symplectic and time reversible. Because they have such similar properties I wondered if Leapfrog and Verlet are actually the same method – after all, how many second order sympletic methods can there be? It turns out that they are indeed equivalent, a fact apparently well known among numerical methods people. Their equivalence is easier to understand with code than with math notation, so that’s what I’ll do here.
I’m assuming that we have a function $$a(x)$$ to compute the forces/accelerations from the positions, and $$x$$ and $$v$$ are initialised to the initial positions and velocities. Here is Verlet:
for(i = 0..n){
x_prev = x
x += v*dt + a(x)*dt/2
v += (a(x_prev) + a(x))*dt/2
}
And here is Leapfrog:
for(i = 0..n){
x += v*dt
v += a(x)*dt
}
At first sight there’s no way that they could be equivalent, because Verlet computes $$x(t)$$ and $$v(t)$$ at multiples of the time step $$t=0,1,2,\dots$$, whereas Leapfrog computes $$x(t)$$ at $$t=0,1,2,\dots$$ and $$v(t)$$ at $$t=0+\frac{1}{2},1+\frac{1}{2},2+\frac{1}{2},\dots$$ shifted one half step from each other (that’s why it’s called Leapfrog: the $$x$$ and $$v$$ values leapfrog over each other). However, at the start of the simulation we’re given $$x(0)$$ and $$v(0)$$, whereas Leapfrog requires $$x(0)$$ and $$v(\frac{1}{2})$$. To get Leapfrog started we must compute $$v(\frac{1}{2})$$ first, which we can do with one step of Euler’s method with $$\Delta t=\frac{1}{2}$$. Similarly, at the end we’ve got $$x(n)$$ and $$v(n+\frac{1}{2})$$ whereas we’d like to know $$v(n)$$, which we can do by doing one step of Euler’s method backward with $$\Delta t=-\frac{1}{2}$$. That gives us the corrected Leapfrog:
v += a(x)*dt/2
for(i = 0..n){
x += v*dt
v += a(x)*dt
}
v -= a(x)*dt/2
Let’s rewrite this in an apparently silly way by splitting the $$v$$ update in half:
v += a(x)*dt/2
for(i = 0..n){
x += v*dt
v += a(x)*dt/2
v += a(x)*dt/2
}
v -= a(x)*dt/2
Think about what this is doing by unrolling this loop in your mind:
v += a(x)*dt/2
x += v*dt
v += a(x)*dt/2
v += a(x)*dt/2
x += v*dt
v += a(x)*dt/2
v += a(x)*dt/2
... n times ...
x += v*dt
v += a(x)*dt/2
v += a(x)*dt/2
v -= a(x)*dt/2
The last two updates cancel each other out, so we can remove both. A different way of looking at it emerges:
v += a(x)*dt/2
x += v*dt
v += a(x)*dt/2
v += a(x)*dt/2
x += v*dt
v += a(x)*dt/2
... n times ...
v += a(x)*dt/2
x += v*dt
v += a(x)*dt/2
In other words, the Leapfrog code is equivalent to this:
for(i = 0..n){
v += a(x)*dt/2
x += v*dt
v += a(x)*dt/2
}
An iteration of this loop is exactly the same as an iteration of Verlet! Can you see why? (Hint: incorporate the first $$v$$ update into the subsequent $$x$$ and $$v$$ updates).
## Second variant
Instead of advancing $$v$$ by half a timestep at the start and end of Leapfrog, we could also advance $$x$$ by half a timestep to get the second variant of Leapfrog:
x += v*dt/2
for(i=0 to n){
v += a(x)*dt
x += v*dt
}
x -= v*dt/2
We can do the same rewrite and move everything into the loop:
for(i = 0..n){
x += v*dt/2
v += a(x)*dt
x += v*dt/2
}
This variant has the advantage that $$a(x)$$ is only computed once per iteration. We don’t really need to compute $$a(x)$$ twice in the previous variant of Leapfrog-Verlet either, because the second call to $$a(x)$$ will be the same as the first call in the subsequent iteration, so we could save that instead of recomputing it. That complicates the code a bit, so I find this second variant nicer. By incorporating the first $$x$$ update into the subsequent $$v$$ and $$x$$ updates we obtain the second variant of Verlet:
for(i = 0..n){
v_prev = v
v += a(x+v*dt/2)*dt
x += (v_prev + v)*dt/2
}
This variant of Verlet also has the advantage of only computing $$a(x)$$ once. For some reason the other variant seems to be more popular.
