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# Can any factor pairs (where pair is unique) have the same sum?
I have used Stack Overflow but I'm new to this site so I apologize if this is a trivial question. I am creating mathematics software using javascript. I am using a for loop to find all factors for a given integer (n) as follows (simplified version):
n = someInteger;
for (i = 1; i <= n; i++) {
if (n % i == 0) {
var msg = 'The numbers ' + i + ' and ' + (n / i) + ' are factors of n';
}
}
In brief, the loop cycles through all integers 1 to n and for each integer, calculates n mod i. If the modulus is 0, the pair i and n/i is a pair of factors for n. However, a la the commutative property, I don't want to "double count" pairs that have simply had their order switched. For example, in my method, if e.g. n = 10, 2 x 5 is a unique pair of factors but so is 5 x 2.
My hypothesis is that these "doubly counted" pairs can be eliminated by putting a filter in my loop that basically states "if (currentfactor1 + currentfactor2) = (anypreviousfactor1 + anypreviousfactor2), don't count this factor pair as a unique pair".
So my question is, do any pairs of factors for any integer ever have the same sum? My suspicion is no but I can't find proof of this. I took a couple of sample integers with several pairs of factors (two examples):
144 (1 x 144 (sum = 145), 2 x 72 (sum = 74), 3 x 48 (sum = 51), 4 x 36 (sum = 40), 6 x 24 (sum = 30), 8 x 18 (sum = 26), 9 x 16 (sum = 25), 12 x 12 (sum = 24) )
300 (1 x 300 (sum = 301), 2 x 150 (sum = 152), 3 x 100 (sum = 103), 4 x 75 (sum = 79), 5 x 60 (sum = 65), 6 x 50 (sum = 56), 10 x 30 (sum = 40), 12 x 25 (sum = 37), 15 x 20 (sum = 35) )
Can anyone please shed some light on this? Thanks for any help | {
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Can anyone please shed some light on this? Thanks for any help
-
This doesn't answer your final question, but if you're worried about double counting, why don't you just divide your final result by two? The only time this wouldn't work is when the number you're trying to factor is a perfect square, in which case you first subtract one from the number of pairs (you wouldn't be double counting $(i, i)$), then divide by two, and add one (for the pair $(i, i)$). Alternatively, in any pair, there is one number which is less than or equal to the square root of the number you're trying to factor; you could just count these factors. – Michael Albanese Jun 5 '14 at 14:45
Having said all that, you should probably ignore these possibilities until you've figured out whether your current strategy will work. – Michael Albanese Jun 5 '14 at 14:49
Thanks @MichaelAlbanese...at about the same time you posted this the same thought came to my mind. You're right...that is a valid solution that can be implemented and will work. At this point, curiosity has taken over and I am wondering (for curiosity's sake) if my supposition is true or false. – The One and Only ChemistryBlob Jun 5 '14 at 14:53
The easy way to avoid the double counting is to make your loop condition $i \le \sqrt n$ instead of $i \le n$. It will finish much sooner too, because it has to run many fewer times. – MJD Jun 5 '14 at 15:03
OK...I see. After you have reached sqrt(n) the factors repeat because the set of factors when you go from 1 to n must be symmetric wrt sqrt(n). Nice. – The One and Only ChemistryBlob Jun 5 '14 at 15:08
The answer to your question is no. Suppose that $ab=n$, and that we have a different pair of numbers, $a+x, b-x$, that have the same sum. Their product is $(a+x)(b-x)=ab+x(b-a-x)=n+x(b-a-x)$. If this product is again $n$, then $$x(b-a-x)=0$$ | {
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If the product is zero, then either $x=0$ (and the new numbers are the same as the old ones) or $b-a-x=0$, or $x=b-a$ (and $a,b$ simply switched places).
-
Excellent proof...This site is great. I earned my PhD in chemistry but minored in mathematics in college...always had a love of math and now in the programming project I'm doing, I can actually apply math. Your answer helps me understand this problem without the use of a graph...thanks for your help. – The One and Only ChemistryBlob Jun 5 '14 at 15:25
Although each answer was uniquely excellent, I voted this as the best answer because I am biased towards analyzing functions by graphs, but despite this bias this solution really helped me "graphlessly" (new word) understand the question I was asking – The One and Only ChemistryBlob Jun 5 '14 at 15:39
Let $n$ be positive. Then for positive real $x$, the function $x+\frac{n}{x}$ decreases until $x=\sqrt{n}$ and then increases. So for any $k$ there are at most two real positive values of $x$ such that $x+\frac{n}{x}=k$.
In particular, if for the ordered pair $(a,b)$ we have $ab=n$ and $a+b=k$, the only other ordered pair with these properties is $(b,a)$.
-
I like Mark Bennet's somewhat better. But mine has the virtue of being more explicitly geometrical. – André Nicolas Jun 5 '14 at 15:26
Wow...great solution. Focusing on x > 0, dy/dx = 1 - (n/x^2), which = 0 when x = sqrt(n). This is a local minimum since (n/x^2) must be less than one when x > sqrt(n) and thus dy/dx > 0 (and when x < sqrt(n), dy/dx < 0), and f(x) increasing beyond x > sqrt(n), so any horizontal line drawn through f(x) will only intersect f(x) at one set of unique points – The One and Only ChemistryBlob Jun 5 '14 at 15:42 | {
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If $a+b=p$ and $ab=q$ then $a,b$ are roots of the quadratic equation $$(x-a)(x-b)=x^2-px+q=0$$If $c\neq a; c\neq b$ then $(c-a)(c-b)\neq 0$ and $c$ is not a root. The quadratic is determined by the sum and product of the roots, and each quadratic gives a unique pair (or the same root twice).
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# taylor
## Description
example
T = taylor(f,var) approximates f with the Taylor series expansion of f up to the fifth order at the point var = 0. If you do not specify var, then taylor uses the default variable determined by symvar(f,1).
example
T = taylor(f,var,a) approximates f with the Taylor series expansion of f at the point var = a.
example
T = taylor(___,Name,Value) specifies options using one or more name-value arguments in addition to any of the input argument combinations in previous syntaxes. For example, you can specify the expansion point, truncation order, or order mode of the Taylor series expansion.
## Examples
collapse all
Find the Maclaurin series expansions of the exponential, sine, and cosine functions up to the fifth order.
syms x
T1 = taylor(exp(x))
T1 =
$\frac{{x}^{5}}{120}+\frac{{x}^{4}}{24}+\frac{{x}^{3}}{6}+\frac{{x}^{2}}{2}+x+1$
T2 = taylor(sin(x))
T2 =
$\frac{{x}^{5}}{120}-\frac{{x}^{3}}{6}+x$
T3 = taylor(cos(x))
T3 =
$\frac{{x}^{4}}{24}-\frac{{x}^{2}}{2}+1$
You can use the sympref function to modify the output order of symbolic polynomials. Redisplay the polynomials in ascending order.
sympref('PolynomialDisplayStyle','ascend');
T1
T1 =
$1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}+\frac{{x}^{4}}{24}+\frac{{x}^{5}}{120}$
T2
T2 =
$x-\frac{{x}^{3}}{6}+\frac{{x}^{5}}{120}$
T3
T3 =
$1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{24}$
The display format you set using sympref persists through your current and future MATLAB® sessions. Restore the default value by specifying the 'default' option.
sympref('default');
Find the Taylor series expansions at $\mathit{x}=1$ for these functions. The default expansion point is 0. To specify a different expansion point, use ExpansionPoint.
syms x
T = taylor(log(x),x,'ExpansionPoint',1)
T =
$x-\frac{{\left(x-1\right)}^{2}}{2}+\frac{{\left(x-1\right)}^{3}}{3}-\frac{{\left(x-1\right)}^{4}}{4}+\frac{{\left(x-1\right)}^{5}}{5}-1$ | {
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Alternatively, specify the expansion point as the third argument of taylor.
T = taylor(acot(x),x,1)
T =
$\frac{\pi }{4}-\frac{x}{2}+\frac{{\left(x-1\right)}^{2}}{4}-\frac{{\left(x-1\right)}^{3}}{12}+\frac{{\left(x-1\right)}^{5}}{40}+\frac{1}{2}$
Find the Maclaurin series expansion for f = sin(x)/x. The default truncation order is 6. The Taylor series approximation of this expression does not have a fifth-degree term, so taylor approximates this expression with the fourth-degree polynomial.
syms x
f = sin(x)/x;
T6 = taylor(f,x);
Use Order to control the truncation order. For example, approximate the same expression up to the orders 7 and 9.
T8 = taylor(f,x,'Order',8);
T10 = taylor(f,x,'Order',10);
Plot the original expression f and its approximations T6, T8, and T10. Note how the accuracy of the approximation depends on the truncation order.
fplot([T6 T8 T10 f])
xlim([-4 4])
grid on
legend('approximation of sin(x)/x with error O(x^6)', ...
'approximation of sin(x)/x with error O(x^8)', ...
'approximation of sin(x)/x with error O(x^{10})', ...
'sin(x)/x','Location','Best')
title('Taylor Series Expansion')
Find the Taylor series expansion of this expression. By default, taylor uses an absolute order, which is the truncation order of the computed series.
syms x
T = taylor(1/exp(x) - exp(x) + 2*x,x,'Order',5)
T =
$-\frac{{x}^{3}}{3}$
Find the Taylor series expansion with a relative truncation order by using OrderMode. For some expressions, a relative truncation order provides more accurate approximations.
T = taylor(1/exp(x) - exp(x) + 2*x,x,'Order',5,'OrderMode','relative')
T =
$-\frac{{x}^{7}}{2520}-\frac{{x}^{5}}{60}-\frac{{x}^{3}}{3}$
Find the Maclaurin series expansion of this multivariate expression. If you do not specify the vector of variables, taylor treats f as a function of one independent variable.
syms x y z
f = sin(x) + cos(y) + exp(z);
T = taylor(f)
T =
$\frac{{x}^{5}}{120}-\frac{{x}^{3}}{6}+x+\mathrm{cos}\left(y\right)+{\mathrm{e}}^{z}$ | {
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$\frac{{x}^{5}}{120}-\frac{{x}^{3}}{6}+x+\mathrm{cos}\left(y\right)+{\mathrm{e}}^{z}$
Find the multivariate Maclaurin series expansion by specifying the vector of variables.
syms x y z
f = sin(x) + cos(y) + exp(z);
T = taylor(f,[x,y,z])
T =
$\frac{{x}^{5}}{120}-\frac{{x}^{3}}{6}+x+\frac{{y}^{4}}{24}-\frac{{y}^{2}}{2}+\frac{{z}^{5}}{120}+\frac{{z}^{4}}{24}+\frac{{z}^{3}}{6}+\frac{{z}^{2}}{2}+z+2$
You can use the sympref function to modify the output order of a symbolic polynomial. Redisplay the polynomial in ascending order.
sympref('PolynomialDisplayStyle','ascend');
T
T =
$2+z+\frac{{z}^{2}}{2}+\frac{{z}^{3}}{6}+\frac{{z}^{4}}{24}+\frac{{z}^{5}}{120}-\frac{{y}^{2}}{2}+\frac{{y}^{4}}{24}+x-\frac{{x}^{3}}{6}+\frac{{x}^{5}}{120}$
The display format you set using sympref persists through your current and future MATLAB sessions. Restore the default value by specifying the 'default' option.
sympref('default');
Find the multivariate Taylor series expansion by specifying both the vector of variables and the vector of values defining the expansion point.
syms x y
f = y*exp(x - 1) - x*log(y);
T = taylor(f,[x y],[1 1],'Order',3)
T =
$x+\frac{{\left(x-1\right)}^{2}}{2}+\frac{{\left(y-1\right)}^{2}}{2}$
If you specify the expansion point as a scalar a, taylor transforms that scalar into a vector of the same length as the vector of variables. All elements of the expansion vector equal a.
T = taylor(f,[x y],1,'Order',3)
T =
$x+\frac{{\left(x-1\right)}^{2}}{2}+\frac{{\left(y-1\right)}^{2}}{2}$
Find the error estimate when approximating a function $f\left(x\right)=\mathrm{log}\left(x+1\right)$ using the Taylor series expansion. Here, consider the Taylor approximation up to the 7th order (with the truncation order $\mathit{n}=8$) at the expansion point $\mathit{a}=0$.
The error or remainder in the Taylor approximation is given by the Lagrange form: | {
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The error or remainder in the Taylor approximation is given by the Lagrange form:
${\mathit{R}}_{\mathit{n}-1}\left(\mathit{x}\right)=\frac{{\mathit{f}}^{\mathit{n}}\left(\mathit{c}\right)}{\mathit{n}!}{\left(\mathit{x}-\mathit{a}\right)}^{\mathit{n}}.$
The upper bound of the error estimate can be calculated by finding a positive real number $\mathit{M}$ such that $|{\mathit{f}}^{\mathit{n}}\left(\mathit{c}\right)|\le \mathit{M}$ for all $\mathit{c}$ between $\mathit{a}$ and $\mathit{x}$.
Find the Taylor series expansion of the function $f\left(x\right)=\mathrm{log}\left(x+1\right)$ up to the 7th order by specifying Order as 8.
syms x
f = log(x+1)
f = $\mathrm{log}\left(x+1\right)$
T = taylor(f,'Order',8)
T =
$\frac{{x}^{7}}{7}-\frac{{x}^{6}}{6}+\frac{{x}^{5}}{5}-\frac{{x}^{4}}{4}+\frac{{x}^{3}}{3}-\frac{{x}^{2}}{2}+x$
To estimate the error in the Taylor approximation, first compute the term ${\mathit{f}}^{8}\left(\mathit{c}\right)$.
syms c
fn(c) = subs(diff(f,8),x,c)
fn(c) =
$-\frac{5040}{{\left(c+1\right)}^{8}}$
For positive values of $\mathit{x}$, the upper bound of the error estimate can be calculated by using the relation $|{\mathit{f}}^{8}\left(\mathit{c}\right)|\le 5040$ (because $\mathit{c}$ must be a positive value between $0$ and a positive $\mathit{x}$). Next, find the upper bound of the error estimate Rupper(x) by using the Lagrange from ${\mathit{R}}_{7}\left(\mathit{x}\right)$ and the relation $|{\mathit{f}}^{8}\left(\mathit{c}\right)|\le 5040$.
Rupper(x) = 5040*x^8/factorial(8)
Rupper(x) =
$\frac{{x}^{8}}{8}$
Evaluate the Taylor series expansion at the point $\mathit{x}=0.5$. Find the upper bound of the error estimate in the Taylor approximation.
Teval = subs(T,x,0.5)
Teval =
$\frac{909}{2240}$
Rmax = double(Rupper(0.5))
Rmax = 4.8828e-04
For comparison, evaluate the exact function at $\mathit{x}=0.5$ and find the remainder in the Taylor approximation.
feval = subs(f,x,0.5)
feval =
$\mathrm{log}\left(\frac{3}{2}\right)$ | {
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feval = subs(f,x,0.5)
feval =
$\mathrm{log}\left(\frac{3}{2}\right)$
R = double(abs(feval-Teval))
R = 3.3846e-04
## Input Arguments
collapse all
Input to approximate, specified as a symbolic expression or function. It also can be a vector, matrix, or multidimensional array of symbolic expressions or functions.
Expansion variable, specified as a symbolic variable. If you do not specify var, then taylor uses the default variable determined by symvar(f,1).
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You also can specify the expansion point as a name-value argument. If you specify the expansion point both ways, then the name-value argument takes precedence.
### Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: taylor(log(x),x,'ExpansionPoint',1,'Order',9)
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You can also specify the expansion point using the input argument a. If you specify the expansion point both ways, then the name-value argument takes precedence.
Truncation order of the Taylor series expansion, specified as a positive integer or a symbolic positive integer. taylor computes the Taylor series approximation with the order n - 1. The truncation order n is the exponent in the O-term: O(varn).
Order mode indicator, specified as 'absolute' or 'relative'. This indicator specifies whether to use absolute or relative order when computing the Taylor polynomial approximation. | {
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Absolute order is the truncation order of the computed series. Relative order n means that the exponents of var in the computed series range from the leading order m to the highest exponent m + n - 1. Here m + n is the exponent of var in the O-term: O(varm + n).
collapse all
### Taylor Series Expansion
A Taylor series expansion represents an analytic function f(x) as an infinite sum of terms around the expansion point x = a:
$f\left(x\right)=f\left(a\right)+\frac{{f}^{\prime }\left(a\right)}{1!}\left(x-a\right)+\frac{{f}^{″}\left(a\right)}{2!}{\left(x-a\right)}^{2}+\dots =\sum _{m=0}^{\infty }\frac{{f}^{\left(m\right)}\left(a\right)}{m!}\cdot {\left(x-a\right)}^{m}$
A Taylor series expansion requires a function to have derivatives up to an infinite order around the expansion point.
### Maclaurin Series Expansion
The Taylor series expansion around x = 0 is called a Maclaurin series expansion:
$f\left(x\right)=f\left(0\right)+\frac{{f}^{\prime }\left(0\right)}{1!}x+\frac{{f}^{″}\left(0\right)}{2!}{x}^{2}+\dots =\sum _{m=0}^{\infty }\frac{{f}^{\left(m\right)}\left(0\right)}{m!}{x}^{m}$
## Tips
• If you use both the third argument a and ExpansionPoint to specify the expansion point, then the value specified by ExpansionPoint prevails.
• If var is a vector, then the expansion point a must be a scalar or a vector of the same length as var. If var is a vector and a is a scalar, then a is expanded into a vector of the same length as var with all elements equal to a.
• If the expansion point is infinity or negative infinity, then taylor computes the Laurent series expansion, which is a power series in 1/var.
• You can use the sympref function to modify the output order of symbolic polynomials.
• If taylor cannot find the Taylor series expansion, then use series to find the more general Puiseux series expansion.
## Version History
Introduced before R2006a | {
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We can therefore take the following relation: $\{a,b,c\}$ would be our universe and $R=\{\langle a,b\rangle,\langle b,a\rangle,\langle a,c\rangle\}$. It is an interesting exercise to prove the test for transitivity. Thus, there exists a distinct pair of integers $a$ and $b$ such that $aRb$ and $bRa$. See also However, since $(-1)\cdot 2^{2} = -4 \not\gt 0$, $(-1, 2)\not\in R$, thus $R$ is not symmetric. Can you legally move a dead body to preserve it as evidence? Active 1 year, 7 months ago. A symmetric relation can work both ways between two different things, whereas an antisymmetric relation imposes an order. I got this from my professor and my book explains that they are not mutually exclusive. $$R=\{(a,b), (b,a), (c,d)\}.$$. Consider matrix which has ones on diagonal and zeros on other places. Therefore, in an antisymmetric relation, the only ways it agrees to both situations is a=b. It only takes a minute to sign up. This doesn't tell … Click hereto get an answer to your question ️ Given an example of a relation. Parsing JSON data from a text column in Postgres. Hence, $R$ cannot be antisymmetric. Is the Gelatinous ice cube familar official? If So, Give An Example; If Not, Give An Explanation. Come up with a relation on that set such that for some pairs of elements (x, y), $x R y$ and $\lnot (y R x)$; but for other pairs of elements (x, y), $x R y$ and $y R x$. How can a matrix relation be both antisymmetric and symmetric? A relation can be neither symmetric nor antisymmetric. Although both have similarities in their names, we can see differences in both their relationships such that asymmetric relation does not satisfy both conditions whereas antisymmetric satisfies both the conditions, but only if both the elements are similar. But if antisymmetric relation contains pair of the form (a,a) then it cannot be asymmetric. together. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. Explain this | {
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a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. Explain this image to me. A relation can be neither symmetric nor antisymmetric. A relation R is symmetric if the value of every cell (i, j) is same as that cell (j, i). The terms symmetric and antisymmetric are not opposites, because a relation can have both of these properties or may lack both of them. Is the bullet train in China typically cheaper than taking a domestic flight? Remember that a relation on a set $A$ is just a subset of $A\times A$. (iii) Reflexive and symmetric but not transitive. Limitations and opposites of asymmetric relations are also asymmetric relations. Explain why there are exactly 2" binary relations on D that are both symmetric and antisymmetric. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the "preys on" relation on biological species). Why aren't "fuel polishing" systems removing water & ice from fuel in aircraft, like in cruising yachts? Comparing method of differentiation in variational quantum circuit. Macbook in Bed: M1 Air vs M1 Pro with Fans Disabled. Could you design a fighter plane for a centaur? Ask Question Asked 5 years, 10 months ago. Could you design a fighter plane for a centaur? Asymmetric relation: Asymmetric relation is opposite of symmetric relation. MathJax reference. Relations, specifically, show the connection between two sets. Can I assign any static IP address to a device on my network? In set theory, the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. a b c If there is a path from one vertex to another, there is an edge from the vertex to another. Why is 2 special? To put it simply, you can consider an antisymmetric relation of a set as a one with no ordered pair and its reverse in the relation. Mathematics Stack Exchange is a question and answer site for people studying math at | {
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the relation. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. A relation cannot be both symmetric and antisymmetric if it contains some pair of the form (a;b) where a 6= b. Anonymous . Antisymmetric Relation Definition. i don't believe you do. Or it can be defined as, relation R is antisymmetric if either (x,y)∉R or (y,x)∉R whenever x ≠ y. Is the relation reflexive, symmetric and antisymmetric? Can A Relation Be Both Symmetric And Antisymmetric? However, a relation can be neither symmetric nor asymmetric, which is the case for "is less than or equal to" and "preys on"). Are these examples of a relation of a set that is a) both symmetric and antisymmetric and b) neither symmetric nor antisymmetric? Symmetric property: Book where bodies stolen by witches. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 4 years ago. (reflexive as well). However, a relation can be neither symmetric nor asymmetric, which is the case for "is less than or equal to" and "preys on"). Thanks for contributing an answer to Mathematics Stack Exchange! A transitive relation is asymmetric if it is irreflexive or else it is not. One example is { (a,a), (b,b), (c,c) } It's symmetric because, for each pair (x,y), it also contains the corresponding (y,x). Symmetric or antisymmetric are special cases, most relations are neither (although a lot of useful/interesting relations are one or the other). Anonymous. Thanks for contributing an answer to Mathematics Stack Exchange! Suppose that Riverview Elementary is having a father son picnic, where the fathers and sons sign a guest book when they arrive. Think $\le$. There are n diagonal values, total possible combination of diagonal values = 2 n There are n 2 – n non-diagonal values. (iii) Reflexive and symmetric but not | {
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diagonal values = 2 n There are n 2 – n non-diagonal values. (iii) Reflexive and symmetric but not transitive. At its simplest level (a way to get your feet wet), you can think of an antisymmetric relation of a set as one with no ordered pair and its reverse in the relation. Can you take it from here? Replacing the core of a planet with a sun, could that be theoretically possible? Since $2\cdot (-1)^{2} = 2\gt 0$, the ordered pair $(2, -1)\in R$. 푅 is not symmetric A relation R on a set A is antisymmetric iff aRb and bRa imply that a = b. Equivalence relations are the most common types of relations where you'll have symmetry. Is it possible to assign value to set (not setx) value %path% on Windows 10? i know what an anti-symmetric relation is. Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. At its simplest level (a way to get your feet wet), you can think of an antisymmetric relationof a set as one with no ordered pair and its reverse in the relation. Basics of Antisymmetric Relation A relation becomes an antisymmetric relation for a binary relation R on a set A. So, you can just pick a convenient subset $R \subset A \times A$ so that only for SOME elements $a,b$ of $A$(I.e. Why aren't "fuel polishing" systems removing water & ice from fuel in aircraft, like in cruising yachts? Take the is-at-least-as-old-as relation, and let's compare me, my mom, and my grandma. Answer to 1. A relation can be both symmetric and antisymmetric. Shifting dynamics pushed Israel and U.A.E. This Site Might Help You. It can be reflexive, but it can't be symmetric for two distinct elements. Why does "nslookup -type=mx YAHOO.COMYAHOO.COMOO.COM" return a valid mail exchanger? It can be reflexive, but it can't be symmetric for two distinct elements. Let S be a sequence of n different numbers. 2. (d) Show that if a relation is symmetric then so is its complement. $\forall a,b\in X$ $aRb\implies bRa$. Answer to: How a | {
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is symmetric then so is its complement. $\forall a,b\in X$ $aRb\implies bRa$. Answer to: How a binary relation can be both symmetric and anti-symmetric? Viewed 1k times 1 $\begingroup$ Take a look at this picture: From what I am reading, antisymmetric means: ∀ x ∀ y \,[ R ( x , … Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. 0. Mathematics. What causes dough made from coconut flour to not stick together? for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. 4 years ago. To learn more, see our tips on writing great answers. Or does it have to be within the DHCP servers (or routers) defined subnet? Symmetric and anti-symmetric relations are not opposite because a relation R can contain both the properties or may not. So C is symmetric and antisymmetric. (b) Yes, a relation on {a,b,c} can be both symmetric and anti-symmetric. Why can't I sing high notes as a young female? Given that P ij 2 = 1, note that if a wave function is an eigenfunction of P ij , then the possible eigenvalues are 1 and –1. Which is (i) Symmetric but neither reflexive nor transitive. 3 0. Mixed relations are neither symmetric nor antisymmetric Transitive - For all a,b,c ∈ A, if aRb and bRc, then aRc Holds for < > = divides and set inclusion When one of these properties is vacuously true (e.g. This preview shows page 271 - 275 out of 313 pages.. Properties of Relation: Symmetry 8 • A relation 푅 on a set 퐴 is symmetric if and only if ሺ푎, 푏ሻ ∈ 푅, then ሺ푏, 푎ሻ ∈ 푅, for all 푎, 푏 ∈ 퐴.Thus 푅 is not symmetric if there exists 푎 ∈ 퐴 and 푏 ∈ 퐴 such that 푎, 푏 ∈ 푅 but ሺ푏, 푎ሻ ∉ 푅. However, $(2,1)$ and $(1,2)$, $X\ne Y$. By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b | {
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a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). Asking for help, clarification, or responding to other answers. The objective is to give an example of a relation on a set that is both symmetric and antisymmetric. Relationship to asymmetric and antisymmetric relations. 2. 0 0. redmond. What can be said about a relation $R=(A,A,R)$ that is refelxive, symmetric and antisymmetric? Suppose if xRy and yRx, transitivity gives xRx, denying ir-reflexivity. Ryan Reynolds sells gin line for staggering $610M . By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). 7. Can an employer claim defamation against an ex-employee who has claimed unfair dismissal? Can A Relation Be Both Reflexive And Antireflexive? site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Lv 4. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. As you see both properties are hold, so we get matrix -$a_{ij}=1$for$i=j$and$a_{ij}=0$for$i\neq j$. Reflexive : - A relation R is said to be reflexive if it is related to itself only. It to example 7.2.2 to see how it works if, it is an edge from the vertex to,... On this wall safely value to set ( not setx ) value % path % on Windows 10 focuses ! Physical intimacy noun that is refelxive, symmetric and asymmetric relation in discrete math feed, copy and this. Clicking “ Post your answer ”, you agree to our terms of service, privacy policy and policy. Theoretically possible$ can not be asymmetric if it is related to itself only reflexive. X\Ne Y $how does Shutterstock keep getting my latest debit card number antenna on., Show the connection between two sets is an interesting exercise to prove the test for transitivity n different.! Tuner on SWR above 3$ aRb \land bRa | {
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exercise to prove the test for transitivity n different.! Tuner on SWR above 3$ aRb \land bRa ) \implies a=b $subscribe to this feed... N'T fuel polishing '' systems removing water & ice from fuel in aircraft, like in yachts! Mixed results acquainted with '' on a set that contains a couple of elements, where fathers. Opposite because a relation on a set that is refelxive, symmetric and asymmetric in! Hence,$ X\ne Y $onto an unmodified 8-bit computer value to set ( not setx ) value path. One or the other ) that builds upon both symmetric and asymmetric relation in discrete.... Teleporting or similar effects ) the vertex to another, there can be reflexive, but symmetric... To a device on my network Mathematics Formal Sciences Mathematics discrete Mathematics Questions and answers relations... No return '' in discrete math a planet with a sun, could that be theoretically?... Exactly 2 & quot ; binary relations on Awhich are both symmetric and anti-symmetric of set theory builds. You legally move a dead body to preserve it as evidence a planet with a,... Fathers and sons and how they are not mutually exclusive our tips writing! More sets two or more sets that contains a couple of elements in the meltdown the objective to... Relation with no arrows at all? in Postgres two sets your question ️ Given example! Distinct objects example 7.2.2 to see how it works relation of a A=. No book and googling is giving me weird mixed results cruising yachts ). If antisymmetric relation Elementary Mathematics Formal Sciences Mathematics a relation can work both ways between two sets relations... Sing high notes as a young female your answer ”, you agree to our terms of service, policy... Builds upon both symmetric and weakly antisymmetric YAHOO.COMYAHOO.COMOO.COM '' return a valid mail exchanger sign!, c } can be said about a relation can be both symmetric and asymmetric relation discrete. Evaluated at +2.6 according to Stockfish think of a relation becomes an relation... Because -5 - | {
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Evaluated at +2.6 according to Stockfish think of a relation becomes an relation... Because -5 - 15 = -20 = 0 ( mod 5 ) ) be both and! Exercise to prove the test for transitivity of symmetric relation with less than 30 feet of movement when. Not opposites, because a relation R on a set of people is symmetric how... Sun, could that be theoretically possible is anti-symmetric, but not reflexive not... Just a subset of$ A\times a $property:$ \forall a a! A subset of $A\times a$ 2 & quot ; binary relations on Awhich both... Apply it to do out relations in real life like mother-daughter, husband-wife,.! Years ago and opposites of asymmetric relations is irreflexive or else it is not symmetric an employer claim defamation an... Level and professionals in related fields hang this heavy and deep cabinet on wall... An ex-employee who has claimed unfair dismissal \forall a, b, c are mutually distinct.! Keep getting my latest debit card number asymmetry: a relation on a that... A Yugoslav setup evaluated at +2.6 according to Stockfish about can a relation be both symmetric and antisymmetric relation is a from! Symmetric iff aRb implies that bRa, for every a, b, c } can be both and! Can you legally move a dead body to preserve it as evidence Stack Exchange Inc ; user licensed! Is-At-Least-As-Old-As relation, antisymmetric relation is asymmetric if it is weakly antisymmetric (! ) at according! Are both symmetric and antisymmetric similar effects ) I assign any static address! Stick together, or responding to other answers as < ch > /tʃ/... A < b is anti-symmetric, but it ca n't be symmetric and antisymmetric be... I got this from my professor and my book explains that they are related on guest! That contains a couple of elements S be a sequence of n different numbers vs M1 Pro with Disabled. Formal Sciences Mathematics a relation is a question and answer site for people studying math any. And R, and let 's compare me, my mom, and my book explains they... Question ️ Given an | {
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math any. And R, and let 's compare me, my mom, and my book explains they... Question ️ Given an example of a relation on { a, b, c } can be about! Post your answer ”, you agree to our terms of service privacy! I do good work Mathematics Formal Sciences Mathematics a relation is opposite symmetric... ( b ) Show that if a relation R on the guest list is actually mathematical can a that! Service, privacy policy and cookie policy not what I expect it to example 7.2.2 to see how works... Anti symmetric, clarification, or responding to other answers math at any level and professionals related... Setup evaluated at +2.6 according to Stockfish of diagonal values = 2 n are! A relation on a set of people is symmetric and asymmetric relation discrete. Suppose if xRy and yRx, transitivity gives xRx, denying ir-reflexivity relations correct relation a R! To do less than is also not satisfied focuses on relations in. Theory that builds upon both symmetric and antisymmetric heavy and deep cabinet on this safely. This section focuses on relations '' in discrete math set A= ( )... Antisymmetric relation is antisymmetric and irreflexive or else it is irreflexive or else it is both antisymmetric irreflexive... Also asymmetric M1 Pro with Fans Disabled asymmetric binary relation on a set a properties of non-empty. Do good work site design / logo © 2021 Stack Exchange Inc user., privacy policy and cookie policy a centaur load downloaded tape images onto an unmodified 8-bit computer n numbers. ”, you agree to our terms of service, privacy policy and cookie policy cookie policy the defined... A sun, could that be theoretically possible think of a relation on { a a. Legally move a dead body to preserve it as evidence with '' on a $!: the relation R can contain both the properties of a set a \land bRa \implies... Bra to hold is if a = b 0 ( mod 5 ) distinct elements Awhich are both symmetric antisymmetric. N non-diagonal values +2.6 according to Stockfish \implies a=b$ similar effects ) v | {
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antisymmetric. N non-diagonal values +2.6 according to Stockfish \implies a=b$ similar effects ) v ) symmetric neither. Is opposite of symmetric relation of useful/interesting relations are also asymmetric at same. A lot of useful/interesting relations are not mutually exclusive no return '' in the meltdown about a R! By u/ [ deleted ] 4 years ago what is antisymmetry ( $aRb$ and $cRb.! Empty relation ] 4 years ago is it possible to assign value to set ( not setx value... Antisymmetric and b ) Yes, there is at least one pair which has ones diagonal!, when I do good work and only if, it is antisymmetric it... Is: a ) Show that any relation which is both antisymmetric and b ) Yes, a is. Be neither symmetric nor antisymmetric movement dash when affected by Symbol 's effect. At least one pair which fails to satisfy that then it is irreflexive or else is... A relation is opposite of symmetric relation can be reflexive, but it ca n't be symmetric for two elements.: the relation R on the guest list is actually mathematical with references or personal experience % %! Just a subset of$ A\times a $to give an example of a set of people is symmetric aRb! The DHCP servers ( or routers ) defined subnet are special cases, relations. Planet with a sun, could that be theoretically possible exactly proportional when line! Equal or exactly proportional when a line is drawn in the Chernobyl series that ended the. M1 Pro with Fans Disabled /tʃ/ ) let Rbe a symmetric relation, and R a... -5 - 15 = -20 = 0 ( mod 5 ) latest debit card number$ a. Preserve it as evidence, a = b, husband-wife, etc are! Symmetric nor antisymmetric symmetric iff aRb implies that bRa, for every a, b, }! Like in cruising yachts ended in the meltdown but not symmetric suppose if xRy yRx. Give an example of a non-empty relation which is neither symmetric nor antisymmetric is a! Sign a guest book when they arrive not transitive does Shutterstock keep my... Being acquainted with '' on a set a is symmetric but not | {
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not transitive does Shutterstock keep my... Being acquainted with '' on a set a is symmetric but not transitive both of these properties may. Answers – relations opposite because a relation be both antisymmetric and irreflexive or else it is to... Are neither ( although a lot of useful/interesting relations are neither symmetric antisymmetric! Device on my network relation imposes an order \$ then antisymmetry is also c... Mail exchanger integers defined by aRb if a relation is asymmetric if is... Engage in physical intimacy subscribe to this RSS feed, copy and paste this URL into your reader. | {
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# Calculating Wattage for Resistor in High Frequency Application?
I am making a MOSFET driving circuit.
Frequency : 400 kHz [50% duty cycle]
Gate voltage: 12 V
Total gate charge : 210 nC as per datasheet IRFP460
Rise time: 100 ns
[Q=I*t]
Current: 2.1 A
Gate resistor: V/I > 12/2.1 > 5.7 ohm
Peak power: I * I * R > 2.1 * 2.1 * 5.7 > 25.1370 W
1 watt resistor is OK ?
• Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts. – Elliot Alderson Mar 23 at 22:29
• I think i should remove Average Power line . – Israr Sayed Mar 24 at 11:23
The figure below shows the Gate Voltage versus Total Gate Charge for the IRFP460 MOSFET:
With a gate drive voltage $$\V_{DR} = 12\,\mathrm{V}\$$, it's possible to estimate a total gate charge of $$\155\,\mathrm{nC}\$$.
If $$\i_g \$$ represents the gate current, $$\Q\$$ the charge going into the gate and $$\tb\$$ (beginning time) and $$\te\$$ (ending time) to represent a time interval, then:
$$Q = \int_{tb}^{te}i_gdt$$
METHOD 1: (a first estimate)
Here the $$\i_g\$$ is considered constant ($$\Ig_{(ON)}\$$) during the charge ($$\tp_{(ON)}\$$) and constant ($$\Ig_{(OFF)}\$$) during discharge time ($$\tp_{(OFF)}\$$); roughly shown in the figure below:
So, the integral above reduces simply to (considering $$\tp_{(ON)}=100\,\mathrm{ns}\$$ and $$\Q_g\$$ as the total gate charge):
$$Q_g = Ig_{(ON)} \times tp_{(ON)}$$ or $$Ig_{(ON)} = \frac{Q_g}{tp_{(ON)}} = \frac{155\,\mathrm{nC}}{100\,\mathrm{ns}}= 1.55\,\mathrm{A}$$
The gate resistor $$\R_G\$$ must be calculated taking in account that, in “flat” part of the switching period (plot above), the gate voltage is constant at about $$\5.2\$$ V:
$$R_G = \frac{12\,\mathrm{V} - 5.2\,\mathrm{V}}{1.55\,\mathrm{A}} = 4.39 \space \Omega \approx 4.7 \space \Omega$$
In order to simplify I consider here $$\Ig_{(OFF)}=-Ig_{(ON)}\$$. So, the root mean square value for $$\i_g\$$ is: | {
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$$I_{RMS}= Ig_{(ON)}\sqrt{2 \times \frac{tp_{(ON)}}{T} } \approx 0.438\,\mathrm{A}$$
Finally, the average power for $$\R_G\$$ is:
$$P = I_{RMS}^2R_G \approx 0.9\,\mathrm{W}$$
METHOD 2:
Here the $$\i_g\$$ is considered as a straight line with maximum value $$\Ig_{pk_{(ON)}}\$$ and decreasing to zero at the end of time $$\tp_{(ON)}\$$ - as an approximation to the actual exponential decay (more realistic). Similar consideration is made for the gate discharge time:
An example of real measurement:
Retaining a $$\R_G = 4.7 \space \Omega\$$, the peak gate current can be calculated as:
$$Ig_{pk_{(ON)}} = \frac{12\,\mathrm{V}}{4.7 \space \Omega} \approx 2.553,\mathrm{A}$$
In order to simplify I consider here $$\Ig_{pk_{(OFF)}}=-Ig_{pk_{(ON)}}\$$. So, the root mean square value for $$\i_g\$$ is:
$$I_{RMS}= Ig_{pk_{(ON)}}\sqrt{\frac{2}{3} \times \frac{tp_{(ON)}}{T} } \approx 0.417\,\mathrm{A}$$
Finally, the average power for $$\R_G\$$ is:
$$P = I_{RMS}^2R_G \approx 0.817\,\mathrm{W}$$
No major differences from the value previously calculated.
THIRD METHOD
Just to mention a more precise (and more laborious) method. Here, $$\i_g\$$ is considered a true exponential decaying function (see figure above):
$$i_g = Ig_{pk_{(ON)}}e^{-\frac{t}{R_GC_{eff}}}$$
where $$\C_{eff}\$$ is the effective gate input capacitance of MOSFET. So:
$$i_g = \frac{V_{DR}}{R_G}e^{-\frac{t}{R_GC_{eff}}}$$
In the time interval $$\0\$$ to $$\t_s\$$, the total gate charge ("consumed") is given by:
$$Q_g = \int_{0}^{t_s} \frac{V_{DR}}{R_G}e^{-\frac{t}{R_GC_{eff}}}dt$$
This integral can be solved for a parameter ($$\R_G\$$ or $$\t_s\$$), when others are known.
CONCLUSION: The average power values were below $$\1\,\mathrm{W}\$$, but a margin of safety can be applied for guarantee. | {
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• I created an account to let you know that this answer is one of the best answers I've seen, on any site, in a long time. Just brilliant. – LogicalBranch Mar 24 at 12:01
• This could become the canonical answer to questions of this kind. Thorough and well written. However, could you please add links or citations to the sources of the graphics? – Elliot Alderson Mar 24 at 13:20
• So much effort for approximate solutions, have a look to the simple and exact Dave Tweed's one below. – carloc Mar 24 at 13:47
• Estimating Total Gate Charge from Graph nice. – Israr Sayed Mar 24 at 13:58
• @carloc: Just AFTER doing the paper/pencil work it's possible to conclude that the solutions are similar. My contributions is aligned with the standard procedures recommended by manufacturers. Also deals with the gate resistor estimation. – Dirceu Rodrigues Jr Mar 24 at 14:09
Dividing the peak power by the frequency is not useful.