## Conclusion
In my opinion, the best way to write Leapfrog-Verlet is this:
for(i = 0..n){
x += v*dt/2
v += a(x)*dt
x += v*dt/2
}
The advantage is that it’s pretty, computes both $$x$$ and $$v$$ at $$t=0,1,2,\dots$$, and doesn’t use any state other than $$(x,v)$$. The disadvantage is that it updates $$x$$ twice per iteration, instead of once as Leapfrog does. This is likely to be of negligible cost compared to computing $$a(x)$$, but if you really care about it then use the second variant of Leapfrog. Just keep in mind Leapfrog computes $$x$$ at shifted time steps, so if you use $$\frac{1}{2}mv^{2}+U(x)$$ to compute the energy you’ll get an incorrect value. | 2021-06-20T03:54:01 | {
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https://www.physicsforums.com/threads/central-limit-theorem-question.752080/ | # Homework Help: Central Limit Theorem Question
1. May 4, 2014
### dirtybiscuit
1. The problem statement, all variables and given/known data
The Rockwell hardness of certain metal pins is known to have a mean of 50 and a standard deviation of 1.5.
a)if the distribution of all such pin hardness measurements is known to be normal, what is the probability that the average hardness for a random sample of 9 pins is at least 52.
b)what if sample is 40 pins?
2. Relevant equations
So I know that μ = 50 and σ(original) = 1.5, and n = 9
3. The attempt at a solution
σ(sample) = σ(original)/$\sqrt{9}$ = 1.5/3 = .5
now find P($\bar{x}$ > 52) by relating it to the standard normal dist
find z = (52 - 50)/.5 = 4.
This seems like a huge z score to have and it pretty much much makes the probability 0. Am I doing something wrong here? It just seems like a pointless question if in both cases the probability comes to zero.
Last edited: May 4, 2014
2. May 4, 2014
### ehild
The probabilities are really very small, but still differ from zero. There are online calculators http://www.solvemymath.com/online_math_calculator/statistics/cdf_calculator.php [Broken], for example.
ehild
Last edited by a moderator: May 6, 2017
3. May 4, 2014
### Zondrina
Your sample size is quite small. Shouldn't you be using a t-estimator instead?
4. May 4, 2014
### Staff: Mentor
We know (or assume) that all pins follow a normal distribution. Then the mean of 9 pins follows a normal distribution as well.
The calculation is fine, the result is just a small number.
5. May 4, 2014
### Ray Vickson
No. When the variance is known we use the normal distribution. When the variance is unknown (and is estimated using the data itself) we use the t-distribution.
6. May 4, 2014
### Ray Vickson
The probability is small, but nowhere near zero. (In fact, the ability to estimate very small probabilities is crucial in reliability and safety studies, etc.) For large $x > 0$ you can find simple, fairly accurate approximations to $G(x) \equiv P(X > x)$ for standard normal $X$.
Let $$\phi(t) = \frac{1}{\sqrt{2 \pi}} e^{-t^2/2}.$$
We have
$$G(x) = \int_x^{\infty} \phi(t) \, dt \\ A_1(x) = \frac{\phi(x)}{x} \;\; \text{approximation 1}\\ A_2(x) = \frac{\phi(x)}{x} - \frac{\phi(x)}{x^3} \;\; \text{approximation 2}$$
The functions $A_1(x)$ and $A_2(x)$ are approximations to $G(x)$, with $A_2$ being a bit better than $A_1$ for large $x > 0$ (but may be worse for small $x > 0$). You can see how they perform from the following table:
Code (Text):
x G(x) A1(x) A2(x)
1.0 1.58655e-01 2.41971e-01 0.00000e+00
1.5 6.68072e-02 8.63451e-02 4.79695e-02
2.0 2.27501e-02 2.69955e-02 2.02466e-02
2.5 6.20967e-03 7.01132e-03 5.88951e-03
3.0 1.34990e-03 1.47728e-03 1.31314e-03
3.5 2.32629e-04 2.49338e-04 2.28984e-04
4.0 3.16712e-05 3.34576e-05 3.13665e-05
These approximations are based in integration by parts, using the fact that $d \phi(t)/dt = - t \phi(t)$:
$$G(x) = \int_x^{\infty} \phi(t) \, dt = \int_x^{\infty} \frac{1}{t} \left(- \frac{d \phi(t)}{dt} \right) \, dt = \frac{\phi(x)}{x} - \int_x^{\infty} \frac{\phi(t)}{t^2} \, dt$$
The approximation $A_1(x)$ is just the first term above, and it should be a pretty good approximation for large $x > 0$ because the magnitude of the neglected term is less than $G(x)/x^2$. The approximation $A_2(x)$ estimates the second term above through another integration by parts.
Last edited: May 4, 2014 | 2018-07-16T01:36:36 | {
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https://proofwiki.org/wiki/Infimum_in_Ordered_Subset | # Infimum in Ordered Subset
## Theorem
Let $L = \left({S, \preceq}\right)$ be an ordered set.
Let $R = \left({T, \preceq'}\right)$ be an ordered subset of $L$.
Let $X \subseteq T$ such that
$X$ admits an infimum in $L$.