Instead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 × 100 ns out of every 2.5 µs. This would be an average power of
$$25 W \cdot\frac{2 \cdot 100 ns}{2.5 \mu s} = 2 W$$
Clearly, your 1W resistor is not going to cut it!
However, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.
For an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents
$$Energy = \frac{1}{2}\cdot Charge \cdot Voltage$$
$$0.5 \cdot 210 nC \cdot 12 V = 1.26 \mu J$$
This is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.
To get the average power, multiply the energy per cycle by the number of cycles per second, giving | {
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To get the average power, multiply the energy per cycle by the number of cycles per second, giving
$$1.26 \mu J \cdot 2 \cdot 400 kHz = 1.088 W$$
Your 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here. | {
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# Intersection is Largest Subset
## Theorem
Let $T_1$ and $T_2$ be sets.
Then $T_1 \cap T_2$ is the largest set contained in both $T_1$ and $T_2$.
That is:
$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$
### General Result
Let $T$ be a set.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$
### Family of Sets
In the context of a family of sets, the result can be presented as follows:
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
$\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\ds \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$.
## Proof
### Sufficient Condition
From Set is Subset of Intersection of Supersets we have that:
$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
$\Box$
### Necessary Condition
Let:
$S \subseteq T_1 \cap T_2$
From Intersection is Subset we have $T_1 \cap T_2 \subseteq T_1$ and $T_1 \cap T_2\subseteq T_2$.
From Subset Relation is Transitive, it follows directly that $S \subseteq T_1$ and $S \subseteq T_2$.
So $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$.
$\Box$
From the above, we have:
$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
$S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$
Thus $S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ from the definition of equivalence.
$\blacksquare$ | {
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# Matrices with eigenvalues 0 and 1
How can you describe all $2\times 2$ matrices whose eigenvalues are 0 and 1?
My attempt: I know that 0 and 1 has to be solutions of the characteristic polynomial. And I've considered some examples like $$\begin{pmatrix}0&0\\0&1\end{pmatrix},\begin{pmatrix}1&0\\0&0\end{pmatrix} \text{ etc.}$$but I haven't found any regularity. But I also know from linear algebra that a matrix $A$ satisfying $A^2=A$ has only eigenvalues 0 and/or 1. Are there any more matrices? What do they have to satisfy?
• Do you mean that "the matrix has two eigenvalues, which are 0 and 1" or "the matrix has some unknown number of eigenvalues, all of which are either 0 or 1"? – Sharkos May 10 '13 at 21:31
• @Sharkos I would be interested in both cases :) – user31035 May 10 '13 at 21:34
Consider a matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ You can easily work out the characteristic polynomial, which is $$X^2 - (a+d)X + (ad-bc)$$ Based on your information, you have $$\begin{cases} a+d=1 \\ ad-bc=0 \end{cases}$$ because in a polynomial $X^2+pX+q$ the sum of the roots is $-p$ and their product is $q$.
This gives $d=1-a$, so $a-a^2-bc=0$ or $bc=a-a^2$. You can thus choose arbitrarily $a$. If $a=0$ or $a=1$, then one among $b$ and $c$ must be zero and the other one is arbitrary. For $a-a^2\ne0$, the matrices you look for have the form $$\begin{bmatrix} a & b \\ \dfrac{a-a^2}{b} & 1-a \end{bmatrix}$$ So the general form is one of these five cases
1. $\begin{bmatrix}0 & b \\ 0 & 1\end{bmatrix}\quad$ (arbitrary $b$)
2. $\begin{bmatrix}0 & 0 \\ c & 1\end{bmatrix}\quad$ (arbitrary $c$)
3. $\begin{bmatrix}1 & b \\ 0 & 0\end{bmatrix}\quad$ (arbitrary $b$)
4. $\begin{bmatrix}1 & 0 \\ c & 0\end{bmatrix}\quad$ (arbitrary $c$)
5. $\begin{bmatrix} a & b \\ \dfrac{a-a^2}{b} & 1-a \end{bmatrix}\quad$ ($a\ne0$, $a\ne1$, $b\ne0$) | {
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5. $\begin{bmatrix} a & b \\ \dfrac{a-a^2}{b} & 1-a \end{bmatrix}\quad$ ($a\ne0$, $a\ne1$, $b\ne0$)
Since the eigenvalues are 0 and 1 (hence distinct), the matrix must be diagonalizable. Thus it must be of the form $P^{-1}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}P$ for some invertible matrix $P$. Since there is a nice explicit formula for the inverse of a $2 \times 2$ matrix, you can work out the form of any matrix of the aforementioned form. Take $P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $\det P = ad-bc \ne 0$, then write out $P^{-1}$, and then compute.
EDIT: I am assuming you mean that the set of eignevalues of your matrix must be $\{0,1\}$, as opposed to being contained in this set, which is what Sharkos assumes. So it depends on what you intended.
Assuming that we exclude matrices from having complex eigenvalues but no real eigenvalues (which seems reasonable given your question) we make use of the Jordan normal form of the matrix.
There is some invertible matrix $P$ (a change of basis) such that
$$A = P^{-1} J P$$
where $J$ is in Jordan normal form with either $0$ or $1$ on the diagonal. In particular,
$$J = \pmatrix{0 & 0 \\ 0 & 0}, \pmatrix{1 & 0 \\ 0 & 1}, \pmatrix{1 & 0 \\ 0 & 0}, \pmatrix{0 & 1 \\ 0 & 0}, \pmatrix{1 & 1 \\ 0 & 1}$$
generate all other forms of these matrices.
• The first case has $A=0$.
• The second case has $A=I$.
• The third case is a change-of-basis from having one of each eigenvalue; it is a projection onto a line.
• The fourth case is a nilpotent matrix such that $A^2=0$. It maps one direction into another, and that direction into oblivion.
• The fifth case generates a skew/shear in some direction.
This is a partial answer that suggests how to parametrize such matrices explicitly. | {
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This is a partial answer that suggests how to parametrize such matrices explicitly.
The other answers are more elegant and tell the story more succinctly, but they construct the answers non-uniquely. (There are different matrices that you can conjugate a given Jordan normal form by and still obtain the same matrices: $P^{-1} J P = Q^{-1} J Q$ is possible with $P \ne Q$.)
First a little result about the characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$ for a $2 \times 2$ matrix $A$: $$p_A(\lambda) = \lambda^2 - (\operatorname{tr} A) \lambda + \det A,$$ where $\operatorname{tr} A = a + d$ is the trace. Check this from the definition if you're not familiar with it.
Now, proceed by cases. Suppose that $\lambda = 1$ is the only eigenvalue. Then, $$p_A(\lambda) = (\lambda - 1)^2 = \lambda^2 - 2\lambda + 1.$$ So, we know that \left\{ \begin{align} \operatorname{tr} A &= 2 \\ \det A &= 1 \end{align} \right. With $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix},$$ \left\{ \begin{align} a + d &= 2 \\ ad - bc &= 1 \end{align} \right.. Taking advantage of the trace equation to reduce the number of parameters, let's make the substitution $a = 1 + t$ and $d = 1 - t$. Now the determinant equation becomes $(1 + t)(1 - t) - bc = 1$ or $$bc = t^2.$$ Either $b = 0$, in which case, $t = 0$, as well, and $c$ is free to take any value: $$\begin{bmatrix} 1 & 0 \\ c & 1 \end{bmatrix}.$$ Or $b \ne 0$, so $c = -\frac{t^2}{b}$, so we have: $$\begin{bmatrix} 1 + t & b \\ -\tfrac{t^2}{b} & 1 + t \end{bmatrix}.$$
The space of possibilities in $(b, c)$-plane consists of both axes (corresponding to having $t = 0$, hence $1$ on the diagonal) together with the second and fourth quadrants (corresponding to $t \ne 0)$.
A similar analysis can give an explicit description of the other cases for the possibilities of the eigenvalues. (Note that this idea generalizes beyond your $\lambda \in \{0, 1\}$ assumption.) | {
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• I think the second instance of $P$ near the top of your answer should be $Q$. – Alistair Savage May 10 '13 at 22:47
• Now it's fixed. – Sammy Black May 10 '13 at 22:49
If a matrix $A\in\mathbb{C}^{n\times n}$ has only eigenvalues $0$ and $1$ then its characteristic polynomial is of the form $$z^{a}\left(z-1\right)^{b}=z^{a}\sum_{k=0}^{b}\left(\begin{array}{c} b\\ k \end{array}\right)z^{k}\left(-1\right)^{n-k}=\sum_{k=0}^{b}\left(\begin{array}{c} b\\ k \end{array}\right)z^{a+k}\left(-1\right)^{n-k}$$ with $a+b=n$. A matrix that has the above characteristic polynomial is $$B=\left[\begin{array}{cccccc} 0 & & & & & -c_{0}\\ 1 & 0 & & & & -c_{1}\\ & 1 & 0 & & & -c_{2}\\ & & 1 & \ddots & & \vdots\\ & & & \ddots & 0 & -c_{n-2}\\ & & & & 1 & -c_{n-1} \end{array}\right]$$ where $$c_{j}=\begin{cases} 0 & j<a\\ \left(\begin{array}{c} b\\ k \end{array}\right)\left(-1\right)^{n-k} & j\geq a \end{cases}.$$ The above is called the Frobenius companion matrix to the characteristic polynomial. Another matrix that will satisfy this is $B^{T}$ since $\det\left(X^{T}\right)=\det\left(X\right)$. Note that $B$ is not a projection (i.e. $B^2\neq B$).
Obviously, not all matrices that have only eigenvalues of $0$ and $1$ are of this form. For example, $I$ only has $1$ as an eigenvalue but cannot be written in the above form.
• This construction gives (up to conjugation) the matrices which have the polynomial as minimal polynomial. – Mariano Suárez-Álvarez May 10 '13 at 22:20 | {
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# Proof that the set of irrational numbers is dense in the reals
The hint I was given was to simply prove that $y=xz$ is irrational given that $x$ is nonzero, $x$ is rational and $z$ is irrational. Here's how I did it:
Claim: $y=xz$ is irrational.
Proof: Assume $x\neq0$, $x$ is rational and $z$ is irrational.
By contradiction assume that $y=xz$ is rational. This means $y$ can be expressed as $m/n$, $m$ and $n$ being integers; $y$ can be expressed similarly as $p/q$, $p$ and $q$ being integers. By substitution, we have that $$p/q=mz/n$$ and $$z=pn/qm, qm \neq 0.$$ Since $pn$ and $qm$ are integers $z$ has to be rational.
In addition to this it seems like there's a part 2 as follows:
Proof: Given an interval $(x,y)$ we will choose a positive irrational number, $z$, say. By density of the rationals there is a rational $p$ in the interval $(x/z, y/z)$ s.t. $$x/z <p< y/z.$$ From this we see that $pz$ is irrational since it is the product of a rational and irrational number.
Is the $pz$ the $xz$ that we proved is irrational in the first proof? So ideally when presenting a full proof like this, should we do part 2 then part 1?
Great proof! You can apply your result about what you called $xz$ in part 1 to $pz$ in part 2 because it obeys your assumptions: $p \ne 0$, $p$ is rational and $z$ is irrational. So from part 1 we know that $pz$ is irrational.
We also know that $pz$ lies in the interval $(x,y)$, so irrationals are dense.
What is ideal to present when presenting a full proof depends on what your reader already knows. It sounds like in this case you want to present both part 1 and part 2, and that will be a complete proof. But in some other case your reader (or lecturer or marker) might not know or want to assume, for example, that rationals are dense in the real numbers, so you might have to include a part 1.5 that proves that. | {
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EDIT: Oops, I think I misunderstood you. Yeah, I would present part 2 then part 1 when presenting this proof; I think the sequence of ideas flows better. But it's up to you!
Also, just to point out a few typos:
• I think you mean 'This means $x$ can be expressed as $m/n$' when you write "This means y can be expressed as m/n".
• I think you mean 'By substitution' when you write "By substation".
Hope that helps! | {
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# How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$?
I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
• Hint: Fermat's Little Theorem. – Arturo Magidin Feb 15 '11 at 6:13
• Hint #2: Chinese Remainder Theorem. – Pete L. Clark Feb 15 '11 at 6:19
• So you have $n^6\equiv 1\pmod{7}$, but this implies $n^7\equiv n\pmod{7}$. So $7|n^7-n$. In general, $n^p\equiv n\pmod{p}$ for a prime $p$. By the same token, $n^7\equiv (n^3)^2\cdot n\equiv n^3\equiv n\pmod{3}$ by FlT, so $3|n^7-n$. It remains to show $2|n^7-n$, and then you'll be done by the two hints above. – yunone Feb 15 '11 at 6:32
• @Yuval: $p=6$ ? :-) – lhf Feb 15 '11 at 12:26
• It works because $42=6(6+1)$ and Little Fermat, but I like to believe it works because it is the answer to the life, the universe and everything. – chubakueno Sep 1 '13 at 22:22
Problems like this appear frequently here. There are a couple of standard approaches. One is to use Fermat's little theorem, which says that if $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$.
Since $42=2\times 3\times 7$, what we need to do is to check that 2, 3, and 7 divide $n^7-n$, no matter what $n$ is.
That 7 does is direct from Fermat's little theorem.
The theorem also ensures that 3 divides $n^3-n$. Now: $$n^7-n=n(n^6-1)=n((n^2)^3-1)=n(n^2-1)(n^4+n^2+1)=(n^3-n)(n^4+n^2+1),$$ so 3 indeed divides $n^7-n$.
Finally, $n$ and $n^7$ always have the same parity, so $n^7-n$ is even.
We can actually argue without Fermat's little theorem in this case. An approach that only requires patience is as follows: The idea is to factor the polynomial $x^7-x$ and then analyze the result when $x=n$ is an integer. (This is a trick that Bill Dubuque suggests sometimes in his solutions.) | {
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We have: $x^7-x=x(x^6-1)=x(x^3+1)(x^3-1)=x(x+1)(x^2-x+1)(x-1)(x^2+x+1)$. When $x=n$, we have $$n^7-n=(n-1)n(n+1)(n^2-n+1)(n^2+n+1).$$ Now we analyze this prime by prime, as before. Note that one of $n$ and $n-1$ is always even, so the product is even. Also, of 3 consecutive numbers, such as $n-1,n,n+1$, one is always divisible by 3, so it only remains to verify divisibility by 7.
We may assume that $n=7k+b$ where $b=\pm2$ or $\pm3$, since otherwise $(n-1)n(n+1)$ is a multiple of 7. In that case, $n^2\equiv 4$ or $2\pmod 7$, and one of $n^2+n$, $n^2-n$ is $\equiv 6\pmod 7$, so $(n^2-n+1)(n^2+n+1)$ is a multiple of 7.
The disadvantage of this approach over the previous one, of course, is the need to analyze different cases. Fermat's little theorem allows us to analyze all cases simultaneously, which typically (as here) results in a much faster approach.
If you are comfortable with the method of induction, this gives us a way of verifying divisibility by 7 which is not without some elegance (divisibility by 2 and 3 is probably best approached as before). Note that $(-n)^7-(-n)=-(n^7-n)$, so we may as well assume that $n\ge 0$. If $n=0$ it is obvious that 7 divides $n^7-n$.
Suppose then that $7|n^7-n$, and argue that $7|(n+1)^7-(n+1)$. For this, actually expand $(n+1)^7$ using the binomial theorem: $$(n+1)^7=n^7+7n^6+{7\choose 2}n^5+{7\choose 3}n^4+\dots+1.$$ The point is that $${7\choose k}=\frac{7!}{k!(7-k)!}$$ is obviously divisible by 7 as long as $k\ne0,7$, so (modulo 7) we have that $(n+1)^7-(n+1)\equiv (n^7+1)-(n+1)=(n^7-n)$. Now we invoke the induction hypothesis, that precisely says that the latter is divisible by 7, and we are done.
Of course, exactly the same inductive argument gives us a proof of Fermat's little theorem: $p|n^p-n$ for any $p$ prime and any integer $n$.
• To prove that $3$ divides, you can observe that $n^7-n=n^7-n^5+n^5-n^3+n^3-n$ – N. S. Jun 29 '12 at 15:42 | {
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It is a special case of the following global-form of little Fermat. For $$\rm\, a,k,n\in\mathbb N$$ with $$\rm\ a,k > 1$$
$$\rm\qquad d\ |\ n^k\! -\! n\$$ for all $$\rm\:n\:$$ $$\rm \iff\ d\:$$ is squarefree, and $$\rm\ p\!-\!1\ |\ k\!-\!1\$$ for all primes $$\rm\:p\:|\:d$$
Hence for $$\rm\: a = 42 = 2\cdot 3\cdot 7\$$ we deduce: $$\rm\ \ 42\ |\ n^k\!-n\$$ for all $$\rm\:n \iff 6\ |\ k\!-\!1$$
For the simple proof and further discussion see Korselt's criterion for Carmichael numbers here (or my 2009/04/10 sci.math post, link is now broken by google)
Here is another useful variation:
Theorem $$\ \ \ n^{\large k+\phi}\equiv n^{\large k}\pmod{p^i q^j}\ \$$ assuming that $$\ \color{#0a0}{\phi(p^i),\phi(q^j)\mid \phi},\,$$ $$\, i,j \le k,\,\ p\ne q.\ \ \$$
Proof $$\ \, p\nmid n\,\Rightarrow\, {\rm mod\ }p^i\!:\ n^{ \phi}\equiv 1\,\Rightarrow\, n^{k + \phi}\equiv n^k,\,$$ by $$\ n^{\Large \color{#0a0}\phi} = (n^{\color{#0a0}{\Large \phi(p^{ i})}})^{\large \color{#0a0}m}\!\overset{\color{blue}{\rm (E)}}\equiv\! 1^{\large m}\!\equiv\! 1$$ by $$\rm\color{blue}{E}$$=Euler
$$\qquad\quad\ \, \color{#c00}{p\mid n}\,\Rightarrow\, {\rm mod\ }p^i\!:\ n^k\equiv 0\,\equiv\, n^{k + \phi}\$$ by $$\ n^k = n^{k-i} \color{#c00}n^i = n^{k-i} (\color{#c00}{mp})^i$$ and $$\,k\ge i.$$
So $$\ p^i\mid n^{k+\phi}\!-n^k.\,$$ By symmetry $$\,q^j$$ divides it too, so their lcm $$= p^iq^j\,$$ divides it too. $$\$$ QED
Remark $$\$$ Obviously the proof immediately extends to an arbitrary number of primes. This leads the way to Carmichael's Lambda function, a generalization of Euler's phi function.
• Very nice application of Fermat's Little Theorem's global-form (I had never seen it). However, the link to your proof takes me nowhere... – Dr. Mathva May 7 at 13:20
• @Dr.Mathva I added a local link (alas Google continues to break links to its usenet newsgroup archive). – Bill Dubuque May 7 at 14:06
There are many equivalent ways of proving it. | {
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There are many equivalent ways of proving it.
First observe that $42$ divides a number iff $2,3$ and $7$ divides the number.
(Since $42 = 2 \times 3 \times 7$ and $\gcd(2,3) = \gcd(3,7) = \gcd(2,7) = 1$)
Divisibility by $2$:
Clearly, $2|(n^7-n)$ since $n^7$ and $n$ are of the same parity.
Equivalently you could argue out that $2|(n^2-n)$ directly from Fermat's little Theorem. (This is an overkill of Fermat's Little Theorem.)
Divisibility by $3$:
$n^7-n = n(n^6-1) = n(n^2-1)(n^4+n^2+1)=n(n+1)(n-1)(n^4+n^2+1)$.
$3|n$ or $3|(n-1)$ or $3|(n+1)$ and hence $3|(n^7-n)$.
Equivalently you could argue out that $3|(n^3-n)$ directly from Fermat's little Theorem.
Divisibility by $7$:
Note that $n$ can be either $7k$ or $7k \pm 1$ or $7k \pm 2$ or $7k \pm 3$.
If $n=7k$ or $n=7k \pm 1$, we are again done since then $7|n$ or $7|(n+1)$ or $7|(n-1)$ and hence $7|(n^7-n)$.
If $n=7k \pm 2$, then $n^2 = (7k \pm 2)^2 = 7m + 4$ and $n^4 = (7m+4)^2 = 7l+2$. Hence $n^4 + n^2 + 1 = 7l+2 + 7m + 4 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.
If $n=7k \pm 3$, then $n^2 = (7k \pm 3)^2 = 7m + 2$ and $n^4 = (7m+2)^2 = 7l+4$. Hence $n^4 + n^2 + 1 = 7l+4 + 7m + 2 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.
Hence, $7|(n^7-n)$.
Equivalently you could argue out that $7|(n^7-n)$ directly from Fermat's little Theorem.
Therefore, we have that $2|(n^7-n)$ and $3|(n^7-n)$ and $7|(n^7-n)$, $\forall n \in \mathbb{N}$.
Hence, $42|(n^7-n)$, $\forall n \in \mathbb{N}$.
As $42=2.3.7$. Therefore,we need to check that $n^7-n$ is divisible by $2,3$ and 7. For divisibility by $2$, by fermat's little theorem, $n^2=n\pmod 2 \implies {(n^2)}^3.n=n^4\pmod 2=n^2\pmod 2=n\pmod 2 \implies n^7-n=0\pmod2$. For divisibility by 3, $n^3=n\pmod 3\implies n^7=n^3\pmod 3=n\pmod 3 \implies n^7-n=0\pmod 3$. For divisibility by 7, $n^7=n\pmod 7 \implies n^7-n=0\pmod 7$. These relations implies that $42|(n^7-n)$.
Combinatorial Polynomial Approach | {
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Combinatorial Polynomial Approach
Since \begin{align} n^7-n &=5040\binom{n}{7}+15120\binom{n}{6}+16800\binom{n}{5}+8400\binom{n}{4}+1806\binom{n}{3}+126\binom{n}{2}\\ &=42\left[120\binom{n}{7}+360\binom{n}{6}+400\binom{n}{5}+200\binom{n}{4}+43\binom{n}{3}+3\binom{n}{2}\right] \end{align} Therefore, we have that for all $n\in\mathbb{Z}$, $$42\mid n^7-n$$
Little Fermat Approach
Little Fermat Theorem says $$n^7\equiv n\pmod7$$ and since $7\equiv1\pmod2$ $$n^7\equiv n\pmod3$$ and since $7\equiv2\pmod1$ $$n^7\equiv n\pmod2$$ Thus, $$n^7\equiv n\pmod{42}$$
• +1. But I have a question: how do you know that $$n^7-n =5040\binom{n}{7}+15120\binom{n}{6}+16800\binom{n}{5}+8400\binom{n}{4}+1806\binom{n}{3}+126\binom{n}{2}$$? Is there some special formula or theorem for this expansion? I saw a similar one for the case $30\mid n^5-n$ – Dr. Mathva May 4 at 15:34
• If we know the values of $f(n)=n^7-n=\sum\limits_{k=0}^7a_k\binom{n}{k}$ for $n=0$ to $n=7$, it is very easy to compute the coefficients $a_k$ since for $n\lt k$, $\binom{n}{k}=0$. Since $f(1)=f(0)=0$, we get $a_1=a_0=0$. Since $f(2)=126$, we get $a_2=126$. Since $f(3)=2184$, $a_3=2184-126\binom{3}{2}=1806$. Since $f(4)=16380$, $a_4=16380-1806\binom{4}{3}-126\binom{4}{2}=8400$. etc. – robjohn May 4 at 16:04
• Oh, see. Thanks ;) – Dr. Mathva May 4 at 16:06
• Sorry for asking again, but I've just noticed there's one last thing I don't understand. You claim "If we know the values of $\displaystyle f(n)=n^7-n=\sum^7_{k=0}a_k\binom{n}{k}$ [...]". How can we, however, prove that $f$ can be represented as this sum? – Dr. Mathva May 11 at 20:52
• If you look at the method for computing $a_n$, you will see that as long as $f(n)$ is integral for $8$ consecutive integers $n$, we get the $8$ coefficients for this sum. – robjohn May 12 at 0:52
F$\ell$T$^*$ (if $p\not\vert n$, then $p\vert n^{p-1}-1$) $\Rightarrow$
if $2\not\vert n$, then $2\not\vert n^6$, and then $2\vert (n^6)^1-1$ | {
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if $2\not\vert n$, then $2\not\vert n^6$, and then $2\vert (n^6)^1-1$
if $3\not\vert n$, then $3\not\vert n^3$, and then $3\vert (n^3)^2-1=n^6-1$
if $7\not\vert n$, then $7\vert n^6-1$
so $2,3,7\vert n^7-n$.
$^*$: $\ell$=little.
Just for completeness, here is induction (just for divisibility by 7) :
Claim : $n^7 - n$ is divisible by 7
Base Case: True for n = 1,2
Induction Step: Assume true for n = k. To prove true for n = k + 1.
Now, $$(k+1)^7 - (k+1) = k^7 + 7k^6 + 21k^5 + 35k^4 + 35k^3 + 21k^2 + 7k + 1 - k - 1 \\= (k^7 - k) + 7(k^6 + 3k^5 + 5k^4 + 5k^3 + 3k^2 + k)$$ We know by our assumption that $k^7 - k$ is divisible by 7. Therefore, $(k + 1)^7 - (k + 1)$ is divisible by 7.
This shows that $n^7 - n$ is divisible by 7.
To show divisibility by 2 and 3, unfortunately, one has to fall back on some of the earlier tricks. $$n^7 - n = n(n^6 - 1) = (n-1)n(n+1)(n^2 - n + 1)(n^2 + n + 1)$$ $(n-1)n(n+1)$ is a product of three consecutive integers. Product of two consecutive integers is divisible by 2 and product of three consecutive integers is divisible by 3. Since, $(2,3) = 1$, product of three consecutive integers is divisible by 6. Since $(6,7) = 1$, $n^7 - n$ is divisible by 42.
QED
• Where is 42?... – lhf Jun 29 '12 at 11:24
• Good catch!! :P – TenaliRaman Jun 29 '12 at 14:58 | {
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# Smallest subgroup and $\langle a \rangle$.
Gallian says in Chapter 3 of Contemporary Abstract Algebra that
For any element $a$ of a group $G$, it is useful to think of $\langle a \rangle$ as the smallest subgroup of $G$ containing $a$.
But wouldn't this mean the set $\langle a \rangle$ is simply $\{ a, a^{-1}, e \}$?
• If it doesn't contain $a^2$, that isn't a subgroup. – Robert Israel Apr 10 '17 at 16:00
• @RobertIsrael Why not? – Airdish Apr 10 '17 at 16:03
• It needs to be closed under the group's operation. – Akiva Weinberger Apr 10 '17 at 16:58
• That isn't a group unless $a = a^{-1}$, in which case it would be $\{a, e\}$ only – q.Then Apr 10 '17 at 17:53
• @q.Then Actually that is a group if $a^2 = a^{-1}$. – fleablood Apr 10 '17 at 18:09
The set $\{a,a^{-1},e\}$ has an identity and is closed under inverses, but it isn't necessarily closed under the group's operation.
If $H$ is a subgroup and $a \in H$, you also need $a^2=aa\in H$ and $a^3=aaa \in H$ etc. Likewise, you need inverses for all of those elements (i.e. $a^{-1}, a^{-2}, \dots \in H$).
So closure under the group's operation and inverses require at least $\langle a \rangle = \{ a^n \;|\; n \in \mathbb{Z} \} \subseteq H$.
Once you know $\langle a \rangle$ is itself a subgroup. This shows that it is the smallest subgroup containing $a$. | {
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• So what you're saying is that once you have $a$ in the subgroup, you automatically need its powers too because $a * a$ and $a *a * a$, etc. ought to be allowed operations under the group operation? – Airdish Apr 10 '17 at 16:05
• A group's operation is associative, has an identity, and inverses...but it also must be closed. If $a,b \in G$, then $ab \in G$ as well. So this means $a \in G$ then $a^2=aa \in G$ etc. So closure is what is forcing this. – Bill Cook Apr 10 '17 at 16:06
• Ah okay so the set $\{ a, a^{-1}, e \}$ isn't actually a group at all? – Airdish Apr 10 '17 at 16:08
• Not in general. For example: let $i=\sqrt{-1}$. Then $\{i,i^{-1},e\} = \{i,-i,1\}$ is not a group since multiplication isn't a closed binary operation operation on this set ($i^2=-1 \not\in \{i,-i,1\}$. Since we don't have a closed operation, it's not a group. – Bill Cook Apr 10 '17 at 16:12
• On the other hand, it could be. Consider $a=-1$ (with multiplication). Then $\{a,a^{-1},e\}=\{-1,(-1)^{-1},1\} = \{1,-1\}$ which is a perfectly good (cyclic) group (under multiplication). So $\{a,a^{-1},e\}$ can be a group, but usually it isn't. :) – Bill Cook Apr 10 '17 at 16:14
Groups are closed under their operations. In other words if $a,b \in G$ then we know that $ab \in G$.
So if $a\in <a>$ then we know that $a*a=a^2 \in <a>$ and $a^2*a = a^3 \in <a>$ and inductively we know that $a^2 \in <a>$ for all $n \in \mathbb N$. And as inverses must exist we know $a^{-n} \in G$.
So in general the group $<a> = \{.......a^{-3},a^{-2},a^{-1}, e,a,a^2,a^3,....\}$. But the $a^i$ need not be distinct. It could be possible for $a^i = a^k;i \ne j$. An example is $\mathbb Z_6=\{0,1,2,3,4,5\}$ under modulo addition. $2^5 = 2+2+2+2+2 = 4 = 2+2 = 2^2$. So $\{......2^{-3}=0,2^{-2}=2, 2^{-1}=4, 0, 2, 2^2 = 4, 2^3 = 0,....\} = \{0,2,4\}$ is not infinite. | {
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Now it is possible that $a^k =e$ for some number $k$. Then $a^{-1} = a^{k-1}$ Example: In $\mathbb Z_6$, $2^3 = 2+2+2= 0$ and $2^{-1} = 4 = 2^{3-1}$. In this case $<a> = \{0, a, a^2, a^3, .... a^{k-1}\}$. Example $<2> \subset \mathbb Z_{6} = \{0,2,2^2=2^{-1}=4\}$
But it's also possible that $a^k \ne e$ for any $k \ne 0$. In this case $<a> = \{a^n|n \in \mathbb Z\}$. Example: If the group is $\mathbb Z$ under addition. Then $<7> = \{0, 7, -7, 14, -14, 21, -21, ....\} = \{7n| n \in \mathbb Z\}$. Which is infinite.
....
In general. If $a^n= e$ and $a^k\ne e; 1\le k < n$ we so "the order of $a$" is $n$ and we write it as $|a| = n$. If no such $n$ exists, we say $|a| = \infty$.
If $|a| = n$ then $<a> =\{0,a, a^2, ....., a^{n-1}\}$.
If $|a| = \infty$ then $<a> = \{a^k|k \in \mathbb Z\}$.
Notice $<a>$ always has $|a|$ units. | {
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# Understanding limits
#### Rido12
##### Well-known member
MHB Math Helper
I know this is probably a dumb question, but I have a question regarding this. My textbook says the following: "A function f is differentiable at a if f'(a) exists."
It then follows with and example regarding if f(x) = |x| is differentiable at x = 0. They prove this by finding the limit of its derivative, and then splits it in two equations: for the limit as h -> 0+ and h -> 0-. Finally concluding that it is not differentiable as the limits are different. Done.
My question is as follows: does the method, as demonstrated above, work for all functions when you're trying to find if a certain point is differentiable? I question it because even if the aforementioned limit DOES exist, it doesn't mean that f'(a) exists. It just means that it has a limit, as you can have a limit of something when f'(a) is undefined. Thus, making the method inadequate to making such a conclusion.
EDIT:
Sorry if this is confusing, and if you want me to clarify, I can provide a hypothetical situation.
Ok, time to clarify what I'm trying to convey with a hypothetical situation (not real)
We are asked to see if f(x) = |x| is differentiable at x = 0. Let's PRETEND that the limit exists at f'(a). That when you took the limit from the positive and negative side of it, it resulted in 5. However, it is actually undefined at 5, but the limit as you approach f'(a) is 5. By using the same logic as above, the logic that the textbook example used, we are to assume that YES, It is differentiable, BECAUSE the limit exists! However, that is actually NOT the case, since the by the definition, f'(a) MUST exists. But in this example, YES, the limit exists, but f'(a) doesn't exist as it is undefined at a.
That being said, would additional steps be taken/required, in order to prove if something is or is not differentiable. As we see in my hypothetical situation, just determining the limit was not enough.
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#### Bacterius
##### Well-known member
MHB Math Helper
[JUSTIFY]Okay, this is hurting my brain so I'll conservatively assert that:
1. If the two limits are the same, you are correct that this says absolutely nothing about the existence of the derivative at $x = a$, so if the limits are the same this is as far as this method will take you.
2. If the two limits are different, then for most well-behaved functions the derivative does not exist at that point, however there are may exist crazy analytical functions which are continuous while their derivative isn't, and vice versa, so it may not be a sufficient condition to show the existence of $f'(a)$.[/JUSTIFY]
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#### Jameson
Staff member
This is a really good question.
[Rest of post deleted due to a mistake. Updated thought can be found in my later post]
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Hey Jameson! I think what you said is not entirely correct.
This is a really good question.
In order for a function of one variable, $f(x)$, to be differentiable at a point $a$, then it must be be continuous at $a$ and the following limit must exist ...
I think the thing in bold is redundant. The continuity of $f$ at $a$ follows from the differentiability of $f$ at $a$.
2) Just knowing that the above limit exists is not enough to say that $f(x)$ is differentiable at $a$. ...
I think knowing that the limit exists equivalent to $f$ being differentiable at $a$. That is by definition of differentiability at a point. Moreover, strictly, we should not say that $f(x)$ is differentiable at $a$ but simply say that $f$ is differentiable at $a$.
Please correct me if I am wrong or if I have misinterpretted your response.
#### caffeinemachine
##### Well-known member
MHB Math Scholar
I know this is probably a dumb question
This is not a dumb question at all. | {
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My textbook says the following: "A function f is differentiable at a if f'(a) exists."
It then follows with and example regarding if f(x) = |x| is differentiable at x = 0. They prove this by finding the limit of its derivative, and then splits it in two equations: for the limit as h -> 0+ and h -> 0-. Finally concluding that it is not differentiable as the limits are different. Done.
My question is as follows: does the method, as demonstrated above, work for all functions when you're trying to find if a certain point is differentiable? I question it because even if the aforementioned limit DOES exist, it doesn't mean that f'(a) exists. It just means that it has a limit, as you can have a limit of something when f'(a) is undefined. Thus, making the method inadequate to making such a conclusion.
I think you are making mistakes here. We write $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ whether the limit exists or doesn't exist. When the limit doesn't exist, $f'(a)$ is not defined. When the limit $\lim_{x\to a}\frac{f(x)-f(a)}{t-a}$ exists, then $f'(a)$ is same as this limit. So $f'(a)$ is a real number and certainly it exists.
Note that when $\lim_{x\to a}\frac{f(x)-f(a)}{t-a}$ exists then this doesn't say that $f'(a)$ has a limit. This would have no meaning. It says that $\frac{f(x)-f(a)}{x-a}$ has a limit as $x$ approaches $a$. For a short hand notation, and for furthering the theory, we denote this limit, whether it exists or not, as $f'(a)$.
I have not read the content in Spoiler box thoroughly yet. Tell me if you have further doubts or you need to discuss.
EDIT: I have implicitly assumed that $f$ is a real function having an interval $I$ in its domain such that $a\in I$.
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#### Bacterius
##### Well-known member
MHB Math Helper
He is taking the limit of the derivative, not the original function itself.
#### Jameson
Staff member
Hey Jameson! I think what you said is not entirely correct.
Yep, I think I need to edit a couple of things. | {
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I think the thing in bold is redundant. The continuity of $f$ at $a$ follows from the differentiability of $f$ at $a$.
Yes I agree that continuity is a necessary condition of differentiability thus differentiability implies continuity, but the converse is not true. That's what I meant.
I think knowing that the limit exists equivalent to $f$ being differentiable at $a$. That is by definition of differentiability at a point. Moreover, strictly, we should not say that $f(x)$ is differentiable at $a$ but simply say that $f$ is differentiable at $a$.
Please correct me if I am wrong or if I have misinterpreted your response.
About the part in bold - yes I agree with you and would normally write that but thought the OP might be used to seeing functions always written in the form "$f(x)$" so wrote it that way. You're right of course.
Now that I think about it more the if the limit exists then it will be continuous. I was thinking of a situation like $f(x)=\frac{x^2-4}{x-2}$ at $x=2$ where the derivative appears to exist by the limit definition but isn't continuous. However now I see that this won't have a limit for the limit definition of a derivative so the extra stipulation of being continuous isn't necessary as it's already implied by the existence of the limit definition of a derivative.
I suppose the main idea I'm expressing is differentiability implies continuity but the opposite is not true.
#### Jameson
Staff member
He is taking the limit of the derivative, not the original function itself.
Are you sure? I think caffeinemachine addressed his terminology in a way that makes sense. It looks to me like the we're talking about the limit definitions of functions.
#### caffeinemachine
##### Well-known member
MHB Math Scholar
He is taking the limit of the derivative, not the original function itself.
Umm.. now I am totally confused. If you understand his doubt then can you please frame it nicely and post it along with your solution?
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#### Bacterius
##### Well-known member
MHB Math Helper
Are you sure? I think caffeinemachine addressed his terminology in a way that makes sense. It looks to me like the we're talking about the limit definitions of functions.
That was my first interpretation as well. But after rereading his example it actually seems that he is talking about taking the two-sided limit of the derivative of $f$ to establish whether the derivative exists at a given point (so the question would become: is the derivative being (non-)continuous at $x = a$ a sufficient condition for the (non-)existence of the derivative at $x = a$?)
At least that's my take on it. I figured the question was deeper than just "being continuous is necessary but not sufficient for differentiability". But it's possible I am just talking nonsense here haven't done any formal calculus in a while.
#### Rido12
##### Well-known member
MHB Math Helper
I haven't read all your responses yet, but it is the limit of the derivatvive. I don't knwo anymore haha, I'm too confused.
The textbook does this:
f'(x) = lim h-> 0 (|0 + h| - |0|)/ h
Then split in two parts to compute the left and right limits separetely.
#### Jameson
Staff member
I'm pretty sure you don't mean "the limit of the derivative" but the "limit definition of the derivative". Your example seems to confirm this.
#### Rido12
##### Well-known member
MHB Math Helper
Yep, I think I need to edit a couple of things.
Yes I agree that continuity is a necessary condition of differentiability thus differentiability implies continuity, but the converse is not true. That's what I meant.
About the part in bold - yes I agree with you and would normally write that but thought the OP might be used to seeing functions always written in the form "$f(x)$" so wrote it that way. You're right of course. | {
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Now that I think about it more the if the limit exists then it will be continuous. I was thinking of a situation like $f(x)=\frac{x^2-4}{x-2}$ at $x=2$ where the derivative appears to exist by the limit definition but isn't continuous. However now I see that this won't have a limit for the limit definition of a derivative so the extra stipulation of being continuous isn't necessary as it's already implied by the existence of the limit definition of a derivative.
I suppose the main idea I'm expressing is differentiability implies continuity but the opposite is not true.
It's quite strange actually. I always do see functions written in teh form "f(x)" rather than just "f"... My textbook uses "f", so now I do that too. But my previous math teachers use "f(x)"
- - - Updated - - -
I'm pretty sure you don't mean "the limit of the derivative" but the "limit definition of a derivative". Your example seems to confirm this.
I'm confused. OH. My example was just finding the derivative of a function right, by taking its limit right? I got confused because I don't usually see the word "limit" when I take derivatives using power rule, chain, etc, and I got mixed up thinking it was the limit of a derivative.
And I haven't finished reading the responses yet from the previous page.
#### Rido12
##### Well-known member
MHB Math Helper
That was my first interpretation as well. But after rereading his example it actually seems that he is talking about taking the two-sided limit of the derivative of $f$ to establish whether the derivative exists at a given point (so the question would become: is the derivative being (non-)continuous at $x = a$ a sufficient condition for the (non-)existence of the derivative at $x = a$?) | {
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At least that's my take on it. I figured the question was deeper than just "being continuous is necessary but not sufficient for differentiability". But it's possible I am just talking nonsense here haven't done any formal calculus in a while.