Then $\inf_L X \in T$ if and only if
$X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$
## Proof
By definition of ordered subset:
$T \subseteq S$
and
$\forall x, y \in T: x \preceq' y \iff x \preceq y$
### Sufficient Condition
Let $\inf_L X \in T$.
By definition of infimum:
$\inf_L X$ is lower bound for $X$ in $L$.
By definition of $\preceq'$:
$\inf_L X$ is lower bound for $X$ in $R$.
We will prove that
$\forall x \in T: x$ is lower bound for $X$ in $R \implies x \preceq' \inf_L X$
Let $x \in T$ such that
$x$ is lower bound for $X$ in $R$.
By definition of $\preceq'$:
$x$ is lower bound for $X$ in $L$.
By definition of infimum:
$x \preceq \inf_L X$
Thus by definition of $\preceq'$:
$x \preceq' \inf_L X$
$\Box$
Hence $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$
$\Box$
### Necessary Condition
assume that
$X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$
By definition of infimum:
$\inf_R X \in T$
Thus by assumption:
$\inf_L X \in T$
$\blacksquare$ | 2019-03-25T10:29:03 | {
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https://byjus.com/question-answer/assertion-the-inverse-of-the-matrix-a-begin-bmatrix-1-1-1-1-2-3/ | Question
# Assertion :The inverse of the matrix $$A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{bmatrix}$$ does not exist. Reason: The matrix $$A$$ is singular
A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct
Solution
## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion$$|A|=\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{vmatrix}=0$$$$A$$ is a singular matrix.Inverse exists iff matrix is non-singular.Both Assertion and Reason are correct and Reason is the correct explanation for AssertionHence, option A. Mathematics
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https://math.stackexchange.com/questions/1052785/consequence-of-mean-value-theorem/1052864 | consequence of mean value theorem?
Suppose that $f:[0,1]\to\mathbb{R}$ is differentiable on (0,1) and continuous on [0,1]. Would like to assert that if f(0)=0, and $|f'(x)|\leq |f(x)|$ for each $x\in (0,1)$, then f is the zero function.
I have tried applying several different variations of the mean value theorem, but nothing useful has come of this. I also have that the derivative has a max and min (based on the compact domain and the inequality condition). What am I missing here to get that $f=0$?
• If $0<x<1$, there is a $c_1\in(0,x)$ with $|f(x)|\le |f(c_1)|x$ by MVT. Apply MVT again to obtain $|f(x)|\le|f(c_2)|c_1 x$ with $0<c_2<c_1$. Keep going... – David Mitra Dec 5 '14 at 9:47
Denote by $M$ the quantity $\max_{t\in [0,1]}|f(t)|$. Using the mean value theorem we get for $x\in (0,1)$, $$|f(x)|\leqslant x M.$$ We can prove by induction that $|f(x)|\leqslant x^nM$ for each $x\in (0,1)$. Indeed, it is true for $n=1$, and if it is true for $n$, then by the mean value theorem, $$f(x)=f'(c)x$$ for some $c\in (0,x)$, hence $|f(x)|\leqslant x|f'(c)|\leqslant x|f(c)|$. Using the induction hypothesis, we get $|f(x)|\leqslant xc^nM\leqslant x^{n+1}M$.
Since $x^n\to 0$ as $n$ goes to infinity, we get the wanted conclusion.
• interesting. does this mean that the hypothesis f(0)=0 is unnecessary? – reluctant mathematician Dec 7 '14 at 16:51
• oh, wait... it gets used in the quotient $|\frac{f(x)-f(0)}{x-0}|\leq M$ – reluctant mathematician Dec 7 '14 at 16:58
$$|f(x)|=\Bigl|\int_0^xf'(t)\,dt\Bigr|\le\int_0^x|f'(t)|\,dt\le\int_0^x|f(t)|\,dt.$$ Let $g(x)=\int_0^x|f(t)|\,dt$. Then $$g(x)\ge0,\quad g(0)=0,\quad g'(x)-g(x)\le0.$$ Multiplynig the last inequality by $e^{-x}$ we get $$\bigl(e^{-x}g(x)\bigr)'\le0\implies e^{-x}g(x)\le g(0)=0.$$
Actualy, this is a particular case of Gronwall's lemma.
• Don't you need to assume $f'$ is integrable here? – David Mitra Dec 5 '14 at 11:59
• $f'$ is defined everywhere and bounded, so it is Lebesgue integrable and the Fundamental theorem of Calculus applies. – Julián Aguirre Dec 5 '14 at 13:54
• Ah, thanks.${}$ – David Mitra Dec 5 '14 at 14:16
• This proof even works for the more general condition $|f'(x)| \le K |f(x)|$ for some positive constant $K$. – Martin R Mar 2 '17 at 10:40 | 2020-10-20T01:04:23 | {
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