This is precisely what I meant with my question; I don't know if that is the limit definition of derivatives, or whatnot. Now if only my original post was as concise as how you explained it
EDIT: second thought, brain hurts, not sure if that's how i meant it but:
It's just that I'm not convinced that f'(a) exists by just taking its limit from both sides. And it must exist for it to be differentiable at that point.
This:
1. If the two limits are the same, you are correct that this says absolutely nothing about the existence of the derivative at $x = a$, so if the limits are the same this is as far as this method will take you.
I think the notation is confusing me; the limit definition of derivatives has both "f'(x)" and "lim" so it makes me want to think that it's taking the LIMIT of the DERIVATIVE, but now that I think about it, it's simply just taking the derivative.
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#### Rido12
##### Well-known member
MHB Math Helper
This leads me to the following question:
When would it be necessary to use the limit definition of a derivative, and when to just solve for the derivative using chain/product, etc rule? It is to my understanding that the limit definition was used to derive the rules mentioned before.
Isn't the limit definition of a derivative the same as finding the derivative and then apply the limit? So the limit of the derivative?
For example, finding the limit (if there is) for f(x) = |x|using the limit definition of a derivative
We compute the two side limit and we get:
f'(x)=lim as h >0+ |x|= |h|/h = 1
f'(x)=lim as h ->0- |x|= |h|/h = -1 | {
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Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0+ x / |x|
= x / x = 1
limit as h >0- x / |x|
= x / -x = -1
#### Bacterius
##### Well-known member
MHB Math Helper
This leads me to the following question:
When would it be necessary to use the limit definition of a derivative, and when to just solve for the derivative using chain/product, etc rule? It is to my understanding that the limit definition was used to derive the rules mentioned before.
Isn't the limit definition of a derivative the same as finding the derivative and then apply the limit? So the limit of the derivative?
For example, finding the limit (if there is) for f(x) = |x|using the limit definition of a derivative
We compute the two side limit and we get:
f'(x)=lim as h >0+ |x|= |h|/h = 1
f'(x)=lim as h ->0- |x|= |h|/h = -1
Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0+ x / |x|
= x / x = 1
limit as h >0- x / |x|
= x / -x = -1
If you already have an expression for the derivative then you don't need to use the limit definition. You can just evaluate it directly (which you did, since you did not use "h" at all)
I am not sure what you mean though - what would be the purpose of calculating such a limit?
The chain, product rules etc.. can all be derived from the limit definition. They are just much more convenient to use, because calculating derivatives from the limit form (the so-called "first principles") gets rather tedious for anything bigger than a quadratic. The idea is that the limit definition gives the most "basic" definition of what a derivative is, which is important in mathematics (where something without a proper definition is useless).
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#### Rido12
##### Well-known member
MHB Math Helper
If you already have an expression for the derivative then you don't need to use the limit definition. You can just evaluate it directly (which you did, since you did not use "h" at all)
I am not sure what you mean though - what would be the purpose of calculating such a limit?
The chain, product rules etc.. can all be derived from the limit definition. They are just much more convenient to use, because deriving from the limit form (the so-called "first principles") gets rather tedious for anything bigger than a quadratic.
Do you also mean to imply that finding the limit here is redundant?
Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0+ x / |x|
= x / x = 1
limit as h >0- x / |x|
= x / -x = -1
So we could use direct substitution instead of finding the limit, meaning that once we get f'(x) = x / |x| we can do f'(0) = 0/ 0 , meaning that it is not possible to differentiate at x= 0? (Instead of doing all the two side limit hassle that was only necessary when utilizing the limit definition of a derivative)
#### Bacterius
##### Well-known member
MHB Math Helper
Do you also mean to imply that finding the limit here is redundant? | {
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So we could use direct substitution instead of finding the limit, meaning that once we get f'(x) = x / |x| we can do f'(0) = 0/ 0 , meaning that it is not possible to differentiate at x= 0?
Well the same holds true for the function itself. Before differentiating your function you should rigorously show that it is in fact continuous at the locations you want to differentiate it at, otherwise the function is not differentiable there and you cannot even bring up the concept of derivative (it is meaningless). In this case, our function $f(x) = |x| / x$ has a discontinuity at $x = 0$ (you can rigorously show that by taking the left and right limits of the function around that point and demonstrating that $\lim_{x \to 0} f(x)$ does not exist as they are not the same ($-1$ and $+1$ respectively).
Once you know that your function is continuous at that point, then you may try to differentiate it. But that won't necessarily mean the derivative exists! Consider $g(x) = |x|$, it is continuous at $x = 0$ (left and right limits are the same) but it is not differentiable at $x = 0$.
You need to be very careful when differentiating. Just because you get a nice expression for your derivative everywhere except at, say, $x = 0$, does not mean that expression is valid for $x = 0$! This is why it's important to keep track of the domain of each function you are working with. If you find that your derivative is valid for all $x \neq 0$, then that's it - end of story. It is meaningless to plug $x = 0$ into it, the results are undefined because the derivative doesn't apply at that point.
Also, limits don't have much to do with derivatives in the grand scheme of things, but they provide a useful framework on top of which to build the concept of derivative and differentiability (as well as a lot of other important stuff). Don't associate them with limits too much. | {
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Here the "left and right limits" are used to show that the two-sided limit exists, and additionally show continuity. The limit expression of the derivative is different and has a geometrical interpretation, which is basically "take two points on a curve and draw a line between them, as the points get closer and closer together that line starts to approximate a tangent to the curve, which happens to also give the rate of change of the height of the curve with horizontal distance and lots of other useful things".
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#### Rido12
##### Well-known member
MHB Math Helper
Once you know that your function is continuous at that point, then you may try to differentiate it. But that won't necessarily mean the derivative exists! Consider $g(x) = |x|$, it is continuous at $x = 0$ (left and right limits are the same) but it is not differentiable at $x = 0$.
Thanks! So I guess the last question is how would I prove that g(x) is not differentiable at x = 0. Say you prove that g(x) is continuous at 0. How would I prove that it is not differentiable at x = 0 .
#### Jameson
Staff member
You already showed that the left and right hand limits at $x=0$ are 1 and -1, which is all you need to do. That shows that the limit doesn't exist thus $g$ is not differentiable at $x=0$. Once again, differentiability implies continuity but continuity does not imply differentiability.
#### Rido12
##### Well-known member
MHB Math Helper
So Step 1 is to see to determine its continuity, if the function is continuous, we can proceed to find the derivative.
Step 2, we find the limit of both sides of the derivative at the point we want to prove, say x = 0 , and we find that the limit doesn't exist
Step 3, we can make a conclusion that because the limit doesn't exist, g'(x) is discontious at the point and impossible to differentiate. for a function to be differentiable, it must be continuous, but the converse is not true.
Is there anythign wrong with the procedure above?
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Is there anythign wrong with the procedure above?
#### Bacterius
##### Well-known member
MHB Math Helper
Thanks! So I guess the last question is how would I prove that g(x) is not differentiable at x = 0. Say you prove that g(x) is continuous at 0. How would I prove that it is not differentiable at x = 0 .
You take the definition of "differentiable":
A function $f(x)$ on $\mathbb{R}$ is differentiable at $x = a$ if and only if $f'(a)$ exists.
From that definition you can choose different approaches depending on your function. For instance, take $f(x) = |x|$. Then, assume $x < 0$, where the function is (probably) differentiable - we'll prove it is shortly. Now apply the definition of the derivative:
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$
And we get:
$$f'(x) = \lim_{h \to 0} \frac{|x + h| - |x|}{h}$$
Now because $x < 0$, $|x| = -x$, and we can rewrite this as:
$$f'(x) = \lim_{h \to 0} \frac{|x + h| + x}{h}$$
Now assume $x + h > 0$. But this means that $h > -x$, and since $x$ is negative then $-x$ is positive so $h$ is greater than some non-zero value.. which is a contradiction since we assumed that $h \to 0$. Therefore $x + h < 0$, so $|x + h| = - (x + h) = -x - h$ and:
$$f'(x) = \lim_{h \to 0} \frac{-x - h + x}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 ~ ~ ~ \text{for} ~ ~ x < 0$$
A similar reasoning for $x > 0$ leads to:
$$f'(x) = +1 ~ ~ ~ \text{for} ~ ~ x > 0$$
So we have:
$$f'(x) = \begin{cases} +1 ~ ~ &\text{for} ~ ~ x > 0 \\ -1 ~ ~ &\text{for} ~ ~ x < 0 \end{cases}$$
What now? Well now the left and right limits around $x = 0$ are easy (note we are never actually touching zero, $f'(x)$ has not been defined at that point) and we clearly see that:
$$\lim_{x \to 0^{-}} f'(x) \ne \lim_{x \to 0^{+}} f'(x)$$
And we may conclude that:
$$\lim_{x \to 0} f'(x) ~ ~ ~ \text{does not exist} ~ ~ ~ \implies ~ ~ ~ f'(0) ~ ~ ~ \text{does not exist}$$
And therefore by the definition of differentiability, $f(x)$ is not differentiable at $x = 0$. | {
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And therefore by the definition of differentiability, $f(x)$ is not differentiable at $x = 0$.
That was the semi-rigorous, long-winded way. In practice you use things like the chain rule and various limit/derivative theorems (or even plain observation) to help you simplify the function instead of starting from scratch each time.
For instance, using the result above, you now know that $f(x) = |x|$ is not differentiable at $x = 0$. From that point on you can easily show that $f(x) = |x - a|$ is not differentiable at $x = a$, for all $a \in \mathbb{R}$, and you can now use that fact to find other, more complicated derivatives involving absolute values, using the chain rule, product rule, and so on.
#### Jameson
Staff member
So Step 1 is to see to determine its continuity, if the function is continuous, we can proceed to find the derivative.
Step 2, we find the limit of both sides of the derivative at the point we want to prove, say x = 0 , and we find that the limit doesn't exist
Step 3, we can make a conclusion that because the limit doesn't exist, g'(x) is discontinuous at the point and impossible to differentiate. for a function to be differentiable, it must be continuous, but the converse is not true.
Is there anything wrong with the procedure above?
I hope someone who is more versed in analysis stops by to comment on this, but I will add what I can.
1) Yes, this is a good start. There are other things to consider as well, such a bend, cusp or vertical tangent - but your step #2 will sort those out anyway. You'll develop an eye for noticing these things quickly. | {
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2) It's always a good idea to be able to use the limit definition of a derivative (left and right hand limits as you said) but of course in practice you have quicker ways of calculating derivatives, as you know. You can probably tell from the problem if the goal is to quickly calculate the derivative and use that as part of a greater problem, or if the differentiability itself is the main focus of the problem. Anyway, this step is ok.
3) If the limit doesn't exist, it doesn't automatically follow that the function is not continuous at that point. $f(x)=|x|$ is a great example of this. It is continuous at $x=0$ but not differentiable.
The only thing that I am not certain of is whether step 1 is implied by step 2. Last night I thought there could be an example of a function and a point where the limit definition of a derivative results in an answer for that point but the function was actually discontinuous at the point, so the derivative there couldn't exist. However, I'm not able to come up with an example of this and doubt that it's possible now. If anyone could that would be interesting.
Put another way, if we have a function of one real variable, $f$, can the limit following limit exist if the point, $a$ is not in the domain of $f$?
$$\displaystyle \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
I think not because $f(a)$ is not defined. So that leads me to say that step 2 implies step 1.
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#### Rido12
##### Well-known member
MHB Math Helper
Last night I thought there could be an example of a function and a point where the limit definition of a derivative results in an answer for that point but the function was actually discontinuous at the point, so the derivative there couldn't exist. However, I'm not able to come up with an example of this and doubt that it's possible now. If anyone could that would be interesting.
This problem was what compelled me to start this thread in the first place. It occurred to me "what if the limit definition of a derivative results in an answer, but that point was actually discontinuous". I'm also interested if there such as a function as well, as it will clear my problems haha. I agree with you on the premise that step 2 implies step 1.
Thanks guys for the help!
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#### MarkFL
$$\displaystyle f(x)=\frac{x}{|x|}$$
It has a jump discontinuity at $x=0$, however, we find:
$$\displaystyle \lim_{h\to0^{-}}\frac{f(x+h)-f(x)}{h}=0=\lim_{h\to0^{+}}\frac{f(x+h)-f(x)}{h}$$ | {
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# Why on average does each bootstrap sample contain roughly two thirds of observations?
I have run across the assertion that each bootstrap sample (or bagged tree) will contain on average approximately $2/3$ of the observations.
I understand that the chance of not being selected in any of $n$ draws from $n$ samples with replacement is $(1- 1/n)^n$, which works out to approximately $1/3$ chance of not being selected.
What is a mathematical explanation for why this formula always gives $\approx 1/3$ ?
-
I believe this is the origin of the $.632$ in the bootstrap 632+ rule. – gung Mar 6 at 2:46
Essentially, the issue is to show that $\lim_{n\to\infty}(1- 1/n)^n=e^{-1}$
(and of course, $e^{-1} =1/e \approx 1/3$, at least very roughly).
It doesn't work at very small $n$ -- e.g. at $n=2$, $(1- 1/n)^n=\frac{1}{4}$. It passes $\frac{1}{3}$ at $n=6$, passes $0.35$ at $n=11$, and $0.366$ by $n=99$. Once you go beyond $n=11$, $\frac{1}{e}$ is a better approximation than $\frac{1}{3}$.
The grey dashed line is at $\frac{1}{3}$; the red and grey line is at $\frac{1}{e}$.
Rather than show a formal derivation (which can easily be found), I'm going to give an outline (that is an intuitive, handwavy argument) of why a (slightly) more general result holds:
$$e^x = \lim_{n\to \infty} \left(1 + x/n \right)^n$$
(Many people take this to be the definition of $\exp(x)$, but you can prove it from simpler results such as defining $e$ as $\lim_{n\to \infty} \left(1 + 1/n \right)^n$.)
Fact 1: $\exp(x/n)^n=\exp(x)\quad$ This follows from basic results about powers and exponentiation
Fact 2: When $n$ is large, $\exp(x/n) \approx 1+x/n\quad$ This follows from the series expansion for $e^x$.
(I can give fuller arguments for each of these but I assume you already know them) | {
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(I can give fuller arguments for each of these but I assume you already know them)
Substitute (2) in (1). Done. (For this to work as a more formal argument would take some work, because you'd have to show that the remaining terms in Fact 2 don't become large enough to cause a problem when taken to the power $n$. But this is intuition rather than formal proof.)
[Alternatively, just take the Taylor series for $\exp(x/n)$ to first order. A second easy approach is to take the binomial expansion of $\left(1 + x/n \right) ^n$ and take the limit term-by-term, showing it gives the terms in the series for $\exp(x/n)$.]
So if $e^x = \lim_{n\to \infty} \left(1 + x/n \right) ^n$, just substitute $x=-1$.
Immediately, we have the result at the top of this answer, $\lim_{n\to\infty}(1- 1/n)^n=e^{-1}$
As gung points out in comments, the result in your question is the origin of the 632 bootstrap rule
e.g. see
Efron, B. and R. Tibshirani (1997),
"Improvements on Cross-Validation: The .632+ Bootstrap Method,"
Journal of the American Statistical Association Vol. 92, No. 438. (Jun), pp. 548-560
-
More precisely, each bootstrap sample (or bagged tree) will contain $1-\frac{1}{e} \approx 0.632$ of the sample.
Let's go over how the bootstrap works. We have an original sample $x_1, x_2, \ldots x_n$ with $n$ items in it. We draw items with replacement from this original set until we have another set of size $n$.
From that, it follows that the probability of choosing any one item (say, $x_1$) on the first draw is $\frac{1}{n}$. Therefore, the probability of not choosing that item is $1 - \frac{1}{n}$. That's just for the first draw; there are a total of $n$ draws, all of which are independent, so the probability of never choosing this item on any of the draws is $(1-\frac{1}{n})^n$. | {
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Now, let's think about what happens when $n$ gets larger and larger. We can take the limit as $n$ goes towards infinity, using the usual calculus tricks (or Wolfram Alpha): $$\lim_{n \rightarrow \infty} \big(1-\frac{1}{n}\big)^n = \frac{1}{e} \approx 0.368$$
That's the probability of an item not being chosen. Subtract it from one to find the probability of the item being chosen, which gives you 0.632.
-
Sampling with replacement can be modeled as a sequence of binomial trials where "success" is an instance being selected. For an original dataset of $n$ instances, the probability of "success" is $1/n$, and the probability of "failure" is $(n-1)/n$. For a sample size of $b$, the odds of selecting an instance exactly $x$ times is given by the binomial distribution:
$$P(x,b,n) = \bigl(\frac{1}{n}\bigr)^{x} \bigl(\frac{n-1}{n}\bigr)^{b-x} {b \choose x}$$
In the specific case of a bootstrap sample, the sample size $b$ equals the number of instances $n$. Letting $n$ approach infinity, we get:
$$\lim_{n \rightarrow \infty} \bigl(\frac{1}{n}\bigr)^{x} \bigl(\frac{n-1}{n}\bigr)^{n-x} {n \choose x} = \frac{1}{ex!}$$
If our original dataset is big, we can use this formula to compute the probability that an instance is selected exactly $x$ times in a bootstrap sample. For $x = 0$, the probability is $1/e$, or roughly $0.368$. The probability of an instance being sampled at least once is thus $1 - 0.368 = 0.632$.
Needless to say, I painstakingly derived this using pen and paper, and did not even consider using Wolfram Alpha.
-
This can be easily seen by counting. How many total possible samples? n^n. How many NOT containing a specific value? (n-1)^n. Probability of a sample not having a specific value - (1-1/n)^n, which is about 1/3 in the limit.
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# Is it OK to evaluate improper integrals this way?
Today in class we learned that when you have an improper integral like this one:
$$\int_{-\infty}^\infty {f(x)} \: dx$$
you must split it before you do the limits (like so):
$$\lim_{a \to \infty} \int_{-a}^n {f(x)} \: dx + \lim_{a \to \infty} \int_{n}^a {f(x)} \: dx$$
I asked if this would work:
$$\lim_{a \to \infty} \int_{-a}^a {f(x)} \: dx$$
And my teacher said it wouldn't work, but he wouldn't explain why. Why wouldn't that work? An example of a function that doesn't work would be especially nice.
• – beep-boop Dec 16 '14 at 19:11
• I tried the same thing, and it didn't work in a case. That's what convinced me I couldn't do this – Justin Dec 16 '14 at 22:23
The problem is that limiting to both infinities at the same time can overlook divergence.
However, the value: $\lim\limits_{a \to \infty} \int_{-a}^a f(x)\,dx$ (if it exists) is important enough to have a name: the principal value.
So why can't you just do the limit like this? Consider $\int_{-\infty}^\infty x\,dx$
Notice that $\int_0^\infty x\,dx$ diverges to $\infty$ and $\int_{-\infty}^0 x\,dx$ diverges to $-\infty$. So we get (from either side) that our integral diverges.
However, $\lim\limits_{a \to \infty} \int_{-a}^a x\,dx = \lim\limits_{a \to\infty} a^2/2 - (-a)^2/2 = \lim\limits_{a \to \infty} 0 = 0$.
So the principal value of $\int_{-\infty}^\infty x\,dx$ is zero while the integral itself diverges.
In general, IF an improper integral converges, you can be very sloppy and get the right answer. On the other hand, if it actually diverges, sloppiness can miss this!
As I tell my students: You must approach ONE side of EACH bad spot by itself. Then put your answer together.
Addendum: Suppose $\int_{-\infty}^c f(x)\,dx$ and $\int_c^\infty f(x)\,dx$ exist. | {
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Addendum: Suppose $\int_{-\infty}^c f(x)\,dx$ and $\int_c^\infty f(x)\,dx$ exist.
Then $\int_{-\infty}^\infty f(x)\,dx = \lim\limits_{a \to -\infty} \int_a^c f(x)\,dx + \lim\limits_{b \to \infty} \int_c^b f(x)\,dx = \lim\limits_{b \to \infty} \int_{-b}^c f(x)\,dx + \lim\limits_{b \to \infty} \int_c^b f(x)\,dx$
Now because those two limits exist, we can combine them. So we get...
$=\lim\limits_{b \to \infty} \int_{-b}^c f(x)\,dx + \int_c^b f(x)\,dx =\lim\limits_{b \to \infty} \int_{-b}^b f(x)\,dx$
...which is the principal value. So the improper integral and principal value match when the integral actually converges.
• And as further evidence that using $\lim\limits_{a \to \infty} \int_{-a}^a x\,dx$ is wrong, consider that $\lim\limits_{a \to \infty} \int_{-a}^{a+1} x\,dx$ obviously diverges, but there's no reason why we couldn't just as well use that to represent $\int_{-\infty}^\infty x\,dx$. Being more symmetrical doesn't make the first one more right ;-) – Steve Jessop Dec 17 '14 at 9:53
• @SteveJessop Exactly. That's what I tried explain in my answer. – Eff Dec 17 '14 at 11:20
(There is already a good answer, but I thought I'd add something.)
We say that a limit exists, if no matter how we approach it approaches the same. For example $$-1 = \lim_{x\to0^-}\frac{|x|}{x}\neq\lim_{x\to 0^+}\frac{|x|}{x} = 1$$ and hence the limit $\lim_{x\to0}|x|/x$ does not exist. In a similar way, we have to consider any way the interval of integration $[a,b]$ grows, where $a\to-\infty$ and $b\to\infty$. | {
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Consider for example \begin{align} &\int_{-n}^{2n} x\,\mathrm{d}x =\frac{3}{2}n^2 &\text{and }\quad &\int_{-n}^n x\,\mathrm{d}x = 0.\end{align} As $n$ approaches $\infty$ the interval is $[-\infty,\infty]$ for both integrals, but approaching in different ways. In this example we have that the first integral goes to infinity, however the second integral seems to converge? As I said, a limit has to be the same no matter how we approach it, so the integral $$\int_{-\infty}^\infty x\,\mathrm{d}x$$ actually diverges (which it does not using "your" method). I hope this helps you see why it is important to consider the end-points of the interval independently.
The only case in which you can not define the integral that way is when $f$ is not integrable. In terms of the Lebesgue integral you can check this for positive functions.
Wikipedia mentions $$f(x) = \left\{ \begin{array} \;1 \; \; x > 0 \\ -1 \;\;x < 0 \end{array} \right. .$$
You can check that this has $0$ at each term in the sequence $\lim\limits_{a \rightarrow \infty} \int\limits_{-a}^a f(x) dx$, but is clearly not integrable by the definition of the Riemann integral or the Lebesgue integral.
If there is a " discontinuity " at the point "n" , then combining the two integrals , will not (may not) give the correct result. | {
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# Integration Techniques [Solved!]
### My question
You state in Section four that all angles are in radians and that the formulas do not work in degress.
Why do they only work in radians?
### Relevant page
4. Integration: Basic Trigonometric Forms
### What I've done so far
Just wanted to know if there are some integration techniques that work in degrees.
X
You state in Section four that all angles are in radians and that the formulas do not work in degress.
Why do they only work in radians?
Relevant page
<a href="https://www.intmath.com/methods-integration/4-integration-trigonometric-forms.php">4. Integration: Basic Trigonometric Forms</a>
What I've done so far
Just wanted to know if there are some integration techniques that work in degrees.
## Re: Integration Techniques
Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).
Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).
Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.
Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):
We know the following for the green curve:
d/dx sin(x) = cos(x)
At x=0, the slope is cos(0) = 1
But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:
At x=0^@, the slope is pi/180 cos(0^@) = pi/180
So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.
In order for the following formula to "work", x needs to be in radians:
d/dx sin(x) = cos(x)
Hope it helps.
X
Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus). | {
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Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).
Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.
Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):
[graph]310,250;-1,10;-1.1,1.1,1.57,1;sin(x),sin(3.14x/180)[/graph]
We know the following for the green curve:
d/dx sin(x) = cos(x)
At x=0, the slope is cos(0) = 1
But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:
At x=0^@, the slope is pi/180 cos(0^@) = pi/180
So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.
In order for the following formula to "work", x needs to be in radians:
d/dx sin(x) = cos(x)
Hope it helps.
## Re: Integration Techniques
It does. A good example to use. Thank you.
X
It does. A good example to use. Thank you.
You need to be logged in to reply.
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System of Equation. (Complicated)
Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt, I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone would provide some different solutions. Thanks in advance.
• Hint: $\;\frac{12}{5x}+\frac{4}{5y}=\cdots$ – dxiv Mar 26 '17 at 3:01
• Can't you equate through $x^2 + y^2$ term and then isolate x (or y) and then sub back in and solve? – 123 Mar 26 '17 at 3:02
As dxiv has given hint:
$$\frac{12}{5x} + \frac{4}{5y} = 2 \implies x = \frac{6y}{5y-2}$$
Now put this in any one of the relation to make the relation full in terms of $y$.
Multiply equation (1) by y and equation (2) by x,
$$5xy(1+\frac{1}{x^2+y^2})=12y$$
$$5xy(1-\frac{1}{x^2+y^2})=4x$$
$$10xy = 12y + 4x$$
You can find y in terms of x or x in terms of y.
• You can try substitution. May be that is better option. – Kanwaljit Singh Mar 26 '17 at 3:55
Just to provide another way.
Use polar coordinates $$x=\rho \cos(\theta) \qquad y=\rho \sin(\theta)$$ to make the equations to be $$\frac{5 \left(\rho ^2+1\right) \cos (\theta )}{\rho }=12$$ $$\frac{5 \left(\rho ^2-1\right) \sin (\theta )}{\rho }=4$$ So,$$\cos(\theta)=\frac{12 \rho }{5 \left(\rho ^2+1\right)}\qquad \sin(\theta)=\frac{4 \rho }{5 \left(\rho ^2-1\right)}$$ Now, using $$\sin^2(\theta)+\cos^2(\theta)=1 \implies \frac{16 \rho ^2}{25 \left(\rho ^2-1\right)^2}+\frac{144 \rho ^2}{25 \left(\rho ^2+1\right)^2}=1$$ Now, let $t=\rho ^2$ to have $$\frac{16 t}{25 \left(t -1\right)^2}+\frac{144 t}{25 \left(t+1\right)^2}=1$$ Reduce to same denominator to get $$25 t^4-160 t^3+206 t^2-160 t+25=0$$ where you can notice the symmetry of coefficients. | {
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Factoring gives $$25 t^4-160 t^3+206 t^2-160 t+25=(5-t) (5 t-1) \left(5 t^2-6 t+5\right)=0$$ The quadratic does not show real roots; this makes the solutions to be $$t_1=\frac 15\qquad t_2=5$$ then $$\rho_1=\frac 1 {\sqrt 5}\qquad \rho_2=\sqrt 5$$ and now compute the corresponding $\theta$'s from $$\tan(\theta)=\frac{\sin(\theta) }{\cos(\theta) }=\frac{t+1}{3 (t-1)}=-\frac 12$$.
• Up for this answer – Mathxx Mar 26 '17 at 6:09
$$5x(1+\frac{1}{x^2+y^2})=12\cdots \cdots (1)$$
$$5y(1-\frac{1}{x^2+y^2})=4\cdots \cdots \cdots \cdots (2) \times i$$
$$5(x+iy)+5\cdot \frac{x-iy}{(x+iy)(x+iy)}=12+4i$$
put $z=x+iy$ and $\bar{z} = x-iy$ and $|z|^2= z \cdot \bar{z} = x^2+y^2$
So $$5z+\frac{5\bar{z}}{z \cdot \bar{z}}=12+4i$$
So $$5z+\frac{5}{z} = 12+4i$$
So $$5z^2-(12+4i)z+5=0$$ after solving $$z=\frac{12+4i\pm \sqrt{(12+4i)^2-100}}{10} = \frac{12+4i\pm (8+6i)}{10} = 2+i\;,\frac{2-i}{5}$$
So $$z= x+iy = 2+i\;,\frac{2-i}{5}$$
So $$(x,y) = \left\{(2,1)\;\;, \left(\frac{2}{5},-\frac{1}{5}\right)\right\}$$
First of all get rid of $x^2+y^2$
$$\frac{12}{4x} +\frac{4}{5y} = 2,\quad 2x+6y = 5xy$$
It is a rectangular hyperbola passing through origin with asymptotes parallel to axes reminding one of $x= t, y= 1/t$ when axes are asymptotes. So solve for $x,y$
$$x= \frac{6y}{5y-2},\, y= \frac{2x}{5x-6}$$
The asymptotes are
$$x= \frac65 ,\, y= \frac{2}{5}$$
We take hyperbola in the form
$$( x-\frac65 ) (y- \frac25) = k$$
Since it goes through the origin, $k$ must be the product $k=\dfrac65 \cdot \dfrac25 = \dfrac{12}{25}$ so as not to leave behind any constant term.
We can now find $x,y$ upto a parameter $t$
$$( x-\frac65 ) (y- \frac25) = \frac{12}{25}$$
$$\rightarrow x= \frac65 + t \sqrt{12}/5,\, y= \frac25 + \sqrt{12}/{5t},\,$$ | {
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# Algebra Question
Hello,
I wanted to confirm that I am understanding something correctly. I have been doing this all the time, and have never had any problems with it. For some reason now I am wanting to confirm it.
So, if you have something such as: x3-6+y2-√z it would be equivalent to write it as: y2+x3-√z-6
Can you always rearrange terms like this, no matter what they are? What I do is write it as an addition problem so I can change the order, since you can change the order in addition.
For example if you have: log(x)-x3 you could write it as log(x)+(-x3) then as -x3 + log(x).
Just as -11+5-3+5=-4 would be the same as 5-11-3+5=-4 or -11-3+5+5=-4
Does my reasoning seem correct?
Thanks
Last edited:
Yes that is correct. I believe that falls under the commutative property which essentially states
x + y = y + x
And this would also go for multiplication, but not division correct?
Correct.
Example using decimal notation:
2*4 = 8 and 4*2 = 8
8 = 8 TRUE
2/4 = .5 and 4/2 = 2
.5 = 2 FALSE
And here's a tip, never be afraid to go back to the most basic of examples to solidify your thoughts if you start question them. Forget the log(x) and y^3 stuff and just use a scratch piece of paper and walk through simple things like 1+2 = 2+1 to convince yourself that the order of addition does not matter.
arildno
Homework Helper
Gold Member
Dearly Missed
Hi, ThomasMagnus!
As the others have said, you are correct.
The two operations multiplication and addition are COMMUTATIVE, i.e, you can change the order in which they appear.
Division and subtraction are NOT commutative, we can call them ANTI-commutative if you like.
But, as you yourself noted, can't we in 5-4 rewrite this as 5+(-4)=(-4)+5? | {
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But, as you yourself noted, can't we in 5-4 rewrite this as 5+(-4)=(-4)+5?
Yes, we can, by using negative numbers in addition, we can think of the operation of subtraction as "redundant", i.e, the info is contained in the use of adding with negative numbers instead.
That makes for easier calculations, because you then can go about commuting as much as you want!
In 5-(-4), for example, we just write 5-(-4)=5+(-(-4))=(-(-4))+5 and so on.
Does something similar exist with division relative to multiplication?
Yes, because division by a non-zero number "a" can always be thought of as multiplication with the reciprocal of "a", i.e, 1/a.
Thus, 2/4=2*(1/4)=(1/4)*2.
----------------
By thinking in terms of "negative numbers" and "reciprocals", you really have only two basic operations to worry about, addition and multiplication, rather than four operations.
(And, if you think of multiplication as repeated addition, there's only one fundamental operation to worry about..)
HallsofIvy
Homework Helper
And this would also go for multiplication, but not division correct?
Yes, and note that it is largely because (well, perhaps even more that addition and multiplication are "associative" while subtraction and division are not) that we do not consider subtraction and division as separate "operations" but just the inverses of additon and multiplication. That is, that "a subtract b" is just "a plus the additive inverse of b" and "a divided by b" is "a multiplied by the multiplicative inverse of b".
arildno
Homework Helper
Gold Member
Dearly Missed
When we think of ONE operation to be performed, commutativity is the only property that comes into the picture.
However, when we have two or more operations to be performed, we must also take into account what we call the associativity of the operations, that is, what relevance it has which operation we perform first. | {
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We have, for example, for numbers "a", "b" and "c" that:
(a+b)+c=a+(b+c), that is, it doesn't matter WHICH addition we perform first. Addition is associative.
But, is:
(a-b)-c=a-(b-c)??
You easily see that subtraction is NOT associative at all;
We have that (a-b)-c=a-(b+c), whereas a-(b-c)=(a-b)+c
Thank you for the info! So the next time I see something in an answer key as something along the lines of x-b+c and I have -b+c+x I can assume they are the exact same, correct? Also, does this go for all terms, even something such as ln|x-1|-ln|x|+6=-ln|x|+ln|x-1|+6 ?
Thanks!
arildno
Homework Helper
Gold Member
Dearly Missed
"Also, does this go for all terms, even something such as ln|x-1|-ln|x|+6=-ln|x|+ln|x-1|+6"
That's right!
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# Is there any other procedure to find the square root?
If no calculator is allowed, and we want to find the square root of a square number if it is large and analyzing to prime factors is hard, how can one proceed? For example, what to do if the number is a square of a large prime number like 5329?
What are methods to perform this and what are pedagogical reasons to prefer one over the other?
• I agree with @BenjaminDickman. This question is more for MSE. I wonder if it could be rewritten to explore any didactical issues or motivations. – Pablo B. Nov 14 '14 at 21:32
• well friends it is just by luck that I wrote the number to be 5329 .. the concept remains the same .. how to find the root of large numers .. I really voted for matin for the link he supplies. thank you all – Abdallah Abusharekh Nov 14 '14 at 21:37
• Related: hsm.stackexchange.com/q/56/72 – Joel Reyes Noche Nov 14 '14 at 23:39
• Following the (justified) remarks by @PabloB. and Benjamin Dickman I made explicit the educational aspect in the question, which is already present in some of the answers. – quid Nov 15 '14 at 13:26
• There're many methods on wikipedia page. – Ruslan Nov 15 '14 at 20:59
You can use (basically) Newton's method.
Take $x_0>0$, and define $$x_{k+1} = \frac{1}{2} (x_k + \frac{n}{x_k}).$$ The seqeunce converges to $\sqrt{n}$.
And if one starts with $x_0= n$ one has, as soon as $|x_{k+1} - x_k|< 1$, that $\lfloor x_{k+1} \rfloor = \lfloor\sqrt{n} \rfloor$.
But the above is not very feasible for computing by hand, and there is better variant, if you just want to do this for perfect squares or test if something is a perfect square, or more generally compute $\lfloor\sqrt{n} \rfloor$. (It seems this is the case for you.)
You can do $$x_{k+1} = \left \lfloor \frac{1}{2} (x_k + \lfloor \frac{n}{x_k} \rfloor) \right \rfloor$$ only doing arithemtic with integers (including euclidean division). This will converge to $\lfloor\sqrt{n} \rfloor$ in a finite number of iterations. | {
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See Integer Square Root on Wikipedia for further references.
I think this could be an interesting method to teach, as one can provide a good geometric motivation for it.
There is an algorithm to calculate square roots with paper and pencil (like the long division algorithm). See How to calculate a square root without a calculator.
• I learned this in grade school. :-) – Joseph O'Rourke Nov 15 '14 at 21:13
• @JosephO'Rourke, I learned this in primary school! – Martín-Blas Pérez Pinilla Nov 19 '14 at 10:08
A fancy way, for those who know a little calculus, is to linearize $f(x)=\sqrt{x}$ at some point near the number you want to square root. This can only be done when you can think of a square rootable number that isn't too far away, so it's not going to work if you're trying to square root a super huge number, but if you can think of one, it's very doable by hand, and very quick. It also has the added benefit of giving you a rational approximation, if that's something you want.
For example, to compute $\sqrt{2}$, we first have to come up with some other number that is reasonably close to $2$ which we do know how to square root. The first thing that comes to mind is $9/4$ (though there are many other choices that are even more accurate, e.g. such as $49/25$). We linearize $\sqrt{x}$ at $9/4$: $$\left.\frac{d}{dx}\sqrt{x}\right|_{x=9/4}=\left.\frac{1}{2\sqrt{x}}\right|_{x=9/4}=\frac{1}{2\sqrt{9/4}}=\frac{1}{3}$$ $$\mathcal{L}_{9/4}(x)=f^\prime\left(\frac{1}{3}\right)\left(x-\frac{9}{4}\right)+f\left(\frac{9}{4}\right)=\frac{1}{3}\left(x-\frac{9}{4}\right)+\frac{3}{2}$$ Now, we plug $2$ in there. $$\mathcal{L}_{9/4}(2)=\frac{1}{3}\left(2-\frac{9}{4}\right)+\frac{3}{2}=\frac{17}{12}$$ From here, if you want decimals, you can simply use long division. Numerically, $17/12$ comes out to be $1.41667$, while the actual value of $\sqrt{2}$ is $1.41421\ldots$. That's pretty close, and likely close enough to do the job for many on the fly calculations. | {
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• An easy nearby perfect square is always obtainable as the next-lowest or next-highest power of 10 (whichever has an even number of zeros). If you know your powers of two, you can do better using the next-lowest or next-highest power of two. – R.. GitHub STOP HELPING ICE Nov 15 '14 at 17:07
Use log tables and the identity $x^\alpha = e^{\alpha \log x}$.
From an educational point of view, that gives evidence of the usefulness of logarithms (you can calculate any power of $x$ using just a log table and multiplication) and teaches about transforming between problems. It also allows all roots to be calculated by the same method.
An intuitive approach to find the square root of A is to draw a square with the number (A) in the middle as the area, and the sides labeled x and y. We know that x times y equal A, and x is the square root of A when x = y. Begin with a rough estimate of x and calculate y as A/x. Now replace x with the average of the original x and y, and repeat.
For example, let's find the square root of 200. As a rough guess, let's begin with x = 15. Solve for y = 200/15 = 13.333. The new value for x, call it x1 = (15 + 13.333)/2 = 14.167. Now the new value y1 = 200/14.167 = 14.117. The square root of 200 is between 14.167 and 14.117. The average is 14.142. If we use x2 = 14.142, then y2 = 200/14.142 = 14.142, verifying that we have the square root of 200 with five significant digits.
This approach has the advantage of showing what a square root is visually, which can give a problem more meaning. The process makes intuitive sense, and it does not require mathematics beyond division.
Here is the method I used to solve the original poster's problem in about three seconds:
Square root (5329):
Steps 1a-d estimate the answer, to one significant figure, with the correct order of magnitude: | {
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Steps 1a-d estimate the answer, to one significant figure, with the correct order of magnitude:
1b) Find the order of magnitude of the answer. Make a copy of the argument (5329). Shift the decimal of the copy of the argument by two places at a time, to get a number between 1 and 100. Shift the decimal of the estimated answer by one place at a time, in the opposite direction. For example, 53.29 and 10.
1c) Assume that the student knows the squares of the integers between 0 and 10 by heart. What is the closest such square? For example, 49.
1d) Multiply the estimated answer by the square root of the closest square. For example, square root (49) = 7, so the new estimated answer is 70.
Step 2 checks the estimate:
2) Square the current estimated answer. For example, 70 * 70 = 4900.
Steps 3a-d refine the estimate using uses (a+b)^2 = a^2 + 2*a*b + b^2. This is a special case of Newton's method:
3a) Subtract the value found in step 2 from the original argument, to yield a remainder. For example, 5329 - 4900 = 429. Keep track of the sign.
3b) Make a copy of the estimated answer. Multiply the copy by 2, to yield a double estimate. For example, 70 * 2 = 140.
3c) Divide the remainder by the double estimate, to yield an adjustment. Round off as convenient. Keep track of the sign. For example, 429 / 140 ~ 3.
3d) Add the adjustment to the estimate (not the double estimate). We kept track of the sign, so that we could correctly choose whether to add or subtract. For example, 70 + 3 = 73. For subsequent calculations, this value replaces estimate.
Repeat steps 2 - 3 until the answer is close enough. For example, 4900 + 420 + 3*3 = 5329, so the square root of 5329 is 73.
• I can do it in my head.
• It only requires keeping track of a few numbers.
• It quickly produces an estimated answer.
• It is a special case of Newton's method.
• Each time through the loop, the student can round off the adjustment to a convenient decimal or fraction. | {
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• Every iteration, the student checks their work.
Three worked examples:
Argu- Estimated Found Double Adjust-
Step ment Answer So Far Remainder Estimate ment Notes
1a 5329 1
1b 10 53.29
1c 7*7=49
1d 70
2 5329 70 4900
3a 5329 70 4900 429
3b 5329 70 4900 429 140
3c 5329 70 4900 429 140 3
2 5329 73 5329 73*73=4900+420+9
Done!
Step ment Answer So Far Remainder Estimate ment Notes
1a 739 1
1b 10 7.39
1c 3*3= 9
1d 30
2 739 30 900
3 739 30 900 -161 60 -3
2-3 739 27 729 10 54 0.2
2-3 739 27.2 739.84 -0.84 54.4 -0.015
2-3 739 27.185 739.024225 -0.024225 54.37 -0.0005
2-3 739 27.1845 738.99704025 54.369 0.00005
2 739 27.18455 738.9997587025
...Stop when satisfied... | {
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Step ment Answer So Far Remainder Estimate ment Notes
1a 4285 1
1b 10 42.85
1c 6*6=36
1d 60
2-3 4285 60 3600 685 120 5
2-3 4285 65 4225 60 130 0.4
2-3 4285 65.4 4277.16 7.84 130.8 0.06
2-3 4285 65.46 4285.0116 -0.0116 130.92 -0.0001
2-3 4285 65.4599 4284.99850801 130.9198 0.00001
2 4285 65.45991 4284.9998171081
...Stop when satisfied... | {
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# trace
### More from my site | {
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• Does the Trace Commute with Matrix Multiplication? Is $\tr (A B) = \tr (A) \tr (B)$? Let $A$ and $B$ be $n \times n$ matrices. Is it always true that $\tr (A B) = \tr (A) \tr (B)$? If it is true, prove it. If not, give a counterexample. Solution. There are many counterexamples. For one, take $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 […] • Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant Let A=\begin{bmatrix} a & b\\ c& d \end{bmatrix} be an 2\times 2 matrix. Express the eigenvalues of A in terms of the trace and the determinant of A. Solution. Recall the definitions of the trace and determinant of A: \[\tr(A)=a+d \text{ and } […] • Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix? Let A be an n \times n matrix. Is it true that \tr ( A^\trans ) = \tr(A)? If it is true, prove it. If not, give a counterexample. Solution. The answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the […] • If Two Matrices are Similar, then their Determinants are the Same Prove that if A and B are similar matrices, then their determinants are the same. Proof. Suppose that A and B are similar. Then there exists a nonsingular matrix S such that \[S^{-1}AS=B$ by definition. Then we […]
• An Example of a Matrix that Cannot Be a Commutator Let $I$ be the $2\times 2$ identity matrix. Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$. Proof. Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies $ABA^{-1}=-B. […] • If 2 by 2 Matrices Satisfy A=AB-BA, then A^2 is Zero Matrix Let A, B be complex 2\times 2 matrices satisfying the relation \[A=AB-BA.$ Prove that $A^2=O$, where $O$ is the $2\times 2$ zero matrix. Hint. Find the trace of $A$. Use the Cayley-Hamilton theorem Proof. We first calculate the […] | {
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• Matrix $XY-YX$ Never Be the Identity Matrix Let $I$ be the $n\times n$ identity matrix, where $n$ is a positive integer. Prove that there are no $n\times n$ matrices $X$ and $Y$ such that $XY-YX=I.$ Hint. Suppose that such matrices exist and consider the trace of the matrix $XY-YX$. Recall that the trace of […]
• Matrices Satisfying $HF-FH=-2F$ Let $F$ and $H$ be an $n\times n$ matrices satisfying the relation $HF-FH=-2F.$ (a) Find the trace of the matrix $F$. (b) Let $\lambda$ be an eigenvalue of $H$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$. Show that there exists an positive integer $N$ […] | {
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# How to solve an ODE of this form $\dot{P}(t)=A^TP(t)+P(t)A$?
Background
If $$\dot{x}(t)=A \,x(t)$$, then we know the solution is $$x(t)=e^{At}x(0)$$.
Question
Now let $$\dot{P}(t)=A^TP(t)+P(t)A$$, what is $$P(t)$$?
Attempt
If we take $$P(t)$$ as common factor (I'm not sure I'm doing this correctly though) ,then we have: $$\dot{P}(t)=A^TP(t)+P(t)A=(A^T+A)P(t)$$ and using the same rationale in the background section, we have $$P(t)=e^{(A^T+A) t}P(0)=e^{A^T t}P(0)e^{At}$$ Is this a correct solution?
Note
I know the answer is $$P(t)=e^{A^T t}P(0)e^{At}$$ but I'm not sure how to find it.
• Counter question: what is A? – Yuriy S Oct 16 '18 at 16:28
• @YuriyS A is a matrix. – Lod Oct 16 '18 at 16:29
• Dimensions are very likely mismatched in the expression $\dot{p}(t)=A^Tp(t)+p(t)A$. – Ennar Oct 16 '18 at 16:30
• What is $p A$ then? – Yuriy S Oct 16 '18 at 16:30
• @Ennar I fixed the question. – Lod Oct 16 '18 at 16:36
Consider the differential equation $$\dot{P}(t)=A^TP(t)+P(t)A$$
We can rewrite it as: $$\dot{P}(\tau)-A^TP(\tau)-P(\tau)A=0$$
Multiply the last differential equation by $$e^{-A^T\tau}$$ from the left and by $$e^{-A\tau}$$ from the right then
$$$$e^{-A^T\tau}\dot{P}(\tau)e^{-A\tau}-e^{-A^T\tau}A^TP(\tau)e^{-A\tau}-e^{-A^T\tau}P(\tau)Ae^{-A\tau}=0 \quad (1)$$$$ We can easily see that equation $$(1)$$ is generated by the derivative of three multiplied functions: $$$$\frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)=e^{-A^T\tau}\dot{P}(\tau)e^{-A\tau}-A^Te^{-A^T\tau}P(\tau)e^{-A\tau}-e^{-A^T\tau}P(\tau)Ae^{-A\tau}=0$$$$
So we come up with:
$$0=\frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)$$
Take the integral of both sides
$$$$0=\int_0^t\frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)d\tau=\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)\bigg]_0^t=e^{-A^Tt}{P} \tau)e^{-At}-e^{0}{P}(0)e^{0}$$$$ \ $$$$0=e^{-A^Tt}{P}(t)e^{-At}-P(0)$$$$
Multiply left and right sides by $$e^{A^Tt}$$ and $$e^{At}$$ respectively, we get: | {
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Multiply left and right sides by $$e^{A^Tt}$$ and $$e^{At}$$ respectively, we get:
$$$$P(t)=e^{A^Tt}{P}(0)e^{At}$$$$
Note: I used the fact that $$e^{-A^T\tau}$$ and $$A^T$$ commute, i.e. $$A^Te^{-A^T\tau}=e^{-A^T\tau}A^T$$
For small time increment you get approximately $$P(t+dt)=P(t)+(A^TP(t)+P(t)A)dt+O(dt^2)=(I+A^T\,dt)P(t)(I+A\,dt)+O(dt^2)$$ which means that from time step to time step you get an accumulation of these factors on both sides, $$P(N\,dt)=(I+A^T\,dt)^NP(t)(I+A\,dt)^N+O(Ndt^2)$$ Now if $$N\,dt=\Delta t$$ we get, using $$(I+B/N)^N=\exp(B)+O(B^2/N)$$ $$P(t+Δt)=e^{A^T\,Δt}P(t)e^{A\,Δt}+O(Δt\,dt)$$ so that for $$dt\to 0$$ one gets exactly the claimed solution form.
By vectorizing $$P$$ and using the Kronecker product, similar to a method which can be used to solve Sylvester equations, then the differential equation can also be written as
$$\frac{d}{dt}\,\text{vec}\,P(t) = \underbrace{\left(I \otimes A^\top + A^\top\otimes I\right)}_{M}\,\text{vec}\,P(t), \tag{1}$$
which has the solution
$$\text{vec}\,P(t) = e^{M\,t}\,\text{vec}\,P(0). \tag{2}$$
By using the mixed-product property of the Kronecker product it can be shown that $$I \otimes A^\top$$ commutes with $$A^\top \otimes I$$. This commuting property allows you to write the matrix exponential as
$$e^{M\,t} = e^{(I\,\otimes\,A^\top)\,t}\,e^{(A^\top\,\otimes\,I)\,t}. \tag{3}$$
By using the definition of a matrix exponential and again the mixed-product property of the Kronecker product then is can also be shown that
$$e^{X\,\otimes\,I} = e^{X} \otimes I, \quad e^{I\,\otimes\,X} = I \otimes e^{X},$$
thus
$$e^{M\,t} = \left(I \otimes e^{A^\top t}\right) \left(e^{A^\top t} \otimes I\right). \tag{4}$$
Substituting $$(4)$$ into $$(2)$$ gives | {
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Substituting $$(4)$$ into $$(2)$$ gives
\begin{align} \text{vec}\,P(t) &= \left(I \otimes e^{A^\top t}\right) \left(e^{A^\top t} \otimes I\right) \text{vec}\,P(0) \\ &= \left(I \otimes e^{A^\top t}\right) \text{vec}\left(P(0)\,e^{A\,t}\right) \\ &= \text{vec}\left(e^{A^\top t}\,P(0)\,e^{A\,t}\right) \end{align}
thus
$$P(t) = e^{A^\top t}\,P(0)\,e^{A\,t}.$$ | {
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# Series convergence
1. Jun 10, 2009
### utopiaNow
1. The problem statement, all variables and given/known data
Use partial fractions to show
$$\displaystyle\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)} = \frac{1}{4}$$
3. The attempt at a solution
I did the partial fraction decomposition to get: $$\displaystyle\sum_{n=1}^\infty \frac{1}{2n} - \frac{1}{n + 1} + \frac{1}{2n + 4}$$
I'm not sure how to proceed from here, in my textbook the example shows how terms in the partials sums overlap and cancel out if you start looking at the terms in the partial fraction decomposition, however I can't see that happening with this particular series.
Any suggestions would be appreciated. Thanks in advance.
2. Jun 10, 2009
### rock.freak667
Write out a few term, n=1,2,3,4,5....N-3,N-2,N-1,N
then check as N->inf.
3. Jun 10, 2009
### utopiaNow
Hi,
I tried doing that originally, but I don't see any patterns that I can exploit when I write the terms of the partial decomposition for n = 1, 2, 3,..., N-2, N-1, N as you stated.
Do you have any further suggestions?
4. Jun 10, 2009
### Dick
Write the terms for successive values of n each on a row 1/(2n+4) -1/(n+1) 1/(2n)
1/6 -1/2 1/2
1/8 -1/3 1/4
1/10 -1/4 1/6
1/12 -1/5 1/8
1/14 -1/6 1/10
Oblique hint: diagonal.
5. Jun 10, 2009
### utopiaNow
Interesting, so everything seems to be canceling out if you keep going long enough diagonally, except the 1/4 in the second row. However is there any way to formalize this? Or is simply noticing this pattern enough for a formal proof?
6. Jun 10, 2009
### Dick
Depends on how formal you want to be. Writing a table like that and scratching out the cancellations is good for me. If you want to do it formally write out the first term from the n case, the second term from the n+1 case and the third term from the n+2 case and show they cancel algebraically. What is important is to realize just because there isn't an obvious cancellation doesn't mean there isn't one.
7. Jun 10, 2009 | {
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7. Jun 10, 2009
### utopiaNow
Thank you for the help!
8. Jun 10, 2009
### txy
why not take out 1/2 from the partial fraction decomposition so that it's more obvious?
$$\sum^{+\infty}_{n=1}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{n+1})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{1}{n+1}) + \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n+2}-\frac{1}{n+1})$$
9. Jun 10, 2009
### Dick
That's a nice way to handle it. There's more than one way to skin a cat. | {
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# Solutions of $x^2-6x-13 \equiv 0 \pmod{127}$
I started learning number theory, specifically polynomial congruences, and need help with the following exercise. Here it is:
Does the congruence $x^2-6x-13 \equiv 0 \pmod{127}$ has solutions?
I tried to follow the method for solving general quadratic congruence but I didn't get really far. Here's what I've done so far:
Since $(4, 127) = 1$, we may complete the square by multiplying by $4$ without having to change the modulus in order to get the following equivalent congruence
$$(2x-6)^2 \equiv 36 - 4(-13) \pmod{127} \iff (2x-6)^2 \equiv 88 \pmod{127}.$$
If I'm heading in the right direction then I don't know how to continue from here. I suspect there is another method for solving this problem since I didn't make use of the fact that $127$ is prime. Also, the problem doesn't require to actually find the solutions but only determine if there are solutions. Possibly we can avoid computations and make use of some theorem/lemma to find if the congruence has solutions or not.
• The quadratic formula, $x = \frac{6\pm\sqrt{(-6)^2 - 4\cdot 1\cdot (-13)}}{2\cdot 1}$ still works, since $127$ is an odd prime. Note, however, that square root and dividing by $2$ has a different meaning from what it does for real numbers (or, it has the same meaning, but potentially a very different result). – Arthur Sep 19 '17 at 11:16
. To find solutions of $$x^2 - 6x - 13 \equiv 0 \mod 127$$, we must write $$x^2-6x - 13$$ as a square of a linear term, plus a constant. It's seen easily that $$x^2 - 6x - 13 = (x-3)^2 - 22$$. Hence, the congruence is equivalent to $$(x-3)^2 \equiv 22 \mod 127$$. | {
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Now, we must check if $$22$$ is a quadratic residue mod $$127$$. For this, we can use the Legendre symbol, whose notation I will keep the same as the binomial, so do not get confused. $$\binom{22}{127} = \color{green}{\binom{11}{127}}\color{red}{\binom{2}{127}} = \color{green}{-\binom{127}{11}} \times \color{red}1$$ where the terms with same color on the LHS and RHS are equal. The green equality comes by quadratic reciprocity and the second comes by the fact that $$\binom{2}{p}$$ is well known by the remainders which $$p$$ leaves when divided by $$8$$.
Now, we can do: $$-\binom{127}{11} = -\binom{6}{11} = -\color{green}{\binom{2}{11}}\color{red}{ \binom{3}{11}} = -\color{green}{(-1)}\color{red}{(1)} = 1$$
Again, the colored terms on the LHS and RHS are equal because the quantities $$\binom 2p$$ and $$\binom 3p$$ are well known.
Since we have obtained that the Legendre symbol is $$1$$, this implies the existence of a solution, and therefore two.
The question, though, is how to compute them. I do not know of any method other than brute force, unfortunately. However, our reward for writing $$(x-3)^2 \equiv 22 \mod 127$$ is that we basically only need to look for squares of the form $$127k + 22$$ to find a solution, rather than having to substitute values of $$x$$ into the expression $$x^2-6x-13$$ each time.
A brute force : the series $$127k + 22$$ goes like : $$22,149,276,403,530,657,\color{blue}{784},...$$
lo and behold, $$784 = 28^2$$, hence this gives $$x = 31$$. Now, note that the congruence actually has two solutions, one given by $$127 - 28 = 99$$. You can check that $$9801 = 99^2 = 22 + 127 \times 77$$.
Hence, we get two solutions of $$x$$, namely $$x= 31,102$$. | {
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Hence, we get two solutions of $$x$$, namely $$x= 31,102$$.
• I can't thank you enough for this answer! I understood everything and it made me see the link that I was missing between quadratic polynomial/residues and Legendre symbol. – user347616 Sep 19 '17 at 12:07
• But of course, you are welcome! I should mention, +1 for your question. – астон вілла олоф мэллбэрг Sep 19 '17 at 12:08
• Since $127$ is one less than a multiple of $4$, you can get the square roots by "root-power conversion". Fermat's Little Theorem gives $22^{126}\equiv 1$, and with $22$ being a quadratic residue then $22^{63}\equiv 1$. So $22^{64}\equiv 22$ assuring that $22^{32}\equiv 99$ will be one square root. Try it! – Oscar Lanzi Sep 19 '17 at 12:26
• That is a much,much better method than the one I have suggested. I thank you for this. – астон вілла олоф мэллбэрг Sep 19 '17 at 12:29
Using the Berlekamp Algorithm for factoring a polynomial over $\mathbb{F}_{127}$ we immediately obtain that $$x^2-6x-13=(x+25)(x+96).$$ So we have two solutions.
I am a congruence beginner like you.
I did in this way
completed the square $(x-3)^2-9\equiv 13 \mod 127$
$(x-3)^2\equiv 22 \mod 127$
Now call $r$ is the number such that $r^2\equiv 22\mod 127$?
Like a "modular" square root
I found here a way to find the roots
As $127\equiv 3 \mod 4$ then we have $r= \pm 22^{\frac{128}{4}} \mod 127$
$22^{32} =2^{32}\cdot 11^{32}$
$2^7\equiv 1 \mod 127\to 2^{28}\equiv 1 \mod 127\to \color{red}{2^{32}\equiv 2^4\mod 127}$
$11^{32}=121^{16}$ as $121\equiv 6 \mod 127$ we have $121^{16}\equiv 6^{16} \mod 127$ and then $6^{16}\equiv 2^{16}\cdot 3^{16} \mod 127$
$2^k \mod 127$ for $k=0,1,2\ldots$ is cyclical and is $1,2,4,8,16,32,64$ then repeats (because $127=2^7-1$) therefore $2^{16}\equiv 4 \mod 127$
$3^{16}\equiv 81^4 \mod 127\to 3^{16}\equiv 71 \mod 127$
Finally $6^{16}\equiv 4\cdot 71 \mod 127\to 6^{16}\equiv 30 \mod 127$ and $\color{red}{11^{32}\equiv 6^{16}\equiv 30 \mod 127}$ | {
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Collecting the parts in red colour we get $22^{32}\equiv 2^4\cdot 30 \mod 127$ that is $22^{32}\equiv 99\mod 127$ and $-22^{32} \equiv 28 \mod 127$
Concluding the two solutions are $x-3\equiv 99 \mod 127\to x\equiv 102 \mod 127$ and $x-3\equiv 28 \mod 127 \to x\equiv 31 \mod 127$
Edit
My solution is all but elegant. Anyway is a solution provided by a beginner like you and I hope it is understandable
• It took me a little time to go through your solution but I made it. Thank you for the effort, it is always enlightening to see different solutions to the same problem. (+1) – user347616 Sep 19 '17 at 14:59
• To compute $22^{32} \bmod127$ just square five successive times. – Oscar Lanzi Sep 19 '17 at 22:33
• @OscarLanzi Why $22^{32}\equiv 22^5 \mod 127$? – Raffaele Sep 20 '17 at 13:11
• I did not say that. Square five successive times means $22^{2^5}$. – Oscar Lanzi Sep 20 '17 at 13:42
I think you can try $$x=31.$$ It is valid.
Finally we have $$x^2-6x-13\equiv(x-31)(x-102) \pmod{127}.$$
This equation has roots modulo $127$ if and only if the reduced discriminant $\Delta'=9+13$ is a square modulo $127$. We'll use the laws of quadratic reciprocity.
$$\biggl(\frac{22}{127}\biggr)=\biggl(\frac{2}{127}\biggr)\biggl(\frac{11}{127}\biggr)=\biggl(\frac{11}{127}\biggr)\quad\text{since }127\equiv -1\mod8\quad (2^{\mathit{nd}}\textit{supplementary law}).$$ Now we have \begin{align}\biggl(\frac{11}{127}\biggr)&=\biggl(\frac{127}{11}\biggr)\bigl(-1\bigr)^{\tfrac{10\cdot126}4}=-\biggl(\frac{6}{11}\biggr)=-\biggl(\frac{2}{11}\biggr)\biggl(\frac{3}{11}\biggr)\\[1ex] &=+\biggl(\frac{3}{11}\biggr)=\biggl(\frac{11}{3}\biggr)\bigl(-1\bigr)^{\tfrac{2\cdot10}4}=-\biggl(\frac{2}{3}\biggr)=+1. \end{align} Thus the equation has roots. | {
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# Relations between p norms
The $$p$$-norm on $$\mathbb R^n$$ is given by $$\|x\|_{p}=\big(\sum_{k=1}^n |x_{k}|^p\big)^{1/p}$$. For $$0 < p < q$$ it can be shown that $$\|x\|_p\geq\|x\|_q$$ (1, 2). It appears that in $$\mathbb{R}^n$$ a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: $$\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$$(3), $$\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $$\mathbb R^n$$. For instance, for $$n=2$$ and $$n=3$$ one can see that for $$0 < p < q$$, the spheres with radius $$\sqrt n$$ with $$\|\cdot\|_p$$ inscribe spheres with radius $$1$$ with $$\|\cdot\|_q$$.
It is not hard to prove the inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For $$n=2$$ it is easily proven (see below), but not for $$n>2$$. So my questions are:
1. How can relation (3) be proven for arbitrary $$n\,$$?
2. Can this be generalized into something of the form $$\|x\|_{p} \leq C \|x\|_{q}$$ for arbitrary $$0?
3. Do any of the relations also hold for infinite-dimensional spaces, i.e. in $$l^p$$ spaces?
Notes:
$$\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2$$, hence $$=2\|x\|_{2}^{2}$$
$$\|x\|_{1} \leq \sqrt 2 \|x\|_{2}$$. This works because $$|x_{1}|^2 + |x_{2}|^2 \geq 2|x_{1}\|x_{2}|$$, but only because $$(|x_{1}| - |x_{2}|)^2 \geq 0$$, while for more than two terms $$\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0$$ gives an inequality that never gives the right signs for the cross terms. | {
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• Your first link (1) has exactly the equation you seek. Actually a version of Norberts excellent answer, generalized to measure spaces. – Thomas Ahle Mar 31 '16 at 20:57
• Probably a dead thread, but norms in finite dimensional spaces are equivalent, so one can always find c ad C such that $c\|x\|_p\leq \|x\|_q\leq C\|x\|_p$. – BigM Dec 1 '19 at 20:23
1. Using Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$ $$\Vert x\Vert_1= \sum\limits_{i=1}^n|x_i|= \sum\limits_{i=1}^n|x_i|\cdot 1\leq \left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}= \sqrt{n}\Vert x\Vert_2$$
2. Such a bound does exist. Recall Hölder's inequality $$\sum\limits_{i=1}^n |a_i||b_i|\leq \left(\sum\limits_{i=1}^n|a_i|^r\right)^{\frac{1}{r}}\left(\sum\limits_{i=1}^n|b_i|^{\frac{r}{r-1}}\right)^{1-\frac{1}{r}}$$ Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$ $$\sum\limits_{i=1}^n |x_i|^p= \sum\limits_{i=1}^n |x_i|^p\cdot 1\leq \left(\sum\limits_{i=1}^n (|x_i|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}} \left(\sum\limits_{i=1}^n 1^{\frac{q}{q-p}}\right)^{1-\frac{p}{q}}= \left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}$$ Then $$\Vert x\Vert_p= \left(\sum\limits_{i=1}^n |x_i|^p\right)^{1/p}\leq \left(\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}\right)^{1/p}= \left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{1}{q}} n^{\frac{1}{p}-\frac{1}{q}}=\\= n^{1/p-1/q}\Vert x\Vert_q$$ In fact $C=n^{1/p-1/q}$ is the best possible constant.
• @Arun, because for $x=(1,1,1,\ldots,1)$ this bound is attained – Norbert Jun 10 '15 at 14:27
• @Norbert: I'm trying to derive a similar bound for the case $p>q$. Any ideas about how to proceed ? – pikachuchameleon Jun 22 '15 at 12:19
• @AshokVardhan what does "similar" mean? Do you want to prove something like $\Vert x\Vert_p \leq C\Vert x\Vert_q$ for $p>q$ ? – Norbert Jun 23 '15 at 14:24 | {
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Question
# A point moves such that the sum of the square of its distance from the sides of a sqaure of side unity is equal to $$9$$. The locus of the point is a circle such that
A
Center of the circle coincides with that of square
B
Center of the circle is (12,12)
C
Radius of the circle is 2
D
All the above are true
Solution
## The correct option is D All the above are trueLet the sides of the square be $$y=0,y=1,x=0$$ and $$x=1.$$Let the moving point be $$(x,y).$$Then,$${ y }^{ 2 }+{ \left( y-1 \right) }^{ 2 }+{ x }^{ 2 }+{ \left( x-1 \right) }^{ 2 }=9$$ is the equation of the locus.$$\Rightarrow 2{ x }^{ 2 }+2{ y }^{ 2 }-2x-2y-7=0,$$ which represents a circle having centre $$\displaystyle \left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right)$$ (which is also the centre of the square) and radius $$2.$$ Mathematics
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# Looking for a proof of an interesting identity
Working on a problem I have encountered an interesting identity:
$$\sum_{k=0}^\infty \left(\frac{x}{2}\right)^{n+2k}\binom{n+2k}{k} =\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n,$$ where $n$ is a non-negative integer number and $x$ is a real number with absolute value less than 1 (probably a similar expression is valid for arbitrary complex numbers $|z|<1$).
Is there any simple proof of this identity?
• How/where did you meet this identity? Is the LHS just the Taylor series for the RHS? – Daniel Robert-Nicoud Dec 14 '17 at 23:38
• Anything to do with a random walk? Why did you call this interesting, btw? – Mathemagical Dec 14 '17 at 23:44
• Reminiscent of the explicit form for Chebyshev polynomials of the first kind, hence they idea of enforcing the substitution $x=\sin\theta$ and applying the residue theorem looks like a promising one ;) – Jack D'Aurizio Dec 14 '17 at 23:56
• Or you may prove that both sides are the terms of an Appell / Fibonacci sequence of polynomials. – Jack D'Aurizio Dec 14 '17 at 23:57
• @tired This is in regard to the hitting time of a random walk (more particularly, the probability generating function of the hitting time of the walk at the barrier x=1. Please see the equations 15,16, and 17 in the linked document. galton.uchicago.edu/~lalley/Courses/312/RW.pdf The details of the computation are not listed here, but they are shown in greater detail in Michael Steele's book, Stochastic Calculus and Financial applications (pages 8 and 9). – Mathemagical Dec 16 '17 at 15:25
Using
$$\binom{n}{k}=\frac{1}{2 \pi i}\oint_C\frac{(1+z)^{n}}{z^{k+1}}dz$$ we get (integration contour is the unit cicrle)
$$2\pi iS_n=\oint dz \sum_{k=0}^{\infty}\frac{(1+z)^{n+2k}x^{n+2k}}{z^{k+1}2^{n+2k}}=\oint dz \frac{(1+z)^n x^n}{z2^n}\sum_{k=0}^{\infty}\frac{(1+z)^{2k}x^{2k}}{2^{2k}z^k}=\\ 4\frac{x^n}{2^n}\oint dz \underbrace{\frac{(1+z)^n}{4z-(1+z)^2x^2}}_{f(z)}$$ | {
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for $|x|<1$ only we have just one pole of $f(z)$ inside the unit circle namely $z_0(x)=\frac2{x^2}-\frac{2\sqrt{1-x^2}}{x^2}-1$ , so
$$S_n=4\frac{x^n}{2^n}\text{res}(f(z),z=z_0(x))=4\frac{x^n}{2^n}\left[ \frac{1}{4 \sqrt{1-x^2}}\left(2\frac{1-\sqrt{1-x^2}}{ x^2}\right)^n\right]$$
or
$$S_n=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{ x}\right)^n$$
• Thanks Jack :-) – tired Dec 15 '17 at 0:55
• @tired: Thank you very much for the nice proof. It contains however a misprint. $4$ shall be replaced with $4^{1/n}$ in the denominator of the residue value. But probably simpler and more correctly were to write it just as $$\text{res}(f(z),z=z_0)=\frac{1}{4\sqrt{1-x^2}} \left(2\frac{1-\sqrt{1-x^2}}{x^2}\right)^n.$$ – user Dec 15 '17 at 11:56
• @user355705 you are absolutly right – tired Dec 16 '17 at 15:07
Extracting coefficients on the RHS we get the integral (coefficient on $x^{n+2k}$)
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+2k+1}} \frac{1}{\sqrt{1-z^2}} \left(\frac{1-\sqrt{1-z^2}}{z}\right)^n \; dz.$$
Now we put $(1-\sqrt{1-z^2})/z = w$ so that $z = 2w/(1+w^2).$ This has $w = \frac{1}{2} z + \cdots$ so the image in $w$ of the contour in $z$ can be deformed to a small circle enclosing the origin in the $w$-plane. (Moreover we see that the exponentiated term starts at $z^n$ which justifies the corresponding offset in the series.) We get $dz = 2/(1+w^2) - 4w^2/(1+w^2)^2 \; dw = 2(1-w^2)/(1+w^2)^2 \; dw.$ We also have $1-z^2 = 1 - 4w^2/(1+w^2)^2 = (1-w^2)^2/(1+w^2)^2.$ All of this yields
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w^2)^{n+2k+1}}{2^{n+2k+1} w^{n+2k+1}} \frac{1}{(1-w^2)/(1+w^2)} w^n \frac{2(1-w^2)}{(1+w^2)^2} \; dw \\ = \frac{1}{2^{n+2k}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w^2)^{n+2k}}{w^{2k+1}} \; dw.$$
This evaluates by inspection to
$$\frac{1}{2^{n+2k}} [w^{2k}] (1+w^2)^{n+2k} = \frac{1}{2^{n+2k}} [w^{k}] (1+w)^{n+2k} \\ = \frac{1}{2^{n+2k}} {n+2k\choose k}$$
which is the claim. | {
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which is the claim.
I plan to insert this approach into the next (2019) version of my notes.
The even powers of $\arcsin(x)$ all have a nice Maclaurin series (see eqs (19),(20),(21),(22)), and this can be seen as a side effect of the Lagrange–Bürmann inversion theorem. I am going to implement the same trick here, by computing the derivatives at the origin of the RHS (the opposite way has already been investigated by tired).
$$[x^m]\left[\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n\right]=\frac{1}{2\pi i}\oint\frac{(1-\sqrt{1-x^2})^n}{x^{n+m+1}\sqrt{1-x^2}}\,dx$$ where the integral is performed along a small circle enclosing the origin. What about enforcing the substitution $x=\sin z$? The sine function is holomorphic and injective in a neighbourhood of the origin, hence the RHS is simply converted into $$\frac{1}{2\pi i}\oint\frac{(1-\cos z)^n}{\left(\sin z\right)^{n+m+1}}\,dz =\frac{1}{2\pi i}\oint\frac{2^n \sin^{2n}\frac{z}{2}}{2^{n+m+1}\sin^{n+m+1}\frac{z}{2}\cos^{n+m+1}\frac{z}{2}}\,dz$$ or $$\frac{1}{2\pi i}\oint\frac{\tan^{n-m-1}z}{2^{m}\cos^{2m}z}\,dz=\frac{1}{2^{m+1}\pi i}\oint u^{n-m-1}(1+u^2)^{m-1}\,du$$ via $z\to 2z$ and $z\to\arctan u$. Now the RHS is trivially given by a binomial coefficient multiplied by $\frac{1}{2^m}$ (provided by $m\geq n$) and the claim is proved. | {
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# fimplicit3
Plot 3-D implicit equation or function
## Syntax
fimplicit3(f)
fimplicit3(f,[min max])
fimplicit3(f,[xmin xmax ymin ymax zmin zmax])
fimplicit3(___,LineSpec)
fimplicit3(___,Name,Value)
fimplicit3(ax,___)
fi = fimplicit3(___)
## Description
example
fimplicit3(f) plots the 3-D implicit equation or function f(x,y,z) over the default interval [-5 5] for x, y, and z.
example
fimplicit3(f,[min max]) plots f(x,y,z) over the interval [min max] for x, y, and z.
example
fimplicit3(f,[xmin xmax ymin ymax zmin zmax]) plots f(x,y,z) over the interval [xmin xmax] for x, [ymin ymax] for y, and [zmin zmax] for z. The fimplicit3 function uses symvar to order the variables and assign intervals.
example
fimplicit3(___,LineSpec) uses LineSpec to set the line style, marker symbol, and face color.
example
fimplicit3(___,Name,Value) specifies line properties using one or more Name,Value pair arguments. Use this option with any of the input argument combinations in the previous syntaxes.
fimplicit3(ax,___) plots into the axes with the object ax instead of the current axes object gca.
example
fi = fimplicit3(___) returns an implicit function surface object. Use the object to query and modify properties of a specific surface. For details, see ImplicitFunctionSurface Properties.
## Examples
collapse all
Plot the hyperboloid ${x}^{2}+{y}^{2}-{z}^{2}=0$ by using fimplicit3. The fimplicit3 function plots over the default interval of $\left[-5,5\right]$ for $x$, $y$, and $z$.
syms x y z fimplicit3(x^2 + y^2 - z^2)
Plot the hyperboloid specified by the function $f\left(x,y,z\right)={x}^{2}+{y}^{2}-{z}^{2}$. The fimplicit3 function plots over the default interval of $\left[-5,5\right]$ for $x$, $y$, and $z$.
syms f(x,y,z) f(x,y,z) = x^2 + y^2 - z^2; fimplicit3(f) | {
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syms f(x,y,z) f(x,y,z) = x^2 + y^2 - z^2; fimplicit3(f)
Specify the plotting interval by specifying the second argument to fimplicit3. Plot the upper half of the hyperboloid ${x}^{2}+{y}^{2}-{z}^{2}=0$ by specifying the interval $0. For $x$ and $y$, use the default interval $\left[-5,5\right]$.
syms x y z f = x^2 + y^2 - z^2; interval = [-5 5 -5 5 0 5]; fimplicit3(f, interval)
Plot the implicit equation $x\mathrm{sin}\left(y\right)+z\mathrm{cos}\left(x\right)=0$ over the interval $\left(-2\pi ,2\pi \right)$ for all axes.
Create the x-axis ticks by spanning the x-axis limits at intervals of pi/2. Convert the axis limits to precise multiples of pi/2 by using round and get the symbolic tick values in S. Display these ticks by using the XTick property. Create x-axis labels by using arrayfun to apply texlabel to S. Display these labels by using the XTickLabel property. Repeat these steps for the y-axis.
To use LaTeX in plots, see latex.
syms x y z eqn = x*sin(y) + z*cos(x); fimplicit3(eqn,[-2*pi 2*pi]) title('xsin(y) + zcos(x) for -2\pi < x < 2\pi and -2\pi < y < 2\pi') xlabel('x') ylabel('y') ax = gca; S = sym(ax.XLim(1):pi/2:ax.XLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); ax.XTick = double(S); ax.XTickLabel = arrayfun(@texlabel,S,'UniformOutput',false); S = sym(ax.YLim(1):pi/2:ax.YLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); ax.YTick = double(S); ax.YTickLabel = arrayfun(@texlabel, S, 'UniformOutput', false);
Plot the implicit surface ${x}^{2}+{y}^{2}-{z}^{2}=0$ with different line styles for different values of $z$. For $-5, use a dashed line with green dot markers. For $-2, use a LineWidth of 1 and a green face color. For $2, turn off the lines by setting EdgeColor to none.
syms x y z f = x^2 + y^2 - z^2; fimplicit3(f,[-5 5 -5 5 -5 -2],'--.','MarkerEdgeColor','g') hold on fimplicit3(f,[-5 5 -5 5 -2 2],'LineWidth',1,'FaceColor','g') fimplicit3(f,[-5 5 -5 5 2 5],'EdgeColor','none') | {
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Plot the implicit surface $1/{x}^{2}-1/{y}^{2}+1/{z}^{2}=0$. Specify an output to make fimplicit3 return the plot object.
syms x y z f = 1/x^2 - 1/y^2 + 1/z^2; fi = fimplicit3(f)
fi = ImplicitFunctionSurface with properties: Function: [1x1 sym] EdgeColor: [0 0 0] LineStyle: '-' FaceColor: 'interp' Show all properties
Show only the positive x-axis by setting the XRange property of fi to [0 5]. Remove the lines by setting the EdgeColor property to 'none'. Visualize the hidden surfaces by making the plot transparent by setting the FaceAlpha property to 0.8.
fi.XRange = [0 5]; fi.EdgeColor = 'none'; fi.FaceAlpha = 0.8;
Control the resolution of an implicit surface plot by using the 'MeshDensity' option. Increasing 'MeshDensity' can make smoother, more accurate plots while decreasing 'MeshDensity' can increase plotting speed.
Divide a figure into two by using subplot. In the first subplot, plot the implicit surface $\mathrm{sin}\left(1/\left(xyz\right)\right)$. The surface has large gaps. Fix this issue by increasing the 'MeshDensity' to 40 in the second subplot. fimplicit3 fills the gaps showing that by increasing 'MeshDensity' you increased the resolution of the plot.
syms x y z f = sin(1/(x*y*z)); subplot(2,1,1) fimplicit3(f) title('Default MeshDensity = 35') subplot(2,1,2) fimplicit3(f,'MeshDensity',40) title('Increased MeshDensity = 40')
Apply rotation and translation to the implicit surface plot of a torus.
A torus can be defined by an implicit equation in Cartesian coordinates as
$\mathit{f}\left(\mathit{x},\mathit{y},\mathit{z}\right)={\left({\mathit{x}}^{2}+{\mathit{y}}^{2}+{\mathit{z}}^{2}+{\mathit{R}}^{2}-{\mathit{a}}^{2}\right)}^{2}-4{\mathit{R}}^{2}\left({\mathit{x}}^{2}+{\mathit{y}}^{2}\right)$
where
• $a$ is the radius of the tube
• $R$ is the distance from the center of the tube to the center of the torus
Define the values for $a$ and $R$ as 1 and 5, respectively. Plot the torus using fimplicit3. | {
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Define the values for $a$ and $R$ as 1 and 5, respectively. Plot the torus using fimplicit3.
syms x y z a = 1; R = 4; f(x,y,z) = (x^2+y^2+z^2+R^2-a^2)^2 - 4*R^2*(x^2+y^2); fimplicit3(f) hold on
Apply rotation to the torus around the $x$-axis. Define the rotation matrix. Rotate the torus by 90 degrees or $\pi /2$ radians. Shift the center of the torus by 5 along the $x$-axis.
alpha = pi/2; Rx = [1 0 0; 0 cos(alpha) sin(alpha); 0 -sin(alpha) cos(alpha)]; r = [x; y; z]; r_90 = Rx*r; g = subs(f,[x,y,z],[r_90(1)-5,r_90(2),r_90(3)]);
Add a second plot of the rotated and translated torus to the existing graph.
fimplicit3(g) axis([-5 10 -5 10 -5 5]) hold off
## Input Arguments
collapse all
3-D implicit equation or function to plot, specified as a symbolic equation, expression, or function. If an expression or function is specified, then fimplicit3 assumes the right-hand size to be 0.
Plotting interval for x-, y- and z- axes, specified as a vector of two numbers. The default is [-5 5].
Plotting interval for x-, y- and z- axes, specified as a vector of six numbers. The default is [-5 5 -5 5 -5 5].
Axes object. If you do not specify an axes object, then fimplicit3 uses the current axes.
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: '--or' is a red dashed line with circle markers
Line StyleDescription
-Solid line
--Dashed line
:Dotted line
-.Dash-dot line
MarkerDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
's'Square
'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'p'Pentagram
'h'Hexagram
ColorDescription
y
yellow
m
magenta
c
cyan
r
red
g | {
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y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
### Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'Marker','o','MarkerFaceColor','red'
The properties listed here are only a subset. For a complete list, see ImplicitFunctionSurface Properties.
Number of evaluation points per direction, specified as a number. The default is 35.
Example: 100
Line color, specified as 'interp', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default RGB triplet value of [0 0 0] corresponds to black. The 'interp' value colors the edges based on the ZData values.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color | {
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'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Line style, specified as one of the options listed in this table.
Line StyleDescriptionResulting Line
'-'Solid line
'--'Dashed line
':'Dotted line
'-.'Dash-dotted line
'none'No lineNo line
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.
ValueDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
'square' or 's'Square
'diamond' or 'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'pentagram' or 'p'Five-pointed star (pentagram)
'hexagram' or 'h'Six-pointed star (hexagram)
'none'No markers
Marker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the EdgeColor property.
For a custom color, specify an RGB triplet or a hexadecimal color code. | {
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For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Example: [0.5 0.5 0.5]
Example: 'blue'
Example: '#D2F9A7'
Marker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' value uses the same color as the MarkerEdgeColor property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7]. | {
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"openwebmath_score": 0.5654040575027466,
"tags": null,
"url": "https://ch.mathworks.com/help/symbolic/fimplicit3.html"
} |
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